Experimental Biochemistry (1)

572 SWI 3rd ed. CONTENTS

Section I Basic Techniques of Experimental Biochemistry 1

Introduction 3

Experiment 1 Photometry 15

Experiment 2 Chromatography 25

Experiment 3 Radioisotope Techniques 45

Experiment 4 Electrophoresis 61

Section II Proteins and Enzymology 79

Introduction 81

Experiment 5 Acid-Base Properties of Amino Acids 105

Experiment 6 Sequence Determination of a Dipetide 111

Experiment 7 Study of the Properties of ß-Galactosidase 123

Experiment 8 Purification of Glutamate Oxaloacetate Transaminase from Pig Heart 135

Experiment 9 Kinetic and Regulatory Properties of Aspartate Transcarbamylase 149

Experiment 10 Affinity Purificaiton of Glutathione-S-Transferase 157

Section III Biomolecules and Biological Systems 163

Introduction 165

Experiment 11 Microanalysis of Carbohydrate Mixtures by Isotopic Enzymatic, and Colorimetric

Methods 195

Experiment 12 Glcose-1-Phosphate: Enzymatic Formation from Starch and Chemical

Characterization 205

Experiment 13 Isolation and Characterization of Erythrocyte Membranes 217

Experiment 14 Electron Transport 227

Experiment 15 Strudy of the Phosphoryl Group Transfer Cascade Goferning Glucose Metabolism

Allosteric and Covalent Regulation of Enzyme Activity 243

Experiment 16 Experiments in Clinical Biochemistry and Metabolism 253

Section IV Immunochemistry 261

Introduction 263

Experiment 17 Partial purification of a Polyclonal Antibody, Determination of Titer, and

Quantitation of an Antigen Using the ELISA 279

Experiment 18 Western Blot to Identify an Antigen 291

Section V Nucleic Acids 301

Introduction 303

Experiment 19 Isolation of Bacterial DNA 333

Experiment 20 Transformation of a genetic Character with Bacterial DNA 339

Experiment 21 Constructing and Characterizing a Recombinant DNA Plasmid 345

Experiment 22 In Vitro Transcription from a Plasmid Carrying a T7 RNA Polymerase-Specific

Promoter 359

Experiment 23 In Vitro Translation: mRNA tRNA, and Ribosomes 369

Experiment 24 Amplification of a DNA Fragment Using Polymerase Chain Recation 385

Section VI Information Science 397

Introduction 399

Experiment 25 Obtaining and Analyzing Genetic and Protein Sequence Information via the World

Wide Web, Lasergene, and RasMol 405

Appendix Supplies and Reagents 409

Index 437

q /

3

Biochemistry is the chemistry of biological sys-tems. Biological systems are complex, potentiallyinvolving a variety of organisms, tissues, cell types,subcellular organelles, and specific types of mole-cules. Consequently, biochemists must separateand simplify these systems to define and interpretthe biochemical process under study. For example,biochemical studies on tissue slices or whole or-ganisms are followed by studies on cellular systems.Populations of cells are disrupted, separated, andtheir subcellular organelles are studied. Biologicalmolecules are studied in terms of their specificmechanisms of action. By dividing the system un-der study and elucidating the action of its compo-nent parts, it is possible to then define the func-tion of a particular biological molecule or systemwith respect to the cell, tissue, and/or organism asa whole.

Biochemical approaches to the simplificationand understanding of biological systems requiretwo types of background. First, biochemists mustbe thoroughly skilled in the basic principles andtechniques of chemistry, such as stoichiometry,photometry, organic chemistry, oxidation and re-duction, chromatography, and kinetics. Second,biochemists must be familiar with the theories andprinciples of a wide variety of biological and phys-ical disciplines often used in biochemical studies,such as genetics, radioisotope tracing, bacteriology,and electronics. This need reflects the biochemists’ready acceptance and use of theories and techniquesfrom allied areas and disciplines.

It is not possible or appropriate for this book tosummarize the many disciplines and principles used

in biochemistry. However, students will find that areview of the basic principles and units used in thequantitative aspects of experimental biochemistry isquite useful. This section is intended to providesuch a review. In addition, it is valuable for studentsto understand the methods often used in experi-mental biochemistry. Experiments 1 to 4 of this section deal specifically with these techniques: spec-trophotometry, chromatography, radioisotope trac-ing, and electrophoresis. Finally, it is imperativethat students understand the intricacies of dataanalysis. The final part of this Introduction dis-cusses the principles underlying the basics of sta-tistical analysis that are critical to the ability to de-termine the precision or error associated withquantitative data obtained in biochemical experi-ments.

Requirements for a Student ofExperimental Biochemistry

This course is aimed at developing your interest inand understanding of modern biochemical andmolecular biological experimentation. This goalnecessitates a careful emphasis on the experimen-tal design, necessary controls, and successful com-pletion of a wide variety of experiments. This goalwill require additional efforts if you are to benefitfully from Experimental Biochemistry. First, youshould familiarize yourself with general back-ground material concerning each experiment.Three elements have been incorporated into thetext to aid you in this effort: (1) Each experiment

SECTION I

Basic Techniques of Experimental Biochemistry

Introduction

4 SECTION I Basic Techniques of Experimental Biochemistry

is preceded by a short introduction designed to aidyou in understanding the various theories and tech-niques underlying the exercises. (2) The experi-ments are divided into sections that deal with a spe-cific class of biological molecules. The introductionpreceding each section will serve as a review to pro-vide you with enough information to understandthe experiments. This material is intended to rein-force and supplement the knowledge you havegained from biochemistry lecture courses and text-books of general biochemistry, which you shouldreview as needed. (3) Each experiment is followedby a set of exercises and related references that willallow you to further develop your interest and un-derstanding of a particular method, technique, ortopic.

Second, you must keep in mind that the abilityto complete the experiments within allocated timesrequires you to be familiar with the protocol of theexperiment before the start of the laboratory session.Each of the experiments contains a detailed, classtested, step-by-step protocol that will enable you toperform, analyze, and interpret the experiments onyour own. Your success will depend on your abilityto organize and understand the experimental pro-cedures, making efficient use of your time.

Third, efficient use of Experimental Biochemistryrequires that you perform and interpret many cal-culations during the course of the laboratory sessions.Specifically, laboratory work for introductory bio-chemistry, unlike many introductory laboratorycourses, frequently requires you to use the resultsof one assay to prepare and perform additional as-says. Thus, you will have to understand fully whatyou are doing at each step and why you are doingit.

Finally, it is imperative that you maintain a com-plete research notebook containing all your data,calculations, graphs, tables, results, and conclu-sions. Your notebook should be so clear and com-plete that anyone can quickly understand what wasdone and what results were obtained. Your instruc-tor may provide additional specific instructions foryour laboratory reports; the following suggestionsmay be helpful:

1. Use a large, bound notebook, preferably onewith gridded pages. Such notebooks permit the direct construction of data tables and allowyou to attach records of primary accessory

data, such as computer-derived graphs, chro-matograms, dried SDS-PAGE gels, and pho-tographs.

2. Never record your data on separate sheets ofpaper. Rather, record all your data directly inyour notebook. You may consider using oneside of the notebook for raw data and calcula-tions and the other side for results and inter-pretation.

3. All graphs and tables must be clearly and un-ambiguously labeled. Be particularly careful tospecify units on the ordinate ( y-axis) and ab-scissa (x-axis) of every graph.

4. The laboratory report for all experimentsshould include:

a. a brief statement of purpose (why you aredoing the experiment and what you wish todetermine);

b. a brief account of the theory and design ofthe experiment, including a summary orflow chart of the principal manipulativesteps;

c. the raw data;d. all calculations (if analysis requires a single,

repetitive calculation, a sample calculationfor one of a series is acceptable);

e. results;f. conclusions and interpretations (the infor-

mation that you can derive from the resultsof the experiment).

As stated earlier, all the experiments in this text-book have been class tested by hundreds of students.The experiments, therefore, show a high rate ofsuccess. Still, there may be times when your exper-imental results are not particularly useful, or whenthey yield unexpected results that require an expla-nation. If this is the case for a particular experiment,discuss in the results section of your laboratory re-port what may have gone wrong. Did you make animproper dilution of one of the reagents? Did youaccidentally omit one of the experimental steps? Inthe conclusion section, discuss what you may haveexpected to see if the experiment had been suc-cessful. Your knowledge of the theory underlyingthe techniques, along with your understanding ofthe experimental protocol, should be sufficient toallow you to determine what type of data you mayhave obtained under ideal conditions. By doing this,you are likely to turn what appears to be a failed

SECTION I Introduction 5

experiment into a valuable learning opportunity. Itis never sufficient to say, “the experiment did notwork.” Attempt to understand why a particular ex-periment may not have worked as expected.

Laboratory Safety

Experimental Biochemistry employs the use of po-tentially hazardous reagents. Strong acids, strongbases, volatile compounds, flammable compounds,mutagenic compounds, corrosive compounds, ra-dioisotopes, electricity, and sharp objects are thetools of the biochemist. Like any other tool, theseare hazardous only when handled improperly. Atthe beginning of each experimental protocol, wedraw your attention to potential hazards that maybe associated with a particular reagent you are aboutto use.

Safety goggles/glasses must be worn in the lab-oratory at all times. The main purpose of eye pro-tection is to prevent chemical damage to the eye.Laboratory eye protection should also be shatter-proof to protect against debris that would be pro-duced from broken glass in the event of an acci-dent. Although you may feel confident that you willnot be the cause of such an accident, it is impossi-ble to ensure that your laboratory partner or neigh-boring groups will not have accidents.

It is advised that you wear latex or vinyl examgloves at all times in the laboratory. Even if a par-ticular experiment does not require the use of haz-ardous chemicals, one can never be sure that thosefrom a previous experiment have been properly dis-posed of. If volatile compounds are used, theyshould be stored under a fume hood at all times. Ifpossible, students should work with these materialsunder the fume hood as well. The large amounts ofmaterials that are often required for a laboratorygroup may soon fill the room with unpleasant andpotentially hazardous vapors. This is particularilyimportant if the reagent vapors are flammable (seeExperiment 6) or radioactive (see Experiment 12).

Laboratory coats may be worn if desired. It is agood idea to wear them when working with ra-dioisotopes, since very small quantities of a ra-dioactive solution can carry a significant amount ofactivity. It is also a good idea to wash your handsthoroughly with soap before leaving the laboratoryto ensure that you do not take any chemicals out-

side the laboratory. When working with radioiso-topes such as 32P, it is necessary to check your handsand shoes with a Geiger counter before leaving thelaboratory.

The laboratory will be equipped with safetyshowers, eyewash stations, emergency exits, sharpscontainers, and fire extinguishers. Take the time tobecome familiar with the location of all of thesesafety components. All “sharps” (razor blades, Pas-teur pipettes, broken glass, etc.) should be placedin the labeled “sharps” containers. Your laboratorysupervisor will instruct you on the proper use anddisposal of all hazardous reagents. If you do becomeinjured or have any questions about your health riskduring the course of the experiment, immediatelynotify the laboratory instructor. Most laboratorysupervisors have had training in dealing with firesand exposure to different chemicals. Have fun withthe experiments, be safe, and always leave a cleanlaboratory workbench for the beginning of the nextlaboratory session.

Units of Biochemistry

Biochemistry employs a decade system of unitsbased on the metric system. Thus, biochemists useunits such as the mole or the liter and various sub-divisions that differ by three orders of magnitude(Table I-1). With knowledge of the molecularweight of a particular molecule and equation I-1, agiven mass of a molecule can be converted to unitsof moles:

(I-1)

Number of moles �

As indicated in Table I-1, grams may be convertedto milligrams and moles can be converted to mil-limoles simply by multiplying each of the appro-priate values by 103. For example, 0.025 mol of amolecule is equal to 25 mmol:

0.025 mol � � 25 mmol

Volume and mole values define the concentrationterms of molar (M), millimolar (mM), and micro-molar (�M) as shown in equation I-2:

103 mmol��

1 mol

number of grams of molecule��molecular weight of molecule


Table I-1 Basic Units Used in Biochemistry

Mole Units Liter Units

1 mole 1 liter

1 millimole (mmol) � 10�3 moles 1 milliliter (ml) � 10�3 liter

1 micromole (�mol) � 10�6 moles 1 microliter (�l) � 10�6 liter

1 nanomole (nmol) � 10�9 moles 1 nanoliter (nl) � 10�9 liter

1 picomole (pmol) � 10�12 moles

Gram Units Equivalent Units

1 gram 1 equivalent (Eq)

1 milligram (mg) � 10�3 g 1 milliequivalent (mEq) � 10�3 Eq

1 microgram (�g) � 10�6 g 1 microequivalent (�Eq) � 10�6 Eq

1 nanogram (ng) � 10�9 g

1 picogram (pg) � 10�12 g

1 femtogram (fg) � 10�15 g

(I-2)

Concentration (molar) �

Concentration (millimolar) �

�

Concentration (micromolar) �

�

Similarly, volume and equivalent values definethe concentration term of normality (N) commonlyused in the expression for acid (H�) or base (OH�)strength, as indicated by equation I-3:

(I-3)

Concentration (normal) �

�

Because these units involve basic metric princi-ples, one can make use of the metric interconver-sions of mass (grams), fluid volumes (liters or mil-liliters), and spatial volumes (cubic centimeters, cc).Specifically, under most laboratory conditions, 1 ml

number of milliequivalents��

volume (in milliliters)

number of equivalents��

volume (in liters)

number of nanomoles��volume (in milliliters)

number of micromoles��

volume (in liters)

number of micromoles��

volume (in milliliters)

number of millimoles��

volume (in liters)

number of moles��volume (in liters)

of water or dilute aqueous solution weighs approx-imately 1 g and occupies 1 cc of volume (1 ml �1.000027 cc).

These simple interrelationships of moles,weights, volumes, and so forth are often covered inintroductory or freshman level college chemistrytextbooks. Yet, practical experience reveals thatthese basic concepts are a major source of difficultyfor many students in their initial exposure to ex-perimental biochemistry. Therefore, we stronglysuggest that students thoroughly review these con-cepts before conducting the experiments describedin this textbook.

Analysis and Interpretation ofExperimental Data

In nearly all of the experiments outlined in this text-book, you will be asked to collect, analyze, and in-terpret experimental data. Whether you are deter-mining the concentration of a molecule in anunknown solution, the activity of an enzyme, theabsorbance of a solution at a particular wavelength,or the activity of a particular isotope in a biologi-cal sample, the exercise will require you to performa quantitative measurement and calculate a specificvalue. There are several questions that frequentlyarise during the analysis of experimental data: How


do you determine the level of precision of a set ofmeasurements? How many data values or trials ofan experiment must you perform before a mea-surement can be deemed precise? If you have a sin-gle value in a data set that is not in agreement withother members of the set, how do you determinewhether it is statistically acceptable to ignore theaberrant value? In the subsections below, each ofthese issues is addressed.

Accuracy, Precision, and Bias of a

Quantitative Measurement

When interpreting laboratory data, it is importantto recognize that individual measurements, such asthe concentration of a biological molecule observedin an assay, are never entirely accurate. For instance,the serum cholesterol measured by a medical labo-ratory from a blood sample is not the exact averageserum cholesterol in the patient’s blood at the timethe sample was drawn. There are a number of rea-sons for this, the most obvious being that choles-terol may not have been quite uniformly mixedthroughout the bloodstream. The patient’s bloodand the sample drawn from it are never totally ho-mogeneous, the reagents used in the test are nevertotally pure or totally stable if repeatedly used, andthe calibration of the autoanalyzer is never exactlycorrect or totally stable. Even such small deviationsfrom the ideal execution of the assay may some-times noticeably affect the results, and additionalundetected errors in execution sometimes producesubstantial errors. For these reasons, carrying outthe same experiment more than once, or even re-peatedly assaying the same sample, is bound to pro-duce somewhat different numerical results eachtime.

Now, although the quantity being measured isa property of the particular sample under study, thedegree of expected fluctuation from one measure-ment to another depends most fundamentally onthe measurement process itself—that is, how theassay is conducted—rather than on the particularsample. Since, depending on the circumstances, theamount of fluctuation among attempts to measurethe same quantity may be trivial or crucially im-portant, we now consider briefly some basic con-cepts that help the biochemist deal with variabilityamong measurements.

In performing an assay, the biochemist aims foraccuracy. An assay method is accurate when thechance is high that its result will be quite close tothe true value being measured. Since individual as-say results invariably fluctuate, an accurate assaymethod must be (1) highly precise (equivalently, re-producible), having little variability when repeated,and (2) nearly unbiased, meaning that almost all ofthe time the average result from a large number ofrepeated assays of the same sample must be veryclose to correct. Conversely, an assay method canbe poor because it is imprecise, biased, or both. Forinstance, a highly reproducible assay based on a verypoorly calibrated instrument may yield almost thesame, but grossly incorrect result, every time it isapplied to a given sample. Another assay may beunbiased but also never close to correct, because itsfrequently large overestimates are balanced out byequally large and frequent underestimates.

The above concepts become more precise whenexpressed mathematically. Let the Greek letter �represent the true characteristic of a sample that weare trying to measure, and suppose the n observa-tions xi, i � 1, . . . , n, represent the results of en-tirely separate executions of an assay procedure.Then (xi � �) is the error of the ith assay. If wesquare these errors and take their arithmetic mean,we obtain the mean squared error � (1/n)�(xi � �)2

of the group of n repetitions. If we could repeat theassay an extremely large number of times, so thatn is very large, the average result x� � (1/n)�xi wouldeventually stabilize at a limiting value X�, the “long-run” average value of the assay for the given sam-ple. The mean squared error would similarly stabi-lize at a value, MSE, that can be used as an indexof the assay’s inherent accuracy. In principle, a per-fect assay method has MSE � 0 for all samples, butno such assay exists. The bias of the assay is (X� ��) and its square, (X� � �)2, is a component of theMSE. The difference MSE � Bias2 � MSE �(X� � �)2 � 2 is a measure of inherent variabilityof the assay, known as its variance. While we cannever determine the variance of an assay exactly, be-cause that would require performing the assay animpossibly large number of times, we can estimateit by

s2 � � (xi � x�)21

�n � 1


known technically as the sample variance. By takingits square root,

s � �� (�x�i �� x��)2�we obtain a measure of variability in the same unitsas the individual assay results. This measure offluctuation among repeated assays is known as thesample standard deviation, abbreviated SD. Themean, x�, and standard deviation, SD, togetherrepresent a compact summary of a group of re-peated assays. For example, if the absorbance val-ues of three identically prepared solutions at aparticular wavelength are determined to be 0.50,0.44, and 0.32, then their mean value is x� � 0.42and their SD is

s � �� 0�.0�0�8�4� � 0.09

It is common to report such results as x� SD, forexample, 0.42 0.09, although this notation canlead to some unfortunate confusion, as we shall seebelow.

Precision of a Replicated Assay

Intuitively, it seems that one way to improve the ac-curacy of an assay is to do it more than once andtake the average of the repetitions as your result. Inprinciple, random overestimation and underesti-mation by the individual results will cancel one an-other out in the average, leaving a more accurateresult than can be obtained from only one mea-surement. This is true if the assay has been prop-erly calibrated, so that its bias is very low. In thatcase, the accuracy of the assay depends almost en-tirely on its precision, which is represented by theSD, with a precise assay having a small SD. Usingthe SD, we may determine how precision improveswith each additional repetition. Suppose we replacethe procedure of a single assay A with standard de-viation s by a new procedure, which involves n rep-etitions of A, with only the average result reported.Call this “improved” assay An. Thus, each single re-sult of the assay An is an average of n results of A.Though it is beyond the scope of this book todemonstrate, it may be shown that the standard de-viation of the assay An is just s/�n�. Thus, the im-

0.082 � 0.022 � 0.102

��2

1�n � 1

provement in precision obtained by replicating anassay n times is a reduction in the SD by a factorof [1 � (1/�n�)], meaning that two repetitions yieldabout a 29% reduction in SD, three repetitions a42% reduction, and four repetitions a 50% reduc-tion. However, the reduction obtained by An�1 rel-ative to that achieved by An is [1 � (�n�/�n� �� 1�)],or 18%, for three versus two repetitions, 13% forfour versus three, and 11% for five versus four.Thus, the relative benefit of each additional repe-tition declines with n, and the comparison of thebenefit to the cost of an additional repetition gen-erally becomes less favorable to additional repeti-tions as n increases. Nevertheless, in principle anydesired level of precision may be achieved by usingenough repetitions, although the extra repetitionswill not change the bias of the assay.

The quantity s2/n, which represents the vari-ability among arithmetic means (i.e., simple aver-ages) of n repetitions, is known as the standard error of the mean, and is abbreviated either as SEMor SE. Results of assays involving n repetitions, such as the absorbance results reported earlier asx� SD � 0.42 0.09 are also frequently reportedas x� SE, which in this case is 0.42 (0.09/�3�) � 0.42 0.05. Confusion can arise if ei-ther this x� SE or the x� SD format is used with-out specific indication of whether it is SD or SEthat follows the reported mean. The latter is prefer-able whenever the purpose is to represent the pre-cision of the assay’s summary result rather than thevariability of the individual measurements that havecontributed to it. This is almost always the case withchemical analyses, and we recommend the x� SEnotation, supplemented by the number of repli-cates, n, for general use. Thus, in the example, wewould report an absorbance of 0.42 0.05 (SE),from three replications.

When faced with a choice between two assaysyou may compare either their SDs, or their stan-dard errors for any fixed n, to determine which as-say is most useful. The more precise assay is gen-erally preferred if biases are similar, assuming costsand other practical considerations are comparableas well. In such circumstances, an assay with SD �s � 0.09 is much preferable to one with s � 0.36;similarly, a triplicate assay procedure with SE �0.02 is greatly preferable to another triplicate pro-cedure with SE � 0.07.


For an unbiased assay the SE may also often beused to form a range, centered on the reported as-say result, that has a known probability of contain-ing the true value � for the sample being studied.This range, known as a confidence interval, sum-marizes the information that the assay providesabout the sample in a manner that incorporates theunderlying fuzziness of our knowledge due to ran-dom variability in the assay process. For instance,x� 4.31 � SE gives a 95% confidence interval forthe true value estimated by a triplicate assay, andx� 3.19 � SE gives a 95% confidence interval foran assay in quadruplicate. For the triplicate ab-sorbance data, we have 0.42 4.31 � 0.05 �0.42 0.22, or 0.20 to 0.64. In the long-run, 19 of20 (95%) of such ranges obtained from triplicateassays using the given method will include the trueabsorbance of the sample, though we cannot say ex-actly where within the interval that true value lies.However, for 1 in 20 assays (5%), the true ab-sorbance will be outside the interval. For a higherconfidence such intervals may be widened, while in-tervals obtained using a smaller multiplier will ex-clude the true concentration more than 5% of thetimes they are employed. The appropriate multi-plier of the SEM depends on both n and the de-sired degree of confidence, here 95%. When theSD of an assay is known to good accuracy (e.g. re-ported by an instrument manufacturer based onconsiderable data on the performance of the in-strument over the range of likely values) confidenceintervals may be constructed using reported ratherthan observed variability. These intervals requiresmaller multipliers, and hence will tend to be nar-rower, than those based only on the observed repli-cations for an individual assay. The choice of mul-tiplier is beyond the scope of this book.

A consequence of the above ideas is a rule ofthumb that to obtain adequate precision with someassurance that gross error has been avoided, at rea-sonable cost, you will often be well served to per-form three or four trials of a single experiment. Theexperiments outlined in this textbook rarely call forsuch replication, due to time and financial con-straints of the educational process. Remember, how-ever, that a properly designed experiment should be per-formed multiple times, and that the data should bepresented with a well-defined statistical analysis to allowthe reader to ascertain the precision of the experiment.

Outlying Data Values

Suppose that you perform an identical assay fourtimes and obtain the following values: 0.47, 0.53,1.53, and 0.45. Within this small data set, the valueof 1.53 stands out as apparently different. If all fourof these values are included in the reported assayresult, we report 0.75 0.26 (SE) from four repli-cates. This measurement does not appear particu-larly precise, since the error associated with themeasurement is about 35% of its value. By ignor-ing the apparently aberrant value of 1.53, however,you may report 0.48 0.02. It is tempting to as-sume that the 1.53 must have resulted from a mis-take, and that the latter result is likely more accu-rate and far more precise than the former.Unfortunately, in many such instances this is falla-cious, and deleting the apparently outlying numberresults in a less accurate assay result with misrep-resentation of its precision. The problem resultsfrom our intuitive inclination to regard the 1.53 asprobably wrong because it is so deviant from theother three values, when an alternative plausible ex-planation is that any of the other three values issomewhat, but not unusually, lower than might beexpected. When three or four numbers are obtainedfrom a situation partially governed by chance, it isnot at all unusual for one of them to depart furtherfrom the others than intuition might suggest. Whensuch numbers are deleted, the resulting report maybe very misleading.

Yet there are circumstances in which singlegross errors contaminate otherwise valid assays, andit is desirable to be able to delete such errors anduse the remaining repetitions of the assay. How canwe tell when to delete such outlying values, andwhen it is necessary to retain them and accept thepossibility that our assay is less precise than we hadhoped?

The first and best approach should be to exam-ine the assay procedure that was used to try to findan explanation (e.g., a technical error or even a datarecording error) for the outlying value. If a sub-stantial technical error is found, then the outlyingvalue may be discarded, and if possible the assayshould be run again to replace it. If no such expla-nation is found, however, we still may wish to dis-card grossly aberrant values rather than repeat theentire set of n repetitions of the assay. One approach


is to assume a particular mathematical model thatdescribes, in terms of probability, the amounts bywhich repetitions of the same assay vary from oneanother, and then to discard an outlying value onlywhen it seems quite discrepant from what thatmodel predicts. The model commonly used is thegaussian (or “normal”) probability distribution,which is a reasonable approximation of the behav-ior of random fluctuations for many assays (thoughby no means all). This model suggests the follow-ing procedure:

1. Calculate the mean and SD for all observationsother than the outlying value.

2. Subtract the calculated mean from the out-lying value, and divide by the calculated SD.For the data above, this yields (1.530 �0.483)/0.024 � 43.6.

3. Discard the outlying value if the absolute valueof this number is too large. In other words, theoutlying value is discarded when it is too farfrom the mean of the other values, in units oftheir SD.

But how many “SD units” is too far? Specifi-cally, is the ratio of 43.6 large enough to discard theoutlying datum? To answer this question, we mustdecide how much good data we are willing to throwaway, each time biasing our result and exaggeratingits precision, in order to protect ourselves againsterror resulting from including genuinely erroneousvalues. Suppose you feel that genuine gross errorsare not unusual, and thus would be willing to throwaway 1 in 20 good results in order to obtain suchprotection. Then a calculation involving probabil-ities indicates that ratios larger than 6.2 justify dis-carding the outlier. However, if you believe thatgenuine gross errors are rarer, and are willing tothrow out only 1% of good results in order to pro-tect against it, the required multiple is 14.1. In ei-ther case, the value of 1.53 in the above examplewould be discarded. The observed ratio of 43.6 isslightly below the criterion ratio of 44.7 that shouldbe used if you are willing to discard only 1 in 1000good data points. Although that policy may well betoo conservative, a biochemist adhering to it wouldretain the value of 1.53, and report the result of0.75 0.26 (SE) from four replications.

The numerical criteria given above (for decid-ing when the calculated ratio is large enough tojustify discarding a datum) are specifically for qua-

druplicate assays such as the example. Their coun-terparts for triplicate assays are higher, respectively,25.5, 127.3, and 1273.3. Therefore, ratios that jus-tify discarding data points from a quadruplicate as-say may well indicate they should be retained in atriplicate assay. Note that many a practicing biochemistuses numbers far lower than any of these, without ap-preciating that a substantial fraction of good results isbeing discarded. Unless true large errors are common,this frequently yields assays that have lower precisionand accuracy than the full data would provide, with farlower precision and accuracy than is claimed for them.

Presentation of ExperimentalData

There are several options available in the presenta-tion of experimental data. Undoubtedly the twomost popular formats are tables and graphs, sinceboth allow a great deal of material to be presentedin a concise manner. Considerations for each ofthese formats are presented below.

Data Tables

Remember that a data table should be designed topresent a set of data as clearly and concisely as pos-sible. All columns and rows within the table shouldbe clearly defined with respect to the identity andunits associated with each value. In addition, thetable should be titled to allow the reader to deter-mine quickly what features of the table are relevantto the study or experiment. Table I-2 presents ex-amples of both a poorly designed and a properlydesigned data table. Note that the poorly designedtable has no title, is very redundant, is cluttered,and is presented in a manner that does not allowthe reader to easily see differences between (to com-pare) trials of the same experiment. The properlydesigned data table, in contrast, features the exper-imental values and quickly draws the reader’s at-tention to differences between the values.

Graphs

As with data tables, graphs are a good tool for pre-senting a large amount of experimental data in aconcise manner. It is probably a better format touse if you wish to draw the reader’s attention quickly


Table I-2 Examples of a Poorly Designed and a Well-Designed Data Table

A poorly designed data table:

Absorbance values for Absorbance values for Absorbance values for

Reaction #1 at 360, 420, Reaction #2 at 360, 420, Reaction #3 at 360, 420,

and 540 nm and 540 nm and 540 nm

0.876 (360 nm) 0.885 (360 nm) 0.823 (360 nm)

0.253 (420 nm) 0.250 (420 nm) 0.244 (420 nm)

0.164 (540 nm) 0.163 (540 nm) 0.157 (540 nm)

A properly designed data table:

Table 1: Absorbance values for three trials of the experiment at 360, 420, and 540 nm.

Trial # A360 A420 A540

1 0.876 0.253 0.164

2 0.885 0.250 0.163

3 0.823 0.244 0.157

to differences in experimental data. The two mostcommonly used types of graphs in experimentalbiochemistry are the bar graph and the line graph.Figure I-1 presents two simple sets of enzyme ki-netic data in the bar graph and the line graph for-mat. In general, line graphs are preferable to bargraphs for the presentation of data in which the x-axis variables are numbers along a continuum, suchas pH, time, wavelength, etc. Line graphs shouldalways be designed with the controlled variable as theabscissa (x-axis) and the experimentally observed vari-able as the ordinate ( y-axis). For example, if you de-termined the activity of an enzyme at several dif-ferent pH values, you should plot units of enzymeactivity on the y-axis versus pH on the x-axis. Aswith the data table, the axes of any graph should beclearly labeled with respect to identity and units.Bar graphs are more frequently used when the con-trolled variable is not numerical, as in the exampleshown in Figure I-1.

Whenever possible, line and bar graphs shouldbe constructed with the use of computer graphingprograms (CricketGraph, Excel, Lotus, etc.). Asidefrom the fact that they produce graphs that are uni-form and visually attractive, they have the added ca-pability of fitting non-ideal data to a “best-fit” lineor polynomial equation. This operation of fitting aset of data to a best-fit line becomes extremely im-

portant when you are attempting to determine theactivity of an enzyme solution or trying to deter-mine a specific value for an unknown sample withthe use of a standard curve. Although it is possibleto determine a best-fit line through a set of non-ideal data manually with the use of the least squaresformula, the process can be quite time consumingand more prone to human error than the computer-based method. Experiment with different softwaregraphing programs until you find one with whichyou are comfortable.

To the Instructor

The third edition of Experimental Biochemistry is anintroductory biochemistry laboratory textbook tar-geted to juniors, seniors, and first-year graduate stu-dents in biochemistry, chemistry, microbiology, bi-ology, and/or all related disciplines. As such, weassume that students using this book will: (1) havecompleted an introductory lecture and laboratorycourse in organic chemistry; (2) be thoroughly fa-miliar with the mathematics of introductory chem-istry; and (3) be enrolled in, or have completed, anintroductory biochemistry lecture course. Educationin biology, bacteriology, and/or physical chemistry,although helpful, is not essential for this course.


1.2

1.0

0.8

0.6

0.4

0.2

0

Line graph

1

A45

0 (n

m)

2 3 4 5 6 7 8

Time of reaction (min)

120

100

80

60

40

20

Bar graph

Co2+

Enz

yme

activ

ity (

�m

ol/m

in/m

g)

Mn2+ Zn2+ Mg2+ Ca2+

Divalent cation in assay mixture at 10 mM

Figure I-1 Examples of line graph and bar graph

format for presentation of experimen-

tal data.

The introductory sections and experiments inthe third edition of Experimental Biochemistry intro-duce many of the basic techniques of modern bio-chemical research. Thus, the material in this bookwill provide students with the theoretical and quan-titative background to understand and conductmeaningful research in biochemistry.

This goal may be achieved without undertakingall of the experiments in this book, or even all ofthe protocols within any particular experiment. Weare intentionally repetitive in an attempt to intro-duce a wide variety of principles, procedures, tissuesources, and logistic requirements. We know from

experience that the majority of students in an introduc-tory biochemistry laboratory course have difficulty intranslating a very generalized set of instructions into adetailed protocol that will allow them to conduct the ex-periment and obtain reasonable results. Students whoare forced to do this waste a great deal of time de-signing and carrying out experiments that are un-successful. Such students may become frustrated,lose interest in the material, and gain little practi-cal experience from the course. To avoid this situ-ation, this edition of Experimental Biochemistry in-cludes very detailed, step-by-step protocols that willguide the students through each of the 25 experi-ments. This by no means reduces the experimentsto a “cookbook” format; there are many portionsthroughout the book that require students todemonstrate the depth of their knowledge on eachindividual topic. This same format is also valuableto instructors, limiting the time needed to modifyand test the experiments to fit their individual needsand requirements.

We urge instructors to tailor any of the individ-ual experiments in Experimental Biochemistry to fitthe structural and time constraints of their course.We at the University of Illinois schedule a 14-week,1-semester course that meets 3 times per week fora 4-hour period. In this time, we can typically per-form 12 to 14 of the individual experiments in theirentirety. The daily-schedule format of the protocolsfor each experiment reflect what we can performwith a class of up to 50 students in a reasonable pe-riod of time. If you are limited by time, you maychoose to perform fewer experiments or selectedsections from a number of individual experiments.

The majority of the daily experiments outlinedin Experimental Biochemistry can be completed by alarge group of students in 2 to 4 hours with rela-tively limited assistance by the instructor. This isespecially true if the bulk reagents are prepared bythe teaching assistants and instructors prior to thebeginning of the laboratory period. As with the sec-ond edition, the third edition of Experimental Bio-chemistry includes a detailed list of reagents requiredto perform the experiments, as well as a detailedAppendix describing how to prepare each reagentand the quantities needed for students to performthe experiment.

We have made an effort to design experimentsthat require a minimum of equipment and, there-fore, afford maximum flexibility. Modern biochem-


istry, however, makes use of increasingly more so-phisticated procedures and equipment. We wouldbe deficient if we avoided contact with sophisticatedmodern techniques such as scintillation counting,high-performance liquid chromatography (HPLC),sodium dodecyl sulfate–polyacrylamide gel elec-trophoresis (SDS-PAGE), agarose gel electro-phoresis, and the like. This book must thereforerepresent compromises between teaching require-ments and budgetary realities. Where appropriate,we offer optional methods for particular applica-tions that yield similar experimental results.

We urge instructors to add slight modificationsto the experiments to fit their needs. If you are ableto find a more inexpensive source for a reagent, wedo not discourage you from using it. If you thinkthat you can perform a similar experiment with aless expensive apparatus or instrument, we encour-age you to try it. The sources for the commerciallyavailable reagents that we suggest for each experi-ment are included simply to aid the instructor inobtaining the materials. We have found, however,that the quality control procedures performed oncommercially available reagents are more reliable,especially in the case of complex reagents.

One final point is worth specific attention. Westrongly encourage that the experiment be per-formed by the instructor and/or teaching assistantsin its entirety prior to the beginning of the class ex-periment. Further, we strongly recommend thatthis “pre-lab” be performed using the exact samebulk reagents that the class will be using. Thisavoids many of the potential problems of variationsin tissue samples, lot-to-lot variation among com-mercially available reagents, and variation in

reagent composition resulting from scaling up thepreparation of solutions. If you find that you mustmake slight variations of the protocol for a partic-ular trial of an experiment due to the performanceof a particular set of reagents (volumes, incubationtimes, etc.), make the change and bring it to theclass’s attention at the beginning of the experiment.The protocols provided for each experiment are the“ideal” conditions that we have established fromyears of class testing.

We are confident that you will enjoy this text-book and find it to be very valuable in your effortsto design and/or update your biochemistry labora-tory course.

REFERENCES

Garret, R. H., and Grisham, C. M. (1995). Biochemistry.Orlando, FL: Saunders.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Principles of Biochemistry, 2nd ed. New York:Worth.

Lockhart, R. S. (1998). Introduction to Statistics and DataAnalysis. New York: Freeman.

Mathews, C. K., and van Holde, K. E. (1990). Biochem-istry. Redwood City, CA: Benjamin/Cummings.

Moore, D. S. (1998). The Basic Practice of Statistics. NewYork: Freeman.

Stryer, L. (1995). Biochemistry, 4th ed. New York: Free-man.

Voet, D., and Voet, J. G. (1990). Biochemistry. NewYork: Wiley.

Wilson, K., and Walker, J. (1995). Principles and Tech-niques of Practical Biochemistry, 4th ed. New York:Cambridge University Press.

q

15

EXPERIMENT 1

Photometry

Theory

Light can be classified according to its wavelength.Light in the short wavelengths of 200 to 400 nm isreferred to as ultraviolet (UV). Light in the longerwavelengths of 700 to 900 nm is referred to as nearinfrared (near-IR). Visible light falls between thewavelengths of 400 and 700 nm. Within this rangeof wavelengths is found all of the colors visible tothe human eye. For example, light at 400 to 500 nmwill appear blue, while light at 600 to 700 nm willappear red. Any solution that contains a compoundthat absorbs light in the range of 400 to 700 nmwill appear colored to the eye. The solution is col-ored because specific wavelengths of light are re-moved (absorbed) as they pass through the solution.The only light that the eye will perceive, then, arethe wavelengths of light that are transmitted (notabsorbed).

This principle is illustrated in the absorptionspectrum for riboflavin (Fig. 1-1). An absorptionspectrum is a plot representing the absorbance of asolution at a number of wavelengths. Why does asolution of riboflavin appear yellow to the eye? Asshown in Figure 1-1, riboflavin absorbs light at450 nm, which is in the blue region of visible light.Because of this, red and yellow light will be trans-mitted through the solution and detected by theeye. Figure 1-1 also shows that riboflavin absorbslight strongly at 260 and 370 nm. Although this willnot influence the apparent color of the solution(since these wavelengths lie outside the range of vis-ible light), these absorption events in the UV rangecan be detected with a spectrophotometer.

Spectrophotometry can be used to identify andquantitate specific compounds in both pure and im-pure solutions. Spectrophotometry is based on twophysical principles: Lambert’s law and Beer’s law.Lambert’s law states that the proportion of light ab-sorbed by a solution is independent of the intensityof light incident upon it. In addition, it states thateach unit layer of a solution will absorb an equalfraction of the light passing through it. For exam-ple, if the intensity of light incident upon a solu-tion is 1.00 and each unit layer absorbs 10% of thelight passing through it, then the light transmittedthrough the solution will be diminished by 10% perunit layer (1.00, 0.90, 0.81, 0.73, 0.66, . . .).

0.7

0.6

0.5

0.4

0.3

0.2

0.1

260

370

300 400

Wavelength (nm)

500

450

Abs

orba

nce

Figure 1-1 Absorption spectrum of riboflavin

(22 �M in 0.1 M sodium phosphate,

pH 7.06, in 1-cm light path).


Lightsource

Wavelengthselector

Bandwidth Light-detecting

photocell orphototube Meter

Slit

I0 I

Sample tubeor cuvette

(�1 � �2)

Figure 1-2 Operational diagram of a photometer or spectrophotometer.

Lambert’s law gives rise to the following equa-tion (see Fig. 1-2):

(1-1)

I � I0 � e��l

where I0 � intensity of the incident light; I � in-tensity of the transmitted light; l � length of lightpath (in centimeters); and � � absorption coeffi-cient of the medium. Converting Equation 1-1 tothe logarithmic form, we have:

(1-2)

ln I0/I � �l

Using logarithms to the base 10, the absorption co-efficient (�) is converted to a proportionality con-stant (K):

(1-3)

� � 2.303K

Thus,

(1-4)

log10 I0/I � Kl

The log10 I0/I is termed “absorbance” (A) or “op-tical density” (OD). “Absorbance” is currently thepreferred term. It is the absorbance of a solution ata particular wavelength that is of use in the disci-pline of spectrophotometry. As is apparent fromthese equations, absorbance is a unitless value.

Beer’s law recognizes the relationship betweenthe absorbance of a solution and the concentrationof light absorbing compound(s) in that solution. Itstates that light absorption is proportional to thenumber of molecules per unit volume of light-absorbing compound through which the lightpasses. Beer’s law is based on the observation thata solution at concentration c and length l absorbstwice as much light as a solution at concentrationc/2 and length l. In other words, as the concentra-tion of the light-absorbing compound doubles, sotoo will the amount of light absorbed over a givenpathlength of light.

Because of the concentration factors � and K,Beer’s law and Lambert’s law can easily be com-bined to create Equation 1-5. Beer’s law states thatthe proportionality constant K is related to the con-centration of the absorbing solute.

(1-5)

εc � K

where ε is the extinction coefficient and c is the con-centration of the light-absorbing compound. Thisequation leads to:

(1-6)

I � I0 � 10�εcl

This equation can then be rearranged to:

(1-7)

log10 I0/I � A � εcl

EXPERIMENT 1 Photometry 17

The extinction coefficient ε is constant at a givenwavelength for a given solute that absorbs light. Asdiscussed later, the extinction coefficient may de-pend strongly on both the ionic strength and thepH of the solution in which the solute resides.

What is the extinction coefficient? This term al-lows you to relate the concentration of the light-absorbing compound and the pathlength of inci-dent light to the absorbance of a solution. Theextinction coefficient for a compound in solution isdependent on both the solvent used and the wave-length of light at which the absorbance is measured.Clearly, riboflavin absorbs more light at 260 nmthan at 450 nm. This fact will be reflected in theextinction coefficients for riboflavin at these twowavelengths (the extinction coefficient will behigher at 260 nm than at 450 nm).

Since absorbance values are unitless, the ex-tinction coefficient is most often expressed in unitsof inverse concentration times inverse pathlength(i.e., M�1 cm�1 or mM�1 cm�1), since ε � A/cl.The molar extinction coefficient for compound Xis equal to the absorbance of a 1 M solution of thatcompound in a particular solvent at a particularwavelength in a pathlength of 1 cm. The millimo-lar extinction coefficient for compound X is equalto the absorbance of a 1 mM solution of that com-pound in a particular solvent at a particular wave-length in a pathlength of 1 cm. Extinction coeffi-cients are also commonly expressed in units of(mg/ml)�1 cm�1. The greater the extinction coef-ficient under a particular condition (solvent andwavelength), the greater the amount of light ab-sorbed. This idea becomes important when con-sidering the sensitivity associated with quantifyinga compound by measuring the absorbance of a so-lution at a particular wavelength. For example, sinceriboflavin has a greater extinction coefficient at260 nm than at 450 nm, an absorbance readingtaken at 260 nm would be much more useful in at-tempting to determine the concentration of a di-lute solution of this compound.

Qualitative Spectrophotometric Assays

Because many compounds of biological interest ab-sorb light in the UV, visible, and/or near-IR wave-lengths, spectrophotometry can be a very usefultechnique in identifying unknown compounds.When given a sample of a pure compound, an ab-

sorption spectrum can be generated by measuringthe absorbance of the compound in solution at avariety of wavelengths. Modern spectrophotome-ters are often equipped with an automatic scanningcapability that allows the investigator to produce afull absorption spectrum easily and rapidly. By com-paring the absorption spectrum of an unknowncompound with a number of spectra produced fromsolutions containing known compounds, insight canbe gained into the identity of the compound(s) inthe unknown solution.

Often, it is not necessary to produce a full ab-sorption spectrum to identify a compound, partic-ularly if the molecule absorbs light at unique or un-usual wavelengths. For example, the nitrogenousbases that comprise nucleic acids are known to ab-sorb strongly at 260 nm. The aromatic rings ontryptophan and tyrosine (amino acids that are foundin proteins) are known to absorb strongly at280 nm. Many different iron-containing (heme)proteins or cytochromes show a distinct absorbancemaxima in the visible range from 500 to 600 nm,as well as at about 400 nm. The profiles of theseabsorption spectra may even provide informationabout the oxidation state of the heme group in theprotein (whether it is in the oxidized or reducedform).

You must keep in mind that all absorption spec-tra and extinction coefficients are presented in thecontext of a defined pH and salt concentration (ionicstrength). As demonstrated in Figure 1-3, changingone of these conditions may have a dramatic effecton the position and height of one or more peakswithin the spectrum. This is particularly true if thecompound has ionizable groups within or adjacentto its chromophoric (light-absorbing) center. Recallthat the ability to absorb light is dependent on thearrangement and energy state of the electrons sur-rounding the atoms that make up the molecule. Ifthe local environment of the chromophoric centeris altered by changes in pH, ionic strength, or sol-vent composition, this could potentially have a greateffect on the molecule’s ability to interact with pho-tons of specific wavelengths, altering the profile ofthe absorption spectrum.

Quantitative Spectrophotometric Assays

As stated earlier, spectrophotometry can also beused to quantitate the amount of a compound in


1.2

0.8

0.4

220

Wavelength (nm)

300

Abs

orba

nce

260

A260

A280

1.2

0.8

0.4

Adenine Uracil

220

Wavelength (nm)

300

Abs

orba

nce

260

1.360.18

7.6At pH 7.0 = =

A260

A280

0.820.15

5.5At pH 7.0 = =

Figure 1-3 Absorption spectra of 0.1 mM adenine and uracil in a 1 cm cuvette (– ––––) � in 6 N HCl.

(———) � pH 7.0. (———) � pH 13.0.

solution. The Beer–Lambert law (A � εcl ) allowsyou to relate the absorbance of a solution to theconcentration of a particular solute in that solutionusing the light pathlength and the extinction coef-ficient for that solute. The extinction coefficient forhundreds of molecules at specified conditions havebeen published. The Beer–Lambert law thereforeallows you to predict the absorbance that will re-sult from a solution containing a known concen-tration of a solute. Similarly, you can determine theconcentration of a molecule in solution, providedthe extinction coefficient is known. If an extinctioncoefficient is not available, a value can be generatedfor a particular condition and wavelength, providedyou know the concentration of the compound andthe pathlength of light through the photocell or cu-vette. Beer’s law predicts that absorbance is directlyproportional to the concentration of the compound.Though this is true for most molecules, it is a goodpractice to measure the absorbance of several con-centrations of the compound when determining theextinction coefficient. Deviations from Beer’s lawcan occur if the chemical nature of the solute

changes with concentration. For instance, adeno-sine is known to form dimers at high concentra-tions. Adenosine dimers absorb less light thanadenosine monomers. Because of this, the extinc-tion coefficient for an adenosine solution at highconcentration will be less than that of a diluteadenosine solution at a given wavelength.

Is it possible to determine the concentration ofa single compound in solution with other com-pounds? This is possible if the molecule of interestabsorbs light at a wavelength where the other mol-ecules present in solution do not absorb light. Sup-pose that you have a mixture of compounds, X, Y,and Z, as well as extinction coefficients for each ata variety of wavelengths. Compound X absorbslight at 230, 280, and 460 nm, while Y and Z ab-sorb light at only 230 and 280 nm. An absorbancereading at 460 nm will be sufficient to allow you tocalculate the concentration of compound X in theimpure solution. Because compound X is the onlymolecule that absorbs light at this wavelength, youcan attribute all of the absorbance at 460 nm tocompound X and accurately quantify this molecule


Amount of reactant

Testsolution

Linear orvalid range

Abs

orba

nce

Nonlinearinvalid range

Figure 1-4 Standard curve for a color-forming

quantitative reaction. The ab-

sorbance of the test solution can be

used with the standard curve to de-

termine the concentration of a com-

pound in the test solution.

in the presence of Y and Z. The same is not truefor absorbance readings taken at 230 or 280 nm,since all three molecules contribute to the ab-sorbance of the solution at these wavelengths.

Suppose you knew the concentrations of Y andZ in the mixture of X, Y, and Z above. In this case,it would be possible to calculate the concentrationof compound X with an absorbance reading mea-sured at 230 or 280 nm. If you know the extinctioncoefficients for Y and Z at these wavelengths, youcould use these and the concentrations of Y and Zto subtract the absorbance contributions from thesetwo molecules. The remainder of the absorbance at230 or 280 nm, along with the extinction coeffi-cients for compound X at these wavelengths, couldthen be used to calculate the concentration of X inthe mixture. Note that this type of analysis is pos-sible only if you know that X, Y, and Z are the onlycompounds present in the mixture that absorb lightat 230 nm or 280 nm.

The principles of quantitative spectrophoto-metric assays that we have discussed involve directphotometric measurements of light-absorbingcompounds. It is not always the case, however, thatcompounds of biological interest absorb light at aunique wavelength. If a molecule does not absorblight, it can often undergo a reaction with othermolecules to produce a compound that does absorblight (Equation 1-8).

(1-8)

Colorless Color proportional

compoundExcess of

to amount

to be� color-forming 88n

of colorless

assayedreagents

compound

You can quantify the amount of the colored com-pound (and, therefore, the amount of the colorlesscompound) if the extinction coefficient and the re-action stoichiometry are known. It is also possibleto quantify the amount of colorless compound inan unknown by preparing a “standard curve.” Here,the color-forming reagents react with a known se-ries of increasing concentrations of the colorlesscompound, and the absorbance at a defined wave-length is measured against a “blank” containingonly the color-forming reagents (no colorless com-pound) (Fig. 1-4). If the same procedure is per-formed on an unknown (test) solution of the col-

orless compound, the absorbance can be used to de-termine the concentration of the compound in so-lution by comparing where the absorbance readinglies in relation to the concentrations of the com-pound used to produce the standard curve.

Two concepts are very important to keep inmind when designing a colorimetric assay in whichthe color forms as the result of a chemical reaction:the color-forming reagents must be present in excess, andthe data used to produce the standard curve must showa linear relationship between absorbance and concentra-tion of the compound (changes in absorbance must be di-rectly proportional to changes in concentration of the com-pound under study). If the color-forming reagents arenot in excess, the amount of colored productformed may be limited by these reagents rather thanby the amount of colorless compound being ana-lyzed. In other words, the reaction system will be-come saturated at high concentrations of the col-orless compound. The result of this is that thechange in absorbance may no longer be directlyproportional to the amount of the colorless com-pound present, but by the amount of the color-forming reagents present. These nonlinear data athigh concentrations of the compound are the greatestsource of student error in quantitative spectrophotome-


try. If the absorbance of the unknown solution is foundto lie outside the linear range of the standard curve, theexperiment must be performed again on a dilution of theoriginal unknown solution. The effect of the dilutioncan then be taken into account when determiningthe concentration of the compound in the originalsolution. It is not acceptable to dilute the completed re-action that formed a colored solution with water or otherreagent to force the absorbance value into the linearrange of the standard curve. If each of the completedcolored solutions used to generate the standardcurve were diluted to the same extent, the unknownsolution would once again be found to have an ab-sorbance value in the nonlinear portion of the stan-dard curve.

Construction and Properties of

Spectrophotometers

Photometers, colorimeters, and spectrophotome-ters employ the basic components indicated in Fig-ure 1-2. Each of these components can have amarked effect on the efficiency of any colorimetricassay. Accordingly, it is essential to consider each ofthe components in turn.

Light Source. The light source must be capable ofemitting a steady amount of light of sufficient en-ergy in the range of wavelengths required for theanalysis of the sample. Most spectrophotometersemploy a constant voltage-regulated tungsten lampfor spectral analyses in the range of 340 to 900 nm.More sophisticated spectrophotometers, which alsohave the capability for analyses in the UV range,employ an additional constant voltage-stabilizedH2-deuterium lamp that emits light in the range of200 to 360 nm.

Wavelength Selector. Spectrophotometry re-quires assay of absorbance at defined wavelengths.Usually, it is not possible to have light representa-tive of a single wavelength. Therefore, when wespeak of monochromatic light of, say, 500 nm, weusually mean a source that has its maximum emis-sion at this wavelength, with progressively less en-ergy at longer and shorter wavelengths. Thus, a to-tally correct description of this light must alsospecify a spectral bandwidth (Fig. 1-2), which is therange of wavelengths represented. For example,95% of the energy from a 500-nm light source falls

between 495 and 505 nm. From any light source,less intensity is obtained with light of greater pu-rity. On the other hand, the greater the spectral pu-rity of monochromatic light, the greater the sensi-tivity and resolution of the measurements.

Spectrophotometers generate the desired wave-length of light by use of a wavelength selector. Thesimplest wavelength selectors are one or more ab-sorption filters that screen out light above and be-low specific wavelengths. Many of the older col-orimeters employ such filters because of theirsimplicity and low cost. However, these filters gen-erally have broad transmission ranges, and the res-olution of absorption spectra is low (i.e., peaks are“smeared” or obscured). Moreover, measurementscan be made only at wavelengths for which filtersare available. Most modern spectrophotometersovercome these deficiencies through use of mono-chrometers containing a prism or diffraction grat-ing. Such monochrometers generate relatively purelight at any wavelength over a wide range.

A wavelength in the region of maximum ab-sorption is usually employed in assays of the con-centration of colored compounds. This is not ab-solutely necessary, however, since Beer’s law may beexpected to hold true at all wavelengths at whichthere is appreciable absorption. Thus, when an in-terfering substance absorbs at the wavelength ofmaximal absorption of the substance being mea-sured, another wavelength at which the compoundof interest absorbs light may be utilized.

Slit. The intensity of light emitted through anyfilter or monochrometer may be too intense or tooweak for a light-sensing device to record. It is there-fore necessary to be able to adjust the intensity ofthe incident light (I0) by placing a pair of baffles inthe light path to form a slit. Simple colorimetersoften have a fixed slit, but more sophisticated spec-trophotometers usually have a variable-slit-widthadjustment mechanism.

Sample Tubes or Cuvettes. It follows from Lam-bert’s law that if absorbance is to be used to mea-sure concentration, the length of the light path tra-versed by light in the sample container (i.e., asample tube if it is test-tube shaped, or sample cu-vette if rectilinear in shape) must be the same as


that in the blank. Thus, the sample tube or cuvettemust have the same internal thickness as the blank.This is generally true for rectilinear cuvettes, but itcannot be assumed that all sample tubes are iden-tical in this respect. You should always check theabsorbance characteristics of sample tubes and ob-tain a standardized set for spectrophotometric mea-surements.

Light-Detecting Phototubes or Photocells. Spec-trophotometers use either a photovoltaic cell or avacuum phototube to detect the transmitted light(I ) in a light system. Photovoltaic cells contain aphotosensitve semiconductor (e.g., selenium) sand-wiched between a transparent metal film and aniron plate. Light falling on the surface of the cellcauses a flow of electrons from the selenium to theiron, thereby generating an electromotive force. Ifthe circuit is closed through an ammeter, the cur-rent induced is proportional, within a certain range,to the intensity of light transmitted on the selenium(Fig. 1-5). The cell is limited by low sensitivity (itdoes not detect light of very low intensity) and isinsensitive to wavelengths shorter than 270 nm andlonger than 700 nm.

Vacuum phototubes have two electrodes thathave a maintained potential difference. The cath-ode (negative electrode) consists of a plate coatedwith a photosensitive metal. Radiation incidentupon the photosensitive cathode causes an emissionof electrons (the photoelectric effect), which arecollected at the anode (the positive electrode). Theresulting photocurrent can be readily amplified andmeasured (Fig. 1-5).

Because of differences among photosensitivecathodes, certain phototubes are more sensitive incertain regions of the light spectrum. Others aremore sensitive when used in combination with pre-liminary filters that screen out specific regions ofthe spectrum. Accordingly, certain spectropho-tometers require an additional filter or a special red-sensitive phototube when they operate in the redor near-IR ranges of the light spectrum.

Photomultiplier tubes are a variation of the con-ventional phototube. Such tubes have several in-termediate electrodes, known as dynodes, in addi-tion to the primary photocathode and anode.Electrons emitted from the cathode strike the firstof these dynodes, whereupon by secondary emis-

sion, several secondary electrons are emitted. Eachof these, in turn, strikes the second dynode. Theprocess is repeated until the original photocurrentis amplified by as much as 100,000 times. Such pho-tomultipliers are very sensitive light detectors.

Supplies and Reagents

Colorimeter or spectrophotometer

Sample tubes and cuvettes (or colorimeter tubes for thecolorimeter)

5 10�5 M riboflavin and 5 10�5 M adenosine or anyother ribonucleoside

1 mg/ml solution of lysozyme

Solution of lysozyme at unknown concentration

1% (wt/vol) CuSO4 � 5H2O

2% (wt/vol) sodium tartrate

2% (wt/vol) Na2CO3 in 0.1 N NaOH

Folin–Ciocalteau reagent (2 N)

Protocol

Examination of an Absorption Spectrum

This exercise can be performed easily and rapidlyusing a Beckman DU-64 spectrophotometer or anyspectrophotometer having a “scanning wavelength”capacity. Place a 1-ml sample of 5 10�5 M ri-boflavin in a 1-cm pathlength quartz cuvette. Mea-sure the absorbance of the sample at wavelengthsranging from 220 to 620 nm. If your spectropho-tometer is equipped with an RS232 outlet, the ab-sorbances at these wavelengths can be relayed to aprinter to obtain the full absorption spectrum. Ifthe spectrophotometer that you are using is not au-tomatically able to scan a range of wavelengths, thesame procedure can be done manually by increas-ing the wavelength at 10-nm intervals over thisrange (220 to 620 nm). Note that each change of wave-length will require you to adjust the slit width or sensi-tivity control for the blank to a new 100% transmission(zero absorbance) setting before you measure the ab-sorbance of the riboflavin solution at that wavelength.In the range of wavelengths at which the absorbancereading is high, measure the absorbance at 5-nmintervals to identify the wavelength of maximum ab-sorbance. Attempt to produce an absorption spec-


Light

Thin metal film

Meter+ –

Light

Selenium

Iron

Microammeter

Photovoltaic cell circuit Phototube circuit

Figure 1-5 Schematic representation of photovoltaic cell and phototube.

trum by plotting absorbance (vertical axis) versuswavelength (horizontal axis) as shown in Figure 1-1. If time permits, perform the same exercise us-ing a 5 10�5 M solution of adenosine. For thisribonucleoside, it is suggested that you measure theabsorbance of the sample over a range of wave-lengths from 220 to 320 nm.

Once you have obtained an absorption spectrumfor riboflavin and/or adenosine under a given set ofconditions, calculate a molar extinction coefficient(ε) for riboflavin at 450 nm and for adenosine at260 nm using the Beer–Lambert equation (A � εcl ).Remember that the calculation of the extinction co-efficient requires that you know the exact pathlengthof light through the cuvette and the exact concen-tration of the molecule of interest in the solution.Most modern spectrophotometers are designed fora rectilinear, 1-cm-pathlength cuvette. Other pho-tometers, however, are designed for a 13 100-mmcolorimeter tube (pathlength of 1.3 cm).

If time permits, dilute the riboflavin and adeno-sine samples twofold and measure the absorbancesof these samples at the same wavelengths used tocalculate the molar extinction coefficients (450 nmfor riboflavin and 260 nm for adenosine). If Beer’slaw holds true for these compounds, each of theseassays should yield the same molar extinction coef-

ficient values as those determined using the 5 10�5 M samples. Is this true for riboflavin andadenosine?

Photometric Assay of Color Developed by a

Chemical Reaction

Safety Precaution: Avoid skin contact with the

Folin–Ciocalteau phenol reagent. Wear gloves

and safety goggles at all times.

The Folin–Ciocalteau Assay of Protein Concen-

tration. The Folin–Ciocalteau assay is one of themost sensitive and most commonly used assays todetermine protein concentration (sensitive to about10 �g/ml protein). This procedure employs twocolor-forming reactions to assay protein concen-tration photometrically. In the first reaction (a bi-uret reaction), compounds with two or more pep-tide bonds form a dark blue-purple color in thepresence of alkaline copper salts. In the second re-action, tryptophan and tyrosine side chains reactwith the Folin solution to produce cuprous ions.This reaction is most efficient under basic condi-


tions (pH 10.0–10.5). The cuprous ions producedin this reaction act to reduce components in theFolin reagent (sodium tungstate, molybdate, andphosphate), yielding an intense blue-green solutionthat absorbs light at 500 nm. Since the combinedlevels of tryptophan and tyrosine are generally con-stant in soluble proteins, the blue-green color ofthe Folin–Ciocalteau reaction is proportional toprotein concentration in a complex mixture of pro-teins. Since the number of aromatic residues can varyfrom protein to protein, one must keep in mind that thisassay may have to be standardized when measuring theconcentration of a purified protein. In such cases, itmay be best to use a protein standard that has thesame relative proportion of aromatic residues as theprotein under study.

This assay to determine the concentration of proteinin a sample should not be used if the sample contains sig-nificant concentrations of detergents (SDS, Tween, Tri-ton, etc.), sulfhydryl-containing reducing agents (dithio-threitol), copper-chelating reagents, carbohydrates(glucosamine and N-acetylglucosamine), glycerol (above10% [vol/vol]), Tris (above 10 mM), tricine, and K�

(above 100 mM). All of the compounds listed above havebeen reported to interfere with this assay. If a sampledoes contain these compounds, you may use ace-tone or trichloroacetic acid (see Day 6 protocol forExperiment 8) in precipitating the proteins fromthe sample and resuspending them in a more suit-able buffer. For a more detailed discussion of the dif-ferent types of quantitative protein assays available, re-fer to the Section II introduction.

1. Prepare a series of tubes containing the fol-lowing:

Volume of

Volume of Unknown

Volume of 1 mg/ml Lysozyme

Tube # Water (ml) Lysozyme (ml) Solution (ml)

1 1.20 — —

2 1.15 0.05 —

3 1.10 0.10 —

4 1.00 0.20 —

5 0.90 0.30 —

6 — — 1.20

7 0.70 — 0.50

8 1.10 — 0.10

9 1.15 — 0.05

10 1.19 — 0.01

2. Separately prepare 100 ml of fresh alkaline cop-per reagent by mixing together in order:

1 ml of 1% CuSO4 � 5H2O

1 ml of 2% sodium tartrate

98 ml of 2% Na2CO3 in 0.1 N NaOH

3. Add 6 ml of this alkaline copper reagent to eachof the 10 tubes and mix immediately after the ad-dition to avoid precipitation.

4. Incubate the tubes at room temperature for10 min. Add (and immediately mix in) 0.3 ml ofFolin–Ciocalteau reagent to each of the 10tubes.

5. Incubate for 30 min at room temperature andread the absorbance of all 10 tubes at 500 nm.Use tube 1 (blank) to zero your spectropho-tometer.

6. Prepare a standard curve of your results byplotting A500 versus micrograms of protein fortubes 2 to 5. Determine whether A500 is pro-portional to the mass of protein from 0 to300 �g. (Does the plot yield a straight linethrough the data points?)

7. Determine which of tubes 6 to 10 give ab-sorbance readings that lie in the linear portionof the standard curve. What mass of protein (inmicrograms) is indicated by these positions onthe standard curve?

8. Based on the volume of the unknown proteinsolution present in tubes 6 to 10, calculate theconcentration of lysozyme in the unknown so-lution (in milligrams per milliliter).

9. If you have more than two data points for theunknown solution that lie in the linear portionof the standard curve, it is possible to calculateseveral independent values for the protein con-centration in the unknown solution. These val-ues can then be averaged and presented alongwith a standard deviation from the mean.

10. If you do not have more than two data pointsfor the unknown solution that lie in the linearportion of the standard curve, the assay can berepeated again, in triplicate, using a volume ofthe unknown solution known to give a value inthe linear portion of the standard curve.

11. It is also important to realize that there is someerror associated with the data points used toproduce the standard curve. If possible, itwould be best to perform the assays used toproduce the standard curve in triplicate so that


each point could be reported with a standarddeviation. Most computer graphing programsallow one to place “error bars” on the standardcurve at each data point and generate a best-fitcurve through these points. After this best-fitcurve is calculated, the absorbance values forthe unknown solutions can be introduced intothe equation that defines the curve and used tocalculate a value for the protein concentration.

There are three points about the Folin–Ciocal-teau assay that deserve additional comment. First,the greatest source of error in this assay stems frominadequate mixing of reagents. Be sure to mix all ofthe components immediately after each reagent isadded. Second, you may find that this assay isslightly nonlinear. Despite this fact, the assay isvalid up to about 300 �g of protein. Finally, thecolor produced by this assay is largely formed dur-ing the first few minutes after mixing. If you findthat the color in your unknown tubes (6–10) is moreintense than that in tube 5 (the greatest mass of pro-tein used in making the standard curve) after 2 or3 minutes, it would be wise to immediately diluteyour unknown protein solution by half and preparetubes 6 to 10 again using this newly made solution.In doing so, you will avoid waiting 30 minutes forexperimental data that you cannot use to obtain in-formation by comparing to your standard curve.

You must also realize that there are some prac-tical limitations associated with making spec-trophotometric measurements. In general, youshould not trust absorbance measurements greaterthan 1.0 and less than 0.1. If A � 1.0, then only10% of the incident light has been transmitted. IfA � 2.0, then only 1% of the incident light has beentransmitted. Since most spectrophotometers do al-low a small amount of ambient light or scatteredlight from the incident light source to reach thephototube and be recorded as transmitted light, ab-sorbance readings above 1.0 are generally not ac-curate. The most accurate absorbance readings arethose that lie in the range of 0.1 to 0.5.

Exercises

1. You are given 500 ml of a solution at pH 7.0containing an unknown concentration of

adenosine-5-triphosphate (ATP, ε � 14,300M�1 cm�1 at 260 nm and pH 7.0). You place4 ml of this solution in a 1-cm-wide, 1-cm-long, 4-cm-high quartz cuvette and determinethat the A260 of the solution is 0.143. What isthe molarity of the ATP solution? How manymoles of ATP are in 500 ml of this solution?How many micromoles of ATP are in the cu-vette?

2. Given the data shown in Figure 1-3, determinethe molar extinction coefficient for uracil at pH7.0 and 260 nm.

3. Most pure proteins have extinction coefficientsat 280 nm in the range of 0.5 to 3.0(mg/ml)�1 cm�1. Explain why this value maydiffer from one protein to another.

4. A solution of bovine serum albumin (BSA) wasprepared by mixing 0.5 ml of a stock solutionwith 9.5 ml of water. Given that the extinctioncoefficient for BSA at 280 nm is0.667 (mg/ml)�1 cm�1 and the concentrationof BSA in the stock solution was 2.0 mg/ml,predict what absorbance would result from thissolution at 280 nm. You may assume that a 1-cm-pathlength cuvette was used.

REFERENCES

Beaven, G. H., Holiday, E. R., and Johnson, E. A.(1955). Optical Properties of Nucleic Acids and TheirComponents (The Nucleic Acids, Vol. 1). New York:Academic Press.

Cresswell, C. J. (1972). Spectral Analysis of Organic Com-pounds: An Introductory Programmed Text, 2nd ed.Minneapolis: Burgess.

Jaffe, H. H. (1962). Theory and Applications of UltravioletSpectroscopy. New York: Wiley.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Amino Acids and Peptides. In: Principles of Biochemistry, 2nd ed. New York: Worth, pp. 111–133.

Lowry, O. H., Rosebrough, N. J., Farr, A. L., and Randall, R. J. (1951). Protein Measurements withthe Folin Phenol Reagent. J Biol Chem 193:265.

Schenk, G. H. (1973). Absorption of Light and UltravioletRadiation, Fluorescence and Phosphorescence Emission.Boston: Allyn and Bacon.

Stryer, L. (1995). Exploring Proteins. In: Biochemistry,4th ed. New York: Freeman, pp. 45–74.

v;

25

EXPERIMENT 2

Chromatography

Theory

Biological molecules can be separated from one an-other by exploiting differences in their size, charge,or affinity for a particular ligand or solid support.Chromatography is the laboratory technique thatallows separation of molecules based on their dif-ferential migration through a porous medium. Al-though there are many different types of chro-matography, the principle behind the separation ofthe molecules is the same: a mixture of compoundswill have different affinities for the stationary phase(solid support or matrix) on which it is adsorbedand the mobile phase (buffer or solvent) passingthrough the stationary phase.

General Theory

Modern preparative and analytical chromatographyis most often performed in a column format. Here,the porous matrix or solid support is enclosed in adurable cylinder or column saturated with aqueousbuffer or organic solvent (Fig. 2-1). The column isloaded with a solution containing a mixture of com-pounds at the top of the column by allowing it toflow into the porous medium. After the compoundshave entered the solid matrix, they can be differ-entially eluted from the column either by continu-ous buffer flow or by changing the nature of themobile phase passing through the porous matrix.This process of eluting compounds from the col-umn is termed “development.” As the differentcompounds emerge from the column, the eluted so-lution can be separated into multiple “fractions” or

“cuts” that can be analyzed for the presence of amolecule of interest.

Partition Coefficient and Relative Mobility. Asstated earlier, molecules adsorbed on a solid sup-port (stationary phase) will partition between it anda mobile phase passing through the stationaryphase. To predict the behavior of different mole-cules during chromatography, one must be able todefine the affinity of a compound for the station-ary and/or mobile phase. The term used to describethe affinity of a compound for the stationary phaseis the partition coefficient (�). It is defined as the frac-tion of the compound that is adsorbed on the sta-tionary phase at any given point in time. The par-tition coefficient can have a value between 0 and 1.The greater the value of �, the greater the affinityof the compound for a particular stationary phase.For example, a molecule with � � 0.4 will be 40%adsorbed on the stationary phase at any given pointin time. It has less affinity for the stationary phasethan another molecule with � � 0.7 (70% adsorbedat any given point in time). Mathematically, the par-tition coefficient can be expressed as:

� �

The affinity of a molecule for the mobile phaseis described in terms of relative mobility (Rf). Thisterm describes the rate of migration of the mole-cule relative to the rate of migration of the mobilephase passing through the solid support. Mathe-matically, Rf is equal to 1 � �. For instance, a mol-ecule with � � 0.4 will be found in the mobile phase

molecules adsorbed on stationary phase��molecules in stationary and mobile phase


Reservoir of developingsolvent

Fraction collector withtest tubes

Small layer of solventabove column

Chromatographic agentin column

Figure 2-1 Typical column chromatography

system.

60% of the time (1 � 0.4), whereas a molecule with� � 0.7 will be found in the mobile phase 30% ofthe time. A comparison of Rf values indicates thatthe molecule with � � 0.4 will migrate through thecolumn twice as fast as a molecule with � � 0.7.The following formula defines the mobility of an

adsorbed molecule on the stationary phase relativeto the flow rate of the mobile phase:

Fx � Fs � Rf

where Fx is the flow rate of the compound with agiven relative mobility Rf, and Fs is the flow rate ofthe mobile phase.

Chromatographic “Plates.” In theory, a chro-matographic “plate” can be described as the largestarea of the column where two molecules with dif-ferent partition coefficients will have the opportu-nity to display different rates of migration. In practice, this will often be determined by the dimensions (volume) of the column relative to thevolume of sample that is loaded. The larger the dif-ference of � between two compounds, the fewerchromatographic plates they will be required to passthrough before achieving separation. Whereas twomolecules with � values of 0.2 and 0.8 may requirepassage through only 5 chromatographic plates toachieve separation, two molecules with � values of0.8 and 0.7 may require passage through as manyas 50 plates to achieve separation.

Zone Spreading. From the concept of chromato-graphic plates just described, it would appear thatyou could separate any two molecules with slightlydifferent partition coefficients provided that thecolumn contained a sufficient number of plates.Due to a phenomenon termed “zone spreading,”this is not always the case. As the number of platetransfers increases, so too does the volume in whicha particular molecule is found (the molecule is di-luted as it passes through the column). A moleculewith � � 0.5 may elute from a column in a sharppeak after 20 plate transfers and a very broad peakafter 50 to 100 plate transfers. Since zone spread-ing will occur independently on two different mol-ecules with similar partition coefficients, you mayfind that the elution peaks of the two will overlapafter a large number of plate transfers. In general,zone spreading increases with increasing partitioncoefficient: a molecule with � � 0.2 will show lesszone spreading after 100 plate transfers than a mol-ecule with � � 0.7. Zone spreading is the conse-quence of simple diffusion: A sample of moleculesintroduced to the top of a column will move from

EXPERIMENT 2 Chromatography 27

Elution volume

d

W1

W2

Con

cent

ratio

n of

com

poun

ds

A B

Figure 2-2 Resolution of two compounds (A and

B) emerging from a chromatography

column.

areas of high concentration to areas of low con-centration as the mobile phase continuously passesthrough the stationary phase.

Resolution. The goal of any chromatographicstep is to maximize resolution or to minimize zonespreading. Resolution is a function of the positionof the maximum elution peak height and the elu-tion peak width (Fig. 2-2). The greater the resolu-tion between two elution peaks, the greater the de-gree of separation between the two molecules.Mathematically, resolution (R) is two times the dis-tance between two elution peak maxima (d) dividedby the sum of the widths of the two elution peaks(W1 � W2):

R �

For any given chromatographic step, you mustachieve a compromise between the number of platetransfers and the resolution. In other words, youmust provide enough plate transfers to allow sepa-ration while preventing excess zone spreading.

Isocratic versus Gradient Elution. The principlesof chromatography just described hold true for iso-cratic elution schemes, in which the nature of themobile phase remains constant throughout the de-

2d��(W1 � W2)

velopment of the column. Many chromatographiccolumns, however, are eluted with some sort of mo-bile phase gradient. Here, the nature of the mobilephase changes continuously throughout develop-ment with respect to ionic strength, pH, ligand con-centration, or organic:aqueous solvent ratio. As oneor more of these parameters changes, so too doesthe partition coefficient of one or more moleculesadsorbed on the stationary phase. Although thisgreatly complicates the calculation of �, it cangreatly enhance the ability of a given sized columnto separate a number of different compounds. Bycontinuously changing the partition coefficients ofmolecules passing through the stationary phase, youcan effectively increase the number of plate trans-fers occurring over a given length of the column.A 20-ml column that provides 10 plate transfers uti-lizing an isocratic elution scheme may provide 100to 200 plate transfers when used in conjunction withan elution gradient.

Remember that one problem encountered withincreasing the number of plate transfers is an in-crease in zone spreading, decreasing the resolutionof two elution peaks containing compounds withsimilar partition coefficients. Gradient elutionschemes often work to minimize zone spreading.Recall that molecules with lower � values show lesszone spreading after a given number of plate trans-fers than molecules with large � values. If a mole-cule that has a relatively large � value under load-ing conditions can be eluted with a gradient thatcontinually decreases its � value as it passes throughthe column, zone spreading can be minimized, andresolution between neighboring elution peaks canbe maintained. In general, mobile phase gradientsare most often used to optimize a specific chro-matographic procedure by maximizing both thenumber of plate transfers and the resolution of theelution peaks.

Types of Chromatography

Gel Filtration or Size Exclusion. Gel filtration isperhaps the easiest type of chromatography to un-derstand conceptually (Fig. 2-3). It relies on theability of a given size molecule to enter the uni-formly sized pores of a solid support or matrix.Imagine that you have a solution containing twomolecules: one of 8000 Da and one of 50,000 Da.If this solution is passed through a porous matrix


Flow route of large molecules excludedfrom gel-filtration granules

Flow route of small molecules includedin gel-filtration granules

Void fluid volumeoutside ofgranules = 30 ml

Total fluid volumein and outside ofgranules = 100 ml

Magnification of a gel-filtrationgranule indicating its spongelikecharacter

To fractionating device such asfraction collector

Figure 2-3 Diagram of the gel-filtration process.

capable of excluding molecules larger than 10,000Da, then the two molecules will be separated fromone another as they pass through the column.While the 8000-Da molecule will be able to enterand occupy the pores within the matrix (total vol-ume of the column), the 50,000-Da molecule willbe excluded from the mobile phase occupying thepores and be forced to occupy only the volume out-side of the matrix (void volume of the column).Since the 8000-Da molecule can occupy a largervolume of the column than the 50,000-Da mole-cule, the migration of the smaller molecule throughthe column will be retarded relative to the rate ofmigration of the mobile phase through the column.

Therefore, the 8000-Da molecule will elute later inthe development than the 50,000-Da molecule,which will elute in the void volume of the column.

In the last example, one molecule was excludedfrom the pores of the matrix while the other wasable to occupy the volume between the pores. Gel-filtration chromatography can also separate mole-cules that are only differentially excluded from thepores of the matrix. Suppose that you pass a solu-tion containing molecules of 10,000 Da and 45,000Da through a porous matrix capable of excludingmolecules larger than 60,000 Da. Although bothmolecules will be able to occupy the volume be-tween the pores of the matrix, the mobility of the


Total column volume

Eluate from column (ml)

Void volume

MW = 100,000

Pro

tein

con

cent

ratio

n (m

g/m

l)

MW = 60,000

MW = 50,000

MW = 40,000

MW = 30,000

Figure 2-4 Gel filtration of excluded, partially included, and fully included proteins. (Partially included,

spherical proteins generally elute in an order directly proportional to the log of their molecu-

lar weights. Gel filtration of a protein of unknown molecular weight along with proteins of

known molecular weights can therefore yield the molecular weight of the “unknown” protein.)

10,000-Da molecule will be retarded more than thatof the 45,000-Da molecule relative to the migra-tion or flow rate of the mobile phase. As a result,the 45,000-Da molecule would elute from the col-umn before the 10,000-Da molecule. The volumeat which both molecules elute would be somewherebetween the void volume and the total volume ofthe column (Fig. 2-4).

Gel filtration is usually performed with an iso-cratic mobile-phase gradient. It is most often usedto separate molecules that differ significantly insize. It can also be used as a method to exchangethe buffer in which a molecule of interest resides.Suppose that you have purified a 50,000-Da pro-tein in a solution containing 50 mM phosphatebuffer at pH 8.0 containing 0.5 M NaCl. TheNaCl could easily be removed from the proteinsolution by passing it through an appropriate gel-filtration column with a small pore size using 50-mM phosphate buffer at pH 8.0 as the mobilephase. The pore size of the matrix should be smallenough to fully exclude the protein but still largeenough to fully include the ions that you wish toremove.

Ion-Exchange Chromatography. Ion exchange isperhaps the most classic and popular type of chro-matography used in biochemistry. It relies on thedifferential electrostatic affinities of molecules car-

rying a surface charge for an inert, charged sta-tionary phase or solid support. There are two typesof ion-exchange chromatography, determined bythe charges present both on the stationary phaseand on the molecules that will be adsorbed to it.Cation exchange refers to a situation in which thestationary phase carries a negative charge. It willhave affinity for molecules in solution that carry anet positive surface charge (i.e., cations). The op-posite is true for anion exchange: the stationary phasecarries a positive charge that will have affinity formolecules that carry a net negative surface charge(i.e., anions).

Anion and cation exchange resins can be classi-fied as being either strong or weak. Strong ion ex-changers maintain a net charge at extremes of pHwhile weak ion exchangers are subject to titrationat these extremes (see Table 2-1). The most com-monly used anion-exchange resin is diethyl-aminoethyl (DEAE) cellulose. It is a weak ion-exchange resin that carries a positive charge belowpH 8.5 (pKa � 10.0). The most common cation ex-change resin in use today is carboxymethyl (CM)cellulose. It is also a weak ion exchanger that car-ries a negative charge above pH 4.5 (pKa � 4.0).Since most enzymes display optimal activity andstability between pH 4.0 and 9.0, these two ion-exchange resins will suffice for most protein purifi-cation schemes. A variety of ion-exchange resins,


Table 2-1 Commercially Available Chromatography Adsorbents

Gel-Filtration Chromatography

Adsorbent MW Fractionation Range (Da)* Matrix Composition Supplier

Superdex 1 � 104 to 6 � 105 Dextran/agarose Pharmacia

Superose 1 � 103 to 5 � 106 Agarose Pharmacia

Sephacryl 1 � 103 to 8 � 106 Acrylamide Pharmacia

Bio-Gel P 1 � 102 to 1 � 106 Acrylamide BIORAD

Sepharose 1 � 104 to 4 � 107 Agarose Pharmacia

Sephadex 7 � 102 to 2.5 � 105 Dextran Pharmacia

Bio-Beads (SX) 600 to 1400 Styrene divinylbenzene BIORAD

(for organic compounds and other hydrophobic molecules)

*Fractionation range depends on the particle size of the gel.

Ion-Exchange Chromatography

Adsorbent Functional Group Application Supplier

Mono Q –CH2N�(CH3)3 Strong anion Pharmacia

Q Sepharose –N�(CH3)3 Strong anion Pharmacia

AG 1 –CH2N�(CH3)3 Strong anion BIORAD

AG 2 CH2N�(CH3)2C2H4OH Strong anion BIORAD

DEAE-Sepharose –CH2N�H(CH3)2 Weak anion Pharmacia

AG 4-X4 –CH2N�H(CH3)2 Weak anion BIORAD

Mono S –CH2SO3� Strong cation Pharmacia

AG 50W –SO3� Strong cation BIORAD

SP Sepharose –SO3� Strong cation Pharmacia

CM-Sepharose –CH2COO� Weak cation Pharmacia

Bio-Rex 70 –COO� Weak cation BIORAD

Hydrophobic Interaction Chromatography

Adsorbent Functional Group Supplier

Methyl HIC –CH3 BIORAD

T-butyl HIC –CH(CH3)3 BIORAD

Phenyl Sepharose benzene ring Pharmacia

Butyl Sepharose –CH2–CH2–CH2–CH3 Pharmacia

Octyl Sepharose –CH2–(CH2)6–CH3 Pharmacia

each highly resolving, sturdy, and able to maintaina high flow rate under pressure are now commer-cially available (see Table 2-1).

How are molecules adsorbed on ion-exchangecolumns eluted? Two options are available: chang-ing the ionic strength or the pH of the mobile phase(Fig. 2-5). Unlike with gel-filtration columns, ion-exchange columns are most often developed using

a mobile-phase gradient that continuously altersone of these two parameters. A molecule contain-ing one or more ionizable groups will be charged(positive or negative) over a given pH range. Sup-pose that a molecule of interest contains a carboxylgroup (COO�, COOH) with a pKa value of 4.2.Above pH 4.2, this group will take on a negativecharge (COO� form). In order for a molecule con-


Table 2-1 Continued

Affinity Chromatography

Adsorbent Functional Group Application Supplier

Arginine-Sepharose L-Arginine Serine proteases Pharmacia

Affi-Gel Blue Cibacron Blue Albumin, interferons, BIORAD

Blue Sepharose NAD�-requiring enzymes Pharmacia

Red Sepharose Procion Red NADP�-requiring enzymes Pharmacia

Carboxy-peptidase G

Calmodulin Sepharose Calmodulin Calmodulin-binding proteins, Pharmacia

neurotransmitters

Glutathione Sepharose Glutathione S-transferases Pharmacia

Affi-Gel Heparin Heparin DNA-binding proteins, BIORAD

Heparin Sepharose growth factors, steroid receptors Pharmacia

Lipases, lipoproteins

Lysine Sepharose Lysine Ribosomal RNA Pharmacia

Affi-Gel Protein A Protein A Immunoglobulin G BIORAD

Protein A Superose Pharmacia

Protein G Superose Pharmacia

Con A Sepharose Pharmacia

Lentil Lectin Sepharose Lectins Membrane proteins, Pharmacia

Wheat Germ Lectin glycoproteins, polysaccharides

Sepharose T lymphocytes Pharmacia

Poly A Sepharose Polyadenylic acid mRNA-binding proteins Pharmacia

Viral RNA, RNA polymerase

Poly U Sepharose Polyuridylic acid mRNA, reverse transcriptase, Pharmacia

interferons

Poly T Sepharose Polydeoxythymidylic mRNA Pharmacia

acid

DNA Agarose Calf thymus DNA DNA/RNA polymerase, Pharmacia

polynucleotide kinase

Endonucleases and exonucleases

5�AMP Sepharose 5�AMP NAD�-requiring enzymes, Pharmacia

kinases

Affi-Gel Polymixin Polymixin Endotoxins BIORAD

Affi-Gel 501 Sulfhydryls Proteins with sulfhydryl BIORAD

groups

Affi-Gel 601 Boronate Sugars, nucleotides, BIORAD

glycopeptides

taining this group to adsorb to a DEAE column,you must ensure that the pH of the mobile phaseis well above 4.2 (say, pH 6.0). If the pH of the mo-bile phase is then lowered to 3.0, the moleculewould no longer be negatively charged (the pH is

below the pKa value for this group) and would nolonger have affinity for this anion-exchange resin.This principle underlies the use of a pH gradientin the elution of molecules from an ion-exchangecolumn: the molecule is adsorbed to the column at

32

Resin

Increase [cations]in solution

Increase the pH until theprotein is no longer positively

charged

Carboxymethyl cellulose(cation exchange)

O�O

O

CCH2

O� O�

� Protein

� Protein

Resin O

O

CCH2 Resin O

O

CCH2Na�

Na�

Na�

Na�

Elutes

� Protein

Elutes

Resin

Increase [anions]in solution

Decrease the pH until theprotein is no longer negatively

charged

Diethylaminoethyl cellulose(anion exchange)

�

O

H

NCH2 CH2 CH3

CH3

� Protein

Resin�

O

H

NCH2 CH2 CH3

CH3

Resin�

O

H

NCH2 CH2 CH3

CH3

Elutes

� Protein� Protein

Elutes

Cl�Cl�

Cl�Cl�

Figure 2-5 Elution of proteins from ion-exchange resins by altering pH or ionic strength.


a pH that gives it a charge opposite that of the resin,and is eluted by a change in pH that causes it tolose its charge or affinity for the resin. Anion-exchange columns are usually developed with a decreasing pH gradient, while cation-exchangecolumns are usually developed with an increasingpH gradient.

Although elution with a pH gradient appearstheoretically feasible, a number of practical aspectscomplicate its use in the laboratory. First, as the pHof the mobile phase is altered, the charges on boththe resin and the adsorbed molecules are subject tochange. This is particularly true if you are using aweak ion-exchange resin near the pKa of the sub-stituent (charged) group. Second, the adsorbedmolecules contribute significantly to the overall pHof the mobile phase, particularly in the microenvi-ronment of the resin where the exchange process istaking place. Finally, the buffering power con-tributed by the adsorbed molecules as they elutefrom the column can turn what appear to be smallchanges in the pH dictated by the introduced gra-dient into large changes in pH that cause the elu-tion of a large number of different molecules fromthe column. For example, the decrease in the pHof 0.5 that is introduced by a gradient may trans-late into a pH decrease of 1.5 in the microenvi-ronment of the resin because of the high concen-tration of titratable molecules trapped within it ascompared with the concentration of bufferingspecies in the bulk mobile phase.

Because of the practical aspects just describedfor pH gradients, ionic strength (salt) gradients aremost often used for the elution of molecules fromion-exchange columns. As the concentration ofcounter-ions in the mobile phase increases, they be-gin to compete electrostatically with adsorbed mol-ecules bound to the charged resin. Suppose that youhave a positively charged molecule bound to CM-cellulose resin. As the concentration of positivelycharged ions in the mobile phase increases (throughan increasing gradient of KCl or NaCl), the Na�

and K� will compete with the positively chargedmolecules for binding to the resin. Molecules withlower � values will elute at a lower salt concentra-tion than molecules with larger � values (in thiscase, the � value will be dictated by the magnitudeof the positive charge on the molecule). Since theconcentration of salt in the mobile phase is contin-

ually increasing, the top of the band of moleculesmoving down the column will always experience aslightly higher salt concentration (lower � value)than the portion of the band toward the bottom ofthe column. This, in effect, compacts the bands orzones of eluting molecules, minimizing zonespreading and maximizing resolution. By using gra-dient elution schemes in ion-exchange chromatog-raphy, one can offset the deleterious effects (mainly,zone spreading) that accompany chromatographicprocedures that require a large number of platetransfers to achieve separation.

Choosing an appropriate buffer for ion-exchange chromatography is an important consid-eration. In general, there is only one rule to keepin mind: choose a buffer that will provide the maximumbuffering power at a particular pH while contributingas little as possible to the overall ionic strength of the mo-bile phase. Recall that a buffering species in solutionwill have its greatest buffering capacity at or verynear its pKa value. Therefore, a buffer with a pKaof 6.5 would be a poor choice for an ion-exchangecolumn equilibrated and loaded at pH 8.0. As a gen-eral rule, it is suggested that the ionic strength ofthe mobile phase be limited to 5 to 10 mM with abuffer that has a pKa 0 to 0.5 pH units away fromthe operating pH of the column. Remember thatthe charged groups on the ion-exchange resin willelectrostatically attract any molecule or ion carry-ing an opposite charge. Therefore, it is preferableto use a buffer whose charged form is of the samesign as that on the resin. For example, Tris is abuffer commonly used in conjunction with anion-exchange resins. Tris can either exist in the proto-nated form (positively charged) or the unproto-nated (neutral) form (Cl� is the counter-ion for Trisin the protonated form). Since the anion exchangecolumn is positively charged, the concentration ofnegatively charged species in the mobile phaseshould be minimized. Since Tris can exist only inthe positively charged or neutral form, the con-centration of Cl� counter-ion is the only speciescontributing to the effective ionic strength of themobile phase. The same is not true for a mobilephase buffered by, say, phosphate or acetate ions.Here, at least one form of the buffer carries a chargeopposite that of the anion-exchange resin, poten-tially affecting the ability of negatively chargedmolecules in the sample to adsorb to the resin (a


decrease in the capacity of the column). Any buffer-ing ion that carries a negative charge would be bet-ter to use in conjunction with a cation-exchangecolumn, where the resin also carries a negativecharge. A list of buffers recommended for use inanion- and cation-exchange chromatography can befound in Table 2-2.

Hydrophobic Interaction Chromatography (HIC).

As many proteins fold into their native, three-dimensional structure, a number of hydrophobic re-gions or “patches” become exposed on their sur-faces. The more of these hydrophobic regions present on the surface, the less water soluble theprotein is likely to be. This difference in hy-drophobicity between proteins can be exploited asa means of purification by a technique termed “hy-drophobic interaction chromatography.” If a sta-tionary phase contains a large number of aliphatic(multicarbon) chains on it, hydrophobic interac-tions may allow proteins with different hydropho-bic surface character to adsorb to it with variablestrength. The principles that underlie this tech-nique are quite similar to those forces at work dur-ing the process of “salting out” of proteins underhigh ionic strength conditions (see Section II, Ex-periment 8, “Purification of Glutamate–Oxaloac-etate Transaminase from Pig Heart”).

In general, the conditions of the mobile phasethat promote the binding of proteins to hydropho-bic adsorbents are exactly opposite those that pro-mote binding of proteins to cation-exchange resins:low pH and high ionic strength. Whereas high ionicstrength serves to weaken protein-adsorbent elec-trostatic interactions in ion-exchange chromatog-raphy, high ionic strength serves to promote hy-drophobic interactions. High ionic strength is amuch more critical parameter than low pH, espe-cially since many proteins are subject to denatura-tion under mildly acidic conditions. The more hydrophobic the protein of interest, the less hydrophobic the aliphatic chain on the adsorbentis required for binding. While a very hydrophobicprotein may bind a C4 column under conditions ofhigh ionic strength, a slightly hydrophobic proteinmay require a C8 or C10 resin (more hydrophobic)for adsorption.

Elution of proteins from hydrophobic adsor-bents can be achieved by altering the mobile phase

in a manner that disrupts the hydrophobic interac-tions between the molecule and the adsorbent.Since high ionic strength conditions promote suchinteractions, elution can often be induced by sim-ply lowering the ionic strength of the mobile phase.In other cases, it may be necessary to include a com-ponent in the mobile phase that actively disruptsthe hydrophobic interactions between the proteinand the resin. Organic solvents, non-ionic (nonde-naturing) detergents such as Triton X-100, andpolyethylene glycol (PEG) are commonly used. Fi-nally, some hydrophobic interactions can also beweakened by increasing the pH of the mobile phase,provided that it is not detrimental to the stabilityof the protein of interest.

Affinity Chromatography. Affinity chromatogra-phy is perhaps the most rapidly advancing type ofchromatography in the field of biochemistry. Con-sidering that the goal of any chromatographic stepis to achieve the highest yield and the highest de-gree of purity possible, affinity chromatographypresents itself as the most promising candidate forthis purpose. If one considers a biological mole-cule such as an enzyme, what distinguishes one en-zyme from the hundreds that are found in the cell?The answer is in the ligand(s) (substrates, ions, co-factors, peptides, etc.) for which the enzyme hasaffinity. Whereas the other types of chromatogra-phy discussed thus far separate biological mole-cules based on the more general characteristics ofsize, surface charge, and hydrophobicity, affinitychromatography attempts to separate individualmolecules by exploiting characteristics specific tothem.

Substrate Affinity Chromatography. A number ofadsorbents designed to isolate specific classes of bi-ological molecules are currently in use in the fieldof biochemistry. DNA-binding proteins are oftenpurified with adsorbents containing heparin (a sul-fonated polysaccharide with high negative charge)or calf-thymus DNA. Oligosaccharides and glyco-proteins can be purified with lectin columns. Forexample, concanavalin A (a lectin) is known to havestrong affinity for both mannose and glucose.Resins derivatized with 5�AMP, NAD�, or NADP�

are often used in place of dye-ligand adsorbents (seebelow) as an alternative in the purification of ki-


Table 2-2 Buffers for Use in Ion-Exchange Chromatography

For Anion Exchange

Buffer pKa

Histidine 6.0

bis-[2-Hydroxyethyl]iminotris[hydroxymethyl]methane (BIS-TRIS) 6.5

1,3-bis[Tris(Hydroxymethyl)methylamine]propane (BIS-TRIS PROPANE) 6.8

Imidazole 7.0

Triethanolamine (TEA) 7.8

Tris[Hydroxymethyl]aminomethane (TRIZMA) 8.1

N-Tris[Hydroxymethyl]methylglycine (TRICINE) 8.1

N,N-bis[2-Hydroxyethyl]glycine (BICINE) 8.3

Diethanolamine 8.9

For Cation Exchange

Buffer pKa

Lactic acid 3.9

Acetic acid 4.8

2-[N-morpholino]ethanesulfonic acid (MES) 6.1

N-[2-Acetamido]-2-iminodiacetic acid (ADA) 6.6

2-[(2-Amino-2-oxoethyl)amino]ethanesulfonic acid (ACES) 6.8

Piperazine-N,N�-bis[2-ethanesulfonic acid] (PIPES) 6.8

3-[N-morpholino]-2-hydroxypropanesulfonic acid (MOPSO) 6.9

N,N-bis[2-Hydroxyethyl]-2-aminoethanesulfonic acid (BES) 7.1

3-[N-Morpholino]propanesulfonic acid (MOPS) 7.2

N-Tris[Hydroxymethyl]methyl-2-aminoethanesulfonic acid (TES) 7.4

N-[2-Hydroxyethyl]piperazine-N�-[2-ethanesulfonic acid] (HEPES) 7.5

3-[N,N-bis(2-Hydroxyethyl)amino]-2-hydroxypropanesulfonic acid (DIPSO) 7.6

3-[N-Tris(Hydroxymethyl)methylamino]-2-hydroxypropanesulfonic acid (TAPSO) 7.6

N-[2-Hydroxyethyl]piperazine-N�-[2-hydroxypropanesulfonic acid] (HEPPSO) 7.8

Piperazine-N,N�-bis[2-hydroxypropanesulfonic acid] (POPSO) 7.8

N-[2-Hydroxyethyl]piperazine-N�-[3-propanesulfonic acid] (EPPS) 8.0

N-Tris[Hydroxymethyl]methyl-3-aminopropanesulfonic acid (TAPS) 8.4

3-[(1,1-Diethyl-2-hydroxyethyl)amino]-2-hydroxypropanesulfonic acid (AMPSO) 9.0

2-[N-Cyclohexylamino]ethanesulfonic acid (CHES) 9.3

3-[Cyclohexylamino]-2-hydroxy-1-propanesulfonic acid (CAPSO) 9.6

3-[Cyclohexylamino]-1-propanesulfonic acid (CAPS) 10.4

nases and dehydrogenases. Poly dT columns are of-ten used to purify messenger RNA from eukaryoticcells that often end in a poly A tail (A will form hy-drogen bonds with T in DNA). In all instances, themolecules of interest are eluted by the addition ofexcess free ligand, change in ionic strength, changein pH, addition of low concentrations of chaotropic

salts or denaturing agents, addition of organic sol-vents, or addition of some type of detergent (seebelow).

In order to perform affinity chromatography,you must have a means of covalently attaching thedesired substrate molecule to the resin. A numberof different techniques can be used to chemically


Resin BrCCH2OH O CCH2

Cyanogen bromide activation

� N

ResinO

OH

Epoxy group activation

Toluene sulfonyl chloride activation

�

Resin

O

OO

CCH2

CH2

NH2

NH

CH2 CH2 CH2 CH2

ResinO

O CH2 CH

OH

CH2 O CH2O CH2CH2CH2CH2

Resin O

Resin OH

CH2 CH

OH

CH2 CH2 CH2 CH2 CH2 CH2 CH

OH

CH2O O O

Resin

N

RNH2

R

Ligand

RHO

RNH2

R

R

Ligand

�

Cl CH3S

O

O

�

�

1,4-Butanediol diglycidyl ether

Toluene sulfonyl chloride

Resin CH3S

O

O

O

Resin CH3S

O

O

HONH

Cyanogen bromide

Figure 2-6 Activation of carbohydrate-based resins for covalent attachment of affinity ligands (R).


NH2

SO3�

SO3�

SO3�

N N

Cl

N

N

N

N

H

H

H

O

O

Figure 2-7 Structure of Cibacron Blue F3GA.

“activate” carbohydrate-based resins for derivatiza-tion with a ligand of interest (Fig. 2-6). Most ofthese chemical reactions take place on the hydroxylgroups of the resin and create intermediates thatare subject to nucleophilic attack by a free aminogroup on the desired ligand.

Dye-Ligand Chromatography. Of all the types ofchromatography presented in this chapter, the prin-ciples governing the technique of dye-ligand chro-matography are perhaps the least well understood.Purely by accident in the 1960s, a blue dye knownas Cibacron Blue was found to have strong affinityfor two enzymes found in yeast: pyruvate kinase and phosphofructokinase. As shown in Fig. 2-7,Cibacron Blue consists of a series of individual andfused aromatic rings containing several conjugateddouble bonds common to molecules that absorblight in the visible range of wavelengths. Ring struc-tures similar to this are present in purine nu-cleotides. This molecule also contains three sul-fonic acid (SO3

�) groups that may be mistaken forphosphate (PO4

�) groups by an enzyme. Based onthese two pieces of evidence, it is currently believedthat Cibacron Blue acts as a substrate analog ofADP-ribose, explaining why the molecule has beenfound to have affinity for many different NAD�-requiring enzymes (dehydrogenases) and kinases(commonly bind purine nucleotides).

Unfortunately, the specificity of this blue dyefor the types of enzymes just described is not asgreat as originally thought. Indeed, Cibacron Bluehas been proven to have great affinity for a num-ber of seemingly unrelated proteins, such as -interferon and blood serum albumin. The nonspe-cific interactions between the protein and the dyemay be due to cation exchange effects (due to theSO3

� groups) or hydrophobic and/or hydrogen-bonding effects between the proteins and the aro-matic rings on the dye.

A number of methods have proven successful ineluting proteins from Cibacron Blue. Most often,the greatest purification is achieved through somesort of affinity elution scheme. Here, a protein ofinterest bound at its nucleotide binding site to theresin is eluted from it through the addition of itstrue substrate or a high concentration of a substrateanalog to the mobile phase. This free ligand willcompete with the “fixed” groups on the resin forbinding to the protein, lowering its � value. Sub-

strates or ligands commonly used to elute proteinsfrom Cibacron Blue columns include: S-adenosyl-methionine (SAM), ATP, GTP, NAD�, NADH,ADP, AMP, and cAMP. If a suitable ligand cannotbe identified that will cause the protein of interestto elute from the column, elution can be achievedusing the same mobile-phase gradients commonlyused with cation-exchange columns: increasing pHor ionic strength. Two commercially available dye-ligand chromatography adsorbents are shown inTable 2-1.

Immunoaffinity Chromatography. As will be de-scribed in Section IV, an antibody is capable of se-lectively recognizing and binding a single proteinor other antigen in the presence of thousands ofother proteins and other biological molecules. Nat-urally, then, immunoaffinity chromatography offersperhaps the greatest potential in attempting to pu-rify a protein in a single step. A purified antibody(polyclonal or monoclonal) is covalently bound toa stationary-phase resin through which a solutioncontaining a protein of interest is passed. As thecolumn is washed with an appropriate buffer, all ofthe proteins not recognized by the antibody willelute from the column. Ideally, only the protein ofinterest will remain adsorbed to the resin. Unlikethe other types of chromatography discussed thusfar, the challenge in immunoaffinity chromatogra-phy is in trying to determine the conditions of themobile phase that will lower the � value of the pro-tein, causing it to elute from the column.

Antigen–antibody interactions rank as some ofthe strongest found among biological molecules,with dissociation constants (KDs) as high as 10

�12M.


Typical dissociation constants for these interactionsrange from about 10�8 to 10�10 M. Because thebinding between an antibody and its antigen is sostrong, eluting the protein from the column with-out denaturing it or the column-bound antibodiesis not a trivial matter. Part of the problem is due tothe variety of chemical forces known to play a rolein antigen–antibody complexes, such as hydropho-bic interactions, electrostatic interactions, and hy-drogen bond formation. A mobile-phase gradientdesigned to alter the pH drastically may elute theprotein, but it may also denature it and the anti-bodies derivatized to the column (a very expensiveand time-consuming column to prepare). A mobile-phase gradient of increasing ionic strength may dis-rupt electrostatic interactions between the proteinand the antibody but increase the affinity of the twocaused by hydrophobic interactions.

Although the use of immunoaffinity chro-matography may seem problematic, there are sev-eral precautions that can be taken to increase theodds of successful, nondenaturing elution. First andforemost, monoclonal antibodies work best for im-munoaffinity chromatography. Remember that amonoclonal antibody will recognize a specific re-gion or epitope on a protein with a defined affin-ity (KD). This situation is much more predictableand subject to control than that using polyclonalantibodies, which contain many different antibod-ies that recognize a number of different epitopeswith different affinities. During the screeningprocess for monoclonal antibodies, you could selectfor a clone producing an antibody with an affinityfor the protein that is great enough to allow for pu-rification but not so great as to present problemsduring the elution step (KD on the order of 10�6

to 10�7). Second, you may need to experiment withdifferent mobile phases to try to determine whichforces predominate in the specific antigen–anti-body complex under study (hydrophobic, electro-static, or hydrogen bonding). Reagents and condi-tions commonly used to elute proteins fromimmunoaffinity columns include introduction of adecreasing pH gradient, low concentrations of de-naturing agents such as urea or guanidine hy-drochloride, low concentrations of chaotropic (de-naturing) salts such as lithium bromide andpotassium thiocyanate, and low concentrations ofnon-ionic (nondenaturing) detergents such as Tri-ton X-100.

Immobilized Metal Affinity Chromatography (IMAC).

Immobilized metal affinity chromatography isgrowing in popularity as a means of purifying pro-teins that contain histidine residues on their surface(Fig. 2-8). A resin derivatized with a metal ion-chelating agent is first equilibrated with a buffercontaining a metal ion. Salts containing a divalentcation are most often used for this (Zn2�, Fe3�,Ni2�, Cu2�). A protein solution is then passedthrough the column and washed extensively withbuffer to remove unbound proteins. Proteins thatbind the column through their exposed histidineresidues can be eluted by one of three methods: (1)lowering the pH of the mobile phase, causing thehistidine residue to become positively charged andlose its affinity for the metal ion, (2) addition of ametal-chelating agent (i.e., EDTA) stronger thanthat on the IMAC resin with which the metal ionis complexed, or (3) increasing the concentration ofa compound in the mobile phase that will competewith the protein for metal binding (i.e., imidazoleor histidine). Though sound in theory, certain prac-tical problems prevent widespread use of this typeof chromatography. First, since the resin carries astrong negative charge due to the presence of themetal chelator, it is possible that cation exchangeeffects may occur if all the sites containing thechelator are not complexed (saturated) with themetal ion in the mobile phase. Second, the longspacer arms commonly used in IMAC resins oftenbind a number of proteins nonspecifically throughhydrophobic interactions. This latter problem iscompounded by the fact that IMAC columns areoften developed under conditions of high ionicstrength in an attempt to minimize the potentialion-exchange effects. Despite these problems,IMAC has proven to be a powerful tool in the pu-rification of a number of different proteins.

Partition Chromatography. Unlike the othertypes of chromatography discussed thus far, thistype is usually used for analytical (rather thanpreparative) applications. In addition, this type ofchromatography is often performed in a thin-layer(rather than a column) format. The stationary phasein partition chromatography is usually a glass plate(rigid) or polyester sheet (flexible) coated with avery thin layer of the desired adsorbent. For mostapplications, the adsorbent is a cellulose, silica,polyamine, or aluminum oxide–based matrix. In


OCH2

CH2

CH CH

OH

C O�

O

CH2 C O�

O

CH2 C O�

N

Fe3�

Fe3�

Fe3�

Solidsupport

OCH2

CH2

CH CH

OH

C O�

O

CH2 C O�

O

CH2 C O�

N

Ni2�

Ni2�

Ni2�

Solidsupport

6 � His Protein

Phospho-protein

Figure 2-8 Immobilized metal affinity chromatography resin (nitrilotriacetic acid).

some cases, these types of adsorbents can be de-rivatized with DEAE, polyethyleneimine, acetyl, orC18 groups.

In normal-phase partition chromatography, thestationary phase is polar and the mobile phase usedin development is nonpolar. In reverse-phase par-tition chromatography, the stationary phase is non-polar and the mobile phase used in development ispolar. In thin-layer partition chromatography, thesamples are spotted in a straight line across the bot-tom of the solid support (at the origin). Care mustbe taken to avoid widespread diffusion of these sam-ples (small aliquots of the total applied sample arespotted and allowed to dry). After all the appliedsamples have dried, the plate is placed, origin sidedown, in a chamber containing a small amount ofthe mobile-phase solvent on the bottom. It is es-sential that the origin lies above the level of the mobile-phase solvent in the chamber (see Fig. 6-8)and that the chamber is saturated with the vaporsof the mobile phase before development is begun.

The mobile phases used in the development ofthin-layer chromatography plates are usually com-binations of organic and aqueous solvents. Ethanol,t-amyl alcohol, acetonitrile, and methanol arecommonly used organic solvents. Acetic acid is themost commonly used aqueous solvent. In normal-phase partition chromatography, the polar station-ary phase is developed with relatively nonpolar mo-

bile phase. Compounds will display different ratesof migration through the stationary phase depend-ing on their polarity. More-polar compounds willhave more affinity for the polar stationary phasethan for the nonpolar mobile phase. As a result, themore-polar compounds will travel less distanceacross the plate than the nonpolar compounds,which have a higher affinity for the nonpolar mo-bile phase. The opposite is true for reverse-phasepartition chromatography: the more polar the com-pound, the greater its affinity for the mobile phase,the further that compound will travel across theplate in a given period of time. The behavior of acompound in thin-layer partition chromatographyis usually described in terms of relative mobility(Rf):

Rf �

The larger the value of Rf, the more affinity thecompound has for a particular mobile phase sol-vent.

The major advantages of thin-layer partitionchromatography are cost and versatility. Comparedto column chromatography and HPLC, thin-layerchromatography is very inexpensive to perform. Experimentation with different adsorbents and mobile-phase compositions will allow the separa-

Distance traveled by sample (cm)��Distance traveled by solvent front (cm)


tion of a wide variety of biological molecules. Thin-layer chromatography has been used successfully inanalysis of proteins, hormones, amino acids, aminosugars, carbohydrates, nucleotides, antibiotics, fattyacids, pesticides, and a number of other drugs andmetabolic intermediates.

A variation of this same technique uses paper asthe stationary phase. Since all commercially avail-able chromatography paper is composed of a poly-saccharide matrix saturated with water or someother polar solvent, paper chromatography is mostoften carried out in the normal-phase format usinga nonpolar mobile phase for development. It is evenpossible to perform reverse-phase paper chro-matography by saturating the paper with some non-polar solvent (such as liquid paraffin) and develop-ing it with a more-polar solvent. The drawback tothis technique, however, is that this type of adsor-bent is not commercially available and must be usedimmediately after it is prepared. The techniques ofthin-layer and paper chromatography are describedin greater detail in Experiment 6.

One final variation of partition chromatographyexploits the affinity of different molecules in the gasstate for a solid adsorbent. This technique is termed“gas chromatography,” and it works particularlywell for the separation of nonpolar compounds. Inthis technique, a liquid sample is vaporized and de-livered to a thin-diameter, heated coil coated witha solid matrix or viscous liquid (grease) possessinga very high boiling point. The vaporized sample isdelivered to the coil using an inert gas, such as ni-trogen, helium, or argon. Volatile molecules thathave affinity for the solid or liquid matrix will beretarded as the mobile-phase gas carries the sam-ple through the coil. A detector linked to the sys-tem at the end of the column will allow you to iden-tify different compounds as they elute from the coil.

High-Performance (Pressure) Liquid Chromatog-

raphy. Successful chromatography relies on theability to maximize the number of chromatographicplates transfers while maintaining good resolution(minimizing zone spreading). Therefore, it wouldbe desirable to minimize the volume of the columnand still provide a reasonable number of chro-matographic plates to allow separation of the mol-ecule of interest. As stated earlier, a chromato-graphic plate can be described as the largest area ofthe column where two molecules with different par-

tition coefficients will have an opportunity to dis-play different rates of migration. The number ofchromatographic plates in a given column volume,then, will be related to the surface area of the ad-sorbent (stationary phase) with which the com-pounds passing through the column can interact. A5-ml column containing a stationary phase matrixwith a 200-M pore size will provide less effectivesurface area for the compounds to interact with thana 5-ml column containing a stationary-phase ma-trix with a 5-M pore size. In other words, the 200-M-pore-size stationary phase will provide fewerchromatographic plates per unit volume than the5-M-pore-size stationary phase.

Although stationary phases with small pore sizesprovide more chromatographic plates per unit vol-ume than those with larger pore sizes, smaller-pore-size matrices provide more resistance to the flow ofthe mobile phase passing through it. As a result,high-pressure mobile-phase delivery systems(pumps) have been developed that are capable ofdriving the mobile phase through small-pore ma-trices at pressures exceeding 7000 psi. Since thepressure on the stationary phase increases with theflow rate of the mobile phase, the flow rate that aparticular HPLC column can accommodate will de-pend on the strength or rigidity of the matrix. Ingeneral, the pressure on the stationary phase in-creases as the column diameter decreases and col-umn length increases. Therefore, a 20-mm-by-100-cm column will experience less pressure at aflow rate of 1 ml/min than will a 5-mm-by-100-cmcolumn. HPLC columns of different dimensionsand different matrix composition are now com-mercially available that can sustain flow rates ashigh as 100 ml/min or as low as 0.5 ml/min.

Another feature of modern HPLC systems thatmakes them desirable for both analytical andpreparative applications is the complex mobile-phase gradients that they are capable of producing.Many systems come equipped with a pump inte-grator or controller (computer) that allow a num-ber of different mobile-phase solvents to be simul-taneously mixed and delivered to the stationaryphase. Since this process is automated, complex gra-dients used for a particular application are quite re-producible.

As with most technologies, HPLC is not with-out some disadvantages. First, and perhaps fore-most, is cost. Modern HPLC systems, along with


the specialized tubing, fittings, and columns neededto operate them, are quite expensive to acquire andmaintain. Second, extreme care must be taken toensure that both the sample and the mobile-phasesolvents are free of any particulates that could clogthe column, increase the internal pressure, andcrush the stationary phase. Third, the mobile-phasesolvents must be degassed to prevent the formationof air bubbles in the system. If this occurs, the in-tegrity and reproducibility of the applied gradientwill be compromised. Finally, the smaller dimen-sions of most HPLC columns as compared withmore traditional gravity-driven columns limits thesize of the sample that can be applied. If HPLCcolumns are “overloaded,” resolution will decreaseand the number of available chromatographic platesper unit volume will not be realized.

Chromatography systems that operate withpressurized mobile-phase gradients are undoubt-edly the wave of the future. Solid supports capableof withstanding the pressures associated withHPLC have already been developed for many ofthe types of chromatography discussed in this sec-tion (normal- and reverse-phase partition, ion ex-change, gel filtration, etc.). In recent years, thetrend in technology has been toward the develop-ment of high-capacity, large-pore-size solid sup-ports that can operate at high flow rates (low pres-sure) and still maintain good resolution. Systemsthat operate under this principle are commonly re-ferred to as fast protein liquid chromatography(FPLC) systems or low-pressure liquid chromatog-raphy (LPLC) systems.

There is no doubt that the technique of chro-matography (particularly affinity-based) will con-tinue to make large advances as our understandingof the molecular mechanisms governing proteinfolding and enzyme–substrate interactions in-

creases. Thousands of proteins and other moleculesof biological interest have been purified using com-binations of the different types of chromatographydescribed in this chapter. Although no type of chro-matography is without problems, each type can pro-vide a great deal of purification when used at thecorrect time under optimal conditions as deter-mined through thoughtful and carefully performedexperiments.


5- to 8-mm (inside diameter) glass tubing, or equivalentchromatography columns

CM-Sephadex (G-50) in 0.1 M potassium acetate, pH 6.0

CM-Sephadex (G-50) in 1.0 M potassium acetate, pH 6.0

0.1 M Potassium acetate, pH 6.0

1.0 M Potassium acetate, pH 6.0

Glass wool

Mixture of blue dextran (1 mg/ml), cytochrome c(2 mg/ml) and DNP-glycine (1 mg/ml)

Tape

Pasteur pipette with dropper bulb

Protocol

Demonstration of Chromatographic

Separation on CM-Sephadex (G-50)

This simple exercise demonstrates the chromato-graphic separation of the components of a mixtureby a procedure that relies on two of the principlesdiscussed above, namely ion-exchange chromatog-raphy and size-exclusion chromatography. Thecomponents to be separated (see Table 2-3) differgreatly from one another both in their molecular

Table 2-3 Compounds in Colored Solution Applied to CM-Sephadex

Name Color Molecular Weight Ionic Character

Blue dextran Blue �500,000 Nonionic

Cytochrome c Red 12,400 Ionic protein, pI � 10.7

(i.e., net cation below pH 10.7)

DNP-glycine Yellow 241 Anion above pH 3.0


Clamp on aring stand

Tape toidentify column

Fluid aboveresin bed

Glass wool plug

Adjustable pinchclamp on rubbertubing tip

Figure 2-9 Setup for demonstration of proper-

ties of CM-Sephadex.

weight (the basis of their separation by size-exclusion chromatography on Sephadex G-50) andin their net charge at pH 6.0 (exploited for theirseparation by ion exchange on CM-Sephadex).Each component also has a distinctive chro-mophore, so that their separation is easily followedby simply observing the color of the componentsas they separate.

1. Heat a 14- to 16-in. piece of 5- to 8-mm in-side diameter glass tubing and pull out a con-striction in the center of the tube. Cut the tubeat the constriction and mount each of the 2 sec-tions as shown in Figure 2-9.

2. Using a small-diameter glass rod or a Pasteurpipette, push a small piece of glass wool intothe throat of each column, being careful not topush the glass wool beyond the beginning ofthe constriction of the column. Next, label twostrips of tape, one “CM-Sephadex in 0.1 Mpotassium acetate, pH 6.0” and the other “CM-Sephadex in 1.0 M potassium acetate pH 6.0.”Attach the two tapes to the separate columns(see Fig. 2-9).

3. Using a Pasteur pipette equipped with a drop-per bulb, begin to add well-mixed slurries ofCM-Sephadex in potassium acetate solutions tothe appropriate columns. Allow the columns todrip while adding increasing increments of theslurried CM-Sephadex until the settled resinbed reaches a height of 4–5 in. Do not allow thecolumns to run dry; that is, do not allow the topfluid surface to penetrate into the settled resinbed. Instead, add small aliquots of 0.1 M potas-sium acetate buffer, pH 6.0, or 1.0 M potas-sium acetate buffer, pH 6.0, to the appropriatecolumns to keep the fluid surfaces above theresin bed.

4. Once a 4- to 5-in. defined resin bed has accu-mulated in each column, pipette off most of thetop fluid with a Pasteur pipette, and then allowthe columns to drip until the fluid surface justreaches the resin bed.

5. Immediately add 0.2 ml of the colored solution(containing each of the components in Table 2-3) so that it forms a thin band on top of theresin. (NOTE: Do not disturb the top of theresin.) Allow the colored solution to penetratethe columns until the fluid surface again meetsthe resin bed, then gently add a few drops of

the appropriate potassium acetate buffer, pH6.0, as designated by the taped label on eachcolumn. Allow this buffer to penetrate into thecolumn, and then slowly fill up the rest of thecolumn with the potassium acetate buffer of theconcentration that matches the column label.Replenish these upper fluid volumes as neces-sary while observing the course of the elution.


2. You have an aqueous solution containing:

alanine (a monoamino, monocarboylic amino acid)

fructose (a non-ionic monosaccharide)

glycogen (a non-ionic large polysaccharide)

ribose-5-phosphate (an anionic monosaccharidephosphate)

tRNA (a polyanionic nucleic acid, MW � 30,000)

Assuming you have a distinctive assay for eachof these compounds, what procedures wouldyou use to obtain gram quantities of each ofthese compounds free of each of the other com-pounds?

REFERENCES

Block, R. J., Durrum, E. L., and Zweig, G. (1955) AManual of Paper Chromatography and Paper Elec-trophoresis. New York: Academic Press.

Brenner, M., and Neiderwieser, A. (1967). Thin-LayerChromatography (TLC) of Amino Acids. EnzymeStructure. Methods Enzymol 11:39.

Fischer, L. (1969). An Introduction to Gel Chromatogra-phy. New York: American Elsevier.

Jakoby, W. B. (1971). Enzyme Purification and RelatedTechniques. Methods Enzymol 22.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). An Introduction to Proteins. In: Principles ofBiochemistry, 2nd ed. New York: Worth.

Robyt, J. F., and White, B. J. (1990). ChromatographicTechniques. In: Biochemical Techniques: Theory andPractice. Prospect Heights, IL: Waveland Press.

Scopes, R. K. (1994). Protein Purification: Principles andPractice. New York: Springer-Verlag.

Wilson, K., and Walker, J. (1995). ChromatographicTechniques. In: Principles and Techniques of PracticalBiochemistry, 4th ed. Hatfield, U.K.: CambridgeUniversity Press.

6. Prepare a description of your results that in-cludes an explanation for the degree of migra-tion of each colored component in bothcolumns. Once the first colored reagent beginsto exit from the low-salt column (i.e., the col-umn containing 0.1 M potassium acetate buffer,pH 6.0), replace this elution fluid with thehigh-salt buffer (i.e., 1.0 M potassium acetate,pH 6.0), and observe and explain the results.Which colored component is blue dextran,which is cytochrome c, and which is DNP-glycine? Explain the order in which they elutein terms of their physical properties and the na-ture of the CM-Sephadex G-50 column. Fi-nally, predict and justify the results that wouldhave occurred if the same experiment had beenrun with a CM-Sephadex column equilibratedin and eluted with 0.l M HCl.

7. After the relative movements of all the coloredcomponents have been observed under the var-ious elution conditions specified, remove thecolumns from their clamps, take them to a cen-tral container for used CM-Sephadex, andempty the used gel-filtration ion exchangerinto the container by shaking or blowing outthe resin. (Discard the glass wool plug; do notleave it in the CM-Sephadex resin. The resinwill be “regenerated” by appropriate washingsand used again.)

Exercises

1. What is the chemical nature of the chro-mophore of each of the three materials sepa-rated on CM-Sephadex in this experiment; thatis, what is the light absorbing component thatgives it a visible color? Describe the nature ofthe linkage of the chromophore to the rest ofthe molecule.

q )

45

the element and atomic number in question. Thesubscript value is often omitted so that 12C, 13C,and 14C depict the three isotopes of carbon.

Certain combinations of nuclear protons andneutrons for an atom are stable and result in mul-tiple stable isotopes of an atom (e.g., 12C and 13C).In contrast, other combinations of protons and neu-trons are unstable, which results in spontaneous nu-clear transformations of nuclear mass and charge(e.g., 14C). There are a variety of nuclear transfor-mation mechanisms by which unstable isotopesachieve a stable state. Many of these mechanismsresult in emission of specific particles. Such unsta-ble, particle-emitting isotopes are called “radioiso-topes,” or radioactive forms of an element. Fortu-nately for biochemists, the emitted particles can bedetected readily. Thus, introduction of a radioac-tive isotope into a specific biologically importantpopulation of molecules allows tracing of the pas-sage or conversion of that radiolabeled molecularspecies through biochemical reactions or biologicalprocesses. It is this ability to follow the rate and ex-tent of flow of specific molecules through biologi-cal processes—radiotracer technology—that makesthe use of radioisotopes important to biochemists.

Any radioisotope decays by one or, at most, onlya few nuclear transformation mechanisms. Thecharacteristics of the emitted particles are suffi-ciently different from one another that the proce-dures for detection of the different particles mustalso be different. Therefore, a fundamental knowl-edge of radioisotope techniques in biochemistry be-gins with a study of the particles produced duringdifferent mechanisms of nuclear transformation.Because the great majority of biochemical applica-

EXPERIMENT 3

Radioisotope Techniques

Theory

Nuclear Transformations

An atom can be conceptualized as a central nucleus,composed of protons and neutrons, surrounded bya specific cloud of electrons, whose energy levelsare defined by the laws of quantum mechanics. Pro-tons carry a mass of 1 and a charge of �1, neutronscarry a mass of 1 but lack a charge, and electronshave a relatively insignificant mass but carry acharge of �1. In a neutral atom the number of elec-trons equals the number of protons in the nucleus.Ions are atoms from which one or more electronshave been added or removed, creating a species witha net charge. The atomic number (i.e., the numberof protons in the nucleus) and the correspondingcloud of electrons largely determine the chemicalproperties of an atom. Neutrons present in any nu-cleus therefore increase the mass of the atom inquestion, but do not influence the charge or atomicnumber of the atom. Atoms with a constant nuclearcharge (constant number of protons) but with var-ied masses (varied number of neutrons) are calleddifferent “isotopes” of an atom. Different isotopesof an atom react alike, having very nearly the samechemical properties.

Chemists and physicists use specific symbols todepict these mass and charge relationships. For anyatom X, superscript values denote the nuclear massof the atom (number of protons plus neutrons),whereas subscript values denote the atom’s atomicnumber (number of protons). Thus 12

6C, 136C, and

146C are three different isotopic forms of the ele-

ment carbon. In this convention, the letters define


Table 3-1 Decay Properties of Some Biochemically Important Isotopes

Isotopic Tracer Average Energy (MeV) Maximum Energy (MeV)

or Label Reaction of �� Particle of �� Particle (Emax) Half-Life

13H 31H 88n

32He � �� 0.0055 0.015 12.3 years

14C 146C 88n

147N � �� 0.05 0.15 5500 years

32P 31

25P 88n

31

26S � �� 0.70 1.71 14.3 days

35S 31

56S 88n

31

57Cl � �� 0.0492 0.167 87.1 days

33P 31

35P 88n

31

36S � �� 0.25 25 days

0.05

3H

Energy (MeV)

Num

ber

of �

� p

artic

les

0.10 0.15

14C

Figure 3-1 �� energy profiles of 3H and 14C.

tions involve only two types of radioactive decay,namely, �� emission and � emission, only these twokinds of nuclear transformation and emitted parti-cles will be discussed in detail.

�� Particle Emission

The process of �� particle emission occurs in thebiologically important radioisotopes indicated inTable 3-1. � particles are negatively charged parti-cles of negligible mass (i.e., electronlike particles).� particles are emitted along with antineutrinosduring the conversion of neutrons to protons.

Neutron 88n proton� � �� antineutrino

Because the newly formed proton remains in thenucleus after discharge of the �� particle, the prod-uct atom has an atomic number increased by one;for example, carbon is transformed to nitrogen andphosphorus is transformed to sulfur (see Table 3-1).Antineutrinos have no charge and essentially nomass, and they are therefore hard to detect. Becauseof this, antineutrinos are of little significance to bio-chemists. �� Particles are readily detected, how-ever, because of their charge and the energy withwhich they are expelled from the decaying nucleus.The total energy available from a neutron-to-pro-ton transformation is shared randomly by the ��

particle and the antineutrino. Thus, the �� particlemay be emitted at a variety of different energies (aspectrum) up to the maximum energy available inthe transformation. The maximum energies and theassociated �� particle energy profiles for a given��-emitting radioisotope are therefore distinct(Fig. 3-1). These differences in �� emission ener-

gies play an important role in radioisotope detec-tion and differentiation in so-called double-label ex-periments, in which more than one radioactive iso-tope is assayed in a reaction or process (seeExperiment 11).

� Particle Emission

Many particle emission processes in radioisotopesof higher nuclear masses follow a more complex,multistep decay process that results in both parti-cle emission (e.g., �� particle emission) and emis-sion of high energy photons called � particles or �rays. Figure 3-2 depicts a typical scheme for the ori-gin of � rays during a nuclear decay event.

� Ray decay systems are different from �� de-cay events in three ways. First, these systems mayproduce more than one � ray. Second, the resultant� rays possess distinct energies rather than a spec-

EXPERIMENT 3 Radioisotope Techniques 47

Unstable intermediate �– Emission, Emax = 0.314 MeV

� Emission, E = 1.173 MeV

� Emission, E = 1.332 MeV

60Co

60Ni

Unstable intermediate

Figure 3-2 Predominant decay scheme for 60Co.

trum of energies. Third, a � ray decay system mayoriginate from unstable isotopes that initially emitparticles other than �� particles (e.g., � particles).� Ray emission usually occurs soon after the emis-sion of the primary particle (�� in Fig. 3-2). Theemission of the initiating primary � or �� particleis often hard to detect, whereas fairly efficient sys-tems exist to detect � rays. � Ray spectroscopy istherefore the best way to detect � ray emitting ra-dioisotopes, such as 125I, 59Fe, or 60Co.

Other Particles

A number of other nuclear transformation mecha-nisms are found in nature. � Particle decay, positronemission, and electron capture events are examplesof such nuclear transformations. Techniques havebeen developed for measuring such decay events.However, the radioisotopic atoms that yield suchdecay products are used relatively infrequently inbiochemistry and biology. If your research requiresanalysis of such atoms, you should consult a prac-tical textbook in modern physics or radioisotopictechniques.

Nuclear Decay Rates

All nuclear transformations proceed spontaneouslyat rates that are not altered by ordinary chemicalor physical processes. For any population of unsta-ble atoms, the rate of nuclear transformation or ra-dioactive decay is first order; that is, proportionalto the number, N, of decomposing nuclei present:

(3-1)

� �N

where t is time and � is a disintegration constantcharacteristic for a particular isotope.

At t � 0, N0 atoms are present. At other times,N atoms are present. Integration of Equation 3-1within the limits of t � 0 and t is:

(3-2)

�N

N0

� �� t

0dt

which yields:

(3-3)

N � N0e��t

or in natural logarithms:

(3-4)

ln � ��t

Scientists frequently express Equation 3-4 interms of half-life (t1/2), which is the time requiredfor one half of the population of unstable atoms todecay. In other words, after one half-life, the rateof particles emitted has decreased to one half therate of the starting population of unstable atoms.Substitution in Equation 3-4 yields Equation 3-5:

(3-5)

ln � ln (1/2) � ��t1/2 or t1/2 �

This equation, in combination with half-lives (seeTable 3-1) and Equation 3-4, allows calculation ofthe quantity of unstable atoms that will be present

0.693

�

1/2 N0

N0

NN0

dNN

dNdt


10,000

1,000

2Error (%)

Cou

nts

(plo

tted

on lo

g 10 s

cale

)

4 6 8 10 12 14 16 17

Figure 3-3 Counts versus percent error relation-

ship.

at any time t in a population of atoms containing aknown quantity of unstable atoms or known ra-dioactive decay rate.

Statistics and Units of Radioactive Decay

Radioactive particle emission is a random process;that is, the timing of emission of individual parti-cles over time is random. You must therefore de-tect a large number of particles or counts to definestatistically significant count rates (e.g., counts perminute, cpm). Figure 3-3 illustrates the relationshipbetween total recorded counts and the statistical ac-curacy of the observed count rate. In general, allradioactive samples should be counted to a valuewithin the 5% error limit of other experimentalerrors (e.g., pipetting error) within an experiment.In some cases, the design of the experiment mayjustify accumulating counts per minute to a higherstatistical accuracy, such as 1 or 2%. These sta-tistically significant data of counts per minute canthen be converted to data of disintegrations perminute with knowledge of the counting efficiencyof the system (see below).

The basic unit of radioactive particle emissionis the curie, Ci. Originally, the curie was defined asthe number of disintegrations per second emittedfrom a gram of impure radium (3.7 � 1010/s or2.2 � 1012/min). This unit and its subdivisions themillicurie, mCi (2.2 � 109 dpm), the microcurie,�Ci (2.2 � 106 dpm), are the fundamental units todefine radioactive particle emission rates. Radioac-

tive compounds display different rates of particleemission depending on the number of unstable, potentially radioactive atoms present and the disin-tegration constant of the radioisotope. These differences in radioactivity among compounds areexpressed as radioactive specific activity; that is, par-ticle emission rates per amount of compound. Themost convenient units of radioactive specific activ-ity are curies and moles and their respective subdi-visions (e.g., curies per mole, millicuries per mil-limole, microcuries per mole), but certain instancesmay require radioactive specific activity units suchas millicuries per milligram or counts per minuteper gram.

You may also encounter the SI unit of radioac-tivity, the Becquerel (Bq), which is defined as 1 dis-integration per second (dps). In this system 1 curieequals 3.7 � 1010 Bq. For interconversion, it is use-ful to note that 1 �Ci � 3.7 � 104 Bq and 106 Bq �27.027 �Ci.

Measurement of Radioactive Decay

Biochemists most frequently make practical use ofradioisotopic compounds that emit �� and � par-ticles. A number of different techniques have beendeveloped for detecting the rates and energies withwhich these particles are released during radioac-tive decay. We will discuss in detail two of the meth-ods most commonly used in biochemical research.These are liquid and solid scintillation counting andradioautography using photographic and phosphorstorage techniques.

Scintillation Counting. Scintillation counting isused to detect both �� and � particles. Differencesexist in the detection process for these different par-ticles, but in both cases, emitted particles cause aseries of brief light flashes that are detected by aphotocell and recorded as counts for quantitativeinterpretation. Solid scintillation counting is pre-ferred for the measurement of � particles. Thedense array of atoms in the detection crystal, whichis usually sodium iodide, optimizes the chances ofabsorption of the � particle, which can be describedas a highly energetic x-ray. On the other hand, liq-uid scintillation counting is preferred for the de-tection of � particles, which are of relatively lowenergy and must be intimately mixed with the mol-ecules used for detection.


3H

Voltage from photomultiplier tube

Num

ber

of e

vent

s

14C

Figure 3-4 Photomultiplier-tube responses from3H and 14C decay events in a scintil-

lation cocktail.

Liquid Scintillation Counting of �� Emissions. Liq-uid scintillation counting is the most efficient tech-nique to detect �� particle emission. The light-emitting process that is the basis of scintillationcounting occurs as a result of a radioactive samplebeing placed in a “scintillation cocktail” containingan excitable solvent and one or more fluorescentcompounds, called “fluors.” Emitted �� particlescontact solvent molecules (S) and transfer some oftheir energy to these molecules. This yields an ex-cited solvent molecule (S*) and a �� particle con-taining less energy (E):

(3-6)

�� S 88n S* � (�� E)

The energy required to excite a solvent moleculein this reaction is often small compared to the en-ergy of a given �� particle. Therefore, the residual�� particle (�� E in Equation 3-6) may excitemany solvent molecules before its energy is de-pleted. The eventual number of excited solventmolecules is therefore proportional to the energyof the �� particle in question.

Excited solvent molecules transfer their energyto other solvent molecules, and eventually to a primary fluor molecule (F*

1), which then emits aphoton (h 1) and decays to its ground state (Equa-tions 3-7 and 3-8). A widely used fluor is 2,5-diphenyloxazole (PPO).

(3-7)

S* � F1 88n F*1 � S

(3-8)

F*1 88n F1 � h 1

Some scintillation cocktails may employ a sec-ondary fluor that absorbs photons from the primaryfluor decay. The excited secondary fluor thenreemits new photons at a wavelength more favor-able for detection of experimental materials in thescintillation cocktail or more favorable to the pho-totubes of the particular scintillation counter. How-ever, most modern scintillation counters no longerrequire the use of a secondary fluor.

The photons emitted by the fluor are detectedby a phototube-photomultiplier system. The life-time of the �� particle passage, solvent excitation,and eventual photon emission from fluors is short

compared with the response time of the photo-multiplier tube system. Therefore, all the events froma single �� emission are scored by the photomultipliertube system as a single “pulse,” the energy (voltage) ofwhich is proportional to the number of photons developedby the �� particle emission event. Summation of a se-ries of such pulses from a given isotope recreates,in electrical terms (voltage), a curve proportional tothe energy profile of an isotope (Fig. 3-4; comparethis to Fig. 3-1). The individual voltage pulses thatgenerate the curves of Figure 3-4 (i.e., the areas un-der the curves) are passed on by the circuitry of thescintillation counter to a scaler system. This systemrecords the pulses as individual counts, and accu-mulation of them over a period of time leads to acounts per minute (cpm) value.

This is the basic liquid scintillation countingprocess. However, a working knowledge of scintil-lation counting requires familiarity with additionalfeatures of modern scintillation counters. Considera modular diagram of a typical scintillation counter(Fig. 3-5). (Frequent design changes and design dif-ferences among manufacturers make it difficult todefine a typical scintillation counter. Figure 3-5 isa generalization.)

The overall energy of the photons from a sin-gle �� emission is often quite small. The photo-multiplier tube used to detect such low-energyevents must therefore be very sensitive, but this sen-sitivity also causes it to detect “photomultipliernoise,” or spurious counts unrelated to actual ��

particle emissions. This noise is greatly reduced by


Photomultipliertube

Photomultipliertube

Sample

Tap or gain voltage

Coincidence side Analysis side

Signalsummation

Signalsummation

Channel 1discriminator

Scaler

Channel 2discriminator

Figure 3-5 Unit diagram of a typical scintillation counter.

use of a coincident circuit that employs two pho-tomultiplier tubes. With such coincidence systems,both photomultiplier tubes must simultaneouslydetect a light flash to yield a count in the scaler.Thus, most noise events, which occur in a singlephototube independently of the other phototube,are screened out. However, the coincident circuitryis also potentially detrimental in that it can excludedetection of low-energy �� emissions. For exam-ple, certain low-energy �� emissions may not gen-erate enough excited solvent molecules to allowphoton detection by both photomultiplier tubes(e.g., only one photon may be formed). This limi-tation prevents coincident circuit scintillation coun-ters from achieving 100% efficiency (i.e., detectionof all of the �� particle decay events), and is themajor limitation in the detection of low-energy ��

particles such as those from 3H.As seen in Figure 3-5, both the analysis side and

the coincidence side of the scintillation countercontain discriminators, which are particularly use-ful in experiments that employ two or more ��

emitting isotopes. Discriminators contain elec-tronic “gates” that limit the portions of an energy

profile observed by each channel. This is best il-lustrated by the example of a sample containingboth 3H and 14C (Fig. 3-6). You can adjust the dis-criminator in Channel 1 to detect only events thatproduce the energies or voltages between A and B,and adjust the Channel 2 discriminator to detectonly events between B and C. Such settings dictatethat Channel 2 will detect 14C independently of anyquantity of 3H present in the sample. Channel 1will detect 3H, but will also detect a significantnumber of counts from low-energy 14C decayevents. Counting a 14C standard (i.e., a 14C samplewith a known quantity of radioactivity) in bothChannel 1 and Channel 2 allows you to determinethe percentage of the total 14C counts detected inChannels 1 and 2. This allows you to correct forthe “overlap” of 14C counts within the 3H channel(Channel 1). Thus, independent channels and ap-propriate discriminator settings provide a usefultechnique for double-label experiments (see Exper-iment 11).

The “gain” voltage applied to the photomulti-pliers of the phototube–photomultipliers definesthe degree of amplification these units provide to


3H

Energy of �– particle or voltage from photomultiplier tube

Channel 2Channel 1

A B C

Num

ber

of e

vent

s

Num

ber

of e

vent

s

14C 3H

Channel 2Channel 1

A B C

14C

Figure 3-6 Discriminator settings for simultaneous detection of 3H and 14C in linear amplification (e.g.,

Packard Inst.) at left, and logarithmic amplification (e.g., Beckman), at right.

3H

Voltage from phototube–photomultiplier

Channel 2Channel 1

A B C

Num

ber

of e

vent

s

14C

Figure 3-7 Effect of gain voltage on energy pro-

files of 3H and 14C.

the voltage pulses from the phototubes. This has adirect bearing on the desired discriminator settingsof the channels. Consider the two gain settings ofFigure 3-7. The solid lines in Figure 3-7 representan optimal gain voltage on the photomultipliersthat cause the energy profiles to fall within the pre-set channels. In contrast, use of a gain setting thatis too high (dotted lines) yields profiles where some3H decay events appear as 14C events, while somehigh-energy 14C events go unrecorded (i.e., areabove the 14C gate). It is important to rememberthat no single gate setting is optimal for all count-ing conditions; the nature of the scintillation fluid

used will determine the photomultiplier gain to beused.

As this example illustrates, gain voltage and dis-criminator or channel settings are interrelated. Dif-ferent makes of scintillation counters vary in thecontrol they allow the operator. Many current mod-els have fixed (preset) channels and variable gainvoltage, which is adjusted to bring the observedcounts into the preset channels. Others have highlyflexible channels and sophisticated amplification ofthe energy profiles. Most manufacturers also offermodels with varying numbers of independent chan-nels and varying counting sequences on these chan-nels. A clear understanding of the interaction of amplification of energy profiles and control of dis-criminators will suffice for all these differences inmakes and models of scintillation counters.

Measurement of radioactive decay can also beaffected by various components present in, or addedto, the scintillation cocktail. These components cancause quenching; that is, they can decrease the ef-ficiency of the scintillation process. Scintillationcounting provides data in counts per minute (cpm).Quenching dictates that the counts per minute de-tected is less than the actual decay rate, or disinte-grations per minute (dpm). Almost every sample en-countered experimentally is quenched to somedegree; for example, O2 picked up by the scintilla-tion fluid from contact with air serves as a quencher.Therefore, researchers frequently count an addi-tional sample containing a standard of known de-


3H

Voltage from photomultiplier tube

A B C

Num

ber

of e

vent

s

Unquenched

Low level of quencher

High level of quencher

Unquenched

Low level of quencher

High level of quencher

14C

Figure 3-8 Effect of quenching on observed ��

energy profiles.

cay rate (dpm) along with the experimental samplevials. Comparison of the observed counts perminute with the known disintegrations per minuteallows you to determine the counting efficiency ofa given scintillation system.

(3-9)

% Counting efficiency � � 100%

Quenching can occur in three ways. Some ofthese are quite simple, as in color quenching, in whichthe color of the sample absorbs photons emittedfrom the fluor before they are detected by the pho-tomultiplier tube. Another type of quenching ispoint quenching, in which the sample is not solubi-lized in the scintillation fluid, and �� particles areabsorbed at their origin before they contact the sol-vent. Color quenching can usually be overcome byrefining the sample before adding the scintillationfluid; point quenching can frequently be eliminatedby adding specific solubilizers or detergents. Themost important and subtle quenching is chemicalquenching. This phenomenon occurs when variouscompounds present in the scintillation vial decreasethe counting efficiency by interacting with excitedsolvent or fluor molecules so as to dissipate the en-ergy without yielding photons. Chemical quench-ing usually cannot be eliminated, so methods havebeen developed to quantify and correct for it.

The methods often used to analyze chemicalquenching make use of the characteristic impactthat chemical quenching has on the apparent en-ergy profile of radioisotopes that emit �� particles(Fig. 3-8). Some radioisotopes are more prone tochemical quenching than others (see Fig. 3-8). Ingeneral, the higher the average energy of the emit-ted �� particles, the less chemical quenching is ob-served. Thus, high-energy �� emitters (such as 32P)are not sensitive to chemical quenching. Also,chemical quenching agents vary in their quenchingefficiency per gram or per mole of quenching agent.Thus, the exact degree of quenching is potentiallyhighly variable, yet the overall effect of chemicalquenching is a reduced counting efficiency.

If we set up arbitrary channels, such as A n B,B n C, and A n C on the 14C energy profiles ofFigure 3-8, we can measure chemical quenching forthis radioisotope. An examination of the ratio of

cpmdpm

counts in the A n B window to the counts in theA n C window shows that increased quenching in-creases the A n B/A n C counts ratio. You canconstruct a plot that relates decreases in countingefficiency (that is, degree of quenching) to the A n

B/A n C counts ratio. Such a plot (Fig. 3-9) formsthe basis for the channels ratio method of quenchanalysis. These plots are obtained from determina-tions of channels ratio values on a series of samplescontaining a known quantity of a known isotopeand increasing amounts of a chemical quencher.

The channels ratio method makes use of exist-ing counts within the sample vial. This method issuitable when large numbers of counts are present,but it becomes very time consuming with samplescontaining few counts, because a long time is re-quired to accumulate sufficient counts for statisti-cal accuracy. Most modern scintillation counterstherefore employ an automatic external standard-ization system of quench analysis to avoid the timerequired for the internal channels ratio method.This method utilizes a specially selected external �radiation source carried in a lead-shielded chamberthat is buried in the instrument. Before the regu-lar counting of the sample, the external standard is


100

0

0

100

AA

BC

Cou

ntin

g ef

ficie

ncy

(%)

Que

nchi

ng (

%)

Figure 3-9 Channels ratio quench correction plot

for 14C. Quench correction data may

curve slightly up or down depending

on the make and model of counter

used.

Sample vialExternal standardmoves up tubeto point nextto sample vial

Lead shielding

External isotopicstandard (e.g., 137Cs)

Figure 3-10 Example of external standard

mechanics.

automatically placed next to the sample vial to beanalyzed for quenching (Fig. 3-10). � Rays from theexternal standard penetrate the sample vial and col-lide with orbital electrons of atoms in moleculeswithin the scintillation cocktail. Because of thechoice of � emitter selected, these collisions resultin a spectrum of Compton electrons that possess anenergy spectrum analogous to, but of higher en-ergy than a �� spectrum. These electrons then in-teract with the solvent and fluors to produce a col-lection of light flashes of varied intensity that aresimilar to, but higher in energy than that producedfrom �� particle interaction with the scintillationcocktail. Chemical quenching agents will quench

spectra that are induced by � rays in a manner iden-tical to chemical quenching in �� particle analysis.Therefore, analysis of quenching in this inducedspectrum in the high-energy range above the ��

particle range provides an accurate analysis of thequenching of a �� spectrum that would occur inthe absence of the internal standard. (The ��

particle-emitting sample within the sample vialdoes not influence the external standard system be-cause the energy of the �� particles do not fall inthe high-energy Compton range.)

Various scintillation counters analyze thequenching of the induced spectrum of the externalstandard differently. Some machines relate decreasesin the count rate of a selected portion of the inducedspectrum with the quenching in any ��-emittingsample spectrum. Others utilize a channels ratioanalysis of selected channels within the external orinduced spectrum. In all instances, the scintillationcounter generates an external standard number thatreflects the method of quenching analysis employedon the external or induced spectrum. This externalstandard number is directly related to the countingefficiency or quenching observed in a �� emissionsystem, as indicated in the quench curve of Figure3-11. Use of the channels ratio method or the au-tomatic external standardization method of quenchanalysis allows the conversion of counts per minutedata to true disintegrations per minute data. Suchdisintegrations per minute data are necessary forconversion of counting data to curies, the basic unitsof radioactivity.

100

0

0

100

Cou

ntin

g ef

ficie

ncy

(%)

Que

nchi

ng (

%)

4.0 3.0 2.0 1.05.0

External standard number

Figure 3-11 Typical quench curve.


Preparation of Samples for Liquid Scintillation Count-

ing. Quenching in its various forms constitutesthe major concern during sample preparation forliquid scintillation counting. The following sectionsdiscuss some of the problems commonly encoun-tered and also present some remedies to overcomethese problems.

1. Color quenching can only be eliminated by re-moval of the colors. This requires refining ofthe sample before counting. If the compoundto be counted is colored, little can be done ex-cept to minimize and equalize the quantities ofcolored material added to the sample vials.

2. Incomplete solubility and associated pointquenching constitute a major problem in scin-tillation counting. Efficient scintillation count-ing requires that the sample be fully soluble inthe excitable organic solvents of the scintilla-tion fluid. However, biological systems, whichare usually aqueous systems or assays, fre-quently contain water or hydrophilic moleculesthat will not dissolve in standard toluene-basedscintillation cocktails.

Several approaches have been used to cir-cumvent these problems. First, scintillationcocktails have been developed that will acceptmore water and associated hydrophilic com-pounds. Bray’s solution (4 g PPO, 0.2 gPOPOP, 60 g naphthalene, 20 ml ethylene gly-col, 100 ml methanol, and dioxane up to 1 liter),Kinard solution (xylene, p-dioxane, ethanol(5�5�3) containing 0.5% PPO, 0.005% �-NPO, and 6% naphthalene), and ethanol sys-tems (e.g., 3 parts ethanol, 4 parts 0.8% PPO,0.01% POPOP in toluene) are notable exam-ples. In these examples, PPO denotes 2,5-diphenyloxazole, POPOP denotes 1,4-bis-2-(5-phenyloxazolyl)-benzene, and �-NPO denotes2-(1-naphthyl)-5-phenyloxazole. (POPOP and�-NPO are secondary fluors and can usually beomitted with modern scintillation counters.)

A second approach has been developed, inpart because of difficulties encountered in dis-posing of scintillation fluids containing naph-thalene, which is insoluble in water. In this ap-proach, solubilizer or detergents are added tothe standard toluene-based scintillation cock-tails. Hyamine hydroxide, NCS, Soluene-100,

Aquasol, and the Bio-solve series are sold com-mercially for this purpose.

Third, difficult samples of materials con-taining 3H or 14C (e.g., tissue slices, etc.) canbe converted to 3H2O or 14CO2 in commer-cially available “sample oxidizers.” These unitsburn samples under controlled conditions andthen separate the 3H2O and the 14CO2 forcounting in an acceptable scintillation cocktail.Sample oxidation also eliminates color-quench-ing problems.

A fourth solution is the “internal standard”method. In this procedure, a sample is countedthat is known to have solubility or point-quenching problems (e.g., aqueous samples, in-soluble materials embedded in paper or cellu-lose derivative filters, etc.). A known amount ofisotope (labeled) is added to the sample (i.e., tothe fluid or dried paper) and the countingprocess is repeated. Evaluation of the countingefficiency of the added isotope (the internalstandard) allows determination of the countingefficiency of the original sample. (NOTE: Al-ternatively, the internal standard method mayemploy addition of internal standard to onepair of duplicate samples).

3. Chemical quenching is always a problem in liq-uid scintillation counting. Certain compounds(e.g., chlorinated hydrocarbons) are notoriouschemical quenchers. Dissolved O2 from the air,water, even the solubilizers mentioned before,chemically quench scintillation counting some-what. Thus, it is unlikely that chemical quench-ing can be avoided completely. The best ap-proach for minimizing chemical quenching isto design experimental protocols that add thesimplest or most refined assay components tothe final scintillation cocktail. Further, whenchemical quenchers are added (including wa-ter, buffers, etc.) to scintillation vials for count-ing, the same amounts should be added to allsample vials to equalize quenching errors.

Solid Scintillation Counting of �� Emissions. Toavoid some of the problems associated with liquidscintillation counting, a method has been developedfor measuring �� radioactivity by scintillation ofsolid samples. Liquid samples containing the ra-dioactive material to be counted are placed in small


sample cups coated on the bottom with a solid ma-trix containing yttrium, which acts as the scintil-lant. The sample is dried and counted in the sameinstrument as used for liquid scintillation counting(with some changes in the counting “window” andcoincidence circuit settings). The volume of liquidthat can be analyzed is limited to 200 �l, and theradioactive material cannot be volatile. Pointquenching can still present a problem for samplesthat contain substantial amounts of solid matter.However, other kinds of quenching are minimized,and use of expensive and toxic liquid scintillationfluid, which is also difficult to dispose of properly,is avoided. Counting efficiencies depend on thecounting equipment used, but can be nearly as highas with liquid scintillation counting under optimalconditions.

Solid Scintillation Counting of � Emissions. Scintil-lation counting is also the most efficient method fordetection of � rays. � Rays are high-energy pho-tons that lack both charge and mass. As such, theyare highly penetrating and require dense materialsto be absorbed or recorded efficiently. The solublescintillation cocktails employed with �� particle as-says in scintillation counters are not sufficientlydense to absorb �-ray emissions effectively. Scintil-lation counting of � rays is usually accomplished byplacing the � ray source in a solid NaI crystal scin-tillation cell (Fig. 3-12). The brief light flashesemitted by the NaI crystal scintillator are measuredas counts by phototube–photomultipliers. They are

amplified and processed by the conventional scin-tillation counter circuitry and then recorded by thescaler element of the unit. Thus, a liquid scintilla-tion counter can be converted to a �-ray counterby substitution of an NaI crystal scintillator count-ing cell into the conventional circuitry.

In contrast to �� particles, � rays are emittedwith distinct or single energy values. Because of thisproperty, the identity of unknown �-emitting ra-dioisotopes can be determined through analysis ofthe energies of emitted � particles. More impor-tant, the progress of biological reactions involvingvarious components labeled with different �-ray-emitting radioisotopes can be followed. Thus, dou-ble- and triple-label experiments are possible with�-emitting radioisotopes if those that emit � raysof rather different energies are selected, and a � rayscintillation counter equipped with an energy chan-nel analyzer is used.

Cerenkov Counting. �� particles of very high en-ergy (above 0.5 MeV ) pass through aqueous mediawith a velocity greater than the speed of light inwater. This causes the emission of light, which isknown as the Cerenkov effect. The amount of lightemitted is proportional to the number of radioac-tive nuclei that emit high energy �� particles, andit can be readily detected in the same scintillationcounters as are used for liquid scintillation count-ing. Because Cerenkov counting does not requireany special solvents or fluors, it is not subject tochemical quenching and is an extremely convenientand inexpensive means of determining the radioac-tivity of samples of suitable isotopes. Of the iso-topes commonly used in biochemical research, only32P emits �� particles of sufficiently high energy tobe counted by the Cerenkov method. Not all of the�� particles emitted by 32P have energies above 0.5MeV, of course, so this isotope can be detected withonly about 40% efficiency. This provides little prac-tical limitation to the use of Cerenkov counting for32P, however. Other isotopes occasionally used inbiological research with �� particles of sufficientenergy for Cerenkov counting include 22Na, 36Cl,and 42K.

Autoradiography. In biochemical research, it isfrequently desired to determine the location of oneor more radioactive components after analysis by

Sample vial

Photomultiplier tube

Wire from photomultiplier tube

Lead shieldingNaI scintillator

Figure 3-12 NaI crystal scintillation cell.


chromatographic or electrophoretic separations.For example, an enzyme reaction may be assayed byincluding a radioactive substrate and separating thereactants and products after a certain reaction timeby paper or thin-layer chromatography. To deter-mine the rate of the enzyme-catalyzed reaction, itis necessary to separate the radioactive product fromthe radioactive substrate and to determine theamount of radioactivity in the product. Examples ofother situations that require the identification of thelocation of radioactive materials after elec-trophoretic or chromatographic analysis include theseparation of radioactive oligonucleotides in DNAsequencing, separation of protein–DNA or pro-tein–RNA complexes from unbound radioactiveDNA or RNA in electrophoretic gel mobility shiftanalysis, and analysis of the protein content of wholecells by radiolabeling and two-dimensional gel sep-aration of the very complex mixtures of radioactiveproteins found in such cells. Autoradiography offersa very convenient solution to such problems.

Photographic Autoradiography. The most com-monly used form of autoradiography involves ex-posure of a paper chromatogram or electrophoreticgel to a sheet of photographic film, such as x-rayfilm. Radioactive emissions are able to activate sil-ver halide granules in photographic film for pho-toreduction in much the same way as visible lightor x-rays do. Thus, after a period of exposure of thefilm to a sheet of paper or gel containing radioac-tive zones, the film can be developed to reveal apattern of dark spots showing the location andshape of the radioactive zones. The location of thespots can then be compared to the locations of stan-dard compounds and used for identification. Themost common and most reliable use of autoradiog-raphy is qualitative analysis. However, the radioac-tive zones can be cut out of the original gel or pa-per and their radioactivity can be quantitated byliquid scintillation counting (although care must betaken to eliminate or correct for point quenchingof the radioactivity by the paper or gel matrix inwhich the radioactive material is embedded).

The efficiency of various radioactive isotopes inproducing autoradiograms is related to the energyof their decay. Thus, isotopes that emit high-energy �� particles cause more darkening of x-rayfilm per decay event than weak �� emitters. Veryweak �� emitters, such as 3H, may require very long

exposure to the film because many of the radioac-tive particles are too weak to escape the chromato-graphic or electrophoretic medium (e.g., paper orgel), or the wrapping needed to protect the filmfrom sticking to the gel. 3H can be detected in gelsby fluorography, where the gel is dehydrated withdimethylsulfoxide, saturated with a fluor, and rehy-drated. When the weak �� particles excite the fluor,light is given off that is more effective at darkeningthe photographic film than the original �� parti-cle. Weak �� emitters are infrequently used for au-toradiography of chromatograms or electrophero-grams, but they are valuable for autoradiography ofvery thin tissue samples, in which you wish to de-termine the localization of the isotope in a biolog-ical sample. In this latter application, a very-high-energy �� emitter, such as 32P, is less suitable forlocalization of radioactivity with high resolution be-cause the high energy decay leaves too long of atrail of darkened silver particles on the film.

The darkness of the spots on photographic filmafter exposure to a radioactive substance will be afunction of several variables. The radioactivity ofthe material (i.e., the disintegrations per minutepresent in the original spot) and the time of expo-sure to the photographic film are obviously impor-tant, because the more radioactive decay events oc-cur during the exposure of the film, the more grainsof silver halide will be reduced to elemental silver.In a crude semiquantitative sense, then, you canconclude that the darker the spots on the autoradi-ogram, the more radioactivity was present in theposition of the spot on the original chromatogramor electropherogram. Eventually, however, all of thesilver halide near the radioactive material will be re-duced, and the spot cannot become darker onlonger exposure or exposure to higher amounts ofradioactivity. In other words, the film may becomesaturated in a particular zone that contains a highlevel of radioactivity. Under very carefully con-trolled conditions, and within the limits of avoid-ing overexposure, the intensity (or darkness) ofspots on autoradiograms can be used as a quantita-tive measure of the amount of radioactive materialon the original chromatogram or electrophero-gram. This technique requires the use of a densi-tometer for quantitative determination of the dark-ness of the spots and careful standardization of thedensitometer response using samples of known ra-dioactivity separated by the same method.


Autoradiography Using Storage Phosphor Technol-

ogy. A powerful new method for detection of ra-dioactive materials on chromatograms or electro-pherograms that also provides a convenient meansfor quantitative determination of the amount of ra-dioactive material in each location has been pro-vided by the use of storage phosphor techniques.Instead of using a photographic film to detect theradioactive decay events on a chromatogram orelectropherogram, this technique uses a screencoated with a thin layer of an inorganic phosphorin which a chemical change is induced by interac-tion with a strong �� or � particle. The changedphosphor is then detected in a second stage in whichthe screen is scanned with a laser beam, which in-duces the local emission of light only from the al-tered phosphor molecules. The emitted light is col-lected by a high-resolution photocell system, so thatits position and intensity can be stored in digitalform, for later analysis.

An example of storage phosphor technology is that used by the PhosphorImager (Molecular Dynamics). This device uses imaging screens that arecoated with a finely crystalline layer of BaFBr:Eu2�.A radioactive particle from a nearby decay eventcauses oxidation of the phosphor to BaFBr:Eu3�.This change is stable, so that the screen eventuallyacquires a chemical image of the adjacent radioac-tive sheet in which the position and intensity of theradioactive zones is recorded on the phosphorscreen. The chemical image is read by placing thescreen in a device in which it is scanned with a finebeam of laser light (633 nm), which is absorbed onlyby the activated phosphor. The absorbed light al-lows reduction and phosphorescent decay of the

back to with the releaseof a photon at 390 nm.

Thus, the pattern of radioactive decay events istransformed into a pattern of localized light emis-sions, which is collected digitally and can be viewedon a computer screen, printed onto ordinary paper,and, most importantly, analyzed quantitatively. Theamount of light released from the phosphor screenis proportional to the number of decay events de-tected over a rather broad range. The technologyis useful for all of the ��-emitting isotopes com-

BaFBr:Eu2�* � h BaFBr:Eu2�

BaFBr:Eu2� BaFBr:Eu3�� h (laser)

BaFBr:Eu3�BaFBr:Eu2�

monly used in biochemical research except 3H. Un-like x-ray film, which has a dynamic range of about10, some phosphor screens have a dynamic rangeon the order of 105.

Safety Precautions for Work with

Radioisotopes

Use of radioisotopes in the laboratory generates po-tential biological hazards for investigators, and forall of society during subsequent radioisotope removaland disposal. Accordingly, those who work with ra-dioisotopes should follow these safety precautionsduring all applications that employ radioisotopes.

1. Do not ingest radioisotopes. Specifically, neverpipette radioactive solutions by mouth. Instead,use a Propipette bulb or a micropipetter towithdraw and dispense radioactive solutions.Use an appropriate ventilated hood whenworking with volatile radioactive compounds.

2. Avoid contact of radioisotopes with your skin;wear disposable gloves and long-sleeved wash-able clothing when possible. Most biochemi-cally important radioisotopes are metabolitesthat are readily incorporated into the body oningestion or penetration.

3. Avoid close contact with high-energy radioiso-topes. The 3H and 14C systems, and even the32P systems presented in this textbook, do notrepresent excessive radiation hazards. Yet, allexperiments with radioisotopes present someradiation hazard. Those who plan experimentswith compounds that emit �-ray and high-energy �� particles should recognize this anduse appropriate Plexiglas and lead shielding.

4. Avoid contamination of the laboratory with radioisotopes. Work over disposable surfaces(e.g., blotter paper) and remove and clean all“spills” in a thorough manner (consult the labo-ratory supervisor). Further, do not leave ra-dioisotope solutions in the laboratory; they maysupport bacterial or mold growth and eventualproduction of 3H2O or 14CO2, which are volatile.

5. Wash all lightly contaminated equipment(glassware, pipettes, hose tubing, etc.) thor-oughly to avoid future contamination of your-self or your experiments.

6. Place all highly contaminated solutions and alldisposable items (planchets, used filters, etc.) in


radioactive waste containers for prompt re-moval and disposal as specified by law.

7. Survey your hands and all work areas with aGeiger-Müller meter after each use of isotopesthat emit � rays or high-energy �� particles.After each use of low-energy �� isotopes, suchas 3H, sample your hands and work areas witha damp cotton swab and place the swab in a vialcontaining an appropriate scintillation fluid toscreen for possible contamination. Survey thebottoms of your shoes as well, to prevent“tracking” of radioisotopes around the labora-tory.

8. If significant levels of high-energy �� emittersor � emitters are used, you may be required towear a radiation badge to monitor your cumu-lative exposure. Persons who regularly workwith 125I should have their throats surveyedroutinely to monitor possible accumulation ofradioiodine in the thyroid gland.


Marking pen3H toluene, 14C toluene unknowns14C toluene standard of known disintegrations per

minute per milliliter

Volumetric dispensing bottles

Scintillation fluid (0.5% PPO in toluene)

Chloroform

Quenched 14C unknown

Scintillation counter

Protocol

Analysis of a Labeled Unknown Containing

Both 3H and 14C by Scintillation Counting

This experiment tests your ability to determine thecounts per minute of both 3H and 14C in a mixturethat contains unknown amounts of 3H and 14Ctoluene. The unknown and a 14C toluene standardemployed in the experiment are both equallyquenched (by dissolved O2, etc.), so quenching dif-ferences do not influence the final result. The ex-periment can therefore employ measurement ofcounting efficiencies of the unknown and the stan-

dard in separate 3H and 14C above 3H channels, fol-lowed by correction for the overlap of 14C countsin the 3H channel.

1. Using a volumetric dispensing bottle, tip 5 or10 ml of scintillation fluid into each of two scin-tillation vials, depending on the size of scintil-lation vials to be used. Label the bottles “A” and“B” at points on the bottle above the fluid levelor on the cap. In the same fashion, label thebottles with your name for later identification.

2. Add the following:

Bottle A: 0.1 ml of unknown radioactive toluene so-lution (unknown quantities of 3H and 14C)

Bottle B: 0.1 ml of standard 14C toluene solution

3. Cap each bottle, invert several times to mix,and count both bottles for 1 min (no automaticexternal standardization) in both the 3H chan-nel (channel 1) and the 14C-over-3H channel(channel 2), which have been designated byyour instructor.

4. Using the following equations, calculate the to-tal number of counts per minute of both 3Hand 14C in your unknown solution:

A. Determine the percent of the total 14C cpmdetected in channel 2 using the followingequation:

%14C cpm in channel 2 �

B. Determine the total number of 14C cpm inthe unknown sample using the followingequation:

Total 14C cpm in unknown �

C. Determine the percent of the total 14C cpmdetected in channel 1 using the followingequation:

% 14C cpm in channel 1 �14C cpm in channel 1 for bottle B

total 14C cpm in bottle B

channel 2 cpm for bottle A

% 14C cpm in channel 2

14C cpm in channel 2 for bottle B

total 14C cpm in bottle B


D. Determine the 14C cpm detected in Chan-nel 1 for the unknown using the followingequation:

14C cpm in channel 1 for unknown �

E. Determine the total 3H cpm in the un-known using the following equation:

3H cpm in unknown � (total cpm in channel

1 for bottle A) � (14C cpm in channel 1 for

unknown)

Determination of a Quench Curve and

Analysis of a Quenched 14C Sample

This experiment demonstrates two methods foranalysis of quenching within samples, namely, thechannels ratio method and the automatic externalstandardization method. Either or both of themethods may be demonstrated with the quench se-ries of bottles described in the following protocol,depending on the capabilities of your particularscintillation counter.

1. Using a volumetric dispensing bottle, tip 5 or10 ml of scintillation fluid into each of five scin-tillation vials. Label the vials 1 to 5 on the out-side, above the level of the fluid in the vial oron the cap. Label each vial with your name forlater identification.

2. Add to the vials as indicated below:

14C toluene standard Quenched

Sample of known dpm/ml Chloroform 14C unknown

1 1.0 ml — —

2 1.0 ml 0.1 ml —

3 1.0 ml 0.2 ml —

4 1.0 ml 0.3 ml —

5 — — 1.0 ml

If the channels ratio method is used, proceed withSteps 3 to 8. If the automatic external standard-ization method is used, proceed with Steps 9 to 13.

3. Determine three different but specific gain andchannels settings, each of which will detect allor a portion of the 14C counts in the samples

channel 1 cpm for bottle A

% 14C cpm in channel 1

described above. This may have already beendone for you by the laboratory instructor (seeFig. 3-8).

4. Using these three prescribed channel settings,count each sample for 1 min in each of thechannels.

5. Using the data obtained from samples 1 to 4,the known number of 14C disintegrations perminute added to each of these samples, andconsidering the counts per minute data ob-tained from the widest window as a reflectionof overall counting efficiency (the total possi-ble number of counts per minute that can bedetected in each sample), construct a channelsratio quench correction curve similar to thatshown in Figure 3-9.

6. What do you notice about the relationship be-tween the chloroform concentration in thesample and the counting efficiency for 14C?

7. Based on the A n B/A n C ratio of sample 5,calculate the number of disintegrations perminute and Ci of 14C present in the unknownquenched 14C sample using the following for-mula:

14C dpm �

8. What is the maximal 14C counts per minutevalue that you would expect to find for this sam-ple in the absence of any of the chloroform-quenching agent?

Proceed with Steps 9 to 13 only if you are using the au-tomatic external standardization method.

9. Count each of the samples in the prescribed14C-above-3H channel for 1 min in the auto-matic external standardization mode. The ma-chine reports an “external standard number”(often termed an “H” number for many ma-chines) for each of the samples.

10. Using the data obtained from samples 1 to 4,the known number of 14C disintegrations perminute added to each of these samples, andconsidering the counts per minute data ob-tained from sample 1 (unquenched sample) asa reflection of overall counting efficiency (thetotal possible number of counts per minute thatcan be detected in each sample), construct a

cpm 14C in A n C channel

counting efficiency


channels ratio quench correction curve similarto that shown in Figure 3-11.

11. What do you notice about the relationship be-tween the chloroform concentration in thesample and the counting efficiency for 14C?What effect does increased quenching have onthe external standard number reported by yourinstrument?

12. Based on the external standard number re-ported for sample 5, calculate the number ofdisintegrations per minute and curies of 14Cpresent in the unknown quenched 14C sampleusing the following formula:

14C dpm �

13. What is the maximal 14C counts per minutevalue that you would expect to find for thissample in the absence of any of the chloroform-quenching agent?

Exercises

1. You obtain a 1.0-ml aqueous sample contain-ing 100 �l of uniformly labeled 14C-glycine(specific activity � 10 mCi/mmol). What sizealiquot of the total sample should you remove,and how should you subsequently dilute thisaliquot with 12C-glycine, to obtain 10 ml of0.01 M 14C-glycine containing 2.2 � 105

dpm/ml?2. Most liquid scintillation counters that count �

particles utilize PPO (2,5-diphenyloxazole) asa fluor. Excited PPO emits a light flash in theblue region of the visible spectrum. Recogniz-ing this fact, would 1 nCi of yellow-coloredDNP-14C-alanine count more or less effi-

cpm 14C in prescribed window

counting efficiency

ciently that 1 nCi of colorless 14C-alanine whenboth are counted in identical PPO scintillationcocktails? Explain your answer.

3. During the course of using automatic externalstandardization to prepare a 14C quench curve,you determine that you have mistakenlycounted your series of quenched samples in awide 14C channel instead of a 14C-above-3Hchannel. Will your data generate a quenchcurve with a greater or lesser slope than thecurve you would obtain from counting the samequenched samples in a narrow 14C-above-3Hchannel? Explain your answer.

4. Is a quench curve determined for 14C applica-ble for analysis of quenching of 3H com-pounds? Why or why not?

REFERENCES

Hawkins, E. F., and Steiner, R. (1994). Scintillation Sup-plies and Sample Preparation Guide. Fullerton, CA:Beckman Instuments.

Horrocks, D. L., and Peng, C. (1971). Organic Scintilla-tion and Liquid Scintillation Counting. New York:Academic Press.

Kobayashi, Y., and Maudsley, D. (1969). Practical Aspects of Liquid Scintillation Counting. MethodsBiochem Anal 17:55.

Johnston, R. F., Pickett, S. C., and Barker, D. L.(1990). Autoradiography Using Storage PhosphorTechnology. Electrophoresis 11:355.

Steiner, R. (1996) Advanced Technology Guide for LS6500 Series Scintillation Counters. Fullerton, CA:Beckman Instruments.

Wang, C. H., and Willis, D. L. (1965). RadiotracerMethodology in Biological Science. Englewood Cliffs,NJ: Prentice-Hall.

Wilson, K., and Walker, J. (1994). Principles and Tech-niques of Practical Biochemistry, 4th ed. Chapter 5.Cambridge, UK: Cambridge University Press.

( F

61

EXPERIMENT 4

Electrophoresis

Theory

Basic Principles

Electrophoresis is the process of migration ofcharged molecules through solutions in an appliedelectric field. Electrophoresis is often classified ac-cording to the presence or absence of a solid sup-porting medium or matrix through which thecharged molecules move in the electrophoreticsystem. Solution electrophoresis systems employaqueous buffers in the absence of a solid supportmedium. Such systems can suffer from samplemixing due to diffusion of the charged molecules,with resultant loss of resolution during sample ap-plication, separation, and removal steps. Thus, so-lution electrophoresis systems must employ somemeans of stabilizing the aqueous solutions in theelectrophoresis cell. For example, soluble-gradientelectrophoresis systems use varying densities of anon-ionic solute (e.g., sucrose or glycerol) to min-imize diffusional mixing of the materials beingseparated during electrophoresis (Fig. 4-1). Evenwith these refinements, solution electrophoresissystems have only limited application, usuallywhen preparative scale electrophoretic separationis required.

Most practical applications of electrophoresis inbiochemistry employ some form of zonal elec-trophoresis, in which the aqueous ionic solution iscarried in a solid support and samples are appliedas spots or bands of material. Paper electrophore-sis, cellulose acetate strip and cellulose nitrate strip,and gel electrophoresis are all examples of zonal

Water jacketfor cooling

(�) Pole

(�) Pole

Sample applied inintermediate densityof sucrose in buffer

Soluble gradient ofincreasing densityin buffer

Dense salt at (�) pole

Stopcock for serialdraining of system

Figure 4-1 A solution electrophoresis system.


Two or more gelcontaining tubes supportedbetween two buffer pools

Buffer-dampened paper orcellulose strip supportedbetween two buffer pools

Buffer pools

Cover

Paper or cellulose strip electrophoresis Gel electrophoresis

(�) Pole (�) Pole

(�) Pole

(�) Pole

Figure 4-2 Two zonal electrophoresis systems.

electrophoresis systems (Fig. 4-2). Such systems aretypically employed for analytical, rather thanpreparative scale, separations.

All types of electrophoresis are governed by thesingle set of general principles illustrated by Equa-tion 4-1:

(4-1)

Mobility of a molecule �

The mobility, or rate of migration, of a moleculeincreases with increased applied voltage and in-creased net charge on the molecule. Conversely, themobility of a molecule decreases with increasedmolecular friction, or resistance to flow through theviscous medium, caused by molecular size andshape. Total actual movement of the molecules in-creases with increased time, since mobility is de-fined as the rate of migration.

An understanding of the relationships describedin Equation 4-1 is essential for many practical as-pects of experimental biochemistry. Most elec-trophoretic systems employ an equal and constantvoltage on all of the cross-sectional areas of the pa-per strips, gels, or solutions employed in the elec-trophoretic separation. These electric fields arebest defined in terms of volts per linear centime-

(applied voltage)(net charge on the molecule)��

(friction of the molecule)

ter. However, Ohm’s law (V � IR) dictates thatvoltage (V ) is a function of current (I ) and resis-tance (R). The nature of the electrophoresis appa-ratus and buffer composition dictates the resistancein the system. Therefore, current (e.g., mA) is of-ten used to define the voltage requirements of anelectrophoretic separation. The resistance of thesystem is important because it will determine theamount of heat generated during electrophoresis.Since electrophoretic mobility is also a function oftemperature, heating of the separation matrix mustbe controlled. If significant heating occurs duringelectrophoresis, it will be necessary to provide somemeans of cooling the apparatus so as to maintain aconstant temperature. The “smiling” pattern oftenseen on slab gel electrophoresis (see below) is theresult of nonuniform heating of the gel.

If the voltage or current applied to an elec-trophoresis system is constant throughout an elec-trophoretic separation, the mobilities of the mole-cules being resolved will reflect the other terms ofEquation 4-1, namely, the net charge and frictionalcharacteristics of the molecules in the sample. Con-sider the paper electrophoretic separation of glu-tamic acid, glutamine, asparagine methyl ester, andglycinamide at pH 6.0. The charges and molecularweights of these compounds at pH 6.0 are indicatedin Figure 4-3. As seen, the equal-sized amino acidsand the asparagine methyl ester separate strictly asa function of their net charges at pH 6.0, whereas

EXPERIMENT 4 Electrophoresis 63

NH3

CH2

CH COO�

COO�

CH2

�

Glutamic acid(MW � 146)

NH3

CH2

CH COO�

CONH2

CH2

�

Glutamine(MW � 146)

NH3

CH2

CH COOCH3

CONH2

�

Asparagine methyl ester(MW � 147)

NH3 CH2 CONH3�

Glycinamide(MW � 75)

(�) (�)

Origin line

Figure 4-3 Paper electrophoretic separation of glutamic acid, glutamine, asparagine, methyl ester, and

glycinamide at pH 6.0. The glycinamide, with a charge of �1 and a molecular weight of 75,

may not necessarily migrate twice the distance migrated by asparagine methyl ester, with the

same charge and �2 times larger size. Specifically, the character of the buffer can influence

the expression of the electrophoretic frictional contribution of small molecules. Higher ionic

strength decreases the frictional contribution.

the glycinamide (which has the same net charge asasparagine methyl ester but has only half its size)migrates proportionally further toward the (�) poleof the system.

Generally, the principles illustrated in Figure 4-3 can be used to predict the relative mobility ofsmall ionic molecules. These principles are particu-larly useful for predicting the potential for elec-trophoretic separation of small molecules contain-ing weakly acidic or weakly basic groups that carryaverage partial charges in the pH ranges associatedwith their titration, as, for example, during the elec-trophoretic separation of nucleotides or amino acids.

The electrophoretic separation of larger macro-molecules follows the general principles of Equa-tion 4-1, but other factors influence the resolutionof macromolecules. The friction experienced bymolecules during electrophoretic migration reflectsboth molecular size and molecular shape. If the elec-trophoresis is carried out in a medium that offerssignificant barriers to the movement of macromol-ecules through it (as is the case with polyacrylamideand agarose, two very commonly used systems), mol-ecular size may prove to be the most important deter-minant of mobility. If the charge to mass ratio on themacromolecules being separated is approximatelyequal (as in several of the cases discussed below),

molecular size becomes the sole determinant ofelectrophoretic mobility. These conditions are ex-ploited for the determination of the molecularweight of protein subunits by electrophoresis inpolyacrylamide gels containing sodium dodecyl sul-fate (SDS-PAGE), and in the electrophoretic sepa-ration of oligonucleotide “ladders” during DNA se-quencing. Molecular shape is not very significant insmall molecules, in which bonds are free to rotate,so size alone defines their friction. However, macro-molecules often have defined shapes with specificaxial ratios (i.e., length to width ratios). As a result,both size and shape influence migration. Moleculeswith high axial ratios demonstrate lower elec-trophoretic mobility than more spherical moleculesthat have equal weight and equal charge. In addi-tion, macromolecules may deviate from the elec-trophoretic principles of Equation 4-1 because ofinteraction with ions or because of charge-depen-dent intermolecular associations.

Specific Forms of Electrophoresis

Commonly Used in Biochemistry

Paper Electrophoresis. Paper electrophoresis is acommonly used electrophoretic method for analy-sis and resolution of small molecules. This method


is not used to resolve macromolecules (e.g., pro-teins) because the adsorption and surface tensionassociated with paper electrophoresis usually alteror denature the macromolecules, causing poor res-olution.

Two different methods are used routinely to ap-ply samples to the electrophoretic paper. In the dryapplication procedure, a sample of solutes dissolvedin distilled water, or a volatile buffer, is applied asa small spot or thin stripe on a penciled “origin line”on the paper. Appropriate standards of known com-pounds are applied at other locations on the originline. If you anticipate electrophoretic migration to-ward both poles of the system, the origin line shouldbe in the center of the paper. If you anticipate mi-gration in only one direction, the origin line shouldbe near one end of the paper. After the solvent con-taining the samples has evaporated, the paper isdampened with the electrophoresis buffer, either byuniform spraying or by dipping and blotting theends of the paper so that wetting of the paper fromboth ends meets at the origin line simultaneously.In the wet application procedure, samples dissolvedas concentrated solutions in distilled water are ap-plied to paper predampened with electrophoresisbuffer. The dry application procedure has the ad-vantage of allowing small initial sample spots andbetter resolution of similarly mobile compounds.However, this method is awkward because the dip-ping or spraying requires considerable skill to avoidspreading the applied samples. The wet applicationprocedure is simpler to perform, but usually yieldslarger spots and poorer resolution because of sam-ple diffusion.

After the sample is applied to the dampenedpiece of paper, the paper is placed in the elec-trophoresis chamber so that both ends are in con-tact with reservoirs of the electrophoresis buffer atthe electrodes (Fig. 4-4). If the origin line is not inthe center of the paper, the paper must be posi-tioned to allow maximum migration toward the cor-rect electrode. After the chamber is covered orclosed to protect against electric shock, an electricfield is applied to the system.

Application of the electric field and the resul-tant resistance to current flow in the buffered pa-per generates heat. This is the greatest source of diffi-culty with paper electrophoresis. Heat dries the paper,which in turn leads to more resistance to currentflow, which causes greater resistance, and so forth.

Even if paper drying is prevented, heating willchange the current flow and resistance propertiesof the system, which will distort the migration ofmolecules.

Because of these difficulties, modern paper elec-trophoresis systems have been designed to com-pensate for potential heating problems. Most low-voltage systems are portable and can be operated incold rooms or refrigerated chambers. In contrast,most high-voltage systems either employ a cooledflat-bed system to dissipate the heat or operate ina cooled bath of inert and nonpolar solvent (e.g.,Varsol, a petroleum distillate). This solvent absorbsthe heat generated by the system without mixingwith the water, buffers, or samples on the paper(Fig. 4-4). After electrophoretic resolution of thesamples on the paper for the time and voltage re-quired for optimal separation, the current is turnedoff, the paper is removed and dried, and the pres-ence and location of the molecules of interest aredetermined.

Capillary Electrophoresis. A new method of ana-lytical electrophoresis that is rapidly finding in-creasing application in biochemical research is cap-illary electrophoresis. As the name suggests, thematerial to be analyzed and the electrophoresismedium (a conducting liquid, usually aqueous) areplaced in a long, fine-bore capillary tube, typically50 to 100 cm long and 25 to 100 �m inside diam-eter. A very small sample (in the nanoliter range) isplaced at one end of the capillary and subjected toelectrophoresis under fields up to 20 to 30 kV. Theanalytes are separated by the principles illustratedin Equation 4-1 and detected as they emerge fromthe other end of the capillary by any of the meth-ods commonly used in high-performance liquidchromatography (see Experiment 2). Capillary elec-trophoresis offers the advantages of extremely highresolution, speed, and high sensitivity for the analy-sis of extremely small samples, but is obviously notuseful as a preparative method. It has proven espe-cially useful in the separation of DNA moleculesthat differ in size by as little as only a single nu-cleotide. Because of its high resolution, capillaryelectrophoresis is the basis of separation of polynu-cleotides in some of the newer designs of DNA se-quencers. Capillary electrophoresis can also beadapted to the separation of uncharged moleculesby including charged micelles of a detergent (such


Cooled bath of inertnonpolar solventsuch as Varsol

Paper heldby centersupport

Center bafflebetween bufferpools

Wireelectrodein bufferpool

Paper restingbetween bufferpools

Cooled bath or tank systemFlat-bed system

(�) Pole

(�) Pole

Top closes(�) Pole

Water-cooledor refrigeratedtop and base

Aqueous bufferbelow Varsol

Wireelectrode

Figure 4-4 Two paper electrophoresis systems.

as SDS) in the aqueous electrophoresis medium. Ifa mixture of solute molecules that partitition be-tween the aqueous medium and the hydrophobicinterior of the micelles is introduced into such asystem, they can be separated by electrophoresis.Capillary electrophoresis is a highly adaptablemethod, and the range of its applications and opti-mal methodology are still being explored.

Gel Electrophoresis. In gel electrophoresis, mol-ecules are separated in aqueous buffers supportedwithin a polymeric gel matrix. Gel electrophoresissystems have several distinct advantages. First, theycan accommodate larger samples than most paperelectrophoresis systems, and so can be used forpreparative scale electrophoresis of macromole-cules. Second, the character of the gel matrix can

be altered at will to fit a particular application. Thisis possible because the gel enhances the friction thatgoverns the electrophoretic mobility (see Equation4-1). Low concentrations of matrix material or alow degree of cross-linking of the monomers inpolymerized gel systems allow them to be usedlargely as a stabilizing or anticonvection device withrelatively low frictional resistance to the migrationof macromolecules. Alternatively, higher concen-trations of matrix material or a higher degree ofcross-linking of monomers are used to generategreater friction, which results in molecular sieving.Molecular sieving is a situation in which viscosityand pore size largely define electrophoretic mobil-ity and migration of solutes. As a result, the mi-gration of macromolecules in the system will besubstantially determined by molecular weight.


Many gel-like agents are used in electrophoreticsystems. Agarose (a polygalactose polymer) gelshave proven quite successful, particularly when ap-plied to very large macromolecules such as nucleicacids, lipoproteins, and others. Polyacrylamide gelsare among the most useful and most versatile in gelelectrophoretic separations because they readily re-solve a wide array of proteins and nucleic acids (seeTables 4-1 and 4-2 for instructions on how to pre-pare SDS-PAGE gels and polyacrylamide gels forlow molecular weight nucleic acid samples).

Polyacrylamide gels are formed as the result ofpolymerization of acrylamide (monomer) andN,N�-methylene-bis-acrylamide (cross-linker) (Fig.4-5). The acrylamide monomer and cross-linker arestable by themselves or mixed in solution, but poly-merize readily in the presence of a free-radical gen-

erating system. Biochemists use either chemical orphotochemical free-radical sources to induce thepolymerization process. In the chemical method(the most commonly used method), the free radi-cal initiator, ammonium persulfate (APS), is addedalong with a N,N,N�,N�-tetramethylethylenedi-amine (TEMED) catalyst. These two components,in the presence of the monomer, cross-linker, andappropriate buffer, generate the free radicalsneeded to induce polymerization. In the photo-chemical method (less widely used), ammoniumpersulfate is replaced by a photosensitive compound(e.g., riboflavin) that will generate free radicalswhen irradiated with UV light. Many modificationscan be made to produce a gel that will be useful fora particular application. If larger pores are required,you could decrease the amount of monomer and/or

Table 4-1 Recipe for Polyacrylamide Gels with Various Percent Acrylamide Monomer for Use with

SDS-PAGE

% Acrylamide in Resolving Gel

Component (ml) 7.5 10 12 15 20

Distilled water 9.6 7.9 6.6 4.6 2.7

30% acrylamide solution 5.0 6.7 8.0 10.0 11.9

1.5 M Tris chloride (pH 8.8) 5.0 5.0 5.0 5.0 5.0

10% (wt/vol) SDS 0.2 0.2 0.2 0.2 0.2

TEMED 0.008 0.008 0.008 0.008 0.008

10% (wt/vol) ammonium persulfate 0.2 0.2 0.2 0.2 0.2

Prepare the ammonium persulfate fresh and add last to induce the polymerization process (polymerization of the

resolving gel will take approximately 30 min).

4% Acrylamide Stacking Gel for SDS-PAGE

Component Volume (ml)

Distilled water 2.7

30% acrylamide 0.67

1.0 M Tris chloride (pH 6.8) 0.5

10% (wt/vol) SDS 0.04

TEMED 0.004

10% (wt/vol) ammonium persulfate 0.04

Prepare the ammonium persulfate fresh and add last to induce the polymerization process (polymerization of the

resolving gel will take approximately 30 min).

30% Acrylamide Solution

Dissolve 29.2 g of acrylamide and 0.8 g of N,N�-methylene-bis-acrylamide in 100 ml of distilled water. Filter through a

0.45-�M-pore-size membrane (to remove undissolved particulate matter) and store in a dark bottle at 4°C.


cross-linker in the polymerization solution. Ifsmaller pores are required, one may increase theconcentration of monomer and/or cross-linker.

The pore size required for a particular elec-trophoretic separation will depend on the differ-ence in size of the compounds that you wish to re-solve. For instance, if you wish to resolve two smallproteins of 8,000 Da and 6,000 Da, you will requirea small-pore-size gel for this application (�15–20%acrylamide). This same percent acrylamide gelwould not permit the resolution of two larger pro-teins of say 150,000 Da and 130,000 Da. A larger-pore-size gel would be required for this application(�7.5–10% acrylamide). A list of components re-quired to produce polyacrylamide gels of varyingpercent acrylamide is shown in Table 4-1. A list ofthe effective separation range of proteins on acryl-amide gels made with various percent acrylamide isshown in Table 4-3.

Most important, any methods used to producepolyacrylamide gels must be followed exactly eachtime they are cast, since reproducible electro-phoretic separations require uniform gel-formingconditions. Current literature contains many elec-trophoresis procedures and applications for poly-acrylamide gels. Most of these employ the poly-acrylamide gel in some sort of “slab” format. Oneof the most commonly used of these procedures isdiscussed below.

Table 4-2 Recipe for a 15% Polyacrylamide Gel

for Use in Separating Nucleic Acids

Smaller Than 500 Bases in Length

Component Volume or Mass

Urea 15 g

40% acrylamide solution 11.3 ml

10� TBE buffer 3.0 ml

Distilled water 4.45 ml

TEMED 30 ml

10% (wt/vol) ammonium persulfate 200 ml

Prepare the ammonium persulfate fresh and add last to

induce the polymerization process. This recipe will be

enough to cast 3 to 4 gels, depending on the size of the

plates that you use. The recipe for a 0.5� TBE solution is

shown in Table 4-5. A 5� concentrated stock can be

prepared and stored at room temperature for a short time

by multiplying all of the masses and volumes of the Tris

base, boric acid, and EDTA components by 5. After several

days at room temperature, a precipitate may begin to

form in the 5� TBE stock. When this occurs, a fresh

solution should be made. This 5� working stock should be

diluted 1�10 with distilled water to produce a 0.5�

running buffer for use with separation of low-molecular-

weight nucleic acids by polyacrylamide gel electrophoresis.

40% Acrylamide Solution

Dissolve 76.0 g of acrylamide and 4.0 g of N,N�-

methylene-bis-acrylamide in 200 ml of distilled water.

Filter through a 0.45-�M-pore-size membrane (to remove

undissolved particulate matter) and store in a dark bottle

at 4°C.

CH2

NH2

CH

OC

CH2

CH2

NH

NH

CH

OC

CH2 CH

OC

CH2

CH2

NH

NH

CH

OC

CH2

CH2CH2 CHCHCH2CH

OC

CH2 CHCH2CH

NH2

OC

NH2

OC

NH

OC

NH2

OC

Acrylamide(monomer)

�Free radicals

TEMEDPolymer

N,N�-Methylene-bis-acrylamide(cross-linker)

Figure 4-5 Formation of polyacrylamide gels.


Sodium Dodecyl Sulfate–Polyacrylamide Gel Elec-

trophoresis (SDS-PAGE). Sodium dodecyl sulfate(SDS) gel electrophoresis systems are used to de-termine the number and size of protein chains orprotein subunit chains in a protein preparation. Ini-tially, the protein preparation is treated with an ex-cess of soluble thiol (usually 2-mercaptoethanol)and SDS. Under these conditions, the thiol reducesall disulfide bonds (–S–S–) present within and/orbetween peptide units, while the SDS (an ionic or

denaturing detergent) binds to all regions of theproteins and disrupts most noncovalent intermo-lecular and intramolecular protein interactions.These two components result in total denaturationof the proteins in the sample, yielding unfolded,highly anionic (negatively charged) polypeptidechains (Fig. 4-6).

The anionic polypeptide chains are then re-solved electrophoretically within a polyacrylamidegel saturated with SDS and the appropriate current-carrying buffer. The excess SDS is included in thegel to maintain the denatured state of the proteinsduring the electrophoretic separation. The SDS-coated polypeptides (carrying approximately oneSDS molecule per two amino acids) creates a situ-ation in which the charge-to-mass ratio of all of theproteins in the sample is approximately the same.At this point, the intrinsic charge on the individualpolypeptide chains (that is, in the absence of SDS)is insignificant as compared with the negativecharge imposed on them by the presence of theSDS. The friction experienced by the population ofmolecules as they migrate through the polyacryl-amide matrix is now the major factor influencingdifferences in their mobility. In addition, the fric-

Table 4-3 The Effective Separation Range of

Polyacrylamide Gels of Various

Percent Acrylamide Monomer for

Use With SDS-PAGE

% Acrylamide in Effective Separation

Resolving Gel Range (Da)

7.5 45,000–200,000

10 20,000–200,000

12 14,000–70,000

15 5,000–70,000

20 5,000–45,000

s

s

ss

Organized nativeprotein with3-dimensional structure

Proteins without Subunits

Proteins with Subunits

subunit

subunit

subunit

subunit

R–SHSDS

R–SHSDS

SH

2

2

SH

SH

SH

SH

SH

SH

Denaturedprotein

Denatured subunits

Denatured subunits

+

Figure 4-6 Disruption of proteins with excess thiol and SDS.


tion experienced by the molecules during the sep-aration is governed by the pore size of the poly-acrylamide matrix: larger polypeptides will experiencegreater friction when passing through a gel of definedpore size (will migrate more slowly) than smallerpolypeptides (which will migrate more rapidly). In sum-mary, SDS-PAGE allows the separation of proteinson the basis of size.

The principles underlying the most commonlyused form of SDS-PAGE are best illustrated by adescription of the step-by-step progress of a pro-tein migrating in the gel. As seen in Figure 4-7 andTable 4-1, SDS-PAGE employs two buffer/poly-acrylamide gel compositions in a single slab. Theseare referred to as the “stacking gel” and the “run-ning (or resolving) gel.” Protein samples are firstintroduced into wells cast within the stacking gelafter they are mixed with a viscous sample buffer(30% glycerol) containing SDS and thiols. After thesamples have been loaded, voltage is applied to thesystem (current is carried through the gel, �3–4V/cm2) between two separated pools of glycinebuffer, pH 8.3 (Table 4-4).

Remember that ion (current) flow must followthe principles dictated in Equation 4-1, namely,

greater charge increases mobility, whereas greatersize leads to greater friction and decreased mobil-ity. Glycine carries an average charge of about �0.1per molecule at pH 8.3 (running buffer) and almostno net negative charge at pH 6.8 (stacking gel pH).In contrast, the SDS-coated proteins in the systemcarry a high negative charge that is essentially in-dependent of the pH of the system. At the low pHin the stacking gel, glycine anions lose negativecharge and display decreased mobility in the sys-tem. In contrast, the chloride ions contained in thestacking gel migrate ahead of the proteins becauseof their small size (low friction) and full negativecharge. Since the negatively charged proteins in thesystem have a larger frictional factor, they migrateinto the stacking gel at a rate that is slower than thatof the chloride population, but faster than that ofthe virtually uncharged glycine anion population.The resultant scale of anion mobilities in the lowpercentage acrylamide stacking gel (Cl� � pro-teins� � glycine�) causes the proteins to accumu-late ahead of the advancing glycine front (Fig. 4-7b)and eventually stack into concentrated, narrowbands at the interface between the stacking gel andthe running gel (Fig. 4-7c). Meanwhile, the chloride

(–)

(+)

(–)

(+)

(–)

(+)

(–)

(+)

b c da

Resolvingproteins

Bands ofprotein

Proteinsbeginningto stack

Glycine buffer,pH 8.3

Sample of mixedproteins, pH 8.3

Stacking gel,pH 6.8

Resolving gel,pH 8.8

Glycine buffer,pH 8.3

Figure 4-7 Migration of proteins through stacking and resolving gels during SDS-PAGE.


Table 4-4 Recipes for Buffers Used in Agarose Gel Electrophoresis of Nucleic Acids and SDS-PAGE

Tris/Borate/EDTA (TBE) Buffer (0.5�, working solution)

Dissolve 5.4 g of Tris base and 2.75 g of boric acid in 700 ml of distilled water. Add 2 ml of 0.5 M EDTA, pH 8.0. Bring

the final volume of the solution to 1 liter with distilled water. NOTE: A 5� concentrated stock can be prepared and

stored at room temperature for a short time by multiplying all of the masses and volumes of the Tris base, boric acid,

and EDTA components by 10. This 5� concentrated stock will be diluted 1�10 with distilled water to give a 0.5�

working solution.

Tris/Acetate/EDTA (TAE) Buffer (1�, working solution)

Dissolve 4.84 g of Tris base in 700 ml of distilled water. Add 1.14 ml of glacial acetic acid and 2 ml of 0.5 M EDTA, pH

8.0. Bring the final volume of the solution to 1 liter with distilled water. NOTE: A 50� concentrated stock can be

prepared and stored at room temperature for long term by multiplying all of the masses and volumes of the Tris base,

glacial acetic acid, and EDTA components by 50. This 50� concentrated stock will be diluted 1�50 with distilled water to

give a 1� working solution.

Electrophoresis Buffer for SDS-PAGE (1X working solution)

Dissolve 3.02 g of Tris base and 18.8 g of glycine in 700 ml of water. Add 10 ml of 10% (wt/vol) SDS and adjust the pH

of the solution to 8.3 with HCl. Bring the final volume of the solution to 1 liter with distilled water. NOTE: A 5�

concentrated stock can be prepared by multiplying the masses and volumes of all of the components by five. This 5�

concentrated stock will be diluted 1�5 with distilled water to give a 1� working solution described above.

High concentration gel

Intermediate concentration gel

Low concentration gel

1009080706050

40

30

20

100.2 0.4 0.6 0.8

Relative mobility (protein movement relative to ion front)

Mol

ecul

ar w

eigh

t (�

10-3

)

Figure 4-8 Relation between molecular weight

and relative mobility of proteins on

SDS gels.

ions in the stacking gel readily migrate into the run-ning gel.

As the advancing front of anions enters the run-ning gel, the high concentration of pH 8.8 bufferand the high concentration of acrylamide (de-creased pore size) in the gel triggers two events.First, the high pH buffer imparts a greater nega-tive charge to the glycine anions, whose migrationwas retarded in the pH 6.8 buffer in the stackinggel. Because of their small size and increased an-ionic character, the glycine ions quickly overtakethe proteins as they migrate through the system.Second, the reduced pore size of the gel imparts asignificant frictional component to the mobility ofeach individual protein present in the sample. Theequal charge-to-mass ratio imparted on the pro-teins by the SDS present in the system (see above)now dictates that all of the proteins in the systemwill migrate through the gel on the basis of size.

As shown in Figure 4-8, the relative mobility ofeach of the anionic polypeptide chains is a functionof the logarithm of its molecular weight. If a set ofpolypeptides of known molecular weight are in-cluded with the sample during the electrophoreticseparation, their relative mobilities may be deter-mined and plotted as log molecular weight versus

relative mobility. The standard curve producedfrom this analysis of the polypeptides of knownmolecular weights can then be used, along with therelative mobilities of the unknown polypeptides inthe sample, to estimate their molecular weight (seeFig. 4-8).


When voltage is applied to the system, the cur-rent flow will be due largely to migration of thecharged ampholyte species present in the gel. Theampholytes migrate toward either pole in a mannerconsistent with their charge distributions; am-pholytes with lower isoelectric points (e.g., thosethat contain more carboxyl groups and have a netnegative charge) migrate toward the anode, whileampholytes with higher isoelectric points (e.g.,those that contain more amine groups and have anet positive charge) migrate toward the cathode.Eventually, each ampholyte in the system will reacha position in the gel that has a pH equal to its iso-electric point. When this occurs, these ampholytescarry no net charge and will no longer migrate inthe gel. The net effect is that, after a sufficient period of electrophoresis, the population of am-pholytes will act as local “buffers” to establish a sta-ble pH gradient in the polyacrylamide gel. This pHgradient forms the basis for the separation ofmacromolecules that are present in the system (orare subsequently introduced into the system).

Proteins present in the system act like am-pholytes in that they migrate as a function of theirnet charge. As the ampholytes establish a pH gra-dient in the gel, the proteins in the system also mi-grate toward their respective poles until they reacha pH in the gel at which they too carry no net neg-ative charge (the pI of the protein). At this point,the attractive forces applied to the protein by theanode and cathode are equal, and the protein nolonger migrates in the system. In effect, the differ-ent proteins in the sample are “focused” to partic-ular portions of the gel where the pH is equal totheir pI values.

How do you determine when the electro-phoretic separation is complete? As stated above,the separation is complete when the ampholytes andproteins reach a pH value in the gel equal to theirpI and no longer migrate. Once the proteins andampholytes no longer migrate, the current (I ) inthe system decreases dramatically. Because isoelec-tric focusing is performed in the presence of fairlylow concentrations of ampholytes (2–4%), highvoltage potential (150 V/cm2) can be applied to thesystem without the generation of excess amounts ofheat caused from high currents.

Since this method is designed to separate pro-teins on the basis of isoelectric point, one must takecare not to impose a significant “friction factor” to

How do you visualize the proteins that havebeen separated following SDS-PAGE? Two visual-ization methods are most often used. The choicebetween them depends largely on the sensitivity ofdetection that is required for your application. Thefirst method involves saturating the gel with a so-lution of acetic acid, methanol, and water contain-ing Coomassie Brilliant Blue R-250 dye. As themethanol and acetic acid in the solution work to“fix” the proteins within the gel matrix, CoomassieBrilliant Blue binds to the proteins in the gel. Theinteraction between the Coomassie dye and pro-teins has been shown to be primarily through argi-nine residues, although weak interactions with tryp-tophan, tyrosine, phenylalanine, histidine, andlysine are also involved. After the gel is “destained”with the same aqueous acetic acid/methanol solu-tion without the dye (to remove the dye from por-tions of the gel that do not contain protein), theproteins on the gel are visible as dark blue “bands”on the polyacrylamide gel. When conducted prop-erly, this method of detection is sufficiently sensi-tive to detect a protein band containing as little as0.1 to 0.5 �g of a polypeptide. An alternative stain-ing method, silver staining, is sensitive to about10 ng of protein contained in a single band on theacrylamide gel. Silver staining begins by saturatingthe gel with a solution of silver nitrate. Next, a re-ducing agent is added to cause the reduction of Ag�

ions to metallic silver (Ag), which precipitate on theproteins in the gel and cause the appearance of pro-tein bands that are black in color. Silver staining istechnically more difficult, and therefore is used onlywhen extreme sensitivity is required.

Isoelectric Focusing. Isoelectric focusing is anelectrophoretic technique that separates macro-molecules on the basis of their isoelectric points (pI,pH values at which they carry no net charge). Aswith SDS-PAGE, this process can be carried out ina “slab” format. A pH gradient is established in thepolyacrylamide gel with the aid of ampholytes,which are small (�5000 Da) polymers containingrandom distributions of weakly acidic and weaklybasic functional groups (e.g., carboxyls, imidazoles,amines, etc.). A polyacrylamide gel containing theseampholytes is connected to an electrophoresis ap-paratus that contains dilute acid solution (H�) inthe anode chamber and a dilute base (OH�) solu-tion in the cathode chamber (Fig. 4-9).


the macromolecules being separated. In otherwords, the pore size of the gel should be largeenough that all of the macromolecules in the sys-tem can migrate freely to their appropriate iso-electric points. If the pore size of the matrix lim-its the rate of migration of the proteins with highermolecular weights, the separation may turn out tobe influenced more, or as much as, by size as byisoelectric point. If polyacrylamide gels are usedfor this purpose, the concentration of acrylamidein the gel should not exceed 8%. If desired, youcan perform isoelectric focusing using protocolsthat employ other “slab” matrix materials, such asagarose, or in a liquid matrix format that uses a su-crose density gradient as the anticonvection agent.You should also be aware that ampholytes are com-mercially available in a wide array of pH rangesthat may be required for separation of a moleculeof interest (e.g., pH 3–5, pH 2–10, pH 7–9, etc.).In the past, isoelectric focusing has been performedin the “slab” format and employed largely as an an-alytical technique. More recently, systems havebeen developed that will allow isoelectric focusing

to be carried out in a preparative (large-scale, non-denaturing) format. Although the principles of theseparation are the same as those described above,preparative isoelectric focusing is carried out us-ing a liquid anticonvection agent rather than a gelslab.

Agarose Gel Electrophoresis. Agarose gel elec-trophoresis is the principal technique used to de-termine the size of high-molecular-weight nucleicacids (DNA and RNA). Agarose is a long polymerof galactose and 3,6-anhydrogalactose linked via (1 n 4) glycosidic bonds. This material is readilyisolated from seaweed. Agarose polymers may con-tain up to 100 monomeric units, with an averagemolecular weight of around 10,000 Da. Agarosegels are cast by dissolving the white agarose pow-der in an aqueous buffer containing EDTA and ei-ther Tris-acetate or Tris-borate as the bufferingspecies (TAE or TBE buffer, respectively, see Table4-4). When the sample is heated to just below boil-ing, the agarose powder dissolves in the buffer toform a clear solution. As the solution slowly cools

(–)

(+)

(–)

(+)

Anticonvection agent(gel system or

sucrose gradient)

Ampholytes

Protein molecules

Diluteacid

Dilutealkali

Protein bandedat isoelectric pH value

Current flowfor short time

a b

Figure 4-9 Steps of an isoelectric focusing procedure.


Table 4-5 The Effective Separation Range of

Agarose Gels of Various

Composition for Separation of

Nucleic Acids

Effective Separation

% Agarose (wt/vol) Range (base pairs)

0.8 700–9000

1.0 500–7000

1.2 400–5000

1.5 200–3000

2.0 100–300

to room temperature, hydrogen bonding within andbetween the polygalactose units in the solution willcause the formation of a rigid gel with a relativelyuniform pore size. The induction of this polymer-ization event involves no chemical reaction, unlikethe polymerization process described above forpolyacrylamide gels.

As with polyacrylamide gels, the pore size of thegel can be controlled by the percentage of theagarose dissolved in the solution. A high percentagarose gel (say, 3% wt/wt) will have a smaller poresize than a lower (0.8% wt/wt) agarose gel. Thepercent of agarose to be cast in the gel will be de-termined by the size of the various molecules to beresolved during electrophoresis; the smaller themolecular weight of the molecules to be resolved,the higher percent agarose (smaller pore size) thegel should contain. A list of the effective separationranges with agarose gels of various percent agaroseis shown in Table 4-5.

Once the solution is heated to dissolve theagarose, the solution is cooled momentarily andpoured into a slab mold fitted at one end with aTeflon or plastic comb. After the solution poly-merizes, the comb is removed to create wells intowhich the desired DNA or RNA samples will beapplied. The gel is then transferred to an elec-trophoresis chamber and is completely covered withthe same TAE or TBE buffer that was used to castthe gel. Next, the nucleic acid sample is mixed witha viscous buffer (30% glycerol) containing one ormore tracking dyes that will be used to monitor theprogress of the electrophoresis. Bromophenol bluedye will migrate at the same rate as a DNA mole-cule of about 500 base pairs, while xylene cyanole

dye will migrate at the same rate as a DNA mole-cule of about 4000 base pairs.

The nucleic acid samples are loaded into thewells of the gel, along with a sample of DNA frag-ments of known molecular weight (number of basepairs) in one of the wells. These standards will beused following the electrophoretic separation to aidin the determination of the size of the nucleic acidsamples present in the unknown sample (see below).Remember that the DNA is negatively charged. Because of this, the cathode (negative electrode)should be connected to the side of the apparatusnearest the wells in the gel, while the anode (posi-tive electrode) is connected to the opposite side ofthe apparatus. A voltage of about 4 to 6 V/cm is ap-plied to the system (�50–70 mA of current), andthe electrophoresis is continued until the bro-mophenol blue dye front reaches the end of the gel.You will notice that agarose gel electrophoresis, un-like SDS-PAGE, does not employ the use of a stack-ing gel. Since the nucleic acids in the sample havea much greater frictional component in the gel thanthey do in the buffer contained in the wells, the nu-cleic acids focus very quickly at the buffer–gel in-terface before entering the matrix.

How do you visualize the nucleic acids follow-ing the electrophoretic separation? Ethidium bro-mide is a fluorescent dye that has the ability to in-tercalate between the stacked bases of nucleic acidduplexes (i.e., the double helix of DNA). After elec-trophoresis, the gel is placed in a solution of TBEor TAE buffer containing ethidium bromide, whichdiffuses into the gel and associates with the nucleicacids. Following a short destaining period in bufferwithout ethidium bromide (to remove it from theareas on the gel that do not contain nucleic acids),the gel is briefly exposed to ultraviolet (UV) light(256–300 nm), revealing the nucleic acid fragmentsas orange or pink fluorescent bands in the gel. Thepattern of fluorescent bands is recorded by pho-tographing the gel in a dark chamber while the gelis exposed to UV light from below. A filter is in-cluded to screen out UV light, so that most of thelight that exposes the film is from the fluorescenceof the nucleic acid bands that contain tightly boundethidium bromide.

The method of determining the molecularweight of an unknown nucleic acid sample is ex-actly the same as that described for the determina-tion of protein molecular weight in SDS-PAGE.


The log molecular weight of the nucleic acid sam-ples of known sizes are plotted against their rela-tive mobilities to produce a standard curve. Fromthe relative mobility of an unknown nucleic acidfragment on the same gel, the molecular weight andnumber of base pairs that the fragment contains canbe readily determined. (A single base has an aver-age molecular weight of approximately 320 Da,while a single base pair has an approximate averagemolecular weight of 640 Da).

In the experiment that follows, you will have anopportunity to gain experience with some of theforegoing principles by using SDS-PAGE to deter-mine the number of polypeptides and its (their)molecular weight(s) in an unknown protein sample.


1.5 M Tris chloride, pH 8.8


30% acrylamide (292 g/liter acrylamide, 8 g/liter N,N�-methylenebisacrylamide) [CAUTION, ACRYL-AMIDE IS TOXIC!]

10% (wt/vol) SDS in water

TEMED (N,N,N�,N�-tetraethylethylenediamine)

10% (wt/vol) ammonium persulfate in water—freshlyprepared

Running buffer (25 mM Tris, 250 mM glycine, 0.1%(wt/vol) SDS, pH to 8.3 with HCl)

4X Sample buffer:


30% (vol/vol) glycerol

10% (vol/vol) 2-mercaptoethanol

8% (wt/vol) SDS

0.001% (wt/vol) bromophenol blue

Coomassie Blue staining solution (40% methanol, 10%glacial acetic acid, 0.25% (wt/vol) Coomassie Bril-liant Blue R-250 in water)

Destaining solution (40% methanol, 10% glacial aceticacid in water)

Power supply and electrophoresis apparatus

Protein molecular weight standards (suggested standards:phosphorylase [97,400 Da], bovine serum albumin[66,200 Da], ovalbumin [45,000 Da], carbonic an-hydrase [31,000 Da], soybean trypsin inhibitor[21,500 Da], and lysozyme [14,400 Da] at 1 mg/mleach in a single mixture)

Unknown protein solutions (approximately 1 mg/ml)

Protocol

NOTE: The quantities of materials to be used anddetails of preparation of the PAGE gels for elec-trophoresis will depend on the number of studentsand the characteristics of the electrophoresis appa-ratus to be used. A typical apparatus for SDS-PAGEin vertical slab gels is shown in Figure 4-10; the in-structions that follow assume that you are usingsuch a system. Your instructor will provide detailedinstructions and will demonstrate the techniques tobe used in preparing the gels.

1. Prepare the running gel. (The volume to beprepared will depend on the size of the elec-trophoresis apparatus you will use and thenumber of gels to be prepared; consult your in-structor. The procedure described here is for20 ml of a 10% monomer running gel.) Mixthe following, in order, in a clean 100-mlbeaker: 7.9 ml distilled H2O, 6.7 ml 30% acryl-amide solution, 5.0 ml 1.5 M TrisHCl, pH 8.8,200 �l 10% SDS, 8 �l TEMED, and 200 �lfreshly dissolved 10% ammonium persulfate.The ammonium persulfate should be addedlast, as it initiates polymerization.

2. Pour the freshly mixed running gel solutioninto the spaces between parallel glass or plas-tic gel-forming sheets of the electrophoresisapparatus, leaving suficient space at the top forthe stacking gel to be added later. (This is usu-ally about 1 cm from the top, but the exact dis-tance depends on the apparatus being used;consult your instructor.) Create a flat surfaceabove the running gel solution by gently addinga water layer on top of the more dense gel so-lution, and allow the gel to polymerize for atleast 30 min.

3. Pour off the water from above the polymerizedrunning gel. This should leave a very sharp flatsurface at the top of the gel.

4. Prepare the stacking gel. (The procedure de-scribed here is for 4 ml of a 4% monomer stack-ing gel; again, consult your instructor to see ifyou should prepare a different amount.) Mixthe following, in order, in a clean test tube: 2.7ml distilled H2O, 0.67 ml 30% acrylamide so-lution, 0.5 ml 1.0 M TrisHCl, pH 6.8, 40 �l10% SDS, 4 �l TEMED, and, added last, 40 �lfreshly dissolved 10% ammonium persulfate.


Clamp to secure gel

Upperelectrode

Lowerelectrode

Supports for bottom of gel

Lower bufferreservoir

Upper bufferreservoir

Spacers

Glass plates

Well forming template

Loading wells forsample application

Gel slab

Figure 4-10 A typical apparatus for vertical slab SDS-PAGE. (From Wilson, K., and Walker, J. (1994)

Principles and Techniques of Practical Biochemistry, 4th ed. Cambridge, UK. Cambridge

University Press, Fig. 9-1. Reprinted with permission.)

Mix the solution gently and proceed immedi-ately to Steps 5 and 6.

5. Pour the stacking gel into the space above therunning gel.

6. Immediately insert the comb to create samplewells in the stacking gel before it polymerizes.Be sure that the comb is clean and free of for-eign material and that no air bubbles formaround the comb. If bubbles do form, tilt thegel slightly and tap the glass near the area ofthe bubble to dislodge it and allow it to rise tothe top. Be sure that you fill the stacking gelslightly over the top of the small plate, as thestacking gel will shrink slightly during poly-merization. Allow the gel to polymerize for atleast 30 min at room temperature.

7. Remove the bottom spacer from between theplates, and place the polymerized gel in theplates in which it was formed into the elec-trophoresis apparatus.

8. Add running buffer to the chambers at the bot-tom and top of the gel. Gently remove thecomb from between the plates. Be sure that nobubbles interfere with uniform contact of theliquid in the running buffer with the gel at thebottom or at the top. The wells in the stackingportion of the gel should be uniformly filledwith buffer with no bubbles or gel debris inthem. If bubbles need to be removed, use a sy-ringe and buffer to gently force them out. Also,be sure that there is no leaking between the twobuffer chambers. Remember that the current


will take the path of least resistance, and it willnot pass through the gel unless the two bufferchambers are separated.

9. Prepare the standard and unknown proteinsamples for analysis by mixing 10 �l of eachwith 5 �l of 4X sample buffer in microcen-trifuge tubes, mixing gently, and heating at100°C for 5 min in a boiling water bath. Thisstep is to ensure that all of the proteins in thesample are completely denatured. Remove thesamples from the water bath and allow them tocool to room temperature.

10. Load the samples onto the gel with a syringe,taking care to place the needle so that the denseblue solution settles gently in a layer at the bot-tom of the sample wells, displacing the runningbuffer upward as it is added. Be careful not totear or distort the soft stacking gel as you addthe samples. Several students can share a gel;for example, if a 10-well gel is used, eight stu-dents can analyze their unknowns and two sam-ples of standards can be analyzed, one in a cen-ter well and one in a well at the edge of the gel.

11. Connect the electrodes of the apparatus to thepower source. The anode (� pole) must be atthe bottom of the apparatus. (Can you explainwhy?)

12. Apply a constant 200-V field to the apparatus,and continue the electrophoresis until the bro-mophenol blue tracking dye reaches the verybottom of the gel. Bromophenol blue is a smallanionic dye, which will migrate faster than anyprotein during electrophoresis.

13. Turn off the power supply, and disconnect theelectrodes from it.

14. Gently remove the gel from the apparatus. Us-ing a spatula or razor blade, separate the twoplates by gently prying up one corner of a sin-gle plate. The gel should adhere to one of theplates. Using the same spatula or razor blade,separate the soft stacking gel from the morerigid running gel. Discard the stacking gel inthe specified container (polyacrylamide is toxicand should be properly disposed of ).

15. Gently transfer the running gel to a pan filledwith approximately 50 ml of Coomassie Bluestaining solution and allow it to soak in the so-lution for 1 hour.

16. Pour off the staining solution (save it, becauseit can be reused many times), and replace it

with destaining solution for 1 hour, replacingit with fresh destaining solution every 15 min.Destaining proceeds more rapidly if the gel isgently agitated in the solution continuouslyduring destaining. Destaining is completewhen the proteins appear as blue bands on atransparent gel background.

17. The gel can now be photographed or placedbetween two sheets of cellophane (using waterto make the gel easier to slide around, then gen-tly squeezing out the excess water, but leavingno bubbles) and dried under a vacuum on acommercial gel drier to keep a permanentrecord of the electropherogram.

Data Analysis

1. For each protein band in the lanes in whichstandards and the unknown sample were ana-lyzed, measure the distance from the bottom ofthe well (in mm) to the center of the band. Alsomeasure the distance from the bottom of thewell to center of the dye front (in millimeters)in the same lane.

2. Calculate the Rf value for each protein band ac-cording to:

Rf �

3. Use the Rf values for the protein standards toprepare a standard curve that plots the log ofthe molecular weight of each standard proteinon the ordinate versus the Rf value for that pro-tein on the abscissa. Do all of the values fall ona straight line? Can you predict what the curvewould look like for proteins that are muchlarger and much smaller than the standards youused?

4. From the Rf value(s) of the protein band(s) inyour unknown sample, determine the molecu-lar weights of these proteins by interpolationon your standard curve. How many bands didyou observe in your unknown? What was/werethe molecular weight(s)? If there was more thanone band in your unknown, what can you sayfrom the intensity of Coomassie staining of thebands about the relative abundance of the pro-teins in the sample?

distance migrated by the protein band (mm)��

distance migrated by the dye front (mm)


Exercises

1. You wish to resolve the compounds in Figure4-11 by paper electrophoresis in a system thatuses a pH 7.0 buffer. At pH 7.0, will all of thesecompounds migrate in one direction, so thatyou can place the origin line at one end of thepaper? If so, would you place the origin linenear the anode (� pole) or cathode (� pole)?What is the predicted relative order of elec-trophoretic migration of these compounds atpH 7.0?

2. Will the three compounds in Figure 4-11 beclearly separated from one another by elec-trophoresis at pH 11? If so, describe the elec-trode toward which they will migrate and theorder of migration. If not, why not?

REFERENCES

Dunbar, B. S. (1987). Two Dimensional Electrophoresisand Immunological Techniques. New York: PlenumPress.

Gersten, D. M. (1996). Gel Electrophoresis (EssentialTechniques Series). New York: Wiley.

Kuhr, W. G. (1990). Capillary Electrophoresis. AnalChem 62:403R–414R.

Laemmli, U. K. (1970). Cleavage of Structural Proteinsduring the Assembly of the Head of BacteriophageT4. Nature 227:680.

Sambrook, J., Fritsch, E. F., and Maniatis, T. (1989).Molecular Cloning: A Laboratory Manual, 2nd ed.Cold Spring Harbor, NY: Cold Spring HarborLaboratory Press.

Tal, M., Silberstein, A., and Nusser, E. (1980). WhyDoes Coomassie Brilliant Blue Interact Differentlywith Different Proteins? J Biol Chem 260:9976–9980.

Wilson, K., and Walker, J. M. (1994). Principles andTechniques of Practical Biochemistry, Chapter 9. Cambridge, UK: Cambridge University Press.

CH2

NH3

CH

CH COOH

COOH

CH

�

CH2OH

NH3 CH2 COOH�

glycine

glutamicacid

pKa2 � 9.6 pKa1

� 2.4

pKa3 � 9.8

pKa1 � 2.1

pKa2 � 3.9

pKa2 � 6.0

pKa1 � 1.0

C OH

OH

OHO

OH

P

-glycerolphosphate

Figure 4-11 Compounds to be separated in the

Exercises, questions 1 and 2.

l (

81

so as to introduce specific amino acid substitutionsinto it, has become a powerful and very widely usedtool for the investigation of structure–function re-lationships in proteins. Whether major or minor,such differences in structure are the consequencesof differences in the total number and chemical na-ture of amino acids in a protein, the primary se-quence of these amino acids, and the manner inwhich the polypeptide chain is folded in space. Tounderstand the way biological function is conferredon proteins it is necessary to study the chemistry ofamino acids and proteins, as well as the methodsfor elucidating the higher levels of structure.

Amino Acids: Identification andQuantitative Determination

The individual amino acids of a protein can be lib-erated by hydrolyzing the peptide (amide) bonds (–CO–NH–) that link them. The usual procedureis to dissolve the peptide or protein in 6 N HCl andheat the solution in a sealed, evacuated tube at100°C for 8 to 72 hr (or for shorter times at highertemperatures). All amide linkages (including theside chain amides of glutamine and asparagine) arecleaved under these conditions. Certain amino acidsare entirely (tryptophan) or partially (serine, thre-onine, tyrosine, cysteine) destroyed, so that specialprecautions are required for the quantitative deter-mination of these amino acids.

The amino acids in a protein hydrolysate can be conveniently separated for qualitative analysis by paper or thin-layer chromatography or by elec-

SECTION II

Proteins and Enzymology

Introduction

Most of the astonishing variety of structural fea-tures and biochemical activities of cells is a conse-quence of an equally extraordinary variety in thestructures of the proteins in those cells. In somecases the relation between the structure and func-tion of a protein is fairly obvious. For example, themajor structural proteins of hair and collagen fi-brils, keratin and tropocollagen, have high molec-ular weights and regularly repeated amino acid se-quences that endow them with the ability to formlong, regularly structured strands. These proteinsare quite different in size and amino acid composi-tion from lower-molecular-weight, soluble pro-teins, such as enzymes and hormones, for which the relationship between protein structure and biochemical function is far less obvious. Anotherexample is provided by integral membrane or transmembrane proteins, which usually containpredictable sequences of amino acids that fold intohelices with hydrophobic surfaces that adapt theseregions of the protein to stable residence in the hy-drophobic interior of the membrane. The varia-tions in chemical structure among other proteinsare small or subtle; yet these small differences maybe associated with major differences in physiologi-cal function. For example, hemoglobin S differsfrom hemoglobin A by only a single amino acidresidue, yet the behavior of these two proteins inred blood cells is very different. Many other casesare known in which replacement of a single aminoacid residue by mutation radically alters the bio-logical function of a protein. In fact, the techniqueof in vitro mutagenesis, the process of engineeringknown changes into the DNA encoding a protein

82 SECTION II Proteins and Enzymology

trophoresis. Qualitative and quantitative analysis ofamino acid mixtures can be achieved with muchgreater convenience and accuracy using automatedion-exchange chromatography. A common meansof detecting amino acids is by determination of theirreaction with ninhydrin (triketohydrindene hy-drate). Ninhydrin decarboxylates � amino acidsproducing an intensely colored blue-purple prod-uct, CO2, water, and aldehydes derived from thecarbon skeleton of the amino acid (see Fig. 6-9, Ex-periment 6). Ninhydrin yields a similar blue-purple product on reaction with primary amines orammonia. The reagent also reacts with imino acidssuch as proline to yield a yellow product. Undercontrolled conditions the color development isquantitative. Amino acids also react with fluo-rescamine or with orthophthalaldehyde and mer-captoethanol to yield fluorescent products that pro-vide a more sensitive means of analyzing amino acidmixtures.

Another widely used method for qualitative andquantitative analysis of amino acid mixtures is high-performance liquid chromatography (HPLC) (seeExperiments 2 and 6). The mixture of amino acids isfirst subjected to reaction with phenylisothiocyanate(PITC) to convert them to the phenylthiocarbamyl–amino acid derivatives, which are then subjected tochromatographic separation. The derivativatizationof the amino acids serves two purposes: it attaches aUV-absorbing “tag,” which makes their quantitativedetermination easy, and it converts them to a morehydrophobic form, which is necessary for good sep-aration on the reverse-phase system commonly usedwith this technique. This method of amino acidanalysis will be used in Experiment 6.

Amino Acids: Ionic Properties

All amino acids contain ionizable groups that act asweak acids or bases, giving off or taking on protonswhen the pH is altered. As is true of all similar ion-izations, these ionizations follow the Henderson–Hasselbalch equation:

(II-1)

pH � pKa � log10

The pKa is the negative log10 of the acid dissocia-tion constant of the ionizable group. Examination

[unprotonated form (base)]��

[protonated form (acid)]

of this equation leads to a further understanding ofpKa. When the concentration of the unprotonatedform equals that of the protonated form, the ratioof their concentrations equals 1, and log10 1 � 0.Hence, pKa can be defined as the pH at which the con-centrations of unprotonated and protonated forms of aparticular ionizable species are equal. The pKa alsoequals the pH at which the ionizable group is at itsbest buffering capacity; that is, the pH at which thesolution resists changes in pH most effectively. ThepKa value of the weak acid or base in a bufferingsystem is therefore an important consideration inchoosing a buffer of a given pH. For example, if youwanted to prepare a buffer at pH 7.0, you mightchoose inorganic phosphate (pKa2

� 6.8). pKa val-ues are derived with unit activity concentrations (ap-proximately 1.0 M) of components. The pKa valuesof many biochemically important ionic compoundsshift slightly on dilution to more physiological con-centrations. Biochemists therefore use the symbolpKa� to designate pKa values in dilute ionic systems.

Consider applying the Henderson–Hasselbalchequation to the titration of glycine with acid andbase. Glycine has two ionizable groups: a carboxylgroup and an amino group, with pKa values of 2.4and 9.6, respectively. In water at pH 6.0, glycine ex-ists as a dipolar ion, or zwitterion, in which the car-boxyl group is unprotonated (–COO�) and theamino group is protonated to give the substitutedammonium ion (–NH3

�). (Verify this, using theHenderson–Hasselbalch equation, pH 6.0, and therespective pKa values.) Addition of acid to the so-lution lowers the pH rapidly at first and then moreslowly as the buffering action of the carboxyl groupis exerted (see Fig. II-1). At pH 2.4 the pKa isreached and half of the acid has been consumed;the carboxyl group is half ionized and effectivelybuffers the solution. Further addition of acid resultsin titration of the remainder of the carboxylate ionsin the solution (e.g., at pH 1.4 only 1 carboxyl in11 is ionized). Titration of the protonated �-aminogroup with base follows a similar curve into the al-kaline region. The concentration of the protonatedand unprotonated species of each ion at any pH canbe calculated from the Henderson–Hasselbalchequation. The intersection between the titration ofthe carboxyl group and the titration of the aminogroup defines the pH at which glycine has no netcharge, and is called the “isoelectric point” (pI, Fig.II-1). Experiment 5 provides the student with an

SECTION II Proteins and Enzymology 83

14

1.0 0.5 0 0.5 1.0

12

10

8

6

4

2

0

pKa1 = 2.4

pKa2 = 9.6

pI

pH

Acid Alkali

Equivalents

Figure II-1 Titration curve of glycine.

opportunity to develop a further understanding ofthe acid–base properties of amino acids as these arestudied by titration.

Most amino acids contain carboxyl and aminogroups having pKa values similar to those of glycine.In addition to these groups, many amino acids con-tain other ionizable groups, which introduce other“steps” or pKa values into their titration curves. Itis largely these side chain groups—those not linkedin peptide bonds to adjacent amino acids—that account for the ionic properties of proteins. TableII-1 lists the ionizable groups of proteins and aminoacids, the nature of their ionizations, and their ap-proximate pKa values. Notice that the pKa valuesfor groups on proteins are given as ranges ratherthan as single values. This is because the ionizationof various groups on proteins is often modified bythe presence of other groups on the protein. Thisis evident from a comparison between the typicalpKa values for the �-amino and �-carboxyl groupsof a free amino acid, 2.3 and 9.5, respectively, andthe corresponding values for these groups whenthey occur at the ends of a peptide, even a peptideas small as a dipeptide, which are about 3.5 and 8.5.The close proximity of the charged –COO� and –NH3

� groups in an amino acid affects the ioniza-tion of the other group by electrostatic attractionand repulsion. These effects are largely lost whenthese groups are farther apart in peptides.

Protein Structure

The elements of protein structure are divided intofour classes: primary, secondary, tertiary, and qua-ternary. Primary structure refers to the linear se-quence of amino acids linked by amide bonds alongprotein chains. These polymer chains vary in lengthfrom a few amino acid residues (oligopeptides) tomolecules containing 2000 or more amino acids.Most proteins are from 100 to 500 amino acidresidues in length. One or more of each of 20 nat-ural amino acids may be present in each proteinmolecule. In some cases the amino acids undergoposttranslational chemical modification, which in-troduces still more variety into protein structure.

Both secondary and tertiary structure refer tothe arrangement of the polypeptide chain in a three-dimensional structure. Secondary structure generallydescribes the arrangement of neighboring aminoacids into more or less regular patterns that are heldtogether by hydrogen bonding (H-bonding) be-tween the carbonyl oxygens and the amide hydro-gens (CPOZH–N) of the peptide backbone. Sev-eral such regular patterns have been discovered byx-ray crystallographic analysis of many proteins.The two patterns of secondary structure that occurmost frequently are the �-helix and the �-pleatedsheet (or simply �-strands) (Fig. II-2). Certain fi-brous proteins may consist almost entirely of oneof these secondary structural elements. In contrast,H-bonded secondary structures account only for aportion of the peptide chain folding in globular pro-teins (such as most enzymes) and tend to occur inshorter segments connected by bends or turns tofold them into a compact structure.

The complete pattern of folding of the polypep-tide chain of a protein, whether regular or irregu-lar, is called the “tertiary structure.” The tertiarystructure of any protein is the sum of many forcesand structural elements, many of which are the re-sult of interactions between the side chain groupsof amino acids in the protein. Some of these inter-actions are described below.

Disulfide Bonds. The sulfhydryl groups of theamino acid cysteine often (but not always) formpartners in intrachain or interchain disulfide bonds,and they thus serve to fix a protein molecule into athree-dimensional structure. While such bonds are,


Table II-1 Dissociation of Ionizable Groups of Amino Acids and Proteins

pKa Range pKa

Group Ionization (on a protein) (free amino acid)

Carboxyl (�) OCOOH uv H� � OCOO� 2.0–4.0 1.8–2.4

Carboxyl OCOOH uv H� � OCOO� 3.0–4.7 3.8–4.3

(, of glu, asp)

Ammonium (�) ONH3�

uv H� � ONH2 7.5–8.5 8.9–9.7

sec-Ammonium 10.2–10.8 10.6

(proline)

Ammonium ONH3�

uv H� � ONH2 9.4–10.6 10.5

(, lysine)

Phenolic

hydroxyl 9.8–10.4 10.1

(tyrosine)

Imidazolium 5.6–7.0 6.0

(histidine)

Guanidinium 11.6–12.6 12.5

(arginine)

Sulfhydryl OSH uv H� � OS� 9.4–10.8 10.4

(cysteine)

H� �NH2

NH2

CNH�

NH2

NH

CNH

N

N�

H

H

N

N

H

H� �

OH O�H� �

NH�NHH�

2 �

of course, part of the primary covalent linkages ofthe protein molecule, it is more useful to think ofthem as contributing to the chain folding.

Hydrogen Bonds. Hydrogen atoms that are at-tached to the electronegative atoms oxygen or nitrogen (e.g., tyrosine-OH, glutamine, and as-paragine amides) frequently interact by H-bondingwith other electronegative atoms (usually oxygen,as in carboxyl oxygen). This noncovalent bondingcan stabilize the interactions of amino acid sidechains with other side chains or with the carboxylsand amides of the protein backbone.

Ionic Attractions and Repulsions. Attractions andrepulsions occur between ionized functional groups.

Hydrophobic and Hydrophilic Interactions. Al-though proteins are generally soluble in water ordilute salt solutions, they contain many amino acidswith aliphatic or aromatic side-chain residues thathave low solubility in water, apparently because of

the highly organized “icelike” structure that mustbe formed around them. Hence, such hydrophobicresidues are frequently found clustered inside theprotein molecule, where there are weak van derWaals associations among them and they are “removed” from energetically unfavorable contactwith water molecules. The net energy gained fromsuch “oil drop” arrangements contributes substan-tially to overall protein folding. Similarly, aminoacids with highly polar or charged side chainresidues tend to be exposed to the solvent, wheretheir interaction with water molecules is energeti-cally favorable.

Proline Residues. Since proline is an imino acid,unlike the other amino acids, it does not fit into aregularly organized helical or pleated sheet struc-ture. Hence, proline residues are likely to be foundin disorganized regions within the polypeptidechain or in “bends” associated with a change in theoverall direction of the peptide chain.


RC

C

O

O

O

C

C

C

C

CO

N

R

R

R

R

R

R

N

N

N

N

N

N

N

N

R

R

N

C

C

C

C

C

C

C

CC

C

C

O

O O

O

O

O

O

O

O

O

O

O

R

N

R

H

H

H

H

HH

H

H

H

H

H

HH

H

H

H

H

H

H

H

H

H

H

H

C

C

C

CC

O

O

N

N

C

�C

�C

Figure II-2 Major elements of secondary structure of proteins. Left, the �-helix; right, representation of

the antiparallel pleated sheet structures for polypeptides. (After Pauling, L., and R. B. Corey

(1951). Proc Natl Acad Sci USA 37:729).

Many of these features are evident in the three-dimensional structures of ribonuclease and myo-globin shown in Figure II-3. To understand thecommon and distinguishing features of the sec-ondary and tertiary structure of various proteins asdetermined by x-ray crystallographic analysis, thestudent should study several examples in standardtextbooks of biochemistry.

Some proteins can be fully described by primary,secondary, and tertiary structure, but many others,

perhaps most, have an additional level of structuralorganization—quaternary structure. Quaternarystructure refers to the organization of identical ordifferent polypeptide chains into noncovalentlylinked aggregates containing a fixed number ofpolypeptide subunits in a single functional protein.Thus, many proteins have been found to consist of2, 3, 4, 6, 8, 12, or other numbers of polypeptidechains, usually bound in regular arrangements.Some examples are shown in Figure II-4.


20 Å 20 Å

Figure II-3 Ribbon diagrams showing the three-dimensional structures of ribonuclease A (left) and myo-

globin (right). Helical segments are presented as coiled ribbons, �-strands as smoothly curv-

ing ribbons, and regions of irregular structure as thin tubes. The four disulfide bonds of ri-

bonuclease are shown as ball and stick models as are the atoms of the heme A prosthetic

group and the side chains of two histidine residues that ligate to the heme in myoglobin.

(From Stryer, L. (1995). Biochemistry, 4th ed. New York: Freeman, Figures 2-45 and 2-43.

Reprinted with permission.)

Determination of ProteinStructure

The primary structure (i.e., the amino acid se-quence) of a protein can be determined by stepwisechemical degradation of the purified protein. By farthe most powerful and commonly used techniquefor doing this is the automated Edman degradation.The amino terminal amino acid residue of thepolypeptide is reacted with Edman’s reagent(phenylisothiocyanate) to form the phenylthiocar-bamyl derivative, which is removed without hy-drolysis of the other peptide bonds by cyclizationin anhydrous acid. The amino acid derivative is con-verted to the more stable phenylthiohydantoin andidentified by HPLC. The process can be repeatedmany times, removing the amino acids from theamino terminus of the polypeptide one residue at atime and identifying them until the entire sequence

is determined. These reactions are shown in Fig. 6-5 (Experiment 6). When long polypeptides aredegraded stepwise, the accumulated effects of smallimperfections in the chemistry of the process even-tually makes it impossible to deduce the sequence.Under favorable conditions, up to 100 residues canbe determined, and as little as 5 pmol of protein canbe analyzed. In the future, it is likely that mass spec-trometry will also provide increasingly powerfulmeans of determining the sequence of polypeptides.

In contemporary biochemistry the amino acidsequence of a protein is much more likely to be de-termined indirectly from the sequence of the DNAthat encodes the protein or, in the case of eukary-otes, from the sequence of cDNA made frommRNA encoding it. This is because DNA se-quences can be determined far more rapidly and ac-curately than protein sequences. Since the geneticcode is known, it is a simple matter to deduce the


45 Å

45 Å

140 Å R

RR

C C

C

C

C

R

R

R

a

c

b

Dihydrolipoyltransacetylase core(each spherical enzyme unitcontains 3 identical subunits)

24 molecules ofdihydrolipoyldehydrogenase stackedabout the dihydrolipoyltransacetylase core

Complete pyruvicdehydrogenase complexcontaining 24 moleculesof pyruvate dehydrogenasearranged on outsideof dihydrolipoyl transacetylase–dihydrolipoyldehydrogenase center

Figure II-4 Examples of the quaternary structure of proteins. (a) A drawing of glutamine synthetase of E.

coli showing the orientation of the 12 identical subunits of the enzyme. (b) A drawing of as-

partate transcarbamylase of E. coli showing the proposed orientation of the 6 catalytic sub-

units (labeled C, each MW � 33,000), and 6 regulatory subunits (labeled R, each MW �17,000), of the enzyme. (c) A drawing of the pyruvate dehydrogenese complex of E. coli

showing the proposed orientation of the various proteins in the multienzyme complex.

amino acid sequence of a protein from the DNAsequence of the gene that specifies it. In fact, com-puter programs have been developed to do this (seeExperiment 25).

The determination of the secondary and ter-tiary structure—that is, the details of the three-dimensional folding of the polypeptide chain of aprotein at high resolution—relies on one of twopowerful techniques: x-ray diffraction analysis ofprotein crystals and multidimensional high-fieldnuclear magnetic resonance (NMR) spectroscopy.Both methods provide very detailed structural in-

formation at the level of atomic resolution, andseveral thousand structures of proteins have nowbeen determined using these techniques. Thestandard repository for such structural informa-tion is the Brookhaven Protein Data Base(http://www.pdb.bnl.gov/). These structures pro-vide a rich source of information about the natureof protein architecture and the relationship be-tween protein structure and biological function. Adetailed description of these methods is beyondthe scope of this book, but a few general commentsshould be made. X-ray diffraction requires that


you be able to prepare suitable crystals of the pureprotein. The method is applicable to both largeand small proteins. Structures of very large pro-tein aggregates, e.g., the yeast proteasome, havebeen determined by x-ray diffraction. NMR spec-troscopy does not require that the protein be incrystalline form; analysis is conducted on concen-trated solutions of the protein, so the method isespecially valuable for proteins that are difficult tocrystallize. At present, structure determination byNMR is limited to proteins with molecularweights lower than about 30,000; this limit maybe raised in the future by advances in the tech-nology.

At a much lower level of resolution, it is possi-ble to estimate the �-helical and �-strand contentof proteins by measuring their circular dichroicspectra. Such determinations give a general idea ofthe secondary structural features of a protein, butdo not determine which segments of the primarystructure possess which secondary structures. Thiscan be predicted by computer analysis of the pri-mary sequence, however, since analysis of manyprotein structures determined from crystallographyand the study of synthetic polypeptides has given agood indication of the propensity of a given se-quence of amino acids to fold into �-helices or �-strands (see Experiment 25). You should be awarethat the predictions of such analyses have beenfound by subsequent structure determination to becorrect only about 70% of the time. Although cir-cular dichroism may be a crude tool for determin-ing secondary structures of proteins, it is quite valu-able for studying changes in the structures ofproteins, particularly their unfolding and refolding,as occurs during protein denaturation (see below).

Determination of MolecularWeights and QuaternaryStructures of Proteins

The quaternary structure of a protein can be largelydetermined by careful molecular weight determi-nation of the native protein, followed by dissocia-tion of the protein into its constituent polypeptidechains with denaturing agents. If nonidentical sub-units are found, they must be separated from eachother. Then the molecular weight of the isolatedsubunits must be determined, usually under dena-

turing conditions. Usually a small, integral (mostcommonly an even number) of subunits comprisethe native species. That is, the molecular weight ofnative species is a simple multiple of the molecularweight(s) of the subunits. Hence, techniques for de-termining the molecular weights of macromole-cules are central to determining quaternary struc-ture.

Determination of Subunit Molecular Weight

The determination of the primary structure (i.e.,the complete amino acid sequence) of a protein au-tomatically allows the calculation of the molecularweight of the polypeptide from the sum of theresidue weights of the amino acids. This is normallydone by the same computer programs that facili-tate the determination of the amino acid sequencefrom the sequence of the gene (see Experiment 25).This is probably the most common way of deter-mining subunit molecular weights in contemporarybiochemistry.

If small amounts of a purified protein are avail-able, but the amino acid sequence is not known, twomethods are generally available for determinationof its subunit molecular weight.

Electrophoresis in Sodium Dodecyl Sulfate (SDS).

(See also, Experiment 4.) Heating a protein at100°C in an SDS-containing solution (usually in thepresence of thiols to reduce any disulfide bonds)will completely dissociate and denature all polypep-tide chains from one another. The polypeptidechains bind a large quantity of SDS (about one SDSmolecule per two amino acid residues), therebygaining a strongly net negative charge. The bind-ing of so much SDS gives all of the proteins ap-proximately the same (negative) charge to mass ra-tio, regardless of what this ratio might be for thenative proteins at a given pH. Thus, the rate atwhich such SDS-polypeptide complexes migrateduring electrophoresis on polyacrylamide gels is inmost instances determined by viscous resistance totheir flow in the gel (i.e., by their frictional coeffi-cients), rather than by net charge. This phenome-non generates an empirical relation between mole-cular weight and electrophoretic mobility: thelogarithm of the molecular weight is proportionalto the relative mobility. Standards of known molec-


ular weight are used to calibrate the results. Obvi-ously, only molecular weights of constituentpolypeptide chains, not native aggregates, can bedetermined with this method. Proteins that bindabnormal amounts of SDS, proteins with very un-usual amino acid compositions, glycoproteins, andintrinsic membrane proteins often fail to migrate as expected on SDS–polyacrylamide gel elec-trophoresis.

Mass Spectrometry. Advances in the technologyof mass spectrometry have made it possible to de-termine the molecular weights of polypeptides ofthe sizes commonly found in proteins. This has re-sulted primarily from the development of noveltechniques for obtaining molecular ions of theselarge, nonvolatile molecules. Two such techniquesare electrospray ionization (ESI) and matrix-assisted laser desorption ionization (MALDI).When these methods for obtaining molecular ionsof proteins are combined with improvements intechnology for resolution and mass determinationof such ions, it has become feasible to determinemolecular weights of protein subunits to within0.01% of their value, which is more precise thanany method other than determination of the com-plete primary structure. In fact, mass spectrometryis now frequently used to check for errors in thedetermination of the sequence of proteins fromDNA sequencing.

Determination of Native Molecular Weight

Gel-Filtration or Size-Exclusion Chromatography.

Protein molecules will diffuse into the pores of appropriately chosen gel-filtration beads (e.g.,Sephadex or Bio-Gel). The molecules partition be-tween the solvent outside and inside of the gel beadsaccording to their ease of diffusion, which is relatedto their molecular weight (see Experiment 2). Formost globular proteins of generally similar shapes,a linear relationship between the logarithm of themolecular weight and the elution volume will beobserved, within certain molecular weight limits.Hence, comparison of the elution volume of a pro-tein to the elution volume of standards of knownmolecular weights on the same gel column will pro-vide a reasonably good estimate of the molecularweight of the protein. The method is unreliable,

however, if the protein in question has a shape thatis significantly different from that of the standardsused or if the protein subunits participate in a rapidequilibrium between states of aggregation contain-ing different numbers of subunits.

Direct Determination of Quaternary Structure.

None of the above methods is capable of revealingthe precise geometric arrangements of subunits ina native protein. Two methods have proven usefulfor this purpose: x-ray crystallographic analysis andelectron microscopy of the protein molecules usingnegative staining. The examples of quaternarystructure shown in Figure II-4 were determined bythese techniques.

Protein Denaturation

Proteins in their natural form are called “native pro-teins.” Any significant change in the structure of aprotein from its native state is called “denatura-tion.” Denaturation results from changes in the sec-ondary, tertiary, and quaternary structures of pro-teins. Generally, denaturation is to be avoided,because you wish to study a protein in as close toits native state as possible. Occasionally, however,you may wish to denature proteins deliberately. Forexample, proteins are often quickly denatured tostop an enzyme reaction when determining therates of enzyme reactions. Proteins are also oftendeliberately denatured so that the detailed natureof the unfolding and refolding of their polypeptidechains can be studied. Denaturation of proteinsusually leads to greatly decreased solubility, so it canbe a useful way to separate them from other classesof biological molecules during purification. Thefollowing are a few agents and techniques com-monly used to denature proteins.

Chaotropic Agents. Compounds such as urea andguanidine hydrochloride usually cause denaturationwhen present in high concentrations (4 to 8 M).The chemical basis of this disruption of proteinstructure is not completely understood. Urea andguanidine are thought to form competing hydro-gen bonds with the amino acid residues of the pep-tide chain, thereby disrupting the internal hydro-gen bonding that stabilizes the native structure.


These agents also alter the solvent properties of wa-ter so that hydrophobic interactions in the proteinare weakened. They probably act by a combinationof these effects. Urea- or guanidine-induced de-naturation is frequently partially or completely re-versed when the concentration of the competitivehydrogen bonding agent in the protein solution islowered by dialysis or dilution.

Detergents. Sodium dodecyl sulfate (SDS) andother ionic detergents are extremely effective pro-tein denaturants. Such agents usually have a highlypolar end and a hydrophobic end. SDS has beenshown to bind tightly to polypeptides, probably byhydrophobic interactions, and to convert them intorodlike structures that have a strong negativecharge. Associations between polypeptide chains(quaternary structure) are also disrupted by deter-gents.

Heat. Heat denatures most dissolved proteinswhen the temperature reaches higher than about50°C. Harsh heat treatment alters the secondaryand tertiary structures of proteins. Heat denatura-tion usually results in eventual protein precipitationas a result of destruction of the secondary structureand formation of random aggregates. Some pro-teins have been found to be heat-stable, especiallywhen ligands are bound to them (i.e., many en-zymes are protected against heat by their sub-strates). This property can be exploited during pu-rification (see Experiment 8).

Acid or Alkali. Proteins are amphoteric polyelec-trolytes; thus, changes in pH would be expected toaffect the existence of salt bridges, which reinforcethe tertiary structure of proteins. Further, when lo-calized areas of a protein acquire a large net posi-tive or negative charge, the ionizable groups repeleach other and thus place the molecule under con-formational strain. For example, denaturation of theproteolytic enzyme, pepsin, occurs at alkaline pHvalues. This may be attributed to internal strainarising from mutual repulsions of the ionized car-boxyls of aspartic and glutamic acids. Moderateconcentrations of certain acids (for example,trichloroacetic, phosphotungstic, or perchloricacid) usually produce complete denaturation and

precipitation. Proteins commonly are treated withthese acids to stop enzymatic reactions and to con-centrate protein samples prior to SDS-PAGEanalysis (see Experiment 8).

Oxidation or Reduction of Sulfhydryl Groups.

Many, but not all, proteins are sensitive to alter-ations in the oxidation–reduction potential of theirenvironment. The effect is caused in part by oxi-dation of sulfhydryl groups or reduction of disul-fide bonds. Not all proteins are equally sensitive tosuch alterations, but when they are, it is critical tobe aware of their sensitivity. The purification or as-say of some proteins can be accomplished only byproviding reducing conditions (reduced glu-tathione, free cysteine, dithiothreitol, or mercap-toethanol) in all buffer solutions.

Enzymatic Action. Proteinases (also called pro-teases) present in crude or unpurified proteinpreparations often catalyze the breakdown of pro-teins by hydrolyzing the peptide bonds. Since theseenzymes act more slowly at low temperatures, crudeprotein solutions are frequently kept cold (0 to 2°C)during the early stages of purification. In some casesit is also necessary to add inhibitors of proteinases,such as p-methylphenylsulfonyl fluoride (PMSF).

Protein Purification

There is no single or simple way to purify all pro-teins. Procedures and conditions used in the pu-rification of one protein may result in the denatu-ration of another. Further, slight chemicalmodifications in a protein may greatly alter itsstructure and thus affect its behavior during purifi-cation. Nevertheless, there are certain fundamentalprinciples of protein purification on which mostfractionation procedures are based. In all cases, theexperimenter takes advantage of small differencesin the physical and/or chemical properties of themany proteins in a crude mixture to bring aboutseparation of the proteins from one another. Sincethe precise nature of these differences is not usu-ally known in advance, the development of a pu-rification procedure necessarily involves consider-able trial and error. A procedure leading to a


homogeneous protein from a natural source usuallyrequires the sequential application of a number ofthe following techniques.

Overexpression of the Recombinant Protein.

Advances in molecular biology have provided an ex-tremely powerful means of purifying proteins byoverexpressing the genes encoding them from re-combinant plasmids in suitable host cells grown inculture. Instead of having to isolate a protein fromthe cell or tissue in which it normally resides, of-ten as a tiny fraction of the total protein, genomicDNA or cDNA specifying the protein is splicedinto appropriately designed plasmids, called “over-expression vectors,” which can be induced to directthe synthesis of very large amounts of the protein(see Section V). It is not unusual for such recom-binant proteins to be made at levels from 10 to 50%of the total cell protein on induction. This greatlysimplifies the task of purifying the protein: insteadof having to devise a procedure for a 1000- to100,000-fold enrichment of the protein, as is typi-cal for isolation of a protein from its natural source,purification of a recombinant protein by 2- to 10-fold yields essentially pure protein.

Although the use of recombinant genes for pro-tein production is often very successful, problemscan arise. Some genes or cDNAs are not expressedwell in the commonly used vector/host systems.Occasionally, the overproduced proteins do not foldproperly and aggregate into insoluble inclusionbodies in which the recombinant protein is dena-tured. Posttranslational modifications, such asphosphorylation or glycosylation, that occur in thenative cells and are important for the function ofthe protein may not occur in the recombinant ex-pression system.

Heat Denaturation. Because all proteins are notequally stable when heated in aqueous solution,fractionation can often be achieved by controlledheating. As noted above, the presence of a substrate,a product, or an inhibitor often stabilizes a proteinagainst heat denaturation.

Protein–Nucleic Acid Complexes. During pro-tein purification it occasionally becomes necessaryto remove basic proteins or contaminating nucleicacids. The controlled addition of sparingly soluble

material of opposite charge (e.g., addition of a ba-sic protein such as protamine for nucleic acid re-moval) results in the coprecipitation of the addedand unwanted material (e.g., as a protamine–nucleic acid complex).

Solubility Properties. The solubility of most pro-teins in aqueous solutions can be attributed to thehydrophilic interaction between the polar mole-cules of water and the ionized groups of the pro-tein molecules. Reagents that change the dielectricconstant or the ionic strength of an aqueous solu-tion therefore would be expected to influence thesolubility of proteins. The addition of ethanol, ace-tone, or certain salts (such as ammonium sulfate),usually at low temperatures to avoid denaturation,frequently results in precipitation of proteins. Suchagents are thought to act by effectively removingwater molecules that normally solvate the surfaceof proteins and favor their solubility. As the watermolecules are removed, nonspecific protein–protein interactions become dominant over pro-tein–water interactions and protein molecules ag-gregate and form insoluble precipitates. Of theseprecipitating agents, fractionation with ammoniumsulfate is by far the most commonly used. Use oforganic solvents is less generally useful, becausemany proteins are denatured by them. Step-by-stepapplication of such techniques to heterogeneousprotein solutions often results in fractionation (pu-rification) because of the differing degrees of solu-bility among the proteins in the solution.

The precipitations noted above usually resultwhen the attractive forces among protein moleculesexceed those between the protein molecules andwater. Proteins are polyelectrolytes with differentnumbers and types of ionizable groups; the netcharge on any given protein molecule and hence itsattraction to another protein molecule are functionsof the pH and ionic strength of the medium. There-fore, any treatment designed to separate proteinsby making use of relative solubility must take intoconsideration the pH of the medium. Often, alter-ation of only pH under a set of conditions is suffi-cient to effect an “isoelectric” precipitation. Choos-ing the correct conditions of pH, ionic strength, ordielectric constant to carry out a fractionation issomewhat arbitrary and is usually determined em-pirically.


Adsorptive Properties. Under certain conditionsof pH and low ionic strength, certain proteins willbe adsorbed by various substances. Calcium phos-phate gel and alumina C� gel, for example, are fre-quently used to adsorb specific proteins from het-erogeneous mixtures. The adsorbed proteins canfrequently be released from the insoluble supporteither by altering the pH or increasing the ionicstrength. Thus, purification can be obtained by re-moving either the extraneous unwanted proteins orthe desired protein with the gel.

Chromatographic Separation. Chromatographicseparation of protein mixtures has become one ofthe most effective and widely used means of puri-fying individual proteins. A more detailed discus-sion of the principles of chromatography and theirapplication to the separation of protein mixtures ispresented in Experiment 2 and is also illustrated inExperiments 8 and 10.

Proteins, as polyelectrolytes, can be separated byion-exchange chromatography in much the sameway as smaller molecules. Because of the size, easeof denaturation, and complex charge distribution ofprotein molecules, certain special techniques are re-quired. The most generally useful ion-exchange ma-terials for proteins are derivatives of cellulose, dex-tran, agarose, and polyacrylamide (e.g., Sephadex,Sepharose, BioGel, etc.). The most commonly usedanion exchangers are substituted with diethyl-aminoethyl (DEAE) or quaternary ammonium ethyl(QAE) groups; useful cation exchangers are substi-tuted with carboxymethyl (CM), phosphoryl (P), orsulfoethyl (SE) groups. Excellent separation of pro-tein mixtures can often be brought about by step-by-step or gradient elution of the sample from a col-umn of exchange material in which the ionicstrength of the eluting buffer is increased, the pHaltered, or both.

Chromatographic separation of proteins on thebasis of molecular size (or, more precisely, ease ofdiffusion) is readily accomplished by size-exclusionchromatography on dextran, agarose, or polyacryl-amide beads.

Affinity Chromatography. Affinity chromatogra-phy is a very useful new technique that takes ad-vantage of the natural affinity of a protein such asan enzyme for specific molecules, for instance, thesubstrate or a very effective inhibitor of that en-

zyme. The molecule that binds to the protein is co-valently attached to an insoluble support, such asagarose. This insoluble material can then be usedto adsorb the appropriate protein selectively fromother proteins that do not bind to the molecule at-tached to the support material. Some remarkablyeffective purifications have been accomplished byuse of this useful technique. A much more detaileddiscussion of the technique of affinity chromatog-raphy is presented in Experiment 2. An alternativeapproach to isolation of proteins by affinity chro-matography is to use genetic engineering to add se-quences coding for extra amino acids or even an en-tire protein to the gene encoding the protein youwish to isolate; the extra polypeptide is chosen toprovide a chemical “tag” that binds tightly andspecifically to an affinity adsorbent. The desiredprotein with its covalently attached tag can then beeasily isolated from all the other proteins in the cellsin which the recombinant protein was produced.The protein is then released from the adsorbent bydissociating the tag from the adsorbent. This ap-proach is illustrated in Experiment 10.

Electrophoretic Separation. The net charge on aparticular protein varies with the pH of the mediumin which it is dissolved. Accordingly, application ofan electric field to a buffered, heterogeneous pro-tein solution often results in their differential mi-gration in free solution or in heterogeneous systemswith an inert supporting material, such as a poly-acrylamide gel. Because of the difficulty of “scalingup” electrophoretic procedures, they are more com-monly used for analytical applications than for thepreparative scale purification of proteins. The useof electrophoresis is discussed in more detail in Ex-periment 4.

Criteria of Protein Purity. There are no tests forprotein purity, only for impurities. Crystallinity, theabsence of contaminating enzymes, and homoge-neous behavior in centrifugal fields (ultracentrifu-gation), in electric fields (electrophoresis), in ion-exchange chromatography, and in immunologicalreactions may all be consistent with purity, but theydo not prove it. We attempt to prove inhomo-geneity with as many tests as are available. Practi-cally, if a protein preparation behaves as a single en-tity in careful studies using these criteria, it may beassumed to have a high degree of purity.


Quantitative Determination of Proteins

In general, there is no completely satisfactory sin-gle method to determine the concentration of pro-tein in any given sample. The choice of the methoddepends on the nature of the protein, the nature ofthe other components in the protein sample, and thedesired speed, accuracy, and sensitivity of assay. Sev-eral of the methods commonly used for protein de-termination are discussed in the sections that follow.

Biuret Test. Compounds containing two or morepeptide bonds (e.g., proteins) take on a character-istic purple color when treated with dilute coppersulfate in alkaline solution. The name of the testcomes from the compound biuret, which gives atypically positive reaction. The color is apparentlycaused by the coordination complex of the copperatom and four nitrogen atoms, two from each oftwo peptide chains (Fig. II-5). The biuret test isfairly reproducible for any protein, but it requiresrelatively large amounts of protein (1 to 20 mg) forcolor formation. Because of its low sensitivity, thebiuret assay is no longer widely used.

Folin–Ciocalteu (Lowry) Assay. The quantitativeFolin–Ciocalteu assay (also often called the “Lowryassay”) can be applied to dried material as well asto solutions. In addition, the method is sensitive;samples containing as little as 5 �g of protein canbe analyzed readily. The color formed by theFolin–Ciocalteu reagent is thought to be caused bythe reaction of protein with the alkaline copper inthe reagent (as in the biuret test) and the reduction

of the Cu2� (cupric) ions in the reagent to Cu1�

(cuprous) by the tyrosine and tryptophan residuesof proteins. The cuprous ions in turn reduce thephosphomolybdate–phosphotungstate salts in thereagent to form an intensely blue complex. Becausethe content of these two amino acids varies sub-stantially within proteins, the color yield per mil-ligram of protein is not constant. It may differ sub-stantially from that of a protein standard.Nevertheless, the method is very useful for follow-ing changes in protein content, as, for example,during purification of a protein. This assay is usedin most of the experiments in this book, because itis convenient and inexpensive.

Bicinchoninic Acid (BCA) Assay. The bicin-choninic acid assay for proteins is based on the samereactions as the Folin–Ciocalteau assay. Proteinsare again reacted with alkaline cupric ions to formthe biuret complex, and these ions are reduced tocuprous ions by the aromatic amino acids in theproteins. In this case, however, the Cu1� ions forma complex with bicinchoninic acid (Fig. II-6), whichhas an intense absorbance maximum at 562 nm.This assay shows the same variation from proteinto protein as the Folin–Ciocalteau assay, but ismore convenient experimentally and can be madesomewhat more sensitive.

Dye-Binding (Bradford) Assay. The binding ofproteins to Coomassie Brilliant Blue 250 causes ashift in the absorbance maximum of the dye from465 nm to an intense band at 595 nm. Determina-tion of the increase in absorbance at 595 nm as afunction of protein added provides a sensitive assay

Figure II-5 The biuret reaction.

C

NH2

NH2

NHO

CO

C

NH2

NH2

NH2

NH2

NH NHO C O

CO C O

2

CNH

NH

O

C

CHR

O

2

CNH

NH

O C O

C

CHR

NH

NH

CHR

O C O

or

Cu2�

OH�

Cu2� Cu2�or

Biurets

Peptide chains Biuret complexes


�OOC N

�OOC N

�OOC N

�OOC N

COO�N

COO�N

�

�

�

Cu1�

Cu2�

Cu1�

Cu1�

Protein

Protein–Cu2� complex

Protein–Cu2� complex (biuret reaction)

Uncharacterized protein oxidation products

BCA BCA–Cu1� complex

Figure II-6 Chemistry of the bicinchoninic acid protein assay.

of protein that is quite constant from protein toprotein and relatively free from interference fromother cellular components, commonly used salts,etc. This assay uses commercially available reagentsand is about four times as sensitive as the Folin–Ciocolteu assay.

Spectrophotometric Assay. The tyrosine andtryptophan residues of proteins exhibit an ultravi-olet absorbance at approximately 275 nm and280 nm, respectively. Because the combined con-tent of these amino acids is roughly constant inmany proteins, the concentration of protein (in theabsence of other ultraviolet-absorbing materials) isgenerally proportional to the absorbance at 280 nm.(Note that because individual proteins can vary sub-stantially in their content of tryptophan and tyro-sine, the absorption spectra of pure proteins andtheir absorbance at 280 nm may vary widely.) Manyprotein solutions, especially solutions containingmixtures of many proteins, at 1 mg of protein permilliliter exhibit an absorbance at 280 nm of about0.8 when viewed through a 1-cm light path. Thus,the concentration of most proteins can rapidly beassayed by measurement of the absorbance at280 nm, if they are free of other materials that ab-sorb light at 280 nm. Such assays are advantageousbecause they are rapid and allow full recovery ofthe protein for subsequent analyses.

Unfortunately, other compounds present in nat-ural materials also exhibit absorbance at 280 nm.Specifically, nucleic acids, which have maximal ab-sorbance at 260 nm, exhibit extensive absorbance at280 nm. Corrections can be made for nucleic acidabsorbance at 280 nm if the ratio of absorbance at280/260 nm is known. Pure proteins have a 280nm/260 nm ratio of approximately 1.75 (variationsare due to differences in number and type of aro-matic amino acids present), whereas nucleic acidshave a 280 nm/260 nm ratio of 0.5. Intermediateratios corresponding to various mixtures of proteinand nucleic acid can be used to compute the con-centration of each.

Dry Weight. Most of the above methods are suit-able for comparisons, such as following the in-creasing specific activity of an enzyme during pu-rification. However, for careful chemical work witha pure protein, in which the mass must be knownaccurately, it is generally necessary to measure di-rectly the dry weight of a salt-free sample of theprotein. This is done after drying a sample of pro-tein, which has been extensively dialyzed againstdistilled water or a volatile buffer (e.g., ammoniumcarbonate) to constant weight in a vacuum at 50 to100°C over a drying agent. This mass can then berelated to one of the more convenient assays, suchas the absorbance of the protein at 280 nm.


Enzymology

Theory of Enzyme Action

Enzymes are proteins that catalyze biochemical re-actions. As is true of other catalysts, enzymes influencethe rate at which equilibrium is obtained, but do notaffect the overall equilibrium of the reaction. The re-action is accelerated by providing a reaction routehaving a lower free energy of activation for thetransition of substrate to products than the uncat-alyzed process. These two fundamental points areillustrated by the highly simplified free energy pro-files of hypothetical enzyme-catalyzed and non–enzyme-catalyzed reactions depicted in Figure II-7. The free energy levels of substrates and prod-ucts are the same in these two systems, and there-fore the net free energy change in the overall re-action, F (also called G in some textbooks); isidentical for both processes. The equilibrium con-stant Keq is directly related to F° as in EquationII-2:

(II-2)

� F° � RT lnKeq

where F° is F at standard states (one molar) ofreactants and products, R � gas constant, and T �absolute temperature. It follows that the equilib-

rium constants for both enzymatic and nonenzy-matic processes are identical as long as much lessthan stoichiometric amounts of catalyst are used.

The rate of a process is determined by the freeenergy levels of the rate-limiting transition state inthe reaction pathway (the higher the free energybarriers the slower the rate). This relationship isdefined by the absolute reaction rate theory as inEquation II-3.

(II-3)

kvel � e� F‡/RT

where kvel � reaction velocity constant, kB � Boltz-mann’s constant, h � Planck’s constant, R � gasconstant, and T � absolute temperature. The freeenergy differences between intermediates and tran-sition states are indicated by F‡; the superscript ‡

emphasizes that the thermodynamic parameterrefers to transition state complexes determining therate of the reaction.

Kinetic Analysis of Enzyme Activity

Unlike most uncatalyzed reactions, the first ob-served rate or initial velocity (v0) of enzyme-catalyzed reactions increases with increasing sub-

kBT�

h

– F

F2++

F1++

E • S

Energy levelof reactants

Activation energy for nonenzymatic reaction

Non–enzyme-catalyzed reaction

Enzyme-catalyzed reaction

Net free energy change

Energy level of products

Progress of reaction

(Activation energyfor enzymatic reaction)

F (

Free

ene

rgy)

Figure II-7 Energy profiles of enzymatic and nonenzymatic catalysis of a reaction.


Figure II-8 The relationship of substrate concen-

tration and initial velocity in enzy-

matic reactions.

Vmax

KM

vmax

v 0 (I

nitia

l rea

ctio

n ve

loci

ty)

2

Substrate concentration [S]

strate concentration only until a substrate level isreached beyond which further additions of substratedo not increase the initial rate. This phenomenonand the maximum initial velocity (Vmax) achieved atsubstrate saturation, are illustrated in Figure II-8.These observations, as they apply to enzymatic re-actions that employ a single substrate or the hy-drolysis of a single substrate, have been explainedby the following postulated reaction scheme (Equa-tion II-4), where E is enzyme, S is substrate, and Pis product:

(II-4)

k1 k3E � S uv ES uv E � Pk2 k4

Thus, the reaction proceeds by means of an oblig-ate intermediate enzyme–substrate complex (ES).When all of the enzyme is in the ES state (i.e., thesystem is saturated with substrate), the observedrate of reaction is maximal and the reaction is atVmax.

Michaelis and Menten, and later Briggs andHaldane, used the scheme shown in Equation II-4to derive a mathematical expression that describesthe relation between initial velocity and substrateconcentration. (Consult a biochemistry textbookfor the step-by-step derivation of this relationship,because it is important to be aware of the assump-

tions that were used.) The Michaelis–Menten equa-tion (Equation II-5) is:

(II-5)

v0 �

In this expression, Vmax and v0 are velocities as de-fined earlier, [S] is the substrate concentration, andKM is a new constant, the Michaelis constant, equalto k2 � k3/k1 where the k values are the specific rateconstants of Equation II-4.

If we rearrange the Michaelis–Menten equationto Equation II-6,

(II-6)

KM � [S] (Vmax/v0 � 1)

we can readily see that KM is numerically equal toa substrate concentration; KM therefore has the di-mensions of moles per liter (or millimolar or mi-cromolar, depending on the units used for the sub-strate concentration). The substrate concentration thatyields a velocity, v0, equal to half the maximum veloc-ity, Vmax, is equivalent to KM (Fig. II-8). The valueof the KM reflects the stability of the enzyme–substrate interaction and is of great practical value.KM is not, however, the true dissociation constantof the enzyme–substrate complex. This dissociationconstant is commonly termed KS. Only when k2 »k3 does

(II-7)

KM � KS � k2/k1 �

The reciprocal of KS is the affinity constant of en-zyme for the substrate. That is, the greater theaffinity of the enzyme for the substrate, the smallerthe KS. It is not correct to consider KS and KM to beequivalent without additional evidence, but this as-sumption is a common error.

Although KM may be determined using datasimilar to that presented in Figure II-8, it is moreaccurate and convenient to use one of the linearforms of the Michaelis–Menten equation to deter-mine KM. Lineweaver and Burk first pointed outthat equation II-8 can be obtained by inversion ofthe Michaelis–Menten equation:

[E] [S]�

[ES]

Vmax [S]��KM � [S]


Figure II-9 Lineweaver–Burk plot.

Vmax

Vmax

KM

KM

v0

1

1

1

1

[S]

Slope =

–

(II-8)

� �

Velocities at different substrate concentrations maybe plotted as reciprocals, as shown in Figure II-9.Values for KM and Vmax are obtained by extrapola-tion of the line, as indicated. Plots derived fromother formulations of the Michaelis–Menten equa-tion may also be used to obtain the same constants(Equations II-9 and II-10).

(II-9)

� [S] � (for a Hanes–Woolf plot)

and

(II-10)

v0 � �KM � Vmax (for an Eadie–Hofstee plot)

Plots derived from these equations (e.g., v0 versusv0/[S]) are often preferred because they spread outpoints that are close together on Lineweaver–Burkplots. In current research, it is preferable to usecomputer programs, such as KinetAsyst (IntelliKi-netics, State College, PA), KINSIM, or numerous

v0�[S]

KM�Vmax

1�Vmax

[S]�v0

1�Vmax

1�[S]

KM�Vmax

1�v0

others, to fit kinetic data for the Michaelis–Mentenequation, or for the other equations shown belowbecause these programs provide a precise estimateof the statistical reliability of the fit of the data andcalculate standard error values for the kinetic con-stants. Graphic representations are still conven-tionally used to display the data, but computer-based statistical analysis is essential for accurateanalysis.

Most enzymes react with two or more sub-strates. For this reason, the Michaelis–Mentenequation is inadequate for a full kinetic analysis ofthese enzyme reactions. Nonetheless, the same gen-eral approach can be used to derive appropriateequations for two or more substrates. For example,most enzymes that react with two substrates, A andB, are found to obey one of two equations if initialvelocity measurements are made as a function ofthe concentration of both A and B (with productconcentrations equal to zero). These are

(II-11)

v0 �

for “sequential” mechanisms, which require inter-mediate formation of an EAB complex (Kia is aproduct inhibition constant) or

(II-12)

v0 �

for “ping-pong” mechanisms in which a product isreleased from the enzyme before substrate B binds.Ka and Kb are Michaelis constants for the substratesA and B, respectively, in both equations. It is read-ily shown by converting these equations to theirdouble reciprocal form, analogous to theLineweaver–Burk equation, that they can be dis-tinguished by experiments in which v0 is measuredas a function of one substrate at several fixed levelsof the other substrate. For sequential mechanisms,both the intercepts at 1/[S] � 0 and the slopes of adouble reciprocal plot (e.g., 1/v0 versus 1/[A]) willbe changed if the concentration of B is changed.That is, a family of intersecting lines will be ob-tained. For ping-pong mechanisms only the inter-cepts are altered by changing the concentration of

Vmax [A] [B]��Kb [A] � Ka [B] � [A] [B]

Vmax [A] [B]��KiaKb � Kb [A] � Ka [B] � [A] [B]


Figure II-10 Lineweaver–Burk plot: competitive

inhibition.

Vmax

Vmax

KM

KM KM

v0

1

1– – 1

KI

[I]

1

1

[S]

Slope =

Vmax

KMSlope =

1 + ( )

KI

[I]1 + ( )

[I] = Inhibitor concentration

[I] = 0

B; that is the double reciprocal plots for a ping-pong kinetic mechanism are parallel lines—theslopes are not changed by altering the concentra-tion of the second substrate.

Vmax is the velocity of an enzyme-catalyzed re-action when the enzyme is saturated with all of itssubstrates and is equal to the product of the rateconstant for the rate-limiting step of the reaction atsubstrate saturation (kcat) times the total enzymeconcentration, ET, expressed as molar concentra-tion of enzyme active sites. For the very simple en-zyme reaction involving only one substrate de-scribed by Equation II-4, kcat � k3. However, morerealistic enzyme reactions involving two or moresubstrates, such as described by Equations II-11 andII-12, require several elementary rate constants todescribe their mechanisms. It is not usually possi-ble to determine by steady-state kinetic analysiswhich elementary rate constant corresponds to kcat.Nonetheless, it is common to calculate kcat valuesfor enzymes by dividing the experimentally deter-mined Vmax, expressed in units of moles per liter ofproduct formed per minute (or second), by the mo-lar concentration of the enzyme active sites at whichthe maximal velocity was determined. The units ofkcat are reciprocal time (min�1 or sec�1) and the rec-iprocal of kcat is the time required for one enzyme-catalyzed reaction to occur. kcat is also sometimescalled the “turnover number” of the enzyme.

Kinetics of Inhibition of Enzymes

Many substances interact with enzymes to lowertheir activity; that is, to inhibit them. Valuable in-formation about the mechanism of action of the in-hibitor can frequently be obtained through a kineticanalysis of its effects. To illustrate, let us consider acase of competitive inhibition, in which an inhibitormolecule, I, combines only with the free enzyme, E,but cannot combine with the enzyme to which thesubstrate is attached, ES. Such a competitive in-hibitor often has a chemical structure similar to thesubstrate, but is not acted on by the enzyme. For ex-ample, malonate (�OOCCH2COO�) is a compet-itive inhibitor of succinate (�OOCCH2CH2COO�)dehydrogenase. If we use the same approach that wasused in deriving the Michaelis–Menten equation to-gether with the additional equilibrium that definesa new constant, an inhibitor constant, KI,

(II-13)

KI �

we will obtain

(II-14)

v0 �

As before, it is most useful to convert this equationto its reciprocal form:

(II-15)

� (1 � [I] / KI) �

Measurement of v0 at various substrate levels andknown inhibitor concentration yields data in theform shown in Figure II-10. The various kineticconstants may be obtained as shown in the figure.A convenient way to obtain KI is to plot the valuesof the slopes of lines such as in Figure II-10 at sev-eral inhibitor concentrations. This replot should bea straight line with an intercept on the x-axis equalto �KI. A statistically reliable determination of in-hibition constants, however, should always involve

1�Vmax

1�[S]

KM�Vmax

1�v0

Vmax [S]��KM(1 � [I] / KI) � [S]

[E] [I]�

[EI]


fitting the data to the appropriate equation using acomputer program.

Biochemists observe other kinds of enzyme in-hibition. Noncompetitive inhibition consists of casesin which an inhibitor combines with either the Eor the ES form of the enzyme. This requires defi-nition of two new inhibitor constants:

(II-16)

KIe � and KIs �

Derivation of a rate expression in a manner similarto that used to derive the Michaelis–Menten equa-tion, and subsequent inversion of this rate expres-sion yields Equation II-17.

(II-7)

� (1 � [I] / KIs) � (1 � [I] / KIe)

In some instances, KIe � KIs; that is, the extent ofcombination of inhibitor with the E and ES formsof the enzyme are identical. The left panel of Fig-ure II-11 illustrates a plot of such data. In othercases of noncompetitive inhibition, KIe is different(usually larger) than KIs as illustrated in the rightpanel of Figure II-11.

1�Vmax

1�[S]

KM�Vmax

1�v0

[ES] [I]�

[ESI]

[E] [I]�

[EI]

In uncompetitive inhibition, the inhibitor com-bines only with the ES form of the enzyme. Thispattern of inhibition is not seen frequently exceptin studies of inhibition by the products of enzymereactions. If you define an inhibitor constant

(II-18)

KI �

and derive a rate expression for uncompetitive in-hibition, its inverted form is

(II-19)

� � (1 � [I] / KI)

as illustrated in Figure II-12. Note that these threekinetic patterns of inhibition can be distinguishedmost easily by obtaining initial velocity data as afunction of substrate and inhibitor concentrationand plotting the results in double-reciprocal form(Figures II-10 to II-12). The pattern will differ inthe effects of the inhibitor on the slopes and inter-cepts of these plots (see Table II-2).

It is important to recognize that the three pat-terns of inhibition just described are observed inmany different situations (e.g., product inhibition)

1�Vmax

1�[S]

KM�Vmax

1�v0

[ES][I]�[ESI]

Figure II-11 Double reciprocal plots of noncompetitive inhibition.

Vmax

KM

a. KIe= KIs= KI b. KIe= 2KIs

v0

KI

[I]

1

1

1

v0

1

[S]

1

[S]

Slopes = 1 + [ ] Vmax

KM

KIs

[I]Slopes = 1 + [ ]

Vmax1

Vmax

–KM1

Vmax

KI

[I]1 + [ ] 1

VmaxKIe

[I]1 + [ ]

[I] = 2 [I] = 2

[I] = 1 [I] = 1

[I] = 0 [I] = 0


v0

1

1

1

[S]

Vmax

KMSlopes =

Vmax

1

Vmax

KI

[I]1 + [ ]

[I] = 1

[I] = 0

Figure II-12 Double reciprocal plot of uncom-

petitive inhibition.

Table II-2 Inhibition Patterns Can Be

Distinguished by the Effects of

Inhibitors on Double Reciprocal

Plots

Effect of Increasing [I] upon:

Intercept

Type of Inhibition Slope �at �v

1

0

� � 0�Competitive increases no change

Noncompetitive increases increases

Uncompetitive no change increases

in addition to the somewhat oversimplified cases de-scribed above. In all cases the type of inhibition canbe identified and analyzed by the equations shown,but the physical meaning of the inhibition constantsand the exact mechanism of inhibition may not bethe same as given in our simple examples.

Other Factors Affecting Enzyme Activity

A variety of factors other than substrate and enzymeconcentration affects the activity of enzymes. Some

of the most important of these are discussed verybriefly below and are studied in the laboratory inExperiment 7.

Temperature. In general, the dependence of en-zyme activity on temperature will be described bythe curve illustrated in Figure II-13. At the lowertemperatures, the temperature dependence of Vmax

will be described by the Arrhenius equation (Equa-tion II-20).

(II-20)

log10 Vmax � � � constant

Thus, a plot of log10 Vmax versus the reciprocal ofthe absolute temperature T is linear with the slopeequal to �Ea/2.3 R, where Ea is an empirical quan-tity called the Arrhenius activation energy and R isthe gas constant, 1.98 cal/deg-mol. At higher tem-peratures the maximal velocity generally is muchlower than predicted; this deviation is the result ofdenaturation and inactivation of the enzyme at el-evated temperatures.

pH. Most enzymes show a strong dependence ofactivity on the pH of the medium. For this reasonit is essential to conduct enzyme assays in bufferedsolutions. The pH dependence of enzyme reactionsis the consequence of changing degrees of ioniza-tion of groups in the enzyme, in the substrate, orin both. In some cases it is possible to make edu-

1�T

Ea�2.3 R

2.3 R

EaSlope =

Log

Vm

ax

Thermalinactivationof enzyme

1/Absolute temperature (T)

Figure II-13 Relation of enzyme activity to tem-

perature. This is a typical Arrhenius

plot.


cated guesses as to functional groups in enzymesthat are involved in enzyme activity by determin-ing pKa values associated with changes in enzymeactivity and comparing them to values for knownprotein groups (Table II-1).

Cofactors. Many enzymes require relativelyloosely bound organic cofactors (coenzymes) ormetal ions for activity. A biochemist must be alertfor indications of such a requirement during pu-rification of an enzyme, because it may be neces-sary to add such cofactors to the assay solution toobtain enzyme activity. The interactions of coen-zyme with enzyme can be analyzed by kinetic andspectral techniques. Similar techniques are used toevaluate interactions with metal ions.

Allosteric Effectors. Many enzymes are subject tometabolic regulation through interaction withmetabolites that often act at allosteric sites, whichare distinct from the active site. The kinetic be-havior of such enzymes is often more complex thanthe behavior we have discussed above, and suchcomplex kinetics may serve as an indication that youare dealing with an allosteric enzyme. Further dis-cussion of this subject is found in Experiments 9and 15.

Enzyme Assay

During the isolation of an enzyme from any source,you must determine the amount of enzyme presentin the preparations. You usually accomplish this in-directly, by measuring the catalytic activity of theenzyme under some fixed conditions of pH, tem-perature, and added ions. Examination of anotherformulation of the Michaelis–Menten equation(Equation II-21) shows us that this is straightfor-ward.

(II-21)

v0 � �

(In this formulation, S0 equals the initial substrateconcentration and ET equals the total enzyme, E �ES). Because k3 and KM are fundamental constantsof the system and S0 is known from the assay con-ditions, the initial velocity v0 is directly propor-

k3 [ET] [S0]��KM � [S0]

Vmax [S0]��KM � [S0]

tional to [ET]. Careful measurement of v0 values,therefore, allows an estimation of the enzyme con-centration. Formally, v0 is expressed as the initialrate of change in concentration of substrate orproduct per unit time (e.g., millimolar per minute),but if you always use the same reaction volume forthe assay, it is satisfactory to express v0 simply asthe amount of substrate used or product formed perunit of time (e.g., micromole per minute). This con-vention is used in Experiments 7, 8, and 9.

The simplest procedure for obtaining v0 is touse such high concentrations of substrate that anegligible loss occurs during the reaction: [S] al-ways � [S0]. In such cases the velocity after 5 min,or even 30 min, may be the same as the initial ve-locity. Eventually, the substrate will become limit-ing or inhibitory concentrations of products will beformed. If a low [S] is used (i.e., [S] � KM), the ve-locity will continually decrease as substrate con-centration decreases, and it may be difficult to es-timate v0.

An accurate determination of v0 requires that anappropriate amount of enzyme be used. You can useso little enzyme that the velocity cannot be accu-rately determined (see Assay 1, Figure II-14). Sim-ilarly, use of too much enzyme makes it difficult todetermine an extremely high velocity in the timerequired to make the assay measurements (see As-says 3 and 4, Figure II-14).

True initial rate (Assay 3)

False initial rate (Assay 3)with fixed-time assay

Limit of fixed-time assay

Equilibrium

(Fixed-time assaygives true initial rate)

Assay 2

Assa

y 3

Assay 1

Time (min)

Ass

ay4

Mol

es o

f sub

stra

te c

onsu

med

,or

mol

es o

f pro

duct

form

ed

Figure II-14 Time course of enzyme-catalyzed

reactions.


–5 –4 –3 –2 –1

Log10 ml of enzyme solution used

Effective range

Enz

yme

activ

ity o

f und

ilute

d so

lutio

n (u

nits

/ml)

Figure II-15 Effect of enzyme dilution on appar-

ent assay.

Valid range

Enzyme concentration

Vo

Figure II-16 Relationship of enzyme activity to

enzyme concentration.

A so-called kinetic assay, in which the reactionrate is followed continuously, is advantageous be-cause it is possible to observe directly the linearityor nonlinearity of the response with respect to time.Many enzyme assays, however, are based on a sin-gle measurement at a defined time, a so-calledfixed-time assay. It is usually not possible to predictthe appropriate amount of enzyme in either kineticor fixed-time assays to obtain an optimum velocitylike that of Assay 2 in Figure II-14. This may beempirically determined by a dilution experiment intwo stages. At first, constant volumes of serial 10-fold dilutions of enzyme are assayed to find therange of dilution in which the calculated activity ismaximal and constant (see Figure II-15).

Such “range finding” may frequently straddlethe optimal enzyme concentration for assay. Thus,a second series of enzyme dilutions is assayed topinpoint the enzyme concentration over a narrowerrange. Usually four to five dilutions over a 10- to20-fold concentration range are adequate.

The prime test of the validity of v0 assays is agraphical test to establish that the rates observedare a linear function of enzyme concentration, as il-lustrated in Figure II-16. This test should be ap-plied to all assays in which reaction rates are usedto measure enzyme concentrations. This procedureis especially important when fixed-time assays areused (see Figure II-14).

Assay Procedure during EnzymePurification

Because enzyme (protein) purification involves theselective removal of other proteins, it is necessaryto assess the amount of enzymatic activity relativeto the amount of protein present. A measure of en-zymatic activity per milligram of protein is usuallyemployed to indicate the degree of purity of the en-zyme in the various fractions obtained during pu-rification.

This quantity, the specific activity, is calculatedfrom a measure of enzymatic activity or units, usu-ally expressed in terms of micromole of productformed per minute, and a protein determination, asin Equation II-22.

(II-22)

Specific activity � units per milligram of protein

� micromole of substrate consumed

per minute per milligram of

protein

� micromole of product formed per

minute per milligram of protein

The efficiency of an enzyme purification is alsoassessed by determining how much of the enzyme


activity is recovered from each purification step.This requires that you calculate the total activity ineach fraction.

Total activity � (specific activity) �(total mg protein in preparation)

or

(II-23)

Total activity � (activity per milliliter) �(volume of the preparation in milliliter)

Once you know these values for any prepara-tion of an enzyme, you can proceed with the pu-rification of the enzyme by applying one or moreof the techniques discussed elsewhere in this sec-tion and in Experiments 2, 8, and 10. It is essentialto calculate specific activity, total activity, and per-cent yield for every fraction obtained during a pu-rification.

(II-24)

Percent yield �

� 100%

The usefulness of a particular step may then beevaluated with reference to the increase in the spe-cific activity of the enzyme and percent yield in thefractions of greatest enrichment. An increase in spe-cific activity indicates a purification. An ideal frac-tionation would provide complete enrichment(pure enzyme) in 100% yield; in practice, few pro-

total activity of a given fraction��total activity of the starting material

cedures are so selective, and various, less ideal frac-tionation steps are combined. Of course, if enrich-ment is great, a lower yield may be allowed in a par-ticular step. Table II-3 shows a convenient and usualway to present the results obtained during a purifi-cation procedure. Because protein concentrationand enzyme activity are usually determined as mil-ligram per milliliter and units per milliliter, re-spectively, it is also always necessary to record thevolume of each enzyme fraction. Experiment 8 pro-vides the student with experience in the applicationof these concepts to the purification of an enzyme.

REFERENCES

Ackers, G. K. (1970). Analytical Gel Chromatographyof Proteins. Adv Protein Chem 24:343.

Andrews, P. (1965). The Gel-Filtration Behavior ofProteins Related to their Molecular Weights over aWide Range. Biochem J 96:595

Banks, J. F., Jr., and Whitehouse, C. M. (1996). Elec-trospray Ionization Mass Spectrometry. Methods En-zymol 270:486.

Beavis, R. C., and Chait, B. T. (1996). Matrix-AssistedLaser Desorption Ionization Mass-Spectrometry ofProteins. Methods Enzymol 270:519.

Benson, J. R., and Hare, P. E. (1975). o-Phthalalde-hyde: Fluorogenic Detection of Primary Amines inthe Picomole Range. Proc Natl Acad Sci 72:619.

Bradford, M. M. (1976). A Rapid and Sensitive Methodfor the Quantitation of Microgram Quantities ofProtein Utilizing the Principle of Protein-DyeBinding. Anal Biochem 72:248.

Carter, C. W., and Sweet, R. M. (1997). Macromolecu-lar Crystallography, Parts A and B. Methods Enzy-mol 276 and 277.

Table II-3 Typical Summary of an Enzyme Purification Procedure

Volume Activity Specific Activity Fold Total Activity Yield

Purification Step (ml) (�mol/min/ml) (�mol/min/mg protein) Purification (�mol/min) (%)

Crude extract 50 20 0.1 (1.0) 1000 (100)

First (NH4)2SO4 20 40 0.7 7.0) 800 80)

precipitate

DEAE-Sephadex 15 30 200 2000 450 45)

chromatography

Affinity 3 100 4500 45000 300 30)

chromatography


Dixon, M., and Webb, E. C. (1964). Enzymes, 2nd ed.New York: Academic Press.

Gordon, J. A., and Jencks, W. P. (1963). The Relation-ship of Structure to the Effectiveness of DenaturingAgents for Proteins. Biochemistry 2:47.

Groll, M., L. Ditzel, Löwe, J., Stock, D., Bochtler, M.,Bartunik, H. D., and Huber, R. (1997). Structure of20S Proteasome from Yeast at 2.4 Å Resolution.Nature 386: 463.

Haschemeyer, R. H. (1970). Electron Microscopy ofEnzymes. Adv Enzymol 33:71.

Hirs, C. H. W., Stein, W. H., and Moore, S. (1954).The Amino Acid Composition of Ribonuclease. JBiol Chem 211:941.

Hunkapillar, M. W., Hewick, R. M., Dreyer, W. J., andHood, L. E. (1983). High-Sensitivity Sequencingwith a Gas-Phase Sequenator. Methods Enzymol91:399.

Layne, E. (1957). Spectrophotometric and Turbidomet-ric Methods for Measuring Proteins. Methods Enzy-mol 3:447.

Lowry, O. H., Rosebrough, N. J., Farr, A. L., and Randall, R. J. (1951). Protein Measurement withthe Folin Phenol Reagent. J Biol Chem 193:265.

Moore, S., and Stein, W. H. (1963). ChromatographicDetermination of Amino Acids by Use of Auto-matic Recording Equipment. Methods Enzymol6:819.

Reynolds, J. A., and Tanford. C. (1970). The GrossConformation of Protein-Sodium Dodecyl SulfateComplexes. J Biol Chem 245:5161.


Segel, I. H. (1975). Enzyme Kinetics. New York: Wiley.Smith, P. K., Krohn, R. I., Hermanson, G. T., Mallia,

A. K., Gartner, F. H., Provenzano, M. D., Fuji-moto, E. K., Goeke, N. M., Olson, B. J., andKlenk, D. C. (1985). Measurement of Protein Using Bicinchoninic Acid. Anal Biochem 150:76.

Woody, R. W. (1995). Circular Dichroism. Methods Enzymol 246:34.

q

105

and proteins is largely due to the additional ioniz-able groups on the side chains.

This experiment is a study of the reaction oftypical amino acids with hydrogen ions. As an acidor base is added to a solution of an amino acid, achange in the pH of the solution is observed. Afterallowances for the dilution of the acid or base dur-ing titration, the results can be predicted by use ofthe general Henderson–Hasselbalch equation. TheHenderson–Hasselbalch equation was presented inthe Introduction to Section II (Equation II-1), andits application to the analysis of acid and base titra-tion of a typical amino acid was described. Be surethat you understand this discussion well before pro-ceeding with this experiment. Test your under-standing further by predicting the titration curvesof other amino acids with ionizable side chaingroups, such as histidine, glutamic acid, and lysine.

Amino acids (and peptides) are characterizednot only by their acid dissociation constants (pKas),but also by an isoelectric point (pI), which is thepH at which the amino acid or peptide has no netcharge. For a simple amino acid with only an �-carboxyl and an �-amino group the pI is determinedby a balance between the tendency of the proto-nated amino group to lose a proton and the un-protonated carboxyl group to bind a proton. It canbe readily shown that this pH is

(5-1)

pI � �pKa1 �

2

pKa2�

However, for an amino acid or peptide with threeor more ionizable groups, you must avoid the trap

EXPERIMENT 5

Acid–Base Properties of Amino Acids

Theory

Most biologically important molecules, large andsmall, are water-soluble and ionic. Their ionic char-acter derives from the association and dissociationof protons to and from weakly acidic or basicgroups in their structure. It is difficult to overesti-mate the importance of these ionizing groups.They confer water solubility on the molecules,which is important because water is the funda-mental intracellular medium. They are also crucialfor the formation of native three-dimensionalstructure, for molecular recognition, and for bio-logical function, as for example, in the catalytic ac-tivity of enzymes. The degree of protonation ofacidic and basic groups, and hence the charge onsuch groups, is a function of the pH of the sur-rounding medium. This explains why pH is so im-portant in biochemical systems.

In this experiment we shall examine the effectsof pH on the ionization (protonation) of aminoacids, but it is important to remember that the sameprinciples and quantitative relationships can be usedto describe the ionization behavior of all biologicalmolecules that contain weak acids or bases. As dis-cussed in the Introduction to Section II, all aminoacids contain at least one acidic (�-carboxyl) groupand one basic (�-amino) functional group. Someamino acids contain other side chain groups thationize readily: amino, carboxyl, p-hydroxyphenyl,sulfhydryl, guanidinium, and imidazole groups. Be-cause the �-amino and �-carboxyl groups of aminoacids all participate in peptide linkages (except forthose at the amino and carboxyl termini of thepolypeptide), the ionic character of polypeptides


12

10

8

6

4

2

0 0.5 1.0 1.5 2.0 2.5 3.0

Equivalents of base

pH

Cl–

COOH

H3+N–CH–COO

CH2

CH2

COOH

H3+N–CH–COO–

Na+

COO–

H3+N–CH–COO–

CH2

CH2

CH2

CH2

2Na+

COO–

H2N–CH–COO–

CH2

CH2

Figure 5-1 Titration of glutamic acid hydrochlo-

ride with base, showing possible

ionic states of the amino acid as a

function of pH.

of thinking that pI is the average of pKa values.Rather, it is always determined by the balance ofonly two pKa values:

(5-2)

pI ��pKn �

2

pKn�1�

pKn and pKn�1 are the two pKa values that describethe ionization of the species with a net zero charge;that is the first ionization that adds a proton to the neu-tral species and gives it a net charge of �1 and the firstionization that removes a proton from the neutral speciesand gives it a net charge of �1. Test your under-standing of this concept by calculating the pI val-ues for aspartic acid, histidine, and lysine.

In this experiment you will determine pKa andpI values for an unknown amino acid by titrating itwith acid and base. You will also determine the mol-ecular weight of the amino acid by determining thenumber of equivalents of standardized acid or basethat were required to titrate one functional group—an �-carboxyl group or an �-amino group. From theknown mass of the amino acid titrated and the equiv-alents required to titrate one group, the molecularweight of the amino acid can be readily calculated.

Note that the amino acid may be given to youin one of several ionic states. For example, it couldbe given to you as a zwitterionic (isoelectric) form,or as a hydrochloride salt or a sodium salt. See Fig-ure 5-1 for an illustration of this concept as appliedto the amino acid glutamic acid. This amino acidcould be isolated as the hydrochloride, the zwitte-rionic form, the monosodium salt, or the disodiumsalt. The initial pH of a solution of glutamic acidgives you information about its ionic form. For ex-ample, the zwitterionic form would yield a solutionwith a pH of about 3, whereas the monosodium saltwould give a solution with a pH near 7. The titra-tion curves for glutamic acid would be the same forall forms, except that the groups titrated by acid orbase depend on the initial state. For example, titra-tion of the hydrochloride would reveal three groupstitrated by base and none with acid. The mono-sodium salt would require one equivalent of baseand two equivalents of acid to be fully titrated.

Your instructor may also give you unknownsthat are monofunctional weak acids or bases, butthat are not amino acids. Careful examination ofyour titration curves should enable you to distin-

guish among these possibilities. If you receive anamino acid as an unknown, you should be able toidentify it from its molecular weight and pKa val-ues. Table 5-1 lists the pKa values and molecularweights for the common amino acids.


Unknown (amino acid, monofunctional weak acidor weak base) in powder form

2 N NaOH

2 N H2SO4

pH meter

Magnetic stirring motor and bar

10-ml burette

Burette holder

Buffers for standardization of pH meter

EXPERIMENT 5 Acid–Base Properties of Amino Acids 107

Table 5-1 Molecular Weights and pKa Values

for Common Amino Acids

pKa of pKa of pKa of

MW �-Carboxyl �-Amino Side Chain

Amino Acid (g/mol) Group Group Group

Glycine .750 2.35 9.78

Alanine .890 2.35 9.87

Valine .117 2.29 9.74

Leucine .131 2.33 9.74

Isoleucine .131 2.32 9.76

Methionine .149 2.13 9.28

Proline .115 2.95 10.65

Phenylalanine .165 2.16 9.18

Tryptophan .204 2.43 9.44

Serine .105 2.19 9.21

Threonine .119 2.09 9.10

Asparagine .133 2.10 8.84

Glutamine .146 2.17 9.13

Tyrosine .181 2.20 9.11 10.13

Cysteine .121 1.92 10.78 8.33

Lysine .147 2.16 9.18 10.79

Arginine .175 1.82 8.99 12.48

Histidine .155 1.80 9.20 6.00

Aspartic acid .132 1.99 9.90 3.90

Glutamic acid .146 2.10 9.47 4.07

Chloride 35.5

Sodium .023

Magneticstirrer

pH meterelectrodes

Burette

Magneticstirring baror paperclip

Figure 5-2 Setup for titration.

Protocol

1. Using a sample of standard buffer, familiarizeyourself with the operation of the pH meterand standardize the instrument. For titrationswith acid, standardize the pH meter with thepH 4.0 standard buffer; for titrations with baseuse the pH 9.0 or 10.0 standard buffer. Be sureto rinse the electrodes with deionized water afterstandardization.

2. After recording your unknown number, weighabout 400 mg of the unknown, recording themass to the nearest milligram, and dissolve itin 20 ml of deionized H2O. If the sample doesnot dissolve readily, gently heat the solution to50 to 60°C and allow it to dissolve. Allow thesolution to cool to room temperature beforebeginning the titration.

3. Set up the electrodes of the pH meter, a mag-netic stirring bar (a small metal paper clip canbe substituted for a stirring bar) and stirringmotor, and a burette as shown in Figure 5-2,so that the sample can be titrated with contin-uous stirring. Take extreme care to ensure thatthe electrodes are not damaged by the stirringbar or by rough treatment.

4. Titrate the dissolved unknown sample with 2 NH2SO4, recording the burette readings (andfrom these, the volume of H2SO4 added) andthe pH at frequent intervals throughout thetitration until the solution reaches pH 1.0.

5. Titrate another 20-ml H2O sample (a waterblank containing no unknown) with 2 NH2SO4 1 drop at a time for the first 10 drops,2 drops at a time for the next 10 drops, and fi-nally 4 drops at a time until the solution reachespH 1.0. Record the pH and volume of acid de-livered after each addition.

6. Dissolve another sample of the unknown (againapproximately 400 mg, weighed to the nearest

Table 5-2 Amounts of Acid Required to

Titrate Sample and Water Blank in

Solution

Volume (ml) of Acid (2 N HCl)

Water plus Water

pH Sample Blank Difference

3.5 0.103 0.003 0.100

3.0 0.335 0.020 0.315

2.5 0.667 0.032 0.635

2.2 1.063 0.063 1.000

2.0 1.425 0.200 1.225


H2O blank

Sample in H2O

Milliliters 2 N HCI

pH

6

5

4

3

2

1

1 2 30

Figure 5-3 Uncorrected titration curves of water

blank and sample.

milligram) in 20 ml of H2O and titrate this sam-ple with 2 N NaOH in the same manner as forthe acid titration until the pH of the solutionsreaches 12.0. As before, record the volume ofNaOH delivered and the pH after each addi-tion.

7. Titrate a 20 ml water blank with 2 N NaOHto pH 12.0 in a manner similar to the first wa-ter blank (see step 5).

Data Analysis

To determine the true titration curve of any sub-stance, you must measure how much acid or baseis consumed in titrating the solvent (water) to eachpH and then subtract this amount from the totalamount of acid or base consumed in reaching thatpH when titrating the sample (amino acid � water).The following example with the acid side of thetitration of an amino acid illustrates the method forcorrecting for such acid or base dilution that youshould use in correcting your data.

1. For both the sample and the water blank, plotthe volume of acid added versus the pH reached(see the example in Fig. 5-3).

2. From the graph (or from the original data), pre-pare a table like Table 5-2, listing the amount

of acid required to attain each pH, determin-ing a value for as many points as needed to de-fine the titration curve. Then, subtract the vol-ume of acid required to bring the water blankto any pH from the volume of acid required tobring the sample to the same pH. This differ-ence represents the amount of acid consumedin the titration of the sample only.

3. Using the data from your table, plot the pHversus the number of equivalents of acid neededto titrate the amino acid sample to any pH (seethe example in Fig. 5-4).

Milliequivalents of acid

pH

6

5

4

3

2

1

0240

pK´a

Figure 5-4 Corrected titration curve of sample.

EXPERIMENT 5 Acid–Base Properties of Amino Acids 109

4. Use the same method to correct for titration ofthe water blank with NaOH. Prepare a com-plete, corrected NaOH titration curve for theunknown that you titrated.

5. Combine the corrected acid and base titrationcurves into a single titration curve for the un-known sample for the entire pH range from 1to 12. If your unknown was a simple amino acidwith no ionizable side chains, this figure willlook like Figure II-1 in the Introduction to Sec-tion II.The number of equivalents of acid orbase consumed in passing through one full in-flection of a curve (in the zone of a single ion-ization, as in Fig. 5-4) represents the quantityof acid required to titrate one ionizable groupin the quantity of amino acid you have assayed.Using this relationship and noting the numberof full inflections of your curve, calculate themolecular weight for your unknown. The endpoint of the titration should be recognizable asthe point at which the pH rises (or falls) sharplywith addition of titrant. This point is usuallymost accurately determined from the titrationof the amino acid with base.

6. Compare the molecular weights obtained fromthe titration of each ionizable group. How welldo they agree?

7. Using the Henderson–Hasselbalch equation,calculate the pKa value for each ionizable grouptitrated.

8. What is the identity of your unknown? Justifyyour conclusion by comparing the observedmolecular weight and pKa values to those forall amino acids that might have been your un-known (Table 5-1). If your unknown was notan amino acid, what can you suggest about itschemical nature? How could you confirm youridentification of your unknown?

Exercises

1. How do the true titration curves of aspartic acidand lysine differ from that of glycine? Explainyour answer.

2. Is the true titration curve of leucyl valine thesame as that of glycine? Explain your answer.

3. The quantity 432 mg of a monoamino mono-carboxylic amino acid is observed to consume

7.39 mEq of acid and base in the titration rangepH 0.8 to 12.0. What is the name of the aminoacid?

4. Give an example of a polypeptide that is not apolyelectrolyte.

5. Calculate the isoelectric point for the follow-ing polyfunctional amino acids: (a) histidine, (b)lysine, and (c) aspartic acid. (See Table 5-1.)

6. A useful method to assay many hydrolytic en-zymes is to follow enzymatic activity by deter-mining production or uptake of hydrogen ionsduring the reaction. For example, consider aprotease that catalyzes cleavage of internal pep-tide bonds in a protein:

a. Assume the enzyme is fully active at pH 8.8,that the pKa of an �-carboxyl group in a pep-tide is 3.5, and that the pKa of an �-aminogroup in a peptide is 8.5. How will the ac-tion of this protease on a protein affect thepH of an unbuffered reaction mixture (start-ing at pH 8.8)? Explain your answer.

b. The protease reaction is run in a pH-stat, adevice that maintains a constant pH in a re-action mixture by continuous, automatic ad-dition of acid or base. The instrument hasa recorder that plots the volume of acid orbase added versus time of reaction. If the re-action in (a) is run in a pH-stat maintainedat pH 8.8, calculate the millimoles of titrantthat must be added per millimole equivalentof peptide bonds hydrolyzed by the enzyme.

REFERENCES

Christensen, H. N. (1963). pH and Dissociation.Philadelphia: Saunders.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Amino Acids and Peptides. In: Principles ofBiochemistry, 2nd ed. New York: Worth.

Stryer, L. (1995). Biochemistry, 4th ed. New York: Free-man, pp. 17–23, 42–43.

–C–NH–CH–C–NH–CH–C–NH–

R2R1O O O

–C–NH–CH–COOH � H2N–CH–C–NH–

R2R1O O

protease

H2O

v w

111

EXPERIMENT 6

Sequence Determination of a Dipeptide

Theory

The compound 1-fluoro-2,4-dinitrobenzene(FDNB) reacts with free amino, imidazole, andphenolic groups at neutral to alkaline pH to yieldthe corresponding, colored dinitrophenyl (DNP)compounds. Thus, FDNB will react with the free,unprotonated �-amino groups on amino acids, aswell as with the side chains of lysine, histidine, andtyrosine (Fig. 6-1). Dansyl chloride is another com-pound that is known to react with the unproto-nated, N-terminal amino groups of peptides. De-rivatization of peptides with this compound yieldsfluorescent products that provide a very sensitivemethod of detection of the amino acid derivatives(Fig. 6-2).

The reaction of FDNB with a peptide proceedsby the attack of an uncharged amino group on theelectron-deficient carbon atom adjacent to the flu-orine atom. The pKa value for the �-amino groupon most peptides is approximately 8.5. For this rea-son, it is critical for the reaction solution to bebuffered at or slightly above pH 9.0. Sodium bi-carbonate (NaHCO3), a weak base, is used to bufferthe reaction, since OH� ions (strong base) will re-act with FDNB to form 2,4-dinitrophenol as a sideproduct (Fig. 6-3). As shown in Figure 6-1, acid(HF) is produced on the reaction of a peptide withFDNB. As a result, additional NaHCO3 may haveto be added as the reaction proceeds to maintain analkaline pH.

If the DNP derivative of a peptide is acid hy-drolyzed and extracted with ether at low pH, allnonpolar (uncharged) compounds will be extractedinto the ether phase while all polar (charged) com-

pounds will remain in the aqueous phase. At lowpH, the N-terminal DNP-amino acids will extractinto the ether phase because their �-carboxylgroups are protonated (uncharged). In contrast, allamino acids that are not N-terminal will remain inthe aqueous phase because their �-amino groupsare protonated (charged) at low pH (see Fig. 6-4).The only DNP-amino acids that will not enter the ether phase at low pH conditions are DNP-arginine and DNP-cysteic acid, since these sidechains continue to carry a charge at low pH. Oncethe N-terminal DNP-amino acid derivatives areseparated from those amino acids or DNP-aminoacid derivatives that are not N-terminal in the pep-tide, the DNP derivatives and free amino acids maybe identified by a variety of chromatographic tech-niques. These data will yield part of the sequenceof the peptide under study.

Although the reaction of peptides with FDNBis useful in identifying the N-terminal amino acid,the acid hydrolysis step destroys the remainder ofthe peptide (hydrolyzes all of the remaining pep-tide bonds). It would be extremely valuable to havea means of cleaving a derivative of the N-terminalamino acid from the peptide without hydrolyzingthe other peptide bonds. With such a technique,you could determine the entire sequence of a pep-tide merely by repeating the derivatization andcleavage cycle, removing and identifying one aminoterminal residue at a time. Such a technique is cur-rently available with the use of Edman’s reagent,phenylisothiocyanate (PITC). This reagent reactsto form a phenylthiocarbamyl derivative of the free�-amino group of a peptide, which is then cyclizedin anhydrous HF to cleave the amino terminal

112

O2N F

NO2

�� O2N N

NO2

H

R

C COO�H2N

H2N

H

RH

C COO�

Amino acid DNP-Amino acidFDNB

HF

O2N F

NO2

�� O2N N

NO2

NO2

NO2

H

C COO�

CH2

CH2

CH2

NH2

CH2

H

C COO�

CH2

CH2

CH2

NH

CH2

H

Lysine

FDNB

2HF2

-DNP-Lysine�

H2NO2N F

NO2

�� O2N N

NO2

NO2

NO2

O2N

NO2

H

C COO�

CH2

H

C COO�

O

CH2

H

Tyrosine

FDNB

2HF2

O-DNP-Tyrosine

O�

H2NO2N F

NO2

��

O2N

N

NO2

H

C COO�

CH2

H

C COO�

CH2

H

Histidine

FDNB

2HF2

Imidazole-DNP-histidine

C

C

N

NC H

H

H

C

C

N

NC H

H

Figure 6-1 Reaction of amino acids with 1-fluoro-2,4-dinitrobenzene (FDNB). DNP � dinitrophenyl;

HF � hydrofluoric acid

EXPERIMENT 6 Sequence Determination of a Dipeptide 113

H3C CH3

��

H

R

C COO�H2N

N

H

H

R

C COO�

Dansyl chloride(Dimethylaminonaphthalene

sulfonyl chloride)

HCI

S

Cl

OO

NH3C CH3

S OO

N

Dansyl-amino acid(fluorescence-sensitive detection)

O2N F

NO2

�� O2N OH

NO2

NaOH

2,4-DinitrophenolFDNB

NaF

O2N N N

NO2

DNP-glycyl alanine at pH 1.0(will extract into ether phase)

DNP-glycine(will extract into ether phase)

C C C COOH

H H

H

O H H

O2N N

NO2

C COOH

H H

H

O2N N N

NO2

CH3CH3

C COOH

H

CH3

DNP-glycyl alanine at pH 9.0(will remain in the aqueous phase)

Alanine(will remain in the aqueous phase)

C C C COO�

H H

H

O H H

Hydrolysis of peptide(6 N HCI, 110�C)

H3N�

Figure 6-2 Reaction of amino acids with dansyl chloride.

Figure 6-3 Reaction of NaOH with 1-fluoro-2,4-dinitrobenzene (FDNB).

Figure 6-4 Partitioning of an acid hydrolyzed DNP-dipeptide between the aqueous and ether phases is

pH-dependent.


N

Phenylisothiocyanate

Anilinothiazolinone

C S

OR2OR1

CH CHNH2 C N C

�

�

�

H

OR2OR1

CH CHC

SC

NN

C

NH

OR2

CHNH3 C

slightly alkaline pH(7.0–10.0)

anhydrousacid

aqueousacid

H

CN

R1

C O

CH

S

NH

H

PTC-peptide(phenylthiocarbamyl)

Product peptide readyfor next reaction cycle

PTH-Amino acid (phenylthiohydantoin)

C

O

S

C

NR1CH

NH

Figure 6-5 The steps of the Edman degradation

of a polypeptide.

residue from the peptide without cleaving the otherpeptide bonds (Fig. 6-5). This derivative of the N-terminal amino acid (the anilinothiazolinone deriv-ative) can be converted to a stable phenylthiohy-

dantoin (PTH-amino acid), which can be identifiedby reverse-phase high-performance liquid chro-matography (HPLC). This procedure is then repeated to determine the identity of the next N-terminal amino acid in the peptide. This techniquehas been automated for step-by-step sequence de-termination of peptides and proteins. As many as60 to 100 amino acids can be identified by this tech-nique under ideal conditions. The first step in thereaction of an amino acid or peptide with PITC isthe formation of the phenylthiocarbamyl (PTC) de-rivative. As in the case of reaction of amino acidswith FDNB, this reaction takes place at unproto-nated �-amino groups (as well as at the ε-aminogroup on lysine) under alkaline conditions. In thisexercise, you will prepare PTC derivatives of theamino acids released by the hydrolysis of the un-known dipeptide and identify them by reverse-phase HPLC and spectrophotometry. The PTCmoiety on the amino acids serves two purposes inthis experiment: first, it provides a method of de-tection (absorbs light at 254 nm). Second, it adds ahydrophobic component to all of the amino acidsthat will make them much easier to separate by reverse-phase HPLC partition chromatography(see Experiment 2).

In this three-period experiment, milligramamounts of an unknown dipeptide will be examinedby the procedures that we have just discussed. InFigure 6-6, a flow diagram of the experiment is pro-vided to aid your understanding of the varioussteps.


P-20, P-200, P-1000 Pipetman with disposable tips

Glass test tubes (13 by 100 mm and 16 by 125 mm)

Dipeptides

DNP-amino acid standards (in acetone)

Amino acid standards (in acetone)

Peroxide-free ether

Whatman 3 MM filter paper (8.5 in. by 8.5 in.)

Hydrolysis vial (4 ml) with Teflon-lined screw caps

6 N HCl

FDNB solution (0.25 ml of FDNB to 4.75 ml of absoluteethanol)

4.2% NaHCO3 (4.2 g of NaHCO3 in 100 ml of water)

Pasteur pipettes


Identify by thin-layer chromatography

N-Terminal DNP-amino acid(ether)

C-Terminalamino acid(aqueous)

DNP-dipeptide(ether)

Dipeptide

FDNB(ether)

FDNB (pH ~9.0)

DNP-dipeptide(aqueous)

DNP-dipeptide+excess FDNB

•Evaporate to dryness (N2)•Dissolve in 6 N HCI•Hydrolyze (110°C)•Ether extract

Salts(aqueous)

Etherextract(Acidic pH)

Etherextract(Alkaline pH)

6 N HCI (100°C)

Amino acids

PITC

Identify by reverse-phase HPLC

PTC-amino acidsPaperchromatography

Figure 6-6 Flow diagram for sequence determi-

nation of a dipeptide.

Ninhydrin spray solution (Dissolve 1 g of ninhydrin in500 ml of n-butanol. Add 0.5 ml of 2,4-collidinejust before use.)

Silica-gel thin-layer chromatography plates (Fisher cata-log 05-713-317)

pH paper (1–11)

Pressurized N2

Heating block

Acetone

Chloroform

t-Amyl alcohol

Glacial acetic acid

n-Butanol

Ethanol

Triethylamine

Phenylisothiocyanate (Sigma catalog P-1034)

PicoTag Eluant A (Waters catalog 88108)

Amino Acid Standard H (Sigma catalog AA-S-18)

HPLC unit

Alltech Adsorbosphere HS C18 reverse-phase column(250 � 4.6 mm)

Spectrophotometer

Chart recorder

Vacuum microcentrifuge (Speed-Vac)

Protocol

Safety Precautions: Avoid getting FDNB on your

skin by wearing rubber or latex gloves at all

times. Ether is very noxious and very flammable.

Keep the ether in the fume hoods at all times

and avoid having open flames in the laboratory.

FDNB can cause burns to the skin. Wear safety

goggles at all times. Ninhydrin vapors are corro-

sive, so use caution. Perform all ether extrac-

tions in glass test tubes, since polypropylene

tubes will dissolve in ether.

Day 1: Acid Hydrolysis of the Peptide

and Preparation of the DNP-Derivatized

Dipeptide

1. Obtain duplicate samples of the unknowndipeptide, each containing about 2 mg. Onesample will be in a microcentrifuge tube andthe other will be in a 13-by-100-mm glass testtube. Record the number of your unknown samplein your notebook.

2. To the sample in the glass test tube, add thefollowing:

200 l distilled water

50 l of 4.2% NaHCO3400 l of FDNB reagent

3. Allow the sample to incubate for 1 hr at roomtemperature, mixing gently every 3 or 4 min.During this hour, check the pH every 10 minby dropping a spot of the reaction solution onpH paper. Maintain the pH at �9.0 by addingdrops of 4.2% NaHCO3 as needed. Proceedwith steps 4 to 5 during this 1-hr incubation.

4. To the sample in the microcentrifuge tube, add200 l of 6 N HCl and mix gently to dissolvethe sample. Using a Pasteur pipette, transfer


N2

Figure 6-7 Setup for ether extract evaporation.

the sample to a 4-ml glass vial and seal with aTeflon-lined screw cap.

5. Label the top of the vial (not the sides) withyour name and place in a 110°C heating blockovernight (�12 hr).

6. To the sample in the glass test tube (after the1-hr incubation), add 1 ml of distilled water and0.5 ml of 4.2% NaHCO3. Make sure that thepH of the solution is above 8.0 before contin-uing.

7. Add an equal volume (�2 ml) of ether and gen-tly swirl to mix for 10 sec. Allow the aqueousand ether phases to separate and remove anddiscard the ether (upper) phase containing ex-cess (unreacted) FDNB.

8. Repeat Step 7 two more times or until the etherphase no longer shows any yellow color.

9. To the resulting aqueous phase (containing theDNP-dipeptide), add 100 l of 6 N HCl. Makesure that the pH is �1.0 by putting a small dropof the reaction on pH paper after mixing. Youmay notice some bubbles forming in the solu-tion as the H� causes the formation of carbondioxide gas and water from the formation ofH2CO3.

10. Extract the acidified, aqueous solution with 2ml of ether as in step 7. After mixing, allow thephases to separate, remove the ether phase(containing the DNP dipeptide) and transfer toa 16-by-125-mm glass test tube.

11. Repeat step 10 two more times. The etherphases from all three extractions can be savedin a single 16-by-125 mm glass test tube.

12. In the fume hood, evaporate the ether con-taining the DNP-dipeptide by placing the tubein a beaker of warm (not hot!) water and di-recting a gentle stream of N2 over the solution(see Fig. 6-7). Continue with this procedureuntil a dry, yellow solid remains at the bottomof the tube.

13. Dissolve the dried DNP dipeptide (yellowsolid) in 0.5 ml of acetone and transfer the sam-ple to a 4-ml glass vial.

14. Evaporate the acetone under N2 in the fumehood and redissolve the dried sample in 0.5 mlof 6 N HCl.

15. Seal the vial with a Teflon-lined screw cap, la-bel the top (not the sides) with your name, andplace it in a 110°C heating block overnight(�12 hr).

Day 2: Thin Layer Chromatography

of the DNP-Amino Acid and Paper

Chromatography of the Acid-Hydrolyzed

Dipeptide

Thin-Layer Chromatography of the

DNP-Amino Acid

1. Transfer the contents of the 4-ml vial contain-ing the acid-hydrolyzed DNP dipeptide (yel-low sample) to a clean 13-by-100-mm glass testtube.

2. Add 2 ml of distilled water and 2 ml of ether.Swirl for 10 sec, allow the phases to separate,and transfer the ether phase to a clean 16-by-125-mm glass test tube. Repeat thisether extraction two more times and combinethe ether phases in the same glass test tube.This final ether extraction is done to separatethe N-terminal DNP-amino acid from theother amino acid (or DNP-amino acid) in thesample. Remember that the N-terminal DNP-amino acid carries no charge at low pH (etherphase), while the C-terminal amino acid has acharged �-amino group at low pH (aqueousphase).

3. Evaporate the ether extracts (containing yourN-terminal DNP-amino acid) to dryness underN2 as before.


Solvent

Origin line

Solvent

Solvent front

Solvent front

PaperTLC plate

Ascendingthin-layer

chromatography

Ascendingpaper

chromatography

Figure 6-8 Ascending thin-layer and paper chromatography.

4. “Activate” a silica-gel thin-layer chromatogra-phy (TLC) plate by heating for 5 min in a100°C oven. This procedure drives off excessmoisture from the plate so that the TLC sep-aration is more reproducible.

5. Make a small scratch on the edge of the TLCplate (short axis) about 2 cm from the bottom.This will indicate the position of the originwhere the samples will be spotted (do not markthe plate with pencil and avoid touching theface of the plate with your fingers). Refer toFig. 6-8 for proper setup of thin-layer chro-matography experiment.

6. Carefully spot 5 l of the DNP-amino acidstandards (provided by the instructor) along theorigin of the plate. To prevent diffusion of thesample, it is a good idea to spot 1 l of the sam-ple, allow it to dry, spot another 1-l aliquot,allow it to dry, and so forth.

7. Dissolve the dried DNP-amino acid (N-terminal) in 100 l of acetone.

8. Using the same technique as described in step6, carefully spot a 5-l aliquot of your samplealong the same origin as the DNP-amino acidstandards. If you did not recover a sufficientamount of the sample in the last ether extrac-tion, you may also want to spot a 10-l samplenext to this. The sample spot should be yellowenough in color that you will be able to iden-tify it once the plate is developed.

9. When all of the spots have dried, place theTLC plate (sample side toward the bottom)in a TLC tank containing 1 cm of the mobile-phase solvent [chloroform:t-amyl alcohol:glacial acetic acid (70:30:3)] (see Fig.6-8). Remember to make a map or drawing of theTLC plate in your notebook so that you will re-member which spot corresponds to which sample.

10. Allow the mobile phase to migrate across (up)the plate until the solvent front is about 1 cmfrom the top of the plate (�2 hr).

11. Remove the plate from the tank and mark theposition of the solvent front with a pencil. Af-ter the plate has dried, circle the position of allof the spots on the plate with a pencil. Also,draw a line across the bottom of the plate inpencil to mark the position of the origin.

12. For each spot on the plate, calculate a relativemobility value (Rf ):

Rf =

If the sample spot showed some diffusion dur-ing development, use the center of the spot indetermining the Rf value. Also, if the solventfront migrated unevenly across the plate, mea-sure the distance traveled by the sample (cm)relative to the distance traveled by the solventfront (cm) in each lane.

distance traveled by sample (cm)distance traveled by solvent front (cm)


N

�

� � � �

COO�

OH

OH

O

O

C

R

H

3 H2O H�

H3N�

2

Ninhydrin

O O�

O O

Amino acid

Blue-purple in color

CO2 HR C

O

Figure 6-9 Reaction of amino acids with ninhydrin.

13. From the Rf value of your unknown samplecompared to the Rf values of the DNP-aminoacid standards, what is the identity of the N-terminal amino acid in your unknown dipep-tide? Discuss any uncertainty that may be in-volved in your identification, as well as howthese uncertainties may be eliminated throughmodifications in the experiment.

Paper Chromatography of

Acid-Hydrolyzed Dipeptide

1. Recover the 4-ml vial containing your acid hy-drolyzed dipeptide (colorless sample).

2. Obtain an 8.5 in. by 8.5 in. piece of 3 MMWhatman filter paper from the instructor anddraw a straight line across the bottom using asoft lead pencil about 1.5 in. from the edge ofthe paper. This line will designate the originwhere your samples will be spotted.

3. On both edges of the paper perpendicular to theorigin, draw two straight lines from the top tothe bottom that intersect the origin 1 in. fromeither edge of the paper. All of the samples willbe spotted on the origin between these two lines.

4. Spot 3 l of each of the amino acid standardsalong the origin. As with the thin-layer chro-matography plate, spot each sample 1 l at atime. Indicate the identity of each sample spotbelow the origin using a soft lead pencil.

5. Next to these standards, spot a 1-l and a 3-lsample of your unknown acid hydrolyzed

dipeptide. Save the remainder of this sample foruse on Day 3.

6. When all of the spots have dried, roll the pa-per into the form of a cylinder so that the spot-ted samples face inward toward the bottom ofthe paper (see Fig. 6-8). Staple the paper inthree places (top, middle, and bottom) avoidingany overlap of the paper on the edges.

7. Place the paper (origin side down) in a large jar(10 in. in height with a 5.5-in. diameter) con-taining 2 cm of mobile-phase solvent on thebottom (butanol:acetic acid:water [60:15:25]).

8. Cap the jar and allow the mobile phase to mi-grate through (up) the paper until it reaches 1in. from the top (�4–6 hr). Remove the paperfrom the jar, mark the position of the solventfront with a pencil, and allow the paper to dryin a fume hood.

Day 3: Analysis of the Paper

Chromatogram and Preparation and

Analysis of PTC-Amino Acids

Analysis of Paper Chromatogram

1. Spray the paper chromatogram lightly withninhydrin reagent and heat for 5 min in a100°C oven. The amino acids will appear asblue or purple spots (except proline, which willappear yellow). Figure 6-9 shows the reactionthat amino acids undergo with ninhydrin.


2. Circle the position of all spots with a pencil andnote the color, which may give some clue as tothe identity of the amino acid.

3. Calculate the Rf values of the amino acids inyour unknown sample as well those of theamino acid standards, using the same procedureand equation as that used on Day 2.

4. Based on the Rf values and colors of the spotspresent in your unknown sample compared tothose produced by the amino acid standards,what two amino acids are present in your un-known dipeptide? Can you now determine thesequence of your unknown dipeptide based onthese results and those from Day 2? As before,discuss any uncertainty that may be involved inyour identification, as well as how these un-certainties may be eliminated through modifi-cations in the experiment.

Preparation of PTC-Amino Acids

1. Transfer 10 l of your acid-hydrolyzed dipep-tide sample (colorless) to a microcentrifugetube and add 20 l of distilled water. In addi-tion, pipette 0.25 ml of Amino Acid StandardH into a separate microcentrifuge tube. Per-form steps 2 to 6 on both of these samples.

2. Dry the sample in the vacuum microcentrifuge(Speed-Vac) under the high heat setting for 30min or until all of the HCl has evaporated.

3. Add 20 l of wash reagent (ethanol:triethyl-amine:water [2:2:1]), mix, and dry the sampledown in the Speed-Vac under the high heat set-ting. Repeat this wash procedure one moretime. The amino acids standards are suppliedin a solution of 0.1 N HCl. The washes are per-formed to remove all traces of acid, which mayinterfere with the coupling reaction.

4. Add 15 l of coupling reagent (ethanol:tri-ethylamine:PITC:water [7:1:1:1]), mix, and in-cubate at room temperature for 20 min.

5. Dry the PTC-amino acids in the Speed-Vacunder the low heat setting.

6. Resuspend the sample in 1 ml of PicoTag Elu-ant A. Centrifuge the sample for 5 min at

12,000 � g to pellet any undissolved material.Transfer the top 0.7 ml of the clarified sampleto a fresh microcentrifuge tube. This samplewill be analyzed by reverse phase HPLC as de-scribed below.

HPLC Analysis of PTC-Amino Acids

1. Fill HPLC Buffer Reservoir A with PicoTagEluant A and Buffer Reservoir B with 60%(vol/vol) acetonitrile in HPLC grade water.

2. Equilibrate the C18 reverse-phase HPLC col-umn with a solvent composition of 98% BufferA and 2% Buffer B (flow rate at 1 ml/min).

3. Inject 100 l of the PTC-amino acid sampleand apply the following mobile-phase gradientduring development (mark the time of injec-tion, or T � 0, on the chart recorder):

Time (min) % Buffer A % Buffer B

0 98 2

12 90 10

13 80 20

23 60 40

25 0 100

31 0 100

4. Detect the PTC-amino acids as they elute witha spectrophotometer set at 254 nm. When thesignal output (10 mV maximum) is relayed toa chart recorder with a speed of 0.5 cm/min,the PTC-amino acid peaks will show goodheight and good resolution.

5. Inject a 100-l sample of the 20 PTC-aminoacid standards and apply the same mobile-phase gradient. The sequence of the PTC-amino acids eluting from the column developedwith this gradient is shown in Table 6-1.

6. Measure the retention times (RT) of the un-known PTC-amino acids and compare these tothose of the PTC-amino acid standards. Theretention time (RT) can be calculated as fol-lows:

RT �distance from injection point (T � 0) to the middle of the elution peak (cm)

0.5 cm/min


Table 6-1 Retention Times of PTC-Amino Acids

on a C18 Reverse-Phase HPLC

Column at a Flow Rate of 1 ml/min

Amino Acid RT (min)

PTC-aspartic acid 7.0

PTC-glutamic acid 7.4

PTC-serine 8.4

PTC-glycine 9.4

PTC-histidine 10.4

PTC-arginine 12.4

PTC-threonine 12.4

PTC-alanine 13.4

PTC-proline 14.0

PTC-tyrosine 22.0

PTC-valine 23.0

PTC-methionine 24.0

PTC-cysteine 25.6

PTC-isoleucine 26.4

PTC-leucine 26.6

PTC-phenylalanine 28.6

PTC-lysine 30.4

7. Based on these results, what two amino acidsare present in your unknown dipeptide? Asshown in Table 6-1, it may be difficult to re-solve some PTC-amino acids (PTC-arginineand PTC-threonine). Still, this mobile-phasegradient will resolve the majority of the PTC-amino acid derivatives. How might the condi-tions of this experiment be changed to resolvemolecules such as PTC-arginine and PTC-threonine?

NOTE: Several amino acids will be modified ordestroyed during the harsh acid hydrolysis step:

Tryptophan is destroyed.

Glutamine is converted to glutamate.

Asparagine is converted to aspartate.

Cysteine is oxidized to cysteic acid.

Because of this, we suggest that you performthe experiment with dipeptides that do not con-tain these amino acids.

NOTE: In addition to the HPLC mobile-phasesystem described above, you can replace Pico-Tag Eluant A with 140 mM KOAc, pH 5.3(sodium acetate is not an acceptable alterna-

tive). If this is done, Resevoir B must contain84% (vol/vol) acetonitrile in HPLC-grade wa-ter. This change in mobile-phase compositionwill require the flow rate to be 2 ml/min (forgood resolution), with the following gradient:

Time (min) % Buffer A % Buffer B

0 94 6

0.1 93 7

2 91 9

8.5 89 11

9 78 22

11 77 23

15 76 24

17 74 26

18.6 64 36

20.5 64 36

21.5 0 100

22.5 0 100

The retention times for each of the PTC-amino acidsdescribed above (Table 6-1) will not apply to a chro-matography experiment done with this mobile-phasesystem. If this alternative gradient is used, these val-ues will have to be determined independently.

Exercises

1. Draw the chemical structures of PTC-valineand PTH-valine. With what biochemical tech-nique is the formation of PTH-amino acid de-rivatives associated?

2. The five dipeptides listed below were treatedas follows: (a) The dipeptide was reacted ex-haustively with FDNB. (b) The reaction wasthen acidified and extracted with ether. (c) Theether phase from this extraction was drieddown, resuspended in 0.5 ml of 6 N HCl, andheated at 110°C for 12 hr. (d) 1 ml of waterand 2 ml of ether were then added to the sam-ple. For each of the dipeptides listed below, listthe peptide-derived compounds that you wouldexpect to find in the final ether phase and thefinal aqueous phase following the series of fourtreatments described above.

Alanyl-asparagine Valyl-arginine

Glutamyl-lysine Arginyl-valine

N-acetyl-alanyl-tryptophan


3. A peptide (Compound A) was acid hydrolyzedand found to contain equimolar amounts ofArg, Val, Tyr, Glu, Lys, Ala, and Gly.

a. Exhaustive treatment of Compound A withtrypsin yields the following compounds:Arg, Ala-Lys, a peptide (Compound B) con-taining Glu, Gly, Tyr, Val. Treatment ofcompound B with chymotrypsin yields Val-Tyr and Glu-Gly.

b. Short-term treatment of Compound A withcarboxypeptidase yields free Gly as the firstdetectable amino acid.

c. N-terminal analysis of Compound A withFDNB yields DNP-alanine in the etherphase.

What is the sequence of the dipeptide?

4. With what groups on amino acids can FDNBreact? Why could an amino acid that is not N-terminal, but still reacts with FDNB on its sidechain, fail to be extracted into the ether phaseat low pH conditions?

REFERENCES

Fraenkel-Conrat, H., Harris, J. I., and Levy, A. L.(1955). Recent Developments in Techniques for Termi-nal and Sequence Studies in Peptides and Proteins.(Methods of Biochemical Analysis, Vol. 2). NewYork: Wiley.

Konigsberg, W. (1972). Subtractive Edman Degrada-tion. Enzyme Structure, Part B. Methods Enzymol25:326.

Laursen, R. A. (1972). Automatic Solid-Phase EdmanDegradation. Enzyme Structure, Part B. MethodsEnzymol 25:344.


Schroeder, W.A. (1972). Degradation of Peptides. En-zyme Structure, Part B. Methods Enzymol 25:298.

Stryer, L. (1995). Exploring Proteins. In: Biochemistry,4th ed. New York: Freeman.

Tarr, G. E. (1985). Microcharacterization of Polypeptides:A Practical Manual (Manual Edman SequencingSystem). Totowa, NJ: Humana Press.

Wilson, K., and Walker, J. (1995). ChromatographicTechniques. In: Principles and Techniques of PracticalBiochemistry, 4th ed. Hatfield, UK: Cambridge University Press.

xb

123

EXPERIMENT 7

Study of the Properties of �-Galactosidase

Theory

The enzyme �-galactosidase allows lactose metab-olism in Escherichia coli. As shown in Figure 7-1, thisenzyme catalyzes the hydrolysis of lactose into itsmonosaccharide units, galactose and glucose. A sideproduct of the �-galactosidase reaction, allolactose,is known to control the expression of the enzymeat the level of transcription.

The gene for �-galactosidase (lacZ) resideswithin a set of genes and regulatory DNA sequencescollectively referred to as the lac operon (Fig. 7-2).

An operon is a set of genes whose expression is co-ordinately regulated. Often, these genes encodeproteins or enzymes related in a particular functionor to a particular biological pathway. The lac operonconsists of a promoter sequence, an operator (reg-ulatory) sequence, and three genes. As mentionedabove, lacZ encodes �-galactosidase. LacY encodesgalactoside permease, a transmembrane transportprotein that allows lactose to enter the cell. LacAencodes thiogalactoside transacetylase, an enzymewhose function is not fully understood. The pro-moter sequence within the operon is the site where

HO

H

H

H

H

H H

H

OH H OH

OH

H

HOH

H

HOH

H

OH

OH

CH2OH CH2OHO O

O�-Galactosidase

Lactose(D-Galactopyranosyl ( , 1 4) D-glucopyranose)�

(D-Galactopyranosyl ( , 1 6) D-glucopyranose)�

HO

HO

HO

H

H

H H

H

OH H OH

OH

CH2OH CH2OHO O

OH

H

HOH H

HOH

HO

H

H

H

H

H

OH

H OH

OH

CH2OH

CH2

O

OO

Galactose Glucose

Allolactose

�

Figure 7-1 Reaction catalyzed by �-galactosidase.


PI lacI

lacI

lacZPlac operatorCAPsite lacY lacA

When lactose is absent

mRNA

mRNA

Translation

RNA polymerase is blockedfrom transcribing lacZ, lacY,and lacA.

PI lacI

lacI


When lactose is present

CAP

Inducer(allolactose or IPTG)

When lacI binds the inducer, it loses its affinity for the operator and dissociates. RNA polymerase is now free to transcribe lacZ, lacY, and lacA. If glucose is not present, the CAP–cAMP complex will form and bind the CAP site. This CAP–cAMP complex bound at the CAP site enhances transcription of the operon.

mRNA

PI lacI

lacI


When lactose and glucose are present

CAP

Inducer(allolactose or IPTG)

Although the inducer causes the lacI repressor protein to dissociate from the operator, the glucose will lead to a low level of cAMP (low adenylate cyclase activity). As the CAP–cAMP complex dissociates, the CAP protein will leave the CAP site. This will repress transcription of lacZ, lac Y, and lac A (catabolite repression).

cAMP

Figure 7-2 Regulation of the expression of the lac operon.

RNA polymerase binds the DNA and initiates tran-scription of lacZ, lacY, and lacA. As we shall see, theoperator sequence adjacent to the promoter is animportant regulatory sequence that will participatein the control of the expression of the lac operon.

Transcription of the lac operon can be induced(turned on) in the presence of lactose and repressedin the absence of lactose when the components re-quired for its metabolism are not needed. This reg-ulatory system ensures that the cell will not waste

EXPERIMENT 7 Study of the Properties of �-Galactosidase 125

valuable energy producing proteins that it does notimmediately require. Expression of the lac operonis repressed in the presence of the lac repressor pro-tein. This regulatory protein, encoded by the lacIgene immediately 5� to the lac operon, is a consti-tutively expressed DNA binding protein that has ahigh affinity for the lac operator sequence. Whenthe lac repressor binds the operator sequence, itblocks RNA polymerase from transcribing the lacoperon. As stated earlier, transcription of the lacoperon is induced in the presence of lactose. Thisphenomenon is due to the effect that allolactose(the inducer) has on the affinity of the lac repres-sor for the operator sequence (Fig. 7-2). As the lacrepressor (LacI ) binds allolactose, it loses its affin-ity for the operator sequence. RNA polymerase isthen able to transcribe the lac operon to producethe proteins required for lactose metabolism. In thepresence of lactose, expression of the lac operon canincrease by up to as much as 1000-fold. Althoughallolactose is the natural inducer of the lac operon,it is not the only compound that will increase its level of expression. Isopropylthiogalactoside(IPTG, Fig. 7-3) is a gratuitous inducer of the lac operon. Although it is not a substrate for �-galactosidase, it too will complex with the lac re-pressor protein and lower its affinity for the oper-ator sequence.

The regulation of the lac operon just describedis a form of negative regulation: LacI represses ex-pression of the operon in the absence of the in-ducer. It turns out, however, that transcription ofthe lac operon is also subject to positive regulation.Immediately 5� to the lac promoter is another reg-

ulatory DNA region called the CAP (catabolitegene activator protein) site. When CAP is boundat this site, transcription of the lac operon can beincreased 20-fold to 50-fold. By itself, CAP has lowaffinity for the CAP site in the operon. The affin-ity of CAP for this site increases dramatically, how-ever, when the protein binds cAMP (Fig. 7-2). Sincethe lac promoter by itself is relatively weak, thecAMP–CAP complex bound at the CAP site is re-quired for significant expression of the lac operon.It is apparent, then, that the expression of the lacoperon is closely tied to the level of intracellularcAMP.

Adenylate cyclase is the enzyme responsible forcatalyzing the conversion of ATP to cAMP (Fig.7-4). This enzyme is inactive in the cell when glu-cose levels are high. As a result, cAMP levels de-crease in the cell when glucose is present. The neteffect of this is that the cAMP–CAP complex is notable to form, and transcription of the lac operon isdecreased. This will be the case even if lactose is also

HO

H

H

H

H

H

OH

OH

CH2OH

CH

CH3

CH3O

�

S

Isopropylthio- , D-galactopyranoside

Figure 7-3 Isopropylthiogalactoside, a gratuitous

inducer of the lac operon.

O

H HH H

OH OH

O OP

O

NH2

N

�NN

N

O

OH H

H

O OH

NH2

PPiCH2

N

NN

N

O P

O

�O

�O �O �O

P

O�

O

P

O

Adenosine triphosphate (ATP)

Adenylatecyclase

Adenosine-3�, 5�-cyclic monophosphate(cAMP)

Figure 7-4 Reaction catalyzed by adenylate cyclase. The activity of adenylate cyclase is decreased when

glucose is present in the growth medium.


HO

H

H

H

H

H

OH

OH

CH2OH NO2

O

�

�O

O-Nitrophenyl- , D-galactopyranoside(ONPG)

HO

H

H

H

H

H

OH

OH

OH

CH2OHNO2O

�O-Galactosidase

�

Galactose

O-Nitrophenol (ONP)(Absorbs light at 420 nm

in alkaline conditions)

Figure 7-5 Colorimetric assay for �-galactosidase activity.

present. The ability of glucose to prevent expres-sion of the lac operon even in the presence of itsinducer is an example of catabolite repression. Sinceglucose and gluconate are the preferred carbonsources for E. coli, these compounds are very ef-fective catabolite repressors of the lac operon.Other carbon sources, such as glycerol, acetate, andsuccinate, are less effective catabolite repressors.Since catabolite repression is due to the effect ofhigh glucose levels on adenylate cyclase activity, theeffect of catabolite repression can be overcome (re-versed) in vivo through the addition of cAMP tothe growth medium.

You will use o-nitrophenyl-�,D-galactopyra-noside (ONPG), rather than lactose, as the sub-strate for the �-galactosidase assays performed inthis experiment. �-galactosidase will hydrolyze theONPG substrate to produce two products, one ofwhich absorbs light at 420 nm under alkaline con-ditions (Fig. 7-5). By monitoring the change in A420

of a solution containing �-galactosidase andONPG, you will be able to continuously measure thechange in concentration of the product (o-nitro-phenol, ONP) over time.

In this four-period experiment, you will studyand characterize the kinetics of �-galactosidase atvarying pH and temperature as well as in the pres-ence of two different inhibitors of the enzyme. Inthese exercises, you will determine the KM and Vmax

for the enzyme with respect to its ONPG substrate,as well as the inhibition constant (KI) for the twoinhibitors. In addition, the temperature studies willallow you to calculate the activation energy for the�-galactosidase catalyzed reaction. Finally, you willutilize a number of different E. coli strains in an invivo �-galactosidase assay to demonstrate the intri-cate regulatory system of the lac operon and the

concept of catabolite repression. Before you beginthe experiment, review the section on enzymologyin the Introduction to Section II.


0.08 M sodium phosphate buffer, pH 7.7





�-galactosidase solution (�10 units/milliliter in 0.08 Msodium phosphate buffer, pH 7.7, supplementedwith 1 mg/milliliter bovine serum albumin)

o-nitrophenyl-�,D-galactopyranoside solutions (2.5 mMand 10 mM)

13 � 100 mm glass test tubes

16 � 125 mm glass test tubes with caps (sterile)

1.5-ml plastic microcentrifuge tubes

P-20, P-200, P-1000 Pipetmen with disposable tips

ice

methyl-�,D-galactopyranoside (MGP) solution (750mM)

methyl-�,D-thiogalactoside (MTG) solution (150 mM)

isopropylthiogalactoside (IPTG) solution (200 mM)—filter-sterilized

cAMP solution (100 mM)—filter sterilized

1 M Na2CO3

0.1% (wt/vol) sodium dodecyl sulfate (SDS) solution

chloroform

YT broth (1% wt/vol bactotryptone, 0.5% wt/vol yeastextract, 0.5% wt/vol NaCl)

20% (wt/vol) glucose solution—filter sterilized

spectrophotometer to read absorbance at 280 nm, 420nm, 600 nm, and 550 nm

heating blocks or temperature-controlled water baths

colorimeter tubes and quartz cuvettes


Protocol

Day 1: Determination of the Activity and

Specific Activity of the �-Galactosidase

Solution

1. Set up the following reactions in six 13-by-100-mm glass test tubes. Keep the enzyme solution onice and add it to the tubes below, which are at roomtemperature.

Volume of 0.08 M Sodium Volume of 2.5 mM

Tube Phosphate, pH 7.7 (ml) ONPG (ml)

1 4.0 0.5

2 4.0 0.5

3 4.0 0.5

4 4.0 0.5

5 4.0 0.5

6 4.0 0.5

2. Add 0.5 ml of sodium phosphate buffer, pH 7.7to tube 1 (blank) and mix gently in a Vortexmixer. Blank your spectrophotometer to readzero absorbance at 420 nm against this solu-tion.

3. Add 0.5 ml of undiluted �-galactosidase so-lution to tube 2. Mix gently with a Vortexmixer for 5 sec and place the solution in yourspectrophotometer. Exactly 30 sec after theaddition of the enzyme, record the A420 of the solution in your notebook. Continue torecord the A420 value at 30-sec intervals un-til the A420 value no longer changes (indicatesthat the reaction is complete). This final A420 value will be used later to determine theactivity of �-galactosidase in each reaction(�A420max).

4. Prepare a twofold dilution of the enzyme byadding 250 �l of the �-galactosidase stock so-lution to 250 �l of 0.08 M sodium phosphatebuffer, pH 7.7. Add this entire 0.5-ml solutionto tube 3. Exactly 30 sec after the addition ofthe enzyme, record the A420 of the solution inyour notebook. Continue to record the A420

value at 30-sec intervals. Unlike with tube 2, itis not necessary that this reaction is followed to com-pletion. Rather, you are looking to obtain a set ofA420 values that are linear with respect to time overabout 2 min.

5. Prepare a fivefold dilution of the enzyme byadding 100 �l of the �-galactosidase stock solu-tion to 400 �l of 0.08 M sodium phosphatebuffer, pH 7.7. Add this entire 0.5-ml solutionto tube 4. Exactly 30 sec after the addition of theenzyme, record the A420 of the solution in yournotebook. Continue to record the A420 value at30-sec intervals. Unlike with tube 2, it is not nec-essary that this reaction is followed to completion.Rather, you are looking to obtain a set of A420 valuesthat are linear with respect to time over about 5 min.

6. Prepare a 10-fold dilution of the enzyme byadding 50 �l of the �-galactosidase stock solu-tion to 450 �l of 0.08 M sodium phosphatebuffer, pH 7.7. Add this entire 0.5-ml solutionto tube 5. Exactly 30-sec after the addition of theenzyme, record the A420 of the solution in yournotebook. Continue to record the A420 value at30-sec intervals. Unlike with tube 2, it is not nec-essary that this reaction is followed to completion.Rather, you are looking to obtain a set of A420 valuesthat are linear with respect to time over about 5 min.

7. Prepare a 20-fold dilution of the enzyme byadding 25 �l of the �-galactosidase stock solu-tion to 475 �l of 0.08 M sodium phosphatebuffer, pH 7.7. Add this entire 0.5-ml solutionto tube 6. Exactly 30 sec after the addition of theenzyme, record the A420 of the solution in yournotebook. Continue to record the A420 value at30-sec intervals. Unlike with tube 2, it is not nec-essary that this reaction is followed to completion.Rather, you are looking to obtain a set of A420 valuesthat are linear with respect to time over about 5 min.

8. Prepare a plot of A420 versus time for the dataobtained from the reactions in tubes 2 to 6. Plotthe data from all five reactions on a single graphfor comparison.

9. Draw a “best-fit” curve through each set of datapoints. For each curve, determine over whattime frame the reaction kinetics appear linear(where �A420/�time yields a straight line). Cal-culate the slope of the linear portion of eachcurve.

10. Use the following equation to determine the �-galactosidase activity (micromoles of ONPGhydrolyzed per minute/per milliliter of en-zyme) in each reaction:

Activity (�A420/�time)(1.25/�A420max)

Vp


where 1.25/�A420max is a conversion factor re-lating the number of micromoles of ONPG hy-drolyzed to the change in absorbance at420 nm, and Vp is the total volume (in milli-liters) of �-galactosidase solution present ineach reaction. Taking the volumes of the stock�-galactosidase solution (undiluted) used ineach reaction into consideration, what is the ac-tivity of �-galactosidase in the stock solution ofthe enzyme? NOTE: You should be able to cal-culate several activity values for the stock �-galactosidase solution using the reactions intubes 2 to 6. These values should be averagedand presented along with a deviation from themean.

11. Obtain two 1.5-ml plastic microcentrifugetubes. To one, add 0.5 ml of distilled water and0.5 ml of the stock �-galactosidase solution. Tothe other, add 0.8 ml of distilled water and0.2 ml of the stock �-galactosidase solution.Cap each tube and invert several times to mix.

12. Read the absorbance of both solutions in aquartz cuvette at 280 nm using water as a blankto zero your spectrophotometer. Record theA280 value of both solutions in your notebook.

13. Based on these values, estimate the total pro-tein concentration (in milligrams per milliliter)of the stock �-galactosidase solution usingBeer’s law (A εcl ). Assume an extinction co-efficient of 0.8 (mg/ml)�1 cm�1. This extinctioncoefficient is based on the average value of aromaticamino acids found in most proteins.

14. Calculate the specific activity of the stock �-galactosidase solution (micromoles of ONPGhydrolyzed per minute per milligram of pro-tein) using the following equation:

Specific activity [prot

a

e

c

in

ti

]

v

(

i

m

ty

g/ml)

Day 2: Determination of KM and Vmax and

the Effect of Inhibitors on �-Galactosidase

Activity

Determination of KM and Vmax

1. From the data obtained on Day 1, determinea dilution of the stock �-galactosidase solu-

tion that showed good linear kinetics over a5-min period (where �A420/�time produceda straight line over 5 min). Use a dilution ofthe enzyme that gives a change in absorbanceat 420 nm of about 0.1 per min. Prepare 9 mlof a dilute �-galactosidase solution by mak-ing the appropriate dilution of the �-galac-tosidase stock solution in sodium phosphatebuffer, pH 7.7. This dilute �-galactosidase so-lution will be used in both of the assays de-scribed below. Keep this solution on ice and addto the tubes below at room temperature when youare ready to perform the assay.

2. Set up the following reactions in six 13-by-100mm glass test tubes:

Volume of Phosphate Volume of

Tube Buffer, pH 7.7 (ml) 2.5 mM ONPG (ml)

1 4.00 0.50

2 4.00 0.50

3 4.20 0.30

4 4.30 0.20

5 4.35 0.15

6 4.40 0.10

3. Add 0.5 ml of sodium phosphate buffer, pH 7.7,to tube 1 (blank) and mix by gently swirling.Blank your spectrophotometer to read zero ab-sorbance at 420 nm against this solution.

4. Add 0.5 ml of the dilute �-galactosidase solu-tion to tube 2. Mix gently with a Vortex mixerfor 5 sec and place the solution in your spec-trophotometer. Exactly 30 sec after the addi-tion of the substrate, record the A420 of the so-lution in your notebook. Continue to recordthe A420 value at 30-sec intervals for 5 min.

5. Follow the exact same procedure as describedin step 3 for tubes 3 to 6. Record the A420 at30-sec intervals for at least 5 min for each re-action.

6. Prepare a plot of A420 versus time for the dataobtained from the reactions in tubes 2 to 6. Plotthe data from all five reactions on a single graphfor comparison.

7. Calculate the slope of each line. If your enzymedilution prepared in step 1 was correct, the�A420 per minute for each reaction should belinear for at least 5 min.


8. Use the following equation to determine theinitial velocity of each �-galactosidase reaction(micromoles of ONPG hydrolyzed per minute):

Initial velocity (V0) (�A420/�time)(1.25/�A420max)

Note how the slope of each line (initial veloc-ity) changes with decreasing ONPG concen-tration.

9. Prepare a plot of 1/V0 versus 1/[ONPG] (aLineweaver–Burk plot) using the data obtainedfrom tubes 2 to 6. Do the data points “fit” toa straight line? From the x-intercept of thisgraph, can you determine the KM (mM) for �-galactosidase with respect to ONPG? From they-intercept of this graph, can you determine theVmax for the �-galactosidase solution (micro-moles of ONPG hydrolyzed per minute)?

Effect of Inhibitors on �-Galactosidase Activity

1. Set up the following reactions in twelve 13-by-100-mm glass test tubes:

Volume of

Phosphate Volume Volume Volume of

Buffer, of MGP of MTG 2.5 mM

Tube pH 7.7 (ml) (ml) (ml) ONPG (ml)

1 3.9 0.1 — 0.50

2 3.9 0.1 — 0.50

3 4.1 0.1 — 0.30

4 4.2 0.1 — 0.20

5 4.25 0.1 — 0.15

6 4.30 0.1 — 0.10

7 3.9 — 0.1 0.50

8 3.9 — 0.1 0.50

9 4.1 — 0.1 0.30

10 4.2 — 0.1 0.20

11 4.25 — 0.1 0.15

12 4.3 — 0.1 0.10

2. Add 0.5 ml of sodium phosphate buffer, pH 7.7,to tube 1 (blank) and mix by gently swirling.Blank your spectrophotometer to read zero ab-sorbance at 420 nm against this solution.

3. Add 0.5 ml of the dilute �-galactosidase solu-tion to tube 2. Swirl gently for 5 sec and placethe solution in your spectrophotometer. Ex-actly 30 sec after the addition of the substrate,

record the A420 of the solution in your note-book. Continue to record the A420 value at 30-sec intervals for 5 min.


5. Add 0.5 ml of sodium phosphate buffer, pH 7.7,to tube 7 (blank) and mix by gently swirling. Zeroyour spectrophotometer at 420 nm against thissolution.

6. Add 0.5 ml of dilute �-galactosidase solution totube 8. Swirl gently for 5 sec and place the so-lution in your spectrophotometer. Exactly 30sec after the addition of the substrate, recordthe A420 of the solution in your notebook. Con-tinue to record the A420 value at 30-sec inter-vals for 5 min.


8. Prepare two plots of A420 versus time: one plotshould include data obtained from the reactionsin tubes 2 to 6, the other should include dataobtained from the reactions in tubes 8 to 12.Plot the data from reactions containing the sameinhibitor on a single graph for comparison.

9. Use the following equation to determine theinitial velocity of each these �-galactosidasereactions (micromoles of ONPG hydrolyzedper minute):

Initial velocity (V0) (�A420/�time)(1.25/�A420max)

10. Prepare a plot of 1/V0 versus 1/[ONPG] (aLineweaver–Burk plot) using the data obtainedfrom tubes 2 to 6. Compare the x-intercept andy-intercept of this graph with those obtained inthe absence of inhibitor (to determine KM andVmax). Are they the same? What type of inhi-bition does methyl-�,D-galactopyranoside dis-play toward �-galactosidase? Can you calculatean inhibition constant (KI) for this inhibitor (re-member that the methyl-�,D-galactopyra-noside stock solution was at a concentration of750 mM)?

11. Prepare a plot of 1/V0 versus 1/[ONPG] (aLineweaver–Burk plot) using the data obtained


from tubes 8 to 12. Compare the x-interceptand y-intercept of this graph with those ob-tained in the absence of inhibitor (to determineKM and Vmax). Are they the same? What typeof inhibition does methyl-�,D-thiogalactosidedisplay toward �-galactosidase? Can you calculate an inhibition constant (KI) for this inhibitor (remember that the methyl-�,D-thiogalactoside stock solution was at a concen-tration of 150 mM)?

12. Based on the KI values for methyl-�,D-galacto-pyranoside and methyl-�,D-thiogalactoside,which compound is the more potent inhibitorof �-galactosidase?

13. Based on the type(s) of inhibition displayed bythese two inhibitors, can you determine the siteon the �-galactosidase enzyme with whichthese two compounds interact? Explain. Wouldyou expect to be able to reverse the effects ofthese inhibitors by changing the concentrationof ONPG used in these assays? Explain.

Day 3: Effect of pH and Temperature

on �-Galactosidase Activity

Effect of pH on �-Galactosidase Activity

1. Prepare two different dilutions of the stock �-galactosidase solution: the first dilution shouldbe the same as that done for the experimentson Day 2 (one that gave a linear change in ab-sorbance at 420 nm per minute of about 0.1).Designate this solution “�-gal 1.” The secondsolution should be twice as dilute as the firstsolution. Designate this solution “�-gal 2.”You will need a total of 1.0 ml of each dilute �-galactosidase solution for the assays describedbelow.

2. Obtain 10 ml of sodium phosphate buffersranging in pH from 6.4 to 8.0:

Buffer 1 pH 6.4

Buffer 2 pH 6.8

Buffer 3 pH 7.2

Buffer 4 pH 7.7

Buffer 5 pH 8.0

3. Set up the following reactions in thirteen 13-by-100-mm glass test tubes. All volumes aregiven in milliliters:

Buffer Buffer Buffer Buffer Buffer �-gal �-gal

Tube 1 2 3 4 5 1 2

1 4.0 — — — — — —

2 3.9 — — — — 0.1 —

3 3.9 — — — — — 0.1

4 — 3.9 — — — 0.1 —

5 — 3.9 — — — — 0.1

6 — — 3.9 — — 0.1 —

7 — — 3.9 — — — 0.1

8 — — — 3.9 — 0.1 —

9 — — — 3.9 — — 0.1

10 — — — — 3.9 0.1 —

11 — — — — 3.9 — 0.1

12 — — — — 4.0 — —

13 — — — 3.5 — — —

4. To tube 1 (blank), add 0.5 ml of 2.5 mM ONPGand 0.5 ml 1 M Na2CO3, mix, and use the so-lution to zero your spectrophotometer at420 nm.

5. Add 0.5 ml of 2.5 mM ONPG to tube 2, mixgently with a Vortex mixer for 5 sec, and incu-bate at room temperature for 4 min.

6. After the 4 min incubation, stop the reaction byadding (and mixing in) 0.5 ml of 1 M Na2CO3.The addition of this weak base will raise the pH ofthe reaction enough to completely stop the reaction.

7. Perform steps 5 and 6 on tubes 3 to 7. Recordthe final A420 of all these solutions in your note-book.

8. To tube 12 (blank), add 0.5 ml of 2.5 mMONPG and 0.5 ml of 1 M Na2CO3, mix, anduse the solution to zero your spectrophotome-ter at 420 nm.

9. Add 0.5 ml of 2.5 mM ONPG to tube 8, mixgently with a Vortex mixer for 5 sec, and incu-bate at room temperature for 4 min.

10. After the 4-min incubation, stop the reaction byadding (and mixing in) 0.5 ml of 1 M Na2CO3.

11. Perform steps 9 and 10 on tubes 9 to 11. Recordthe final A420 of all these solutions in your note-book.

12. To tube 13, add 0.5 ml of the undiluted �-galactosidase stock solution. Add 0.5 ml of2.5 mM ONPG, mix gently with a Vortexmixer for 5 sec, and incubate at room temper-ature for 10 min, or until the �A420/�time isequal to zero (indicates that the reaction iscomplete).


13. Add 0.5 ml of 1 M Na2CO3, mix, and recordthe final A420 value of this solution in yournotebook. This value (A420max2) will be used tocalculate the initial velocity of each of these �-galactosidase reactions. It is necessary to calculatethis new conversion factor since these fixed time-point assays are being stopped by increasing the pH.Recall from Experiment 1 that the absorbance of asolute is strongly dependent on the pH and ionicstrength of the solution in which it resides. Since thepH at which you are measuring the absorbance ofthese solutions is not the same as the pH at whichyou were measuring the absorbance of the solutionson Day 1, a new conversion factor must be calcu-lated.

14. Calculate the initial velocity of each �-galacto-sidase reaction using the following equation(micromoles of ONPG hydrolyzed per minute):

Initial velocity (V0) (�A420/4 min)(1.25/�A420max2)

NOTE: Two dilutions of the enzyme were usedat each pH because this is a fixed time-point as-say. Rather than continuously measuring thechange in absorbance at 420 nm over time, youstopped each reaction at 4 min by increasing thepH. If the enzyme dilution that showed linearkinetics over 4 min at pH 7.7 did not show lin-ear kinetics at a different pH, the solution witha lower concentration of the enzyme (�-gal 2)will produce linear kinetics over 4 min. If �-gal1 and �-gal 2 both displayed linear kinetics over4 min at a particular pH, then the initial veloc-ity of the solution containing �-gal 1 will betwice that of the solution containing �-gal 2.

15. Prepare a plot of V0 versus pH based on theresults of this assay. From this plot, can you de-termine the pH optimum for the reaction cat-alyzed by �-galactosidase?

16. Based on the pH optimum for �-galactosidaseand your knowledge of the pKa values of thevarious amino acid side chains, can you hy-pothesize which amino acids might be presentor important in the active site of the enzyme?

Effect of Temperature on �-Galactosidase

Activity

1. Prepare the following reactions in nine 13-by-100 mm glass test tubes:

Volume of

Phosphate Buffer, Volume of 2.5 mM

Tube pH 7.7 (ml) ONPG (ml)

1 4.0 0.5

2 3.9 0.5

3 3.9 0.5

4 3.9 0.5

5 3.9 0.5

6 3.9 0.5

7 3.9 0.5

8 3.9 0.5

9 3.9 0.5

2. Place tubes 2 and 6 in a 25°C water bath. Placetubes 3 and 7 in a 30°C water bath. Place tubes4 and 8 in a 37°C water bath. Place tubes 5 and9 in a 48°C water bath. Incubate all of thesetubes at the indicated temperatures for 10 min.

3. Add and mix 0.1 ml of �-gal 1 to tubes 2 to 5.Note the exact time that the dilute �-galactosidasesolution was added to each of the tubes. After ex-actly 4 min, add 0.5 ml of 1 M Na2CO3 to eachof tubes 2 to 5.

4. Add and mix 0.1 ml of �-gal 2 to tubes 6 to 9.Note the exact time that the dilute �-galactosidasesolution was added to each of the tubes. After ex-actly 4 min, add 0.5 ml of 1 M Na2CO3 to eachof tubes 6 to 9.

5. To tube 1 (blank), add 0.5 ml of 1 M Na2CO3,mix, and use the solution to zero your spec-trophotometer at 420 nm.

6. Measure and record the final A420 values of thesolutions in tubes 2 to 9 in your notebook.

7. Calculate the initial velocity of each �-galacto-sidase reaction using the following equation(micromoles of ONPG hydrolyzed per minute):

Initial velocity (V0) (�A420/4 min)(1.25/�A420max 2)

NOTE: Two dilutions of the enzyme were usedat each temperature because this is a fixed time-point assay. Rather than continuously measur-ing the change in absorbance at 420 nm overtime, you stopped each reaction at 4 min by in-creasing the pH. If the enzyme dilution thatshowed linear kinetics over 4 min at a particu-lar temperature did not show linear kinetics ata different temperature, the solution with alower concentration of the enzyme (�-gal 2)


will produce linear kinetics over 4 min. If �-gal1 and �-gal 2 both displayed linear kinetics over4 min at a particular temperature, then the ini-tial velocity of the solution containing �-gal 1will be twice that of the solution containing �-gal 2.

8. Construct a plot of log V0 versus 1/absolutetemperature (°K). Determine the slope of theline through the linear portion of this curve.From this slope, determine the Arrhenius acti-vation energy (Ea) for the reaction catalyzed by�-galactosidase using the following equation:

Slope � 2

E

.3

a

R

where R is the gas constant (1.98 cal/deg.-mol).9. What do you notice about the activity of the

enzyme at higher temperatures? Explain whatyou think is happening to the enzyme at highertemperatures.

Day 4: Catabolite Repression and

Regulation of the lac Operon in Vivo

We suggest that steps 1 through 3 be performed by theinstructor prior to the beginning of the experiment.

1. The night before the experiment, grow 5-mlovernight cultures of each of the following E.coli strains in YT broth with shaking at 37°C.

EM1: E. coli K-12 ( lacI�,cya�,crp�,thi�)

EM1327: E. coli K-12 ( lacI�,cya�,�crp,thi�)

EM1328: E. coli K-12 ( lacI�,�cya,crp�,thi�)

EM1329: E. coli K-12 (�lacI,cya�,crp�,thi�)

Recall that the lacI, crp, and cya genes encodethe LacI repressor protein, the catabolite geneactivator protein, and adenylate cyclase, re-spectively.

2. The next morning (�3–4 hr before the exper-iment), dilute the cultures 1:10 to 1:200 in 50ml of fresh YT broth. We have found that thesestrains display different growth rates. It is very im-portant that you perform growth trials beforehandto determine the correct dilution for each strain toensure that the cultures will grow to the same op-tical density prior to the beginning of the experi-ment.

3. Grow these cultures at 37°C with shaking tolate log phase (A600 �1.0). This will take ap-proximately 3 to 4 hr.

4. Set up the following inoculations in 12 sterile(16 � 125 mm) glass test tubes:

YT Broth 200 mM 100 mM 20% Glucose

Tube (ml) IPTG (ml) cAMP (ml) (ml)

1 5.0 — — 0.05

2 5.0 0.05 — 0.05

3 5.0 0.05 0.05 0.05

4 5.0 — — 0.05

5 5.0 0.05 — 0.05

6 5.0 0.05 0.05 0.05

7 5.0 — — 0.05

8 5.0 0.05 — 0.05

9 5.0 0.05 0.05 0.05

10 5.0 — — 0.05

11 5.0 0.05 — 0.05

12 5.0 0.05 0.05 0.05

5. Using sterile technique (consult the instructor),add 100 �l of late-log-phase strain EM1 cul-ture to tubes 1 to 3. Add 100 �l of late-log-phase strain EM1327 culture to tubes 4 to 6.Add 100 �l of late-log-phase strain EM1328culture to tubes 7 to 9. Add 100 �l of late-log-phase strain EM1329 culture to tubes 10 to 12.

6. Grow all of the cultures with shaking at 37°Cfor 2.5 to 3.0 hr.

7. Set up 12 16 � 125 mm glass test tubes (onefor each culture) containing the following:

3.75 ml of sodium phosphate buffer, pH 7.7

50 �l of chloroform

50 �l of 0.1% (wt/vol) sodium dodecyl sulfate (SDS)solution

Number the tubes 1 to 12. The reaction foreach numbered culture will be done in the cor-responding numbered reaction tube.

8. Add 0.25 ml of each culture to the corre-sponding reaction tube. Mix vigorously in a Vor-tex mixer for 5 sec. The chloroform and SDSwill disrupt the integrity of (solubilize) the cellmembrane, allowing the ONPG substrate tocome into contact with the �-galactosidasepresent in the cell.

9. Add 0.5 ml of 10 mM ONPG to each reactiontube. Mix and incubate at room temperature


for 15 to 20 min or until a yellow color beginsto appear (be sure that some yellow color begins toappear in at least some of the reaction tubes. If itdoes not appear after 20 min, continue the reactionuntil it does. If you see yellow color appear after 2or 3 min, decrease the incubation time to 5 min).It is important that you note the exact time that theONPG is added to each tube.

10. Stop each of the 12 reaction tubes by addingand mixing in 0.5 ml of 1 M Na2CO3. It is im-portant that you note the exact time that theNa2CO3 is added to each tube to stop the reaction.

11. Prepare a blank solution in a large glass testtube containing the following:

3.75 ml of sodium phosphate buffer, pH 7.7

50 �l of chloroform

50 �l of 0.1% (wt/vol) SDS solution

0.25 ml of YT broth

0.5 ml of 10 mM ONPG

0.5 ml of 1M Na2CO3

Use this solution to zero your spectropho-tometer at both 550 nm and 420 nm in the ab-sorbance determinations described below.

12. Measure the A420 and A550 of the reaction so-lutions in each of the 12 reaction tubes. Recordthese values in your notebook.

13. Measure the A600 of each of the 12 cultures (notthe reaction tubes). For the absorbance measure-ments taken at 600 nm, the spectrophotometershould be blanked to read zero absorbance ofa solution containing only YT broth.

14. Calculate the activity of �-galactosidase (nano-moles of ONPG hydrolyzed per minute/A600

of culture) using the following equation:

Activity � 0.0

1

045 � 5 ml

where t time of reaction (min), 0.0045 extinction coefficient (�M�1 cm�1) for ONPunder these conditions, (1.75 � A550) cor-rection factor for the light scattering of theculture at 420 nm, and 5 ml total reactionvolume.

15. Based on your knowledge of the regulation ofthe lac operon and catabolite repression, dis-cuss whether or not each strain displayed a levelof �-galactosidase activity that was expectedunder the various growth conditions.

A420 � (1.75 � A550)

(A600)(t)

Exercises

1. During the purification of �-galactosidase, astudent found that 1 ml of a 1:10,000 dilutionof crude extract gave A420 readings as follows:

Time (sec) A420

15 0.059

30 0.084

45 0.109

60 0.134

75 0.159

The total reaction volume was 10 ml, contain-ing 8.0 ml of buffer (pH 7.5), 1.0 ml of 7.5 mMONPG solution, and 1.0 ml of the dilute �-galactosidase solution. A standard 10-mM so-lution of ONP, pH 7.5, has an A420 of 0.50 inthe 1-cm pathlength colorimeter cuvettes usedfor these assays.

a. One unit of �-galactosidase under theseconditions should correspond to what in-crease in A420 per minute (1 unit one mi-cromole of ONPG hydrolyzed per minute)?

b. If the protein concentration of the undilutedcrude extract was 13.3 mg/ml, what is thespecific activity of the extract in units permilligram?

2. Should the presence of IPTG in the growthmedium have any effect on the expression ofthe lac operon in the lacI� strain? Explain.Should the presence of cAMP in the growthmedium have any effect on the expression ofthe lac operon in the lacI� strain? Explain.

3. Should the presence of IPTG in the growthmedium have any effect on the expression ofthe lac operon in the cya� strain? Explain.Should the presence of cAMP in the growthmedium have any effect on the expression ofthe lac operon in the cya� strain? Explain.

4. Should the presence of IPTG in the growthmedium have any effect on the expression ofthe lac operon in the crp� strain? Explain.Should the presence of cAMP in the growthmedium have any effect on the expression ofthe lac operon in the crp� strain? Explain.

5. Can you think of a possible mutation in the op-erator sequence of the lac operon that would


produce a strain with the same level of expres-sion of the lac operon as the lacI� strain in thepresence of IPTG and/or cAMP?

6. Explain how the activity of adenylate cyclaseaffects expression of the lac operon. Describesituations (growth conditions) when the activ-ity of adenylate cyclase would be high and whenit would be low.

7. Explain how a mutation in the lacY gene couldaffect the regulation of the lac operon.

8. Why was it necessary to calculate a new con-version factor relating the �A420 to the num-ber of micromoles of ONPG hydrolyzed forthe kinetics experiments done on Day 3, de-spite the fact that the concentration of ONPGstock solution was the same as that on Day 1and Day 2?

REFERENCES

Beckwith, J. R., and Zipser, D., eds. (1970). The LactoseOperon. Cold Spring Harbor, NY: Cold SpringHarbor Laboratories.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Regulation of Gene Expression. In Principlesof Biochemistry, 2nd ed. New York: Worth.

Miller, J. H. (1972). Experiments in Molecular Genetics.Cold Spring Harbor, NY: Cold Spring HarborLaboratories.

Perlman, R., Chen, B., DeCrombrugghe, B., Emmer,M., Gottesman, M., Varmus, H., and Pastan, I.(1970). The regulation of lac Operon transcriptionby cyclic adenosine-3�,5�-monophosphate. ColdSpring Harbor Symp Quant Biol 35:419.

Stryer, L. (1995). Enzymes: Basic Concepts and Kinet-ics. In: Biochemistry, 4th ed. New York: Freeman.

Zubay, G., Schwartz, D., and Beckwith, J. (1970) Themechanism of activation of catabolite-sensitivegenes. Cold Spring Harbor Symp Quant Biol 35:433.

O O

135

EXPERIMENT 8

Purification of Glutamate–OxaloacetateTransaminase from Pig Heart

Theory

The detailed study of enzyme mechanisms requiresthe use of purified if not homogeneous enzymes.This experiment presents three procedures com-monly used in protein purification: ammonium sulfate precipitation, heat denaturation, and ion-exchange chromatography. Although the purifica-tion procedure outlined in this experiment is use-ful in the isolation of glutamate–oxaloacetatetransaminase (GOT), the same techniques can bemodified to aid in the purification of many otherproteins of interest.One of the most important requirements for

successful protein purification is the availability ofan accurate, rapid, and quantitative assay that isspecific for the protein of interest. If the proteinhas no known enzymatic activity, the amount ofthe protein present after each purification stepmust be analyzed by some general method such assodium dodecyl sulfate–polyacrylamide gel elec-trophoresis (SDS-PAGE), Western blotting, or anenzyme-linked immunosorbent assay (ELISA). If

the protein is known to bind a particular molecule,it may be possible to design a quantitative bind-ing assay for the protein. If the protein of inter-est has enzymatic activity, an assay that is specificfor the enzyme can be designed, and the specific ac-tivity of the enzyme can be monitored to assess theeffectiveness of each purification step. Althoughyou may find that the total enzyme activity decreasesthroughout the purification (percent yield lessthan 100), the specific activity of the enzymeshould increase with each purification step. Recallthat it is the specific activity of an enzyme that isthe true measure of its purity (see Introduction toSection II).GOT is a 90-kDa, homodimeric enzyme that

catalyzes the reversible transfer of an amino groupfrom aspartate to �-ketoglutarate (Fig. 8-1). Ox-aloacetate and glutamate are the products of thisreaction. GOT is one member of the family oftransaminase enzymes. All members of this familycatalyze the transfer of an amino group from anamino acid to an �-keto acid, which is nearly al-ways �-ketoglutarate or pyruvate. In the process,

CH2

C

Oxaloacetate

COO�

COO�

O

CH2

H3N C

Aspartate

COO�

COO�

H

CH2

CH2

C

COO�

COO�

�

�

CH2

H3N C

Glutamate

CH2

COO�

COO�

H

�

�

O

-Ketoglutarate�

Glutamate–oxaloacetatetransaminase

Figure 8-1 Reaction catalyzed by GOT.


glutamate (or alanine) and the corresponding �-keto acid of the amino acid substrate are produced.The specificity of each transaminase is determinedby the amino acid it recognizes as a substrate. Thetransaminase enzymes provide an early and criticalstep in amino acid catabolism, funneling aminogroups from different amino acids into the pro-duction of glutamate. The glutamate that is formedfrom these transaminase reactions has several fates:it can be converted to a neurotransmitter (�-aminobutyric acid, GABA), it can be used in the produc-tion of glutathione, it can be deamidated to pro-duce NH4

� for the urea cycle, it can be convertedto glutamine (a method of NH4

� assimilation inprokaryotes), or it can act as the precursor in thebiosynthesis of other amino acids.All of the transaminase enzymes rely on a com-

mon coenzyme, pyridoxal phosphate (PLP, Fig. 8-2). The PLP coenzyme is linked to the enzymethrough the formation of a Schiff base between theε-amino group of a lysine residue at the active siteand the aldehyde carbon of PLP. The pyridoxalform of GOT is heat-stable, a property that can beexploited in its purification. As aspartate enters theactive site, the PLP will form a new Schiff base withthe �-amino group forming an aldimine interme-diate. Addition of a water molecule will cleave thebond, producing oxaloacetate (the corresponding�-keto acid of aspartate) and pyridoxamine phos-phate. This reaction will be reversed as �-ketoglu-tarate enters the active site and removes the aminogroup from pyridoxamine phosphate, forming glu-

tamate and the pyridoxal form of the enzyme thatis able to accept another aspartate molecule. Themechanism of the reaction catalyzed by GOT isoutlined in Figure 8-3.How do you measure GOT activity? The ox-

aloacetate produced by the enzyme can exist in ei-ther the keto or the enol form. At basic pHs favoredby GOT (pH 8.3), the keto–enol equilibrium is dis-placed in favor of the enol form (Fig. 8-4). The enolform of oxaloacetate absorbs light at 256 nm, pro-viding a direct means of measuring GOT activity(the �A256/�time is directly proportional to thechange in concentration of oxaloacetate over time).In this experiment, you will use a different methodto measure GOT activity. You will quantify the rateof production of oxaloacetate (catalyzed by GOT)through the use of a coupled enzyme assay. In thepresence of malic dehydrogenase (MDH), oxaloac-etate and NADH will be converted to malate andNAD� (Fig. 8-5). Since NADH absorbs light at 340nm, you can continuously measure the rate of oxi-dation of NADH to NAD� by following the de-crease in absorbance at 340 nm over time. As longas the components of the coupled assay (MDH andNADH) are not rate-limiting, the �A340/�timewill be directly proportional to the change in con-centration of oxaloacetate over time, providing anaccurate measure of GOT activity. Since inexpen-sive spectrophotometers are not designed to giveaccurate absorbance readings in the short UV rangeof light, 340 nm is a much more convenient wave-length to use in the GOT assay.

C

N�

NH2

Enzyme with free �-amino group

Pyridoxal phosphate

Enzyme

�

O

O

�O

H

HO

H3C

H2O

CH2

H

O O�P

CH

N�

�NH

Pyridoxal form of enzyme

Enzyme

O

�O

HO

H3C

CH2

H

O O�P

Schiff base

Figure 8-2 PLP is a common coenzyme involved in transamination reactions.

EXPERIMENT 8 Purification of Glutamate–Oxaloacetate Transaminase from Pig Heart 137

CH

N�

NH

O

�O

HO

H3C

CH

2

H

OO

�P

�N

H3

CC

OO

�H

CH

2

CO

O�

�H

CC

OO

�H

CH

2

CO

O�

�

�

�

Pyr

idox

al fo

rm o

f GO

T

Asp

arta

te

Lys

GO

T

N�

O

�O

HO

H3C

CH

2

CH

2

H

OO

�P

CC

OO

�O

CH

2

CO

O�

C CO

O�

CH

2

CH

2

CO

O�

� �

Pyr

idox

amin

e ph

osph

ate

(bou

nd to

GO

T)

Oxa

loac

etat

e

C N�

O

�O

HO

H3C

CH

2

H

OO

�P

NH

3Ly

sG

OT

NH H

CC

OO

�C

H2

CO

O�

�

�

N�

O

�O

HO

H3C

CH

2

CH

2

H

OO

�P

NH

3

� NH

3

Lys

GO

T� N

H3

Lys

GO

T

NH

CH

N�

NH

O

�O

HO

H3C

CH

2

H

OO

�P

�N

H3

CC

OO

�H

CH

2C

H2

CO

O�

CC

OO

�H

H

CH

2C

H2

CO

O�

CH

2

CO

O�

�

�

�

Pyr

idox

al fo

rm o

f GO

T

Glu

tam

ate

Lys

GO

T

C N�

O

�O

HO

H3C

CH

2

H

OO

�P

NH

3Ly

sG

OT

NH H

CC

H2

CO

O� �

C N�

O O�

H�

HO

H3C

CH

2

H

OO

�P

NH

3Ly

sG

OT

NH H

H2O

H2O

O

�-K

etog

luta

rate

Figure 8-3 Mechanism of the reaction catalyzed by GOT.


CH2

C

C

CO

O

O�

O�

O

C

C

C

CO

O

O�

O�

HO

H

Keto form Enol form(absorbs at 256 nm)

Figure 8-4 Keto–enol equilibrium of oxaloac-

etate.

Unfortunately, there is another enzyme in pigheart mitochondria, glutamate dehydrogenase(GDH), that also recognizes �-ketoglutarate andNADH as substrates. Glutamate dehydrogenasewill convert �-ketoglutarate to glutamate, with thesubsequent oxidation of an NADH molecule (Fig.8-6). Since the change in concentration of NADHis what is being measured in the coupled GOT as-

say, GDH has the potential to interfere with theGOT activity assay, producing an artificially highvalue. Although �-ketoglutarate is a common sub-strate for both GOT and GDH, only GOT utilizesaspartate. Because of this, aspartate can be omittedfrom the GOT assay to measure GDH activity di-rectly and accurately. Once GDH activity has beenquantified, it can be subtracted from the apparentGOT activity to give a correct activity value for thetransaminase enzyme.In this experiment, you will perform a three-

step purification of glutamate–oxaloacetate trans-aminase from pig heart. With each step in the pu-rification, you will assay different fractions forGOT activity and total protein concentration.These values will then be used to calculate the to-tal activity, specific activity, and percent yield ofGOT following each purification step. The goal ofthe experiment is to achieve 80- to 100-fold purifi-cation of the enzyme with �5% yield. The overalleffectiveness of the purification procedure will beanalyzed by SDS-PAGE on the last day of the ex-periment. Be sure that you determine three different

CH2

C

Oxaloacetate

COO�

COO�

O

CH2

C

Malate

COO�

COO�

OHH

CH2

H3N C

Aspartate

COO�

COO�

H

CH2

CH2

C

COO�

COO�

�

�

CH2

H3N C

Glutamate

CH2

COO�

COO�

H

�

�O

-Ketoglutarate�

Glutamate–oxaloacetatetransaminase

NADH(absorbs at 340nm)

Malic dehydrogenase (MDH)

NAD�

Figure 8-5 Coupled enzymatic assay to measure GOT activity.


O

H HH H

OH OH

O

O

O

P

ONH2

NH4

N

NN

O

H H

H H

H H

OH OH

N

N�O

OP

C

O

�O

Nicotinamide adenine dinucleotide (NADH)(reduced form)

CH2

CH2

C

COO�

COO�

��

H3N�

�

O

-Ketoglutarate�

NH2

O

H HH H

OH OH

O

O

O

P

ONH2

H2O

N

NN

O

H H

H

H H

OH OH

N�

N�O

OP

C

O

�O

Nicotinamide adenine dinucleotide (NAD�)(oxidized form)

CH2

CH2

C

COO�

COO�

� �

Glutamate

NH2

H

Glutamatedehydrogenase

Figure 8-6 Reaction catalyzed by glutamate dehydrogenase (GDH).

values for each fraction following each purification step:The total volume of each fraction (in milliliters), theGOT activity in each fraction (in micromoles of ox-

aloacetate per minute per milliliter of enzyme), and theprotein concentration in each fraction (in milligrams permilliliter).




0.5 M potassium malate buffer (pH 6.0) containing 5 mMEDTA

0.04 M �-ketoglutarate (adjusted to pH 6.0 with NaOH)

Saturated ammonium sulfate (reagent grade) solution

0.03 M sodium acetate buffer (pH 5.4)



Carboxymethylcellulose (equilibrated in 0.03 M sodiumacetate, pH 5.0)

0.1 M potassium phosphate buffer (pH 7.5)

Standard bovine serum albumin solution (1 mg/ml)

Solutions for Folin-Ciocalteau protein assay:

2 N Folin–Ciocalteau reagent

2% Na2CO3 in 0.1 N NaOH

1% CuSO4 5H2O

2% sodium tartrate

250-ml plastic centrifuge bottles

50-ml plastic centrifuge tubes

Bunsen burner

Pasteur pipettes

Spectrophotometer to read absorbance at 340 nm and500 nm

Cuvettes or colorimeter tubes for spectrophotometer

Small and large glass test tubes

Cheesecloth

Glass wool

Chromatography column (20 ml)

Fresh pig hearts

Meat grinder

Blender

0.1 M potassium phosphate and 0.12 M aspartate buffer(pH 7.5)

3 mM NADH in 0.1 M potassium phosphate buffer(pH 7.5)—prepared fresh daily

Malic dehydrogenase (MDH), 200 units/ml in 0.1 Mpotassium phosphate buffer (pH 7.5)

0.1 M �-ketoglutarate solution (pH 7.5)

Dialysis tubing (8,000–12,000 Da molecular weight cut-off )

Reagents and supplies for SDS-PAGE (see Experiment 4):

10 Buffer A (1.5 M Tris HCl, pH 8.8)

10 Buffer B (1.0 M Tris HCl, pH 6.8)

30% acrylamide (292 g/liter acrylamide, 8 g/literN,N�-methylenebisacrylamide)

10% (wt/vol) SDS in water

TEMED (N,N,N�,N�-tetramethylethylenediamine)

10% (wt/vol) ammonium persulfate in water—prepared fresh

SDS-PAGE running buffer (25 mM Tris, 250 mMglycine, 0.1% (wt/vol) SDS, pH 8.3)

4 SDS sample buffer:

0.25 M Tris, pH 7.0

30% (vol/vol) glycerol

10% (vol/vol) 2-mercaptoethanol

8% (wt/vol) SDS

0.001% (wt/vol) bromophenol blue

Coomassie Blue staining solution (40% methanol, 10%glacial acetic acid, 0.25% (wt/vol) Coomassie Bril-liant Blue R-250, 50% distilled water)

Protein molecular weight standards (see Experiment 4)

Destaining solution (40% methanol, 10% glacial aceticacid, 50% distilled water)

Power supply

GOT Activity Assay

The following procedure will be used for all of theGOT activity assays performed in this experiment.

1. Add the following, in order, to a small glass testtube:

Reagent Volume (ml)

0.1 M potassium phosphate and

0.12 M aspartate buffer (pH 7.5) 2.5

3 mM NADH solution 0.1

MDH (200 units/ml) 0.1

GOT dilution (in 0.1 M potassium

phosphate buffer, pH 7.5) 0.1

2. Blank your spectrophotometer to read zero ab-sorbance at 340 nm against water.

3. Add 0.2 ml of 0.1 M �-ketoglutarate solutionto the reaction tube, mix, and place the solu-tion in the spectrophotometer.

4. Record the A340 of the solution at 30-sec in-tervals for 4 min. Determine whether the ini-tial velocity of the reaction is linear over 4 min(�A340/�time yields a straight line). If the ki-netics are not linear over 4 min, dilute the en-zyme twofold in 0.1 M potassium phosphatebuffer, pH 7.5, and repeat the assay. Prepare aplot of A340 versus time and calculate the ab-solute value of the slope of the line.


5. Determine the activity of GOT (micromoles ofoxaloacetate produced per minute per milliliterof enzyme) using the following equation:

Activity � (3 ml)

where (�A340/�time) is the absolute value ofthe slope of the line produced from a plot ofA340 versus time, 6.22 mM

�1cm�1 is the mil-limolar extinction coefficient for NADH at340 nm, 3 ml is the total reaction volume, folddilution is the extent to which the GOT sam-ple was diluted prior to the assay, and 0.1 ml ofenzyme is the volume of the dilute enzyme so-lution used in the assay.

GDH Activity Assay

The following procedure will be used for all GDHactivity assays performed in the experiment. It is es-sential that GDH activity be assayed for and sub-tracted from the apparent GOT activity until youare sure that the GOT and GDH have been sepa-rated by one of the purification steps. Once this hasoccurred, you are no longer required to assay forGDH activity.

1. Add the following, in order, to a 13 100 mmglass test tube:

Reagent Volume (ml)

0.1 M potassium phosphate

(pH 7.5)—no aspartate 2.5

3 mM NADH solution 0.1

MDH (200 units/ml) 0.1

GOT dilution (in 0.1 M potassium

phosphate buffer, pH 7.5) 0.1

2. Blank your spectrophotometer to read zero ab-sorbance at 340 nm against water.

3. Add 0.2 ml of 0.1 M �-ketoglutarate solutionto the reaction tube, mix, and place the solu-tion in the spectrophotometer.

4. Record the A340 of the solution at 30-sec in-tervals for 4 min. Determine whether the ini-

fold dilution 0.1 ml of enzyme

1 6.22 mM�1cm�1

�A340 �time

tial velocity of the reaction is linear over 4 min(�A340/�time yields a straight line). If the ki-netics are not linear over 4 min, dilute the en-zyme twofold in 0.1 M potassium phosphatebuffer, pH 7.5, and repeat the assay. Prepare aplot of A340 versus time and calculate the ab-solute value of the slope of the line.

5. Determine the activity of GDH (micromolesof NADH oxidized per minute per milliliter ofenzyme) using the following equation:

Activity � (3 ml)

where (�A340/�time) is the absolute value ofthe slope of the line produced from a plot ofA340 versus time, 6.22 mM

�1cm�1 is the mil-limolar extinction coefficient for NADH at340 nm, 3 ml is the total reaction volume, folddilution is the extent to which the GOT sam-ple was diluted prior to the assay, and 0.1 ml ofenzyme is the volume of the dilute enzyme so-lution used in the assay.

Folin–Ciocalteau Assay

The following procedure will be used for all pro-tein assays performed in this experiment. Becauseof day-to-day variation in the assay, you must pre-pare a standard curve each day that you performthe Folin–Ciocalteau assay.

1. Set up the following reactions in 16 125 mmglass test tubes:

Volume of

1 mg/ml Bovine Volume of GOT Volume of

Tube Serum Albumin (ml) Sample (ml) Water (ml)

1 — — 1.20

2 0.02 — 1.18

3 0.05 — 1.15

4 0.10 — 1.10

5 0.20 — 1.00

6 0.30 — 0.90

7 — 0.10 1.10

8 — 0.20 1.00

fold dilution 0.1 ml of enzyme

1 6.22 mM�1cm�1

�A340 �time


2. Prepare 100 ml of fresh alkaline copper reagentby mixing, in order:

98 ml of 2% Na2CO3 in 0.1 M NaOH

1 ml of 1% CuSO4 5H2O

1 ml of 2% sodium tartrate

Add (and immediately mix in) 6 ml of thisreagent to each tube and incubate at room tem-perature for 10 min.

3. Add (and immediately mix in) 0.3 ml of Folin–Ciocalteau reagent to each tube and incubateat room temperature for 30 min.

4. Blank your spectrophotometer to read zero ab-sorbance at 500 nm against the solution in tube1 (blank).

5. Read and record the A500 of the solutions intubes 2 to 8.

6. Prepare a standard curve using the data fromtubes 2 to 6 by plotting A500 versus milligramsof protein. Use the standard curve to determinethe concentration of protein in the GOT sam-ple (tubes 7 and 8). If the A500 of tubes 7 and8 lies outside the range of the standard curve,dilute the GOT sample an additional twofoldand repeat the assay.

Protocol

Day 1: Preparation of Pig Heart

Homogenate and Heat Denaturation

We suggest that steps 1 to 3 be performed by the in-structor prior to the beginning of the experiment.

1. In the cold room (4°C), trim the fresh pighearts free of fat and auricles. Mince small por-tions of the heart tissue with a meat grinder.This minced heart tissue can be frozen at�20°C for up to 4 weeks.

2. Add 300 ml of ice cold 0.5 M potassium malatebuffer (pH 6.0) containing 5 mM EDTA toeach 200 ml portion of minced tissue.

3. Homogenize in a blender until no large tissueportions are present (about 1 min). Store thehomogenate on ice until the beginning of theexperiment.

4. Obtain 75 ml (record the exact volume in yournotebook) of pig heart homogenate from the in-structor and place in a 250-ml plastic centrifuge

bottle. Although you will only be assaying a smallportion of the crude pig heart homogenate for GOTactivity (see step 5 below), the total volume of thissample should be used in the calculations for the pu-rification table at the end of the experiment.

5. Remove 5 ml of the homogenate, place it in aclean 30-ml plastic centrifuge tube, and cen-trifuge at 8000 g for 10 min. The supernatantis Fraction I (FR-I). Store it on ice. The pelletproduced in this step may be discarded.

6. Prepare a water bath in a large beaker by heat-ing over a Bunsen burner. Submerge the 250-ml centrifuge bottle (containing 70 ml of pigheart homogenate) into the water bath andbring the temperature of the homogenate to60°C. Place the clean thermometer directly into thehomogenate (not the water bath) to monitor its tem-perature.

7. Add 15 ml of 0.04 M �-ketoglutarate to thehomogenate to convert the GOT to the heat-stable pyridoxal form.

8. Increase the temperature of the homogenate to72°C and incubate it for exactly 20 min. Do notexceed 75°C or incubate for longer than 20 min. Ifthe time of the incubation is too long or the tem-perature is too high, the GOT will denature.

9. Remove the homogenate from the water bathand chill on ice for 20 min.

10. Filter the heat-treated homogenate throughcheesecloth into a clean 250-ml plastic cen-trifuge bottle. The solid material contains thedenatured proteins and cell debris that precip-itated out of solution at high temperature. Thesupernatant contains GOT, small molecules,and other heat-stable proteins.

11. Clarify the supernatant fraction further by cen-trifugation at 8000 g for 10 min. The super-natant is Fraction II (FR-II). Record its exactvolume and store it on ice. The pellet producedin this step (cell debris and denatured proteins)may be discarded.

12. Perform GOT and GDH assays on FR-I andFR-II using the following suggested dilutionsin 0.1 M potassium phosphate buffer (pH 7.5):

1:160 for FR-I

1:80 for FR-II

Calculate the GOT activity present in eachfraction (micromoles of oxaloacetate producedper minute per milliliter of enzyme).


13. Place 0.5 ml of FR-I and FR-II in separate 1.5ml microcentrifuge tubes labeled with yourname. Return these to the instructor, who willstore them at �20°C for SDS-PAGE analysison the last day of the experiment.

14. Store the remainder of FR-I and FR-II on iceat 4°C for use on Day 2.

Day 2: Development of an Ammonium

Sulfate Fractionation

1. Place 0.4 ml of FR-II into each of seven 1.5-mlmicrocentrifuge tubes. Place the tubes on ice.

2. Slowly (one drop at a time with gentle mixingin between) add the following volumes of sat-urated ammonium sulfate solution (SAS) toeach tube:

Volume of % Ammonium

Tube SAS (ml) Sulfate Saturation

1 — 0

2 0.40 50

3 0.49 55

4 0.60 60

5 0.74 65

6 0.93 70

7 1.20 75

Incubate all the tubes on ice for 20 min. NOTE:It is very important that the protein solutionbe brought up to the desired ammonium sul-fate saturation slowly. If this step is not donecorrectly, it will be difficult to reproduce theseresults when the same procedure is carried outon the bulk preparation. Also, avoid frothing inthe solution (a sign that some of the proteinsare being denatured) by gently mixing the so-lution as the ammonium sulfate is added.

3. Centrifuge tubes 2 to 7 for 5 min at 4°C in amicrocentrifuge. Remove and discard the su-pernatant from the precipitated protein pellet.

4. Resuspend the precipitated protein pellet fromtubes 2 to 7 in 1.0 ml of 0.03 M sodium acetatebuffer, pH 5.4. Add 0.6 ml of this same bufferto tube 1 (untreated) to bring the final volumeof all of the samples to 1.0 ml.

5. Perform GOT assays (and GDH assays, if nec-essary) on all seven tubes using the following

suggested dilutions in 0.1 M potassium phos-phate buffer (pH 7.5):

Tube Dilution

1 1:20

2 Undiluted

3 1:20

4 1:50

5 1:10

6 1:40

7 1:20


6. Perform Folin–Ciocalteau assays on FR-I, FR-II, and each of the solutions in tubes 1 to 7 (24assays in all, including those for the standardcurve) using the following suggested dilutionsin water (not phosphate buffer):

1:20 for FR-I

1:5 for FR-II

1:2 for tubes 1 to 7

Calculate the concentration of protein (in mil-ligrams per milliliter) present in FR-I, FR-II,and each of the ammonium sulfate saturationtrials.

7. Store FR-I and FR-II on ice at 4°C for use onDay 3.

8. Prepare a plot of percent GOT recovered ver-sus percent ammonium sulfate saturation usingthe data from tubes 2 to 7 with the use of thefollowing equation:

% GOT recovered �

100

where total GOT activity is the activity ofGOT (micromoles of oxaloacetate producedper minute per milliliter of enzyme) times thetotal volume of the sample (1 ml).

9. On the same graph, prepare a plot of percentprotein recovered versus percent ammoniumsulfate saturation using the following equation.

% Protein recovered �

100mass of protein in sample (mg) mass of protein in tube 1 (mg)

total GOT activity in sample total GOT activity in Tube 1


where mass of protein is the concentration (inmilligrams per milliliter) of protein in the sam-ple (determined from the standard curve) timesthe total volume of the sample (1 ml).

10. Determine the highest percent ammonium sul-fate saturation that gave less than 20% GOTrecovered. This will be a good percent ammo-nium sulfate saturation to use for the first cuton Day 3 (when the bulk of the GOT is stillsoluble). Next, determine the lowest percentammonium sulfate saturation that gave morethan 80% GOT recovered. This will be a goodpercent ammonium sulfate saturation to use forthe second cut on Day 3 (when the bulk of theGOT will precipitate out of solution). This in-formation will be critical for the experiment per-formed on Day 3, so these calculations must be com-pleted prior to the beginning of the next experiment.

Day 3: Ammonium Sulfate Fractionation of

FR-II (Bulk Preparation)

1. Based on your results from the ammonium sul-fate precipitation trials performed on Day 2,determine what volume of SAS you will needto add to the remainder of FR-II to precipitateless than 20% of the total GOT in solution us-ing the following formula:

Volume of SAS �

where S1% is the initial percent ammoniumsulfate saturation (zero at this point for FR-II)and S2% is the desired final percent ammo-nium sulfate saturation.

2. Slowly (one drop at a time with gentle mixingin between) add the correct volume of saturatedammonium sulfate solution to the remainder ofFR-II. Incubate on ice for 25 min.

3. Centrifuge the sample at 8000 g for 20 min.Decant the supernatant into a clean 250-mlplastic centrifuge bottle.

4. Redissolve the precipitated protein pellet pro-duced in step 3 in 3 ml of 0.03 M sodium ac-etate buffer (pH 5.0). This is ammonium sul-fate fraction I (AS-I). Record the volume of this

[(volume of sample at S1%)(S2% � S1%)]/100

[(1 � (S2% / 100)]

fraction and store AS-I on ice at 4°C for useon Day 4.

5. Determine what volume of SAS you will needto add to the remainder of the supernatant fromthe first cut to precipitate more than 80% ofthe total GOT in solution. Use the same for-mula shown in step 1, but remember that S1%is now greater than zero and that the volume ofFR-II has changed.

6. Slowly (one drop at a time with gentle mixingin between) add the correct volume of SAS so-lution to the supernatant fraction from the firstsalt cut. Incubate on ice for 25 min.

7. Centrifuge the sample at 8000 g for 20 min.Decant the supernatant into a clean, 250-mlplastic centrifuge bottle. This is AS-III. StoreAS-III on ice at 4°C for use on Day 4 andrecord its volume.

8. Redissolve the precipitated protein pellet pro-duced in step 7 in 3 ml of 0.03 M sodium ac-etate (pH 5.0). This is AS-II. Store this frac-tion on ice and record its exact volume.

9. Perform GOT assays (and GDH assays, if nec-essary) on AS-I, AS-II, and AS-III using the fol-lowing suggested dilutions in 0.1 M potassiumphosphate buffer (pH 7.5):

1:100 for AS-I

1:200 for AS-II

1:10 for AS-III

Calculate the GOT activity present in eachfraction (micromoles of oxaloacetate producedper minute per milliliter of enzyme). Also, cal-culate the total GOT activity in each of thesefractions (GOT activity times the total fractionvolume). Note that the sum of the total GOT ac-tivity in AS-I, AS-II, and AS-III should be roughlyequal to the total GOT activity in FR-II.

10. Obtain a 12-in. length of prepared dialysis tub-ing from the instructor (always wear gloveswhen handling the dialysis tubing and do notallow it to dry out). Tie two knots in one endof the tubing and fill the tubing with the entireAS-II fraction (�3 ml). Tie two more knots inthe other end of the dialysis tubing and dialyzeovernight against 6 liters of 0.03 M sodium ac-etate buffer (pH 5.0) to remove the residual am-monium sulfate from the sample. This is doneto ensure that the ionic strength of the solution


(and the pH) will be low enough to allow GOTto bind the carboxymethylcellulose column. Itis also necessary to remove the residual ammo-nium sulfate so that the NH4

� will not inter-fere with the Folin–Ciocalteau assay performedon AS-II.

Day 4: Folin–Ciocalteau Assay on Fractions

and Preparation of Carboxymethylcellulose

Column

1. Remove the dialysis bag containing AS-II andcentrifuge at 8000 g for 10 min. Remove andsave the supernatant on ice. This is ammoniumsulfate fraction II dialyzed (AS-IId). Remove0.5 ml and place in a 1.5 ml microcentrifugetube labeled with your name. Return this to theinstructor, who will store it at �20°C for SDS-PAGE analysis at the end of the experiment.

2. Perform Folin–Ciocalteau assays on AS-I, AS-II, and AS-IId using the following sug-gested dilutions in water:

1:20 for AS-I

1:10 for AS-II

1:10 for AS-IId

Calculate the concentration of protein in (mil-ligrams per milliliter) present in each fraction.

3. Obtain a 10-ml aliquot of carboxymethylcellu-lose (CMC) resin saturated with 0.03 M sodiumacetate (pH 5.0). Close the outlet valve at thebottom of the column. Slowly pour the slurryinto the chromatography column containing asmall plug of glass wool at the bottom (consultthe instructor). Allow the resin bed to settlewith each small addition of the slurry. Whenthe resin has settled to half of the columnheight (�5 ml), open the output valve and ad-just the flow rate to �1 ml/min. Equilibrate thecolumn in 0.03 M sodium acetate (pH 5.0) bypassing three column volumes (30 ml) of thisbuffer through the column. Leave about 10 mlof buffer above the resin bed and close the out-let valve at the bottom of the column.

4. Label the column with your name and store itat 4°C for use on Day 5.

5. Store the AS-IId fraction on ice at 4°C for useon Day 5.

Day 5: Carboxymethylcellulose

Chromatography

1. Open the outlet valve on the CMC column andallow the level of the buffer to fall even withthat of the resin bed (but not below it). Dur-ing this time, adjust the flow rate to 1 ml/min.

2. Gently load all but 0.1 ml of AS-IId on the topof the bed and allow it to flow into the resin.Carefully layer (without disturbing the resinbed) the top of the resin bed with 20 ml of 0.03M sodium acetate buffer (pH 5.0). Immediatelybegin collecting 2 ml fractions as they elutefrom the column. When this is complete, closethe outlet valve at the bottom of the column.

3. Perform a quick GOT assay on every otherfraction to ensure that the GOT adsorbed tothe column. These rapid assays do not requireany sample dilution. Simply add 0.1 ml of everyother fraction to a GOT assay tube and deter-mine whether you see any significant decreasein A340 after 2 min.

4. Elute GOT from the CMC column by passing40 ml of 0.08 M sodium acetate buffer (pH 5.0)through the column. Immediately begin col-lecting 2-ml fractions as they elute.

5. Perform rapid GOT assays on every other frac-tion (see step 3) as well as absorbance readingsat 280 nm for each fraction.

6. Prepare an elution profile by plotting relativeGOT activity (�A340/2 min) versus fractionnumber. Combine the three fractions that showthe highest relative GOT activity.

7. Accurately assay the precolumn sample andpooled eluted samples for GOT activity usingthe following suggested dilutions in 0.1 Mpotassium phosphate buffer (pH 7.5):

1:100 for pre-column fraction

1:20 for pooled eluted fraction


8. Perform a Folin–Ciocalteau assay on thepooled eluted fraction. No dilutions are neededat this point, since the total protein concentra-tion is likely to be low. Calculate the concen-tration of protein (in milligrams per milliliter)present.


Volume [Protein] Total GOT Activity Specific Activity Fold

(ml) (mg/ml) (�mol/min) (�mol/min/mg) % Yield Purification

Fraction I

(crude homogenate)

Fraction II

(heat-treated)

AS-II

(predialysis)

AS-IId

(postdialysis)

Pooled CMC

fraction

9. Label the pooled eluted fraction with yourname and return it to the instructor, who willstore it at �20°C for SDS-PAGE analysis onDay 6.

10. Prepare a purification table for GOT:

specifically what the changes would be and howthe new procedure would be performed (buffercomposition, etc.).

Day 6: SDS-PAGE Analysis (See the Section

II Introduction and Experiment 4)

1. Determine the concentration of protein in theFR-I, FR-II, AS-IId, and pooled CMC frac-tions from the Folin–Ciocalteau assays per-formed throughout the experiment. Rememberthat 0.5 ml of each of these fractions was savedfor SDS-PAGE analysis.

2. For fractions that had a protein concentrationgreater than 1 mg/ml, determine how mucheach one would have to be diluted with waterto bring the final concentration to 1 mg/ml.For example, if FR-I were at a concentrationof 10 mg/ml, it could be diluted 10-fold withwater to bring it to 1 mg/ml.

3. For fractions that had a protein concentrationless than 1 mg/ml, determine what volume ofthat fraction would contain 10 �g of protein.For example, if AS-IId is at a concentration of0.1 mg/ml, 100 �l of AS-IId contains 10 �g ofprotein. Add the appropriate volume of thesefractions to a microcentrifuge tube and add

The total GOT activity of a fraction can be cal-culated by multiplying the GOT activity of thefraction (micromoles of oxaloacetate producedper minute per milliliter of enzyme) by the to-tal fraction volume.The specific activity of GOT of a fraction

can be calculated by dividing the GOT activ-ity of the fraction by the protein concentration(in milligrams per milliliter) for that fraction.The percent yield for each fraction can be

calculated by dividing the total GOT activityin a fraction by the total GOT activity in frac-tion I (crude homogenate) and multiplying thevalue by 100.The fold purification for each fraction can

be calculated by dividing the specific activity ofGOT in a fraction by the specific activity ofGOT in fraction I (crude homogenate).

11. What is the overall fold purification of GOTin your experiment?

12. What is the percent yield of GOT followingthis purification procedure?

13. Which step in the purification procedure af-forded the greatest increase in the purity ofGOT?

14. Which step in the purification procedure af-forded the least increase in the purity of GOT?

15. Discuss the overall effectiveness of this GOTpurification protocol in terms of fold purifica-tion and percent yield. Suggest possible modi-fications in the experiment that might improvethe percent yield or fold purification. Describe


water to bring the final volume of each of thesesamples to 0.2 ml. Add 0.2 ml of 20% trichlor-oacetic acid to each of the samples, mix, andincubate on ice for 10 min. Centrifuge the sam-ples at 4°C for 5 min and remove the super-natant (there should be a precipitated proteinpellet at the bottom of the tube). Wash the pro-tein pellet with 100 �l of acetone, centrifugefor 1 min at 4°C, and remove the supernatant.Allow the precipitated protein pellet to air dryand resuspend it in 10 �l of water.

4. Prepare the following samples for SDS-PAGE:

Volume of Sample Volume 4� SDS

Tube at 1 mg/ml Sample Buffer

1 10 �l (FR-I) 5 �l

2 10 �l (FR-II) 5 �l

3 10 �l (AS-IId) 5 �l

4 10 �l (pooled CMC fraction) 5 �l

5 10 �l (protein molecular 5 �l

weight standard)

The molecular weight protein standards(tube 5) contain 1 mg/ml each of phosphory-lase (97,400 Da), bovine serum albumin(66,200 Da), ovalbumin (45,000 Da), carbonicanhydrase (31,000 Da), soybean trypsin in-hibitor (21,500 Da), and lysozyme (14,400 Da)and will be provided by the instructor.

5. Incubate the samples for 5 min in a boiling wa-ter bath and centrifuge briefly to collect thesample at the bottom of the tube.

6. Load the samples on a 10-well, 12% SDS poly-acrylamide gel with a 4% polyacrylamide stack-ing gel. Two groups can share a single SDS-PAGE gel. Apply 200 V (constant voltage), andallow the electrophoresis to continue until thebromophenol blue dye front reaches the bot-tom of the gel.

7. Remove the gel from the plates and place intoa Coomassie Blue staining solution (50 ml pergel) for 1 hr. Pour off the staining solution andplace in destaining solution for 1 hr, changingthe destain solution every 15 min.

8. The proteins should appear as bright bluebands on the gel as the destain process pro-gresses. At this point, the gel can be pho-tographed or dried to serve as a permanentrecord.

9. Calculate the Rf values of the molecular weightprotein standards using the following formula:

Rf �

10. Prepare a standard curve by plotting the log ofmolecular weight versus Rf for each of the sixprotein molecular weight standards (see Fig. 4-8). Based on the Rf value of the proteinband(s) in the lane containing the pooled CMCfraction, what is the molecular weight of theprotein(s) present at the end of the purifica-tion? Can you identify GOT (GOT is a90,000-Da homodimer composed of two iden-tical subunits)? Are there any other proteinsthat copurified with GOT? What are theirmolecular weight(s)? Has the intensity of theprotein band corresponding to GOT changedduring the course of the purification? Explainwhat this means.

Exercises

1. What was being removed from FR-II duringdialysis? Give two reasons why it was necessaryto remove this component prior to continuingwith the assay.

2. The pH of the buffer used to load FR-II on theCMC column was the same as that of the bufferused to elute GOT from the CMC column.How could two buffers of the same pH affectthe ability of GOT to adsorb to the CMCresin?

3. Describe the difference between the activityand the specific activity of an enzyme. Howis it possible that the total activity of an en-zyme in a particular fraction could decreasewhile the specific activity of that same frac-tion increases?

4. Using the data shown below, answer questionsa and b: The GOT assay was performed in atotal volume of 3 ml. The value in parenthesesindicates how much the fraction was diluted be-fore using 0.1 ml in the assay. The protein as-say was performed by diluting the samples asindicated in parentheses and determining anA280 value in a 1-cm pathlength cuvette. Assume

distance traveled by protein band (cm)

distance traveled by dye front (cm)


that the proteins in the solution have an ex-tinction coefficient of 0.7 (mg/ml)�1 cm�1. Themillimolar extinction coefficient for NADH is6.22 mM�1 cm�1 at 340 nm.

Volume GOT assay

Fraction (ml) (� A340/min) A280

Crude cell 100 0.57 (1:100) 0.464 (1:1000)

lysate

FR-II 080 0.40 (1:100) 0.186 (1:1000)

AS-IId 003 0.81 (1:1000) 0.440 (1:100)

CMC pooled 006 0.10 (1:1000) 0.200 (1:10)

a. Which single step in the purification proto-col resulted in the greatest purification ofGOT? Use specific numbers to justify youranswer.

b. Calculate the overall fold purification andpercent yield of GOT following this com-plete purification protocol.

5. Propose an experiment that would allow you topurify GOT with the use of a DEAE (diethy-laminoethyl) cellulose column rather than witha CMC (carboxymethylcellulose) column. ThepI of GOT is�6.0. Include in your answer adescription of the buffer system that you woulduse (buffering ions, ionic strength, pH) forloading GOT on the column and for elutingGOT from the column.

REFERENCES

Banks, B. E. C., Doonan, S., Lawrence, A. J., and Ver-non, C. A. (1969). The Molecular Weight andOther Properties of Aspartate Aminotransferasefrom Pig Heart Muscle. Eur J Biochem 5:528.

Christen, P., and Metzler, D. E. (1985). Transaminases.New York: Wiley-Interscience.

Jansonius, J. N., and Vincent, M. G. (1987). StructuralBasis for Catalysis by Aspartate Aminotransferase.(Biological Macromolecules and Assemblies, Vol.3.) New York: Wiley.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Amino Acid Oxidation and the Productionof Urea. In: Principles of Biochemistry, 2nd ed. NewYork: Worth.

Nielsen, T. B., and Reynolds, J. A. (1978). Measure-ment of Molecular Weights by Gel Electrophoresis.Methods Enzymol, 48:3.



Sizer, I. W., and Jenkins, W. T. (1962). Glutamic As-partic Transaminase from Pig Ventricles: Prepara-tion and Assay of Enzymes. Methods Enzymol 5:677.

Stryer, L. (1995). Amino Acid Degradation and theUrea Cycle. In: Biochemistry, 4th ed. New York:Freeman.

Voet, D., and Voet, J. G. (1990). Amino Acid Metabo-lism. In: Biochemistry. New York: Wiley.

: P

149

EXPERIMENT 9

Kinetic and Regulatory Properties of Aspartate Transcarbamylase

Theory

The efficient and balanced functioning of the me-tabolism of an organism requires that various meta-bolic pathways be regulated. A biosynthetic pathway,for example, should be most active when the organ-ism has a high demand for the product of that path-way and inactive when the organism has an outsidesource for the product. In biosynthetic pathways ofbacteria, such regulation is accomplished by twokinds of mechanisms: control of the rate of synthe-sis of enzymes (generally referred to as repression)and control of the catalytic efficiency of enzymes byinhibition or activation. The latter mode of controlwill concern us in this experiment. It is commonlyobserved that the first enzyme in an unbranchedpathway leading to the biosynthesis of a compoundis subject to feedback inhibition by the end productof the pathway. Thus, the end product exerts feed-back control over the role of its own synthesis.

In this experiment we will examine some of theproperties of the aspartate transcarbamylase of Escherichia coli, which is typical of many enzymes subject to feedback inhibition and which has beenstudied extensively. Aspartate transcarbamylase(ATCase) catalyzes the first reaction unique to thebiosynthesis of pyrimidine nucleotides. ATCase issubject to specific inhibition by quite low concen-trations of one of its end products, cytidine 5�-triphosphate (CTP). This relationship and twoother regulatory interactions important to the con-trol of pyrimidine biosynthesis are summarized inFigure 9-1.

The study of the kinetic properties of the ATCase reaction and of its inhibition by CTP, some

of which will be repeated in this experiment, revealsseveral unusual characteristics that cannot be ex-plained by the simple Michaelis–Menten kinetictheory, presented in the Section II introduction.The inhibition of ATCase by CTP can be over-come by increased concentrations of the substrateaspartate, which suggests that these molecules com-pete for the active site of the enzyme. Yet the chem-ical structures of CTP and aspartate are very dif-ferent and would not be expected to bind to thesame site. Furthermore, inhibition by CTP is veryspecific; UTP does not inhibit. Another nucleotide,ATP, is a stimulator rather than an inhibitor of ATCase. Treatment of ATCase with mercurials re-sults in an enzyme that is fully active (actually moreactive than the native enzyme) but which is insen-sitive to inhibition by CTP. These observations ledto the proposal that aspartate and CTP bind at en-tirely separate sites—aspartate at the active or cat-alytic site and CTP at a separate allosteric site. Theinteractions between ligands at the active site andthe allosteric site are negative; that is, binding ofCTP lessens the ability of aspartate to bind. Hence,the effect of CTP can be overcome by high con-centrations of aspartate. Clearly, if aspartate doesnot bind to ATCase, no product (carbamyl aspar-tate) can be formed, so CTP acts as an inhibitor.The allosteric nature of the CTP binding site onATCase was proven by the demonstration that theactive site and the CTP binding site actually resideon two different polypeptide chains (Fig. 9-2).

ATCase and many (but not all) enzymes subjectto metabolic regulation have a second interestingproperty: their reaction velocity is not a hyperbolicfunction of substrate (aspartate) concentration, as

150

The source of the enzyme to be used in thesestudies is a mutant strain of E. coli, strain EK1104,in which the gene encoding ATCase (pyrB) has beendeleted and has a defect in another enzyme ofpyrimidine biosynthesis, orotidylic decarboxylase(encoded by the pyrF gene) (see Nowlan andKantrowitz, 1985). The pyrB gene is supplied to theorganism in many copies per cell on the pEK2 plas-mid. However, because of the “leaky” defect in theorotidylate decarboxylase gene, the strain makespyrimidine nucleotides very slowly. Thus, the poolof UTP, which is the repressing metabolite for thepyrBI that encodes ATCase, is very low, and thebiosynthesis of ATCase is markedly derepressed.These observations underscore the fact that controlof ATCase activity occurs in E. coli cells both by re-pression of enzyme synthesis and by inhibition ofenzyme activity.


E. coli strain EK1104 bearing plasmid pEK2

Materials for growth medium and culturing bacteria (seeAppendix)

would be predicted from simple steady-state kinet-ics (see Fig. II-8). Rather, the aspartate saturationcurve is sigmoid (S-shaped). You can rationalize theresults by suggesting that aspartate binds to morethan one site on the enzyme and that once aspar-tate is bound to one of these sites, it binds morereadily to a subsequent site. This is called coopera-tive binding. Such binding is similar to the interac-tions between aspartate and CTP binding, exceptthat in this case interactions between aspartate sitesare positive rather than negative. Simple enzymekinetics generally assume single ligand sites or theabsence of interactions between multiple sites.However, you must consider both positive and neg-ative interactions between substrate and allostericsites in order to understand many regulatory en-zymes. Biochemists are now trying to learn the pre-cise physical nature of such site–site interactions.

Some of the kinetic properties of ATCase de-scribed above will be examined in this experiment.The assay to be used is a fixed-time, colorimetricprocedure. Carbamyl aspartate accumulated in thefirst step of the procedure is assayed in a secondstep (Fig. 9-3).

SECTION II Proteins and Enzymology

Inh ib i t ion

CO2

NH2

C

O�

O�

O

O P

O

ATP�

Glutamine

ADP�

Glutamate

CH

CH2

NH2 COO�

COO�

CH

CH2NH2

CN COO�

COO�

Carbamyl phosphatesynthetase

Aspartatetranscabamylase

Carbamylphosphate

O�

O�HO P

O

Aspartate

O

H

Carbamylaspartate

UMP UTP CTPCTP synthetase

Various reactionsthat consumecarbamyl phosphate

Inhibition

Pyrimidine compounds

Inhibition

Figure 9-1 Sites of feedback inhibition in carbamyl phosphate metabolism of E. coli. Note that aspartate

trascarbamylase is the first enzyme on the unique pathway to pyrimidine compounds.

EXPERIMENT 9 Kinetic and Regulatory Properties of Aspartate Transcarbamylase 151

N

NC

C

C

C

N

N

N

N

C

C

20Å

Catalyticsite

CTP and ATPsite

Figure 9-2 Three-dimensional structure of E. coli aspartate transcarbamylase. Half of the native c6r6

molecule (see Fig. II-4) is shown. The catalytic (c) submit, which binds the substrates aspar-

tate and carbamyl phosphate is shown in light shading. The regulatory (r) subunit, which

binds the allosteric effectors CTP and ATP, is shown in dark shading. From Kantrowitz, E.R.,

et al. (1980). E. coli Aspartate Transcarbamylase. Part II: Structure and Allosteric Interactions.

Trends Biochem Sci 5:150; and Stryer, L. (1995). Biochemistry, 4th ed. New York: Freeman,

Figure 10-5, p. 240. Reprinted by permission.

NH2C NH CH COO�

COO�

CH3

CH3

CH3

CHC NOH

C ON

N OO

Diacetylmonoxime

� �

Carbamyl aspartate

Antipyrine

Yellowproduct

of undeterminedstructure

Step 1: Aspartate � Carbamyl phosphateATCase

Carbamyl aspartate � Pi

Step 2:

Figure 9-3 Colorimetric assay of aspartate transcarbamylase.


Sonicator


0.125 M Na-L-aspartate

0.036 M carbamyl phosphate, freshly prepared (keepcold)

1.0 mM carbamyl aspartate standard solution

Antipyrine-H2SO4 reagent

Diacetylmonoxime (butanedione monoxime) reagent (re-frigerate in foil covered bottle)

2% Perchloric acid

40 mM ATP (keep cold)

10 mM CTP (keep cold)

5 mM p-Mercuribenzoate

30 and 45°C water baths

Centrifuge

0.l M Tris-hydrochloride buffer, pH 9.2

Protocol

Growth of E. coli EK1104/pEK2

(1-liter procedure)

(This portion of the experiment is usually per-formed in advance by an assistant.)

1. Inoculate 10 ml of sterile growth medium con-taining 36 �g/ml uracil and 100 �g/ml ampi-cillin (see Appendix) with strain EK1104 car-rying the pEK2 plasmid, and grow for 16 to 18hr (typically overnight) in a shaking incubatorat 200 rpm and 37°C. The culture should beturbid with bacterial cell growth at this time(absorbance at 566 nm of about 1 or above).

2. Prepare 1 liter of sterile growth medium con-taining 12 �g/ml uracil and 100 �g/ml ampi-cillin as described in the Appendix. Cool to37°C.

3. Using sterile technique, add 10 ml of the in-oculum from step 1 to the 1 liter of culturemedium. Cultivate the bacteria with shaking(300 rpm) at 37°C for 20 to 24 hr, whereuponthe absorbance at 566 nm should be about 0.9.

Harvesting of Cells and Preparation of the

Cell Extract

(This portion of the experiment is usually per-formed in advance by an assistant.)

1. Weigh two large centrifuge bottles and recordtheir weight.

2. Pour the medium from the cell culture in step3 of the growth procedure into the large bot-tles and centrifuge the cell-containing mediumat 5000 � g for 10 min. Following centrifuga-tion, decant the supernatant and discard.

3. The cell yield of the pellet, after allowing excessliquid to drain off, is determined by weighingthe pellet-containing bottles and subtracting theweight of the empty bottles obtained in step 1.Record the cell yield, which should be about 5to 6 g of wet cell paste per liter of culture.

4. Resuspend the cell pellets in ice-cold 0.1 M Trischloride buffer, pH 9.2, using 4 ml of bufferper gram of cells. Transfer the resuspendedcells to a small metal sonication container andpack in an ice-water bath. (The cells may bestored frozen at �20 or �70°C at this pointfor later use, if desired.)

5. To disrupt the cell walls, sonicate the coldbreakage buffer-cell suspension for 30-sec intervals for a total of 20 to 30 min (30 sec sonication/ 30 sec off ). Do not operate the soni-cator without immersing the tip.

6. Pour the sonicated mixture into the medium-sized centrifuge tubes and centrifuge the mix-ture for 30 min at 10,000 � g and discard thepellet. The cell extract may be stored frozen at�20°C in small aliquots (0.5 ml) for subsequentuse by the class.

Before it is used by the class, the extract shouldbe tested for overproduction of ATCase by assay;see “General Procedure for the Assay of AspartateTranscarbamylase Activity,” step 9 “Range Finding”below. The extract can also be examined by SDS-PAGE for overproduction of the ATCase catalyticand regulatory subunits (34 kDa and 19 kDa, re-spectively). To do this, use the procedure describedin Experiment 4 for 15% gels.

General Procedure for the Assay of

Aspartate Transcarbamylase Activity

The standard assay procedure for ATCase is as fol-lows. In the subsequent procedures, modificationsand additions will be made as indicated. Read theentire experiment before planning your procedure.To obtain good results in the following experi-ment, it is essential to pipette enzyme and substratesolutions accurately and reproducibly and to mea-sure absorbance values carefully. When you are


comparing two sets of experimental conditions,you should assay the sets at the same time with thesame enzyme dilution.

1. Add the following to each of a series of assaytubes: 0.5 ml 0.08 M sodium phosphate buffer,pH 7.0, varying amounts of 0.125 M aspartate(0.2 ml when range finding), 0.10 ml of an ap-propriate enzyme dilution (kept cold), and wa-ter to a final total volume of 0.90 ml.

2. Prepare a blank tube from which enzyme isomitted.

3. Incubate the tubes at 30°C for about 5 min tobring the contents to 30°C. At timed intervals(30 sec is convenient) add 100 �l of 0.036 Mcarbamyl phosphate to initiate the reaction andmix.

4. Allow the reaction to proceed at 30°C for30 min. Then terminate each reaction at thesame timed intervals as used in initiating thereactions by adding 1 ml 2% perchloric acid.

5. Set the tubes on ice until all are ready for thecolor test. (If necessary, remove any precipitateby centrifuging the cold tubes for 5 min in theclinical centrifuge, followed by decanting.) De-velop the color in these tubes at the same timeas the series of carbamyl aspartate standards de-scribed in the next step.

6. Prepare a series of carbamyl aspartate standardscontaining 0 (blank), 25, 50, 100, 150, 200, 250,and 375 �l of 1.0 mM carbamyl aspartate andwater to a final volume of 2 ml.

7. Immediately before use, prepare the colorreagent by mixing well two parts of antipyrine-H2SO4 reagent with one part of diacetyl-monoxime reagent.

Warning: Sulfuric acid can burn clothing and

skin. Wear protective gloves.

Prepare the amount of color reagent you need—3mlto be added to each tube containing 2 ml of carbamylaspartate standard or enzyme assay mixture afteraddition of perchloric acid. Add 3 ml to each assayand standard tube and mix thoroughly. Cap thetubes with marbles or parafilm, cover them withaluminum foil, and store in a dark place at roomtemperature until the next lab period (15–48 hr issatisfactory).

8. After the incubation in the dark, place the tubesin a 60°C water bath exposed to room light,but not direct sunlight, for 70 to 75 min. Coolthe tubes in cold water, and read the absorbanceof each at 466 nm versus the blank containingall reagents except carbamyl aspartate. Read thetubes promptly, because the color will slowlyfade.

9. Range finding. (If time is limited, this portionof the experiment should be performed in ad-vance by an assistant.) In order to conduct thesubsequent experiments under conditions thatwill yield valid assays, it is necessary to deter-mine the amount of enzyme that will give a re-sponse proportional to the enzyme activity andwithin the range of the carbamyl aspartate colortest. This corresponds to an amount of enzymethat forms about 0.05 to 0.10 �mol of carbamylaspartate when 100 �l of a suitable dilution isassayed under standard assay conditions. Be-cause the activity of your extract will dependon a number of variables, you must assay 25-,50-, 75-, and 100-�l samples of a series of di-lutions of the extract in 0.08 M sodium phos-phate buffer, pH 7.0, to determine the appro-priate enzyme level for use in the subsequentexperiments. Suggested dilutions are: 1 to5,000, 1 to 8,000, and 1 to 10,000.

Dependence of ATCase Activity on

Aspartate Concentration in the Absence

and Presence of Allosteric Effectors

1. Prepare a series of tubes that contain 0.5 ml0.08 M sodium phosphate buffer in each and10, 20, 40, 60, 100, 150, and 200 �l of 0.125M aspartate. To each add 0.1 ml of an enzymedilution shown to yield about 0.1 �mol of car-bamyl aspartate under standard assay condi-tions above and shown to respond linearly toenzyme concentration (i.e., 50 �l gives half asmuch product as 100 �l). Add water sufficientto bring each to a volume of 0.9 ml. This isSeries A. Use Table 9-1 to assist in preparationof the tubes.

2. Prepare two other series of tubes identical tothe first series except that one series (Series B)also contains 50 �l 10 mM CTP, and the otherseries (Series C) contains 50 �l 40 mM ATP.Tables 9-2 and 9-3 describe these series.


Table 9-1 Series A: Activity of ATCase as a Function of Aspartate Concentration

(All Volumes in Microliters)

Component Tube Number

0 1 2 3 4 5 6 7

0.125 M Asparate — 010 020 040 060 100 150 200

Water 300 290 280 260 240 200 150 100

0.08 M Na-Pi buffer, pH 7 500 500 500 500 500 500 500 500

Enzyme 100 100 100 100 100 100 100 100

Table 9-2 Series B: Activity of ATCase as a Function of Aspartate Concentration, with CTP Added

(All Volumes in �l)


0 1 2 3 4 5 6 7

0.125 M Asparate 100 010 020 040 060 100 150 200

Water 150 230 240 210 190 150 100 050

0.08 M Na-Pi buffer, pH 7 500 500 500 500 500 500 500 500

Enzyme — 100 100 100 100 100 100 100

10 mM CTP 050 050 050 050 050 050 050 050

Table 9-3 Series C: Activity of ATCase as a Function of Asparate Concentration, with ATP Added (All

Volumes in �l)


0 1 2 3 4 5 6 7

0.125 M Aspartate 100 010 020 040 060 100 150 200

Water 150 240 230 210 190 150 100 050

0.08 M Na-Pi buffer, pH 7 500 500 500 500 500 500 500 500

Enzyme — 100 100 100 100 100 100 100

40 mM ATP 050 050 050 050 050 050 050 050

3. Conduct the assays for Series A, B, and C, ini-tiating each reaction with 100 �l 0.036 M car-bamyl phosphate, terminating after 30 min ofreaction with 1 ml of 2% perchloric acid, anddeveloping the color by adding 3.0 ml of freshlyprepared color reagent as described above. In-clude a series of carbamyl aspartate standardsalso as described above.

4. Using your carbamyl aspartate standard curve,convert your A466 readings to micromoles ofcarbamyl aspartate formed per hour per milli-

liter of diluted enzyme. Plot the rate of car-bamyl aspartate formation as a function of as-partate concentration. Do the results obeyMichaelis–Menten kinetics? What is the effectof adding CTP? Of adding ATP?

5. Prepare a Lineweaver–Burk plot of the data.Can you determine an accurate KM value foraspartate? Prepare Hill plots of the data andcalculate Hill coefficients (nH) for each exper-imental series. (The Hill plot is a plot of log(v/Vmax) versus log aspartate concentration,


Table 9-4 Series D: Activity of ATCase as a Function of Aspartate Concentration, after Treatment

with Mercuribenzoate (All Volumes in �l)


0 1 2 3 4 5 6 7

0.125 M Asparate 100 010 020 040 060 100 150 200

Water 290 280 270 250 230 190 140 090

0.08 M Na-Pi buffer, pH 7 500 500 500 500 500 500 500 500

Enzyme — 100 100 100 100 100 100 100

5 mM Mercuribenzoate 010 010 010 010 010 010 010 010

where v is the reaction velocity at any aspartateconcentration and Vmax is the maximal velocityat saturating aspartate concentration (obtainedby extrapolation on the Lineweaver–Burkplot). The slope of this plot is the Hill coeffi-cient, nH, and is a measure of the “sigmoidic-ity” of a velocity versus substrate concentrationcurve.)

Desensitization of the Allosteric Site of

ATCase with Mercuribenzoate

1. Prepare two series of tubes similar to those de-scribed above. Series D should contain no nu-cleotides, and Series E should contain 50 �l of10 mM CTP in each tube. Each tube shouldalso contain 10 �l of 5 mM p-mercuribenzoate.You do not need to prepare a series containingATP. The amount of enzyme added to eachtube should be the same as that used in the pre-vious series. Follow Tables 9-4 and 9-5.

2. Incubate the tubes at 30°C for 15 min. Theninitiate the reactions with 100 �l of 0.036 Mcarbamyl phosphate. After 30 min of reaction,terminate each reaction (in the order they wereinitiated) with 1 ml 2% perchloric acid, and de-velop the color in the assay tubes (along witha set of carbamyl asparate standards) by adding3.0 ml of freshly prepared color reagent as de-scribed above.

3. Analyze the data as in steps 4 and 5 above.What is the effect of 50 �M mercuribenzoateon the activity of ATCase and on its sensitivityto allosteric inhibition by CTP?

Exercises

1. Describe the current view of the molecular ar-chitecture of E. coli aspartate transcarbamylase.What techniques were used in arriving at thisview?

Table 9-5 Series E: Activity of ATCase as a Function of Asparate Concentration, with CTP Added

after Treatment with Mercuribenzoate (All Volumes in �l)


0 1 2 3 4 5 6 7

0.125 M Aspartate 100 010 020 040 060 100 150 200

Water 240 230 220 200 180 140 090 040

0.08 M Na-Pi buffer, pH 7 500 500 500 500 500 500 500 500

Enzyme — 100 100 100 100 100 100 100

10 mM CTP 050 050 050 050 050 050 050 050

5 mM Mercuribenzoate 010 010 010 010 010 010 010 010


2. In this experiment a sigmoid dependence of re-action velocity on substrate concentration wasinterpreted as evidence for cooperativity ofbinding of the substrate to the enzyme. Whatkinds of experiments would be necessary toconfirm this proposal? Describe other possiblesituations, especially in a crude enzyme prepa-ration, that would result in such a sigmoidcurve; that is, what kinds of artifacts might leadyou to conclude falsely that a sigmoid curve re-sults from cooperativity in substrate binding?

3. How is aspartate transcarbamylase regulated inother bacteria? In fungi and yeasts? In mam-malian systems?

4. List three other examples of enzymes that aremade up of nonidentical subunits. List threeenzymes that contain specific allosteric sites forregulatory molecules but that are made of onetype of subunit only.

5. Is it possible to observe allosteric interactionsin an enzyme made up of only one subunit? Is

it possible to observe cooperativity in substratebinding in an enzyme made up of only one sub-unit? If so, under what circumstances; if not,why not?

REFERENCES

Monod, J., Changeux, J.-P., and Jacob, F. (1963). Al-losteric Proteins and Cellular Control Systems. JMol Biol 6:306.

Nowlan, S. F., and Kantrowitz, E. R. (1985). Superpro-duction and Rapid Purification of Escherichia coliAspartate Transcarbamylase and Its Catalytic Sub-unit under Extreme Derepression of the PyrimidinePathway. J Biol Chem 260:14712.

Prescott, L. M., and Jones, M. E. (1969). ModifiedMethods for the Determination of Carbamyl As-partate. Anal Biochem 32:408.

Stadtman, E. R. (1966). Allosteric Regulation of En-zymic Activity. Adv Enzymol 28:41.

P z

157

EXPERIMENT 10

Affinity Purification of Glutathione-S-Transferase

Theory

As discussed in the introduction to Experiment 2,affinity chromatography is one of the most rapidlygrowing, new, and effective techniques in modernprotein purification. A single protein or family ofproteins that bind a particular ligand can often bepurified in a single step, provided that the ligand ofinterest can be coupled to some type of solid sup-port or matrix. As with other types of chromatog-raphy, affinity chromatography is carried out in athree-step process: a binding step that allows theprotein of interest to interact with the immobilizedligand, a wash step to remove unbound proteins,and an elution step to recover the purified protein.

In this experiment, you will purify the C-terminal domain of the enzyme glutathione-S-transferase (GST) from an E. coli cell extract byexploiting the affinity that the enzyme has for its natural substrate, glutathione. Although thechromatography experiment used in the purifica-tion will not be performed in the column format(see Experiment 2), the “batch wash” format usedhere could easily be adapted for use in the col-umn format.

Glutathione is a tripeptide that is involved in a number of different aspects of metabolism (Fig.10-1). Found in nearly all cell types, the moleculecan exist in either the oxidized or the reduced form.In the reduced form, it acts as a general reducingagent, protecting the cell from harmful peroxides,radicals produced from exposure to ionizing radia-tion, and maintaining the general sulfhydryl groupstatus of other proteins in the cell. Some reductaseenzymes utilize the sulfhydryl group on glutathione

as an electron donor. Glutathione-S-transferase isan enzyme that catalyzes condensation reactions be-tween glutathione and a number of other molecules.This reaction is often a necessary step in the detox-ification of foreign organic compounds (xenobi-otics) and toxic compounds produced by the cy-tochrome P-450 oxidase enzymes. Condensation ofsuch compounds with glutathione converts theminto harmless, water-soluble glutathione conjugatesthat can be filtered out of the blood in the kidneys.

The E. coli strain used in this experiment con-tains the pGEX-2T plasmid (Fig. 10-2). This plas-mid contains the bla gene, which allows the cell toproduce �-lactamase and grow in media containingampicillin (a �-lactam antibiotic). It also containsthe coding sequence for the C-terminal domain ofGST from Schistosoma japonicum. This GST gene isunder the control of the tac promoter, a modifiedform of the lac promoter that produces a high levelof transcription of genes under its control when notbound by LacI (the lac repressor). Cells harboringthis plasmid will not express GST until the inducer(isopropylthio-�,D-galactopyranoside, IPTG) isadded, causing the lac repressor (LacI) to lose itsaffinity for the promoter sequence.

Following induction with IPTG and an incu-bation period to allow for the expression of GST,the cells are harvested and lysed by sonication inphosphate buffer containing 150 mM NaCl. TritonX-100 (a non-ionic or nondenaturing detergent) isadded to help solubilize the membranes and releaseproteins that may be peripherally associated withthem. The clarified cell extract is then incubatedwith the glutathione Sepharose 4B resin to allowGST to bind the immobilized ligand. The Triton

158

(10-1)

The dissociation constant (Ks) � k�1/k1, while theequilibrium constant (Keq) � [ES]/[E][S]. k1 isequal to the rate of formation of the ES complex,while k�1 is equal to the rate of dissociation of theES complex.

If a large concentration of free (non-immobi-lized) ligand is introduced, the enzyme will estab-lish a new equilibrium with the immobilized ligand,releasing the protein from the resin into the buffer.The free glutathione in solution with GST follow-ing elution from the resin can be removed eitherby dialysis or by gel filtration chromatography. Asstated in Experiment 2 (affinity chromatography),it may also be possible to elute GST from the solidsupport by altering the pH and/or the ionicstrength of the elution buffer. Keep in mind, how-ever, that the most specificmethod for eluting a pro-tein of interest from an affinity-based chromato-graphic support is with the application of excess freeligand. If other proteins besides GST are bound tothe glutathione resin after the numerous wash steps

E � S ESk1

k�1

X-100 added earlier may help prevent nonspecifichydrophobic interactions between other proteins inthe extract and the solid support matrix (or the GSTbound to it), which could decrease the effectivenessof the purification. In the same fashion, the NaClpresent in the binding buffer may prevent the for-mation of nonspecific electrostatic interactions be-tween other proteins in the extract and the immo-bilized glutathione (remember that glutathione willcarry a negative charge at pH 7.3 and may displaysome general cation exchange effects with otherproteins in the extract). The glutathione resin isthen washed several times in the binding buffer toremove loosely associated proteins.

To recover the purified GST from the resin, abuffer containing a high concentration of the lig-and (free ligand) is introduced. This free ligand willcompete with the immobilized ligand for the GST,eventually drawing the protein from the resin intothe buffer. Remember that the enzyme–substrateinteraction is a dynamic one. The enzyme does notbind the substrate irreversibly. Rather, the enzymeis in a rapid equilibrium between the free form andthe ligand bound form, as determined by the dis-sociation constant (Equation 10-1).

SECTION II Proteins and Enzymology

COO�

SH

H3N CH CH2 CH2 CH2

CH2

C NH CH

COO� O

C NH

O�

COO�

S

S

H3N CH CH2 CH2 CH2

CH2

CH2

C NH CH

COO� O

C NH

O�

Reduced form (GSH)

Oxidized form (GSSG)

COO�H3N CH CH2 CH2 CH2C NH CH

COO� O

C NH

O�

Figure 10-1 The structure and forms of glutathione (�-glutamyl cysteinyl glycine).

this vector in the same reading frame as the GSTgene, a fusion protein will be expressed as a hybridincluding the GST domain and the protein encodedby the gene of interest. This protein “tagged” withthe GST domain may then be purified using thesame procedure described in this experiment. In

through some sort of nonspecific electrostatic or hy-drophobic interactions, the addition of free glu-tathione will only cause the elution of the GST.

Notice that the pGEX-2T plasmid contains amultiple cloning site immediately 3� to the GSTcoding sequence. If a gene of interest is cloned into

EXPERIMENT 10 Affinity Purification of Glutathione-S-Transferase 159

P tac

Glutath

ione-S-transferase

lacI

ori

bla

pGEX - 2T

~5000 bp

CTG GTT CCG CGT GGA TCC CCG GGA ATT CATGAC CAA GGC GCA CCT AGG GGC CCT TAA GTA

Bam HISma I

Eco RI

Thrombin recognition sequence encoded by these codons

Figure 10-2 The pGEX-2T plasmid. Multiple-cloning site allows genes of interest to be inserted in frame

with the GST coding sequence to express the gene as a GST fusion protein that is easily pu-

rified with affinity chromatography. The GST tag can be removed after purification by treat-

ment with thrombin.


Table 10–1 Some Commonly Used Affinity

“Tags” to Aid in Protein

Purification

“Tag” Affinity Ligand

Glutathione-S-transferase Glutathione

Maltose binding protein Amylose

Polyhistidine Ni2�

“Flag” epitope “Flag” antibody

(antibody recognizes

the peptide sequence

DYKDDDDK)

theory, a protein of interest could be “tagged” withany protein domain or peptide that has affinity fora particular ligand for use in affinity chromatogra-phy. Some examples of commercially available fu-sion protein expression vectors are shown in Table10-1. In many cases, these vectors also contain aprotease recognition sequence between the affinitytag coding sequence and the multiple-cloning site.This cleavage site (thrombin, factor Xa) will allowyou to remove the affinity tag domain from the pu-rified protein once it has been isolated. A graphi-cal description outlining the technique of affinitychromatography used in this experiment is pre-sented in Figure 10-3.


Cell extract from induced E. coli strain (TG1 or others;see Appendix) carrying pGEX-2T

Isopropylthio-�,D-galactoside (IPTG)

Triton X-100

YT broth (1% wt/vol bactotryptone, 0.5% (wt/vol) yeastextract, 0.5% (wt/vol) NaCl)

Glutathione sepharose 4B (Pharmacia Biotech, Inc.)

PBS buffer (150 mM NaCl, 16 mM Na2HPO4, 4 mMNaH2PO4, pH 7.3)

50 mM Tris, 10 mM reduced glutathione (pH 8.0)

4� SDS sample buffer (0.25 M Tris, pH 6.8, 8% SDS,20% wt/wt glycerol, 10% �-mercaptoethanol, 0.1% bromophenol blue)

Purified glutathione-S-transferase (prepared by instruc-tor)



Protocol

We suggest that steps 1 through 6 be performedby the instructor prior to the beginning of the ex-periment. We perform this experiment in con-junction with Experiment 18 (“Western Blot toIdentify an Antigen”). If this option is chosen, theprotocol described above can be followed exactly.If the Western blot is not to be performed, theSDS-PAGE experiment described in Experiment18 can be performed on a single set of samples (seestep 16).

1. Prepare a 50-ml overnight culture of E. coli(strain TG1) containing pGEX-2T in YTbroth with 100 g/ml ampicillin. Incubate withshaking at 37°C overnight (�20 hr).

2. The next morning, inoculate 2 liters of freshYT broth containing 100 g/ml ampicillin withthe 50 ml overnight culture. Incubate at 37°Cwith shaking for 1 hr.

3. Add IPTG to a final concentration of 0.5 mMand incubate at 37°C with shaking for 4 hr.

4. Harvest cells by centrifugation at 5000 � g for10 min.

5. Resuspend the cell pellet in 20 ml of PBS bufferand sonicate to lyse the cells (four separate 30-sec bursts, with 30-sec incubations on ice inbetween each burst). Add Triton X-100 to 1%final concentration and incubate on ice for30 min.

6. Centrifuge the lysate at 13,000 � g for 30 minand collect the supernatant (containing GSTand other cytoplasmic proteins). This clarifiedcell extract will be used in the GST purifica-tion experiment.

7. Obtain 0.1 ml of a 50% slurry of Glutathionesepharose 4B resin in PBS buffer. Centrifugethe slurry for 30 sec in a 1.5 ml plastic micro-centrifuge tube and discard the supernatant.

8. Obtain 0.5 ml of the E. coli cell extract fromthe instructor. Save 15 l of the crude cell extractfor SDS-PAGE analysis. Add the remainder ofthe cell extract to the resin. Flick the bottomof the tube gently to resuspend the resin in theextract.

9. Incubate for 20 min at room temperature.Make sure that the resin remains fully sus-pended in the cell extract by inverting the tubefrequently during the incubation.

EXPERIMENT 10 Affinity Purification of Glutathione-S-Transferase 161

= Glutathione Sepharose 4B resin

= Glutathione-S-transferase

= Reduced glutathione

= Other E. coli proteins

Collect resin bycentrifugation

Remove unboundfraction

Removesupernatantand add reducedglutathione to theresin; collect theresin by centrifugation

Collect the supernatantcontainingfree glutathione andglutathione-S-transferase

Wash repeatedlyto removeunbound proteinand collect resinby centrifugation

Mix resin withcell extract

Figure 10-3 Purification of glutathione-S-transferase with glutathione Sepharose 4B resin.

10. Centrifuge the tube for 30 sec and collect thesupernatant. Label the supernatant, “unboundfraction.” Store this fraction on ice.

11. Add 0.5 ml of PBS buffer to the resin and in-vert the tube several times to completely re-suspend the resin in the buffer. Centrifuge for30 sec to collect the resin; discard the super-natant.

12. Repeat step 11 four more times.13. Add 0.1 ml of 50 mM Tris/10 mM reduced glu-

tathione (pH 8.0) to the resin and flick the bot-

tom of the tube gently several times until theresin is fully resuspended.

14. Incubate the tube for 10 min at room temper-ature, inverting the tube frequently to keep theresin suspended in the buffer.

15. Centrifuge the tube for 30 sec to collect theresin. Save the supernatant and label it, “elutedfraction.” Store this fraction on ice.

16. Prepare duplicate samples for SDS-PAGE andWestern blot analysis (prepare in four tubesmarked A to D):


Sample A: 4 l of crude cell lysate � 11 l of water � 5 l of 4� SDS sample buffer

Sample B: 4 l of “unbound fraction” � 11 l of wa-ter � 5 l of 4� SDS sample buffer

Sample C: 15 l of “eluted fraction” � 5 l of 4�SDS sample buffer

Sample D: 5 l of purified GST (prepared by theinstructor) � 10 l of water � 5 l of 4� SDSsample buffer

Proceed with SDS-PAGE (Experiment 18,Day 1).

Exercises

1. What characteristics of glutathione-S-trans-ferase make it an attractive affinity “tag” for usein bacterial expression systems?

2. Is there another method that could be used toelute GST from the glutathione sepharose 4Bresin besides the addition of reduced glu-tathione? Are there any advantages to elutingthe enzyme with the addition of free substrateas opposed to these other methods?

3. Describe how the protocol used in this exercisewould change if the experiment were per-formed in the column format rather than in the“batch wash” format.

4. How does the substrate affinity chromatogra-phy used in this experiment differ from theother types of affinity chromatography dis-cussed in Experiment 2?

5. Why is it that fusion protein expression vectorscontain thrombin and factor Xa recognition se-quences rather than trypsin or chymotrypsinrecognition sequences? Why would it some-times be desirable to remove the affinity tagfrom the fusion protein?

6. You have isolated a protein of interest taggedwith the GST domain. At this point, the fusionprotein is bound to the glutathione Sepharose4B resin and the unbound proteins have beenwashed away. Describe, in detail, two differenttechniques that you could use to isolate the pro-tein without the affinity tag using only throm-bin, glutathione Sepharose 4B, and reducedglutathione.

REFERENCES

Garrett, R. H., and Grisham, C. M. (1995). Recombi-nant DNA: Cloning and Creation of ChimericGenes. In: Biochemistry. Orlando, FL: Saunders.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Biosynthesis of Amino Acids, Nucleotides,and Related Molecules. In: Principles of Biochemistry,2nd ed. New York: Worth.

Mathews, C. K., and van Holde, K. E. (1990). Metabo-lism of Sulfur-Containing Amino Acids. In: Bio-chemistry. Redwood City, CA: Benjamin/Cummings.

Scopes, R. K. (1994). Purification of Special Types ofProteins. In: Protein Purification: Principles and Prac-tice, 3rd ed. New York: Springer-Verlag.

Smith, D. B., and Johnson, K. S. (1988). Single-StepPurification of Polypeptides Expressed in Es-cherichia coli as Fusion Proteins with Glutathione-S-Transferase. Gene 67:31.

Stryer, L. (1995). Pentose Phosphate Pathway and Glu-coneogenesis: Biosynthesis of Amino Acids andHeme. In: Biochemistry, 4th ed. New York: Free-man.

Uhlen, M., and Moks, T. (1990). Gene Fusions forPurpose of Expression: An Introduction. MethodsEnzymol 185:129.

v

165

Carbohydrates

Carbohydrates are the most abundant of all organiccompounds in the biosphere. Many members of thecarbohydrate class have the empirical formulaCx(H2O)y, and are literally “hydrates of carbon.”The fundamental units of the carbohydrate class,the monosaccharides, are polyhydroxy aldehydes orketones and certain of their derivatives. As withother classes of biologically important compounds,much of the function of the carbohydrates derivesfrom the ability of the monosaccharides to com-bine, with loss of water, to form polymers: oligosac-charides and polysaccharides. The chemistry of car-bohydrates is, at its core, the chemistry of carbonyland hydroxyl functional groups, but these func-tional groups, when found in the same compound,sometimes exhibit atypical properties. The discus-sion that follows is designed to review the aspectsof carbohydrate chemistry that are especially im-portant for isolation, analysis, and structure deter-mination of biologically important carbohydrates.

Structure and Stereochemistry of

Monosaccharides

Drawing Monosaccharide Structures: The Fischer

Projection. From the stereochemical point ofview, monosaccharides are considered to be derivedfrom the two trioses, D- and L-glyceraldehyde (seeFig. III-1). These two parent compounds differ onlyin the steric arrangement of the atoms about thecentral asymmetric carbon; they are mirror images(enantiomorphs, optical antipodes) of one another.

The two compounds differ physically only in theorientation of their crystals and the direction inwhich their solutions rotate polarized light. Thesemay seem like trivial differences, but when such op-tical isomers interact with other asymmetric mole-cules (especially enzymes) they may acquire very dif-ferent chemical reactivities. This principle of opticalselectivity is central to the biochemistry of carbo-hydrates (and other classes of naturally occurringcompounds as well). The manner of drawing the twoisomers as shown in the top part of Figure III-I iscalled the “Fischer projection.” In this projection,the most oxidized (aldehyde) end of the molecule isdrawn up and is considered to be behind the planeof the paper. The –CH2OH end is drawn down andis also behind the plane. Because the groups are

SECTION III

Biomolecules and Biological Systems

Introduction

CH2OH

CHO

CHO

H

H

C

OH HO

OH

D-Glyceraldehyde

CH2OH

CHO

H

L-Glyceraldehyde

CH2OH

CH2OH

CHO

HCHO

Figure III-1 The two forms of glyceraldehyde.

The asymmetric carbon is shown in

boldface.

166 SECTION III Biomolecules and Biological Systems

arranged about the central carbon atom (which is inthe plane of the paper) as at the corners of a tetra-hedron, the H and OH group must project in frontof the plane of the paper. The drawings at the bot-tom should make the arrangement of atoms clear.Although the arrangements of atoms in the D and Lforms of glyceraldehyde were originally arbitrarilyassigned by Emil Fischer, we now know from x-raycrystallographic analysis that the actual arrangementof atoms in these structures is the same as was orig-inally assigned. You must always remember thethree-dimensional meaning of the two-dimensionalFischer projection, especially when consideringstructures drawn in other arrangements.

If you have access to ball-and-stick models usedfor instruction in organic chemistry, use them toconstruct models corresponding to the drawings inFigure III-1 and determine the consequences ofdrawing flat projections of the structure in arrange-ments other than the Fischer convention. In thisway, you will see why it is essential to use a gener-ally accepted convention for drawing asymmetricstructures in two dimensions.

When the carbohydrate molecule contains ad-ditional CHOH groups, the number of possible iso-mers increases (it will be equal to 2n, where n is thenumber of asymmetric carbon atoms). By conven-tion, all such sugars are considered to be membersof two families, the D family and the L family. Mem-bers of the D family have the same arrangement atthe next to last carbon atom (the penultimate car-bon, the first asymmetric center from the bottom)as does D-glyceraldehyde. Likewise, the L sugarshave the configuration of L-glyceraldehyde aboutthe penultimate carbon. As before, the most oxi-dized end of the sugar (–CHO for aldo sugars and

for keto sugars) is always drawn at the top.

The name of the sugar is determined by thearrangement of all the asymmetric centers; the fam-ily (D or L) is determined by the penultimate car-bon only. (Note that D and L have no other mean-ing; they do not describe the direction of rotationof polarized light of solutions of the sugar.) D andL isomers of the same sugar (enantiomorphs or enan-tiomers) are mirror images. Isomers that are not mir-ror images, but differ in configuration about one ormore carbon atom are called diastereoisomers. Twodisastereoisomers differing in configuration aboutonly one carbon atom are called epimers. Studentsshould test their understanding of these conceptsby stating how various pairs of the sugars in FigureIII-2 are related.

A further point should be made about the three-dimensional meaning of the Fischer projection ofsugars with more than three carbon atoms. The car-bon atoms are arranged in a line (oxidized end up,as before) such that all H and OH groups projectto the front and the C–C bonds curve back. Youview the groups as they would project onto a planecurved toward you. The carbon chain is curved con-tinuously in a single direction rather than staggered.This concept is illustrated in Figure III-3.

Monosaccharides Form Rings as a Consequence

of Internal Hemiacetal or Hemiketal Formation.

There is yet another level of isomerism of sugars,because aldehydes (and ketones) form freely re-versible adducts with alcohols (hemiacetals andhemiketals) in aqueous solutions. In the presence of

CH2OH

C O

CH2OH

H C

CHO

OH

H C OH

HO C H

H C OH

CH2OH

C

CHO

H C OH

HO C H

HO H

H C OH

CH2OH

C

CHO

CHO

C

H

HO H

HO H

H

C

OH

CH2OH

C

CHO

C

HO C H

HO H

H

C

OH

H OH

D-Glucose L-Glucose D-Mannose L-Idose

Figure III-2 Which pairs of sugars are enantiomers? Diastereoisomers? Epimers? The stereochemistry of

the penultimate carbon (boldface) determines whether the sugar belongs to the D family or

the L family.

SECTION III Introduction 167

CH2OH CH2OH

H C

CHO

OH

H C OH

HO C H

H C

O

C

HCC

C

C

OH

Front view Side view

OHOH

OH

HOHH

HH

Figure III-3 Physical meaning of the Fischer pro-

jection of a hexose.

an acid catalyst, the hemiacetal can react with an-other alcohol molecule to form an acetal (or ketal)(see Fig. III-4a). Acetal formation is favored underdehydrating conditions (strong acid, anhydrousconditions); acetal hydrolysis is favored in diluteaqueous acid. Most sugars exist primarily as inter-nal hemiacetals, in which intramolecular addition ofa hydroxyl occurs to form a five-member (furanose)or six-member (pyranose) ring (see Fig. III-4b).

Note that formation of the hemiacetal ring formof a sugar creates a new optically active center. Sug-

ars that differ in symmetry at this center are calledanomers, and the aldehyde (or ketone) carbon in-volved is often called the “anomeric carbon.”Anomers are named as derivatives of the parentaldehyde (or keto) sugar and are designated �- and�-. The best definition of these two designationsrelates them to the Fischer projection of the par-ent sugar. An �-form is the one in which the –OHon the anomeric carbon is on the same side of thecarbon backbone in the Fischer projection as the–OH on the penultimate carbon. The examples inFigure III-5 illustrate this principle.

Drawing Monosaccharide Structures: The Haworth

Projection. In an attempt to represent sugar struc-tures in a manner more closely resembling the ac-tual geometric arrangement of the atoms, W.N. Ha-worth proposed the system of perspective drawingsgenerally used by biochemists. The Haworth for-mula is a natural consequence of turning the Fis-cher projection on its side and drawing the hemi-acetal ring oxygen with more realistic bonddimensions as part of a ring. This idea should bemade clear by the following step-by-step procedurefor converting Fischer projections illustrated here with �-D-glucopyranose and �-D-fructofuranose.

H

O

R OR�

H

C

OH

R OR�

H

C

OR�reversibleequilibrium

HO R�

HO R�

HO R�CR �

AcetalHemiacetalH2O

�H2O

acidhydrolysis

a

CH2OH

H C OH

H C OH

HO C H

H

H C

C

OH

CH2OH

C

CHO

C

H

HO H

H

C

OH

CH O

H OH

O

O

CH2OH

C

CHO

C

H

H

C

OH

H OH

CH

H OH

O�

b

Figure III-4 (a) Hemiacetal and acetal formation and acetyl hydrolysis. (b) Internal hemiacetal formation

with the OH on carbon 5 of hexoses yields two six-membered rings.


CH2OH

C

CHO

C

H

HO H

H

C

OH

CH

H OHO

CH2OH

C

C

HO C H

HO H

HO H

H

C

OH

C H

O

CH2OH

C

CHO

C

H

H

C

OH

H OH

CH

H OHO

-D-Glucopyranose� -D-Glucopyranose� -L-Glucopyranose�

Figure III-5 �- and �-anomers of D- and L-glucose in Fischer projection. The anomeric carbons are shown

in boldface.

1. Place the carbonyl carbon on the right anddraw a potential five- or six-member ring clock-wise (Fig. III-6). This rotation places hydrox-yls that lie to the right on the Fischer projec-tion down in Haworth representation; hydroxylsto the left will project up. (This is like placingthe Fischer projection on its side.)

2. Draw the hemiacetal or hemiketal form of thesugar by making ring closure with the appro-priate hydroxyl added to the carbonyl to form

a furanose (five-member) or pyranose (six-member) ring with the ring O in the upperright-hand corner of the pyranose or at the topof a furanose (Fig. III-7). Ring closure will re-quire a rotation of the carbon that bears the ringO so that remaining carbons of the “tail” aredrawn opposite to the original direction (step1) of the OH group involved in ring closure.

3. Establish the orientation of the asymmetriccarbon created by hemiacetal or hemiketal for-mation. By definition, an �-anomeric OH is onthe same side of the Fischer projection as theO

CH2OH

CH2OH

H C OH

H C

O

C

C

C

OHOH

OHOH

HOHO

C H

HH

H

H

H

HH

C OH

C

D-Glucose

C C

CH2OH

CH2OH

CH2OH

CH2OH

H C OH

H C O

CC

OH

OH

H OH

OH

HO C H

H

HC O

D-Fructose

C C

Figure III-6 Converting Fischer to Haworth

projection—step 1.

HO

H

H

H

H

HH

H OH

OH

OH

OH

CH2OH

CH2OH

O

O

(H, OH)

(OH, CH2OH)

“Tail” (CH2OH) drawn upbecause ring closure OHwould have been down

Orientationunassigned

as yet

Orientationunassigned

as yet

D-Glucopyranose(partial structure)

D-Fructofuranose(partial structure)

Figure III-7 Converting Fischer to Haworth

projection—step 2.


HO

H H

H

H

HH

H

H OH

OHOH

OH

OH

OH

CH2OHCH2OH

CH2OH

O O

-D-Fructofuranose�-D-Glucopyranose�

Figure III-8 Representation of � and � anomers

of D sugars in Haworth projection.

hydroxyl on the penultimate carbon and will bedown in D sugars (Fig. III-8). It follows that a�-anomeric hydroxyl will project up. It also fol-lows that the opposite will be true for L sugars.

4. In cases in which ring formation involves anOH on a carbon other than the penultimatecarbon (e.g., aldohexofuranoses), the remain-ing carbon atoms of the “tail” (e.g., )

are drawn in the Fischer projection convention(Fig. III-9).

Note that this last step becomes particularly con-fusing in cases in which the “tail” projects up,whereas the Fischer convention features the car-bons of the “tail” down (e.g., �-D-glucofuranose).Correct application of step 4 to these cases requiresthat you rotate the plane of the paper to place thetail down and indicate the Fischer convention OHassignments (Fig. III-10). However, there still is un-

CH2OH

CHOH

fortunate confusion over the correct way to drawsuch asymmetric centers. Consequently, it is a goodpractice to indicate in words the intended steroiso-meric configuration on Haworth projections of sug-ars with such centers. It will be valuable to draw theHaworth formulae of �-D-galactofuranose and �-L-galactofuranose to test your ability to apply theserules.

Any sugar in solution should be considered amixture of four ring forms and the straight-chain(aldehydo- or keto-) forms (see Fig. III-11). Themixture of forms existing in solution differs fromsugar to sugar. Glucose, for example, has beenshown to exist primarily as a mixture of �- and �-pyranose forms with only traces of the other formspresent. Pure samples of single forms can often beobtained by crystallization under specific condi-tions.

C

H

H

H

H

OH

HO

OHOH

CH2OH

O

H

H

CH

H

H

H

OH

OH

OH

OH

CH2OH

O

-D-Glucofuranose�

-D-Galactofuranose�

Figure III-9 Converting Fischer to Haworth pro-

jection—step 4.

C

H

H

H

H

OH

OHOH

CH2OH

O

-D-Glucofuranose�

C

H

H

H

H

H

OH

HO

OHOH

CH2OH

O

-D-Glucofuranose�

C

H

H

H

H

H

OH

HO

OHOH

CH2OH

O

-D-Glucofuranose �

rotate

(partial struture)

(partial structure)

rotate andcompletestructure

now makeOH assignment with

carbon 6 down

(H, OH?)

Figure III-10 Haworth projections of some sugar hemiacetals.


Three-Dimensional Representation of Monosac-

charide Rings. Even the Haworth projection issomewhat of an oversimplification. The most ra-tional approach to carbohydrate chemistry requiresus to consider the true conformation of pyranoserings. An examination of models using normal C–C bond angles makes it clear that six-memberedcarbon rings cannot assume a planar conformation.Two conformations with no bond strain are possi-ble (Fig. III-12). The chair formation is favored overthe boat conformation by about 1000 to 1 for pyra-nose rings, so conformational analysis of these car-bohydrates is essentially limited to consideration oftwo different chair conformations (see Fig. III-13).To be certain that you have converted the Haworthprojection into chair forms properly, imagine flat-tening the rings. When you do this, constituentsthat are above the ring in Haworth will be abovethe ring in a flattened chair. Consideration of space-filling models of such chair forms makes it clear thatbulky groups in axial positions (perpendicular to theplane of the ring) will be crowded to distancessmaller than their van der Waals radii, whereas suchnegative steric interactions are largely absent when

such groups are in equatorial positions (projectingoutward from the ring). Hence, the most stableconformation will favor equatorial arrangements of–OH and CH2OH groups. In �-D-glucopyranose,shown in Figure III-13, it is clear that the C4

1 con-formation will be greatly favored over the C1

4 con-formation, which has five bulky axial groups. Eventhe difference in steric crowding between one axialOH group (�-D-glucopyranose in C4

1 conforma-tion) and no axial OH groups (�-D-glucopyranosein C4

1 conformation) is sufficient to cause the � formof glucopyranose to predominate in solution overthe � form by two to one.

HO

H

H

HH

H OH

OH

OH

CH2OHO

HO

H HH

H

H OH

OHOH

CH2OHO

HO

O

HH

H

H

OH

OH

C

H

HHH

H

OH

HO

OH

OH

CH2OH

O

-D-Glucofuranose�

C

H

H

H

H

H

OH

HO

OHOH

CH2OH

O

-D-Glucofuranose�

C OH

CH2OH

C C

CC

D-Aldehydoglucose-D-Glucopyranose� -D-Glucopyranose�

H

Figure III-11 Interconversion of hemiacetal forms of D-glucose.

ChairBoat

Figure III-12 Conformations of six-membered

carbon rings.


Like pyranose rings, furanose rings are not ac-tually planar as shown in Haworth projections, butexist in two conformations, called envelope (E) andtwist (T): In the envelope conformation, four mem-bers of the ring lie in a plane with the fifth mem-ber bent above or below the plane. Only threemembers of the ring lie in a plane in the twist con-formation; the other two members are twistedabove and below the plane of the ring (see Fig.III-14). The differences in stability between con-formations in furanose rings are not as great as inpyranose rings.

The use of conformational analysis is essentialto a rational approach to carbohydrate chemistry,but for simply representing carbohydrate structuresmost biochemists continue to use the Haworth for-mulae.

Biologically Important Monosaccharides. A verylarge number of monosaccharides have been foundin biological systems, but a relatively smaller num-ber of monosaccharides is found very frequently invirtually all kinds of cells. Most biochemists find itconvenient to memorize the structures of thissmaller group. Consult a standard biochemistrytextbook for the structures of the following mono-saccharides. The hexoses, D-glucose and D-fructose

(a 2-ketohexose), are especially important becauseof their central roles in metabolism, as are the pen-toses, D-ribose and 2-deoxyribose, and the trioses,D-glyceraldehyde and dihydroxyacetone.

The rich variety of carbohydrate structuresmade possible by the presence of multiple chiralcenters is further expanded in biology by the in-troduction of a number of modifications to the ba-sic Cx(H2O)y structure. These include formation ofamino (usually 2-amino) sugars and N-acetylationof the amino group, oxidation of the terminal car-bon atom to a COOH group (yielding uronic acids),and attachment of O-linked sulfate groups. Reduc-tion of CHOH centers to CH2 is also important,because it leads to deoxysugars; the formation of 2-deoxyribose, the constituent sugar of DNA, fromribose, the constituent sugar of RNA, is the mostobvious example.

In addition to the monosaccharides listed above,the most commonly occurring monosaccharides innature are mannose, galactose, glucosamine and N-acetylglucosamine, galactosamine and N-acetylgalac-tosamine, sialic acid (see Fig. III-41), glucuronic acid,and fucose (a 6-deoxyhexose). Most of these mono-saccharides are found as components of oligosaccha-rides attached to proteins or lipids and polysaccha-ride polymers, which will be discussed next.

H OH

H

H

HH H

HH

H

H

H

HH

H OH

HOH

OH

HO

OH

CH2OH

C14

C41

O

-D-Glucopyranose�in Haworth projection

OH

OH

OH

OH

OHCH2OH

CH 2OH

HO

O

O

Bulky groups areequatorial

Bulky groups are axial

Figure III-13 Three-dimensional chair projections of �-D-glucopyranose.


Structure of Oligosaccharides

and Polysaccharides

The Joining of Monosaccharide Units into Poly-

mers. The fundamental linkage between mono-saccharide units in oligosaccharides or polysaccha-rides (analogous to the peptide bond betweenamino acids in proteins) is the acetal or ketal link-age. You will recall that this linkage is formed bycondensation of an alcoholic hydroxyl (usually the–OH of another monosaccharide) with the hemi-acetal (or hemiketal) form of the carbonyl carbonof a sugar, accompanied by loss of water. Acetal orketal linkages between the anomeric carbons of sug-ars and alcoholic hydroxyls (of sugars or other al-cohols) are called glycosidic linkages. An example ofsuch a linkage is shown in the disaccharide maltose(Fig. III-15). Such glycosidic linkages are quite sta-ble at neutral and alkaline pH values, but are hy-drolyzed by aqueous acid.

Oligosaccharides contain only a few monosaccha-rides. Most abundant natural oligosaccharides aredisaccharides, which contain two monosaccharideunits linked together. Biologically important disac-charides include the milk sugar, lactose; ordinarytable sugar, sucrose; a derivative of starch commonin plants, maltose; and the major circulating sugarin insects, trehalose. Oligosaccharides are extremelyimportant in biology, where they are usually found

H

H

-D-Ribofuranose�in Haworthprojection

OHOHH

H

OHHOCH2

O

Others can be drawn, but T2 or T3

are probably favored for mostfuranosides.

T2

3

3

2

H

H

H

H

H

H H

H

OHOH

HO

OHOH

OH

CH2OHO

E3

2 1

3

4

5

2

1

3

4

5

CH2OH

O

E3

2

3

1

4

5

Similaryl E1, E1, E2 and E2 can allbe drawn; ribofuranosides areprobably E3.

HH H

OH OH

HHOCH2

OH

O

Figure III-14 Twist and envelope projections of D-ribofuranose.

HO

H HH

H

H OH

OH

CH2OHO

O

H HH

H

H OH

OHOH

CH2OHO

Figure III-15 �-D-Maltose (�-4-[�-D-glucopyra-

nosyl]-D-glucopyranoside).


covalently attached to proteins or lipids. Polysaccha-rides, which are enormously abundant in nature,contain many monosaccharide units and may havemolecular weights in the millions. Polysaccharidesmay contain only a single type of monosaccharideunit (homopolysaccharides, such as starch, glyco-gen, cellulose, chitin) or several different monosac-charide types (heteropolysaccharides).

The monosaccharide units of oligosaccharidesand polysaccharides may be linked in a variety ofisomeric arrangements. That is, glycosidic bondsmay be formed by condensations between �- or �-OHs at the anomeric carbon of one sugar and al-coholic hydroxyl groups at carbon 2, 3, 4, or 6 of asecond sugar (e.g., l n 2, 1 n 3, 1 n 4, and l n 6for a hexopyranose). Furthermore, the polymer maybe branched so that two sugars have glycosidic link-ages to different alcoholic hydroxyls of a third unit(e.g., at a “branch point” in glycogen, one glucoseresidue is condensed with the C-4 hydroxyl of thebranching residue and a second glucose is con-densed with the C-6 hydroxyl of the same residue).Each glycosidic (acetal or ketal) linkage employs thereducing carbon (anomeric carbon) of one monosac-charide. Thus, all oligo- and polysaccharides haveonly a single reducing end (analogous to the C-terminal of a peptide). The only exceptions to thisgeneralization are oligosaccharides, such as sucroseand trehalose, in which a sugar unit is linked by gly-cosidic linkage to the anomeric hydroxyl of anothersugar. Such a sugar has no aldehyde or ketone inhemiacetal linkage and is consequently a nonreduc-ing sugar. A linear polysaccharide also has only a sin-gle nonreducing end (analogous to the N-terminalof a peptide), but a branched polymer has one ad-ditional nonreducing end for each branch point.These elements of linkage isomerism, which are notfound in proteins or nucleic acids, complicate thedetermination of the structure of polysaccharides,as will be discussed in following paragraphs.

Biological Functions of Oligosaccharides and

Polysaccharides. Polysaccharides serve two im-portant biological roles. Glycogen and starch arepolymers of glucose units linked in �(1 n 4) link-ages that serve as carbohydrate reserves for ani-mals, bacteria, and plants. Because these polymersare readily converted to intermediates for pathwaysthat yield metabolic energy, they can also be

thought of as “energy” storage materials. Themechanisms by which these reserves are mobilizedby plants and animals are explored in Experiments12 and 15, respectively. A second important rolefor polysaccharides is the formation of structuralpolymers for cell walls. Cellulose, a polymer of glucose units in �(1 n 4) linkages, is a major constituent of plant cell walls. Chitin, a polymer of �(1 n 4)-linked N-acetylglucosamine units, isabundant in cell walls of yeasts and fungi, and inthe exoskeletons of insects and crustaceans. A morecomplex class of carbohydrate-containing poly-mers are the peptidoglycans (muramylpeptides) andteichoic acids of bacterial cell walls, which containpeptides or phosphate residues, respectively, linkedin polysaccharide structures.

As noted above, oligosaccharides are usuallyfound linked to noncarbohydrate molecules, espe-cially proteins and lipids. Glycoproteins, in whicholigo- or polysaccharide chains are linked topolypeptide chains by N-glycosidic linkages to theamide nitrogen of asparagine, or by O-glycosidiclinkage to the side chain OH of serine or threo-nine, are particularly abundant in blood and on cellsurfaces. It is difficult to exaggerate the biologicalimportance of the oligosaccharide components ofthese proteins. They affect the folding and stabil-ity of the proteins. They are important in cell–cellrecognition, modulation of immunological and in-flammatory responses, the invasion of host cells bypathogenic microorganisms, the biology and diag-nosis of tumors, and a host of other phenomena.Modification of proteins with oligosaccharides gov-erns their compartmentation into specific mem-branes or organelles within cells and their excretionfrom cells. Because these phenomena are highlysensitive to the precise structures of the oligosac-charides involved, it is crucial to develop methodsfor determining and modifying their complex struc-tures. Characterization of the carbohydrate portionof a glycoprotein requires careful cleavage of theintact polysaccharide from the protein by digestionwith proteolytic enzymes or glycosidases (or bymild alkaline degradation in the case of O-glyco-sides linked to serine) prior to structure determi-nation. Two enzymes frequently used for release of oligosaccharides from glycoproteins are endo-�-N-acetylglucosaminidase H and peptide-N 4-(N-acetyl-�-D-glucosaminyl) asparagine amidase.


Glycolipids are a diverse class of compoundsfound throughout the living world. These com-pounds are classified according to the nature of thelipid portion of the molecule; for example, sphin-golipids, abundant in nervous tissue and brain;glycero-glycolipids, found in plants and animals,and the very complex lipopolysaccharides of bacte-rial cell surfaces. Some glycolipids are conjugatedto proteins, as for example, in glycosyl phos-phatidylinositol structures that serve to anchor pro-teins to cell membranes. Glycolipids are discussedfurther in a subsequent section. Determination ofthe structure of the carbohydrate portion of thesevarious complex polymers is obviously very impor-tant for understanding their biochemistry and bio-logical functions. Such determination relies on theprinciples and techniques described in the follow-ing paragraphs.

Isolation of Carbohydrates

The isolation of polysaccharides presents problemslike those encountered in the isolation of other bi-ological polymers. Carbohydrates, with the excep-tion of glycolipids, are insoluble in most organicsolvents. Most are soluble in water, although a few,such as cellulose, are relatively insoluble. In con-trast to proteins and nucleic acids, most polysac-charides are soluble at low pH, so proteins and nu-cleic acids can often be removed from them byprecipitation with acids such as trichloroacetic acid.Proteins are denatured much more readily thanpolysaccharides and can be removed by heating orby treatment with denaturing solvents (such aschloroform�isoamyl alcohol [25�1]). Nucleic acidsare sometimes removed by degradation with DNaseor RNase, and proteins by digestion with proteases.Once other macromolecules have been removed or

enzymatically degraded to low-molecular-weightfragments, most polysaccharides can be precipitatedwith water-miscible organic solvents (ethanol, ace-tone). Acidic polysaccharides or neutral polysac-charides rendered acidic when combined with bo-rate can also be precipitated by the formation ofwater-insoluble complexes with cationic detergents.Other methods used in purification of polysaccha-rides resemble those used for purification of pro-teins, especially when the polysaccharide containsionizable residues. Ion-exchange, gel-filtration, adsorption chromatography, and density-gradientcentrifugation have all proven valuable.

Determination of the Structure of

Polysaccharides

The problem of determining the structure of apolysaccharide is similar to that of determining theamino acid sequence of a protein, but it is compli-cated by the additional structural features unique tocarbohydrates: branching and linkage isomerism.The elements of carbohydrate chemistry discussedin following paragraphs are those relevant to theproblems of determination of carbohydrate com-position and the structure of polysaccharides.

Determination of Carbohydrate Composition:

Cleavage of Glycosidic Linkages. To analyze apolysaccharide, you must first determine the num-ber and nature of monosaccharide units in thestructure. For most oligo- and polysaccharidescomposed of neutral sugars, hydrolysis of the gly-cosidic linkages between sugars is complete in 3 to6 hr in 0.5 to 1 N HCl or H2SO4 at 100°C (Fig.III-16). Furanoside linkages tend to be more labileto acid than pyranoside linkages. Glycosides ofacidic sugar derivatives and amino sugars tend to

HO

H H

HH

H OH

OH

CH2OHO

HO

H H

HH

H OH

OH

OH

CH2OHO

HO

H HHH

H OH

OHOH

CH2OH

CH3OH

OHH

O CH3

HH

H OH

OH

CH2OH

2H2O

OO

-Methyl-D-lactoside�

H�

� �

Methanol

D-Galactose D-Glucose( form written)� ( form written)�

Figure III-16 Acid-catalyzed hydrolysis of acetal linkages in �-methyl-D-lactoside.


O

R

H

H

C OH

C

R

H

H

C O

OHC

R

H

C OH

OHC

D-FructoseCommon enolD-Glucose

O

R

HO

H

C H

C

D-Mannose

Figure III-17 Alkaline interconversion of glucose,

mannose, and fructose.

resist hydrolysis more strongly than correspondingneutral sugars. Only a few monosaccharides un-dergo degradative reactions under these mild acidconditions. Glycosidic linkages are generally stableunder mild alkaline conditions.

Quantitative Determination of Monosaccharides

by Reactions of Their Aldehyde or Ketone Func-

tions: “Reducing Sugars.” The quantitative de-termination of monosaccharides is often based onthe reactivity of the hemiacetal (potential aldehydeor ketone) of the free sugar. (NOTE: This group isnot present when the carbonyl is in acetal linkage.)The hemiacetal or “reducing group” is very reac-tive with dilute alkali (e.g., 0.035 N NaOH or sat-urated aqueous Ca(OH)2), where the carbohydrateundergoes isomerization at room temperature.This interconversion, which probably involves anenolic intermediate (see Fig. III-17), is referred toas the Lobry de Bruyn–Alberda van Eckenstein re-arrangement. Clearly, if the identity of individualsugars is important, you should avoid alkaline con-ditions during polysaccharide analysis. In thecourse of migration of double bonds in alkali,monosaccharides become very susceptible to oxi-dation by O2 or by other electron acceptors (hence,the term reducing sugar for a free monosaccharide).Such oxidations are very complex and usually do

not proceed in a strictly stoichiometric fashion.However, if the conditions (temperature, alkali con-centration, oxidant concentration, and especially time)are carefully controlled, oxidation of monosaccha-rides by various oxidizing agents (e.g., Cu2� orFe(CN)6

�3) can be used for quantitative analysis ofsugar solutions. This is the basis of such methodsas Nelson’s and the Park and Johnson determina-tions, as well as many others not widely used to-day. The nonstoichiometric aspects of these oxida-tions show up as differences in the “reducingpower” of sugars with very similar structures,which after a fixed time reduce different amountsof Cu2� under identical conditions. Hence, differ-ent standard curves are obtained for different sug-ars. When you analyze an unknown sugar, you canexpect substantial uncertainty in the analytical re-sult. Any saccharide that has a reducing end re-duces Cu2�, Fe(CN)6

3�, O2, and like substances.Per unit weight, however, monosaccharides con-tain more reducing ends than polysaccharides. Bymeasuring increases in reducing equivalents youcan follow the cleavage of a polysaccharide tomonosaccharides or oligosaccharides.

Quantitative Determination of Monosaccharides

by Conversion to Furfurals. As described above,polysaccharides are cleaved in mild acid to mono-saccharides, which are generally stable to acid.Strong mineral acids (4–6 N at 100°C, lower tem-peratures with more concentrated acids) not onlycleave glycosidic bonds but cause dehydration ofmonosaccharides to yield furfurals and, subse-quently, levulinic acid (Fig. III-18). Once again, thedehydration reaction is not stoichiometric, and dif-ferent yields of furfurals and other products are ob-tained from very similar carbohydrates under iden-tical conditions. Further, the yield of products isstrongly dependent on the nature and strength ofacid, temperature, and other variables. It is impor-tant to bear these complexities in mind when usingone of many methods that utilize furfural forma-tion for quantitative analysis of carbohydrates. Thefurfurals formed by acid dehydration from carbo-hydrates react in acid with any one of a variety ofphenols (phenol, orcinol, carbazole, anthrone, etc.)or aromatic amines to give highly conjugated, col-ored products. The amount of color formation is afunction of the amount and nature of the carbo-hyrate being analyzed. Because the rate of furfural


formation is so dependent on the nature of the car-bohydrate and the reaction conditions, careful stan-dardization is required.

Chromatographic Analysis of Carbohydrate Mix-

tures. Much effort has been devoted in recentyears to devising methods for analyzing carbohy-drate mixtures quantitatively and qualitatively in asingle process (as is done for amino acid analysis).Four methods that offer most promise of this goalare described here.

Gas–liquid chromatography of trimethyl-silyl(TMS) or acetyl derivatives of monosaccharides, oftheir methyl glycosides, or of their correspondingalditols (monosaccharides that have their carbonylfunction reduced to the corresponding alcohol) isthe most widely used. Some sample reactions illus-trating the formation of these derivatives are shownin Figure III-19. An advantage of methanolysis isthat it simultaneously cleaves the glycosidic bondsof a polysaccharide and forms the methyl glycosidederivatives. However, it suffers from the disadvan-tage that a single monosaccharide may give as manyas four glycosidic derivatives (�- and �-pyranosidesand �- and �-furanosides), a disadvantage not

shared by the borohydride reduction–acetylationprocedure. The volatile derivatives are identifiedand quantitated from the position and size of theelution peak on the gas chromatogram. They mayalso be analyzed by mass spectrometry when con-firmation of identity is needed.

A second method of analyzing unknown carbo-hydrate mixtures is ion-exchange chromatographyof the monosaccharides as their borate complexes.This method takes advantage of the fact that in al-kaline solution, boric acid forms complexes withvicinal hydroxyl groups (Fig. III-20). The numberand stability of the complexes that form on anygiven monosaccharide are a function of the struc-ture and conformation of the carbohydrate, espe-cially the spatial orientations of adjacent hydroxyls.Thus, the net negative charge of the borate com-plexes of various carbohydrates is different. This al-lows for their separation on anion-exchange resinsand subsequent analysis by the colorimetric meth-ods already described.

High-pH anion-exchange chromatography(HPAEC) of monosaccharides and oligosaccharideshas been more recently developed as a useful ana-lytical tool. At sufficiently alkaline pH (12 to 14)

O

D-Ribose

Furfural

D-Glucose 5-Hydroxymethylfurfural

O

CH2OH

H C OH

H C OH

H

H

C OH

C

O

CH

CH

CH

H

C

C

O

O

C

CH

CH

H

C

C

–3H2O

–3H2O 2H2O

� C HO

O

O

CH2OH CH2OH

H C OH

H C OH

HO C H

H

H

C OH

Levulinic acid

CH3

C O

CH2

CH2

COOH

HCOOH

C

Acid

Acid

�

Figure III-18 Acid-catalyzed dehydration of monosaccharides.


anhydrousCH3OH

100C, 3 hours

orTMS—Cl

�

(CH3)3SiSi(CH3)3

HCl

HH

H

H OH

OHO

O

O

CH2OH

HH

H

H OH

OHOH

H

OCH3

O

O

CH2OH

H

HH

H OH

OH

OH H

OH

O

CH2OH

HH

H OTMS

HTMSO

OTMS

H

OCH3

O

CH2OTMS(TMS)2N—CCF3

TMS derivativeof �-Methyl-D-Glucose

METHANOLYSIS-TMS PROCEDURE

BOROHYDRIDE REDUCTION—ACETYLATION PROCEDURE

CH2OH

CH2OH

H C OH

C

HO C H

HO H

H C OH

D-Galactose

Ac2O

CH2OAc

CH2OAc

H C OAc

C

AcO C H

AcO H

H C OAc

drypyridine

Acetyl derivativeof D-Galactitol

NaBH4

Figure III-19 Sugar derivatives commonly formed before gas-liquid chromatography.

the anomeric hydroxyl group of a monosaccharideloses its proton, and the carbohydrate acquires anegative charge. The acidity of the anomeric hy-droxyl groups is a sensitive function of the struc-ture of the monosaccharide from which it is formed,so many mono- and oligosaccharides can be re-solved by gradient elution from an anion-exchangecolumn (see Experiment 2). When this chromato-graphic separation is combined with pulsed amper-ometric detection, it provides a very sensitivemethod for analysis of the separated carbohydrateswithout derivatization or other chemical reactions.

A fourth method, analysis of sugar alcohols af-ter reduction by [3H]-NaBH4 and separation by pa-per chromatography, is described in detail in Ex-periment 11.

Determination of Sequence and Linkage Iso-

merism of Polysaccharides. Structural character-ization of polysaccharides is similar to protein char-acterization in that it is frequently necessary todegrade polysaccharides into more easily charac-terized subunits (oligosaccharides). This approachis, in fact, quite powerful, because very many poly-saccharides are constructed of repeating or alter-nating oligosaccharide sequences rather than a verylong unique sequence. Fragmentation can be car-ried out by partial acid hydrolysis or by use of spe-cific glycosidases. Enzymes are especially useful be-cause they are usually specific with respect to themonosaccharide units (e.g., glucosamine vs. glu-curonic acid), position of the bond (e.g., 1 n 4 vs.1 n 6), and anomeric configuration (� vs. �) of the

C OH

OHH3BO3 B O� H��

C

C O

OC� �2H2O

Figure III-20 Reaction of borate with vicinal hydroxyl groups.


residues cleaved. Hence, cleavage yields structuralinformation about the nature of the linkage be-tween the fragments, as well as releasing smallerfragments that can be analyzed further. In recentyears, the availability of highly purified recombi-nant endoglycosidases with well-characterizedspecificity for their sites of cleavage has provided avery valuable set of tools for analyzing oligosac-charide structure by enzymatic digestion. In addi-tion to the high specificity of such enzymes, theyoffer the advantage of cleavage under very mildconditions, eliminating the undesired side reactionsthat accompany chemical degradation.

Determination of the structure of oligosaccha-rides usually requires characterization of endgroups, especially the reducing terminus. The mostcommon means to determine reducing termini isreduction by [3H]-NaBH4 followed by total hy-drolysis of the oligosaccharide. Only the reducingterminal is converted to the corresponding sugar al-cohol, which can be identified by paper or gas chro-matography (see Experiment 11). The nonreducingtermini are more difficult to identify, but are oftenrevealed by methylation. Exhaustive methylation ofoligosaccharides or polysaccharides (conversion ofall hydroxyl groups to the corresponding methylethers), followed by hydrolysis and identification ofthe constituent methyl sugars by chromatographicmethods is a general means of determining struc-ture. Only those OH groups on anomeric carbonsand the OH groups that participate in glycosidiclinkages will not be methylated (see Fig. III-21).(Note that in the example shown in Fig. III-21, the

reducing terminus was identified by mild oxidationwith hypoiodite, which converts the reducing ter-minus to the corresponding sugar acid.)

Recent advances in techniques for the determi-nation of the structures of oligosaccharides by high-resolution mass spectroscopy make it likely that theclassic chemical techniques discussed above may belargely replaced by mass spectroscopy. When suffi-cient quantities of an oligosaccharide of unknownstructure are available, nuclear magnetic resonance(NMR) spectroscopy provides another valuable toolfor structure analysis.

Two methods are generally used to determinethe anomeric configuration of glycosidic linkages:cleavage by enzymes of known specificity and NMRspectroscopy. In the latter method, the chemicalshift of the anomeric hydrogen and the coupling ofproton spins (spin–spin coupling) between theanomeric hydrogen and the hydrogen on an adja-cent carbon allow the identification of configura-tion.

Chemistry of Biological Phosphate

Compounds

The chemistry of phosphate esters and anhydridesis so central to biochemistry that the subject mightalso be logically discussed in the sections on nucleicacids, lipids and membranes, or intermediary me-tabolism. However, because of the importance ofsugar phosphates and their derivatives in Experi-ment 12, the subject is presented in this section.The background is important to more than Exper-

HO

H HH

H

H OH

OH

CH2OHO

O

O

H HH

H

H OH

OHOH

CH2OHO

HO

H HH

H

H OH

OH

CH2OHO

O

HCH

H

H OH

OH

OH

OH

CH2OH

ONaOl (CH3)2SO4

NaOH

H2O

H�

CH3O HOOH

H HH

H

H OCH3

OCH3

CH2OCH3

OH

CHH

H OCH3

OCH3

OCH3

OH

CH2OCH3

O

�

CH3O

H HH

H

H OCH3

OCH3

CH2OCH3

OH

CHH

H OCH3

OCH3

OCH3

O�

CH2OCH3

O

Figure III-21 Characterization of maltose by mild oxidation followed by exhaustive methylation and acetal

hydrolysis.


R OH OH

OH

O�H2O

HO P OH

OH

O

OR P�

Figure III-22 Formation of a phosphomonoester.

HH H

H

OH OH

OH

CH2 OO

OH

O

HO P

-Ribose-5-phosphate�

OH

H

OH

HH

H

H OH

OH

O

-Glucose-6-phosphate�

CH2O

OH

O

HO P

Figure III-23 Examples of biologically important

phosphomonoesters.

R R�O

OH

O

O P

A phosphodiester A phosphotriester

R R�

R�

O

O

O

O P

Figure III-24

iment 12, however, and you should keep these prin-ciples in mind when dealing with the other subjectsmentioned in this paragraph.

Phosphate Esters. An ester is formed by elimi-nation of H2O and formation of a linkage betweenan acid and an alcohol (or phenol) (Fig. III-22).Phosphomonoesters, especially of monosaccharides,are very common (Fig. III-23). Because phosphoricacid is a tribasic acid, it can also form di- and tri-esters (Fig. III-24). Phosphotriesters are rarelyfound in nature, but diesters are extremely impor-tant, particularly as the fundamental linkage of thenucleic acid polymers, which are sequences of ri-bose (or deoxyribose) units linked by 3� n 5� phos-phodiester bonds (see Fig. III-25). Like phosphoricacid, which has three dissociable protons (Fig.III-26), phosphomono- and phosphodiesters areacidic and typically ionize as shown in Fig. III-27.Note the similarities between the pKa values for

phosphoric acid and the “primary” and “secondary”ionizations of phosphate esters. The ionization ofphosphodiesters accounts for the very acidic prop-erties of nucleic acids. Knowledge of the ionizationof phosphate compounds and the consequent de-pendence of net charge on pH are essential to theuse of ion-exchange chromatography and elec-trophoresis for the separation of these compounds(see Experiments 4 and 12).

H

H or OH

H H

O

Base1

OH

CH2

H

O

H

H or OH

H H

O

Base2CH2

H

O

O

P

O

O 5�

3�

Figure III-25 Pentose phosphodiester polymer

as found in nucleic acids.

H3PO4

H2PO4�

HPO42�

PO43�

pKa1 � 1.9

pKa2 � 6.7

pKa3 � 12.4

Figure III-26 Ionization of phosphoric acid as a

function of pH.


(pKa � 0.8–2.1)

(pKa � 5.9–6.8)

(pKa � 1.0–2.0)

R R�O

OH

O

O P

R OH

OH

O

O P

R O�

OH

O

O P

R O�

O�

O�

O

O P

R R�O

O

O P

Figure III-27 Ionization of phosphomonoesters

and phosphodiesters as a function

of pH.

Other Biological Phosphate Compounds. Elimi-nation of water between phosphoric acid and cer-tain other types of compounds results in formationof a variety of phosphate compounds that haveproperties that are different from simple phospho-monoesters. A phosphate ester of a monosaccharidein which phosphate is linked to the anomeric hy-droxyl is called a phosphoacetal. An example is �-D-glucopyranosyl-1-phosphate (glucose-1-phos-phate) (Fig. III-28). A related group of compounds

might be called pyrophosphoacetals, shown by the ex-amples in Figure III-29. Condensation between twomolecules of an acid with elimination of water re-sults in the formation of an acid anhydride. Exam-ples of such anhydride bonds are found in inorganicpyrophosphate and in the many pyrophosphoryland tripolyphosphoryl esters of biological systems(Fig. III-30). Condensation between two different

OH

H

O

HH

H

H OH

OH

OCH2OH

OH

OH

O

P

Figure III-28 An example of a phosphoacetal,

glucose-1-phosphate.

OH

H

O

HH

H

H OH

OH

O

Uridine diphosphoglucose(UDPG)

CH2OH

OH

O

O

P

OH

O

O

P

O

OH

O

O

P

OH

OH

O

PHH H

H

OH OH

CH2 OO

OH

O

HO P

Phosphoribosylpyrophosphate(PRPP)

ribose uracil

Figure III-29 Biologically important pyrophos-

phoacetals.

HH H

adenine

H

OH OH

CH2 OO

OH

O

PO

OH

O

PO

OH

HO

O

P

OH

OH

O

PO

OH

HO

O

P

Adenosine-5�-triphosphate (ATP)

Anhydride bonds Phosphoester bond

Pyrophosphoricacid

�H2O

�H2O2 H3PO4

Figure III-30 Acid anhydride bonds in pyrophos-

phate and ATP.


adenineribose

CH3

O

OH

O

PO

O

C

NH3

R

H

C

OH

OH

O

PO

O

C

Aminoacyl-AMP

Acetyl phosphate

�

Figure III-31 Two biochemically important com-

pounds that contain mixed anhy-

dride bonds.

acids with elimination of water results in formationof mixed anhydrides, which are important as inter-mediates in biochemical reactions (Fig. III-31).

The chemical properties of an enol phosphate es-ter are quite different from simple phosphate esters.The only important example is the glycolytic in-termediate, phosphoenolpyruvic acid (Fig. III-32).

A final group of biologically important phos-phate compounds is the phosphoramidates, whichare characterized by the structure shown in FigureIII-33. Phosphocreatine, an important energy stor-age compound in muscle, and phosphohistidine, anintermediate in several enzyme reactions, are twoexamples (Fig. III-34).

Phosphorylation of Proteins. Phosphorylation ofproteins is a central element in a large number ofbiological regulatory processes. Enzymes that cat-

alyze this posttranslational modification are calledprotein kinases. They typically transfer the terminal() phosphate group of ATP to a substrate proteinor to a residue on their own polypeptide chain (au-tophosphorylating kinases) to yield phosphoserine,phosphothreonine, phosphotyrosine, phosphohisti-dine, or phosphoaspartate residues. Phosphoryla-tion usually results in dramatic changes in the func-tional properties of the target proteins, which areoften enzymes, receptors, or transcription factors.Removal of phosphoryl groups may be catalyzed byseparate phosphoprotein phosphatases or may be cat-alyzed by the target proteins themselves. The com-bined action of protein kinases and phosphoproteinphosphatases renders protein phosphorylationhighly reversible and susceptible to exquisitely sen-sitive regulation. Hundreds of examples are nowknown in which such processes play important rolesin metabolic regulation and cell biology. An exam-ple is studied in Experiment 15.

Hydrolysis of Phosphate Compounds. From thepoint of view of the biochemist, the most impor-tant single reaction undergone by phosphate com-pounds found in biological systems is hydrolysis. Anunderstanding of hydrolysis of phosphates is perti-nent to the degradation and identification of com-pounds containing such bonds. It also provides aninsight into the biological function of such mole-cules.

Although all of the compounds described hereundergo hydrolysis with the release of inorganic

CH2

OH

OH

OH

O

PO

O

C

C

Figure III-32 Phosphoenolpyruvic acid.

N

H

OH

O

POR R�

Figure III-33 A phosphoramidate.

OHCHHO

OHCH3 H

O

PNN CC

NHO

OH

CH CH

OH

NHC

NN

CH2 O

P

CC

H

OPhosphocreatine

3-Phosphohistidine(in protein linkage)

Figure III-34 Examples of biologically important

phosphoramidates.


Table III-1 Rates of Hydrolysis of Phosphate Compounds in Dilute Acid (100°C, 1 N acid)

Percent of Reaction

Hydrolysis Reaction t1/2 (time in minutes) in 7 min

2140 000

1300 000

0002.8 070

00 01.05 100

0008.3 040

0008 (0.1 N acid) 100

0 011 (0.5 N acid, 40°) 100

0004 (0.5 N acid, 25°) 100

0020 (0.1 N acid) 100

0010 (pH 4, 65°) 100

3-Phosphoglyceric acid

Glucose-6-phosphate

Fructose-1-phosphate

Glucose-1-phosphate

Phosphoenolpyruate

ATP

Acetylphosphate

Phosphocreatine

UDPG

PRPP

glyceric acid � Pi

H2O

glucose � Pi

H2O

fructose � Pi

H2O

glucose � Pi

H2O

pyruvate � Pi

H2O

ADP � Pi

H2O

acetate � Pi

H2O

creatine � Pi

H2O

UDP � glucoseH

2O

ribose-5-P � PPi

H2O

phosphate during treatment with strong acid or al-kali at elevated temperatures, they differ markedlyin their rates of hydrolysis under milder conditions,especially in dilute acid (0.l–l N, 20–100°C). Thesedifferences are often exploited for analysis or spe-cific degradation (see Experiment 12). Table III-1lists some examples from each class and comparestheir relative lability to acid hydrolysis. The com-pounds vary widely in their stability in mild acid. Auseful generalization, based on the behavior of typ-ical phosphate compounds in 1 N HCl at 100°Cfor 7 min is that most phosphomonoesters and di-esters of alkyl alcohols are stable (no hydrolysis in7 min at 100°C), whereas (with a few exceptions)members of the other classes are completely hy-drolyzed under these conditions. This principleserves as the basis for the assays of Experiment 12.

Most of the compounds under discussion arequite labile in acid, but with two important excep-tions they are resistant to mild alkaline hydrolysis(0.1–1 N base, 25°C). One exception is the mixedanhydrides of phosphate and carboxylic acids, suchas acetyl phosphate. They are very susceptible toalkaline cleavage or cleavage by nucleophiles suchas hydroxylamine. The other relatively alkali-labilephosphate compounds are phosphodiesters, inwhich the phosphate is attached to one of two vic-inal hydroxyl groups (as in RNA or derivatives ofphosphatidic acid). In these instances, the neigh-

boring OH group participates in the hydrolysis ofsuch compounds with intermediate formation of acyclic diester. The resultant phosphomonoester isthen generally very resistant to alkaline hydrolysis.

The susceptibility of different types of phos-phoprotein linkages to chemical cleavage is oftenused to diagnose the nature of the attachment ofphosphate to the protein in a case in which this isunknown. For example, O-phosphorylserine andO-phosphorylthreonine are very stable to acid, butare very rapidly hydrolyzed at alkaline pH. The re-verse is true for most phosphoramidates (phos-phorylhistidine, phosphoryllysine, phosphorylargi-nine), which are extremely acid-labile, but relativelystable to base. Acyl phosphates (phosphorylaspar-tate, phosphorylglutamate), as noted above, arequite labile to both acid and base. Finally, O-phos-phoryltyrosine is extremely stable in alkaline. Thephosphate of O-phosphoryltyrosine is not removedby hydrolysis in 5 N KOH at 155°C for 30 min,conditions that hydryolyze all the other types ofphosphorylamino acid linkages.

The free energies of hydrolysis (phosphate grouptransfer potential) of the compounds in the classeswe have been discussing are important to their rolesas phosphate donors and acceptors, and as carriersof chemical energy in metabolism. Table III-2 listssome of these free energies under standard condi-tions (pH 7, 25°C, 1 M reactants and products). A


compound with a higher free energy of hydrolysiscan spontaneously transfer its phosphate group tothe conjugate acceptor of any phosphate compoundwith a lower free energy of hydrolysis under stan-dard conditions and in the presence of an appropri-ate catalyst (enzyme). In this way, a table of grouptransfer potentials is analogous to a table of redoxpotentials; it can be used to predict the direction ofmany reactions and to analyze the bioenergetics ofmetabolic sequences. There is a general coincidencebetween the acid lability of these compounds andtheir free energies of hydrolysis, but you should notbe deceived. There is no predictable relationship be-tween the free energy change of a reaction and therate at which that reaction will proceed.

Lipids

Because the universal liquid milieu of living systemsis water, you might expect that the chemical com-ponents of cells would be polar substances that are

readily soluble in water. This is generally the case,but there exists a rather heterogeneous class of bi-ological molecules that are soluble in nonpolar solvents, such as benzene, ether, chloroform, orslightly more polar solvents, such as methanol oracetone. This class is called lipids. It is perhaps mostuseful to consider the lipids by their functions. Amajor function of the triglycerides (fats and oils) isstorage and transport of metabolic energy. Waxes,hydrocarbons, and fatty alcohols, the most nonpo-lar lipids, function primarily in forming protectivesurfaces of plant and animal tissues, although waxesare used in place of triglycerides as energy storageforms in some aquatic animals. The most impor-tant lipids are those that serve as structural com-ponents of biological membranes. All of the mem-bers of this group (phospholipids, glycolipids,sphingolipids, and sterols—the latter two are usu-ally absent from bacterial membranes) are amphi-pathic. That is, they possess a nonpolar tail, oftencontaining long-chain fatty acyl groups, and a po-lar head group. Considerable attention will be paidin this section to these lipids and their role in mem-brane structure and function. Finally, there is alarge group of lipids of relatively low molecularweight (steroids, terpenes, prostaglandins, andquinones) that have diverse and very importantfunctions as hormones, vitamins, and photopig-ments, as well as other more specialized roles. Asimportant as this group is, it will not be further dis-cussed in this section. We will focus on lipids thatare important in energy metabolism and membraneformation.

Structure and Classification of Lipids

Triacylglycerols and Related Simple Lipids. Tri-acylglycerols (triglycerides)—that is, glycerylesters offatty acids (e.g., triolein and �-palmitoyl-��,�-diolein [Fig. III-35])—are by far the most abundantlipids in higher plants and animals. These lipids ac-cumulate in fatty depots and seeds as a storage formof food energy. Natural fats generally contain a va-riety of fatty acids, and are best represented as“mixed” fats, as in the case of 1-palmitoyl-2,3-diolein in Fig. III-35. Diglycerides and monoglyc-erides exist in various tissues, usually in smallamounts. Diglycerides play an important role in thebiosynthesis of some of the more complex lipids andin signal transduction in eukaryotic cells.

Table III-2 Free Energy of Hydrolysis of

Phosphate Compounds at pH 7,

25°C. (Most of these free energy

values are strongly dependent on

pH and Mg2+ concentration.)

�F°� of Hydrolysis

(Phosphate Transfer

Reaction Potential) kcal/mol

�14.8

�10.5

�10.1

�8.2

�7.3

�7.0

�5.0

�3.3

�3.1

�3.1

Phosphoenolpyruvate

Creatine phosphate

Acetylphosphate

PRPP

ATP

UDPG

Glucose-1-P

Glucose-6-P

Fructose-1-P

3-Phosphoglycerate

pyruvate � Pi

H2O

creatine � Pi

H2O

acetate � Pi

H2O

ribose-5-P � PPi

H2O

ADP � Pi

H2O

UDP � glucoseH

2O

glucose � Pi

H2O

glucose � Pi

H2O

fructose � Pi

H2O

glycerate � Pi

H2O


Sterols and Sterol Esters. The sterols are hy-droxylated derivatives of the cyclopentanoperhy-drophenanthrene (steroid) nucleus (Fig. III-36). Allsterols are capable of forming esters with fatty acids,but in general, the free sterols are more abundant

than their esters. The most abundant sterol is cho-lesterol, which is found in the membranes of mostanimal cells (Fig. III-36). When cholesterol occursin the esterified forms, it usually is combined withcommon fatty acids, such as palmitic, oleic, orlinoleic acids. Plant sterols are occasionally esteri-fied with acids, such as acetic and cinnamic acids,or are incorporated into glycosides (saponins). Cho-lesterol is the precursor to many important steroidhormones that regulate metabolism and reproduc-tion. Well-known examples include cortisol, aldos-terone, progesterone, testosterone, and estradiol(see Fig. III-36), but there are many others.

Glycerophosphatides. These lipids are mainlyacyl derivatives of �-glycerophosphoric acid and of-ten are called “phospholipids.” The simplest glyc-erophosphatides are the phosphatidic acids, whichcontain �-glycerophosphoric acid esterified withtwo fatty acids (Fig. III-37). Small quantities ofphosphatidic acids have been isolated from a widevariety of plant and animal tissues. It is doubtfulthat these compounds exist in large amounts in tis-sues, because more complex glycerophosphatidesare readily hydrolyzed by enzymes that are widelydistributed in such tissues, yielding phosphatidicacids. Phosphatidic acid is a crucial intermediate inthe biosynthesis of phospholipids.

oleyl

oleyl

oleyl

CH2O

CH2O

CH2O

Tri-o-oleylglycerol(triolein)

HO

O

oleyl

oleyl

C

CH2

CH2O O palmityl

1-o-palmityl-2,3-di-o-oleyl-L-glycerol(1-palmityl-2,3-diolein)

CH CH (CH2)7 C

O

Oleyl � CH3(CH2)7

C

O

Palmityl � CH3(CH2)14

1 (or �)

2 (or �)

3 (or ��)

Figure III-35 Structures of two triglycerides.

HO

CH3

CH3

HO

CH3

OH

Cholesterol Estradiol

O

CH3

CH2OH

CH3

OHHO

Cortisol

HO

CH3

CH3

Side chain

Typical sterol

C O

Figure III-36 Structures of sterols.


CH

O

COO

CR O

R

O

PO O�

O�

CH2

CH2

Figure III-37 A phosphatidic acid.

The major glycerophosphatides are esters ofphosphatidic acid that have the general formulashown in Figure III-38, in which R represents along-chain alkyl group and R� an amino alcohol orinositol. Additional phosphoryl groups are esteri-

fied to inositol in some phosphatidyl inositols.Phosphatidyl inositol diphosphate (PIP2) and otherphosphoinositols are important molecules in signaltransduction.

Additional residues of glycerol are incorporatedin the more complex structures of phosphatidylglycerol, phosphatidyl 3�-o-aminoacylglycerol (lip-amino acid), and cardiolipin (see Fig. III-38).

The plasmalogens represent an unusual type of glycerophosphatide that has been isolated fromanimal and plant tissues. Acid hydrolysis of thesematerials yields long-chain aldehydes and mono-acylated �-glycerophosphate. The long-chain al-dehydes originate from vinyl ether structures. Thevinyl ether group appears usually (if not always) in

CH

O

COO

CR O

R

R�

O

PO O

O�

O

C CHO

R

CH2

CH2

N�

R� � CH2CH2 CH3

CH3

CH3

In phosphatidyl choline (lecithin)

R� � CHCH2 CH2

OH

R� � CHCH2 CH2OH

OH

In phosphatidyl glycerol

In cardiolipin

In phosphatidyl 3 �-o -aminoacylglycerol(lipamino acid)

NH3�

NH3�

O

P OOR� � CCH2 CH2

O�OH

O CCH R1

H

CH2

R� �

R� �

CH2CH2

In phosphatidyl ethanolamine

NH3�

R� �

O�

OCH CCH2

In phosphatidyl serine

In phosphatidyl inositol (phosphoinositide)

H OH

HHH

HOH

HO

OH

H

OH

CCH2O R2

O

O

Figure III-38 Structures of various glycerophosphatides. The generic structure is shown at the top of the

figure, and the classes differ in the structure of the R� group as shown.


the � position. The R� phosphomonoester groupmay be either choline or ethanolamine, althoughethanolamine usually predominates. In the struc-ture of a plasmalogen shown in Figure III-39, R isan aliphatic chain and R� is choline or ethanol-amine.

Sphingolipids. The sphingolipids contain C18 hy-droxyamino alcohols. Four such long-chain hy-droxyamino alcohols have been isolated from nat-ural lipids. Sphingosine and dihydrosphingosine aretypical of animal tissues.

For simplicity we will concentrate on sphingo-sine, which is rarely seen in the free form, and itsderivatives found in animal tissues. Sphingosine isformed biosynthetically from palmitoyl coenzymeA and serine (Fig. III-40). Attachment of a fatty acidto the amino group of sphingosine in amide link-age yields the ceramides (Fig. III-40). The mostabundant sphingolipids are structures consisting ofceramides linked to a phosphate ester, as in the caseof sphingomyelin or in glycosidic linkage to one ormore carbohydrate residues to form cerebrosides organgliosides (Fig. III-40). The linkage of the ce-ramide to other residues always involves the termi-nal hydroxyl group of sphingosine. The fatty acidcomponents of the sphingolipids are usually satu-rated with an unusually high proportion of long-chain C24 acids such as lignoceric (n-tetracosanoic)and cerebronic (�-hydroxylignoceric) acid. Thecarbohydrate component of brain cerebrosides isgalactose. Glucose is present in small amounts inspleen cerebrosides. Spleen also contains a disac-charide derivative made up of both glucose andgalactose (ceramide lactoside).

The gangliosides, a more complicated group ofceramide oligosaccharides, contain (in addition togalactose and glucose) both N-acetylgalactosamine

and sialic acid (N-acetylneuraminic acid) (Fig. III-41). Sialic acid is also frequently found in other protein- and lipid-linked oligosaccharides.

Isolation, Separation, and Analysis of Lipids

Solvent Extraction. The first step in the prepara-tion of a lipid is extraction from the tissue. Sincelipids are defined by their solubility in organic sol-vents (see above), whereas proteins, nucleic acids,carbohydrates and most of the low-molecular-weight metabolites of the cell are generally water-soluble, the isolation of lipids is, in principle, quitesimple. Extraction of a biological sample with anorganic solvent that is immiscible with water, fol-lowed by separation of the phases should yield thelipid components in the organic phase. The vari-ous classes of lipids can then be separated by thechromatographic techniques described below. Al-though such procedures are subject to the poten-tial complications listed below, they are satisfactoryif you wish to analyze the “free” lipids in a cell typeor tissue—that is, lipids that are not tightly associ-ated with proteins or oligosaccharides. The extrac-tion procedures can be modified by inclusion of de-tergents or high concentrations of inorganic saltsto disrupt noncovalent, but strong associations be-tween lipids and proteins so that the lipids can beextracted and analyzed. Lipids that are covalently at-tached to proteins or carbohydrates are not neces-sarily extracted by such procedures, however, andspecialized methods of separation must be devel-oped for each case.

In many cases, biological function is dependenton specific associations between proteins and lipids,as in the case of serum lipoproteins and many in-tegral membrane proteins, such as the enzymes ofthe electron transport chain, proteins involved intransmembrane signaling, or proteins involved inthe transport of ions and metabolites. Characteri-zation of such proteins in their native state requirestheir isolation in complexes with lipids. Purificationprocedures in such cases often require modifica-tions of the usual methods used for isolation of pro-teins (see the Introductory Chapter to Section II).Such modifications might include use of nonionicdetergents, differential centrifugal sedimentation,and chromatographic separations using lipophilicmaterials. Harsh extraction of the lipid components

O RCH CH

O

O�

O

CCH R

CH2

PCH2 R�

O

O

O

Figure III-39 Structure of plasmalogens.


from such complexes using organic solvents wouldallow their separation for analysis, but usually re-sults in loss of biological activity and denaturationof the protein components.

Even in the case of extraction of “free” lipids byorganic solvents, the experimenter must be awareof a number of potential problems.

1. Lipid mixtures have remarkable ability to carrynonlipid materials into organic solvents, partlyby way of ionic interactions. Thus, hexane ex-tracts of blood serum contain urea, sodiumchloride, and amino acids, and hexane extractsof soybeans contain the sugars raffinose andstachyose.

O�

O P CH2 CH3

CH3�

�

CH3

CH2

O

O NCeramideOCeramide Hexose

OCeramide Glucose

Galactose

Hexosamine

Acetyl neuramininc acid

Sphingomyelin

CH3(CH2)12CH CH

OH

O

NH

C

R

C C CH2OH

A ceramide

A cerebroside

A ganglioside

H H

CH3(CH2)12CH CH2CH CH CH

OH OHNH2

Sphingosine

(CH2)14 COO�

COO�

CH3 H3N C H

CH2OH

RCOO�

CO2

�

Figure III-40 Origin and structure of the major sphingolipids.


2. The glycerophosphatides usually contain sub-stantial amounts of highly unsaturated fattyacids that undergo rapid autoxidation catalyzedby Cu2� or Fe3� salts. These salts tend to becarried along by lipids into the organic phase.Thus, lipid extractions may have to be made ina nitrogen atmosphere in the cold and may re-quire the use of peroxide-free solvents or an-tioxidants.

3. Extraction of wet tissues often produces in-tractable emulsions; hence prior dehydration ofthe sample is often helpful. Still, you must re-alize that dehydration of certain tissues causesirreversible binding of some of the lipid com-ponents to proteins and consequently leads todecreases in the lipid yield.

4. Solvents vary widely in their ability to extractdifferent types of lipids, and a series of solventsmay be required for complete extraction.

5. Enzymes that hydrolyze lipids are present invarious tissues. Certain of these are activatedby organic solvents, and the rate of activationincreases as the solvent is warmed. At least onesuch enzyme is active in an ether dispersion.Thus, consideration must be given to the pos-sibility of enzymatic degradation, particularlywhen dealing with plant sources.

No single satisfactory procedure has been dis-covered that avoids all these complications. Thus,extraction conditions must be adapted to the tis-sue under study and to the quantity and type oflipid desired. As a point of departure, the totallipids of a wet tissue can usually be extracted withchloroform–methanol mixtures according to theprocedure of Folch (or one of several variants of

this procedure). Often, you can remove nonlipidmaterials present in the chloroform–methanol ex-tract by extracting the organic phase with water.

Separation of Lipid Classes. Biochemists fre-quently obtain preparative scale separations of lipidmixtures by using step-by-step elution from silicicacid or Florosil columns with solvents of graduallyincreasing polarity. This is particularly applicablefor mixtures of neutral lipids. Polar lipids also maybe separated in this way, although ion-exchangeseparations on DEAE- or TEAE-cellulose are morepowerful for large-scale separations of polar lipids.Lipophilic Sephadex (derivatives of the normal dextran-bearing alkyl groups) can be used to sepa-rate gross lipid classes and individual componentswithin a class, such as various cholesterol esters.

Thin-layer chromatography (TLC), high-performance liquid chromatography (HPLC) andgas–liquid chromatography (GLC) are sensitive andpowerful techniques for analytical scale separationsof lipid mixtures into classes. These methods are alsovaluable for further resolution of the members ofeach class into single components or chemically verysimilar subgroups. TLC of lipids generally utilizessilica gel (silicic acid) as the adsorbant and the samenonpolar solvents used in preparative silicic acidchromatography. Like most TLC procedures, TLCresolution of lipids provides speed, sensitivity, andexcellent resolving power. Many methods have beendeveloped for detection of lipids on thin-layer chro-matograms. Spraying the chromatogram with 50%H2SO4, followed by heating, reveals lipids (and allnonvolatile organic components) as charred spots.Most lipids contain unsaturated fatty acids, whichcan be visualized after reaction with I2 vapor. Othermethods for detection of specific lipid classes are alsoavailable. Incorporation of AgNO3 into the silica gelof TLC or preparative column systems is a usefulway of separating very similar lipids that containfatty acids of varying degrees of unsaturation (i.e.,differing numbers of double bonds). The Ag� ionsform complexes with double bonds. The migrationof lipids on the gel is therefore inversely related tothe number of double bonds (i.e., Ag� ion com-plexes) per molecule of lipid.

GLC and HPLC are powerful techniques foranalyzing lipids. GLC is particularly useful foranalysis of fatty acid mixtures or fatty acid methylesters derived from isolated lipid classes or specific

H

H

HN COO�

H

H

H

OH

CH2OH

OH

OH

H

CH3

OC

C

H OHC

O

Figure III-41 Sialic acid (N-acetylneuraminic acid).


lipids. GLC of intact nonpolar lipid classes, such astriglycerides or cholesterol esters, is also practical.In contrast, polar lipids must be converted to morevolatile derivatives or be analyzed after degradationand derivatization. For this reason HPLC is moregenerally applicable to the analysis of polar lipids.The general principles of operation and detectionused in GLC and HPLC have been described inExperiment 2. Identification and determination ofunknown structures of components separated bychromatography can be greatly aided by use of massspectrometry and NMR spectroscopy.

Degradation of Lipids and Analysis of Lipid Com-

ponents. To determine the complete structure ofan isolated lipid, usually you must degrade the lipidinto its components, separate the components, andanalyze them. This section reviews chemical andenzymic techniques for degradation of lipids, withan emphasis on common phospholipids.

Alkaline hydrolysis (e.g., the Schmidt–Thann-hauser procedure, which employs 1 N KOH at37°C for 24 hr) cleaves most ester bonds but leavesthe amide groups in sphingolipids intact. Alcoholicalkali may be necessary with lipid mixtures that donot disperse readily into aqueous solutions. Potas-sium hydroxide is preferable because it is more sol-uble in alcohol than NaOH; similarly, potassiumsalts of fatty acids are more soluble than their equiv-alent sodium salts.

Acid hydrolyses are usually carried out by re-fluxing in 6 N aqueous hydrochloric acid (constantboiling) or 5 to 10% solutions of HCl in methanol(to promote solubility) for 4 to 30 hr, depending onthe lipid in question. Most glycerophosphatides arehydrolyzed by acid to fatty acids, glycerophosphate,and the free base, just as with alkali. However, ino-sitol phosphatides initially yield inositol phosphateand diglycerides on acid hydrolysis. Hydrochloricacid is easily removed by vacuum, which makeschromatographic examination of the hydrolysisproducts easier.

Because the sites of cleavage are often quite spe-cific, enzymatic degradation of triglycerides andglycerophosphatides can provide useful structuralinformation. Lipases generally cleave fatty acidresidues from the 1 and 3 positions of glycerol, leav-ing the acyl groups on the secondary hydroxyl in-tact. The acyl groups can be removed subsequentlyby alkaline hydrolysis. The phospholipases (A1, A2,

C, and D), which are abundant in bee and snakevenoms, are valuable for structure determinationbecause they selectively hydrolyze various phos-phoglyceride components. As seen in Figure III-42,phospholipase A2 removes the central fatty acid,phospholipase A1 removes the terminal fatty acid,and phospholipase B (which is a mixture of phos-pholipase A1 and phospholipase A2 activities) re-moves both the central and terminal fatty acids.Phospholipases C and D cleave on either side of thephosphate, yielding a diacyl glycerol or a phospha-tidic acid, respectively. The specificity of the vari-ous phospholipases for the site of cleavage is quitehigh. These enzymes have some specificity for thenature of the acyl group and (particularly the Xgroup) of the phosphatide, placing some limitationson the use of these enzymes in structural studies ofunknown phospholipids.

Membranes

The most obvious function of a membrane is toserve as a physical barrier or boundary between theinterior of a cell or subcellular organelle (nucleus,mitochondrion, chloroplast, etc.) and the sur-rounding medium. It has become evident, however,that this is far too static a view of the membrane.It is true that membranes act as envelopes and con-tain contents of cells or organelles, but all mem-branes possess machinery for selective movementof molecules across them, and frequently pump

O RC

R O

O�

O

C CH2

CH2

PCH2 X

O

O

O

OPhospholipase A2

Phospholipase A1

Phospholipase D

Phospholipase C

Figure III-42 Sites of hydrolytic cleavage by

phospholipases.


such molecules against a concentration gradient(active transport). Furthermore, the membrane isoften the site of intense metabolic activity, particu-larly electron transport, oxidative and degradativereactions, biosynthesis of cell-surface components,and transformations of metabolic energy. Mem-branes are also important sites of cell-cell and organelle-cell interactions, such as hormone-cellinteractions, contact inhibition, neurotransmitteraction, recognition by the immune system and soon. In addition to being a physical barrier, the mem-brane is also a site of intense biochemical activity.

Most of these activities require both protein andlipid components of the membrane, as well as vary-ing degrees of structural integrity of the membraneor cell. To obtain a chemical and physical under-standing of membrane phenomena, it is necessaryto isolate membranes and study their compositionand structural arrangements. Ideally, you would re-construct the phenomena of interest in isolatedmembranes or reconstituted systems made frommembrane components. In some cases, this is dif-ficult to carry out. An example of the study of bio-chemical functions in isolated membrane fractionsis the characterization of electron transport per-formed in Experiment 14.

Isolation

One of the major problems encountered when iso-lating membranes is that they usually do not occurin sharply defined molecular sizes, so that differen-tial centrifugation, while valuable, does not usuallyyield homogeneous preparations. Isopycnic densitygradient centrifugation, which takes advantage ofthe generally low density of membranes or mem-brane-rich organelles, is therefore an invaluable ad-junct to membrane isolation. When the membraneof a specific organelle is under study, it is usuallydesirable to isolate the organelle before separatingthe membrane. Contamination of plasma mem-branes with cellular organelles and organelle mem-branes is a serious problem in membrane isolation.The mammalian erythrocyte (red blood cell) is anespecially useful source of homogeneous plasmamembranes because these cells have no other or-ganelles (see Experiment 13).

It is essential to have a set of criteria for assess-ing the purity of isolated membrane fractions. Twomethods are most useful. Phase-contrast or electron

microscopy gives a visual indication of the size andappearance of an isolated membrane preparation andreveals contamination by organelles and other mate-rial. If the membrane fraction in question is knownto contain a specific chemical component or enzymicactivity (such as Na�,K�-ATPase or 5�-nucleotidasefor animal plasma membranes, or succinic dehydro-genase for mitochondrial membranes), assay for suchis a valuable criterion of membrane homogeneity.Other criteria for identity of membranes include thepresence of specific antigens, the phospholipid:pro-tein ratio, and the specific protein composition re-vealed by sodium dodecyl sulfate–polyacrylamide gelelectrophoresis (SDS-PAGE).

Composition of Membranes

The primary constituents of membranes are lipidand protein; carbohydrate residues are found in rel-atively small amounts attached to members of bothmajor classes. Any organism has fairly constantmembrane composition, but membranes varywidely from organism to organism and among var-ious kinds of cells. For example, membranes varyfrom 18% protein, 79% lipid, and 3% other com-pounds in the myelin sheath, to 75% protein and25% lipid in the cell membrane of gram-positivebacteria. A “typical” figure would be about 60%protein and 40% lipid.

The lipid composition of membranes can varyin the same manner. Major components of prokary-otic membrane lipids are phospholipids and gly-colipids; eukaryotic membrane lipids typically con-tain these two classes and, in addition, sterols suchas cholesterol, and sphingolipids. Table III-3 givesthe lipid content of some membranes from varioussources. Table III-3 does not list some of the un-usual and less abundant lipids found in membranes,such as lipamino acids, glycolipids, and phos-phatides containing unusual fatty acid residues.

The protein composition of a membrane is dif-ficult to define precisely because proteins are boundto the membrane with highly varying degrees ofavidity. Some proteins can be removed from a mem-brane by fairly mild treatments, such as washingwith water or salt solutions. Such proteins are usu-ally referred to as peripheral membrane proteins. Insuch cases, however, it may be difficult to say withcertainty whether the protein in question is really a“membrane protein” or merely happens to associ-


ate with the membrane during isolation. Completerelease of protein from membrane preparations re-quires use of agents that disrupt the associations be-tween lipids and proteins. Such procedures releaseintegral membrane proteins that are deeply embeddedin the membrane or have one or more transmem-brane segments. Disruption of such protein associ-ations with membranes requires agents rarely usedin purifying soluble proteins, such as organic sol-vents, chaotropic agents, and detergents. Manymembrane proteins are soluble in such agents andmay require them (or phospholipids) for activity.Fractionation of membrane proteins follows con-ventional procedures (see Section II introduction),except that inclusion of detergents is frequently nec-essary. Analytical electrophoresis of membranes onpolyacrylamide gels containing sodium dodecyl sul-fate has proven to be a powerful tool. As we havediscussed, many enzymes have been found to be as-sociated with membranes. Glycoproteins are par-ticularly abundant in animal-cell membranes, al-though they also occur in bacterial membranes.

Structure

The appreciation of the many important biologicalfunctions of membranes has led to intense research

activity into the structural arrangement of lipids andproteins in membranes, the forces involved in form-ing such structures, and the relation between mem-brane structure and function. Our current view ofmembrane structure is based primarily on physicalstudies of natural membranes and of synthetic lipidmonomolecular and bimolecular layers. Amphi-pathic molecules, particularly phospholipids, have astrong tendency to self-associate and to form lay-ered structures when they are dispersed into aque-ous medium. (Amphipathic molecules are mole-cules in which both highly polar, hydrophilicelements and nonpolar, hydrophobic elements arejoined in the same molecule.) Monomolecular lay-ers (monolayers) of amphipathic molecules tend toform at air–water interfaces. Bilayers form acrossan opening between two water-containing cham-bers. When a phospholipid is dispersed into waterby ultrasonic oscillation, the lipids spontaneouslyform bilayered aggregates such as micelles and water-containing vesicles called “liposomes.” Theseself-assembling systems provide models for study-ing the interactions of amphipathic lipids (phos-phatidylcholine is a commonly used lipid) in aque-ous environments. Such bilayers are thought to bemodels for the simplest or most primitive mem-brane structures (see Fig. III-43). The bilayer is

Table III-3 Lipid Composition of Membranes from Various Sources

Animal Bacteria

Lipids

(given in percent Bacillus

of total lipid) Myelin Erythrocyte Mitochondria Microsome E. coli megaterium

Cholesterol 25 25 5 6 0 0

Phosphatides

Phosphatidylethanolamine 14 20 28 17 100 45

Phosphatidylserine 7 11 0 0 0 0

Phosphatidylcholine 11 23 48 64 0 0

Phosphatidylinositol 0 2 8 11 0 0

Phosphatidylglycerol 0 0 1 2 0 45

Cardiolipin 0 0 11 0 0 0

Sphingolipids

Sphingomyelin 6 18 0 0 0 0

Ceramide 1 0 0 0 0 0

Cerebrosides 25 0 0 0 0 0

Others 11 1 0 0 0 10

Source: Modified from Wold, F. (1971). Macromolecules: Structure and Function. Englewood Cliffs, NJ: Prentice-Hall.


formed by hydrophobic interactions that lead to as-sociation of the nonpolar tails of the lipid moleculeand interaction of the polar heads with water. Pro-teins are believed to associate with lipids in mem-branes by hydrophobic interactions as well as in-teractions with polar head groups, such as hydrogenbonds and electrostatic attractions. Synthetic lipidbilayer systems possess many of the physical prop-erties of natural membranes.

The basic tenets of the fluid mosaic model ofmembrane structure shown in Figure III-44 arewidely accepted. The phospholipid bilayer is thebasic structural feature of the membrane, and pro-teins or multiprotein complexes are viewed as is-land embedded in a sea made up by the lipid bi-layer. Membranes are fluid in the sense that lipidsand proteins can diffuse freely in the plane of the

membrane. Flip-flop of phospholipids, or othercharged molecules, across the bilayer is unfavor-able because of the energy required to move acharged or polar molecule through the hydro-phobic interior of the membrane. There is alsoabundant evidence that membranes are asymmet-ric; that is, they have “inside” and “outside” sur-faces with different proteins and lipids found oneach side.

The relationship between the physical structureof membranes and their biological function is notyet well defined. Clearly, the interposition of a non-polar layer between the spaces separated by a mem-brane prevents translocation of polar biologicalmolecules, as a cell envelope must. Furthermore,the positioning of specific proteins at the surface ofthe membrane allows many specific “recognition”

CH2—CH—CH2

O

C—O

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH3

—

—O

C—O

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH3

—

O

O—P—O–

O

CH2—CH2—+N—CH3

CH3

CH3

Symbol for suchamphipathic lipidmolecules

Nonpolar tailsformed byfatty acid chains

Polar head group(phosphate andassociated group ofa glycerophosphatide)

Disperse into

aqueous solution

Introduce lipid into aqueous

environment between 2 dividing

baffles. Then pull baffles apart.

Monolayer at watersurface, polar headgroups associatewith polar water

Spherical micelle,nonpolar tailsassociate away frompolar water

Syntheticbilayer ormodel membrane

Movable baffle

Air

n

Figure III-43 Formation of micelles and monolayers by amphipathic lipid molecules in aqueous solution.


sites to be presented to other cells, hormones, andmetabolites inside and outside the cell.

Studies have suggested that membrane proteinsdo not always undergo free diffusion in the mem-brane, and that they may be anchored by proteinsof the cytoskelton, by the extracellular matrix or inother ways. Also, membranes are not uniform intheir distribution of components, even on one sideof the bilayer. They contain many specialized re-gions or domains, such as clathrin-coated pits,synapses in nerve cells, microvillae, and focal con-tacts. The lipid and protein composition of the do-mains are distinct and have been associated withspecialized biological functions of the membrane.

The use of artificial lipid membranes and iso-lated membrane fragments and vesicles has been ofgreat value in the study of membrane function, par-ticularly transport and membrane-bound enzymereactions. Current research in this area uses so-phisticated techniques for determining the molec-

ular architecture of these organelles at higher res-olutions, including NMR, electron paramagneticresonance, and fluorescence spectroscopy. Othermethods for obtaining images of membranes andmembrane components are x-ray crystallography ofmembrane proteins, Fourier analysis of electronmicroscopic images, and atomic force microscopy.Methods for isolating native membrane compo-nents in the functional state have also become muchmore effective. Important phenomena at the cellsurface involve not only the cell membrane, but cellwalls, capsular materials, and enzyme activities lo-cated outside the membranes of gram-negative bac-teria (periplasmic proteins).

REFERENCES

Barker, R. (1971). Organic Chemistry of Biological Com-pounds. Englewood Cliffs, NJ: Prentice-Hall.

Figure III-44 The fluid mosaic model of membrane structure.


Casey, P. J., and Buss, J. E. (eds.) (1995). Lipid Modifi-cation of Proteins. Methods Enzymol, 250.

Cummings, R. D., Merkle, R. A., and Stults, N. L.(1989). Separation and Analysis of GlycoproteinOligosaccharides. Methods Cell Biol 32:141.

Fukuda, M., and Kobata, A. (1994). Glycobiology, APractical Approach. New York: Oxford UniversityPress.

Gennis, R. B. (1989). Biomembranes: Molecular Structureand Function. New York: Springer Verlag.

Hardy, M. R., and Townsend, R. R. (1994). High-pHAnion-Exchange Chromatography of Glycoprotein-Derived Carbohydrates. Methods Enzymol 230:208.

Hounsell, E. F. (ed.) (1998). Glycoanalysis Protocols. To-towa, NJ: Humana Press.

Lennarz, W. J., and Hart, G. W. (eds.) (1994). Guideto Techniques in Glycobiology. Methods Enzymol,230.

Kobata, A. (1992). Structures and Functions of the SugarChains of Glycoproteins. Eur J Biochem 209:483.

Stryer, L. (1995). Biochemistry, 4th ed. chapters 11 and18. W. H. New York: Freeman.

Vance, D. E., and Vance, J. E. (eds.) (1985). Biochem-istry of Lipids and Membranes. Menlo Park, CA:Benjamin/Cummings.

Varki, A. (1993). Biological Roles of Oligosaccharides:All of the Theories Are Correct. Glycobiology 3:97.

Wold, F., and Moldave, K. (1984). PosttranslationalModifications, Part B Methods Enzymol, 107:3–23;23–26.

, A

195

EXPERIMENT 11

Microanalysis of Carbohydrate Mixtures by Isotopic, Enzymatic, and Colorimetric Methods

Theory

Naturally occurring polysaccharides are made up ofa variety of monosaccharides connected by glyco-sidic bonds. Often, these polysaccharides are linkedto proteins and lipids. The function of such poly-saccharides is often critically dependent on the com-position and sequence of their monomeric units.Determination of the structure of polysaccharides isa multistep process. In the early stages, it is crucialto ascertain both the relative amount and the chem-ical identity of the monomers. After a polysaccha-ride has been broken down to its monomeric unitsby hydrolytic or enzymatic digestion, a variety of an-alytical techniques can be employed to identify andquantify each monomer. Three of these techniqueswill be demonstrated in this laboratory exercise.

In this experiment, you will be given a samplecontaining D-glucose, D-ribose, and D-glucosamine.Your objective will be to determine the concentra-tion of each of these carbohydrates in the sampleusing three independent techniques:

1. Reduction with tritiated NaBH4 (NaB3H4),followed by analysis of the radioactive sugar al-cohols by paper chromatography and liquidscintillation counting.

2. Quantification of D-glucose with the use of acoupled enzyme assay.

3. Quantification of D-glucosamine with the useof the Elson–Morgan colorimetric reaction.

Reaction of Reducing Sugars with Sodium

Borohydride

A characteristic property of reducing sugars is thatthey can be reduced to their corresponding sugar al-

cohols with sodium borohydride (Fig. 11-1). This isone of the few analytical reactions of carbohydratesthat is known to proceed stoichiometrically to givea quantitative yield of product. Notice that one moleof NaB3H4 can reduce four moles of sugar to its cor-responding sugar alcohol. If the specific activity ofthe NaB3H4 used in the reaction and the amount ofcarbohydrate in the sample are known, you can pre-dict the amount of radioactivity that will be presentin the product. In the same sense, the amount ofproduct in a sample can be determined if the spe-cific activity of the NaB3H4 used in the reaction andthe amount of radioactivity present in the productare known. For example, suppose you perform a car-bohydrate reduction reaction with 40 �Ci/�molNaB3H4, and you determine that the sugar alcoholcontains 10 �Ci of 3H. Based on the stoichiometryof the reaction, you know that you have then reduced1 �mol of the sugar to its sugar alcohol. If you knowthe volume of the carbohydrate solution used in thereaction, you can then determine the concentrationof carbohydrate present in the unknown sample.

When utilized in conjunction with paper chro-matography, this technique can be effective in iden-tifying and quantifying a number of different car-bohydrates in an impure solution. Commonly,polysaccharides are hydrolyzed in 1 N acid to pro-duce their constituent monomers. The solution isthen adjusted to pH 8.5 with the addition of 1 vol-ume of 2 N sodium carbonate and reduced withNaB3H4. Following the completion of the reduc-tion reaction, the excess (unreacted) NaB3H4 is re-moved from the solution through the addition of astrong acid:

NaB3H4 � 3H2O � H� Na3H2BO3 � 43H2

196

1. Whether the reduction reaction with NaB3H4

went to completion.2. How much of the total reaction mixture wastransferred to the chromatogram.

3. Which of the 3H-containing peaks on the chro-matogram corresponds to glucitol, the productof the reduction of glucose.

If the reduction reaction was complete, there willbe a single 14C peak superimposed on the 3H-glucitol peak. If the reduction was incomplete, therewill be an additional 14C peak that corresponds tounreduced 14C-glucose. The 14C-glucose internalcontrol will also indicate what percent of the totalreaction was analyzed on the chromatogram. If the reaction contained 1 �Ci of 14C-glucose, and

It is very important to perform this reaction uncappedin a fume hood to prevent the room from filling withtritium gas. The mixture of tritiated sugar alcoholsis then separated by paper chromatography on thinstrips. These strips are divided into half-inch seg-ments and analyzed by liquid scintillation counting.The total counts per minute of 3H in each sugar al-cohol peak can then be used to calculate the amountof each carbohydrate present in the sample.

In addition to the tritium isotope present in thesodium borohydride, the carbohydrate sample thatyou are given will also contain a small amount ofhigh specific activity 14C-glucose, which will serveas an internal control or an internal standard for thereaction. Inclusion of 14C-glucose in the reactionwill allow you to verify three important parameters:

Figure 11-1 Reaction of reducing carbohydrates with sodium borohydride.

OH

CH2OH

NaB[3H]4 NaH2BO3

H C OH

H C OH

HO C H

H C OH

H C 3H

O

CH2OH

H C OH

H C OH

HO C H

H

H

C OH

C

�� 44

D-GlucitolD-Glucose

[3H] sodium borohydride

OH

CH2OH

NaB[3H]4 NaH2BO3

H C OHH C OH

C

H C OH

H OH

H C 3H

O

CH2OH

H C OH

C

H

H

C OH

H OH

C

�� 44

D-RibitolD-Ribose


OH

CH2OH

NaB[3H]4 NaH2BO3

H C OH

H C OH

HO C H

H C

H C 3H

O

CH2OH

H C OH

H C OH

HO C H

H

H

C NH2 NH2

C

�� 44

D-GlucosaminitolD-Glucosamine


ent in the sample (it will add identifiable counts tothe reaction while contributing negligible mass).

Elson–Morgan Determination

of Hexosamines

The most specific and frequently used assay to quan-tify 2-deoxy-2-amino sugars employs the condensa-tion of the carbohydrate with 2,4-pentanedione in ba-sic solution. The product (a pyrrole derivative) is thenreacted with p-dimethylaminobenzaldehyde to forma chromogen that has a maximum absorbance at530 nm (Fig. 11-2). Although both 6-deoxy-6-aminoand 3-deoxy-3-amino sugars can also be analyzed using the Elson–Morgan reaction, the chromogens

0.25 �Ci total are present on the chromatogram,then you know that only 25% of the sample wasanalyzed. Finally, the 14C-glucose internal controlwill allow you to identify the glucitol peak. SinceD-glucose is the only carbohydrate in the unknownsample that is labeled with 14C, the glucitol peakwill be the only sugar alcohol peak that containsthis isotope. It will be important to identify the glu-citol peak, since the migration of all of the othersugar alcohols will be measured relative to glucitol(an Rglucitol value will be calculated for each sugaralcohol). By using high specific activity 14C-glucoseas the internal control component, one can followthe fate of D-glucose in the reaction without sig-nificantly affecting the total mass of glucose pres-

EXPERIMENT 11 Microanalysis of Carbohydrate Mixtures by Isotopic, Enzymatic, and Colorimetric Methods 197

H H

O

CH2OH

H C OH

H C OH

CH2OH

CH3

H C OH

C CH3

H

H C OH

H C OHHO C

H

H

C

H

NH2

H�

C

�

Glucosamine adduct(pyrrole derivative)

4-Dimethylaminobenzaldehyde

D-Glucosamine

2,4-pentanedione

CH3 CH3CH2C

O

C

O

O

heatN

CH2OH

CH3

H C OH

C CH3

CH3H3C

H

H C OH

H C OH

O

C

H3C

H3C O

H

N

N�

Chromogen (A530)

Figure 11-2 The Elson–Morgan reaction.

198

Formic acid

Scintillation fluid (5 g of 2,5-diphenyloxazole/L-toluene)

Scintillation vials


Glucose reagent (For large laboratory groups, thisreagent can be purchased in a mixed dry form fromSigma Chemical Co.)

NAD� (1.5 mM)

ATP (1 mM)

Yeast hexokinase (1 unit/ml)

Glucose-6-phosphate dehydrogenase (1 unit/ml)

MgCl2 (2.1 mM)

Buffer at pH 7.5

Spectrophotometer

4-ml glass vials with Teflon-lined screw caps

Glucosamine standard solution (0.5 �mol/ml)

Acetylacetone reagent (prepared fresh daily); 2 ml of 2,4-pentanedione mixed with 98 ml of 1 M Na2CO3

95% ethanol

p-Dimethylaminobenzaldehyde reagent (dissolve 677.5 mgof p-dimethylaminobenzaldehyde in 25 ml of a 1�1mixture of ethanol:concentrated HCl)

Protocol

Day 1: Reduction of Carbohydrates with

Sodium Borohydride and Enzymatic

Determination of Glucose

Safety Precaution: This laboratory exercise uti-

lizes radioisotopes (3H and 14C). Use caution

when handling these compounds. Avoid ingest-

ing and skin contact with these radioisotopes.

Wear gloves and safety glasses at all times.

Consult the instructor for information on how to

properly dispose of the radioactive material

produced in this experiment.

Reduction of Carbohydrates with Sodium

Borohydride

1. Obtain a numbered 1.5-ml microcentrifugetube containing 1 ml of your unknown carbo-hydrate mixture. Record the number of your un-known solution in your notebook. This solutionwill be used on Day 1 and Day 2.

that they produce absorb light maximally at differentwavelengths.

Determination of Glucose with a Coupled

Enzyme Assay

In the presence of excess ATP and NAD�, and hex-okinase plus glucose-6-phosphate dehydrogenase, theconcentration of glucose in an unknown sample canbe determined in an end-point assay that follows thereduction of NAD� to NADH�H� (Fig. 11-3). Inthe presence of ATP, hexokinase will convert glu-cose to glucose-6-phosphate. This reaction is notspecific for glucose in that hexokinase will recognizemany different six-carbon sugars as a substrate. It isthe second reaction that makes this overall assayspecific for glucose. Glucose-6-phosphate dehydro-genase will convert the glucose-6-phosphate pro-duced in the first reaction to 6-phosphogluconate.During this process, one molecule of NAD� willbe reduced to NADH. This enzyme will only recognize glucose-6-phosphate as a substrate, ig-noring all other six-carbon sugar phosphates thatmay have been produced in the first reaction. Sincethe production of NADH can be followed spec-trophotometrically at 340 nm, you can determinethe number of moles of NADH produced. Thisnumber will be equal to the number of moles ofglucose present in the reaction. This informationcan then be used to determine the concentration ofglucose in the unknown solution, provided youknow the volume of the unknown sample that waspresent in the assay.



Heating block

Microcentrifuge tubes (0.5 ml and 1.5 ml)14C-glucose, 200 mCi/mmol (2,000 dpm/�l)

0.5 M NaB3H4, 15–25 mCi/mmol

Solutions of unknown concentration containing D-glucose,D-ribose, and/or D-glucosamine

0.75 N H2SO4

Fume hood

Whatman 3 MM chromatography paper

Descending chromatography tank

Ethyl acetate

Acetic acid

SECTION III Biomolecules and Biological Systems

199

O

HH

HH

OH

OH

OO

PO

NH

2 N

NNN

OPO

�O

�O

�O

�O

PO

Ade

nosi

ne tr

ipho

spha

te (

ATP

)

O

CH

2O

H

HC

OH

HC

OH

HO

CH

H

H

CO

H

C

�

D-G

luco

se

O

HH

HH

OH

OH

OO

PO

NH

2 N

NNN

PO

�O

�O

�O

O�

O�

PO

O

Ade

nosi

ne d

ipho

spha

te (

AD

P)

O

CH

2O

HC

OH

HC

OH

HO

CH

H

H

CO

H

C

�

D-G

luco

se-6

-pho

spha

te

O�

�OPO

O

O HC

HC

OH

HC

OH

HO

CH

H

H

CO

H

C

D-G

luco

se-6

-pho

spha

te

Hex

okin

ase

O

H

H

O�

O�

O�

POH

C

HC

OH

HC

OH

HO

CH

HC

OH

C

6-P

hosp

hogl

ucon

ate

H

HH

H

OH

OH

O

OO PON

H2 N

NN

O

HHH

HH

OH

OH

N N�

O

OP

CO

�O

Nic

otin

amid

e ad

enin

e di

nucl

eotid

e (N

AD

�)

(NA

D�

, oxid

ized

form

)

�

NH

2

O

HH

HH

OH

OH

O

OO PON

H2 N

NN

O

HH

HH

HH

OH

OH

N N�

O

OP

CO

�O N

icot

inam

ide

aden

ine

dinu

cleo

tide

(NA

DH

; red

uced

form

)

��

��

NH

2

Glu

cose

-6-p

hosp

hate

dehy

drog

enas

e

�

O

Figure 11-3 Coupled enzyme assay to determine glucose concentration.

200

carbohydrate sample is radioactive. Wear gloves andsafety goggles at all times when handling the sam-ple. Be sure that all radioactive waste is disposed ofproperly.

8. Apply a 2-�l aliquot of the sample on the ori-gin and allow the sample to dry. Repeat thisprocedure until a total of 8 �l of the sample hasbeen applied. Successive 2-�l aliquots shouldbe spotted on top of one another to give a sin-gle, 8-�l sample at the origin.

9. Using a soft lead pencil, label the strip at theend opposite the origin with your name.

10. Place the strip at the top of the descending chro-matography tank so that the origin is just abovethe level of the mobile phase (Fig. 11-4). Themobile phase for this experiment is ethyl ac-etate:acetic acid:formic acid:water (18�3�1�4).Allow the chromatogram to develop for �18 to20 hr.

11. When the development is complete, removethe strip from the chromatography tank and al-low it to air dry in the fume hood.

2. Obtain a 0.5-ml microcentrifuge tube con-taining 5 �l of 14C-glucose. The reduction reac-tion will be done in this tube, so label it with yourname.

3. Add 5 �l of your unknown carbohydrate solu-tion to this tube and mix thoroughly.

4. Bring this reaction tube to the instructor, whowill add 5 �l of NaB3H4. Mix all componentsthoroughly.

5. Incubate the reaction in a 50°C oven for30 min. During this incubation, you may pro-ceed with the “Enzymatic Determination ofGlucose,” below.

6. In the fume hood, add 10 �l of 0.75 N H2SO4 tothe reaction. Mix thoroughly and allow the re-action to stand for 30 min uncapped in the fumehood to remove all 3H2 gas.

7. Obtain a strip of Whatman 3 MM chromatog-raphy paper (1 in. by 22 in). Draw a line witha soft lead pencil about 7 cm from one end ofthe strip. This line will mark the origin wherethe sample will be spotted. Keep in mind that the


Equilibration solventin chamber

Paper

Solvent front

Glass trough with solvent

Glass rod to hold paper

Supporting rods

Figure 11-4 Descending paper chromatography.

2. Set up eight 4-ml glass vials containing the fol-lowing:

Volume of Volume

0.5 �mol/ml of Dilute

Glucosamine Carbohydrate Volume of

Vial Standard (�l) Solution (�l) Water (�l)

1 — — 500

2 050 — 450

3 100 — 400

4 150 — 350

5 250 — 250

6 — 050 450

7 — 150 350

8 — 250 250

3. Add 0.5 ml of acetylacetone reagent to all eightvials. Mix thoroughly and loosely seal the vialwith a Teflon-lined screw cap to allow thefumes to escape.

4. Place the vials in a 90°C heating block and in-cubate for 45 min. While you are waiting, youmay complete “Analysis of Paper Chro-matogram,” below.

5. Allow the vials to cool to room temperature(�5 min).

6. Add 2 ml of 95% ethanol to each vial. Slowlyadd 0.5 ml of p-dimethylaminobenzaldehydereagent to each vial. The reaction may bubble vig-orously as this is done, so use caution so as not to loseyour samples.

7. When the solutions have stopped bubbling, capthe vials and incubate for 1 hr at room tem-perature.

8. Blank your spectrophotometer to read zero ab-sorbance at 540 nm against the solution in vial1 (blank).

9. Measure the absorbances of vials 2 to 8 at540 nm and record these values in your note-book.

10. Prepare a standard curve by plotting the num-ber of micromoles of glucosamine versus A540using the data obtained from vials 2 to 5 (thesecontain known amounts of the glucosaminestandard solution). Does this plot yield astraight line through the data points?

11. Determine which of vials 6 to 8 had absorbancereadings that lie in the linear portion of the stan-dard curve. Based on these absorbance values,

Enzymatic Determination of Glucose

1. Dilute your unknown carbohydrate sample1�20 with distilled water to give a total of 40 �lof dilute carbohydrate solution. This dilutedsample will be used in the following assay.

2. Set up the following reactions in two 1.5-mlmicrocentrifuge tubes:

Volume of Volume of Diluted

Glucose Carbohydrate Volume of

Tube Reagent (ml) Solution (�l) Water (�l)

1 1.0 — 20

2 1.0 20 —

3. Mix thoroughly and allow the tubes to incu-bate at room temperature for 15 min.

4. Read the absorbance of both samples at 340 nmin the spectrophotometer. Set the instrumentto read zero absorbance using water as a blank.

5. Subtract the A340 reading of tube 1 (control)from the A340 reading from tube 2 (sample).This corrected value is the true A340 readingfor your carbohydrate sample.

6. Using Beer’s law, calculate the concentration ofNADH in this 1.02-ml sample. The millimo-lar extinction coefficient at 340 nm for NADHis 6.22 mM�1 cm�1.

7. Using this information, calculate the numberof moles of NADH that are present in the 1.02-ml sample.

8. Using the information obtained above, calcu-late the concentration of glucose present in theunknown carbohydrate solution in units of mi-cromoles per milliliter and milligrams per mil-liliter. Remember that the original unknown car-bohydrate solution was diluted for use in this assay.

Day 2: Elson–Morgan Determination of

Glucosamine and Analysis of Paper

Chromatogram

Elson–Morgan Determination of

Glucosamine

1. Dilute your original unknown carbohydrate so-lution 1�100 with distilled water to give a totalof 1 ml of dilute carbohydrate solution. This di-luted sample will be used in the following assay.


202

g. Calculate the Rglucitol values for the other 3H

sugar alcohol peaks on the chromatogram:

Rglucitol �

Using this mobile phase for development,Rglucitol values for the various compoundsare typically:

Ribitol �1.4Glucose �0.85Glucosaminitol �0.3

Which sugar alcohols can you identify onyour chromatogram? Was the reduction re-action with sodium borohydride complete?Explain.

h. Determine the true number of 3H cpm and14C cpm in the glucitol peak:

3H cpm � (channel 1 cpm) � (channel 1/

channel 2 ratio) (channel 2 cpm)

14C cpm � (channel 2 cpm) � (channel 1/

channel 2 ratio) (channel 2 cpm)

The channel 1/channel 2 ratio is a value thatwill be provided by the instructor. It is a cor-rection factor that takes into account the factthat some 14C counts per minute werecounted in channel 1. The instructor deter-mined this ratio by scintillation counting ofa sample containing a known number of 14Cdisintegrations per minute in channel 1 andchannel 2 under conditions identical to yourexperiment. Refer to Section I, Experiment 3for further discussion of the channel ratio methodof quantifying radioactivity.

i. Calculate the number of micromoles of glu-cose, ribose, and glucosamine in your car-bohydrate sample:

Micromoles of sugar �

where

Fraction recovered �

total 14C cpm in glucitol peaktotal 14C cpm in the reaction

(3H cpm in peak)(�mol sugar/3H cpm)

fraction recovered

distance from 3H peak to the origin (cm)distance from glucitol peak to the origin (cm)

determine the number of micromoles of glu-cosamine present in these vials.

12. Taking into account the dilution of the originalunknown carbohydrate sample and the volumeof this diluted sample present in vials 6 to 8, de-termine the concentration of glucosamine in theoriginal unknown carbohydrate solution. Ex-press the concentration in units of micromolesper milliliter and milligrams per milliliter.

Analysis of Paper Chromatogram

1. Using a soft lead pencil, mark off half-inchsegments on the chromatography strip, start-ing from the origin and extending to the endof the strip. As before, remember that the chro-matogram is radioactive. Wear gloves and safetyglasses at all times.

2. Cut along each line with a scissors and placethe half-inch segments into scintillation vialslabeled 1 to 30 (with 1 corresponding to thesegment nearest the origin). Be sure that thesegments are placed in the vials in the correctorder so that you will be able to identify eachpeak by its Rglucitol value.

3. Fill each scintillation vial with enough scintil-lation fluid to completely cover each paper seg-ment.

4. Place the vials (in order) into a scintillationcounter rack and count each vial in the 3Hchannel (channel 1) and the 14C over 3H chan-nel (channel 2). Consult the instructor for theproper scintillation counter channel settings.

5. Data analysis

a. Identify the segment that has the lowest cpmvalue for channel 1 (3H, background countsper minute for channel 1).

b. Subtract this value from all other channel 1cpm values for the segments.

c. Identify the segment that has the lowest cpmvalue for channel 2 (14C, background cpmfor channel 2).

d. Subtract this value from all other channel 2cpm values for the segments.

e. Prepare a plot of 14C counts per minute ver-sus segment number and a plot of 3H countsper minute versus segment number.

f. Identify the glucitol peak, which is (are) thesegment(s) with 14C and 3H cpm.


Questionsa. Calculate the counting efficiencies for 3Hand 14C both in channel 1 and channel 2.

b. Determine how many 14C disintegrationsper minute are in the unknown sample.

c. Determine how many 3H disintegrationsper minute are in the unknown sample.

d. If the specific activity of NaB3H4 used inthis reaction was 15 mCi/mmol, and 6 �l ofthe original carbohydrate solution was usedin the reaction, calculate the concentrationof the carbohydrate in the unknown solu-tion in micromoles per milliliter. Assumethat the fraction recovered was 40%.

REFERENCES

Conrad, H. E., Varboncouer, E., and James, M. E.(1973). Qualitative and Quantitative Analysis of Reducing Carbohydrates. Anal Biochem 51:486.

Elson, L. A., and Morgan, W. T. J. (1933). A Colori-metric Method for Determination of Glucosamineand Chondrosamine. Biochem J 27:1824.

Iyer, R. S., Kobierski, M. E., and Salomon, R. G.(1994). Generation of Pyrroles in the Reaction of Levuglandin E2 with Proteins. J Org Chem59:6038.

Lehninger, A. L., Nelson, D. N., and Cox, M. M.(1993). Structure and Catalysis of Carbohydrates.In: Principles of Biochemistry, 2nd ed. New York:Worth.

Mathews, C. K., and van Holde, K. E. (1990). Carbohydrates. In: Biochemistry. Redwood City, CA: Benjamin/Cumming.

Morrison, R. T., and Boyd, R. N. (1987). OrganicChemistry, 5th ed. Boston: Allyn & Bacon.


Spiro, R. G. (1966). Analysis of Sugars Found in Glycoproteins. Methods in Enzymol 8.

Stryer, L. (1995). Carbohydrates. In: Biochemistry,4th ed. New York: Freeman.

Wheat, R. W. (1966). Analysis of Hexosamines. Methods Enzymol 8:60.

The number of micromoles of sugar/3H cpm orthe specific activity of sodium borohydride arevalues that will be provided by the instructor.The total number of 14C counts per minute pre-sent in the reaction will also be provided by theinstructor.

j. From the number of micromoles of eachcarbohydrate on the chromatogram and thevolume of the original unknown carbohy-drate solution used in the reduction reac-tion, can you determine the concentrations ofall three carbohydrates in your unknown?Express your answer in units of micromolesper milliliter and milligrams per milliliter.

6. How do the concentrations of D-glucose andD-glucosamine determined in this sodiumborohydride reduction experiment comparewith the values that you obtained for these twocarbohydrates in the hexokinase and Elson–Morgan assays, respectively? Discuss the quan-titative differences as well as possible explana-tions for the differences arising from the vari-ous assays.

Exercises

1. Acetyl derivatives of amino sugars (e.g., N-acetyl glucosamine) will undergo an Elson–Morgan reaction in basic solution. Write anequation to describe this reaction and draw thestructure of the likely intermediate.

2. The following data were obtained in a sodiumborohydride reduction reaction as performedin this experiment. The reaction contained asingle carbohydrate and the 14C-labeled inter-nal control:

Data

Measured Measured

Channel 1 Channel 2

Sample cpm cpm

3H (10,000 dpm) 5,800 014C (10,000 dpm) 4,700 3,600

Unknown (14C and 3H) 49,000 7,500


E

205

erning use of a major energy reserve, this enzyme(and the corresponding synthetic enzyme, glycogensynthetase) is under complex metabolic and hor-monal control. The regulation of glycogen phos-phorylase by covalent modification (reversiblephosphorylation) and by allosteric effectors will bestudied further in Experiment 15.

In this experiment, soluble starch is incubatedwith inorganic phosphate (Pi) and a phosphorylasepreparation from potatoes. After 24 to 48 hr, the re-action is stopped by heating the mixture to destroythe enzyme. After removal of the denatured enzyme,the unreacted inorganic phosphate is removed fromthe filtrate by precipitation as magnesium ammo-nium phosphate. The glucose-1-phosphate is thenisolated by ion-exchange chromatography and issubsequently crystallized. The ion-exchange purifi-cation consists of two steps. First, cations in the so-lution (Mg2�, NH4

�, K�) are adsorbed to the cationexhange resin, Dowex 50, with displacement of theH� initially adsorbed to the resin. The H� ions re-leased into the solution lower its pH and convertglucose-1-phosphate to the monanionic form (G-1-P�1). Then, the glucose-1-phosphate is adsorbedonto an anion exchange resin, Amberlite IR-45,which allows uncharged materials (starch, aceticacid) to pass through. The glucose-1-phosphate isthen eluted as the dipotassium salt from the IR-45column with the addition of a concentrated solutionof OH� ions.

Glucose-1-phosphate is determined by mea-surement of the inorganic phosphate released fromthe molecule following acid hydrolysis. Phospho-acetals of all sugars can be hydrolyzed by treat-ment with 1 N acid for 7 min at 100°C, but most

EXPERIMENT 12

Glucose-1-Phosphate: EnzymaticFormation from Starch and Chemical Characterization

Theory

Starch, a complex carbohydrate found in mostplants, is a mixture of two polysaccharides: amylose,a straight-chain polymer of glucose units joined by�-1,4 linkages, and amylopectin, a branched-chainglucose polymer that differs from glycogen pri-marily in its larger number of �-1,4-linked glucoseunits between the �-1,6 branch points. Both glyco-gen (which is similar to amylopectin and is foundin animals and bacteria) and starch are substratesfor the enzyme phosphorylase. Phosphorylase cat-alyzes the phosphorolysis of glycogen or starch to glu-cose-1-phosphate (Fig. 12-1).

The primary function of starch is to serve as astorage form for carbohydrate. Mobilization ofstarch for metabolism or synthesis of other poly-saccharides requires cleavage of monosaccharideunits from the polymer and formation of phospho-rylated sugar derivatives, which are the substratesfor glycolysis or for formation of nucleoside diphos-phosugars that are used for biosynthetic reactions.Phosphorolysis of starch fulfills both requirementsin a single step. If starch were cleaved by hydrolyticreactions (e.g., by amylases), subsequent formationof the sugar phosphate from the released glucosewould require the expenditure of metabolic energyin the form of ATP. Thus, the phosphorylase reac-tion increases the efficiency of starch or glycogenutilization by conserving the energy of the acetalbonds between glucose units through formation ofthe phosphoacetal bond of glucose-1-phosphate.

The regulation of glycogen phosphorylase inmammalian muscle and liver has been studied in-tensively. As might be expected for a pathway gov-

206

total phosphate is greater than 1, the results mayindicate the presence of phosphate-containing con-taminants. Second, it is necessary to distinguish be-tween the known isomers of glucose phosphate.This step is particularly pertinent in this experi-ment because we know that glucose-1-phosphatemay be converted to glucose-6-phosphate by theenzyme phosphoglucomutase, another enzymepresent in the crude potato extracts used for syn-thesis of glucose-1-phosphate from starch.

The characterization of glucose-1-phosphate isbased on the relative stability of various sugar phos-phates in acid solution. Because sugar phosphoac-etals are nonreducing sugars, they release equalamounts of reducing sugar and phosphate on 7-minhydrolysis in 1 N acid at 100°C. In contrast, glu-cose-6-phosphate and other phosphate esters re-quire more concentrated acid or more prolongedheating for complete hydrolysis. Therefore, it ispossible to characterize glucose-1-phosphate byqualitative (chromatographic) and quantitativeanalysis of materials present before and after 7-minhydrolysis. It is also possible to estimate the purityof the product by careful evaluation of the results.

Chromatographic characterization of sugarphosphates can be achieved by two different meth-ods. First, you can usually detect the sugar portionof sugar phosphates by using sugar-specificreagents, such as the aniline–acid–oxalate spray

other sugar phosphate esters (such as glucose-6-phosphate) ordinarily require longer heating timesor more concentrated acid for complete hydroly-sis. The inorganic phosphate released by acid hy-drolysis is measured by the colorimetric method ofFiske and Subbarow, which is specific for inorganicphosphate; that is, inorganic phosphate can be an-alyzed with no interference from organic phos-phate compounds such as phosphate esters or phos-phoacetals that may also be present in the solution.Hence, the increase in inorganic phosphate after7-min hydrolysis of a sample (compared to an un-hydrolyzed sample) is a measure of the phospho-acetal content of the sample. Stability of variousother biologically important phosphate compoundsto acid hydrolysis has been discussed in the Sec-tion III introduction. The color developed in theFiske–Subbarow reaction is dependent on the for-mation of a phosphomolybdic acid complex, whichforms an intense blue color when reduced by a mix-ture of bisulfite and p-methylaminophenol. (Notethat a variant of this assay is used in Experiment13.)

It is necessary to establish both the identity andpurity of the newly isolated glucose-1-phosphate.In this experiment, the first step consists of show-ing that the product presumed to be glucose-1-phosphate is, in fact, composed of equal quantitiesof glucose and phosphate. If the ratio of glucose to


O�HO

O�

P

O

O�

O�

P

O

O

OOH

OH

OH OH

�

�

H

HH

H

H

CH2OH

O

OH

OHH

HH

H

H

CH2OH

OOH

H

HH

CH2OH

O

OH

OHH

HH

H

H

CH2OH

OOH

H

HH

CH2OH

Nonreducing end

phosphorylase

O

OOH

OH

OH

H

HH

H

H

CH2OH

Figure 12-1 Phosphorylase-catalyzed phosphorolysis of glycogen or starch to glucose-1-phosphate.

Centrifuges

300-ml Polypropylene centrifuge bottles

Nelson’s reagents A and B

Arsenomolybdate reagent (see Appendix)

1.0 mM glucose standard (1.0 mM glucose in 1.0 MNaCl)

Whatman No. 1 paper

1% Glucose-1-phosphate

1% Glucose-6-phosphate

1% Glucose

1% Inorganic phosphate

10% Mg(NO3)2 in ethanol

Methanol

88% Formic acid

Aniline–acid–oxalate spray reagent (see Appendix)

Modified Hanes–Ischerwood spray reagent [(4%(NH4)6MoO24 � 4 H2O�1 N HCl�70% HClO4�

H2O (5�2�1�12)]

Acid–FeCl3 in acetone (Dissolve 150 mg FeCl3 � 6 H2Oin 3 ml 0.3 N HCl and mix with 97 ml acetone—prepare fresh before use.)

1.25% Sulfosalicylic acid in acetone

UV lamp

100°C oven

Protocol

The isolation and characterization of glucose-1-phosphate can be completed within four labora-tory periods. The enzyme incubation is begun inthe first period. The removal of cations withDowex 50 should be completed in the second pe-riod, and the product should be stored at 0 to 5°C.The ion-exchange chromatography can then becompleted and the crystallization begun in thethird period. Characterization of the isolated prod-uct will be conducted in the fourth period, aftercollecting and drying the crystals of dipotassiumglucose-1-phosphate dihydrate in between thethird and fourth periods. Store all solutions at 0 to5°C between laboratory periods to avoid bacterialor chemical degradation. Use distilled or deionizedwater throughout the experiment.

Day 1: Incubation of Starch and Pi with

Potato Phosphorylase, Preparation of the

IR-45 Column

1. Using about 50 ml of H2O, make a smoothslurry of 10 g of soluble starch. Add this slowly

reagent. Detection of sugars with this reagent in-volves the formation of colored glycosylamineswhen reducing sugars react with aniline. (Note thatglucose-1-phosphate is a nonreducing sugar, but itis readily hydrolyzed to form glucose under theacidic conditions used to develop the color.) Or-ganic phosphates, including sugar phosphates, canbe detected with phosphate-specific methods. Themodified Hanes–Ischerwood spray reagent used inthis experiment hydrolyzes any organic phosphateso that inorganic phosphate is released. The inor-ganic phosphate then combines with the molybdicacid and is reduced to yield a blue spot (phospho-molybdous acid). The alternate iron–sulfosalicylicacid complex assay involves binding of Fe3� by or-ganic phosphates. The bound Fe3� cannot thenform a red-brown complex with sulfosalicylic acid,so phosphate-containing areas appear as white spotson a colored background.


Phenylmercuric nitrate

0.8 M Potassium phosphate buffer, pH 6.7

Soluble starch

Filter aid

14% NH4OH

2 N NaOH

1 N NaOH

2 N HCl

1 N HCl

Potatoes (fresh)

Blender

Cheesecloth

Mg(OAc)2 � 4 H2O

Dowex 50 (H� form)

IR-45 (OH� form)

Absolute methanol

Cylindrical glass tubes, or equivalent chromatographycolumns

Glass wool

Reducing reagent (3% NaHSO3, 1% p-methylamino-phenol)

Acid molybdate reagent (see Appendix)

Phosphate standard (1 mM)

5% KOH

0.01 M KI, 0.01 M I2Charcoal (Norit)

Vacuum desiccator

P2O5

EXPERIMENT 12 Glucose-1-Phosphate: Enzymatic Formation from Starch and Chemical Characterization 207


to 100 ml of vigorously boiling H2O, and stircontinuously with a glass rod until the solutionis nearly clear. The solution may be cloudy, butshould not be milky. Further heating may berequired to dissolve the starch, but avoid pro-longed heating. Add 100 ml of cold H2O, stir-ring continuously, to help cool the solution toroom temperature. Do not add the enzyme untilthe solution has cooled to room temperature becausehigh temperatures inactivate the enzyme. Step2 can be conducted while the solution is cool-ing.

2. Prepare a 500-ml filter flask to receive thepotato extract by pouring 250 ml of H2O intoit and marking the position of the top of thefluid in the filtration reservoir with a felt-tippedpen. Pour out the H2O. Suspend about 100 mgof phenylmercuric nitrate in a few milliliters of H2O and pour it into the filter flask.Phenylmercuric nitrate is added to inhibitother enzymes and to prevent bacterial growthduring the incubation of potato phosphorylasewith starch.

3. When the starch solution prepared in step 1has cooled, add the solution (250 ml) to 250 mlof 0.8 M phosphate buffer in a 1-liter Erlen-meyer flask.

4. Cut a fresh, medium-sized potato (precooledfor 24 hr at l to 5°C; you need not peel thepotato, but it should be washed to remove dirtand dust) into half-inch cubes. Blend 150 g ofthese cubes, added over a 30-sec period, with150 ml of H2O for 2 min in a blender. Thenquickly pour the resultant slurry onto a Buch-ner funnel lined with two to four layers ofcheesecloth, and filter with vacuum, collectingthe filtrate in the filter flask you prepared instep 2. Wash the crude pulp with 50 ml of H2Oto ensure thorough enzyme extraction; that is,add the water to the pulp, stir the mixture, andpass the mixture through the filter as before.Failure to complete these operations within 2 minof blending may result in loss of enzyme activity.Adjust the extract to a volume of 250 ml withH2O.

5. Immediately add the 250 ml of enzyme solution(potato filtrate) to 500 ml of the solution ofstarch plus phosphate buffer prepared in step3. Record the total volume, and store the solu-tion in a stoppered Erlenmeyer flask at room

temperature in your desk. During the incuba-tion period (24–48 hr), the reaction mixturewill turn red and then dark blue or purple be-cause of the action of tyrosinases present in thecrude extract that catalyze melanin formationfrom proteins. These colored materials will beremoved during later procedures.

6. Prepare the IR-45 column for use on Day 3 asfollows: Prepare a column about 4 cm in di-ameter and 20 to 30 cm long using a rubberstopper, screw clamp, and glass wool plug (Fig.12-2). Before assembling the column, rinse allparts thoroughly with distilled H2O. Insert therubber stopper tightly at the bottom, and tapethe edges to prevent leaks. Fill the columnabout one-third full with H2O, and insert theglass wool plug, pushing it gently down to therubber stopper at the bottom. Mix 250 ml of

Figure 12-2 Anion-exchange column.


H2O with 250 ml of moist IR-45 (OH�) (pre-

pared by the instructor), and pour the resultantslurry into the column so that there are no airpockets in the settled resin. For best results,pour the slurry of resin completely into the col-umn before allowing the resin to settle. Then,wash the resin with H2O until the effluent isabout pH 9 or lower (use a pH meter). Drainor pipette off the excess fluid until the fluid sur-face just covers the top of the resin bed. Coverthe surface of the resin with a layer of glasswool and cover the top of the column withParafilm. The column can stand at room tem-perature until use on Day 3, but be sure that itdoes not leak.

Day 2: Precipitation of Pi, Cation Exchange

with Dowex 50, and Initial Glucose-1-

Phosphate Assays

1. After 24 to 48 hr of incubation, stop the enzy-matic reaction by rapidly heating the potato ex-tract/starch solution to 95°C and then slowlycooling it to room temperature over a 30-minperiod or longer. If possible, come to the labora-tory 4 to 6 hr before the usual laboratory period tocarry out the heating, so that the solution will slowlycool before the next step. Rapid cooling of this so-lution (e.g., in an ice bath) may cause the resid-ual starch to form a gel and should be avoided.

2. Remove the coagulated protein by centrifugingthe fluid in large (300 ml) bottles at room tem-perature for 15 min at 4000 � g. Decant thesupernatant fluid into a large beaker.

3. Remove the excess inorganic phosphate fromthe solution by dissolving 0.2 mol of magne-sium acetate (44 g of Mg(OAc)2 � 4 H2O) in thesolution and adjusting the pH to 8.5 with 14%NH4OH (use pH paper first, then a pH meter;about 25 to 30 ml of 14% NH4OH will be re-quired, but avoid adding excess NH4OH). Coolthe solution in a salted ice bath for 10 min, andremove the precipitated magnesium ammo-nium phosphate by suction filtration. UseWhatman No. 1 filter paper covered with a thinlayer of Celite filter aid. If filtration is very slow,use more than one Buchner funnel or changethe filter as frequently as needed. Filter the so-lution a second time using two layers of What-man No. 1 filter paper.

4. Collect the filtrate and record its volume. Re-move duplicate 0.05, 0.1, 0.2, and 0.5 ml sam-ples of the filtered solution for the inorganicphosphate and 7-min phosphate assays de-scribed in steps 5 through 9. Place these du-plicate samples in small, labeled glass test tubes.

NOTE: Most laboratory detergents containlarge amounts of phosphate, which may con-taminate your glassware. All tubes used inphosphate analyses should be thoroughlycleaned with a sodium dodecyl sulfate deter-gent that does not contain phosphate andrinsed with deionized water.

5. Assay for inorganic phosphate and glucose-1-phosphate:

In the Fiske–Subbarow colorimetricmethod for the determination of phosphate, thecolor yield is directly proportional to the amount ofinorganic phosphate only when the sample analyzedcontains between 0.1 and 1.0 �mol of phosphate.The efficiency of the enzymatic formation ofglucose-1-phosphate on Day 1 will vary some-what depending on the sources of the enzymeand the starch, as well as the length of the in-cubation period. Accordingly, the sample sizessuggested may not lie within the range in whichthe assay is valid. In this event, use the resultsyou obtained to decide on several differentsample sizes to analyze until you find at leastone that falls within the accurate range of theassay. For example, if the suggested samplesizes gave too little color for accurate mea-surement, use larger sample sizes in the nextanalysis. In this experiment, as in all isolationprocedures, you must obtain an accurate mea-surement of the amount of the desired com-pound in each fraction. Therefore, you willneed to determine and use appropriate aliquotranges for analysis before proceeding to thenext steps in the experiment. Further, you mustkeep accurate records of aliquot sizes and pro-tocols in order to evaluate your data correctly.

6. Carry out the 7-min hydrolysis on one of theset of duplicate samples of the magnesium am-monium phosphate supernatant fluid from step4 by adding an equal volume of 2 N HCl toeach and heating at 100°C for 7 min. That is,set up tubes 1 through 4, which contain 0.05, 0.1,0.2, and 0.5 ml of the supernatant fluid, re-spectively. Add 0.05, 0.1, 0.2, and 0.5 ml 2 N


HCl, respectively, to tubes 1 through 4. Placethe tubes in a boiling water bath for exactly7 min. Remove the tubes, cool them in coldwater, and add 0.05, 0.1, 0.2, and 0.5 ml of 2 NNaOH, respectively, to tubes 1 through 4. It isimportant that the solutions be nearly neutral;use pH paper to check them. Bring the volumein each tube to 3 ml by adding 2.85, 2.7, 2.4,and 1.5 ml of H2O, respectively, to tubes 1through 4.

7. Prepare the other (unhydrolyzed) set of dupli-cate samples of the magnesium ammoniumphosphate supernatant fluid from step 4 bybringing their volumes to 3.0 ml with H2O.That is, set up tubes 5 through 8, which contain0.05, 0.1, 0.2, and 0.5 ml of the supernatantfluid, respectively. Add 2.95, 2.9, 2.8, and 2.5 mlof H2O, respectively to tubes 5 through 8.

8. Prepare a phosphate standard curve and blankas follows. Set up tubes 9 through 15, which con-tain 0, 0.1, 0.2, 0.4, 0.6, 0.8, and 1.0 ml of the1 mM (1 �mol/ml) inorganic phosphate stan-dard, respectively. Add 3.0, 2.9, 2.8., 2.6, 2.4,2.2, and 2.0 ml of H2O, respectively, to tubes9 through 15.

9. Add, in order, 1 ml of acid molybdate reagent,1 ml of reducing reagent (3% NaHSO3, l% p-methylaminophenol), and 5 ml of H2O to allof tubes 1 through 15. Mix all of the solutionsby inverting the tubes, and allow the color todevelop for 20 min before reading the ab-sorbance at 660 nm using the solution in tube 9as the blank. Calculate the quantity of inorganicphosphate and glucose-1-phosphate in eachsample and in the entire reaction mixture as fol-lows. Prepare a standard curve that plots ab-sorbance at 660 nm versus number of micro-moles of Pi based on the data obtained fromtubes 10 through 15. From the absorbance at660 nm of your hydrolyzed and unhydrolyzedsamples, determine the number of micromolesof Pi and the concentration of glucose-1-phos-phate in your original sample. Remember thatthe number of micromoles of glucose-1-phosphate inthe sample is equal to the total Pi found in the 7-min hydrolyzed sample minus the number of mi-cromoles of Pi found in the unhydrolyzed sample ofthe same volume. You have made these determi-nations in four different volumes of the mag-nesium ammonium phosphate supernatant

fluid from step 4. How do the different deter-minations agree? Which determinations aremost reliable? Why?

In subsequent phosphate analyses on Days 3and 4, you should prepare a fresh standard curveaccording to step 8 along with the samples be-ing analyzed for that day’s experimental work.

10. If the phosphate assay reveals that an excess ofinorganic phosphate is still present in the in-cubation filtrate (i.e., if an intense blue colorforms in the unhydrolyzed 0.1 ml sample), add1 g of Mg(OAc)2 � 4 H2O, adjust to pH 8.5 with14% ammonia, cool the solution, and filter itagain. Then repeat the phosphate assays. (Step10 is not usually necessary.)

11. Calculate the number of micromoles of Pi and7-min phosphate in the entire volume of fil-tered solution. If the inorganic phosphate (inthe unhydrolyzed sample) in the filtrate is lessthan 15% of the phosphate found after 7-minhydrolysis, you may proceed with the rest ofthe experiment. (NOTE: When students areworking in pairs, some instructors prefer tohave one student proceed directly with steps 12through 14 while the other student conductsthe assays of steps 5 through 9. This assumesthat the magnesium ammonium phosphate pre-cipitation effectively removes most of the in-organic phosphate from your preparation, asoccurs in most (but not all) cases. Consult yourinstructor.)

12. Decolorize the solution of glucose-1-phos-phate (separated from excess inorganic phos-phate) by stirring 2 g of charcoal into the so-lution and then removing the charcoal byvacuum filtration using filter aid. This proce-dure should yield a clear or yellowish solutioncontaining glucose-1-phosphate, unreactedstarch, and many salts that were present eitherin the original potato extract or in the reagentsadded during the course of the experiment.

13. Remove the cations by treatment with Dowex50 in the following manner: Add 350 ml ofmoist Dowex 50 in the H� form (prepared bythe instructor) to the decolorized solution andstir gently for 5 min before separating by vac-uum filtration. Do not use filter aid when remov-ing the Dowex; use only Whatman filter paper.

14. Determine the pH of the filtrate produced instep 13. If the pH of the filtrate is not acidic


(pH 1 to 3), add 100 ml of moist Dowex 50 inthe H� form, then stir and filter the solutionas before. Repeat this procedure until the pHof the solution is between 1 and 3. Record thevolume of the resultant solution, and removeduplicate 0.05-, 0.1-, 0.2-, and 0.5-ml samples.Place these samples in four small, phosphate-free, labeled glass test tubes.

15. Assay the samples from the Dowex supernatantsolution for inorganic phosphate and 7-minphosphate exactly as described under steps 5 to9. If necessary, the remaining solution fromstep 14 may now be stored at 0 to 5°C for sev-eral days. There may be small losses of glucose-1-phosphate during storage in acidic solution,but these are minimized at 0 to 5°C. Alterna-tively, if time is sufficient, proceed directly tothe protocol for Day 3.

16. At the end of the period return the Dowex tothe container provided by your instructor.(NOTE: Do not allow the Dowex to become mixedwith the Amberlite resin used in Day 3.)

Day 3: Isolation of Glucose-1-Phosphate by

Anion Exchange

The next step in the purification procedure is thecolumn chromatography of the Dowex 50 filtrate onAmberlite IR-45 in the OH� form. This involvesremoving the anions in the acidic solution from allother contaminating materials. Thus, when theacidic filtrate (pH 1 to 3) contacts the IR-45 in OH�

form, the glucose-1-phosphate (G-1-P�1) is elec-trostatically adsorbed on the resin, whereas un-ion-ized materials in the solution (such as acetic acidfrom Mg(OAc)2 and unreacted starch) pass throughthe column. When the resin is eluted with strong al-kali (5% KOH), the adhering anions, including glu-cose-1-phosphate, are displaced by the excess OH�

ions and are obtained in the eluate from the column.

1. To cause adsorption of the glucose-1-phosphateby the IR-45, gently pour the acidic solutiontreated with Dowex 50 (steps 13 and 14 of theprotocol for Day 2) onto the column, avoidingthe introduction of air bubbles into the resin bed.Open the screw clamp and adjust the flow rateto about 15 ml/min. Pass the entire solutionthrough the IR-45 resin without permitting airto enter the column.

2. Collect the entire effluent in one fraction, de-termine its volume, and assay duplicate 0.5-mlsamples for inorganic phosphate and 7-minphosphate as described for tubes 4 and 8 understeps 5 to 9 of Day 2.

3. If appreciable 7-min phosphate appears (�30%of total glucose-1-phosphate present in theDowex 50 filtrate), and if time permits, add anadditional 50 g of IR-45 resin to the column,and pass the effluent through the column again.

4. After the original solution has passed into thecolumn and the effluent has been satisfactorilyfreed of the 7-min phosphate (i.e., glucose-1-phosphate should have bound to the column),wash the column with deionized H2O until theeffluent no longer gives a positive test for starch(i.e., a blue color forms on mixing a drop of ef-fluent with a drop of 0.01 M I2 in 0.01 M KI).About 500 ml of water will be required to re-move all of the starch from the column.

5. To elute the glucose-1-phosphate from theresin, pass 5% KOH (Remember: KOH is causticand can cause painful skin burns) through the col-umn, adjusting the flow rate to 20 ml/min. Col-lect separate, successive 100 ml fractions in clean250 ml Erlenmyer flasks. Continue collectingfractions for about 500 ml after the pH of theeffluent becomes markedly alkaline (pH 11 to13) as observed with pH paper. (The usual to-tal volume of all fractions equals 700�800 ml.)

6. Test 0.1-ml samples from each 100-ml fractionfor 7-min phosphate as described for tube 2 un-der steps 5 to 9 of Day 2. Some fractions mayrequire smaller samples to fall within the lin-ear range of the standard curve, but 0.1-mlsamples will serve to find the “peak” for glu-cose-1-phosphate. It is not necessary to repeatanalyses for samples that give too much color to fallon the standard curve.

7. Combine the fractions containing 80 to 90%of the recovered glucose-1-phosphate, and ad-just the solution to pH 8 (or higher) by addinga few drops of 5% KOH, if necessary. Do notadd KOH if the pH is already above 8. Determinethe volume of the combined KOH effluent(glucose-1-phosphate “peak”) fractions. Saveduplicate 0.05-, 0.1-, 0.2-, and 0.5-ml samplesfor inorganic phosphate and 7-min phosphateanalysis in clean, phosphate-free, labeled smallglass test tubes.


8. Add 3 volumes of absolute methanol to thecombined fractions, and leave them at 0 to 5°Cfor at least 12 hr for crystallization of thedipotassium glucose-1-phosphate dihydrate. Inthis temperature range, this compound will re-main stable indefinitely.

9. Collect the crystals by centrifugation, afterpouring off the bulk of the clear supernatantfluid. A clinical centrifuge or refrigerated lab-oratory centrifuge at low speed may be used.Wash the crystals with 5 ml of absolute ethanoland centrifuge again. Then dry the crystals ina tared container in a vacuum desiccator overP2O5 at 5°C. Weigh the crystals to determinethe final yield of glucose-1-phosphate and savethem for analysis on Day 4 of this experiment.It will save much time on Day 4 if you return tothe laboratory and carry out step 9 on the day pre-ceding the experiments of Day 4.

Report of Results for Days 1 through 3

1. Prepare a flow sheet of the steps in the isolationprocedure, indicating the purpose of each step.

2. Prepare a table (see Table 12-1) showing thepercentage of the glucose-1-phosphate recov-ered at each step in the procedure. Account forany poor recoveries.

3. Prepare a graph of the elution pattern of theglucose-1-phosphate from IR-45, plotting mi-cromoles of glucose-1-phosphate/100 ml as theordinate and the milliters of effluent as the ab-scissa.

4. Calculate the maximal yield of glucose-1-phosphate that would be expected if the phos-

phorylase reaction had reached equilibrium.The quantity of starch is altered only slightlyduring the incubation; therefore, the starchconcentrations in the numerator and denom-inator are roughly equivalent and cancel out.Use the following equilibrium constant foryour calculation:

Keq 0.088

Remember that the initial concentration of in-organic phosphate (0.8 M � 250 ml/750 ml 0.267 M) has been decreased at equilibrium bythe amount of glucose-1-phosphate formed.Compare your yield (glucose-1-phosphatefound in the MgNH4PO4 supernatant fluid)with the theoretical yield expected. Suggestreasons for any discrepancy between your re-sults and the theoretical value calculated fromthe phosphorylase equilibrium constant.

Day 4: Chemical Characterization of the

Isolated Glucose-1-Phosphate

You should begin this day’s experiments by prepar-ing tubes 1 through 6 and Tubes 1 through 5 asdescribed in steps 1 through 6 below. Then youshould prepare the chromatograms and start theirdevelopment (steps 7 and 8). Next you should per-form the phosphate and reducing sugar analyses asdescribed in steps 12 and 13. Finally, spray the de-veloped chromatograms to detect the separatedproducts (steps 9 through 11).

glucose-1-phosphateinorganic phosphate

Table 12-1 Percentage of Glucose-1-Phosphate Recovered during Purification

�mol Total

Vol of of 7-min �mol of 7-min Percent

Step Solution Phosphate/ml Phosphate Recovered

MgNH4PO4 supernatant 100

Dowex 50 supernatant

Original IR-45 (OH�) washes

(before eluting with H2O or KOH)

Combined KOH effluent fractions

Glucose-1-phosphate crystals* — —

*Assume that the crystals are dipotassium glucose-1-phosphate dihydrate (MW 372).


1. You will evaluate your sample of isolated glucose-1-phosphate for purity by determin-ing the following quantities (tubes with primevalues are duplicates). Instructions for prepar-ing these tubes follow the list below.

a. Reducing sugar equivalents before 7-minhydrolysis (tubes 1 and 1)

b. Reducing sugar equivalents after 7-min hy-drolysis (tubes 2 and 2)

c. Inorganic phosphate present before 7-minhydrolysis (tubes 3 and 3)

d. Inorganic phosphate present after 7-min hy-drolysis (tubes 4 and 4)

e. Phosphate present after total hydrolysis(tubes 5 and 5)

f. Chromatographic characterization after 7-min hydrolysis (tube 6)

2. Weigh out a 20- to 40-mg sample of your iso-lated glucose-1-phosphate with an analyticalbalance, determining the weight of the sampleto the nearest milligram, and dissolve the sam-ple in 10 ml of H2O. Place 0.1-ml samples ofthis solution in each of 10 small glass tubes,which should be numbered tubes 1, 1, 2, 2, 3,3, 4, 4, 5, and 5. Save the remaining solutionfor chromatographic characterization (step 7,below).

3. To another tube (tube 6), add 0.1 ml of a sep-arately prepared 10 mg/ml solution of your iso-lated glucose-1-phosphate and add 0.1 ml 2 NHCl.

4. Add 1.0 ml of 1 N HCl to tubes 2, 2, 4, and4. Heat them for 7 min in a boiling water bath;then cool them at once in a beaker of cold wa-ter. Finally, neutralize the contents by adding1 ml of 1 N NaOH to each of the tubes. At thesame time as tubes 2, 2, 4, and 4 are heated,heat tube 6 at 100°C, but do not add additionalHCl or NaOH to this tube. Cool tube 6 withthe other tubes and save the contents for chro-matographic analysis (step 7 below).

5. Add 1 drop of 10% Mg(NO3)2 in ethanol totubes 5 and 5. Cautiously evaporate the fluidto dryness over a flame until brown fumes dis-appear and only a white ash remains in thetubes. Cool these two tubes, then add 1.0 ml1 N HCl to each and heat them in a boilingwater bath for 15 min to hydrolyze any py-rophosphates that may have formed during the

heating with Mg(NO3)2. Cool the tubes, neu-tralize with 1 ml of 1 N NaOH, and analyzefor inorganic phosphate as with the other tubes(step 13).

6. Add 2 ml of H2O to tubes 1, 1, 3, and 3 andsave them for the assay of unhydrolyzed prod-uct (steps 12 and 13).

7. Prepare two identical chromatograms usingWhatman No. 1 paper squares, 20 cm on a side,which carry the following spots, 5 to 7 mm indiameter, spotted at 1-in. intervals on a linedrawn lightly with a pencil 1 in. from the bot-tom of the paper:

20 �l of the hydrolyzed material (from tube 6)

10 �l of 1% glucose-1-phosphate standard

10 �l of 1% glucose-6-phosphate standard

20 �l of the isolated, suspected glucose-1-phosphate(from step 2)

10 �l of 1% glucose standard

20 �l of 1% inorganic phosphate standard

Keep the spots small by spotting about 3 �l ata time and allowing drying of the spots betweenapplications.

8. Develop each chromatogram in ascending fash-ion (see Fig. 6-8, Experiment 6), using freshlymixed solvent, which consists of 80 ml absolutemethanol, 15 ml 88% formic acid, and 5 mlH2O. (If time permits, you can improve theseparation by suspending the chromatogram inthe chromatography jar with a fine wire for 2 hrbefore allowing the paper to dip into the sol-vent.) Since the chromatography requires about3 hr for development, you should carry out steps 12and 13 during this time and complete Steps 9through 11 afterward.

9. When the chromatograms have developed, drythem, and spray one of them with aniline–acid–oxalate spray to detect sugars. Spray the chro-matogram with the reagent until it is fullydampened, but avoid excessive application ofthe spray, as this may cause diffusion of thespots on the paper. Heat the paper at 100 to105°C for 5 to 15 min to develop the coloredspots. Note that this spray reagent requires re-ducing sugars to develop color. Explain how itcan be used to detect glucose-1-phosphate,which is not a reducing sugar.

10. Subject the other dried chromatogram to oneof the following phosphate-detection proce-


dures. Either the modified Hanes–Ischerwoodprocedure or the iron-sulfosalicylic acid com-plex method may be used to detect organicphosphate compounds on chromatograms.

a. Modified Hanes–Ischerwood Spray Procedure.Spray the chromatogram lightly with freshlyprepared modified Hanes–Isherwood reagent(25 ml 4% (NH4)6Mo7O24 � 4 H2O, 10 ml 1N HCl, 5 ml 70% perchloric acid, 60 mlH2O). Dry the paper in a 100°C oven for afew minutes, but do not allow the paper todarken. The paper will become fragile, sohandle it very carefully. Irradiate the paperwith a UV lamp held at a distance of 10 cmfor 1 to 10 min. Phosphate-containing com-pounds will appear as blue spots on a lightbackground.

UV will damage eyes: wear glasses and avoid di-

rect irradiation.

b. Iron–Sulfosalicylic Acid Detection of Phosphates.Dip the chromatogram paper containingmore than 0.05 �mol phosphate per spot ina bath of acidic FeCl3 in acetone and im-mediately hang it up in a fume hood to dry.When dry, dip the paper in 1.25% sulfosal-icylic acid in acetone and again immediatelyhang it up in a fume hood to dry. Organicphosphates and inorganic phosphate appearon the dry chromatogram as white spots ina red-brown field.

11. Determine the Rf values for the various stan-dard and experimental spots on both chro-matograms and record them in a table. Whatconclusions can you draw from your observa-tions?

12. Nelson’s Test for Equivalents of Reducing Sugar.(NOTE: A simple and specific alternativemethod to assay the glucose present in tubes 1,1, 2, and 2 is the enzymatic assay for glucosedescribed in Experiment 11. This assay is spe-cific for glucose (i.e., will not detect glucose-6-phosphate or other reducing sugars) and is op-timal in the range from 0.01 to 0.10 �mol ofglucose. If you choose this option, follow thesteps in the section on “Enzymatic Determi-nation of Glucose” in Experiment 11, except

that 100- and 200-�l samples from tubes 1 and2 should be assayed, and equivalent volumes ofH2O should be added to the analysis of the glu-cose standard. In calculating the amount of glu-cose determined, remember that the final vol-ume in the assay will be 1.1 or 1.2 ml.) Analyzetubes 1 and 1 (untreated) and tubes 2 and 2(hydrolyzed 7 min) for their reducing sugarcontent as follows: Prepare a blank sample con-taining 2 ml of H2O and standards containing0.1, 0.2, 0.4, 0.6, and 0.8 �mol of glucose in 2-ml final volume (i.e., 0.1, 0.2, 0.4, 0.6, and0.8 ml of a 1.0 mM glucose standard plus 1.9,1.8, 1.6, 1.4, and 1.2 ml of H2O, respectively).Mix 0.5 ml of Nelson’s reagent B with 12.5 mlof Nelson’s reagent A. Add 1 ml of the com-bined reagent to each tube (i.e., the blank, thefive standards, and tubes 1, 1, 2, and 2). Placethe tubes simultaneously in a vigorously boil-ing water bath (500-ml beaker or larger), andheat for exactly 20 min. Remove the tubes si-multaneously and place them in a beaker ofcold water to cool. When the tubes are cool(25°C), add 1 ml of arsenomolybdate reagentto each and shake well occasionally during a5 min period to dissolve the precipitated Cu2Oand to reduce the arsenomolybdate. Dilute thecontents of each tube to a final volume of 10ml with H2O. Read the absorbance of all of thesolutions at 540 nm. Calculate the number ofmicromoles of glucose-reducing equivalents intubes 1, 1, 2, and 2 from your glucose stan-dard curve.

13. Determination of Inorganic Phosphate. Using themodified Fiske–Subbarow method as describedabove, assay the following for inorganic phos-phate: tubes 3 and 3 (no hydrolysis), tubes 4and 4 (7-min hydrolysis), tubes 5 and 5 (totalhydrolysis), five standards (0.1, 0.3, 0.5, 0.7, and0.9 ml of 1.0 mM standard phosphate plus 1.9,1.7, 1.5, 1.3, and 1.1 ml H2O, respectively), anda 2.0-ml H2O blank. Add 1 ml of acid molyb-date reagent and 1 ml of reducing reagent toall of the tubes, and bring the final volume ofeach tube to 10 ml with H2O. Allow them tostand for 20 min, then read the absorbance ofeach solution at 660 nm against the blank. De-termine the number of micromoles of inor-ganic phosphate in tubes 3, 3, 4, 4, 5, and 5from your standard curve.


Table 12-2 Analysis of Glucose-1-Phosphate

Isolated from Potato

Phosphorylase Reaction

�mol of Reducing �mol of Inorganic

Equivalents/0.1 ml Phosphate/0.1 ml

Sample Theory Observed Theory Observed

Initial sample

(�g/0.1 ml)

7-min hydrolyzed

sample

Total hydrolysis

sample

Report of Results for Day 4

1. Prepare a table similar to Table 12-2. From theweight of isolated glucose-1-phosphate dis-solved in H2O for analysis, calculate the theo-retical values for micromoles per milliliter re-ducing equivalents and inorganic phosphate ineach sample, assuming a molecular weight of372 and 100% purity. Then, determine the ob-served values from your analyses in steps 12and 13. In all cases, report the values as mi-cromoles per milliliter of the original glucose-1-phosphate solution analyzed.

2. From the analysis of tubes 3 and 3 in step 13,determine how many micromoles per milli-liter of inorganic phosphate contaminate yoursample. How many micromoles per milliliterof glucose-6-phosphate contaminate yoursample? (You have two independent determina-tions of this value. What are they? Hint: Onedetermination uses the determinations withtubes 1 and 1 in step 12, and the other re-quires you to compare the determinations intubes 4 and 4 to those of tubes 5 and 5 instep 13.) What do you assume in using thesevalues as assays for glucose-6-phosphate? Dothe two determinations agree? If not, suggestpossible reasons why not. (Note that if youused the enzymatic determination of glucose,you have only a single determination of glu-cose-6-phosphate. Explain why.) Do yourchromatographic results agree with the con-clusions drawn from the chemical analyses?

3. Determine the extent of hydration of your iso-lated compound by performing the followingoperations:

a. Calculate the percentage of 7-min phospho-rus (not phosphate) in your sample (e.g., mi-crograms of phosphorus per 100-�g sam-ple), assuming that the sample is pure.

b. Calculate the expected percentage of 7-minphosphorus from dipotassium glucose-1-phosphate (no water of hydration).

c. Calculate the expected percentage of 7-minphosphorus from dipotassium glucose-1-phosphate dihydrate.

d. Compare your value for the percentage ofphosphorus (a) with the two theoretical val-ues (b and c), and decide which formula bestfits your data. Comment on the reliabilityof your conclusion.

4. Weigh and label the remaining glucose-1-phosphate, and turn it in to the instructor.

Exercises

1. Consider the following set of data for inorganicphosphate determinations on unhydrolyzedand 7-min hydrolyzed aliquots taken from a700-ml volume of the MgNH4PO4 supernatantobtained in this experiment.

Absorbance at 660 nm

Hydrolyzed for

Volume of Sample Unhydrolyzed 7 min

0.1 ml 0.004 0.225

0.2 ml 0.010 0.435

0.5 ml 0.022 0.860

0.4 �mol Pi standard 0.150

0.8 �mol Pi standard 0.300

a. Which absorbance values can be used forfurther calculations?

b. Assume 100% recovery of the glucose-1-phosphate and inorganic phosphate at thebeginning of the anion exchange step. Howmany milliequivalents of anion exchangeresin must be used to adsorb a solution ofthese anions at pH 3.0? (pKal 1, andpKa2 6 for glucose-1-phosphate.)


c. Assume 80% recovery of the glucose-1-phosphate as crystals. How many grams ofdipotassium glucose-1-phosphate dihydratewould be isolated?

2. Propose a series of steps using ion-exchangeresins for the isolation of glucose-6-phosphatefrom a solution containing methylamine, sodiumacetate, glucose, and glucose-6-phosphate.

3. What steps would you perform to convert therather insoluble barium salt of glucose-6-phosphate to the dipotassium salt?

4. Describe chemical tests for determiningwhether a given pure sample of unknown is (a) glucose-6-phosphate, (b) glucose-1-phosphate, (c) �-methylglucoside, (d) glucose,(e) fructose-1,6-diphosphate, or (f ) sorbitol.

5. Which of the following compounds yield inor-ganic phosphate on 7-min hydrolysis: acetylphosphate, 3-phosphoglyceraldehyde, ribose-5-phosphate, adenosine-5-phosphate, ribose-1-phosphate, and pyrophosphate?

6. Point out the chemical similarities betweenNelson’s test and the Fiske–Subbarow test forphosphate, as well as between these tests andthe Folin–Ciocalteau (Lowry) protein deter-mination.

REFERENCES

Bandurski, R. S., and Axelrod, B. (1951). The Chro-matographic Identification of Some BiologicallyImportant Phosphate Esters. J Biol Chem 193:405.

Cori, C. F., Colowick, S. P., and Cori, G. T. (1937).The Isolation and Synthesis of Glucose-1-Phosphoric Acid. J Biol Chem. 121:465.

Fiske, C. H., and Subbarow, Y. (1925). The Colorimet-ric Determination of Phosphorus. J Biol Chem66:375.

Hanes, C. S., and Isherwood, F. A. (1949). Separationof the Phosphoric Esters. Nature 164:1107.

McCready, R. M., and Hassid, W. Z. (1957). Prepara-tion of �-D-Glucose-1-Phosphate by Means ofPotato Phosphorylase. Methods Enzymol 3:137.

Nelson, N. (1944). A Photometric Adaptation of theSomogyi Method for the Determination of Glu-cose. J Biol Chem 153:375.

Runeckles, V. C. and Krotkov, G. (1957). The Separa-tion of Phosphate Esters and Other Metabolites.Arch Biochem Biophys 70:442.

(F

217

EXPERIMENT 13

Isolation and Characterization ofErythrocyte Membranes

Theory

In modern biochemical research it is often neces-sary to isolate and characterize biological mem-branes. Erythrocytes (red blood cells) are a popu-lar cell type for these studies since they areabundant, they contain no organelles, they conductrelatively little metabolism, and they offer a homo-geneous source of easily obtainable membranes.This experiment features a three-stage isolation ofplasma membranes from porcine (pig) erythrocytes.First, the red blood cells will be separated from theplasma by low-speed centrifugation and subsequentwashing. Second, the isolated red blood cells willbe lysed by osmotic shock following transfer froman isotonic to a hypotonic buffer. In the final stage,the membranes will be isolated following a seriesof washes to remove hemoglobin and other cyto-plasmic contaminants. After the erythrocyte mem-branes have been purified, you will perform a se-ries of assays to characterize and quantify thedifferent protein and lipid components present inthese membranes.

Remember that biological membranes containa great number of different types of proteins, whichtypically account for about 60% of their total mass.There are about 20 prominent proteins found inerythrocyte membranes. The chloride–bicarbonateanion exchanger accounts for about 30% of the to-tal protein content of the red blood cell membrane.As carbonic anhydrase converts CO2 to the morewater-soluble form of HCO3

�, this membrane pro-tein will transfer the ion across the erythrocytemembrane. At the same time, chloride ions aretransported in the opposite direction to prevent the

formation of an electrochemical gradient. Anotherprotein that is abundant in the erythrocyte mem-brane is the Na�/K� pump. This transport proteinuses the energy of ATP to import K� ions and ex-port Na� ions. The net effect of this transport sys-tem is to establish an electrochemical gradient thatcan act as an energy source to drive the transportof other molecules against a concentration gradi-ent. Glycophorin is another protein abundant inerythrocyte membranes. This highly charged trans-membrane glycoprotein is the target (docking site)of the influenza virus. It is believed that this pro-tein plays a role in preventing the aggregation ofred blood cells in narrow capillaries. Finally, redblood cells contain a significant concentration of atransport protein called glucose permease. Thistransport protein spans the membrane 12 times andallows passive transport of glucose into the cell ata rate 50,000 times faster than would occur throughthe lipid bilayer in the absence of this transport pro-tein.

All of the proteins just described are examplesof integral membrane proteins. As such, they arefirmly embedded in the phospholipid bilayer andactually span the membrane; one side of the pro-tein faces the cytoplasmic side of the membrane,the other side faces the outer surface of the cell.The erythrocyte membrane also contains a numberof peripheral-membrane proteins. Unlike the membrane-spanning integral-membrane proteins,these peripheral-membrane proteins are tightly as-sociated only with the cytoplasmic side of the phospholipid bilayer. Together, these peripheral-membrane proteins form a meshlike matrix orskeleton on the inner surface of the membrane that


gives the erythrocyte its distinctive and unusual bi-concave (disklike) shape.

The majority of this matrix is composed of a pro-tein called spectrin, a heterodimeric protein con-taining a 220-kDa � subunit and a similar but slightlylarger � subunit. The highly repetitive amino acidsequences of both the � and � subunits give them afilamentous three-dimensional structure. As the �and � subunits of spectrin associate with one an-other, they form the flexible monomeric units thatare used to create the membrane skeleton.

Two additional peripheral membrane proteinsanchor the spectrin filaments to the cytoplasmicside of the erythrocyte membrane. One of thesepolypeptides, the 210-kDa ankyrin protein, bindsboth a single spectrin molecule and the chloride–bicarbonate anion-exchange protein discussed pre-viously. The second of these polypeptides, actin, iscapable of binding several molecules of spectrin.Since actin is able to associate with more than a sin-gle spectrin monomer, it acts as a branch point forthe spectrin protein as the membrane skeleton ormatrix is assembled (see Fig. 13-1). In this experi-ment, you will determine the concentration of to-tal protein in the erythrocyte membrane throughthe use of the Folin–Ciocalteau assay.

Lipids account for about 40% of the total massof the typical mammalian cell membrane. One classof lipids, the sterols, acts as the precursor to steroid

hormones, affects membrane fluidity, and consti-tutes a major component of bile salts. In this ex-periment, you will quantify the amount of choles-terol (a sterol) using a coupled enzymatic assaycontaining cholesterol esterase, cholesterol oxidase,peroxidase, and a chromophore that absorbs lightat 500 nm in the oxidized form (see Fig. 13-2).

Phospholipids, a second class of lipids, are the major components of the lipid bilayer of biological membranes. In this experiment, you will separate and identify phosphatidylcholine, phosphatidylethanolamine, phosphatidylserine, andphosphatidylinositol using thin-layer chromatogra-phy. Cholesterol and sphingomyelin (a sphingolipid)will also be identified in this part of the experiment.In addition, you will quantify the amount of totalphospholipid present in the erythrocyte membraneusing a colorimetric assay that produces a reducedphosphomolybdate complex that absorbs light at660 nm (see Fig. 13-3). Note the relationship be-tween the phosphate assay and the Folin–Ciocalteauprotein assay (Experiment 1). What is the reducingagent in each case? What is the limiting reagent ineach assay? A flow diagram for this three-period ex-periment is presented in Figure 13-4 to help guideyou through the experiment.


Pig blood (if you do not have access to blood from thissource, out of date whole human blood from ablood bank can also be used)

Isotonic phosphate buffer (310 mM sodium phosphate)

Hypotonic phosphate buffer (20 mM sodium phosphate)

250-ml plastic centrifuge bottles

30- to 50-ml plastic centrifuge tubes with screw caps

P-20, P-200, and P-1000 Pipetmen with disposable tips

15-ml plastic conical tubes with screw caps

Microcentrifuge tubes

Heparin sulfate

Pasteur pipettes

Reagents for Folin–Ciocalteau protein assay:

2% sodium tartrate solution

1% CuSO4 � 5 H2O solution

2% Na2CO3 in 0.1 N NaOH

Folin–Ciocalteau Reagent (2 N)

Lysozyme solution (1 mg/ml) in water

4-ml glass vials with Teflon-lined screw caps

Chloroform

Junctional complex(actin)

Spectrin

Ankyrin

Plasmamembrane

Chloride–bicarbonateexchange proteins

Outside

Inside

Figure 13-1 The membrane skeleton of erythro-

cytes.

EXPERIMENT 13 Isolation and Characterization of Erythrocyte Membranes 219

R

O

CO

H2O

CH3

CH3 NH2

CH3

CH3

�

2 H2O2 � � �

OH

R C

O

OHCH3

CH3

�

cholesterol

esterase

H2O2�

Fatty acidCholesterol

Cholesterol ester

HO

O2

CH3

CH3

�

O

CH3

CH3

cholesterol

oxidase

peroxidase

Cholest-4-ene-3-one

Cholesterol

O

OH

�O

OO

NN

S4-Aminoantipyrene

CH3

CH3 N

ONN

p-Hydroxybenzenesulfonate

O

Quinoneimine dye � A500

HSO3� � 4 H2O

Figure 13-2 Quantitative assay for total cholesterol.


Figure 13-3 Quantitative phosphate assay to quantify phopholipids.

CH

O

COO

CR O O

PO O

�O

H2C

H2C

HC

OH

OH

OH

H2C

H2C

O

CR OH

R

X

X

Phospholipid

Perchloric acid

(160�C)Fatty acids

� �PO4�2

�

�PO4�2

Glycerol

Head group

molybdate Phosphomolybdate complex(oxidized)

Phosphomolybdate complex(reduced � A660)

Phosphomolybdate complex(oxidized)

H2N OH

NH2

2,4-Diaminophenol(reducing agent)

2 �

Methanol

Glacial acetic acid

Silica gel thin-layer chromatography plates (Fisher cata-log #05–713–317)

16 125-mm glass test tubes

Phospholipid standard solutions in CHCl3 (10–50 mg/mlof sphingomyelin, phosphatidylcholine, phos-phatidylethanolamine, phosphatidylserine, phos-phatidylinositol, and cholesterol)

I2 chamber

Cholesterol standard solutions (1 mg/ml, 2 mg/ml,4 mg/ml) in isopropanol

Cholesterol reagent (buffered to pH 6.5) (This reagentis supplied in a mixed dry form by Sigma Chemi-cal Company. It is convenient for use with largelaboratory groups.)

Cholesterol esterase (0.1 unit/ml)

Cholesterol oxidase (0.3 unit/ml)

Peroxidase (1 unit/ml)

Chromophore (4-aminoantipyrine, 0.3 mM)

p-Hydroxybenzenesulfonate (30 mM)

Phosphate standard solution (1 mol/ml)

Pressurized N2 tank

Perchloric acid (70%)

5% ammonium molybdate solution in water

2,4-diaminophenol reagent (1 g of sodium bisulfite and40 mg of 2,4-diaminophenol per 4 ml of water)

Heating block

Preparative centrifuge

Clinical tabletop centrifuge (low speed)

Spectrophotometer to read absorbance at 500 and660 nm with cuvettes

100°C oven

Protocol

Day 1: Isolation of Erythrocyte Membranes

and Lipid Extraction.

We suggest that steps 1 to 6 be performed by theinstructor prior to the beginning of the exercise.

Isolation of Erythrocyte Membranes

1. Obtain 3 liters of fresh pig blood (see “Suppliesand Reagents” section for an alternative supplyof blood, if necessary) and immediatelymix withheparin sulfate (30 mg/liter final concentration)to prevent clotting. Place the blood in an icebucket and proceed. All additional steps throughstep 6 should be performed at 4°C. We suggestthat steps 1 to 6 be performed by the instruc-tor prior to the beginning of the experiment.If you plan to perform Experiment 16 (Exper-iments in “Clinical Biochemistry and Metabo-lism”) in conjunction with this experiment, a


Whole blood (30 ml)

• Centrifuge at low speed• Wash twice in isotonic buffer• Discard supernatant

Erythrocytes (5 ml)

Lipids

Cholesterol assay(quantitative)

Phosphate assay(quantitate phospholipids)

Thin-layer chromatography(identify lipids)

• Lyse in hypotonic buffer• Centrifuge at high speed (collect pellet)• Wash three times in hypotonic buffer• Discard supernatant

Membranes (2–4 ml)

• Add chloroform:methanol:water• Centrifuge at low speed• Save organic phase• Evaporate lipids to dryness and record mass• Resuspend dried lipids in chloroform

Folin–Ciocalteau assay(determine protein concentration)

Figure 13-4 Flow diagram for erythrocyte membrane isolation and characterization.

portion of the pig blood can be centrifuged at400 g for 10 min to obtain the plasma (nophosphate buffer added).

2. Add 32 ml of isotonic phosphate buffer to anumber of 250-ml plastic centrifuge bottles.Add 180 ml of whole pig blood to each bottle,cap, and invert several times to mix.

3. Centrifuge at 1000 g for 10 min. Using aPasteur pipette, carefully remove the plasmaand “buffy coat” from the pelleted red bloodcells.

4. Gently resuspend the erythrocyte pellet fromeach bottle in 120 ml of isotonic phosphatebuffer. Centrifuge at 1000 g for 10 min andagain remove the supernatant.

5. Repeat step 4 one more time.6. Resuspend the red blood cell pellet from each

centrifuge bottle in isotonic buffer to a finalvolume of 30 ml. If these erythrocytes are pre-pared in advance, they can be stored on ice at

4°C for several days. DO NOT FREEZE THECELLS AT THIS POINT. Each 30 ml suspen-sion of erythrocytes (obtained from 180 ml of wholeblood) will be enough material for six groups of stu-dents to perform the experiment. We always collectextra blood (3 liters total) in case problems arise inthe preparation of the erythrocytes.

7. Add 5 ml of the erythrocyte suspension to 25 mlof hypotonic phosphate buffer in a 50-ml plasticcentrifuge tube. Cap the tube and invert severaltimes to mix (do not use a Vortex mixer).

8. Centrifuge at 20,000 g for 40 min. Themembrane pellet at the bottom of the tube willbe dark red in color and loosely packed. Care-fully remove the supernatant (hemoglobin, cy-toplasmic contaminants) with a Pasteur pipette.

9. Gently resuspend the membrane pellet in thecentrifuge tube in 25 ml of hypotonic phos-phate buffer. Mix well by inversion and cen-trifuge at 20,000 g for 20 min. Carefully


remove and discard the supernatant above themembrane pellet with a Pasteur pipette.

10. Repeat step 9 two more times. Note the changein color of both the membrane pellet and thesupernatant with each wash.

11. After the last wash step, remove as much of thesupernatant as possible and gently agitate thetube to resuspend the loosely packed, milky-looking membrane pellet in the residual bufferat the bottom of the tube. At this point, youshould have 2 to 4 ml of concentrated mem-brane solution. Transfer this solution to a 15-ml plastic conical tube with a screw cap. Labelthe tube with your name and record the exactvolume of the membrane solution in your note-book. This membrane solution can be storedat 4°C.

Lipid Extraction

1. Transfer 0.8 ml of the membrane suspensioninto two separate 15-ml plastic conical tubeswith screw caps. Avoid polystyrene and any othertypes of plastics that are not compatible with this sol-vent. Add 3 ml of chloroform:methanol (1�2)to each tube. Cap and swirl each tube vigor-ously with a Vortex mixer for 30 sec.

2. Add 1 ml of chloroform to each tube, cap thetube, and swirl vigorously for 30 sec.

3. Add 1 ml of water to each tube, cap the tube,and swirl vigorously for 30 sec.

4. Separate the methanol:water phase (top) fromthe chloroform phase (bottom) in each tube bycentrifugation at low speed (�500 g, 5 min)in a tabletop centrifuge. You will see a smalldisk of white precipitate (denatured protein) atthe interface between these two phases. If thisdisk extends down too far into the chloroformlayer (more than halfway into the chloroformphase), add 0.1 ml of 0.1 M HCl, mix, and re-peat the last low-speed centrifugation step.

5. Remove as much of the upper methanol:waterphase as possible with a Pasteur pipette and dis-card.

6. With a clean Pasteur pipette, carefully punc-ture through the denatured protein disk and re-move the chloroform layer at the bottom ofeach tube. Transfer the chloroform layer fromboth tubes to a single, preweighed 4-ml glass vialwith a Teflon-lined screw cap. Take care not to

transfer any of the denatured protein layer (whiteprecipitate) to these glass vials.

7. Evaporate the combined chloroform phasecontaining the erythrocyte membrane lipids todryness by passing a stream of N2 over the topof the vial in a fume hood (see Experiment 6,“Sequence Determination of a Dipeptide” Fig.6-7).

8. When the lipids have been dried completely,weigh the vial again and determine the mass oflipid that you have extracted from the erythro-cyte membrane. Record the mass of lipids in yournotebook.

9. Based on the mass of lipid present in 1.6 ml ofthe membrane solution, determine the numberof milligrams of lipid present in the entire ery-throcyte membrane solution (determined instep 11 of “Isolation of Erythrocyte Mem-branes”).

10. Resuspend your dried lipids in 1 ml of chloro-form. Cap the vial tightly (chloroform is veryvolatile), label it with your name, and store itat room temperature.

Day 2: Thin-Layer Chromatography of

Lipids, Quantitative Cholesterol Assay,

Folin–Ciocalteau Assay, and Preparation

for Phosphate Assay

Thin-Layer Chromatography of Lipids

1. Obtain a silica-gel thin-layer chromatographyplate that has been activated in a 100°C ovenfor 5 to 10 min. On either side of the plate,make a small scratch in the silica matrix about2 cm from the bottom of the plate. This willmark the position of the origin, along whichthe samples will be spotted. Do not mark onthe plate with pencil.

2. Remove 0.3 ml of your extracted lipids in chlo-roform prepared on Day 1 and place in a 1.5-ml microcentrifuge tube. Dry the sample un-der a stream of N2 and resuspend the lipids in40 l of chloroform.

3. Using your P-20 Pipetman or a thin capillarytube, apply a total of 10 l of this lipid solu-tion on the origin. Keep the area of applicationsmall by repeated spotting and drying of 2-laliquots. Try to keep each sample spot as con-


centrated as possible. This will give you a muchless diffuse lipid sample following develop-ment, which will make the identification of thedifferent phospholipids much easier.

4. Using the same technique, spot 10 l of eachof the lipid standard solutions in chloroform(phosphatidylcholine, phosphatidylserine,phosphatidylethanolamine, phosphatidylinosi-tol, cholesterol, and sphingomyelin) on differ-ent positions along the origin. Spot an addi-tional 20 l of your extracted lipid sample usingthe same technique (in case the concentrationof lipid in your prepared sample is low).

5. When all the spots have dried, place the plate(origin side toward the bottom) in an ascend-ing chromatography tank containing 1 cm ofmobile-phase solvent on the bottom of thechamber. The mobile phase for this experimentis chloroform:methanol:acetic acid:water(25�15�4�1). Be sure that the mobile phase isin contact with the plate, but that the originalong which your samples are spotted lies abovethe level of the mobile phase (See Fig. 6-8).While your thin-layer chromatography plate is de-veloping, you may complete the quantitative choles-terol assay and the Folin–Ciocalteau protein assay.

6. Allow the mobile phase to travel through (up)the stationary phase until the solvent front isabout 1 in. from the top of the plate (�1.5–2.0 hr). Remove the plate from the tank, markthe position of the solvent front with a pencil,and allow the plate to air dry in the fume hoodfor about 5 min.

7. Place the plate in a sealed chamber saturatedwith I2 vapor and incubate for 5 min. The I2 willreact with unsaturated carbon bonds present in thelipids and produce yellow spots on the plate.

8. Remove the plate from the I2 chamber and cir-cle the position of the spots with a pencil (theyellow spots will disappear in about 1 hr).

9. Measure the Rf values of the lipid standards:

Rf �

Based on the Rf values of the lipids present inyour sample, what lipids appear to be presentin the erythrocyte membrane? Based on thesize and intensity of the spots present in yoursample compared to those of the lipid stan-dards, which lipid(s) are most abundant in thepig erythrocyte membrane?

Quantitative Cholesterol Assay

1. Set up the reactions described in the table be-low in 1.5-ml microcentrifuge tubes: The ex-tracted lipids that are currently dissolved in chloro-form cannot be used directly in this assay, since thechloroform will denature the enzymes that will even-tually produce a colored product (see Fig. 13-1). Toprepare the lipids in your sample for this assay, addthe indicated volume of lipids in CHCl3 to tubes 4and 5, dry them under a stream of N2, and resus-pend each sample in 10 l of isopropanol. The stan-dard cholesterol solutions (tubes 1–3) can be used di-rectly in this assay by adding the indicated volumesto the appropriate tubes. (See chart below.)

2. Incubate the tubes at room temperature for15 min.

3. Read the absorbance of the solution in eachtube at 500 nm within 30 min of the comple-tion of the 15-min incubation. Use the solution in the blank tube to zero your spectrophotometer at 500 nm. Record the ab-sorbance values in your notebook.

4. Prepare a standard curve by plotting A500 ver-sus milligrams of cholesterol for tubes 1 to 3.

5. Based on the absorbance readings of tubes 4and 5 at 500 nm, determine the mass of cho-lesterol (in milligrams) present in these tubes.

Distance traveled by sample spot (cm)��Distance traveled by solvent front (cm)

Component Blank Tube 1 Tube 2 Tube 3 Tube 4 Tube 5

Water 100 l 90 l 90 l 90 l 90 l 90 l

1 mg/ml cholesterol std. — 10 l — — — —

2 mg/ml cholesterol std. — — 10 l — — —

4 mg/ml cholesterol std. — — — 10 l — —

Lipids (in CHCl3) — — — — 20 l 100 l

Cholesterol reagent 1.00 ml 1.00 ml 1.00 ml 1.00 ml 1.00 ml 1.00 ml


6. Based on the volume of lipid sample (inCHCl3) present in tubes 4 and 5, determinethe concentration of cholesterol present inyour lipid sample in units of milligrams permilliliter.

7. Using the molecular weight of cholesterol(386.7), calculate the concentration of choles-terol in your entire membrane sample in unitsof micromoles per milliliter.

Folin–Ciocalteau Assay to Determine

Membrane Protein Concentration

1. Dilute your membrane solution prepared on Day1 (not the extracted lipid solution) by adding0.2 ml of the membrane solution to 0.8 ml ofhypotonic phosphate buffer. This diluted ery-throcyte membrane solution will be used in theFolin–Ciocalteau assay described below.

2. Set up the following reactions in 16 125-mmglass test tubes:

Volume of Volume of Volume of

Hypotonic 1 mg/ml Lysozyme Diluted Membrane

Tube Buffer (ml) Solution (�l) Solution (�l)

1 1.20 — —

2 1.18 20 —

3 1.15 50 —

4 1.10 100 —

5 1.00 200 —

6 0.90 300 —

7 1.15 — 50

8 1.10 — 100

9 0.90 — 300

3. Separately prepare 100 ml of fresh alkaline cop-per reagent by mixing, in order, 1 ml of 1%CuSO4 � 5H2O and 1 ml of 2% sodium tartrateinto 98 ml of 2% Na2CO3 in 1 N NaOH.

4. Add 6 ml of this alkaline copper reagent toeach of the nine tubes and mix immediatelyafter each addition to avoid precipitation. In-cubate the tubes at room temperature for10 min.

5. Add (and immediately mix in) 0.3 ml of Folin–Ciocalteau reagent to each of the nine tubes.Incubate for 30 min at room temperature.

6. Read the absorbance of the solution in eachtube at 500 nm. Record the absorbance values inyour notebook. Use the solution in tube 1 (blank)to zero your spectrophotometer at 500 nm.

7. Prepare a standard curve by plotting A500 ver-sus milligrams of protein for tubes 2 to 6.

8. Determine which of tubes 7 to 9 gave an A500

that lies in the linear portion of the standardcurve. What mass of protein (in milligrams) isindicated by its position on the standard curve?

9. Based on the volume of the membrane solutionpresent in tubes 7 to 9 (remember the dilutionat the beginning of the assay), calculate the con-centration of protein present in the membranesolution in units of milligrams per milliliter.

10. Based on the total volume of the entire mem-brane solution isolated on Day 1, how manymilligrams of protein are present in the totalmembrane fraction?

11. Using the total mass of lipid in the membranesolution (calculated at the end of Day 1), whatis the mass ratio of protein to lipid in the ery-throcyte membrane?

Preparation for Phosphate Assay

The glass vials used in this part of the experiment shouldbe cleaned with a 0.5% solution of sodium dodecyl sul-fate (SDS), since many commercially available deter-gents contain a significant concentration of phosphate.

1. Transfer 20 l, 50 l, and 100 l of your ex-tracted lipid sample in CHCl3 to three 4-mlglass vials with Teflon-lined screw caps. Trans-fer 0.30 ml of a 1 mol/ml phosphate standardsolution to a fourth 4-ml glass vial.

2. Dry each of the samples under a stream of N2

in the fume hood.3. Add 0.5 ml of 70% perchloric acid to each vial

and resuspend each of the dried lipid sampleswith gentle agitation.

4. Cap the vials tightly with a Teflon-lined screwcap and incubate in a 160°C oven or heat blockovernight (�12 hr). After cooling to room tem-perature, these solutions will be used on Day 3. The 1 mol/ml phosphate solution in per-chloric acid does not have to be heated overnight.This standard solution can simply be stored at roomtemperature for use on Day 3.


Day 3: Completion of Phosphate Assay

1. Prepare a series of phosphate standard solu-tions in 4-ml glass vials using the 0.3-ml sam-ple of phosphate (1 mol/ml) prepared on Day 2.

Volume of Phosphate Volume of Perchloric

Tube Solution in Perchloric Acid (�l) Acid (ml)

1 — 0.50

2 20 0.48

3 50 0.45

4 100 0.40

5 150 0.35

2. Add and mix the following reagents, in order,to these five standard vials as well as to the threevials containing your lipid sample in perchlo-ric acid that were hydrolyzed on Day 2:

2.4 ml of water

0.1 ml of 5% ammonium molybdate in water

0.1 ml of 2,4-diaminophenol (amidol) reagent

3. Seal all eight vials with a Teflon-lined screw capand incubate at 110°C for 30 min in the heatblock.

4. Allow the vials to cool to room temperature andread the absorbance of all of the solutions at660 nm. Use the solution in vial 1 (blank) tozero your spectrophotometer.

5. Prepare a plot of A660 versus micromoles ofphosphate. Based on the A660 of your threelipid samples, how many micromoles of phos-phate are present in each vial?

6. Based on the fact that there is a 1�1 ratio ofphosphate to phospholipid, and considering thetotal volume of the lipid fraction recovered onDay 1, how many micromoles of phospholipidare present in your extracted lipid sample?

7. Based on the average molecular weight of thefour phospholipids (�780 g/mol), how manymilligrams of phospholipid are present in yourtotal extracted lipid sample?

8. Based on the number of milligrams and thenumber of micromoles of cholesterol presentin your total extracted lipid sample (calculatedon Day 2), what is the mass ratio and mole ra-tio of cholesterol:phospholipid in the erythro-cyte membrane?

Exercises

1. Prior to the beginning of this experiment, theerythrocytes were washed twice in isotonicphosphate buffer. What was the purpose ofthese washes?

2. The erythrocytes were then washed severaltimes in hypotonic phosphate buffer and sub-jected to centrifugation. What purpose did thehypotonic phosphate buffer serve in thesewashes? What components were present in thepellet and supernatant fractions following eachwash step?

3. State what components were present in the wa-ter:methanol phase and the chloroform phasefollowing the water:methanol:chloroform ex-traction of the erythrocyte membranes.

4. The lipid sample that was used in the choles-terol assay underwent a special preparationprocedure before it could be used in the assay.What was this procedure and why was it nec-essary?

5. List the four phospholipids that are commonlyfound in biological membranes. Explain howthe single assay performed in this experimentcould be used to quantify this entire group oflipids in the erythrocyte membrane.

6. Explain why the protein:lipid ratio may varybetween different cell types. Explain why theremay be variations in the relative amounts ofeach type of lipid from one cell type to another.

7. Why are erythrocytes commonly used as asource of membranes for biochemical analysis?

8. Analyze your thin-layer chromatography plate.What lipids displayed the lowest and highest Rfvalues? Explain the reason for this result interms of the chemical nature of the stationaryand mobile phases used in this experiment andthe nature of the different molecules being sep-arated in this experiment (which lipids wouldyou expect to show the greatest and the leastmobility on the plate and why).

REFERENCES

Allain, C. C., Poon, L. S., Chan, C. S. G., Richmond,W., and Fu, P. C. (1974). Enzymatic Determinationof Total Serum Cholesterol. Clin Chem 20:470.


Bennett, V. (1985). The Membrane Skeleton of HumanErythrocytes and Its Implications for More Com-plex Cells. Annu Rev Biochem 54:273.

Dittmer, J. C., and Wells, M. A. (1969). Quantitativeand Qualitative Analysis of Lipids and Lipid Com-ponents. Methods Enzymol 14:482.

Garrett, R. H., and Grisham, C. M. (1995). Lipids andMembranes. In: Biochemistry. Orlando, FL: Saunders.

Gurr, M. I., and Harwood, J. L. (1990). Lipid Biochem-istry. An Introduction, 4th ed. London: Chapman &Hall.

Kates, M. (1986). Techniques of Lipidology: Isolation,Analysis, and Identification, 2nd ed. (Laboratory

Techniques in Biochemistry and Molecular Biology,Vol. 3.) New York: Elsevier Science Publishing Co.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Lipids. In: Principles of Biochemistry, 2nd ed.New York: Worth.

Op den Kamp, J. A. F. (1979). Lipid Asymmetry inMembranes. Annu Rev Biochem 48:47.

Skipski, V. P., and Barclay, M. (1969). Thin-LayerChromatography of Lipids. Methods Enzymol14:530.

Stryer, L. (1995). Membrane Structure and Dynamics.In: Biochemistry, 4th ed. New York: Freeman.

. v

227

EXPERIMENT 14

Electron Transport

Theory

Most of the metabolites that serve as energy sourcesfor aerobic organisms are broken down (catabo-lized) by various pathways to yield substrates for thecentral, energy-yielding citric acid cycle (alsoknown as the tricarboxylic acid cycle or the Krebscycle). The citric cycle is a sequence of enzymaticsteps whose net reaction consists of the oxidationof acetate to carbon dioxide.

(14-1)

CH3COOH � 2 O2 88n 2 CO2 � 2 H2O

Oxygen is not reduced directly during oxidation ofacetate. Rather, pairs of electrons and pairs of pro-tons are removed from the citric acid cycle sub-strates and transferred to intermediate redox carri-ers, as shown by the half reactions of Equation 14-2.

(14-2)

AH2 88n A � 2H� � 2 e�

2 e� � 2 H� � B 88n BH2

Here AH2 represents the reduced substrate, A is theoxidized (dehydrogenated) substrate, B is the oxi-dized carrier and BH2 is the reduced carrier. That is,the oxidation of the substrate A is coupled with thereduction of electron carrier B (Equation 14-3).

(14-3)

AH2 � B 88n A � BH2

Several electron carriers are found in nature; themost important examples include pyridine nu-

cleotides, flavins, quinones, and cytochromes. Elec-tron transport consists of a series of reactions inwhich electrons and protons are passed through acascade of these electron carriers.

Natural electron transport involves consecutivereduction and oxidation of a series of electron carri-ers, which must include electron transfer, but doesnot always include transfer of the protons. In thesecases, the protons become a part of the aqueousmedium (e.g., in the reduction of ferric iron associ-ated with a cytochrome, as shown in Equation 14-4).

(14-4)

CH2 � 2 Fe3�88n C � 2 Fe2� � 2 H�

(in protein) (in protein)

Eventually, the electrons are passed to molecularoxygen, and protons are reclaimed from the aque-ous medium through the formation of water (halfreaction shown in Equation 14-5).

(14-5)

1/2 O2 � 2 H� � 2 e�88n H2O

This final reaction accounts for most of the knownoxygen consumption by aerobic organisms. Thecascade of redox reactions that couples the oxida-tion of organic substrates to reduction of molecu-lar oxygen in biological systems is called electrontransport, and is often presented schematically asshown in Figure 14-1. When substrates are oxidizedby such a system, the rate and extent of substrateoxidation is directly dependent on, and can be mea-sured by, the decrease in concentration of molecu-lar oxygen, as will be done in this experiment.


AH2

A

B

BH2

D

DH2C

CH2

Fe3�

Fe2�

Fe3� H2O

Fe2�

Fe2�

Fe3�2H� 2H�

1O22–

Carriers transferringboth e� and H�

Carriers transferringonly e�

Figure 14-1 Schematic representation of electron transport.

The general features of electron transport aresimilar throughout nature. The processes take placein highly structured environments within the cellmembranes of bacteria and within specialized sub-cellular particles—the mitochondria—of higherplants and animals. The electron carrier moleculesare also similar throughout nature: the diphospho-pyridine nucleotide NAD�, proteins that containflavin (FMN and/or FAD) and iron–sulfur (Fe-S)clusters, quinones that are soluble in the lipid com-ponent of membranes, and several heme-contain-ing proteins called cytochromes. Each of these elec-tron-carrying molecules has a characteristic opticalabsorption spectrum that differs between the oxi-dized and reduced states of the component. Youhave seen the absorption spectra of reduced and ox-idized NAD in Experiment 8 and the spectrum ofoxidized riboflavin, the chromophore of FAD, inExperiment 1. Several cytochromes are involved inthe electron transport scheme. These cytochromesall possess an iron-containing heme, which is Fe-protoporphyrin IX in the case of the various mem-bers of the cytochrome b family (Fig. 14-2). Theheme is covalently linked to the protein via addi-tion of cysteinyl residues to the vinyl groups of Fe-protoporphyrin IX in the cytochromes of the c class(Fig. 14-2). Further modifications of the heme leadto heme a, which is found in cytochromes of the aclass (Fig. 14-2). In all of the cytochromes, electrontransfer reactions involve the reversible oxidationand reduction of the heme Fe between the Fe2� andFe3� states. In their reduced forms cytochromesdemonstrate a characteristic three-peak heme ab-sorption spectrum with �, �, and � (or Soret) bandssimilar to those shown in Figure 14-3.

Biochemists use the changes in the ultraviolet-visible absorption spectra of the individual compo-nents of the electron transport chain to follow the

order of the steps in electron transport and the ratesof individual steps in the electron transport chain.This has been a particularly useful approach in an-alyzing the cytochrome components of the chain.Small differences in the heme structures of differ-ent cytochromes, together with major differencesin their protein components, create characteristicspectral differences among them. These spectraldifferences, as observed in the visible spectra of re-duced cytochromes, originally served as the primarytool of cytochrome classification (see Table 14-1).Functional assays also serve to classify cytochromes.For example, the protein conformation of most cy-tochromes buries the cytochrome’s heme within theprotein and makes it inaccessible to molecular oxy-gen. Those cytochromes therefore cannot reactwith O2 or its analogs (CO, CN�, and azide). Incontrast, the cytochromes of the terminal oxidasecontain a heme iron that is accessible to O2 and itsanalogs. Thus, cytochrome a3 of Table 14-1, for ex-ample, forms a spectrally distinct complex with CO,CN�, and azide, whereas the other cytochromes inTable 14-1 do not.

The components of the electron transport chainare organized into a series of very complex, highlyordered integral-membrane–multiprotein com-plexes. It is essential that the electron-transport ap-paratus be integrated into an enclosed membranesystem that enables the large negative free energyof the oxidation steps to be captured in the form ofan electrochemical gradient. That is, the electron-transport components are arranged so that the pro-tons which are released during oxidation of the hy-drogen-carrying electron carriers (see, for example,Equation 14-4) are released on the outside of themembrane, creating a gradient of protons. The re-verse flow of these protons to the interior of themembrane-enclosed cell or organelle discharges the

EXPERIMENT 14 Electron Transport 229

FeN N

N

NH3C

CH3

CH3

CH2CH2

CH2 CH3

CH2

CH2�OOC

COO�

CH2

HC

HC CH

CH

CHCH

FeN N

N

NH3C

CH3

CH3

CH3

CH2

CH2 CH3

CH2

CH2�OOC

COO�

CH3

HC

HC CH

CH

CH2S

CH2S

CHCH

Heme(Fe-protoporphyrin IX)

FeN N

N

NC

CH3

CH3

(CH2 CH2 CH CH2)3

CH2CH2

CH2 CH3

CH2

CH2�OOC

COO�

CH2

CH3

HC

HC CH

CH OH

CHCH

Heme a

Heme covalently linkedto cytochrome c

Protein

O

H

H

Figure 14-2 Structures of the heme prosthetic groups of the cytochromes. In all cytochromes the heme

Fe undergoes reversible oxidation and reduction between Fe2� and Fe3�.


Nanometers (nm)

400

% R

elat

ive

abso

rptio

n

500

Reduced cytochrome cOxidized cytochrome c

600

100

50

0

� or Soret peak or band

� Peakor band

� Peakor band

Figure 14-3 Absorption spectra of cytochrome c.

gradient and drives the synthesis of ATP from ADPand Pi by an integral membrane enzyme known asthe F0F1-ATPase or ATP synthase. Although elec-tron transport can occur in membrane fragmentswithout an enclosed membrane space, such a sys-tem cannot use oxidation reactions to generate ATP.Such a system is said to be uncoupled.

In higher plants and animals, electron transportoccurs within the mitochondria. Biochemists havedevised methods to isolate intact mitochondria con-taining all of the functional citric-acid-cycle en-zymes and electron-transport components. Such in-

tact mitochondria provide useful information aboutthe overall process. Yet, such intact mitochondriaare too complex to allow the intricate details of elec-tron transport to be studied. For this reason, theintegral-membrane–multienzyme complexes ofelectron transport have been carefully dissected intoseparate functional components that catalyze a por-tion of the overall electron-transport chain, andthese components have been characterized sepa-rately. For example, the electron transport chain ofmammalian mitochondria has been resolved intothree fundamental multienzyme complexes:NADH-ubiquinone reductase, cytochrome c re-ductase, and cytochrome c oxidase (see Fig. 14-4and Table 14-2). Two components of the electron-transport chain, ubiquinone (Fig. 14-5) and cy-tochrome c, are not tightly associated with thesecomplexes. Oxidation of the citric-acid-cycle inter-mediate succinate involves yet a fourth integralmembrane enzyme complex, succinate–ubiquinonereductase, which bypasses NADH–ubiquinone re-ductase and contributes reducing equivalents (elec-trons) from succinate oxidation to ubiquinone,which is dissolved in the membrane (Fig. 14-4 andTable 14-2). An exciting area of current research isthe determination of the detailed structure of themultienzyme complexes of electron transport; thestructure of the 13-subunit cytochrome c oxidasefrom bovine mitochondria has been determined atbetter than 3 Å resolution, for example.

Table 14-1 Absorption Bands (�max) of Cytochromes (position of

bands in nanometers, cytochromes in reduced state)

Absorption Maxima of CO Complex

Maxima of Pigment Formed on Addition of CO

Cytochrome Name of Band in Reduced Form to Reduced Cytochrome

a � 605 —

� 452 —

a3 � 600 590

� 445 432

b � 564 —

� 530 —

� 431 —

c � 550 —

� 520 —

� 415 —


NAD+–ubiquinonereductase

NADH

NAD+

Oxidizedsubstrates

Reducedsubstrates

Ubiquinone Cytochrome c

Ascorbate Dehydroascorbate

Cytochrome creductase

Cytochromeoxidase

Succinate-ubiquinonereductase

O2

2H2O

Succinate

Fumarate

Inhibition byamytal androtenone

Inhibition bymalonate

Inhibition byantimycin A

Inhibition byCN–, azide, andCO

DPIP

Methylene blue O2

Methylene blue

DPIP

O2

Figure 14-4 Schematic representation of electron transport in mitochondria.

Table 14-2 Multienzyme Complexes of the Electron-Transport Chain

in Mammalian Mitochondria

Number of

Complex Mass (kDa) Subunits Prosthetic Groups

NADH-ubiquinone reductase 880 34 FMN, Fe-S centers

Succinate-ubiquinone reductase 140 4 FAD, Fe-S centers

Cytochrome c reductase 250 10 Heme b-562

Heme b-566

Heme c1

Fe-S centers

Cytochrome oxidase 160 13 Heme a

Heme a3

CuA and CuB

SOURCE: Stryer, L. (1995). Biochemistry, 4th ed. p. 534 (Table 21-2), New York: Freeman.

Reprinted by permission.

The sum of a large number of studies with in-tact mitochondria, submitochondrial membranefractions, and purified electron-transport–multien-zyme complexes has led to the somewhat simpli-fied scheme for the functional organization of the

electron-transport system in mammalian mito-chondria shown in Figure 14-4. You should use thisscheme to aid in your interpretation of your resultsfrom this experiment. Note that different substratesare oxidized by different components of the system


CH C HCH2

CH3

OH

CH3O

CH3O

O�CH2

CH3

IO

CH C HCH2

CH3

O

CH3O

CH3O

OCH2

CH3

IO

CH C HCH2

CH3

OH

CH3O

CH3O

OHCH2

CH3

IO

Flavin-H2

Flavin

H� � e�

H� � e�

Figure 14-5 Oxidation–reduction reactions of ubiquinone.

and thus feed their reducing equivalents into thechain at different points. Most citric-acid-cyclecomponents are oxidized by soluble dehydroge-nases of the mitochondria matrix with the con-comitant reduction of NAD� to NADH. NADHis then oxidized by the NADH–ubiquinone reduc-tase. Succinate, on the other hand, is oxidized by amembrane-bound succinate dehydrogenase com-plex (a flavoprotein containing iron–sulfur centers),which reduces ubiquinone without involvement ofpyridine nucleotides. Ascorbate (vitamin C) is nota normal substrate of the electron-transport chain;it readily reduces cytochrome c, but not the elec-tron carriers upstream (such as flavoproteins,ubiquinone, or cytochrome c reductase), so it canbe used to study the properties of the portion ofthe electron-transport chain from cytochrome c toO2. As noted below, artificial electron-acceptingdyes can accept electrons from the flavin-contain-ing components of the electron transport chain (seeFig. 14-4) and can thus be used to study the por-tions of the chain upstream from the flavins. Fi-nally, various inhibitors have been shown to blockthe electron-transport chain at specific points; useof such inhibitors played an important role in de-veloping the scheme shown in Figure 14-4. For ex-ample, malonate is a structural analog of succinateand competitively inhibits succinate dehydroge-nase. CN�, azide ions, and CO, as noted above,bind specifically to the heme a3 of the terminal cy-tochrome c oxidase. The antibiotic antimycin Aspecifically inhibits cytochrome c reductase. The

use of such specific inhibitors is an important partof this experiment.

For less sophisticated studies, preparations ofelectron-transport components in mitochondrialmembranes are useful. Keilin–Hartree heart parti-cles, a preparation of mitochondrial fragments anddamaged mitochondria used in this experiment,provide a useful submitochondrial system for the as-say of many general properties of electron trans-port. These particles have lost most of the citric-acid-cycle enzymes, which are soluble in themitochondrial matrix and are washed away from themembrane fragments during purification. Keilin–Hartree particles are also not capable of oxidativephosphorylation, because the integrity of an en-closed membrane system has been lost. However,the membrane fragments still contain functionalsuccinate–ubiquinone reductase, ubiquinone, cy-tochrome c reductase, cytochrome c, and cyto-chrome c oxidase complexes. Thus, these particleswill catalyze oxidation of succinate and normal elec-tron transport to oxygen. In this experiment, someproperties of the electron-transport chain from suc-cinate to O2 will be characterized using Keilin–Hartree particles from pig heart mitochondria.

Two methods will be used to assay electrontransport catalyzed by these Keilin–Hartree parti-cles. In most of the experiments, an oxygen elec-trode will be used to measure the rate of oxygenconsumption. Oxygen electrodes are described inthe next paragraph. A spectrophotometric assay ofelectron transport will also be used. This assay re-


places O2 with an organic dye that acts as the ter-minal electron acceptor. Some reducible dyes, suchas methylene blue, are “autoxidizable.” That is, theyare reduced as the result of accepting electrons froma reduced electron transport component and they,in turn, pass on electrons by reducing O2 to H2O.Such autoxidizable dyes therefore provide a shunt,or second means of electron transfer to oxygen, butthey are not reduced in a stoichiometric manner.Because of this, they are of little use in aerobic spec-trophotometric assays (Fig. 14-6). In contrast, a dyethat is not autoxidizable, such as 2,6-dichlorophe-nol-indophenol (DPIP), will accept electrons froma biological electron transport system but will notpass these electrons on to O2. Thus, the reductionof DPIP, which is accompanied by a loss of visible

color, can be used to assay biological oxidation re-actions (Fig. 14-6).

Commercially available electrodes that measureO2 in aqueous systems are called oxygen electrodes,dissolved oxygen meters, or oxygen polarographs(although few of these systems involve true oxygenpolarography; i.e., a dropping Hg electrode undercontrolled diffusion conditions). All of the systemsare based on an oxygen detector called an oxygenelectrode similar to that illustrated in Figure 14-7.Oxygen electrode units contain two chambers—asample chamber and an electrode chamber—sepa-rated by a synthetic membrane. The electrodechamber is filled with an electrolyte system thatminimizes artifacts at the electrode surfaces. Themembrane is placed over the electrode chamber,

N

N

X

S

(oxidized)Blue

(oxidized)Blue

CH3

CH3

XH2 H2O

O2

Electrondonor N CH3

CH3

N

N O

Cl

Cl

O�

N S

(reduced)Colorless

CH3

CH3

N CH3

CH3

(reduced)Colorless

N

HCl

Cl

O� O�

H�

H

�

�

X

No reactionwith O2

XH2Electrondonor

Nonautoxidizable DPIP

Autoxidizable methylene blue (MB)

12

Figure 14-6 Comparison of autoxidizable and nonautoxidizable dyes.


and the electrode chamber is placed in contact withthe sample chamber. Next, the sample chamber isfilled with an aqueous solution containing dissolvedO2 such that the fluid is in complete contact withthe membrane. The laws governing diffusion dic-tate that dissolved oxygen from the aqueous systemin the sample chamber will diffuse through themembrane into the electrode chamber at a rate pro-portional to the concentration of dissolved O2 inthe sample. Thus, the system will establish an ini-tial rate of diffusion of O2 into the electrode cham-ber that is defined by the geometry of the samplechamber and the initial concentration of dissolvedO2. The electrode system of the electrode chambermeasures the O2 as it enters the chamber in one oftwo ways. Amperometric systems use a constantvoltage between electrodes. This voltage causes thereduction of O2 at the cathode according to Equa-tions 14-6 through 14-8, which results in a flow ofcurrent in the electrolyte solution.

(14-6)

O2 � 2 H2O � 2 e�88n H2O2 � 2 OH�

(14-7)

H2O2 � 2 e�88n 2 OH�

(14-8)

Sum: O2 � 2 H2O � 4 e�88n 4 OH�

Potentiometric systems, which are less commonlyused, measure changes in electric potential causedby interaction of O2 with the cathode. Appropriateelectronic equipment transduces the amperometricoutput into a voltage, which is amplified andrecorded on a conventional strip chart recorder.

As long as the solution is adequately mixed sothat the solution at the membrane surface has a dis-solved O2 content that is representative of the bulksolution, the voltage output from an oxygen elec-trode is proportional to the rate of diffusion of O2

across the membrane, which is in turn proportionalto the O2 concentration in the bulk solution.Changes in amounts of dissolved O2 in the samplechamber, due to O2 reduction by electron trans-port, alter the rate of O2 diffusion into the elec-

Cathode tip exposed at end of glass core

Adjustable supportring

Anode

Injection port

Metal or clear plastichousing that holdssample and maintainstemperature

Stirring bar

O-ring tohold membrane

Electrode orelectrodechamber

Sample chamber

Figure 14-7 Features of a typical oxygen electrode system.


trode chamber and result in a proportional alter-ation of the recorder readings. Thus, you can eas-ily measure relative rates of O2 consumption (orproduction, as in studies of photosynthesis) with anoxygen electrode. If you have previously calibratedthe oxygen electrode with an O2 standard, as willbe done in this experiment, you can interpret theserates and overall extents of reaction in units of mi-croliters or micromoles of O2 (22.4 �l of a gas 1 �mol under standard conditions).


Oxygen electrode with recorder, electrolyte solution,membranes, magnetic stirrers and sample injectionsyringes (see Appendix)

Fresh pig heart

Meat grinder

1 mM EDTA (pH 7.4)

Blender

0.02 M potassium phosphate containing 1 mM EDTA(pH 7.4)

High-speed centrifuge

0.25 M potassium phosphate containing 1 mM EDTA(pH 7.4)

1 M acetic acid

Oxygen electrode system (e.g., Yellow Springs Instru-ments Model 53)

1 mM NAD�

0.3 M potassium phosphate buffer (pH 7.4)

0.05 M succinate

0.5 M succinate

0.25 M malate

0.25 M �-hydroxybutyrate

0.6 mM cytochrome c

Sodium ascorbate (198 mg sodium ascorbate in 10 mlH2O, or 175 mg ascorbic acid plus 1 ml 1 N NaOHand diluted to 10 ml with H2O—freshly preparedimmediately before use)

0.5 M malonate

0.l M KCN—Poison!

0.01 M methylene blue

Antimycin A (180 �g/ml in 95% ethanol)

95% Ethanol

6 M KOH

2,6-Dichlorophenol indophenol (DPIP) solution (13.5mg dissolved in 50 ml H2O)

Spectrophotometer

Protocol

Preparation of Keilin–Hartree Particles

from Heart Muscle

We suggest that steps 1 through 5 of the experiment beperformed by the instructor or teaching assistant a dayor two before the beginning of the experiment.

1. Working at 0 to 4°C (cold room), free a pigheart from a freshly slaughtered animal of visiblefat and fibrous tissue. Pass the muscle twicethrough a precooled (2°C) meat grinder fittedwith a fine-grind plate. (The minced musclemay be stored at �20°C for months withoutrisking loss of activity.)

2. Wash the minced heart tissue by stirring in 10volumes of cool (less or equal to 20°C) 1 mMEDTA (pH 7.4) for about 4 hr, replacing thebuffer six to eight times during this interval.Squeeze excess fluid from the minced musclewith four layers of cheesecloth after eachwashing.

3. Homogenize the washed tissue in a chilledblender for 7 min at 0 to 4°C with 0.02 Mpotassium phosphate and 1 mM EDTA(pH 7.4), using 500 ml per 200 g of original,trimmed heart muscle. Centrifuge the ho-mogenate at 800 g for 15 to 20 min. Decantthe cloudy supernatant solution into a freshcentrifuge bottle, and save it on ice. Homoge-nize the precipitate again in a chilled blenderfor 2 min with 300 ml of cold 0.02 M potas-sium phosphate and 1 mM EDTA (pH 7.4),and centrifuge as before. Add this supernatantliquid to that from the first centrifugation.(Usually 500 to 600 ml of suspension is ob-tained from 200 g of heart.)

4. Reduce the pH of the cold (�5°C) combinedsupernatant suspension to 5.6 by cautiouslyadding (with stirring) cold 1.0 M acetic acid.Centrifuge at 4000 to 5000 g for 15 min.The clear, reddish supernatant from the cen-trifugation contains cytochrome c and hemo-globin and can be discarded. Wash the precip-itate once by suspending it in an equal volumeof ice water and then centrifuge for 5 min at1500 g. Save the precipitate and discard thesupernatant fluid.


5. Suspend the precipitate in an equal volume of0.25 M potassium phosphate and 1 mM EDTA(pH 7.4). The final volume should be about70 ml from 200 g of heart. Disperse any lumpsof precipitate by shaking gently in a flask or bygentle homogenization in a hand-operatedglass homogenizer. Store this preparation at lto 4°C until used. (It is stable for about 1 week,although some losses in activity may be ob-served by the end of a week.)

Calibration of the Oxygen Electrode

1. Slowly remove your O2 electrode from thesample chamber and check to see that the mem-brane is intact and that the electrolyte solutionwithin the electrode is free of gas bubbles. Ifnecessary, remove the membrane holder, refillthe electrode with electrolyte so a bulging poolof electrolyte is left, and then slip on and re-seal a membrane over the electrode so that noair bubbles are trapped inside. Finally, slowlyplace the O2-electrode into the sample holderso that the electrode contacts the desired quan-tity of fluid to be assayed (�3 ml). Leave thesample chamber empty.

2. Turn on the O2-electrode system and allowthe circuitry to warm up or stabilize for 5 min.Check the operation of the recorder duringthis interval. Then set the recorder on “pen”and set the zero point adjustment control ofthe recorder to achieve a zero setting on therecorder during zero output from the O2-electrode system. This usually requires thatyou disconnect the O2-electrode system fromthe recorder or switch a zero output controlon the O2-electrode system.

3. Set the O2-electrode system to “amplifierzero,” and adjust the amplifier zero control un-til the recorder registers zero. This establishesthe reading that occurs in the absence of O2.You can confirm this setting by later using anO2-free water solution, such as water that hasbeen boiled vigorously and cooled or spargedwith an inert gas.

4. Establish the other end of the recorder scale bycarrying out the following steps in order. First,open the sample chamber of the O2 electrodeand place a small stirring bar into the samplechamber of the O2-electrode system. Second,

using the sample injection syringe containing aflexible end of tubing, carefully fill the samplechamber with 2.9 ml (or more, no remainingair bubbles) of air-saturated water. (Saturate thewater with air by squirting a water sample re-peatedly into a beaker.) Third, set the samplechamber and O2 electrode over a magnetic stir-rer, turn on the stirrer, and adjust to a slow(�100 rpm) stirring rate. Last, set the O2-electrode system to “air” or “air-saturatedH2O” and adjust the 100% adjust control ofthe O2-electrode system so that the recorderregisters 85 units, at which full deflectionequals 100 units. This may require a minute ortwo to reach a stable reading. Since the samplechamber contains 3 ml, and air-saturated H2Ocontains 28.3 �l/ml of O2, this setting cali-brates the recorder so that each unit or divisionon the strip chart equals 1 �l of O2.

5. These steps will give you a fully adjusted andcalibrated system. To proceed directly with theO2 measurements in this experiment, returnthe O2-electrode system to the amplifier zerostate, turn off the magnetic stirrer, replace theair-saturated water sample with a magneticallystirred (�100 rpm) experimental sample fromone of the series tables as described below, re-assemble the system (no air bubbles!) and makeappropriate readings with the O2-electrode sys-tem at the “air” or “air saturated H2O” state.At the conclusion of your O2 measurements,turn off the O2-electrode system, the recorder,and the magnetic stirrer. Empty the samplechamber and rinse all components gently withdistilled water. Repeat this brief calibrationprocedure before performing an experiment ina following lab period.

Characterization of Electron Transport

using Keilin–Hartree Particles and the

Oxygen Electrode

The following tables describe four series of assays.Series I and II characterize the substrate specificityof Keilin–Hartree particles, whereas Series III andIV employ the autoxidizable dye, methylene blue,and various inhibitors to characterize the electron-transport system in these particles. It may be prac-tical to divide the class into groups that will only


perform one or two of the series of experiments andfor the groups to share their results to obtain a fulldata set. Consult your instructor about the distri-bution of the assays among the class members.

1. Prepare separate tubes containing all of the“sample chamber” components of the desiredassays of Series I through IV except for the ad-dition of KCN solution where it is called for in Se-ries III and IV. Add the components to the tubesin the order in which they are listed in the ta-bles. Prevent plugging of micropipette tipsused to transfer the heart particle suspensionby cutting about 2 mm off the tips.

2. Using an injection syringe tipped with a flexi-ble tubing end, load the “sample chamber” con-tents of an individual assay tube into the sam-ple chamber of a 2.9-ml calibrated oxygenelectrode. When the assay tube contents in-clude KCN, add it as the last component justbefore mixing them and adding them to thesample chamber with the injection syringe.

KCN releases dangerous HCN at the pH of the

system and must be added just before starting

the system. Do not pipette by mouth! Use a

propipetter.

Set the oxygen electrode control to “air” or“air-saturated H2O” and turn on the recorder.

3. Two areas to which you should pay particularattention are sample volume and potential in-terfering air bubbles. Most of the assays involveinitiating the reaction by addition of only 50 or100 �l of substrate. Thus, the contents of thesample chamber in these assays essentially fillsthe O2-electrode sample chamber before thesubstrate is added. This allows you to establisha base value of O2 concentration in the mainvessel contents before adding potential sub-strate. Therefore, if the assay is initiated by ad-dition of 100 �l or less of substrate, turn on themagnetic stirrer and the recorder and recordbase-line value readings for 2 min.

4. During this interval, load an injection syringewith the substrate to be used in initiating thereaction, withdrawing the syringe barrel justenough to take up the desired volume of sub-strate. Avoid excess air in the syringe because this

will introduce bubbles into the O2-electrode samplechamber.

5. After recording base-line O2 levels for 2 min,lower the flexible tip of the injection syringeinto the sample chamber, inject the substrate(no air bubbles, let any excess fluid overflowout of the system), and withdraw the syringe.Continue recording data over 2 to 4 min; thatis, as long as a linear reaction occurs. Then,turn off the system, disassemble the O2

electrode, and rinse out the sample chamberand injection syringe with distilled H2O priorto further assays or storage.

6. If the total volume of the main vessel compo-nents of an assay does not equal or exceed2.9 ml—that is, if the substrate volume exceeds100 �l—you cannot establish a good base-linereading before adding substrate because airbubbles in the sample chamber will distort thebase-line values. In such instances, transfer thesample chamber components into the O2-electrode sample chamber, but do not turn onthe recorder or the magnetic stirrer—do notattempt to establish a base-line value. Instead,load the injection syringe with just the desiredvolume of substrate (no air bubbles), and injectthe substrate into the bottom of the O2-electrode sample chamber. Immediately turnon the magnetic stirrer and recorder. With theO2-electrode set on “air” or “air saturatedH2O,” begin to monitor the reaction with therecorder. Continue readings for 2 to 4 min, orfor as long as a linear reaction occurs. Then,turn off the system, disassemble the O2 elec-trode, and rinse out the sample chamber andinjection syringe with distilled H2O prior tofurther assays or storage.

7. Using the linear assay range from each assaystudied, calculate a rate of O2 consumption inunits of microliters of O2 consumed perminute. Explain your results for each tube inall four series in terms of our current knowl-edge of electron transport (see Fig. 14-4). Es-timate a KM for succinate from the data of Se-ries I, tubes 1 through 3. (If you do notremember how to do this, see Experiment 7.)Which tubes represent controls for othertubes? Are the sites of entry of the various elec-tron donors and the modes of action of the in-hibitors depicted in Figure 14-4 consistent with


your findings? Discuss possible reasons for anyunexpected results.

Spectrophotometric Assay of Electron

Transport

1. Prepare a series of tubes containing the com-ponents outlined in Series V below, but withoutthe specified amounts of heart particles and succi-nate. Add the components in the order listed inthe table. Add the KCN carefully to tube 5 just

before adding the heart particle preparation to tube5—not before.

2. Add the heart-particle suspension to tube 1 (theblank) and to tube 2. Mix the contents of eachof these two tubes, and read the optical densityof tube 2 against the blank (tube 1) at 600 nmat 30-sec intervals for 2 min to check for oxi-dation of any endogenous substrate.

3. Add the succinate to tube 2. Mix the contents,and read the optical density at 600 nm againstthe blank at 30-sec intervals for 5 min.

Series I Assays of Citric-Acid-Cycle Substrates

Assay Tube Number

Addition

(all volumes in �l) 1 2 3 4 5 6

Sample chamber

H2O 950 990 950 650 500 500

0.3 M K-phosphate, pH 7.4 1500 1500 1500 1500 1500 1500

1 mM NAD� — — — 300 300 300

Heart-particle suspension 500 500 500 500 500 500

Substrate (added last)

0.05 M succinate 50 — — — — —

0.5 M succinate — 10 50 50 — —

0.25 M malate — — — — 200 —

0.25 M �-hydroxybutyrate — — — — — 200

Series II Assays with Succinate and Ascorbate as Substrates

Assay Tube Number

Addition

(all volumes in �l) 1 2 3 4 5

Sample chamber

H2O 950 650 400 700 400

0.3 M K-phosphate, pH 7.4 1500 1500 1500 1500 1500

0.6 mM cytochrome c — 300 300 — 300

Heart-particle suspension 500 500 500* 500 500


0.5 M succinate 50 50 — — —

0.1 M ascorbate† — — 300 300 300

*For tube 3 use heart particles that have been heated at 100°C for 10 min.

†198 mg sodium ascorbate/10 ml or 175 mg ascorbic acid plus 1 ml 1 N NaOH made up to 10 ml with H2O. Prepare immediately before

use.


Series III Sites of Action of Electron Transport Inhibitors

Assay Tube Number

Addition

(all volumes in �l) 1 2 3 4 5 6 7

Sample chamber

H2O 950 650 350 650 850 500 550

0.3 M K-phosphate, pH 7.4 1500 1500 1500 1500 1500 1500 1500

0.5 M malonate — — — — 100 100 100

10 mM methylene blue — — 300 300 — — 300

Heart-particle suspension 500 500 500 500 500 500 500

0.1 M KCN — 300 300 — — — —


0.5 M succinate 50 50 50 50 50 400 50

4. Repeat steps 2 and 3 with each of the remain-ing 4 tubes (tubes 3, 4, 5, and 6), one at a time,reading the absorbance at 600 nm of each so-lution against the tube 1 blank.

5. Prepare a plot of the decrease in absorbance at600 nm versus time in minutes for the resultsobtained from tubes 2 through 6. Explain the

results for each reaction in terms of our cur-rent knowledge of electron transport, the sitein the electron transport chain at which reduc-ing equivalents are “shunted” to the DPIP dye,and the proposed sites of action of the in-hibitors used (see Fig. 14-4).

Series IV Sites of Inhibitor Action and Ascorbate Oxidation

Assay Tube Number

Addition

(all volumes in �l) 1 2 3 4 5 6

Sample chamber

H2O 920 920 890 370 370 70

0.3 M K-phosphate, pH 7.4 1500 1500 1500 1500 1500 1500

0.6 mM cytochrome c — — — 300 300 300

10 mM methylene blue — — 30 — — —

Antimycin A in 95% ethanol (180 �g/ml) — 30 30 — 30 —

95% Ethanol 30 — — 30 — 30

Heart-particle suspension 500 500 500 500 500 500

0.1 M KCN — — — — — 300


0.5 M succinate 50 50 50 — — —

0.1 M ascorbate* — — — 300 300 300

*198 mg sodium ascorbate/10 ml or 175 mg ascorbic acid plus 1 ml 1 N NaOH made up to 10 ml with H2O. Prepare immediately before

use.


Exercises

1. Define: (a) dehydrogenase, (b) oxidase, (c) cy-tochrome.

2. Ascorbic acid reduces cytochrome c but has lit-tle impact on cytochrome b. Explain this ob-servation in terms of the reduction potentials(redox potentials) of these substances.

3. The following parts describe a series of sepa-rate assays that use the heart particle systemdescribed in this experiment. Assuming eachcomponent (substrate, dye, or inhibitor) ispresent in sufficient concentration to assert itsfunction, predict whether O2 will be con-sumed or not consumed in each of these heart-particle assays.

a. Succinate, DPIP, and cyanideb. Succinate, antimycin A, cyanide, and meth-

ylene bluec. Ascorbate, antimycin A, DPIP, and mal-

onated. Succinate, methylene blue, cytochrome c,

cyanide, and ascorbate

4. Four electron carriers (A, B, C, and D) are re-quired in the electron transport system of a re-cently discovered aerobic bacterium. You findthat in the presence of substrate and O2, fourinhibitors (I, II, III, and IV) block respirationat four different sites. From differential spec-trophotometry of the electron carriers, you findthat these inhibitors yield the patterns of oxi-dation state of the carriers shown below. What

is the order of electron transfer to the electroncarriers in the respiratory chain of this bac-terium? Explain your answer by writing the re-action sequence from reduced substrate to O2

and indicate the site of action of each inhibitor.

Oxidation State of Electron Carrier in the

Presence of Substrate and Inhibitor

(� means fully oxidized, � means fully reduced)

Inhibitor A B C D

I � � � �

II � � � �

III � � � �

IV � � � �

REFERENCES

Iwata, S., et al. (1998). Complete Structure of the 11-Subunit Bovine Mitochondrial Cytochrome bc1

Complex. Science 281:64.Keilin, D., and Hartree, E. F. (1947). Activity of the

Cytochrome System in Heart Muscle Preparations.Biochem J 41:500.

Lehninger, A. L. (1964). The Mitochondrion: MolecularBasis of Structure and Function. Menlo Park, CA:Benjamin.

Stryer, L. (1995). Biochemistry, 4th ed. chapters 20 and21. New York: Freeman.

Trumpower, B. L., and Gennis, R. B. (1994). EnergyTransduction by Cytochrome Complexes in Mito-chondrial and Bacterial Respiration: The Enzymol-

Series V Spectrophotometric Assays of Electron Transport

Assay Tube Number

Addition

(all volumes in ml) 1 (blank) 2 3 4 5 6

H2O 4.4 3.9 3.8 3.4 3.0 3.9

0.3 M K-phosphate, pH 7.4 4.5 4.5 4.5 4.5 4.5 4.5

DPIP solution — 0.4 0.4 0.4 0.4 0.4

0.5 M malonate — — 0.1 0.1 — —

Antimycin A (180 �g/ml in 95% ethanol) — — — — — 0.01

0.1 M KCN — — — — 0.9 —

Heart-particle suspension (next to last) 0.1 0.1 0.1 0.1 0.1 0.1

0.5 M succinate (added last) — 0.1 0.l 0.5 0.1 0.1


ogy of Coupling Electron Transfer Reactions toTransmembrane Proton Translocation. Annu RevBiochem 63:675.

Tsukihara, T. et al. (1995). Structures of Metal Sites ofOxidized Bovine Heart Cytochrome c Oxidase at2.8 Å. Science 269:1069.

Xia, D. et al. (1997). Crystal Structure of the Cyto-chrome bc1 Complex from Bovine Heart Mito-chondria. Science 277:60.

Yoshikawa, S. et al. (1998). Redox-Coupled CrystalStructural Changes in Bovine Heart Cytochrome cOxidase. Science 280:1723.

: )

243

EXPERIMENT 15

Study of the Phosphoryl Group TransferCascade Governing Glucose Metabolism:Allosteric and Covalent Regulation ofEnzyme Activity

Theory

A living cell contains hundreds of enzymes involvedin many very different yet equally important meta-bolic processes. As the environment of the cell andthe demands on it change, so too must the activityof many of these biological pathways to allow thecell to cope with each new challenge. This generalconcept is easy to understand if you consider howcellular activity following a large meal is differentthan that during strenuous physical activity. In theresting state, when energy supplies are plentiful, thecell will begin to switch from a catabolic (energy-yielding metabolism) mode to a more anabolic (en-ergy-storing or biosynthetic metabolism) mode.This situation will be reversed with increases inphysical activity; energy stores replenished duringthe resting state will be metabolized to producelarge quantities of the cell’s energy currency, ATP.

How does the cell regulate the enzymes in-volved in catabolic and anabolic metabolism? Thecell has several options: it can control the expressionof enzymes involved in these processes, or it cancontrol the activity of the enzymes involved in theseprocesses. In Experiment 7 (“Study of the Proper-ties of �-Galactosidase”), you investigated a com-mon strategy utilized by prokaryotes to regulatemetabolic enzyme activity at the level of transcrip-tion. Expression of the gene for �-galactosidase(lacZ) is induced in the presence of the enzyme’s sub-strate, lactose. The cell will repress expression of theenzymes involved in lactose metabolism until itfinds itself in the presence of this carbohydrate. Insome cases, however, the cell is required to respondrapidly to its changing metabolic needs. It may not

have time to produce the enzymes required to per-form a specific reaction. Therefore, cells are alsocapable of regulating the activity of enzymes thatare expressed or maintained at a constant level. Thenet result of either form of regulation is the same:enzymes can be regulated according to the ever-changing metabolic needs of the cell.

Two mechanisms that are commonly employedin altering enzyme activity are covalent modifica-tion and allosteric regulation. Covalent modificationis an enzymatically catalyzed reaction that involvesthe reversible formation of a covalent bond betweena small molecule and a specific amino acid sidechain(s) on an enzyme that affects its activity. Al-losteric regulation of an enzyme’s activity involvesnoncovalent binding of a small molecule at a siteother than the active site that alters the enzyme’sactivity. Unlike the limited examples of covalentmodification that have been discovered (see Table15-1), a wide variety of small molecules have beenfound to regulate the activity of particular enzymesallosterically.

Glycogen phosphorylase was one of the first en-zymes shown to be allosterically regulated. It is alsoone of the first enzymes shown to be covalentlymodified by what has been shown to be a very com-mon mechanism in enzyme regulation, phosphory-lation. Glycogen phosphorylase is the enzyme re-sponsible for cleaving the �(1 n 4) linkage betweenthe glucose subunits that comprise glycogen, acommon energy store for some mammalian cells(Fig. 15-1). The glucose-1-phosphate units that re-sult from this phosphorolysis reaction can eventuallyenter the glycolytic pathway and produce ATP.Glycogen phosphorylase can exist in the a form or


Table 15-1 Established Forms of Covalent Modification in the Regulation of Enzyme Activity

Group Attached Modified

Modification Donor Group Byproduct Residues Examples

Phosphorylation ATP PO4�2 ADP Ser Phosphorylase b,

Thr glycogen synthetase,

His EGF receptor,

Tyr insulin receptor,

protein kinase C,

protein kinase G,

protein kinase A

ADP-ribosylation NAD� ADP- Nicotinamide Cys � subunit of G protein

ribose Arg (cholera toxin),

Gln i subunit of G protein

(pertussis toxin)

Adenylylation ATP AMP PPi Tyr glutamine synthetase

Uridylylation UTP UMP PPi Tyr PII protein

(another form of

glutamine synthetase

regulation)

Methylation S-adenosyl- CH3 S-adenosyl- Glu Methyl-accepting

methionine homocysteine chemotaxis proteins

Glycogen (n glucose subunits)

Glycogen phosphorylaseP

CH2OH

H H H H HH

H

HHOOH

OH

OH

O

O�

�O

O

CH2OH

H

H

HOH

OH

O

O O

...................

Glycogen (n-1 glucose subunits)

CH2OH

H H H H HH

H

HHOOH

OH

O

CH2OH

H

H

HOH

OH

O

O

Glucose-1-phosphate

CH2OH

H HH

H

HHOOH

OH

O

OO

...........

P

O

O�

O�

�

Figure 15-1 Reaction catalyzed by glycogen phosphorylase.

EXPERIMENT 15 Study of the Phosphoryl Group Transfer Cascade Governing Glucose Metabolism: Allosteric and Covalent Regulation of Enzyme Activity

245

the b form, each made up of a dimer of identical94.5-kDa subunits. Glycogen phosphorylase b is theless active form of the enzyme. Glycogen phos-phorylase a is the active form of the enzyme thathas covalently bound phosphoryl groups on theSer-14 side chain of each subunit. The phosphorylgroup transfer from ATP to phosphorylase b to pro-duce phosphorylase a is catalyzed by an enzymecalled glycogen phosphorylase kinase.

Like glycogen phosphorylase, phosphorylase ki-nase is also subject to covalent modification (regu-lation) by phosphorylation. In response to hor-mones that signal an increase in physical activity,adenylate cyclase is activated to produce an increasein the intracellular cAMP concentration. In turn, acAMP-dependent protein kinase is activated tocause an increase in the level of phosphorylatedglycogen phosphorylase kinase. This activated(phosphorylated) kinase will convert phosphorylaseb to phosphorylase a, causing an increase in the rateof glycogen breakdown to glucose-1-phosphate forthe production of ATP. Similar types of phospho-rylation cascades have been shown to be critical inmany signal transduction pathways.

In addition to being activated by phosphoryla-tion, glycogen phosphorylase b is also subject to al-losteric activation by AMP. As physical activity in-creases, so too does the intracellular AMP�ATPratio. This increase in AMP concentration will al-low immediate activation of glycogen phosphory-lase b prior to its hormonally mediated conversionto phosphorylase a by cAMP-dependent protein ki-nase and glycogen phosphorylase kinase. This dualregulation of glycogen phosphorylase ensures thatactive muscle cells will have an adequate supply ofglucose for the production of ATP. A schematic di-agram of the signal transduction process just de-scribed is presented in Figure 15-2.

In this laboratory exercise, you will study the ef-fects of phosphorylation and allosteric regulation onthe activity of glycogen phosphorylase. In the firstexperiment, you will phosphorylate glycogen phos-phorylase b in vitro using �-[32P]ATP and glycogenphosphorylase kinase. In the second experiment, youwill study the effect of phosphorylation on glycogenphosphorylase activity, as well as the effect of AMPon glycogen phosphorylase b activity with the use ofa coupled enzymatic (kinetic) assay (Fig. 15-3).

Although phosphorylation is the type of cova-lent modification that will be investigated in this

experiment, it is not the only form of covalent mod-ification that has been adopted by cells as a meansof regulating enzyme activity. Table 15-1 lists anumber of other types of covalent modificationsthat have been confirmed in a number of differentbiological systems and pathways.


P-20, P-200, and P-1000 Pipetmen with sterile tips

1.5-ml Microcentrifuge tubes

Stock solution of 1 M Tris, pH 8.0

Stock solution of 1 M glycerophosphate, pH 8.0

Stock solution of 60 mM Mg acetate, pH 7.3

Concentrated enzyme storage buffer:

50 mM Tris, pH 8.0

50 mM glycerophosphate, pH 8.0

40% vol/vol glycerol

2 mM EDTA, pH 8.0

0.1% �-mercaptoethanol

Reaction buffer:

50 mM Tris, pH 8.0

50 mM glycerophosphate, pH 8.0

2 mM ATP solution (2 mM ATP prepared in 60 mM Mgacetate, pH 7.3)

Glycogen phosphorylase b stock solution (Sigma catalog#P-6635):

Dissolve lyophilized enzyme to �1000 units/ml inconcentrated enzyme storage buffer. Divide into20-�l aliquots and store at �20°C.

Glycogen phosphorylase kinase stock solution (Sigmacatalog #P-2014):

Dissolve lyophilized enzyme to �2,500 units/ml inconcentrated enzyme storage buffer. Divide into20-�l aliquots and store at �20°C.

4X SDS/EDTA buffer:

0.25 M Tris, pH 7.0

30% vol/vol glycerol

10% vol/vol �-mercaptoethanol

8% wt/vol sodium dodecyl sulfate

250 mM EDTA

0.01% Bromophenol blue (tracking dye)

�-[32P]ATP (4000 Ci/mmol)—ICN Biochemicals, cata-log #35001X. Dilute this solution 1:80 for use in thein vitro phosphorylation assay

Prestained high-molecular-weight protein markers (GibcoBRL cat no. 26041-020)

246

Ser 14 Ser 14

Glycogen phosphorylase a(more active)

O–

P O H2C

2H2O

2Pi

2ADP

2ATP

–O

O

O–

P O O–CH2

O

Ser 14 Ser 14 O–

P O–O

O

–O

P–O O

O

Ser 14 Ser 14

OHHO

O–

P O–OOH

O

Ser 14 Ser 14

Glycogen phosphorylase b(less active)

Glycogen phosphorylasephosphatase

Glycogen phosphorylasekinase

HOH2C CH2OH

Ser 14 Ser 14

Glycogen phosphorylase b(more active)

In liver(glucagon binds to receptor)

adenylate cyclase

Allostericregulation

Covalentmodification

AMP AMP

2AMP (

cAMP + PPi

ADPATP

ATP

)( ) 2AMP

cAMP-dependent proteinkinase

Glycogen phosphorylase a(more active)

Glycogen phosphorylase b(less active)

Glycogen(n)

Glycogen + glucose –1–P(n –1)

Phosphorylasekinase

Phosphorylasekinase

(more active)(less active)

2ADP2ATP

Pi

O

Figure 15-2 Allosteric and covalent regulation of glycogen phosphorylase.


247

phosphoglucomutase

Glucose-1-phosphate

(n glucose subunits)

Nicotinamide adenine dinucleotidephoshate (NADP�-oxidized form)

CH2OH

H

H

HH

H

HHOOH

OH

O

O P

O

O�

Glucose-6-phosphate

CH2

NH2

CH2

NH2

H H

H

HHO

HO

H

OH

OH

OHOH

O

OH

P

O

O�O

�O

�O

C

O

P

O

O�O

P

O

O�O

O P

O

O N

N N

N

O�

O�

HH H

H

H

O N CH2O

CH2O CH2O

H

H H

H H

H H

Nicotinamide adenine dinucleotidephoshate (NADPH-reduced form)

(absorbs light at 340 nm)

H

NH2

NH2

H

OH

OH

OH

C

O

P

O

O�O

P O�O

O P

O

O N

N N

N

O�

O�

P

O

O O�

COO�

HH

H

O N

Glucose-6-phosphate 6-Phosphogluconate

CH2

H HH

H

HHOOH

OH

O

OH

P

O

O�O

�O

�OH

H C

OHH C

OHH C

HHO C

OHH C

glucose-6-phosphate dehydrogenase

glycogen phosphorylase

Pi

Glycogen

(n-1 glucose subunits)

Glycogen Glucose-1-phosphate�

�

Figure 15-3 Coupled enzyme assay to determine glycogen phosphorylase activity. As long as the activity

of phosphoglucomutase and glucose-6-phosphate dehydrogenase are not rate-limiting, the

�A340/�t is directly proportional to the �[glucose-1-phosphate]/�time.


Reagents and supplies for SDS-PAGE (see Experiment 4)

Glass or plastic plates for casting SDS–polyacrylamidegel with a 10-well comb

Plastic wrap

Whatman 3 MM filter paper

Film cassette

Kodak X-omat film

Dark room

Film developer or Kodak developing solutions


Glass test tubes (13 100 mm)

10-ml Pipette with bulb

Spectrophotometer to read absorbance at 340 nm

30 mM cysteine, pH 7.0 (prepare using L-cysteine hy-drochloride, monohydrate, and pH to 7.0 usingsodium bicarbonate)

80 mM cysteine, pH 6.8 (prepare using L-cysteine hy-drochloride, monohydrate, and pH to 6.8 usingsodium bicarbonate)

250 mM Tris buffer, pH 7.7

250 mM glycerophosphate (prepare in 250 mM Tris,pH 7.7)

40 mM glycerophosphate (prepare in 80 mM cysteine,pH 6.8 and adjust pH to 6.8 with 1 M HCl or 1 MNaOH if necessary)

Glycogen phosphorylase b solution (100 units/ml)

Dilute enzyme stock solution (see Day 1 reagents)1:10 in 30 mM cysteine, pH 7.0

30 mM ATP solution (prepare in deionized water and pHto 7.7 with 1 M NaOH)

Glycogen phosphorylase kinase solution

Dilute enzyme stock solution (see Day 1 reagents)1:125 in 30 mM cysteine, pH 7.0

100 mM Mg acetate solution (prepare in deionized wa-ter with Mg acetate tetrahydrate and adjust pH to7.7 with 1 M NaOH)

500 mM potassium phosphate buffer, pH 6.8

4% (wt/vol) glycogen solution (Sigma catalog #G-8876)

300 mM MgCl2100 mM EDTA, pH 8.0

6.5 mM �-nicotinamide adenine dinucleotide phosphatesolution (�-NADP)

Prepare in deionized water immediately before use

0.1% (wt/vol) �-D-glucose-1,6-diphosphate solution(prepare in deionized water)

Phosphoglucomutase solution (10 units/ml)—Sigma cat-alog #P-3397


Glucose-6-phosphate dehydrogenase solution (10 units/ml)—Sigma catalog #G-6378


Glycogen phosphorylase a enzyme solution (2 units/ml)—Sigma catalog #P-1261

Dilute a 1000-units/ml stock in concentrated enzymestorage buffer 1:500 in 40 mM glycerophosphatesolution

Glycogen phosphorylase b control solution (2 units/ml)

Dilute enzyme stock 1:500 in 40 mM glycerophos-phate solution

Glycogen phosphorylase b control solution (2 units/ml)with AMP

Dilute enzyme stock 1:500 in 40 mM glycerophos-phate solution containing 60 mM AMP

Remember to keep all enzyme solutions on ice.

Protocol

Caution: This experiment requires the use of

significant amounts of 32P-ATP. Wear gloves and

safety goggles at all times during the experi-

ment. Plexiglas shields should be used to screen

students from excessive exposure to the ra-

dioisotope. Consult the instructor for proper

handling and disposal of radioactive waste.

Day 1: In Vitro Phosphorylation of

Glycogen Phosphorylase b

1. Prepare an �250-units/ml phosphorylase ki-nase solution by diluting the stock enzyme 1:10with reaction buffer.

2. Set up the following reactions in three 1.5-mlmicrocentrifuge tubes:

Volume of ��250 U/ml Volume of ��1,000

Tube Phosphorylase Kinase U/ml Phosphorylase b Volume of Reaction Buffer

1 — 10 �l 34 �l

2 10 �l — 34 �l

3 10 �l 10 �l 24 �l


249

3. To tubes 1 and 2, add 6 �l of an ATP solutionthat contains 20 �l of 2 mM ATP in 60 mMMg acetate, pH 7.3 and 4 �l of �-[32P]ATP(provided by the instructor). Pipette up anddown rapidly several times to mix. After exactly10 min, remove a 10-�l aliquot from each tubeand add it to 10 �l of 4X SDS/EDTA bufferto stop the reaction.

4. To tube 3, add 6 �l of the same ATP solutiondescribed above and pipette up and downrapidly several times to mix. At exactly 5 minand 10 min, remove a 10-�l aliquot and add itto 10 �l of 4X SDS/EDTA buffer to stop thereaction. Separate 5-min and 10-min samplesshould now have been taken from tube 3.

5. Load each 20-�l sample in the following orderon a 12% SDS–PAGE gel (prepared by the in-structor). Two groups will share one SDS–PAGE gel. Obtain the prestained protein mo-lecular weight standards from the instructor.

Tube 1—10 min

Tube 2—10 min

Tube 3—5 min

Tube 3—10 min

Standards

Standards

Tube 1—10 min

Tube 2—10 min

Tube 3—5 min

Tube 3—10 min

6. Connect the negative electrode (cathode) tothe upper buffer chamber and the positiveelectrode (anode) to the lower buffer cham-ber. The SDS bound to the proteins will givethem a negative charge, causing them to mi-grate toward the positive electrode. Apply 200V (constant voltage) and continue the elec-trophoresis until the bromophenol blue dyefront has migrated off the bottom of the gel.The unreacted 32P-ATP present in the sampleswill migrate off the gel with the dye front. At thispoint, the solution in lower buffer chamber is ra-dioactive. Wear gloves and safety goggles at alltimes.

7. When the electrophoresis is complete, turn offthe power supply and disconnect the elec-trodes. Remove from the unit the glass plates,which contain the gel. Carefully separate theplates (the gel will adhere to one of the twoplates) and carefully transfer the gel to a small

piece of Whatman 3 MM filter paper, avoidingany wrinkles in the gel.

8. Enclose the gel in a piece of plastic wrap andplace it in a film cassette. Using a felt-tippedpen, mark the film in a manner that will allowyou to determine position of gel on the film af-ter it is developed. Expose the gel to a piece ofKodak X-omat film for �18 hr at �70°C. Byexposing the film at low temperature, radioac-tive scattering will be minimized, giving youwell-defined radioactive protein bands on theautoradiogram.

9. Develop the film the next morning. Align theautoradiogram with the gel and mark the po-sitions of the wells and the prestained proteinmolecular weight standards on the film.

10. Prepare a plot of log molecular weight versusdistance migrated on the gel using the molec-ular weight protein standards. Are there any ra-diolabeled protein bands present on the au-toradiogram? What is the molecular weight ofthis protein? What is the identity of this pro-tein? Support your answer in terms of what youknow about the molecular weight of the pro-teins present in the assay. Are the radiolabeledprotein bands of the same intensity on the au-toradiogram? Describe and discuss your resultsin terms of the kinetics of the kinase present inthis assay.

11. Would you expect to see any radiolabeled bandspresent in the lanes containing samples fromreaction tubes 1 and 2? Why or why not?

Day 2: In Vitro Assay for Phosphorylase

Kinase and Glycogen Phosphorylase

Activity

1. Set up the reactions shown in the table on thenext page in 1.5-ml microcentrifuge tubes.The stock glycogen phosphorylase and thestock glycogen phosphorylase kinase solutionsshould be diluted to the proper concentrationswith 30 mM cysteine buffer, pH 7.0.

To start the phosphorylase kinase assay, add25 �l of 30 mM ATP solution to each tube andpipette rapidly up and down several times tomix. All three reactions can be run simultane-ously, staggering the start of each reaction byabout 30 sec. At exactly 5 min and 10 min, re-move 100 �l from each reaction and stop it byadding it to 1.9 ml of 40 mM glycerophosphate,


Volume of 30 mM Volume of 250 mM Volume of Volume of 100 units/ml Volume of 20 units/ml

Tube Cysteine, pH 7.0 Glycerophosphate 100 mM Mg Acetate Phosphorylase b Phosphorylase Kinase

1 122.5 �l 62.5 �l 25 �l — 15 �l

2 37.5 �l 62.5 �l 25 �l 100 �l —

3 22.5 �l 62.5 �l 25 �l 100 �l 15 �l

pH 6.8 in a 13-by-100-mm test tube. Each ofthese “stop” tubes (six of them) should be set up inadvance of starting the reaction and each tube shouldbe labeled with the reaction tube number and thetime (e.g., tube 1—5 min). These stopped, 2.0-ml reactions will be used in the glycogen phos-phorylase a assay described below. Place all ofthese tubes on ice until you are ready to proceed withthe next assay.

3. Prepare a reaction cocktail containing the fol-lowing stock components (this quantity will besufficient for at least four groups):

Stock Solution Volume (ml)

Deionized water 099.5

500 mM potassium phosphate buffer, pH 6.8 15.00

4% (wt/vol) glycogen 07.50

300 mM MgCl2 00.67

100 mM EDTA 00.15

6.5 mM �-NADP 010.0

0.1% (wt/vol) glucose-1,6-diphosphate 00.50

4. Dispense 2.7-ml of this reaction cocktail in 10small glass test tubes (13 100 mm), labeled Athrough J. To each of the 10 test tubes, add100 �l of the 10-units/ml glucose-6-phosphatedehydrogenase solution and 100 �l of the 10-units/ml phosphoglucomutase solution. Thetotal volume in each test tube after this step willbe 2.9 ml. Refer to Table 15-2 to aid you in set-ting up the rest of the assay tubes.

5. To tube A, add 100 �l of water and shake thetest tube gently for about 10 sec to mix. At 1-min intervals, measure and record the A340 ofthe sample. Continue taking absorbance read-ings until the reaction achieves a linear changein A340 versus time (at least 5 min). This is anegative control reaction that contains noglycogen phosphorylase or glycogen phospho-rylase kinase.

6. Perform the same procedure using tubes B toG by adding 100 �l of your different phos-phorylase kinase assay reactions to these tubes.Be sure that you know which phosphorylase kinasereaction corresponds to which test tube (e.g., tube Bis phosphorylase kinase reaction 1—5 min, tube Cis phosphorylase kinase reaction 1—10 min, etc.).Measure and record the A340 of each reactionat 30-sec intervals for at least 5 min, or untileach reaction achieves a linear change in A340

versus time.7. Perform the same procedure using tube H by

adding 100 �l of the 2-units/ml phosphory-lase a solution (this is a positive control forthe glycogen phosphorylase a assay). Measureand record the A340 of the reaction at 30-secintervals for at least 5 min, or until the reac-tion achieves a linear change in A340 versustime.

8. Perform the same procedure using tube I byadding 100 �l of the 2-units/ml phosphorylaseb solution (this is a negative control for theglycogen phosphorylase a assay). Measure andrecord the A340 of the reaction at 30-sec inter-vals for at least 5 min, or until the reactionachieves a linear change in A340 versus time.

9. Perform the same procedure using tube J byadding 100 �l of the 2-units/ml phosphorylaseb solution in 60 mM AMP. This assay is de-signed to show allosteric activation of glycogenphosphorylase b by AMP. Measure and recordthe A340 of the reaction at 30-sec intervals forat least 5 min, or until the reaction achieves alinear change in A340 versus time.

10. Construct a plot of A340 versus time for eachof the 10 glycogen phosphorylase reactions (A–J). Determine the slope of the line for each plot(�A340/min). It may be helpful to constructfour plots for tubes B–J: one graph will con-tain the data for tubes B and C, another willcontain the data for tubes D and E, another will


251

Table 15-2 Assay for Glycogen Phosphorylase Activity

Components Added (ml) Tubes

A B C D E F G H I J

Reaction cocktail 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7

Glucose-6-phosphate

Dehydrogenase (10 units/ml) 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

Phosphoglucomutase (10 units/ml) 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

H2O 0.1 — — — — — — — — —

From Tube Tube Tube Tube Tube Tube

Step 2: — 1, 5 1, 10 2, 5 2, 10 3, 5 3, 10 — — —

Volume from step 2 (ml) — 0.1 0.1 0.1 0.1 0.1 0.1 — — —

Phosphorylase a (2 units/ml) — — — — — — — 0.1 — —

Phosphorylase b (2 units/ml) — — — — — — — — 0.1 —

Phosphorylase b

(2 units/ml in 60 mM AMP) — — — — — — — — — 0.1

contain the data for tubes F and G, and a fourthwill contain the data for tubes H to J.

11. The goal of the analysis is to determine theconcentration (units per milliliter) of phospho-rylase kinase used in these reactions as well asthe concentration of glycogen phosphorylase a(units per milliliter) present in each of the re-actions after 5 min and 10 min. One unit ofphosphorylase kinase is defined as the amountof the enzyme that will convert 1 �mol of phos-phorylase b to phosphorylase a per min. Oneunit of phosphorylase a is defined as the amountof the enzyme that will produce 1 �mol of glu-cose-1-phosphate from glycogen per minute.

12. Determine the glycogen phosphorylase activ-ity present in tubes D to J using the followingequation:

units/ml enzyme

where test tubes other than A to C whose ac-tivity is being tested; blank the blank for thereactions in tubes H to J is tube A (no glyco-

(�A340/min test � �A340/min blank)(20)(3 ml)

(6.22 mM�1 cm�1)(l)(0.1 ml enzyme)

gen phosphorylase). The blank for the reac-tions in tubes D and F is tube B (5-min timepoint, no glycogen phosphorylase b). The blankfor the reactions in tubes E and G is tube C(10-min time point, no glycogen phosphorylaseb); 20 dilution factor (how much the glyco-gen phosphorylase enzyme solution beingtested was diluted before being assayed); 3 ml total volume of glycogen phosphorylase assay;6.22 mM�1 cm�1 millimolar extinction co-efficient for �-NADPH at 340 nm; l path-length through the spectrophotometer (1.3 cmfor benchtop spectrophotometers); and 0.1 mlenzyme volume of the phosphorylase kinaseassay or enzyme solution used in the glycogenphosphorylase assay.

NOTE: No dilution factor (20) is required for thecalculation of glycogen phosphorylase activity intubes H to J, since the enzyme stock solution wasadded directly to the assay. Otherwise, the same for-mula can be used for tubes H to J.

13. Determine the concentration of glycogenphosphorylase kinase present in reaction tube3 (from step 2) using the following equation:

Units/ml enzyme (units glycogen phosphorylase test/ml) � (units glycogen phosphorylase blank/ml)(20)��

(time)(0.015 ml)


where test the 5-min and 10-min time pointsfor phosphorylase kinase reaction 3 (tubes Fand G, respectively); blank the 5-min and10-min time points for phosphorylase kinasereaction 2 (no glycogen phosphorylase kinase;tubes D and E, respectively); 20 dilution fac-tor (how much the glycogen phosphorylase en-zyme solution being tested was diluted beforebeing assayed); time number of minutes (5or 10) of the phosphorylase kinase assay; and0.015 ml Volume (ml) of phosphorylase ki-nase used in reactions 1 and 3.

14. After you have completed these calculations,you will have obtained two values for the con-centration of glycogen phosphorylase kinasepresent in this assay, one based on the 5-mintime point and one based on the 10-min timepoint. From these data, determine the averagevalue of glycogen phosphorylase kinase presentin this assay (units/ml of enzyme).

15. Compare the glycogen phosphorylase activitybetween the reactions in tubes I and J. What ef-fect does AMP have on the activity of glycogenphosphorylase b? Can you estimate how muchthe activity of the enzyme is increased or de-creased in the presence of this allosteric effector?

16. Compare the glycogen phosphorylase activitybetween the reactions in tubes H and I. Whateffect does phosphorylation of Ser 14 have onthe activity of glycogen phosphorylase b? Canyou estimate how much the activity of the en-zyme is increased or decreased with this typeof covalent modification?

Exercises

1. What do you think is the advantage of havingan enzyme such as glycogen phosphorylase bcontrolled both by an allosteric effector and bycovalent modification?

2. Would you expect the glycogen phosphorylaseassay data from phosphorylase kinase reaction3 to be any different if the concentration of thekinase used in this reaction was increased by afactor of 5? Explain.

3. Would you expect the glycogen phosphorylaseassay data from phosphorylase kinase reaction3 to be any different if the concentration ofATP used in this reaction was increased by afactor of 10? Explain.

4. What must be true about the concentrationsof glucose-6-phosphate dehydrogenase andphosphoglucomutase used in the glycogenphosphorylase assay for the change in A340 tobe directly proportional to the change in con-centration of glucose-1-phosphate? Explain.

5. You have obtained a liver cell line that is mu-tant in glycogen metabolism. The mutation issuch that glycogen phosphorylase activity is al-ways high (glycogen phosphorylase activity isno longer regulated). This phenotype could bethe result of a mutation in a number of differ-ent genes involved in glycogen metabolism. Foreach of the enzymes listed below, state whetherthe phenotype of this cell line is consistent withthe activity of the enzyme being constitutivelyhigh or constitutively low. Support your answerusing your knowledge of the control of glyco-gen phosphorylase activity.

a. Adenylate cyclaseb. Glycogen phosphorylase kinasec. Glycogen phosphorylase phosphatased. cAMP-dependent protein kinase

REFERENCES

Antoniw, J. F., Nimmo, H. G., Yeaman, S. J., and Co-hen, P. (1977). Comparison of the Substrate Speci-ficities of Protein Phosphatases Involved in theRegulation of Glycogen Metabolism in RabbitSkeletal Muscle. Biochem J 162:423.

Cohen, P. (1983). Phosphorylase Kinase from RabbitSkeletal Muscle. Methods Enzymol. 99:243.

Garrett R. H. and Grisham, C. M. (1995). Gluconeo-genesis, Glycogen Metabolism, and the PentosePhosphate Pathway. In: Biochemistry. Orlando, FL:Saunders.

Hemmings, B. A., and Cohen, P. (1983). GlycogenSynthase Kinase-3 from Rabbit Skeletal Muscle.Methods Enzymol. 99:337.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Glycolysis and Catabolism of Hexoses. In:Principles of Biochemistry, 2nd ed. New York: Worth.

Sigma Quality Control Test Procedure (EC 2.7.1.38).St. Louis: Sigma Chemical Co.

Stryer, L. (1995). Glycogen Metabolism. In: Biochem-istry, 4th ed. New York: Freeman.

Voet, D. and Voet, J. G. (1990). Glycogen Metabolism.In: Biochemistry. New York: Wiley.

? O

253

EXPERIMENT 16

Experiments in Clinical Biochemistry and Metabolism

Theory

Medical technology has made tremendous advancesin the past several decades, due in part to the ad-vances made in clinical chemistry. Before the de-velopment of clinical chemistry, doctors wereforced to make biochemistry diagnoses based onmacroscopic observations and symptoms. Today,medical professionals have an extensive arsenal ofbiochemical tests available to aid in the early de-tection of a particular disease or condition. Thesetests were made possible through improved knowl-edge of how individual cells and tissues function atthe molecular level.

How do you determine whether a particular or-gan, such as the liver, heart, lung, or kidney is func-tioning properly? Since tissues consist of specializedcells that perform specific biological processes, in-formation about the integrity of a particular tissuecan be obtained by monitoring the activity of en-zymes present in that cell type, or the level of sub-strates or products used or produced by these en-zymes. Since it is not always feasible or desirable toinvade an organ and remove cells for analysis, an alternative means of quantifying the level of biological molecules in an individual had to be developed.

What is the next best thing to studying the ac-tivity present in a particular cell type directly? Agreat deal of information can be obtained fromanalysis of the fluid surrounding a particular tissueor cell type. Extracellular body fluid accounts fornearly 20% of the total weight of an individual. Ofthis, about 75% is found around and between tis-

sues and cells (interstitial fluid). The remainder isin the form of vascular bound fluid, such as bloodand lymph. Cerebrospinal fluid (in the spinal cordand around the brain) and the fluid between jointsaccount for only a trace of the total body fluid ofan individual.

Undoubtedly, blood plasma or serum is the mostcommon source of material for experiments in clin-ical biochemistry. This makes logical sense if oneconsiders the role of the blood in the body. Nearlyevery cell comes into contact with capillary bloodfor the purpose of acquiring oxygen, obtaining nu-trients, receiving signals via hormones, and ex-pelling the products of metabolism. If a particularcell is undergoing some unusual type of metabo-lism due to infection, exposure to a toxin, or othercondition, evidence for this will be present in theidentity and/or quantity of the various moleculesthat the cell secretes into the bloodstream. In thesame sense, dying cells associated with tissue dam-age or necrosis will lyse and secrete enzymes intothe bloodstream. Identifying and quantifying thelevel of particular enzymes in the blood has becomea useful tool in the diagnosis of specific conditionsand diseases. In this laboratory exercise, you willbecome familiar with several assays that are cur-rently used in the field of clinical biochemistry.

Glucose

Although high blood glucose levels are commonlyassociated with diabetes, it has also been found tobe associated with individuals possessing a hyper-active thyroid or adrenal gland. Conversely, a low


blood glucose level (hypoglycemia) is usually asso-ciated with cases of insulin overdose, an underac-tive endocrine system, and some forms of liver dis-ease. In general, blood glucose levels are useful inassessing the overall integrity of carbohydrate me-tabolism in an individual.

The assay used in the quantitation of blood glu-cose is one that you have become familiar with froman earlier experiment. It is a coupled assay that re-lates the reduction of NAD� (�A340) to the con-centration of glucose in the sample (see Experiment11, Figure 11-3).

Lactate

Lactate is formed from the degradation of carbohy-drates via glycolysis. During times of oxygen limita-tion, cells will turn toward anaerobic metabolism. Inthis process, lactate dehydrogenase will convertpyruvate to lactate, with the concomitant oxidationof NADH to NAD�. This, in turn, will lead to anincrease in the concentration of blood lactate. Sincelactate is metabolized by the liver to produce glucosevia the gluconeogenesis cycle, high blood lactate lev-els are usually found in individuals with poor liverfunction. The liver acts as the target organ for manydrugs and toxins; therefore, the assay is also usefulin diagnosing a person’s exposure to these agents.

The assay used to quantify lactate in the bloodallows you to relate the production of a colored dye(chromophore) to the level of lactate in the sample.This chromophore will absorb light at 540 nm, allowing its concentration to be measured spec-trophotometrically (Fig. 16-1).

Urea Nitrogen

One of the major products of amino acid metabo-lism is ammonia (NH3), a molecule known to behighly toxic to higher organisms. In the liver, am-monia and carbon dioxide are used to produce a water-soluble form of nitrogen, urea, via the ureacycle. The liver passes this urea to the blood, whichcarries it to the kidneys to be filtered out and ex-creted in the urine. Since one function of the kid-ney is to collect and excrete urea, increases in theconcentration of this compound in the blood are anindicator of poor kidney function. Since urea isformed in the liver, low blood urea nitrogen is of-ten the consequence of impaired liver function dueto disease or as the result of infection (hepatitis).

The assay used to quantify blood urea nitrogenrelies on coupling the production of ammonia andcarbon dioxide from urea to the transamination of�-ketoglutarate to glutamate and the subsequentoxidation of NADH (A340) (Fig. 16-2).

N N

S

N N

S

CH3C COO�

OH

H

H2O2

Lactate

CH3C

CH3

CH3NH2

CH3

COO�

O

Pyruvate

N,N-dimethylaniline

3-Methyl-2-Benzothiazolinonehydrazone

lactate oxidase

horseradish peroxidase

�

H2O2 � � N

CH3

CH3CH3

NN

Indamine dye(broad absorption maximum at 590 nm)

�

Figure 16-1 Quantitative assay for serum lactate.

EXPERIMENT 16 Experiments in Clinical Biochemistry and Metabolism 255

Creatine Kinase (CK)

Phosphocreatine is a high-energy compound thatcan be used by cells to drive energy-requiring re-actions. This compound is formed from creatine

and ATP by the action of creatine kinase (CK). Thisenzyme is also often called creatine phosphokinase(CPK). The reaction performed by this enzyme isreadily reversible: ADP and phosphocreatine can be

Urease

glutamatedehydrogenase

H2OCH2N NH2

O

Urea

�

� �

CO22 NH3 �

O

H HH H

OH OH

O

O

O

P

ONH2

H2O

N

NN

O

H H

H

H H

OH OH

N

N�O

OP

C

O

�O

NH2

�

O

H HH H

OH OH

O

O

O

P

ONH2

NH3

N

NN

O

H

H H

H H

OH OH

N

N�O

OP

C

O

�O

NH2

O

COO�

COO�

CH2

CH2

C

Nicotinamide adenine dinucleotide(NADH—reduced form; has A340)

-Ketoglutarate�

��

H

COO�

COO�

CH2

NH3

CH2

C

Glutamate

�

Nicotinamide adenine dinucleotide(NAD�—oxidized form)

Figure 16-2 Quantitative assay for serum urea nitrogen.


converted to creatine and ATP. Surprisingly, in-creased levels of this enzyme in the blood have beenfound to be associated with a number of serious yetseemingly unrelated conditions including heart attacks, alcoholism, muscular dystrophy, stroke,epilepsy, and other neurological and endocrine dis-orders. In general, elevated serum CK levels are as-sociated with conditions leading to tissue damageand cell death (necrosis).

Creatine kinase functions as a dimer. The dimercan consist of combinations of two different sub-units, M and B. Different cell types produce or ex-press the MM, MB, or BB forms of CK. Althoughthe amino acid composition of the M and B sub-unit differ, all three dimeric forms can catalyze thereaction described above. Different forms of an en-zyme that catalyze the same reaction (such as theMM, MB, and BB forms of CK) are referred to asisoenzymes.

It has been determined that different tissuesproduce or express different CK isoenzymes. Skele-tal muscle cells express the CK-MM isoenzyme al-most exclusively, while brain cells and cells of thegastrointestinal tract express the CK-BB isoenzymealmost exclusively. Cardiac muscle cells, however,express an approximate 1�3 ratio of the CK-MBand CK-MM isoenzymes, respectively. If you hada means of determining which isoenzyme is presentat elevated levels in the serum, you might be ableto determine what type of tissue the enzyme orig-inated from.

One common method of determining the iden-tity of the three isoenzymes makes use of the dif-ference in charge between the two subunits thatcomprise the dimeric protein. While the M subunitcarries a slightly positive charge at neutral toslightly alkaline pH, the B subunit carries a largenegative charge. When subjected to an electric field(electrophoresis, see Experiment 4), the CK-BBisoenzyme will migrate rapidly toward the anode(positive electrode), while the CK-MM isoenzymewill display a slow migration toward the cathode(negative electrode). The CK-MB isoenzyme, car-rying less of a net negative charge than the CK-BBisoenzyme, will migrate toward the anode at aslower rate.

Another method of determining the identity ofthe various CK isoenzymes that is gaining wide-spread popularity involves immunochemical tech-niques such as immunoprecipitation and Western

blotting (see Section IV). It is now possible to pro-duce and screen for monoclonal antibodies that rec-ognize the M subunit or B subunit exclusively.

The assay used to quantify blood concentrationsof this enzyme is nearly identical to that used in theassay described for determining blood glucose con-centrations (Fig. 16-3). Whereas the ATP in the lat-ter assay is supplied in excess, the ATP in the cre-atine kinase assay is provided by the action of theenzyme (creatine kinase) in the presence of ADPand creatine phosphate. In both assays, you aremeasuring the change in absorbance of the sampleat 340 nm. Since glucose is the limiting reagent inthe glucose assay, the A340 will be directly propor-tional to the glucose concentration. In the case ofthe assay for creatine kinase, glucose is in excess,and the rate of change in absorbance at 340 nm willbe limited by the ATP that is being produced bythe activity of creatine kinase. This demonstratesan important feature common to all coupled assaysystems: for the assay to be valid, the molecule or en-zyme under study must be the limiting reagent in theassay.


Plasma or serum from pig

P-20, P-200, and P-1000 Pipetmen with sterile tips

5-ml and 10-ml pipettes

13 � 100-mm glass test tubes

Heparin sulfate solution (1000 units/ml)

0.5 M EDTA, pH 8.0

B/L spectrophotometer tubes or cuvettes

Spectrophotometer to read absorbance at 340 nm and540 nm

Glucose assay reagent* consisting of:

50 mM Tris, pH 7.5

1.5 mM NAD�

1.0 mM ATP

1.0 units/ml hexokinase

1.0 units/ml glucose-6-phosphate dehydrogenase

5 mM MgCl2

D-Glucose standard solutions in water: 3.00 mM,11.10 mM, and 16.65 mM

Lactate assay reagent* consisting of:

50 mM Tris, pH 7.2

0.5 units/ml lactate oxidase


gluc

ose-

6-ph

osph

ate

dehy

roge

nase

hexo

kina

se

O

HH

HH

OH

O

O

OO PON

H2 N

NN

O

H

H HH

HH

OH

OH

N N�

O

OP

CO

�O

NH

2

�

O

HH

HH

OH

O

O

OO PON

H2 N

NN

O

H

HH

HH

OH

OH

N N

O P

�O

OP

CO

�O

NH

2

Nic

otin

amid

e ad

enin

e di

nucl

eotid

e ph

osph

ate

(NA

DP

H—

redu

ced

form

)N

icot

inam

ide

aden

ine

dinu

cleo

tide

phos

phat

e(N

AD

P�

—ox

idiz

ed fo

rm)

O H

HCO

H

HC

O

O P

HC

OH

OH

HO

CH

H

O�

�O

O�

O�

O�

O P

�O

�O

O PO

�

O�

CC

�

6-P

hosp

hogl

ucon

ate

O H

HCO

H

HC

O

O P

H

H

CO

H

OH

HO

C

H

H

�O

O�

CC

�

D-G

luco

se-6

-pho

spha

te

O H

HCO

H

HC

O

O P

HC

OH

OH

HO

CH

H

H

�O

O�

CC

�

D-G

luco

se-6

-pho

spha

te

O

CH

2O

H

HC

OH

HC

OH

OH

HO

CH

H

H

CC

�

�

D-G

luco

se

O

HH

HH

OH

OH

O

O P

�O

O

NH

2 N

NNN

Ade

nosi

ne d

ipho

spha

te (

AD

P)

crea

tine

kina

seO P

�O

CO

O�

NC

�

Cre

atin

e

O

HH

HH

OH

OH

O

O P

�O

O�

O

O P

�O

O

NH

2N

H2

CH

3

CH

2

H2N

N

NNN

Ade

nosi

ne tr

ipho

spha

te (

ATP

)

O P

�O

CO

O�

NC�

Pho

spho

crea

tine

O

HH

HH

OH

OH

O

O P

�O

O�

O

�O

O

O�

NH

2

NH

CH

3

CH

2

H2N

N

NNN

Ade

nosi

ne d

ipho

spha

te (

AD

P)

O P

�O

OO

HH

HH

OH

OH

O

O P

�O

�O

O P

�O

O

NH

2 N

NNN

Ade

nosi

ne tr

ipho

spha

te (

ATP

)

�

P

Figure 16-3 Assay for serum creatine kinase (CK) activity.


2.5 units/ml horseradish peroxidase

5 mM 3-methyl-2-benzothiozolinone hydrazone

5 mM N,N-dimethylaniline

Urea nitrogen assay reagent* consisting of:

50 mM Tris, pH 8.0

0.25 mM NADH

5.0 mM �-ketoglutarate

15 units/ml urease

25 units/ml glutamate dehydrogenase

Urea standard solutions in water: 3.57 mM, 8.92 mM,10.71 mM, and 35.70 mM

Creatine kinase assay reagent consisting of:

50 mM Tris, pH 7.5

10 mM phosphocreatine

1.5 mM ADP

1 mM NADP

5.0 mM D-glucose

1.5 units/ml hexokinase

0.5 units/ml glucose-6-phosphate dehydrogenase

20 mM dithiothreitol

*(Reagent is sold in a mixed dry form by SigmaChemical Co. This is convenient for use with alarge laboratory class.)

Protocol

Preparation of Blood Plasma

Pig Blood. Collect a 100–200-ml blood sample ina sterile container with cap containing 750 units ofheparin sulfate (anticoagulant) and 25–50-�l of0.5M EDTA, pH 8.0 (preservative). As soon as pos-sible after collection, centrifuge the sample at400 � g for 10 min to separate the plasma or serumfrom the blood cells (erythrocytes, etc.). Removethe clear plasma from the blood cells and store ina fresh, sterile container at 4°C. For best results, theassays described below should be performed within 36 hrafter the serum is isolated.

Blood Glucose Assay

1. Add 2.0 ml of glucose reagent to five 13-by-100-mm test tubes labeled G1 through G5.

2. Add 20 �l of the following to each of thesetubes and shake gently to mix:

G1 Water

G2 Plasma sample

G3 Glucose standard (3.00 mM)



3. Incubate the tubes at room temperature for 10min. It is not critical that the incubation timebe exact. Rather, it is important that the reac-tion be allowed to run to completion (nochange in A340). When the reaction is com-plete, add 1 ml of water to each of the five tubesbefore taking a final A340 reading.

4. Blank your spectrophotometer at 340 nmagainst a water reference. Record the absor-bance of all of your samples, G1 through G5.

5. Subtract the absorbance value of G1 (blank)from the rest of the samples. Construct a stan-dard curve of A340 versus micromoles of glu-cose for your three glucose standard solutions(G3, G4, and G5).

6. Based on the A340 of your plasma sample and thevolume of plasma used in the assay, determine theconcentration of glucose in your plasma sample.

NOTE: The normal or expected concentrationof blood glucose in an adult human is �4.0 mMto 6.5 mM. How does the concentration of glu-cose in the pig serum sample compare to thatexpected in human serum?

Blood Lactate Assay

1. Add 2.0 ml of lactate reagent to three 13-by-100-mm test tubes labeled L1 through L3.


L1 Water

L2 Plasma sample

L3 Lactate standard (4.44 mM)

3. Incubate the tubes at room temperature for 10min. It is not critical that the incubation timebe exact. Rather, it is important that the reac-tion be allowed to run to completion (nochange in A540). When the reaction is com-plete, add 1 ml of water to each of the threetubes before taking a final A540 reading.

4. Blank your spectrophotometer at 540 nmagainst the sample blank (L1). Record the ab-sorbance of your samples, L2 and L3.


5. Use the following equation to determine theconcentration of lactate in your plasma sample:

[Lactate] (mM)

� � 4.44 mM

NOTE: The normal or expected concentrationof blood lactate in an adult human is �0.3 mMto 1.3 mM. The assay is reported to be accu-rate for plasma samples containing as much as�13.0 mM lactate. How does the concentra-tion of lactate in the pig serum sample com-pare to that expected in human serum?

Blood Urea Nitrogen Assay

1. Add 2.0 ml of urea nitrogen reagent to five 13-by-100-mm test tubes labeled U1 through U5.


U1 Water

U2 Plasma sample

U3 Urea standard (3.57 mM)




3. Incubate the tubes at room temperature for 5min. It is not critical that the incubation timebe exact. Rather, it is important that the reac-tion be allowed to run to completion (nochange in A340). When the reaction is com-plete, add 1 ml of water to each of the five tubesbefore taking a final A340 reading.

4. Blank your spectrophotometer at 340 nmagainst a water reference. Record the absor-bance of all of your samples, U1 through U6.

5. Subtract the absorbance value of U1 (blank)from the rest of the samples. Construct a stan-dard curve by plotting the absolute value of A340

versus micromoles of urea nitrogen for your foururea standard solutions (U3, U4, U5, and U6).

6. Based on the A340 of your plasma sample andthe volume of the plasma sample used in the as-say, determine the concentration of urea nitro-gen in your plasma sample.

NOTE: The normal or expected concentrationof blood urea nitrogen in an adult human is

A540 plasma sample (L2)A540 lactate standard (L3)

�2.0 mM to 6.0 mM. How does the concen-tration of urea nitrogen in the pig serum sam-ple compare to that expected in human serum?

Blood Creatine Kinase Assay

This assay is different from that of the glucose, lac-tate, and urea nitrogen assays in that you are at-tempting to measure the activity of an enzymerather than the concentration of a molecule. Re-member that in an enzymatic activity assay, it is criti-cal that the time between absorbance readings be recordedexactly.

1. Add 2.5 ml of creatine kinase reagent to two 13-by-100-mm test tubes labeled CK1 and CK2.

2. Dilute your plasma sample 1�5 in distilled wa-ter. This dilute plasma sample will be used inthe assay described below.

3. Add 100 �l of distilled water to CK1 and 100 �lof the dilute plasma sample to CK2. Shake gen-tly to mix and allow the reaction to incubate atroom temperature for 5 min.

4. Blank your spectrophotometer at 340 nmagainst a water reference. Record the ab-sorbance of CK1 and CK2 at exactly 5 min and10 min.

5. Determine the �A340 of your plasma sample(CK2) and the blank (CK1) by subtracting theA340 for the 5-min reading from the A340 forthe 10-min reading for each of these tubes.

6. To determine the concentration of creatine ki-nase in your plasma sample (units/liter), use thefollowing equation:

[Creatine kinase] (units/liter)

�

where 2.6 � total volume of reaction (ml); 5min � total reaction time; 1000 � conversionfactor to convert micromoles per milliliter tomicromoles per liter; 6.22 mM�1 cm�1 � mil-limolar extinction coefficient for NADPH at340 nm; l � pathlength of spectrophotometer(1.3 cm for benchtop colorimeters); and 0.02 �volume of plasma sample used in the assay (inmilliliters), taking into account the dilution.

NOTE: The normal or expected concentrationof blood creatine kinase in an adult human is

([�A340 CK2 � �A340 CK1] per 5 min)(2.6)(1000)

(5 min)(6.22 mM�1 cm�1)(l)(0.02)


�10 to 50 units/liter. One unit of creatine ki-nase is defined as the amount of the enzymethat will produce 1 �mol of ATP from phos-phocreatine and ADP per minute under thesereaction conditions. How does the activity ofcreatine kinase in the pig serum sample com-pare to that expected in human serum?

Exercises

1. You are presented with two individuals: one ofwhom completed a marathon earlier that dayand one who has been watching television allafternoon. What test would you use to deter-mine which individual had completed amarathon? Describe in your answer for whatmolecule you would assay, the relevant featuresof the test, and how the results for each indi-vidual would be different.

2. The assays for serum glucose and creatine ki-nase were similar in many respects. List thechemical reactions that were common to bothassays. Also, describe how these two similar as-say systems could be used to quantify these twovery different biomolecules.

3. Two of the molecules that were assayed for inpig serum are often used to assess overall liverfunction. Name these two molecules. Usingyour knowledge of the gluconeogenesis path-way and the urea cycle, explain how increasedor decreased levels of these two moleculescould be used to assess the overall function ofthe liver.

4. Why do you think that serum creatine kinaselevels might be elevated in patients with seem-ingly unrelated conditions such as alcoholismand epilepsy?

5. Why was it necessary to know the exact incu-bation times for the creatine kinase assays per-

formed in this experiment, while it was less im-portant in the blood urea nitrogen, glucose, andlactate assays?

REFERENCES

Barhan, D., and Trinder, P. (1972). An Improved ColorReagent for the Determination of Blood Glucoseby the Oxidase System. Analyst 97:142.

Bondar, R. J. L., and Mead, D. C. (1974). Evaluationof Glucose-6-Phosphate Dehydrogenase from Leu-conostoc mesenteroides in the Hexokinase Method forDetermining Glucose in Serum. Clin Chem 20::586.

Gotchman, N., and Schmitz, J. M. (1972). Applicationof a New Peroxide Indicator Reaction to the Spe-cific, Automated Determination of Glucose withGlucose Oxidase. Clin Chem 18:943.

Hess, J. W., Murdock, K. J., and Natho, G. J. W.(1968). Creatine Phosphokinase—A Spectrophoto-metric Method with Improved Sensitivity. Am JClin Pathol 50:89.

Kaplan, A., Szabo, L. L., and Opheim, K. E. (1988).Clinical Chemistry: Interpretation and Techniques, 3rded. Philadelphia: Lea & Febiger.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Bioenergetics and Metabolism. In: Principlesof Biochemistry, 2nd ed. New York: Worth.

Martinek, R. G. (1969). Review of Methods for Deter-mining Urea Nitrogen in Biological Fluids. J AmMed Technol 31:678.

Rosalki, S. B. (1967). An Improved Procedure forSerum Creatine Phosphokinase Determination. JLab Clin Med 69:696.

Sigma Chemical Co., Procedure No. 16-UV (1997).Sigma Chemical Co., Procedure No. 735 (1997).Sigma Chemical Co., Procedure No. 66-UV (1997).Sigma Chemical Co., Procedure No. 45-UV (1997).Stryer, L. (1995). Metabolic Energy. In: Biochemistry,

4th ed. New York: Freeman.Trinder, P. (1969). Determination of Glucose in Blood

Using Glucose Oxidase with an Alternative OxygenAcceptor. Ann Clin Biochem 6:24.

q z

263

Antibodies, or immunoglobulins (Ig), are proteins pro-duced by cells of the immune system that allow thebody to resist viral and bacterial infection, as wellas cells that have undergone malignant transforma-tion. Antibodies act by binding with high specificityto foreign molecules, called antigens. The cells thatproduce and secrete antibodies are called B cells.An individual B cell can differentiate to produce asingle type of antibody that will recognize a singleepitope (region) on a single antigen molecule. Be-cause of their high degree of specificity and oftenhigh affinities for antigens (KD � 10�6 to 10�10 M),antibodies have proven to be powerful tools for bio-chemical studies that require detection and/orquantitation of a specific antigen.

The structural characteristics common to allimmunoglobulins is shown in Figure IV-1. An an-tibody is comprised of two identical heavy chainand two identical light chain subunits that arelinked by several disulfide bonds. Both the heavyand the light chains possess a constant region at theirC-termini and a variable region at their N-termini.Depending on the type of antibody (see below),one or more N-linked carbohydrates may be foundat different positions along the heavy chains. It isthe N-termini of the heavy and light chains that arevariable in amino acid sequence. Together, these vari-able (V) regions form a binding site that interacts withthe antigen and thereby determines the specificity of theantibody (see Fig. IV-1). Because each “arm” of theimmunoglobulin is made up of identical heavy andlight chain subunits, a single antibody molecule hasthe ability to bind the identical epitope or antigenicdeterminant on two individual molecules of the

antigen. In other words, a single antibody can bindtwo molecules of the antigen.

Once the epitope of a single molecule of anti-gen is bound by an antibody molecule, other anti-bodies that bind the same epitope are unable to bind(Fig. IV-2). Antibodies that bind to different epi-topes on the same antigen may still be able to bind.These properties underlie the basis of competitionand capture assays, as developed by many diagnos-tic testing laboratories. Such assays are frequentlyused to determine the amount of a specific antigenthat is present in clinical samples (e.g., shed tumor-associated antigens that might be diagnostic for aparticular type of cancer).

Types of Antibodies

The process of stem-cell differentiation into amature, antibody-secreting B cell is complex. Thefirst part of this process, which takes place in thebone marrow, occurs before the B-cell precursoris exposed to any potential antigens, and is there-fore termed antigen independent differentiation. Asthe B-cell precursor migrates to the lymph andcontinues to mature, it eventually reaches a stagein its differentiation at which further developmentwill be guided by a particular antigen that it hap-pens to encounter (antigen-dependent differentia-tion). The terminal step in differentiation willgive rise to a plasma B cell that secretes a singleimmunoglobulin. The secreted immunoglobulinis classified as one of the five isotypes describedbelow.

SECTION IV

Immunochemistry

Introduction

264

cular bound fluid, and is the major component re-quired for activation of the complement system.IgM antibodies typically have low intrinsic bindingaffinities (i.e., at a single antigen-binding site) buthigh avidities (i.e., multiple sites on the IgM bindvery well to multivalent ligands).

Immunoglobulin D (IgD)

Although the role of IgD in the immune system hasnot yet been fully elucidated, it appears to be in-volved in determining if a particular B-cell precur-sor will continue to develop and differentiate.While an immature B cell expresses only IgM onthe cell surface, a mature B cell expresses both IgMand IgD on the cell surface. A B cell that recog-nizes any soluble self-antigen with its surface IgM orIgD will quickly halt production of IgM and in-crease expression of surface IgD. This high con-centration of surface IgD will place the B cell in a

Immunoglobulin M (IgM)

Immunoglobulin M is the first antibody isotype tobe expressed on the surface of an immature B-cellprecursor as the cell leaves the bone marrow andenters the lymph to undergo antigen-dependentdifferentiation. Unlike IgG (see Fig. IV-1), IgM hasno hinge region between the heavy- and light-chainjunction, and it is heavily glycosylated on the mostC-terminal regions of the heavy chain. Immature Bcells that recognize self-antigens on tissue surfaceswith their membrane-bound IgM will quickly un-dergo programmed cell death (apoptosis), prevent-ing the onset of various autoimmune diseases. Bcells that do continue to differentiate into IgM-secreting cells (those that do not recognize surfaceself-antigens) will produce a pentameric form of theantibody (5 immunoglobulins with 10 identicalantigen binding sites, Fig. IV-3). This pentamericform of IgM resides almost exclusively in the vas-

SECTION IV Immunochemistry

"Hinge"region

Heavy chain (Fc portion of IgG)

Light chain

Antigen-binding sites

C

C

V

C

V

C

C

V

C

V

V = variable regionsC = constant regions

= disulfide bridges

Figure IV-1 Structure of immunoglobulin G.

self-antigens with their surface IgD prevents theonset of various autoimmune diseases. Structurally,IgD is similar to IgG (see Fig. IV-1), but it containsa number of N-linked carbohydrate groups near the

state of anergy, in which it will no longer be allowedto develop into a fully mature, antibody-secretingB cell. As is the case with surface IgM, selectionagainst B-cell precursors that recognize soluble

SECTION IV Introduction 265

IgG1

IgG1

IgG1

a

b

Antigen

Antigen

Antigen

IgG2

IgG2

Figure IV-2 Interaction of an antibody with an antigen. (a) One molecule of antibody can bind the same

epitope or region on two individual molecules of the antigen. (b) Once a single epitope of an

antigen is bound by a single molecule of the antibody (IgG1), another molecule of the same

antibody will not be able to bind the same molecule of antigen. A different antibody (IgG2)

may still be able to bind the same molecule of antigen at a different site or epitope.

266

a more indirect role in protecting the body from in-fection, particularly by parasitic organisms. WhileIgE has affinity for surface antigens found on for-eign pathogens that may enter these tissues, it alsohas high affinity for receptors present on the mastcells. As IgE binds foreign antigens, it signals themast cells to secrete powerful compounds that in-duce the contraction of these tissues. The vomit-ing, coughing, and sneezing that are induced bythese contractions function to rid the body of theinvading pathogen or agent. The IgE-mediated re-sponse can also give rise to allergic reactions towarddust, pollen, mold, spores, and other environmen-tal agents that the immune system becomes sensi-tive to through the development of IgE antibodiesspecific for these antigens (allergens). Like IgM,IgE is larger than the other isotypes and lacks thehinge region between the heavy- and light-chainjunction. Like IgM, IgD, and IgA, IgE is heavilyglycosylated throughout the heavy-chain subunit.

Immunoglobulin G (IgG)

Immunoglobulin G is the most abundant isotypefound in the blood. Unlike the pentameric form ofIgM, which provides protection due largely to itsability to activate the complement system, IgG of-fers protection through opsonization of the invadingpathogen or agent. After IgG binds the antigen, itcan induce macrophages, neutrophils, and otherphagocytes to engulf and destroy it. This processoccurs after the macrophage or other cell binds theantigen–antibody complex through IgG receptors(Fc receptors) on their cell surface. IgG is the ma-jor immunoglobulin isotype induced with the re-peated introduction of a foreign antigen into thebloodstream (hyperimmunization), and generallyexhibits the highest intrinsic binding affinities forantigens. Thus, IgG is the most common isotypeused in biochemical studies (see below).

Polyclonal and MonoclonalAntibodies

An immunogen is defined as an agent that can elicitan immune response. An antigen is defined as anymolecule that a particular antibody has affinity for,or recognizes. While it is possible to raise antibod-ies against virtually any protein, hormone, or chem-

junction between the variable and constant regionsof the heavy chain.

Immunoglobulin A (IgA)

Immunoglobulin A is the isotype responsible forconferring protective immunity to the mucousalmembranes that line the intestinal, respiratory, andurogenital tracts. IgA acts by neutralizing foreignbacteria and viruses that enter the body via thesepathways. Possessing a high affinity (and avidity) forsurface antigens on these types of pathogens, IgAbinds them and prevents their attachment to, andsubsequent infection of, the epithelial cells that linethese tissues. Unlike IgD and IgM, IgA functionsmainly as a dimer in these areas of the body, althoughvery low levels of the monomeric form of IgA may befound in vascular bound fluids. The structure of theIgA isoform most closely resembles that of the IgDisoform, containing a heavily glycosylated stretch ofamino acids at the junction between the variable andconstant regions of the heavy-chain peptides.

Immunoglobulin E (IgE)

The immunoglobulin E isotype is associated withmast cells of the skin, mucosal membranes, and ves-sels that line connective tissue. IgE appears to play


10

1 2

3

4

5

67

8

9

AS

S

B

CD

E

SS

S S

SS

SS

Figure IV-3 The pentameric form of IgM: five

disulfide-linked immunoglobulins

(A to E) with 10 identical antigen-

binding sites.


ical, not all potential antigens are themselves im-munogenic. There are two methods that can be em-ployed to elicit an immune response to an inher-ently nonimmunogenic compound in an attempt toraise antibodies that will recognize it.

The first method involves the use of an adju-vant, an immunostimulant that will increase the im-munogenicity of other compounds that are mixedwith it. If a protein is mixed with heat-killed my-cobacteria in an oil/water emulsion (Freund’s adju-vant) and injected into a host animal, the host’s im-mune system will mount a response to the proteinin the mixture and produce antibodies that will haveaffinity for it. By delivering the protein antigen inan oil emulsion, the protein is slowly released intothe bloodstream so that it can be engulfed bymacrophages and recognized by mature B cells ex-pressing surface IgM and IgD. The componentspresent in the mycobacteria are also crucial in thatthey stimulate macrophages and proliferation ofother immune system cell types that will eventuallybe required for B cell activation. Another adjuvantcontaining aluminum hydroxide and heat-killedBordetella pertussis (the pathogenic bacteria thatcauses whooping cough) is becoming more fre-quently used because of its ability to enhance theimmune response to immunogenic antigens.

The second method involves chemically conju-gating a small molecule (hapten) to an immunogenicprotein carrier. This procedure is often used to raiseantibodies against small organic molecules such asnitrophenols, antibiotics, and hormones, which arenormally not able to elicit an immune response ontheir own. Although this method will lead to theproduction of antibodies that recognize both theprotein carrier and the hapten–carrier conjugate, asubpopulation of antibodies will also be able to rec-ognize only the hapten.

As the antigen of interest is injected into thebloodstream of the host animal, it will encounter Bcells expressing both surface IgD and IgM. Threeevents will take place that will ultimately lead to theactivation and proliferation of B cells that recog-nize the injected antigen. First, the surface IgM onthe B cell will become cross-linked as it binds theantigen, triggering a phosphorylation cascade thatwill release intracellular Ca2�, activate Ca2�-de-pendent kinases, and eventually lead to the activa-tion (phosphorylation) of various transcription fac-tors that promote cell division (proliferation).

Second, some of the IgM-bound antigen will be en-gulfed by the B cell, degraded, and presented againon the surface of the B cell as small peptides boundby the major histocompatibilty complex class II (MHCII). Eventually, this MHC II complex is bound byT helper cells that secrete interleukins-4, 5, and 6.The net effect of T-helper-cell binding and inter-leukin secretion is that the B cell will undergo rapiddifferentiation and proliferation. Third, folliculardendritic cells in the lymph nodes will bind the B celland stimulate its growth and development via a sep-arate phosphorylation cascade. Remember that eachB cell produces a single antibody that recognizes a singleepitope on the antigen with a defined affinity. A popu-lation of antibodies derived from many different Bcells will therefore recognize different epitopes onthe antigen with different affinities and with po-tentially different attractive forces (electrostatic,hydrogen bonding, hydrophobic, etc.).

Polyclonal Antibodies

Polyclonal antibodies are a population of differentimmunoglobulins derived from different B cellsthat recognize an antigen. As such, they frequentlybind to different epitopes on the antigen with dif-ferent affinities. The process of raising polyclonalantibodies against an antigen of interest is relativelysimple (Fig. IV-4). The antigen–adjuvant mixtureis first injected into the bloodstream of the host an-imal. About 7 days after this primary injection, theconcentration of serum IgG that is able to recog-nize the antigen will begin to rise. About 13 to 17days following the primary immunization, theserum level of IgG that recognizes the antigen ofinterest will be at a maximum. The host organismwill then begin to “clear” the antigen from its sys-tem, causing the concentration of antigen-specificserum IgG to decrease back to pre-immunizationlevels.

If the same host animal is again immunizedwith the antigen (a “booster” injection), the im-mune response mounted against it will be bothgreater in magnitude and more rapid than the re-sponse seen following the primary immunization.Approximately 4 days after the secondary immu-nization, the serum levels of antigen-specific IgGwill be at a maximum (as compared to 13 to 17 daysfollowing the primary immunization). In additionto being more rapid, the response to the secondary

268 SECTION IV Immunochemistry

Immunize the animal with an adjuvant–antigen mixture.After several days, the concentration of antibodies in the serum that recognize the antigen will begin to rise.

Analyze the serum for the presence of antibodies thatinteract with or recognize the antigen using a Westernblot or ELISA.

Red blood cells, fibrin, platelets, etc.in the clot

Collect blood from the animal,allow it to clot, and isolate the serum.

Serum

Y

Y

Y

Y

YY Y

Y

YY

Y

Y

Figure IV-4 Raising polyclonal antibodies

against an antigen of interest.

immunization will give rise to a higher concentra-tion of antigen-specific IgG in the serum. Thispopulation of IgG molecules may also have ahigher intrinsic binding affinity for antigen. Theimmune response mounted against the antigen af-ter repeated injection is more intense and morerapid due to the fact that the primary immuniza-tion gave rise to a number of memory B cells thatwill allow the host organism to recognize the anti-gen more quickly and efficiently upon future ex-posures to it. This process of hyperimmunization(multiple rounds of antigen injection), blood col-lection, and serum isolation may be repeated manytimes on an individual animal to obtain large quan-tities of antibodies that will recognize the antigenof interest.

There are several things you must keep in mindwhen attempting to raise a polyclonal antibody

against an antigen. First, you must consider whathost animal should be used in raising the antibody.The immune system naturally selects against B cellsthat produce surface immunoglobulins that recog-nize “self ” antigens in their early stages of devel-opment. Therefore, it may be difficult to raise an-tibodies against a mouse protein by injecting it intoanother mouse. This same mouse antigen, however,may prove to be quite immunogenic when intro-duced into another species, such as a rat, horse,goat, guinea pig, or rabbit. In general, it is best touse a host animal for antibody production fromwhich the antigen of interest was not derived.

Second, you must remember that the serumfrom an immunized animal contains thousands ofdifferent IgG molecules, of which only a small frac-tion (perhaps less than 1%) actually recognize theantigen of interest. The IgG present in the serumprior to the introduction of the antigen is still pres-ent. The immunization process merely attempts to increase the relative concentration of antigen-specific antibodies in the serum with respect to thenumerous other nonspecific antibodies that are alsopresent.

Finally, the heterogeneity of the population ofantibodies in the serum that do recognize the anti-gen make polyclonal antibodies very useful in somebiochemical applications and less useful in others.This stems from the fact that the different anti-bodies can interact with different epitopes on theantigen with a wide variety of prevailing attractiveforces (electrostatic, hydrophobic, hydrogen bond-ing, etc.). This topic will be discussed more thor-oughly later in this introduction.

Monoclonal Antibodies

Monoclonal antibodies are a population of identicalimmunoglobulins all derived from a single B cell.Because of this, they all recognize a single epitopeon the antigen with the same affinity. The processof producing monoclonal antibodies is more com-plex than that used to produce polyclonal antibod-ies (Fig. IV-5). Following hyperimmunization andseveral days of incubation, the animal is killed toobtain its spleen, which contains a large populationof B cells. These B cells are then fused withmyeloma cells (transformed or immortalized cellsof the immune system) with the addition of poly-ethylene glycol to the cell mixture. To select for re-


Immunize mouse with antigen–adjuvant mixture and incubate.Remove spleen cells.

Each hybridoma secretes a single IgG

B cells (mortal) Myeloma cells (immortal)

Polyethylene glycol

Hybridoma(B cell fused withmyeloma cell)

Unfused myeloma cell

Unfused B cell

Limiting dilution

Microtiter platecontaining HATmedia

Mortal B cells willnot grow in culture

Will grow in cultureand express HGPRT

Will not express HGPRT and will die

B B

B

BB

B

B

B

B

B B

B

Y YYY YY Y Y YYYY

Figure IV-5 Production of a hybridoma cell line that produces a monoclonal antibody against an antigen

of interest.


sultant hybridoma cells in culture (mortal B cellsfused with immortal myeloma cells), the mixture isadded to a medium that contains hypoxanthine-aminopterin-thymidine (HAT). Aminopterin is anantifolate that blocks de novo purine nucleotidebiosynthesis, making cells dependent on hypoxan-thine and thymidine for growth. B cells normallyexpress an enzyme called hypoxanthine guanosinephosphoribosyl transferase (HGPRT), which allowscells to take up hypoxanthine. The myeloma fusionpartner does not express this enzyme. A cell thatdoes not express HGPRT, therefore, will not beable to grow in the presence of HAT. Thus, any Bcell that has fused with a myeloma cell (a hy-bridoma) will be immortal, will produce HGPRTderived from the B cell, and will be able to grow inthe HAT-containing medium. Myeloma cells thatdid not fuse with a B cell will not express HGPRTand will die in HAT medium. Although unfused Bcells do express HGPRT, their lifespan in cultureis sufficiently short that they too will be unable togrow effectively in the HAT-containing medium. Ineffect, only hybrid cells that are immortal and whichexpress HGPRT will grow effectively.

Once a collection of hybridoma cells has beenisolated, a long screening and cloning process liesahead. After identification of cultures that containthe desired antibody (the hybridoma cell will se-crete antibody into the growth medium), limitingdilutions of the hybridoma population are made inHAT media in a 96-well microtiter plate. This lim-iting dilution is performed in an effort to obtainone hybridoma cell per well. These hybridomas areallowed to grow, and the medium is again analyzedfor the presence of antibody that binds to the anti-gen of interest. After one or several positive clonesare identified, an additional round of limiting dilu-tion and screening is performed. Each hybridomaclone represents a single, immortalized B cell that willproduce a single (monoclonal) antibody. A number ofdifferent selective criteria may be employed duringthe screening and cloning process. For instance,you may wish to clone a hybridoma that producesan antibody with a defined affinity (KD � 10�8,10�6, etc.). You may wish to raise an antibody thatrecognizes a particular epitope on the antigen of in-terest. You may wish to raise an antibody that bindsthe antigen with a certain type of prevailing attrac-tive force (e.g., electrostatic as opposed to hy-drophobic). If the screening process is designed and

performed properly, it is possible to isolate a hy-bridoma that produces an antibody that fits any se-lective criterion.

Purification of Antibodies

A hybridoma cell line will secrete a monoclonal an-tibody into the culture medium in relatively pureform. Because of this, monoclonal antibodies re-quire less purification than polyclonal antibodies. Aserum sample containing polyclonal antibodiescontains both antigen-specific and nonspecific im-munoglobulins, as well as multiple other serum pro-teins. For example, a large percentage of the serum(80–90%) is comprised of albumin. Unlike theserum immunoglobulins, serum albumin is solublein a solution at 50% ammonium sulfate saturation.Thus, if a serum sample is slowly brought to 50%saturation with ammonium sulfate, the im-munoglobulins will selectively precipitate out of so-lution, leaving the albumin in the supernatant. Thisvery simple procedure, completed in less than anhour, can increase the purity of a polyclonal anti-body solution by an order of magnitude (see Ex-periment 17).

If you wish to separate the antibodies in a poly-clonal antibody solution that recognize the antigenof interest from those that do not, the specificity ofthe antigen–antibody interaction must be exploited.Because all IgGs possess the same general structure,differing only in amino acid composition within thevariable region, you cannot effectively isolate spe-cific IgGs within the population by differences insize, shape, and surface charge on which generaltechniques in protein purification are based (see in-troduction to Experiment 2). The one and perhapsonly property that will distinguish one IgG fromanother is the specificity of the antigen with whichit is able to interact.

If you are able to chemically conjugate the anti-gen of interest to an inert support or matrix, theantigen will “capture” antibodies in the polyclonalsolution that recognize it as it passes through thematrix (Fig. IV-6). After extensive washing, a pop-ulation of immunoglobulins that recognize epitopeson the antigen will remain bound to the antigen onthe matrix. To isolate the antigen-specific antibod-ies, you have to experiment with different mobile-phase compositions (alter pH, ionic strength, de-


Pass a solution of polyclonalantibodies over an inert matrixconjugated with the antigen.Antibodies that have affinity for the antigen will be retained bythe column. Antibodies thatdo not have affinity forthe antigen will wash through.

IgGs with no affinityfor the antigen

Alter the composition of the wash buffer(pH, ionic strength, etc.) to weaken the antigen–antibody interaction enoughto allow the antibodies to elute fromthe column.

IgGs with a very strongaffinity for the antigenmay not elute.

IgGs with affinity for the antigen.

YY

Y

YY

Y

Y

Y

Y

Y

Y

Y

YY Y

Y

Y

Y

Y

Y

Y

Y

YY

Y

Y

YY

Y

Y

Y

Y

Y

Y

Y

Y

Y Y

Y

Y

Figure IV-6 Immunoaffinity chromatography to

purify antigen-specific antibodies.

tergent concentration, etc.) that will weaken theantigen–antibody interaction enough to allow theantibodies to elute from the column. Ideally, the

mobile-phase composition change will disrupt theantigen–antibody interaction without denaturingthe antibody. Typical elution procedures involvelow pH buffers (0.1 M glycine, pH � 2.0) or 5 MMgCl2. Fortunately, IgG molecules are quite sta-ble, and can often withstand the relatively harshconditions that are often required to elute themfrom affinity columns. Although this type of im-munoaffinity chromatography may be difficult tooptimize, it has been used successfully in the pu-rification of antigen-specific antibodies from poly-clonal antibody solutions. Many of the secondaryantibody reagents that are commercially available(e.g., used in Experiments 17 and 18) have been pu-rified in this manner.

Use of Antibodies in BiochemicalStudies

Western Blotting

Western blotting is one of the most widely used im-munochemical techniques in modern biochem-istry. In general, it is used to detect the presenceof a particular protein antigen in an impure mix-ture. Western blots can provide information thatis (1) qualitative: Is the protein present? (2) quan-titative: How much of the protein is present? and(3) structural: What is the size of the protein? Thegeneral protocol for Western blotting is as follows(see Fig. 18-1):

1. A mixture of proteins is separated bySDS–PAGE.

2. The proteins are electrophoretically trans-ferred from the gel to a membrane and incu-bated with a high concentration of nonspecificprotein to “block” or “mask” the areas of themembrane not already bound with protein.

3. A primary antibody specific for the antigen ofinterest is applied to the membrane to allow theantigen–antibody interaction to take place.

4. The membrane is washed extensively to re-move unbound or nonspecifically bound pri-mary antibody.

5. An enzyme-conjugated (alkaline phosphataseor horseradish peroxidase) secondary antibody,which has a strong affinity for the primary an-tibody, is incubated with the membrane.


6. The unbound and nonspecifically bound sec-ondary antibody is removed with extensivewashing.

7. A substrate for the enzyme-conjugated sec-ondary antibody is applied to the membrane.The product of this reaction, a colored precip-itate or light, is detected on the membrane toreveal the location and relative amount of theantigen of interest.

The technique of Western blotting is demon-strated and described more thoroughly in Exper-iment 18.

Enzyme-Linked Immunosorbent Assay

(ELISA)

Like Western blotting, ELISA is a very populartool used in biochemistry. This technique relieson the ability of a particular antibody to recog-nize or bind a specific antigen that is adsorbed tothe wells of a microtiter plate. Like the Westernblot, an enzyme-conjugated secondary antibodythat has affinity for the primary antibody is uti-lized to detect the presence and relative quantityof primary antibody from one well to the next (seeFig. 17-2). When the enzyme is supplied with theappropriate substrate, it will convert it to a solu-ble colored product, the relative concentration ofwhich can be determined easily from well to wellwith the use of a specialized, 96-well plate spec-trophotometer. By monitoring the change in ab-sorbance of a collection of wells containing dif-ferent dilutions of the antibody, the titer of anantibody solution (the inverse dilution of the an-tibody required to give a half-maximum signal)can be determined.

The main advantage of the ELISA over West-ern blotting is its ability to analyze many samplesat one time. In addition, modifications to the gen-eral ELISA protocol can be used to quantify the ex-act concentration of antigen in an unknown solu-tion, the concentration of a specific antibody in anunknown solution, the dissociation constant for aparticular antigen–antibody interaction, or the rel-ative affinity of two different antibodies for thesame antigen. An example of an indirect ELISA,and its use in the determination of the concentra-tion of an antigen in an unknown solution, is thesubject of Experiment 17.

Radioimmunoassay (RIA)

In many respects, the radioimmunoassay is similar tothe ELISA. The two techniques differ only in theirmethod of detection of antibody when bound to theantigen. While the traditional ELISA detects thebinding of the antibody by monitoring the appear-ance of a soluble colored product after the enzyme-conjugated antibody acts on the appropriate sub-strate, the radioimmunoassay detects the antigen–antibody complex with a radioactive atom cova-lently attached to either of the two components.For instance, if the antibody is labeled with 125I andadded to a microtiter plate well coated with theantigen, the amount of radioactivity present in thewell after washing will be directly proportional tothe amount of antibody that bound the antigen.This assay can be altered by coating the well withthe antibody of interest and incubating it with a125I-labeled antigen. The amount of antigen “cap-tured” by the antibody, then, is directly propor-tional to the amount of antigen in the unknown so-lution (Fig. IV-7).

In addition to using radiolabeled antibodies inthe ELISA format, you can also use them in the

Adsorb the antigen to the wells of a microtiter plate and "block"the unoccupied areas of the platewith bovine serum albumin.

Add the antibody labeled with 125I (*).Wash extensively to remove unboundor non-specifically bound antibody.

The amount of 125I retainedon the well is proportional tothe amount of antibody thatbound the antigen.

Figure IV-7 The radioimmunoassay (RIA).


Western blot format. In general, radiolabeled anti-bodies alleviate the need for enzyme-conjugatedsecondary antibodies in both the ELISA and theWestern blot. The major disadvantage to using 125I-labeled antibodies is the hazard associated with thisisotope (125I is volatile in its unconjugated form andemits very high-energy � rays).

Immunoprecipitation

As discussed earlier, a single IgG antibody has twoidentical antigen-binding sites or “arms.” In thecase of polyclonal antibodies, the antigen-binding

sites and the epitopes on the antigen that they rec-ognize differ among members of the population. Ifthe antigen and antibody are mixed together in cor-rect proportions, the different polyclonal antibod-ies can bind to different regions on a number of dif-ferent antigen molecules to create a large complex(Fig. IV-8a). These large protein aggregates are of-ten insoluble, precipitating out of solution. If a fixedamount of either the antigen or antibody is incu-bated with increasing concentrations of the othercomponent, you will be able to determine an opti-mal antigen:antibody ratio for use in future im-munoprecipitation experiments.

Antigen is cross linked by the antibodies to produce an insoluble aggregate.

SDS-PAGE

Aggregate Pure antigenIgG heavy chain

(~50kDa)

IgG light chain(~25kDa)

Any molecule associated strongly with the antigen of interest in solution will precipitate with the insoluble antigen-antibody complex.

SDS-PAGE

Aggregate Pure antigenIgG heavy chain

IgG light chain

b

a

Protein that interacts strongly with the antigen

Figure IV-8 Use of immunoprecipitation to purify antigens and antigen associated proteins.


Although immunoprecipitation is not used aswidely in modern biochemistry, it has been used inthe past to study protein–protein interactions. If anantigen of interest has a strong affinity for anothermolecule present in the solution, the associated mol-ecule will precipitate in the complex along with theantigen that is recognized directly by the antibody(Fig. IV-8b). This technique is often used to iden-tify new members of physiologically important mul-tiprotein complexes when one component of thecomplex has been purified, but others are unknown.

If the antibody is unable to cross-link or aggre-gate the antigen of interest to the extent needed toinduce precipitation of the complex (as is often thecase with monoclonal antibodies), an alternativemethod may be employed. Protein A from Staphy-lococcus aureus is known to have a high affinity forimmunoglobulins. If a soluble antigen–antibodycomplex is incubated with an insoluble matrix co-valently conjugated with Protein A, the complexwill be captured by the Protein A covalently at-tached to the matrix (Fig. IV-9). Because Protein A

Add antibody specific for antigen of interest to impure solution

Add an insoluble matrixconjugated with protein A ( )

Centrifuge

Remove supernatant

Components in solution not bound or recognized by the antibody

Resuspend matrix in elution buffer that will weaken the affinity of protein A for the immunoglobulin

Purified antigen–antibody complex

Figure IV-9 Using Protein A to purify antigen–antibody complexes.


has affinity for most immunoglobulins (not justthose that recognize the antigen of interest), it is oflittle use in purifying antigen-specific antibodiesfrom a polyclonal antibody solution. ImmobilizedProtein A is used, however, to purify immunoglob-ulins from serum samples.

Immunofluorescence Microscopy and

Immunohistochemistry

These two techniques are used to identify the loca-tion of a particular antigen within a cell or tissue.In addition to covalent coupling of antibodies toenzymes for Western blots and ELISAs, antibodiescan also be conjugated to fluorescent dyes. After a tis-sue section or cell has been isolated, ruptured, andfixed to a coverslip, it can be labeled with a fluo-rescently “tagged” antibody that is specific for aparticular antigen believed to reside in it. As thesample is subjected to incident light, the antibod-ies bound to the antigen of interest will emit lightin the visible spectrum and be detectable with a flu-orescence microscope. Because there are a numberof different dyes available that excite at differentwavelengths and emit light of different colorswithin the visible range of wavelengths, a numberof different fluorescent antibodies can be applied to a single cell or tissue section simultaneously. The results of this type of experiment can provideinsight into the cellular location of numerous pro-teins and small molecules, providing the investiga-tor with an understanding of the in vivo environ-ment of the molecule under study.

The technique of immunohistochemistry is verysimilar to fluorescence microscopy. This techniquediffers only in the method of detection or localiza-tion of the antibody and can be performed with aconventional light microscope. As with the ELISAand Western blot, the antibody used in this exper-iment is covalently conjugated to an enzyme, suchas horseradish peroxidase. This enzyme is then in-cubated with a substrate that is converted to an in-soluble colored product that will precipitate or de-posit at the site of enzyme activity. The distributionand location of the colored product is readily de-tected with an ordinary light microscope.

Another related technique utilizes antibodiesthat are conjugated to large, electron-dense atoms,such as gold. When used in conjunction with elec-tron microscopy, these probes can be used to de-

termine the subcellular localization of an antigen ofinterest with high resolution.

Flow Cytometry

In addition to the technique of fluorescence mi-croscopy, fluorescently labeled antibodies are cur-rently used in flow cytometry. This technique allowsthe identification and isolation of individual cells ina complex mixture according to the antigens they ex-press on their surface. For example, among a diversepopulation of cells, some may express an antigen, X,on their surface. After incubating the cells with afluorescently labeled anti-X antibody, the cells arepassed in a narrow tubing through a laser beam. Asthe laser hits the surface of a single cell, it will ex-cite the fluorescently labeled antibody bound toantigen X on the surface of the cells that express it.Each time a fluorescently “tagged” cell passesthrough the laser, a sensitive photomultiplier tube(PMT) will detect and count that event (Fig. IV-10).More than 100,000 cells can be passed through thelaser beam in less than a minute. Cells that do notexpress antigen X on their surface will not fluoresce(although they will scatter the incident light to someextent) and will not be detected by the PMT. Thisapproach can determine the exact number or per-centage of cells in a diverse population that expressantigen X on their surface. In addition, the PMTcan distinguish between differences in fluorescentintensity from one cell to another, providing insightinto variations in the level of expression of antigenX from cell to cell. For example, if cell 1 is five timesgreater in fluorescent intensity than cell 2 as it passesthrough the laser, then the concentration of antigenX on the surface of cell 1 is five times greater thanthat on cell 2.

A flow cytometer may be equipped with morethan one PMT, each capable of detecting fluo-rescence of a particular wavelength. Thus, a cellpopulation can be incubated with more than oneantibody, provided that the fluorescent dyes con-jugated to them emit light of different wave-lengths. This “double label” experiment will al-low you to distinguish cells expressing bothantigens from those expressing either of the twoantigens individually.

A fluorescence-activated cell sorter (FACS) is a flowcytometer capable of isolating individual cellswithin a diverse population that are expressing a


surface antigen of interest. The detection mecha-nism of the FACS is identical to that of the con-ventional flow cytometer. The difference lies in thecomputer-generated feedback system of the FACSthat is absent in the flow cytometer. As the PMTof the FACS identifies a fluorescent cell, the in-strument divides the population into a fine mist thatcarries a single cell per droplet. Each fluorescentdroplet is charged by the PMT and passed throughan electrode that carries a charge opposite that of thedroplet. Any cell that expresses antigen X (fluores-cently labeled) can thereby be separated from theremaining cell population as the mist passes acrossthe charged plate (Fig. IV-11). Cell populations iso-

lated by this technique can be propagated in cul-ture and utilized in future studies.

Advantages and Disadvantagesof Polyclonal and MonoclonalAntibodies

Both polyclonal and monoclonal antibodies haveproven to be very effective in a number of differ-ent biological applications, but there are distinct ad-vantages and disadvantages to employing each typeof antibody for a particular study. First, polyclonalantibodies are generally more stable than mono-

Thin stream of solvent carryinga diverse population of cells, some of which are labeled on acell-surface antigen with a fluorescent anibody ( ).

Cells labeled only with antibody 1

Cells labeled only with antibody 2

Cells labeled with both antibodies

Cells are not labeled

PMT 3 PMT 2

PMT 1

A series of photomultiplier tubes (PMTs) will detect different wavelengths of light emitted by excited, fluorescently labeled antibodies

on the surface of each cell as it passes through the laser.

Intensity of fluorescence fromantibody 2

Inte

nsity

of f

luor

esce

nce

from

antib

ody

1

Laser

Figure IV-10 Schematic diagram of a flow cytometer.


clonal antibodies. While monoclonal antibodiesmay begin to denature after several weeks of stor-age at �80°C, polyclonal antibody serum can oftenbe stored at 4°C for years before any significant lossof titer is observed. While polyclonal antibodies aremore stable, they differ from monoclonal antibod-ies in that their supply is limited. Once the host an-imal dies, the source of a particular preparation ofpolyclonal antibody is gone. While the immuniza-tion process can be repeated on another host ani-mal, a different polyclonal antibody will be pro-duced that requires its own quality control andtesting. In contrast, monoclonal antibodies are pro-duced from a single immortalized B cell. The cellline can be stored indefinitely (as a frozen culturestock) and grown whenever more of the antibodyis needed.

Polyclonal antibody production requires the useof relatively large quantities of pure antigen for im-munization. Remember that the goal of polyclonalantibody production is to increase the relative con-

centration of antigen-specific serum IgG. If an im-pure antigen is used for immunization, the relativeconcentration of serum antibodies that recognizethe impurities in the antigen mixture will be in-creased along with the relative concentration ofantigen-specific antibodies. In contrast, mono-clonal antibodies can often be raised with the in-jection of lower concentrations of impure antigen.Since the thousands of hybridomas produced willbe screened for antigen-specific antibody produc-tion, the B cells that differentiated to produce an-tibodies that recognize antigens other than the oneof interest will eventually be eliminated during thecloning process. In effect, you must find the fewhybridomas that are producing the antibodies of in-terest among the thousands that are screened.

Since polyclonal antibodies recognize a varietyof epitopes on the antigen, they generally work wellin the technique of immunoprecipitation (see be-low). Many polyclonal antibody preparations havealso proven to be effective in the techniques of

Figure IV-11 The fluorescence-activated cell sorter (FACS).

PMT

Cells are labeled on their cell surface with the fluorescent antibody are endowed with a charge

after the stream of solvent is dispersed in a fine mist. Any charged droplet will be

separated from the diverse populationof cells as they pass across a plate

carrying a charge opposite that of the droplets.

Laser


Western blotting and ELISA. The main concernwith the use of polyclonal antibodies for these lat-ter techniques is the potential for cross-reactivityof some of the antibodies with proteins other thanthe antigen of interest. The problem of cross-reac-tivity is easier to control in Western blotting thanin ELISAs, since a Western blot (see Experiment18) will identify the molecular weight of any cross-reacting protein and allow you to evaluate which ofthe signals may be due to nonspecific interactions.Because an ELISA analyzes the extent of interac-tion between all of the antibodies in a solution withall of the potential antigens present in the mi-crotiter plate (see Experiment 17), it is more diffi-cult to determine which of the potentially numer-ous interactions are specific for the antigen ofinterest.

Since monoclonal antibodies recognize a par-ticular epitope on the antigen, they tend to be lesscross-reactive than polyclonal antibodies in West-ern blotting and ELISAs. Also, because the inter-action takes place with a defined affinity (dissocia-tion constant), monoclonal antibodies arefrequently preferred in the technique of im-munoaffinity chromatography (see Introduction toExperiment 2). The wide variety of antigen–anti-body interactions that are present in a polyclonalantibody preparation makes it difficult to define amobile-phase buffer that will promote the elutionof the antigen of interest in good yield from an im-munoaffinity column prepared with polyclonal an-tibodies.

Regardless of the type of antibody you decideto produce, it is likely that you will be able to de-velop effective applications for their use in the studyof a particular biochemical process or pathway. Asurvey of scientific literature indicates that at least25% of all published research articles in the disci-pline of biochemistry utilize antibodies and im-munochemistry in one form or another. In additionto providing scientists with a powerful tool in the

study of biochemistry, natural and recombinant an-tibody technology promises to be of great use inthe treatment and prevention of numerous diseases.

REFERENCES

Alzari, P. M., Lanscombe, M., and Poljak, R. J. (1988).Three-Dimensional Structure of Antibodies. AnnuRev Immunol 6:555.

Burkhardt, A. L., Brunswick, M., Bolen, J. B., andMond, J. J. (1991). Anti-immunoglobulin Stimula-tion of B Lymphocytes Activates src-Related Pro-tein-Tyrosine Kinases. Proc Natl Acad Sci USA88:7410.

Chapus, R. M., and Koshland, M. E. (1974). Mecha-nisms of IgM Polymerization. Proc Natl Acad SciUSA 71:657.

Harlow, E., and Lane, D. (1988). Antibodies, A Labora-tory Manual. Cold Spring Harbor, NY: Cold SpringHarbor Laboratory Press.

Janeway, C. A., Rosen, F. S., Merler, E., and Alper,C. A. (1969). The Gamma Globulins, 2nd ed.Boston: Little, Brown.

Janeway, C. A., Jr., and Travers, P. (1996). Immunobiol-ogy: The Immune System in Health and Disease, 2nded. New York: Garland.


Natvig, J. B., and Kunkel, H. G. (1973). Human Im-munoglobulin: Classes, Subclasses, Genetic Vari-ants, and Idiotypes. Adv Immunol 16:1.

Nemazee, D., and Burki, K. (1989). Clonal Deletion ofB Lymphocytes in a Transgenic Mouse BearingAnti-MHC Class I Antibody Genes. Nature337:562.

Paul, W. E. (1993). Fundamental Immunology, 3rd ed.New York: Raven Press.


Tedder, T. F., Zhou, L. J., and Engel, P. (1994). TheCD19/CD21 Signal Transduction Complex of BLymphocytes. Immunol Today 15:437.

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279

EXPERIMENT 17

Partial Purification of a PolyclonalAntibody, Determination of Titer, andQuantitation of an Antigen Using theELISA

Theory

As discussed in the introduction to Section IV, theenzyme-linked immunosorbent assay (ELISA) hasbecome a widely used and powerful technique inthe field of medical diagnostics. If an individual hasbeen exposed to a pathogenic organism (virus, bac-teria), antibodies that recognize particular antigensspecific to these organisms will be present in theblood plasma. Because of this, a blood test can of-ten reveal the source of an infection and dictate aproper course of treatment.

In addition to being used as a method of iden-tifying and quantifying protein antigens in a sam-ple, the ELISA is also gaining widespread popular-ity as a tool for quantifying low-molecular-weightcompounds (haptens) such as drugs, chemical tox-ins, and steroid hormones. Although many of thesehaptens are not themselves immunogenic (able toelicit an immune response), antibodies specific forthem can often be raised by coupling them to a for-eign protein prior to being injected into the hostanimal. The following is an example of how ELISAtechnology is currently being used: the humanchorionic gonadotropic hormone is produced bythe placenta several days after the onset of preg-nancy. A modified ELISA using an antibody againstthis hormone is used to determine whether an in-dividual is pregnant based on the quantity of thishormone in the urine. The assay is a sensitive, re-liable, an immediate method to verify pregnancy inits very early stages.

In this experiment, you will purify polyclonal an-tibodies against �-galactosidase from the serum ofrabbits immunized with this antigen. Using the

ELISA, you will determine the titer of this anti-body, as well as that of a monoclonal antibody against�-galactosidase, called J1, that is derived from amouse hybridoma cell line. Based on the titer of theJ1 antibody, you will perform a competitive ELISAto determine the concentration of �-galactosidasein an unknown solution.

In many respects, the ELISA performed in thisexperiment is similar to the Western blot describedin Experiment 18. The antigen of interest (�-galactosidase) is first adsorbed to the bottom of a96-well polyvinyl chloride (PVC) or polystyreneplate. Next, the sites on the bottom of each wellnot already bound with protein are “blocked” witha buffered solution containing a high concentra-tion of bovine serum albumin (BSA). This block-ing step is done to ensure that the primary and sec-ondary antibodies added later in the assay will bindonly to the antigen of interest, rather than to thebottom of the well. As with the Western blot, ifthe blocking step is not performed, the backgroundwould be too high to obtain any useful informa-tion (see Experiment 18). Although you will useBSA as the blocking protein in this assay, any pro-tein that does not have affinity for the antibodiescould be used.

At this point, the anti-�-galactosidase antibody(primary antibody) diluted in blocking buffer isadded to the well. During the incubation, it willbind to the �-galactosidase adsorbed to the bottomof the well. Several washes are then performed toremove primary antibody that may have boundnonspecifically to proteins on the bottom of thewell other than �-galactosidase. As in the Westernblot experiment, the wash solution contains a small


amount of the non-ionic detergent, Tween 20, andNaCl to disrupt any weak ionic or hydrophobic in-teractions that may be involved in these nonspecificinteractions. A second antibody specific for the anti-�-galactosidase antibody is then added (secondaryantibody) to allow it to bind the primary antibody.As with the Western blot experiment, the secondaryantibody is covalently conjugated to the enzymehorseradish peroxidase (HRP). This enzyme willeventually take part in a color-producing reactionthat will allow you to determine the titer of the pri-mary antibody. Remember that you will be usingtwo different anti-�-galactosidase primary antibod-ies in this experiment. For assays performed usingrabbit polyclonal antibodies, HRP-goat-antirabbitIgG must be used as the secondary antibody. Forassays performed using the mouse monoclonal anti-body as the primary antibody, HRP-goat-antimouseIgG must be used as the secondary antibody.

After a series of wash steps to remove any sec-ondary antibody bound nonspecifically to proteinson the bottom of the well other than the primaryantibody, a mixture of 3,3�,5,5�-tetramethylbenzi-dene (TMB) and hydrogen peroxide is added. TheHRP enzyme conjugated to the secondary antibodywill use these substrates to produce a soluble, blue-

colored product. The reaction is stopped with theaddition of 4 N H2SO4, which will also convert theblue-colored product to a form that absorbs lightat 450 nm (yellow in color, Fig. 17-1). The inten-sity of the yellow color in each well (A450) will beproportional to the amount of primary antibodythat has bound to the antigen at the bottom of thewell. The ELISA procedure just described is de-picted in Figure 17-2.

Once the titer of the mouse-anti-�-galactosi-dase monoclonal antibody has been determined, acompetitive ELISA can be performed to determinethe concentration of the antigen in an unknown so-lution. The only variation in the competitive ELISAfrom that of the standard ELISA is that the primaryantibody incubation will be performed in the pres-ence of soluble �-galactosidase (antigen that is notadsorbed to the bottom of the well). The theory un-derlying the competitive ELISA is shown in Figure17-3. If the primary antibody-binding step is per-formed in the presence of soluble antigen, the sol-uble antigen will compete with the plate-bound anti-gen for antibody binding. The more soluble antigenthat is present during the primary antibody incu-bation, the less antibody will bind to the antigen onthe plate. The lower the concentration of antibody

Figure 17-1 Oxidation of 3,3�,5,5�-tetramethylbenzidene by horseradish peroxidase.

CH3 CH3

CH3 CH3

3,3�,5,5�-Tetramethylbenzidene

horseradish peroxidaseH2N NH2

CH3 CH3

CH3 CH3

H2N NH2

CH3 CH3

CH3 CH3

H2N NH2

Di-imine product (absorbs light at 450 nm and

appears yellow)

��

H�

(H2O2)

Cation radical intermediate withelectron delocalized over the aromatic

groups. Absorbs light at 652 nm(appears blue)

� �

EXPERIMENT 17 Partial Purification of a Polyclonal Antibody, Determination of Titer, and Quantitation of an Antigen Using the ELISA

281

bound to the plate, the lower the A450 value thateach well will display at the end of the assay. As theconcentration of soluble antigen decreases, moreantibody will be free to bind the antigen on theplate, increasing the A450 value for each well.

It is important that the competitive ELISA ex-periment be performed at a primary antibody con-centration (dilution) that is most sensitive to changesin the concentration of soluble �-galactosidase.This dilution can be determined from a plot of A450

versus the log of the inverse dilution of the primaryantibody as described in the experimental protocol(inverse dilution is the same as the degree to whichthe sample was diluted; for example, if the samplewas diluted 1:1000, the inverse dilution is 1000). Ifthe primary antibody concentration is too high,there will be enough antibody to bind both the sol-uble and plate-bound antigen, and changes in theconcentration of soluble �-galactosidase may go

undetected. If the antibody concentration is toolow, addition of any soluble �-galactosidase will pre-vent binding of the antibody to antigen on the plate,and the A450 values will never increase above thebackground value. As with Western blotting, theideal dilutions of the primary and secondary anti-bodies must often be determined empirically (opti-mized) for each experimental situation.

Horseradish peroxidase is one of several en-zymes that are commonly used for the detection ofantigens in the ELISA. Like HRP-conjugated an-tibodies, secondary antibodies conjugated withthese other enzymes are commercially available. Alist of ELISA detection enzymes, the substrates thatthey use, and the wavelengths of detection for thesoluble colored products that they produce is pre-sented in Table 17-1.

Still another variation of the ELISA exploits thestrong interaction between biotin and streptavidin,

Figure 17-2 Principle of two-antibody (indirect) ELISA.

Adsorb antigen to the bottomof the microtiter plate wells.

Add bovine serum albumin toblock sites on well not boundwith antigen.

TMB + H2O2

Blueproduct

H2SO4Yellowproduct

Primary antibody

Primaryantibody

Incubate with secondary antibodyand allow it to bind to the primaryantibody. Wash away unboundsecondary antibody.

Incubate with primary antibodyand allow it to bind the antigenbound to the bottom of the well.Wash away unbound antibody.

HRP HRPconjugatedsecondary antibody


one of the strongest noncovalent interactions foundin nature (KD � 10�15 M). Many commerciallyavailable reagents allow investigators to cross-link(conjugate) biotin to antibodies. If a biotinylated an-

tibody is used as the primary antibody in an ELISA,it is detected with the use of streptavidin that hasbeen conjugated to one of the enzymes listed inTable 17-1. For more information on biotin, strep-

Figure 17-3 The competitive ELISA. If soluble antigen is included during the primary antibody incubation,

the soluble antigen will compete for antibody binding with the plate-bound antigen.

No competition. Use anantibody dilution thatgives a half–maximumA450 in the ELISA Decreasing competition with soluble antigen

0

++++

50

++

25

+++

100

–

A450

% inhibition

Table 17-1 Soluble Colorimetric Enzyme/Substrate Systems for

Detection of Antigens in the ELISA

Wavelength

Enzyme Substrate for Detection (nm)

Alkaline phosphatase p-Nitrophenyl phosphate 405

(PNPP)

Horseradish peroxidase 2,2�-Azino[3-ethylbenzo- 410 or 650

thiazoline-6-sulfonic acid]-

diammonium salt (ABTS)

o-Phenylenediamine (OPD) 490

3,3�,5,5�-Tetramethyl- 450

benzidene (TMB)

�-Galactosidase o-Nitrophenyl-�,D- 405

galactopyranoside (ONPG)


283

tavidin, and their use in immunochemical assays, seeFigure 18-4 and the introduction to Experiment 18.



Pasteur pipettes

Parafilm

Saturated ammonium sulfate solution

1.5-ml microcentrifuge tubes

10� Phosphate-buffered saline (10X PBS, pH 7.3): Thissolution should be diluted 1:10 with distilled water toproduce the 1X PBS (working solution) described in theprotocol.

2.56 g/liter NaH2PO4 � H2O

22.48 g/liter Na2HPO4 � 7H2O

87.6 g/liter NaCl

Rabbit serum (normal or pre-immune rabbit serum andanti-�-galactosidase serum)*

Spectrophotometer to read absorbances at 280 nm

Cuvettes for spectrophotometer

Polyoxyethylenesorbitan monolaurate (Tween 20, SigmaCatalog #P-5927)

�-Galactosidase solutions (40 g/ml and 10 g/ml) in 1XPBS (Sigma product #G-6008)

Blocking solution (1X PBS containing 1.0% wt/volbovine serum albumin (BSA) and 0.05% vol/volpolyoxyethylenesorbitan monolaurate (Tween 20))

Wash solution (1X PBS containing 0.05% vol/vol Tween20)

Mouse-anti-�-galactosidase monoclonal antibody ( J1)(Sigma Product #G-8021)*

Secondary antibodies:

Goat-antirabbit IgG (HRP conjugated) (Kirkegaard& Perry product #074–1506)

Goat-antimouse IgG (HRP conjugated) (Kirkegaard& Perry product #074–1802)

Horseradish peroxidase (HRP) substrate (3,3�5,5�-tetramethylbenzidene and H2O2) (Kirkegaard &Perry product #50–7600)

Immulon-2 96 well flat-bottomed microtiter plate (Fisherproduct #1424561)

96-Well microplate spectrophotometer

�-galactosidase solution (unknown concentration) in 1XPBS

4 N H2SO4

Protocol

Day 1: Partial Purification of Antibodies

(IgG) from Normal Rabbit Serum and Anti-

�-Galactosidase Rabbit Serum

1. Obtain a 0.2-ml sample of normal or pre-immune rabbit serum (NS) and anti-�-galactosidase rabbit serum (immune serum,IS) from the instructor.

2. Transfer 0.1 ml of each sample to two separate1.5-ml microcentrifuge tubes labeled NS andIS.

3. Slowly add 0.1 ml of saturated ammonium sul-fate solution to each tube, cap, and invert thetubes several times to mix (do not use a Vortexmixer).

4. Incubate the tubes for 15 min on ice. Cen-trifuge the samples at 4°C for 10 min in themicrocentrifuge.

5. Remove the supernatant from the precipitatedIgG pellet in each tube and discard. The serumconsists of 80 to 90% albumin, which is solu-ble at 50% ammonium sulfate saturation. TheIgG is insoluble at 50% ammonium sulfate sat-uration and will precipitate.

6. Resuspend the protein (IgG) pellet in 0.1 ml of1X PBS, pH 7.3, and repeat steps 3 to 5.

7. Resuspend the protein pellet in 1.0 ml of PBS,pH 7.3. Make the following dilutions of the NSand IS samples in 1X PBS as follows:

Volume of Volume of

Sample PBS (ml) Sample (ml)

NS (untreated) 0.990 0.010

IS (untreated) 0.990 0.010

NS [(NH4)2SO4 purified] 0.667 0.333

IS [(NH4)2SO4 purified] 0.667 0.333

8. Read the absorbance of each 1.0-ml solution at280 nm, using 1X PBS to zero your spec-trophotometer. Estimate the total protein con-centration in each sample using Beer’s law

*The dilutions of the anti-�-galactosidase antibodies described in this

experiment are approximate. The titers of particular antibody samples

(even if they are from commercial sources) may vary significantly. Be-

cause of this, we strongly recommend that the instructor test different

dilutions of the primary antibodies that are to be used in the experi-

ment beforehand (using this same general protocol) to optimize the

results. We find that there is less variability with the secondary anti-

body dilutions than with the primary antibody dilutions from semester

to semester.


(ε280 � 0.80 (mg/ml)�1 cm�1). This milligramper milliliter extinction coefficient is based onthe average value of aromatic amino acids foundin most proteins. Remember to take into ac-count the dilution of each sample in your cal-culation when determining the concentrationof protein in each sample. How does the totalprotein concentration compare between theuntreated and ammonium sulfate–treated NSand IS samples? Explain your results. Store allfour of these rabbit serum samples at 4°C foruse on Day 2.

9. Obtain a 96-well Immulon-2 flat-bottomedELISA plate. Cover rows F, G, and H tightly witha piece of parafilm or with tape. To each well inrows A through E, add 50 l of the 10 g/ml�-galactosidase solution in 1X PBS. Tap theplate gently to ensure that the 50-l samplecompletely covers the bottom of each well. Storethe plate at 4°C covered loosely with a pieceof plastic wrap for use on Day 2. During theovernight incubation, the �-galactosidase inthe solution will absorb noncovalently to thebottom of the wells.

Day 2: Determination of Titer

1. Remove the ELISA plate from the cold roomand vigorously shake out the �-galactosidasesolution added to rows A to E. Fill each wellin row A through E completely (to the top ofeach well) with wash buffer and flick the solu-tion out as before. Invert the plate and tap vig-orously several times on a piece of paper towelto remove all of the wash buffer.

2. Fill the well in rows A through E completelywith blocking buffer. Incubate for 30 min atroom temperature. During this incubation, theBSA in the blocking buffer will bind to all ofthe sites on the bottom of the well not alreadybound with �-galactosidase. During the 30-min incubation, prepare the antibody dilutionsdescribed in step 4.

3. Remove the blocking buffer as before and tapextensively on a piece of paper towel to removeall of the excess blocking solution.

4. Dilute the various rabbit serum antibody solu-tions prepared on Day 1 in blocking buffer as fol-lows:

Sample Dilution

NS (untreated) 1:20

IS (untreated) 1:1000

NS [(NH4)2SO4 purified] 1:5

IS [(NH4)2SO4 purified] 1:1000

Prepare at least 0.5 ml of each antibody sam-ple at the appropriate dilution. Also, obtain analiquot of mouse-anti-�-galactosidase mono-clonal antibody ( J1) diluted 1:1000 in blockingbuffer (prepared by the instructor).

5. Add 75 l of blocking buffer to all of the wellsin rows A through E except for 2A, 2B, 2C, 2D,and 2E. Once this is done, you are ready tomake serial, twofold dilutions of the differentprimary antibody solutions across each row (seeFig. 17-4).

6. Add 75 l of normal rabbit serum (NS-untreated) to wells 2A and 3A. Remove 75 lfrom well 3A and place into well 4A. Pipetteup and down several times to mix.

7. Remove 75 l from well 4A and place into well5A. Pipette up and down several times to mix.Continue the twofold serial dilutions of the NSsample all the way down row A to well 12A.Remove 75 l from well 12A and discard.

8. Using a fresh micropipette tip, follow the sameprocedure (steps 6 and 7) to do twofold serialdilutions of the untreated IS sample across rowB. Remember that well 1B will contain no pri-mary antibody.

9. Using a fresh micropipette tip, follow the sameprocedure (steps 6 and 7) to do twofold serialdilutions of the ammonium sulfate–purified NSsample across row C. Remember that well 1Cwill contain no primary antibody.

10. Using a fresh micropipette tip, follow the sameprocedure (steps 6 and 7) to do twofold serialdilutions of the ammonium sulfate–purified ISsample across row D. Remember that well 1Dwill contain no primary antibody.

11. Using a fresh micropipette tip, follow the sameprocedure (steps 6 and 7) to do twofold serialdilutions of the J1 monoclonal antibody acrossrow E. Remember that well 1E will contain noprimary antibody. NOTE: At this point, all of thewells in rows A through E should contain 75 l ofeither blocking buffer or blocking buffer plus a par-ticular antibody dilution.

285

Sam

ples

(D

ay 2

)

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Nor

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A

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(am

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ium

-su

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une

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ck(a

mm

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te –

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ck)

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1:2

1:4

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1:16

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1:51

21:

1024

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lutio

nof

sto

ck1:

21:

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81:

161:

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1:25

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1:10

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1:4

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Figure 17-4 96-Well microtiter plate ELISA. Rows A to E will be used to determine the titer of antibody

samples (dilute antibody across each row). Rows F to H will be used to determine the con-

centration of soluble �-galactosidase in an unknown solution (dilute antigen across each row).


12. Incubate the primary antibody solutions atroom temperature for 30 min. During the in-cubation, the anti-�-galactosidase antibodies inthe various samples will bind to the antigen onthe bottom of the wells.

13. Invert and shake the plate to remove the pri-mary antibody solutions from rows A throughE. Fill each well completely with wash buffer.Discard the wash buffer as before and repeatthe wash step three more times. Remove thelast traces of wash buffer from the wells afterthe final wash by inverting the plate and tap-ping the plate extensively on a piece of papertowel.

14. Dilute the HRP conjugated secondary anti-bodies as follows in blocking buffer:

Secondary Antibody Dilution

HRP-goat-anti-rabbit IgG 1:9000 (2.5 ml total)

HRP-goat-anti-mouse IgG 1:10,000 (0.8 ml total)

15. Add 50 l of the HRP-goat-antirabbit IgG an-tibody to all of the wells in rows A to D exceptwell 1A. Add 50 l of blocking buffer to well1A to prevent it from drying out.

16. Add 50 l of the HRP-goat-antimouse IgG an-tibody to all of the wells in row E.

17. Incubate the secondary antibody solutions atroom temperature for 30 min. During this in-cubation, the secondary antibody will bind theprimary antibody bound to the antigen at thebottom of the wells.

18. Remove the secondary antibody solutions fromrows A to E and wash each well four times withwash buffer as described in step 13. Be sure toremove all of the wash buffer from these wellsfollowing the last wash step.

19. Prepare a fresh 1:1 mixture of the 3,3�,5,5�-tetramethylbenzidene (TMB) solution and theH2O2 solution (5 ml total, 2.5 ml of each so-lution mixed together). Add 100 l to all of thewells in rows A to E. Note the time that the TMBsolution is added to the first well.

20. Carefully monitor the wells visually for the de-velopment of blue color. When you begin tosee the first signs of blue color in any of thewells in columns 11 or 12 (�2 min), the reac-tion in that row is ready to be stopped. Con-sult the instructor if you are not sure when to

stop the reaction. NOTE: It is important thatthe wells within a particular row be incubatedwith the substrate for the same period of time.It is not necessary to have the incubation timebe the same between each row. For instance, thewells in row A may require a 5-min incubationtime before color development, while the wellsin row E may require only a 2-min incubationfor color development. Before you begin tostop the reactions in a particular row, the in-tensity of blue color in wells 2 and 3 of thatrow should appear to be about equal. If the re-actions in a particular row are stopped tooquickly, the determination of the titer will bedifficult (see step 24 and Fig. 17-5).

21. Stop the reaction in each well by adding 100l of 4 N H2SO4. The solution will immedi-ately change from a blue to yellow color. It isimportant that the reactions are stopped in the sameorder in which the TMB substrate was added tostart the reaction. For instance, if row A was started1A to 12A, then row A should be stopped 1A to12A.

22. Using well 1A as a blank to zero the instrumentat 450 nm (no primary or secondary antibody),read the A450 of the wells in rows A to E in the

Half-maximum A450

Titer = 316

Antibody concentrationbecomes too low to

detect the antigen

Plate-bound antigen is saturatedwith antibody at low dilutions

Log of inverse of antibody dilution

1

1.0

0.5

0

A45

0

2 3 4

10Fold dilution: 100 1,000 10,000

Figure 17-5 Sigmoidal titration curves will allow

you to determine the titer of an an-

tibody.


287

microplate spectrophotometer. Consult the in-structor for information on how to operate themicroplate spectrophotometer.

23. Subtract the A450 value of well 1C from all ofthe other A450 values in rows A and C. Sub-tract the A450 value of well 1B from all of theother A450 values in row B. Subtract the A450

value of well 1D from all of the other A450 val-ues in row D. Subtract the A450 value of well1E from all the other A450 values in row E.These corrected absorbance values will be usedto produce the graphs described below. This isdone to account for the fact that the secondary an-tibodies may have bound nonspecifically to other pro-teins on the bottom of the wells besides the differ-ent IgG samples (background). The value in well1 for each row represents the background A450 valueproduced from the secondary antibody bound non-specifically to the bottom of the well (no primaryantibody present).

24. Prepare a plot of A450 versus the log of the in-verse antibody dilution (a 1:10 dilution � 1, a1:100 dilution � 2, etc.). Plot the data from allfive primary antibody samples on a single graphfor comparison. If the assay was successful, eachantibody solution should produce a sigmoidalcurve in this type of plot (see Fig. 17-5). Thevarious primary antibody stock solutions were di-luted prior to performing twofold serial dilutionsacross each row (see step 4). Keep this in mind whenyou construct your graph.

25. Identify the position on each curve where theA450 value is exactly half of the maximum A450

value for each primary antibody sample. Fromthese points on the graph, determine the titerof each antibody sample: For purposes of this ex-periment, the titer will be defined as the inverse ofthe dilution of the antibody that yields a half-max-imum A450 value in the ELISA. For example, anantibody that displayed a half maximum A450

value at a 1:1000 dilution has a titer of 1000.26. Remove the parafilm from rows F to H. Add

50 l of the 10 g/ml �-galactosidase solutionin 1� PBS to all the wells in rows F and G, aswell as wells 1H through 4H. Cover the plateloosely with plastic wrap and store it at 4°C foruse on Day 3.

27. Compare the titer of the untreated NS and ISsamples. Which is larger? Explain your obser-vation.

28. Compare the titers of the untreated NS andIS samples with those of their correspondingammonium-sulfate–treated samples. Has thetiter of NS or IS changed at all followingtreatment with ammounium sulfate? Explain.

29. Using the total protein concentrations (in mil-ligrams per milliliter) for each primary anti-body sample determined on Day 1, calculatethe “specific activity” of the four rabbit serumsamples using the following equation:

“Specific activity” (titer � ml/mg)

�

(Note that this differs from the traditional defini-tion of specific activity as used in the discussion ofenzymes.)

Is this value the same for the untreatedNS and IS samples and their correspondingammonium-sulfate–treated samples? Explainyour results.

30. Determine the titer of the J1 monoclonal an-tibody. The half maximum A450 value shouldbe somewhere between 0.5 and 0.6 if the reac-tions in the wells were stopped at the appro-priate time. This information will be used onDay 3 in the competitive ELISA assay to de-termine the concentration of �-galactosidase inan unknown solution.

Day 3: Determination of the Concentration

of �-Galactosidase in an Unknown Solution

1. Follow steps 1 to 3 from the protocol describedfor Day 2 (blocking steps) for all of the wellsin rows F to H (see Fig. 17-4).

2. Prepare 1 ml of a J1 monoclonal antibody so-lution in blocking buffer that is twice as con-centrated as the dilution that yielded a halfmaximum A450 value in the ELISA performedon Day 2 (A450 � 0.5 to 0.6). For example, ifa 1�10,000 dilution yielded a half-maximumA450 value on Day 2, prepare a solution thatis diluted 1:5000 in blocking buffer. Becauseyou will be adding an equal volume of soluble �-galactosidase to the wells during the primary an-tibody incubation, you will need a primary anti-body dilution that is half that used on Day 2. This

titer of antibody sampletotal protein concentration (mg/ml)


will bring the working dilution of the J1 mono-clonal antibody in the well to its titer value, whereit will be most sensitive to changes in the concen-tration of soluble �-galactosidase.

3. Fill all the wells in rows F and G with 37.5 lof blocking buffer except wells 1F, 2F, 1G, and2G. Also, add 37.5 l of blocking buffer to wells2H, 3H, and 4H. Add 75 l of blocking bufferto wells 1F, 1G, and 1H.

4. Add 37.5 l of the 40 g/ml �-galactosidase so-lution in PBS to wells 2F and 3F. Remove37.5 l from well 3F and add it to well 4F.Pipette up and down several times to mix.

5. Remove 37.5 l from well 4F and add it to well5F. Pipette up and down several times to mix.Continue the twofold serial dilution of the40 g/ml �-galactosidase solution across rowF until you reach 12F. Remove 37.5 l fromwell 12F and discard.

6. Add 37.5 l of the �-galactosidase solution at anunknown concentration to wells 2G and 3G. Per-form twofold serial dilutions of this unknown�-galactosidase solution across row G from 4Gto 12G. Remove 37.5 l from well 12G anddiscard.

7. Add 37.5 l of the J1 monoclonal antibody so-lution prepared in step 2 to all the wells in rowsF and G except for 1F and 1G. Also, add 37.5 lof the J1 monoclonal antibody solution pre-pared in step 2 to wells 2H, 3H, and 4H. Wells2H, 3H, and 4H contain no soluble �-galactosidase.The A450 values of these wells will be important inthe calculations performed at the end of the assay.

8. Incubate the primary antibody for 30 min atroom temperature. Remove the primary anti-body solution from the wells and wash the wellsfour times with wash buffer as described in theprotocol for Day 2.

9. Add 50 l of the HRP-conjugated goat-anti-mouse IgG antibody to all of the wells in rowsF and G except for 1F (using the same dilutionof the secondary antibody as used on Day 2).Also, add 50 l of the secondary antibody so-lution to wells 1H to 4H.

10. Incubate the secondary antibody for 30 min atroom temperature. Wash the wells four timeswith wash buffer as described in the protocolfor Day 2.

11. Add 100 l of a 1:1 mixture of 3,3�,5,5�-tetra-methylbenzidene (TMB) and H2O2 to all of the

wells in rows F and G, as well as wells 1H to4H. Note the time at which you add the substrateto the first well. It is critical for the success of thisexperiment that the total reaction time be the samefor all wells in the competitive ELISA. Allow theblue color to develop until you begin to see anyblue color in wells 2F or 3F (�2–4 min). Stopthe reactions by adding 100 l of 4 N H2SO4.Again, it is very important that the reaction timefor all of the wells is the same in this experiment.

12. Using well 1F as a blank to zero the instrumentat 450 nm (no primary or secondary antibody),read the A450 of all of the wells in rows F to Hin the microplate spectrophotometer.

13. Determine the average A450 value of wells 1Gand 1H and subtract this value from all of theother A450 values in rows F to H.

14. Prepare a plot of percent inhibition versus thelog of the inverse of the dilution of the 40 g/mland unknown �-galactosidase solutions. Usethe following formula to determine the percentinhibition for each well in rows F and G:

% inhibition

� � 100

where A450 of J1 is the average absorbancevalue of wells 2H, 3H, and 4H and A450 of J1 ��-gal are the values of each individual well inrows F and G. Plot the data for rows F and Gon the same graph for comparison. NOTE: Apercent inhibition value can be calculated foreach well in rows F and G.

15. Identify the dilution of each �-galactosidase so-lution (40 g/ml solution and the unknown so-lution) that gave 50% inhibition. If the unknownsolution showed 50% inhibition at a larger di-lution than the 40 g/ml �-galactosidase solu-tion, then the unknown solution is more con-centrated than 40 g/ml. If the unknownsolution showed 50% inhibition at a lower dilu-tion than the 40 g/ml �-galactosidase solution,then the unknown solution is less concentratedthan 40 g/ml. Based on the dilution requiredfor the unknown solution to show 50% inhibi-tion compared to the 40 g/ml solution (0.25times, 0.5 times, 2 times, 4 times, etc.), deter-mine the concentration of �-galactosidase in theunknown solution. For instance, if the 40 g/ml

(A450 of J1) � (A450 of J1 � �-gal)

A450 of J1


289

�-galactosidase solution showed 50% inhibitionat a 1�8 dilution (well 5F), and the unknown �-galactosidase solution showed 50% inhibition ata 1�16 dilution (well 6G), then the unknown so-lution is twice as concentrated as the 40 g/mlsolution (80 g/ml �-galactosidase).

Exercises

1. Explain why it was necessary to incubate theTMB and H2O2 with the HRP-conjugated sec-ondary antibody for the same time in each wellin the competitive ELISA performed on Day3. Why was the incubation time between rowsless of a consideration in the ELISA experimentperformed on Day 2 (to determine the titer ofeach antibody sample)?

2. Discuss the advantages and disadvantages of us-ing the ELISA to determine the concentrationof an antigen in an unknown solution as op-posed to performing Western blots (see Ex-periment 18).

3. You have obtained several monoclonal anti-bodies that all recognize Protein X. You wouldlike to determine whether any of these anti-bodies recognize Protein X only in the nativeform as opposed to the denatured form. De-scribe an ELISA experiment that would allow

you to do this, as well as the predicted resultsof the experiment.

4. Why is it a good idea to test the serum of ananimal for the presence of antibodies against aprotein of interest before the animal is injectedwith the antigen?

5. The production of polyclonal antibodies usu-ally requires a relatively large amount of a pureprotein, whereas the production of monoclonalantibodies may require smaller amounts of animpure protein. Explain this in terms of themethod of production of these two types of an-tibodies.

REFERENCES

Harlow, E., and Lane, D. (1988). Antibodies, a Labora-tory Manual. Cold Spring Harbor, NY: Cold SpringHarbor Laboratory.

Kaplan, A., Szabo, L. L., and Opheim, K. E. (1988).Immunochemical Techniques. In: Clinical Chemistry:Interpretation and Techniques. Philadelphia: Lea &Febiger.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). An Introduction to Proteins. In: Principles ofBiochemistry. New York: Worth.

Mathews, C. K., and van Holde, K. E. (1990). Tools ofBiochemistry. In: Biochemistry. Redwood City, CA:Benjamin/Cummings.

( k

291

EXPERIMENT 18

Western Blot to Identify an Antigen

Theory

A protein has been purified using the latest tech-niques in chromatography. How can you prove thatthe protein band visualized on a Coomassie Blue–stained sodium dodecyl sulfate-(SDS)-polyacry-lamide gel is the protein of interest? If the proteinis an enzyme, it can be subjected to an activity as-say and total protein determination to assess its pu-rity (specific activity). If the protein has no enzy-matic activity, immunological studies may beemployed. Western blotting is a technique com-monly used to verify the presence of a protein ofinterest in a crude mixture. In addition to its use asa qualitative tool (to determine if a particular pro-tein is present), Western blotting can also be usedto quantify the amount of protein present in a pureor impure solution. Western blotting can be usedto detect any protein of interest, provided that anantibody raised against (that recognizes) the pro-tein of interest is available.

In this experiment, you will identify the C-terminal domain of glutathione-S-transferase(GST) from Schistosoma japonicum (purified from anEscherichia coli cell extract in Experiment 10)through the use of a Western blot. After the pro-teins present in the crude cell extract, unboundfraction, and the fraction eluted from the glu-tathione Sepharose resin are separated by SDS–polyacrylamide gel electrophoresis (SDS-PAGE),the proteins will be electrophoretically transferredto a polyvinylidene fluoride (PVDF) membrane(Fig. 18-1). Although membranes composed of ni-trocellulose or nylon can also be used for Westernblotting, PVDF membranes are particularly effec-

tive for use with the chemiluminescent detectionsystem employed in this experiment.

After the proteins have been transferred to thePVDF membrane, a blocking step is performed withpowdered milk to saturate all of the sites on the mem-brane not already bound with protein. This blockingstep is critical to the success of the technique: if theunoccupied sites are not blocked with protein, theprimary and secondary antibodies will bind non-specifically to sites throughout the membrane, mak-ing it very difficult to localize the protein of interestto a single location (protein band) on the membrane.In addition to using powdered milk as a source ofprotein for the blocking step, bovine serum albuminand casein are commonly used as blocking agents. Ifan avidin/biotin system is being utilized for detection of theantigen (see below), nonfat powdered milk or casein shouldnot be used as the blocking agent. The biotin in the milkmay adhere to the membrane and cause high back-ground.

After the proteins have been transferred to thePVDF membrane and the unoccupied sites havebeen blocked with protein, the antibody directedagainst the protein of interest (primary antibody) isadded. During the incubation, the primary antibodywill bind to the protein of interest. The amount ofprimary antibody to be added (the dilution) must bedetermined empirically for each experimental situ-ation. If the antibody concentration is too high, itis likely to bind to other proteins on the membranein addition to the protein of interest. If the primaryantibody concentration is too low, it may not be ableto bind the protein of interest effectively. A com-promise must be reached between the sensitivity ofdetection of the protein of interest (signal) and the

292

sides the protein of interest) that formed during theincubation. The membrane is then incubated with asecondary antibody that will recognize and bind tothe primary antibody. The anti-GST antibody (pri-mary) used in this experiment was raised in a mouse.The secondary antibody used in this experiment is anantimouse IgG antibody that was raised in a goat(goat-antimouse IgG). This secondary antibody isconjugated to an enzyme called horseradish peroxi-

extent of nonspecific binding of the antibody toother proteins on the membrane (“noise”).

After the membrane has been incubated with theprimary antibody (diluted in blocking buffer), themembrane is washed in blocking buffer containingpolyoxyethylenesorbitan monolaurate (Tween 20).This non-ionic (nondenaturing) detergent will dis-rupt hydrophobic interactions between the primaryantibody and other proteins on the membrane (be-


1. Sample preparation 4. Nonspecific binding of unoccupied sites

Membrane

Protein

Blocking agent

2. Separation by gel electrophoresis

Membrane

5. Addition of the primary antibody

Primary antibody

HRP-coupledsecondary antibody

Addition of chemiluminescentsubstrate andexposure to film

3. Transfer to a membrane support 6. Detection

Developedfilm

Figure 18-1 The technique of Western blotting.

EXPERIMENT 18 Western Blot to Identify an Antigen 293

dase (HRP), the critical component necessary for thelight-producing (luminescent) chemical reaction. Aswith the primary antibody, the dilution of the sec-ondary antibody must be determined empirically foreach experimental situation to optimize the results.

The PVDF membrane is again washed withblocking buffer containing Tween 20 to remove anysecondary antibody that has bound nonspecificallyto proteins on the membrane other than the mouseIgG. At this point, the PVDF membrane is readyto be treated with the SuperSignal reagents. Oneof these reagents contains hydrogen peroxide,which will convert the HRP to its oxidized form.The second reagent contains a component calledluminol. As HRP converts luminol from the reducedform to the oxidized form (Fig. 18-2), 428-nm lightis given off. If the PVDF membrane is exposed tophotographic film during the course of this reac-tion, a dark spot will appear on the film where theprimary and secondary antibodies are localized. Ifcare is taken to optimize the system under study, asingle protein of interest can be identified amongthousands of others.

Before the introduction of chemiluminescence tothe technique of Western blotting, the secondary an-tibodies were localized on the membrane with theuse of compounds and chemical reactions thatformed colored precipitates on the membrane. For in-stance, if the membrane used in this experiment weretreated with a solution of 3,3�-diaminobenzidenetetrahydrochloride dihydrate (Fig. 18-3) and H2O2,a brown precipitate would form at the site of actionof the HRP conjugated to the secondary antibody.If the secondary antibody used in this experimentwere conjugated to alkaline phosphatase rather thanHRP, the membrane could be treated with a solu-tion containing 5-bromo-4-chloro-3-indolyl phos-phate and nitroblue tetrazolium chloride (Fig. 18-3)to form a blue precipitate at the site of alkaline phos-phatase action on the membrane. A list of enzymesused for detection of antigens in Western blotting,their substrates, and the color of the insoluble prod-ucts that they produce is shown in Table 18-1. Al-though these latter methods are still in wide use to-day, they are less sensitive than the chemiluminescenttechnique demonstrated in this experiment. For lab-

Figure 18-2 Reaction of luminol and H2O2 to produce light in the presence of horseradish peroxidase.

NH

NH

NH2

H2O2

O�

O�

NH2

O

O

Luminol

O

O

Oxidized form of HRP

�� N2 h�(428 nm)

HRP

GST

Goat antimouse IgG Ab (HRP coupled)

Mouse-anti-Gst Ab

Photographic film(on membrane)


oratories that do not have access to darkroomsand/or developing solutions, we offer an alternativeprotocol that uses a precipitating HRP substrate forthe detection of GST at the end of the experiment.

Why has the “two-antibody” (primary and sec-ondary) technique become so popular for use in im-munochemistry experiments such as the Westernblot? Enzyme-conjugated antibodies that recognize

N

3,3�-Diaminobenzidene tetrahydrochloride

H2N NH2

NH2NH2

N

NN

N

O

Cl

Br

� 4 HCl

For use with Antibodies Coupled to Horseradish Peroxidase

For use with Antibodies Coupled to Alkaline Phosphatase

H

O

PO O�Na

O�Na

�

�

�

5-Bromo-4-chloro-3-indolyl phosphateO

CH3

NO2

NO2CH3

Nitroblue tetrazolium chloride

Cl� Cl� N

NN

N�

Figure 18-3 Other commonly used substrates that form colored precipitates on the membrane when

acted on by enzymes conjugated to secondary antibodies.

Table 18-1 Precipitating Colorimetric Enzyme/Substrate Systems for

Detection of Antigens in Western Blots

Enzyme Substrate Color of Precipitate

Alkaline phosphatase 5-Bromo-4-chloro-3�-indolyl-

phosphate with nitroblue Black/purple

tetrazolium chloride (BCIP/NBT)

Naphthol AS-MX phosphate with

Fast Red TR Bright red

Horseradish peroxidase 4-Chloro-1-naphthol (CN) Blue/purple

3-Amino-9-ethyl carbazole (AEC) Red/brown

3,3�-Diaminobenzidene

tetrahydrochloride (DAB) Brown

(carcinogenic)

Glucose oxidase Phenazine methosulfate with

nitroblue tetrazolium chloride Black/purple

�-Galactosidase 5-Bromo-4-chloro-3-indolyl-

�,D-galactopyranoside (X-Gal) Blue


species-specific IgG are commercially available andrelatively inexpensive. For instance, an HRP-conju-gated goat-antirabbit IgG antibody will recognizeany antibody that has been raised against any anti-gen, so long as the latter antibody has been raisedin a rabbit. It is far more cost effective for biotech-nology companies to develop and conjugate sec-ondary antibodies that recognize antibodies obtainedfrom a limited number of hosts than it is for themto develop and conjugate primary antibodies thatrecognize the seemingly unlimited number of anti-gens that are currently under study. Still, you shouldrealize that it is possible to conjugate a particularprimary antibody of interest to an enzyme and alle-viate the need for a secondary antibody in Westernblotting and in the ELISA (see Experiment 17).

Yet another variation of the Western blot ex-ploits the strong interaction between biotin andstreptavidin. Biotin is a 244-Da vitamin found in allcells (Fig. 18-4). Streptavidin is a 75-kDa, tetramericprotein isolated from Streptomyces avidinii. Eachstreptavidin monomer is capable of binding a singlemolecule of biotin. The dissociation constant of thestreptavidin/biotin complex is on the order of 10�15

M, one of the strongest noncovalent interactionsfound in nature. Many reagents are commercially

Figure 18-4 The chemical structure of biotin.

O

O

CH2 CH2 CH2 CH2 OHC

S

N NHH

O

O O

CH2

CH2

CH2

CH2 NH

CH2

CH2

CH2 H

NHCH2

C C

C

S

N NHH

Free Biotin

Biotin is a coenzyme of CO2-transferring enzymes.In its biochemically active form, biotin is alwayscovalently linked to enzymes via amide linkage

to �-amino groups of lysine residues.

available that allow investigators to cross-link (con-jugate) antibodies with biotin. Antibodies can beconjugated to biotin through a number of differentreactive groups on the protein, including primaryamine groups (lysines), sulfhydryl groups (producedafter reduction of the antibody to break disulfidebonds), carboxyl groups (glutamate and aspartate),carbohydrate groups (after oxidation of cis-diols toreactive aldehyde groups), etc. As is the case withthe enzyme-conjugated secondary antibodies de-scribed above, horseradish peroxidase and alkaline-phosphatase–conjugated streptavidin are also com-mercially available. When the enzyme-conjugatedstreptavidin is incubated with a membrane contain-ing a biotinylated antibody bound to the antigen ofinterest, this antigen can be localized to a particu-lar portion of the membrane with the use of lumi-nol or any other of the substrates described in Table18-1 (Fig. 18-5).


Reagents and apparatus for SDS–PAGE (See Exper-iment 4)

P-20 Pipetman with disposable tips


1. Sample preparation 4. Nonspecific binding of unoccupied sites

Membrane

Antigen

Blocking agent

2. Separation by gel electrophoresis

Membrane

5. Addition of the biotinylated antibody

= biotin

StreptavidinHRP

3. Transfer to a membrane support 6. Detection

Addition of chemiluminescentsubstrate andexposure to film

Developedfilm

Figure 18-5 Use of streptavidin–enzyme conjugates and biotinylated antibodies in the detection of anti-

gens on Western blots.

Prestained high molecular-weight protein standard(Gibco BRL, Catalog #26041-020)

Coomassie Blue staining solution (40% methanol, 10%glacial acetic acid, 0.25% wt/vol Coomassie Bril-liant Blue R-250)

Destaining solution (40% methanol, 10% glacial aceticacid)

Methanol

Transfer buffer (48 mM Tris, 39 mM glycine, 20%methanol, 0.004% SDS, pH 9.2)

Immobilon-P PVDF membrane, 0.45-m pore size (Mil-lipore, Catalog #IPVH 000 10)


Tris/NaCl/powdered milk solution (20 mM Tris–HCl,pH 8, 14.62 g/liter NaCl, 86.22 g/liter nonfat pow-dered milk)


Boiling water bath

Razor blade

Aluminum foil

Small pans or containers for membrane incubations andwashes

Power supply

Tween 20 (polyoxyethylenesorbitan monolaurate, SigmaCatalog #P-5927)

Mouse–anti-glutathione-S-transferase monoclonal anti-body (Sigma Product #G-1160)

SuperSignal West Pico chemiluminescent substrate forWestern blotting (Pierce Catalog #34080)

HRP-conjugated goat-antimouse IgG antibody (PierceCatalog #31430)

Plastic wrap

Kodak X-omat film

Film cassette

Dark room

Transfer apparatus (such as BIORAD Transblot SD®)

Reagents required for the colorimetric detection of HRP (op-tional):

10 mM Tris-buffered saline (TBS buffer), pH 7.4(Dissolve 1.8 g of Trizma base and 13 g of NaClin 1.5 liters of distilled water. Adjust the pH to 7.4with 1 N HCl.)

4-Chloro-1-naphthol (Pierce Catalog #34010)

Absolute ethanol

3% (vol/vol) Hydrogen peroxide in distilled water—prepare immediately before use

Protocol

Day 1: SDS-PAGE and Preparation of Blot

for Western Blot Analysis

1. Heat the eight samples prepared for SDS–PAGE (performed in step 16 of Experiment 10)in a boiling water bath for 5 min, along with a40 l sample of molecular weight protein stan-dards in 4 SDS sample buffer (prepared bythe instructor).

2. Centrifuge the samples for 10 sec to collect thematerial at the bottom of the tubes.

3. Load a 10-well, 12% SDS polyacrylamide gelwith 20 l per well of each sample as follows:

Lane Sample

1 Crude cell extract

2 Unbound fraction

3 Eluted fraction

4 Purified GST

5 Molecular weight protein standards

6 Crude cell extract

7 Unbound fraction

8 Eluted fraction

9 Purified GST

10 Molecular weight protein standards

4. Apply current (�120–180 V, constant voltage,�30 mA) and continue electrophoresis untilthe bromophenol blue dye front has migratedto the bottom of the gel.

5. Turn off the power supply and remove the gelfrom the electrophoresis apparatus. Separatethe plates (the gel will adhere to one of the twoplates), and remove the stacking gel. Be care-ful not to tear the resolving gel during thisprocess.

6. Cut the gel in half between lanes 5 and 6 witha razor blade. The gel will tear if the razorblade is dragged across it, so it is better tomake several cuts holding the blade perpen-dicular to the plate and pressing straight downto cut.

7. Cut a notch in the upper corner of each half ofthe gel, next to lane 1 and lane 6. This will helpyou determine the orientation of the gel fol-lowing Coomassie staining and transfer to thePVDF membrane.

8. Place the portion of the gel containing lanes 1to 5 in a small pan. Add enough CoomassieBlue staining solution to completely cover thegel and incubate at room temperature for 1 hr.

9. Place the portion of the gel containing lanes 6to 10 in a small pan containing transfer buffer.Incubate the gel for 10 to 15 min to allow thegel to equilibrate in the transfer buffer.

10. Cut six pieces of Whatman 3 MM filter paperto the exact size of the gel portion containinglanes 6 to 10.

11. Cut a piece of PVDF membrane (wear gloveswhen handling the membrane) to the exact sizeof the gel portion containing lanes 6 to 10. Wetthe membrane completely by dipping it in asmall pan containing methanol. It is critical thatthe PVDF membrane be fully wetted inmethanol prior to being equilibrated in trans-fer buffer (see below). It is important that the fil-ter paper and membrane be cut to the exact size of


the membrane so that the current applied to thetransfer unit (see below) be forced through the gelrather than around it. If the top and bottom layersof filter paper on either side of the gel come in con-tact with one another, the current will take the pathof least resistance and flow around rather thanthrough the gel. The result of this is that the pro-teins will not transfer to the PVDF membrane, andthe subsequent Western blot will not be successful.

12. Place the methanol-treated PVDF membraneand the four pieces of Whatman 3 MM filterpaper in a small pan containing transfer bufferand incubate for 5 to 10 min.

13. Assemble the blot as described below. Makesure that all components are stacked directly ontop of one another, with no overhanging edgesand no air bubbles in between the layers:

ANODE (�)




PVDF membrane

Gel




CATHODE (�)

14. Connect the electrodes and apply 15 V (con-stant voltage) to the unit for 45 min.

15. Disconnect the power supply from the transferunit, disassemble the layers, and cut a notch onthe PVDF membrane to mark the position oflane 6. Stain the gel following the transfer withCoomassie Blue as described in step 8. This isdone to ensure that the proteins on the gel weretransferred to the PVDF membrane.

16. Place the PVDF membrane in a small pan con-taining 100 ml of Tris/NaCl/powdered milk so-lution (blocking buffer). Incubate at 4°Covernight with gentle shaking. Recall that thisstep is performed to block any portion of themembrane that does not already contain protein.

17. After each half of the gel has been incubatedwith the Coomassie Blue staining solution for1 hr, they may be destained with destain solu-tion by incubating for 1 hr at room tempera-ture. The destain solution should be changedfour times during the course of the 1-hr incu-bation.

18. Calculate the Rf values of the molecular weightprotein standards using the following equation:

Rf �

Prepare a standard curve by plotting log mol-ecular weight vs. Rf.

19. What is the molecular weight of the protein(s)in the eluted fraction? Do you believe that thepurification of GST was successful? Explain.

Day 2: Western Blot

1. Drain the 100 ml of blocking buffer solutionfrom the pan containing the PVDF membrane.Rinse the membrane with a small amount ofdistilled water to remove the last traces ofblocking buffer.

2. Add 20 ml of fresh blocking buffer to thePVDF membrane, along with 2.5 l of themouse–anti-GST monoclonal antibody (a1:8000 dilution of the antibody). Swirl the pancontinuously and gently for 45 min at roomtemperature.

3. Pour out the antibody solution and rinse thePVDF membrane with a small amount of dis-tilled water.

4. Add 100 ml of fresh blocking buffer contain-ing 0.05% Tween 20 (wash buffer) and incu-bate at room temperature for 10 min with con-tinuous gentle swirling. Pour out the washsolution and repeat step 4 two more times.

5. Add 20 ml of fresh blocking buffer to the PVDFmembrane. Add 5 l of a 1:10 dilution (in wa-ter) of the goat-antimouse IgG (HRP-conju-gated). The final dilution of this antibody is1:40,000. Cover the pan with aluminum foil andincubate at room temperature for 45 min withcontinuous gentle swirling. NOTE: If you will bedeveloping the Western blot with a colorimetric sub-strate (see below), the final dilution of the HRP-goat-antimouse IgG should be between 1:5000 and1:20,000. This results from the fact that the 4-chloro-1-naphthol substrate is much less sensitivethan the chemiluminescent substrate (i.e., you willneed more HRP at the site of the antigen to obtaina dark precipitate). Consult the instructor for infor-mation concerning the final dilution of the secondaryantibody if the colorimetric detection option is chosen.

distance traveled by protein band (cm)��

distance traveled by dye front (cm)


6. Pour out the antibody solution and rinse thePVDF membrane with a small amount of dis-tilled water.

7. Add 100 ml of wash buffer and incubate atroom temperature for 10 min with continuousgentle swirling. Pour out the wash solution andrepeat step 7 two more times. If you will be us-ing colorimetric detection with a precipitating HRPsubstrate, proceed to the section entitled “Colori-metric Detection of HRP.”

8. Mix 5 ml of each of the two SuperSignalreagents (peroxide solution and luminol/en-hancer solution) and add them to the top of thePVDF membrane (the side containing the pro-teins). Incubate at room temperature for 5 min.

9. Remove the PVDF membrane from the pan,allow the excess SuperSignal substrate to dripfrom the membrane, and wrap the membranein plastic wrap.

10. Expose the PVDF membrane (protein side up)to a piece of film for 1 to 5 min. Mark the filmwith a notch corresponding to the notch madenext to lane 6 on the SDS gel to aid in orien-tation. Remove the film from the cassette anddevelop. The time of the exposure will depend onhow quickly the blot is exposed to film after thechemiluminescence reaction has been performed.The longer the time between these two events, thelonger the exposure time will be needed to visualizethe bands.

11. Using the notch made on the film, align it withthe notch on the PVDF membrane. Based onthe position of the prestained molecular weightprotein standards present on the PVDF mem-brane, determine the molecular weight of theprotein band(s) present on the film. Do you be-lieve that it is GST? Explain. Are there anyother protein bands besides GST that are pres-ent on the film (in the crude cell lysate and un-bound fraction lanes)? If so, explain what thismeans. Propose how you could alter the con-ditions of the experiment to prevent these otherprotein bands from cross-reacting with the an-tibodies used in the Western blot.

Colorimetric Detection of HRP

1. Dissolve 150 mg of 4-chloro-1-napthol (CN)in 50 ml of ethanol. The stock solution can beprepared in advance and stored at �20°C. This

will be enough reagent for 50 groups to per-form the experiment.

2. Add 1 ml of the 3 mg/ml CN solution preparedin step 1 to 10 ml of Tris-buffered saline, pH7.4.

3. Immediately before use, add 100 l of a 3%(vol/vol) hydrogen peroxide solution to the 10ml of CN/H2O2 reagent prepared in step 2,and pour the entire solution over the side ofthe membrane containing your protein.

4. Incubate the membrane with the CN/H2O2

reagent for approximately 30 min at room tem-perature. Monitor the membrane visually dur-ing this incubation for the appearance of blueor purple protein band(s). If the membranedoes not contain a large amount of GST and/orHRP-conjugated goat-antimouse secondaryantibody, an incubation time longer than 30min may be required.

5. When you are satisfied with the color devel-opment on your Western blot, rinse the mem-brane gently several times with distilled wa-ter to stop the reaction. NOTE: The blue colorwill begin to fade several days after development.To obtain a permanent record of the experiment,we advise that you take a picture or prepare a pho-tocopy of the membrane after it has been allowedto dry.

Exercises

1. Where does the term “Western” blot comefrom?

2. You are performing a Western blot on a puri-fied sample of protein X. You know that pro-tein X is composed of three subunits, one of 50kDa, one of 30 kDa, and one of 15 kDa. Youhave two monoclonal antibody samples raisedagainst native protein that you will use as theprimary antibodies in two separate experi-ments. Describe what you could expect to seein two different Western blots performed withthese two antibodies. Explain your answer.

3. Describe how monoclonal and polyclonal anti-bodies are made.

4. Why do you think that it is more cost effec-tive for companies to sell enzyme-conjugatedantibodies that recognize antibodies from anumber of animals rather than to sell enzyme-


conjugated antibodies that recognize the anti-gens themselves (why is the two-antibody sys-tem commonly used rather than a primary an-tibody that is already enzyme conjugated)?

5. Describe how a chemiluminescent Westernblot could be used to quantify the amount ofan antigen in an unknown sample.

6. How would a Western blot be affected if theblocking step were omitted? How would aWestern blot be affected if the blocking stepwere carried out, but the wash steps followingincubation with the primary and secondary an-tibodies were omitted?

REFERENCES

Harlow, E., and Lane, D. (1988). Antibodies, A Labora-tory Manual. Cold Spring Harbor, NY: Cold SpringHarbor Laboratory.

Hines, K. (1995). Immunoperoxidase Assay Systems: Visu-alization of Immunoblots with SuperSignal® CL-HRPSubstrate System. Rockford, IL: Pierce Chemical.



Thorpe, G. H. G., and Kricka, L. J. (1986). EnhancedChemiluminescence Reactions Catalyzed by Horse-radish Peroxidase. Methods Enzymol 133:331.

Voet, D., and Voet, J. G. (1990). Techniques of ProteinPurification. In: Biochemistry. New York: Wiley.

Weir, D. M. (1986). Handbook of Experimental Immunol-ogy. London: Oxford University Press.

�

303

Structural Features of Nucleic Acids

Nucleic acids are polymers containing nitrogenousbases attached to sugar–phosphate backbones. Thecommon nitrogenous bases of nucleic acids are thebicyclic purines, adenine and guanine, and themonocyclic pyrimidines, cytosine, uracil, andthymine (Fig. V-1).

These purines and pyrimidines join to thesugar–phosphate backbones of nucleic acidsthrough repeating b-linked N-glycosidic bonds in-volving the N 9 position of purines and the N1 po-sition of pyrimidines. There are two classes of nucleic acids: ribonucleic acid (RNA) and deoxyri-bonucleic acid (DNA). DNA and RNA differ in oneof their nitrogenous base components (uracil inRNA, thymine in DNA) and in their sugar (ribose)moiety, as indicated in Fig. V-2.

Repeating 39,59-phosphodiester bonds join theD-ribose units of RNA and the 2-deoxy-D-riboseunits of DNA. A nucleoside is a nitrogenous base plusa sugar, while a nucleotide is a nitrogenous base, plus

a sugar, plus a phosphate. Shorter polymers of 2 to50 nucleotides are referred to as oligonucleotides,while larger polymers of nucleotides are referred toas polynucleotides.

Because nucleic acids contain a large number ofnucleotides, biochemists have devised an abbrevia-tion system to indicate the nucleotide sequence ofnucleic acids (see Fig. V-2). This system assumesthe 39,59-phosphodiester linkages and lists, in a 59to 39 order, the nucleotide sequence of the nucleicacid with p (phosphate) between each nucleotide.Using this latter system, the RNA nucleotide se-quence shown in Figure V-2 is . . . ApGpCpUp . . . .You may also encounter an even simpler abbrevia-tion system where the phosphate (p) between eachnucleotide is omitted (e.g., . . . AGCU . . . ).

Many of the physical and biological propertiesof nucleic acids arise from secondary structures cre-ated by a regular stacking of hydrogen-bonded pairsof nitrogenous bases. Only certain pairs of bases arecapable of hydrogen bonding: Adenine hydrogenbonds with thymine and guanine hydrogen bondswith cytosine in DNA, while adenine hydrogen

SECTION V

Nucleic Acids

Introduction

NH2

N

N N

NH

HH

Adenine Guanine

2

1

3 4

56

78

9

O

N

N N

NH

H

HNumberingsystem

NH2

H2N

N

NO H

H

2

3

1 6

54 N

NO H

H

HH

H

N

NO

OO

H

CH3

H

HNumbering

system

Cytosine Uracil Thymine

Figure V-1 The most common nitrogenous bases of nucleic acids.

304

bitals of the stacked planar base pairs add to the sta-bility of the polynucleotide (Figs. V-3, and V-4).

Most DNAs contain extensive secondarystructure, existing as large double-stranded he-lixes containing two separate, hydrogen-bonded

bonds with uracil and guanine hydrogen bonds withcytosine in RNA and in DNA:RNA duplexes. Itshould be noted, however, that noncanonical basepairing (e.g., GPU) is also commonly found instretches of RNA. The overlapping p electron or-

SECTION V Nucleic Acids

NH2

NH2

CH2

N

N

H HHH

N

N

O

OH

O

P

O

OO2

OO

CH2

NH NH

N

H HHH

N

N

O

OH

P

O

OO2

ONH2

CH2

N

N

H HHH

O

OH

P

O

OO2

O

O

O

O

CH2

NH

N

H HHH

O

OH

P

O

OO2

O

NH2

NH2

CH2

N

N

H HHH

N

N

O

H

H

H

H

O

P

O

OO2

OO

CH2N

H HHH

N

N

O

P

O

OO2

ONH2

CH2

N

N

H HHH

O

P

O

OO2

O

O

O

O

CH2

CH3N

N

H HHH

O

P

O

OO2

O

DNA, unlike RNA, contains2-deoxy-D-ribose and thyminein place of uracil.

or

or

A A

GG

C

C

U

U

A nucleotide: anycombination of a base,a sugar, and a phosphate

A nucleoside: anycombination of a baseand a sugar

Another nucleotide (thisnucleotide has a 59phosphate, whereas the nucleotide above has a39 phosphate)

Figure V-2 Primary structures of RNA and DNA.

acteristics of DNA: extensive viscosity in solution,the potential to rupture or “shear” on agitation,ease of fiber formation for x-ray analysis, and soforth. Double-stranded DNA will denature or“melt” (uncoil to form single strands) when sub-jected to high temperatures, strong bases, and de-naturants such as urea and formamide. Secondarystructure also plays a role in the function of var-ious RNAs. Hydrogen-bonding and base pairing,

complementary sequences. These separate pri-mary strands exist together in opposite or an-tiparallel orientation (one strand running 59 to 39and the other running 39 to 59), while being sta-bilized by intramolecular base stacking and by in-termolecular hydrogen-bonded base pairing. Thisyields the long, fibrous “double helix” character-istic of most DNAs (Fig. V-4). This lengthy struc-ture contributes to the pronounced physical char-

SECTION V Introduction 305

Supporting sugar–phosphate backbone

OverlappingorbitalsCH3 O

ON N

N

N

N

H

H

N

H

N

Planar H-bondedbase pairs

Orbitals

Note opposite or antiparallelorientation of the two sugar–phosphate backbones.

Guanine:Cytosine

Thymine:Adenine

Resultant base stacking

Possible complementaryWatson–Crick base pairing

N

ONN

N

N

N

H

H

O

N

HN

H

H

Figure V-3 Factors governing secondary structural features of nucleic acids.

306

stacking, and interactions with divalent cations. Asis the case for secondary structure, tertiary struc-tural elements are often critical to the function ofa particular RNA molecule, which may include ac-tivating or “charging” amino acids for translation,proper ribosome function, or messenger RNAsplicing (see below).

Cellular Localization and Roles of Nucleic Acids

DNA contains the basic genetic information of allliving cells. As such, cellular DNA is located at the

where present in RNA, are generally intramolec-ular and generate helical hairpin loops within aprimary nucleotide sequence (see Fig. V-4). Thus,precipitated RNAs are more amorphous than fi-brous. Accordingly, most RNAs do not markedlyincrease solution viscosity and can be difficult tocrystallize for subsequent x-ray analysis.

Certain RNAs also possess substantial amountsof tertiary structure. Tertiary structure in RNArefers to the folding of secondary structural ele-ments, such as double helical regions or hairpinstem-loops, into discrete three-dimensional struc-tures. Forces involved in stabilizing such interac-tions are diverse, involving hydrogen-bonding, base


oo

ooo

o

o

oooo

o

oo

oo

oo

ooooo

o o o

o

o ooo

oooo

o

oo o o

o

o

o

oo

o

oo

o

o

o

o

o

o

o

oo

o

o

o

o

o

o

ooCH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2CH2

CH2

N

NN

N

N

N

NN

NN

N NN

N

N

NN

NN

NN

N

NN

N

NN

H

H

H

H

H

H

HH

HHH

P

P

P

P

P

P

P

P

P

Sugar–phosphatebackbones

H-bonded base pairing and

base stacking

A

A

A

Y

Cm

UGm

Unpaired “free”bases in anticodonintramolecularloop

Anticodon arm of tRNA phe (Yeast)

DNA double helix

Figure V-4 Examples of secondary structure in nucleic acids.


Twochromatids(2 x 10 coils)

One coil(30 rosettes)

Nuclearscaffold

One rosette(6 loops)

One loop(50 x 106 bp)

30 nm Fiber

“Beads-on-a-string”form ofchromatin

DNA

Histones

Figure V-5 The highly ordered structure of DNA

in chromatin.

site of primary genetic activity within cells. Inprokaryotic cells (i.e., cells lacking a nucleus) geneticactivity occurs throughout the cytoplasm. Thus, thevarious molecules of circular DNA (chromosomeand plasmids) residing in prokaryotic cells are notlocalized to a specific compartment of the cell. Incontrast, the DNA of eukaryotic cells (i.e., cells withdefined nuclei) lies within discrete, membrane-enclosed organelles. The majority of the DNA of aeukaryotic cell exists in the nucleus. In the “rest-ing” cell not undergoing DNA replication and celldivision, the nuclear DNA is complexed with anumber of histone proteins. This compact form ofDNA is referred to as chromatin (Fig. V-5). Duringmitosis, the nuclear DNA will “unwind” from itscompact structure into individual chromosomesthat can be replicated by the action of DNA poly-merase. The remaining DNA of eukaryotic cells ex-ists in the partially self-duplicating chloroplasts andmitochondria. These nonnuclear DNAs have den-sities different from the nuclear DNA, and are dis-persed throughout these membrane-bound or-ganelles.

There are three types of RNAs found withincells that participate in some aspect of the transferof genetic information from DNA to protein: ri-bosomal RNAs, messenger RNAs, and transferRNAs. Ribosomal RNAs (rRNAs) participate di-rectly in this genetic information transfer processin that they constitute approximately 60% of themass of the ribosome. Ribosomes differ slightly intheir physical properties and component rRNAs de-pending on their prokaryotic or eukaryotic origin(Fig. V-6). Because cells contain a large number ofthese ribosomes, rRNAs account for about 70% ofthe total cellular RNA. Because of their high con-centration and uniform size, rRNAs are also easyto detect and isolate.

Messenger RNAs (mRNAs) and transfer RNAs(tRNAs) participate directly in protein synthesis.The nucleotide sequence of one of the two DNAstrands in the double-helix is transcribed by RNApolymerase to produce a mRNA that is comple-mentary to it (see Experiment 22). A portion or allof this mRNA nucleotide sequence will then serveas the template for the ribosome, which will trans-late the mRNA molecule into a polypeptide or pro-tein (see Experiment 23). Messenger RNAs, then,carry information contained in the DNA into theproduction of protein.

308 SECTION V Nucleic Acids

50S subunitMW = 1.8 3 106

30S subunitMW = 1 3 106


70S overall

80S overall

Eukaryotic ribosome

Prokaryotic ribosome


Contains 1 molecule 23S RNA,1 molecule 5S RNA, and 34 proteins

Contains 1 molecule 16S RNA,and 20–21 proteins

Contains 1 molecule 28S RNA, 1 molecule5S RNA, 5.8S RNA, and approx. 50 proteins

Contains 1 molecule 18S RNAand 33 proteins

( )

( )

( )

( )

Figure V-6 Component parts of natural ribosomes.

Messenger RNAs constitute approximately10% of the total cellular RNA. This mRNA can beeither monocistronic (governing the synthesis of asingle polypeptide) or polycistronic (governing thesynthesis of two or more polypeptide chains). Sincethe ribosomes bind the mRNA and initiate trans-lation at defined positions (the ribosome-bindingsites), translation may begin from more than one siteon a single molecule of mRNA. As suggested above,mRNAs can be quite variable in size. mRNAs arealso quite variable in their stability. While somemRNAs may have a half-life of several hours, oth-ers may have a half-life on the order of minutes. Ingeneral, prokaryotic mRNA is nearly one hundredtimes less stable than the average eukaryotic mRNAin terms of half-life. This difference in mRNA sta-bility may be a function of the faster growth rate ofprokaryotic cells compared to eukaryotic cells (thetypical prokaryotic cell can divide every 20 min,while the typical eukaryotic cell divides once every30 hr). The ribonuclease activity of both types ofcells ensures that the composition and concentra-tion of the total mRNA “pool” within the cell willbe constantly changing to meet the changing needs

of the cell. Eukaryotic cells have developed specificmechanisms to control the stability of individualmRNAs as a means of regulating gene expression.

Transfer RNAs (tRNAs) participate in transla-tion by transferring free amino acids into the grow-ing peptide chain. Unlike mRNAs, tRNAs are uni-form in size (25 to 30 kDa) and constituteapproximately 20% of the total cellular RNA. Nor-mal cells will have a variety of different tRNAs, oneor more that will recognize each of the 20 aminoacids and deliver them to the growing polypeptidechain at the appropriate time. Although there areonly 20 amino acids, the degeneracy of the geneticcode dictates that a single cell must contain at least32 different tRNAs with the capability of recog-nizing different codons (three-base sequence) onthe mRNA.

One final class of eukaryotic RNA that deservesmention are the small nuclear RNAs (snRNAs). Astheir name suggests, this class of RNA is found inthe nucleus and participates in the splicing of in-trons from certain types of eukaryotic mRNAs. FivesnRNAs (U1, U2, U4, U5, and U6) take part inthese splicing reactions when combined with a


number of nuclear proteins. Together, the snRNAsand the proteins form a functional splicing unit re-ferred to as small nuclear ribonuclear proteins(snRNPs or “snurps”). Although the snRNAs arevery important in eukaryotic RNA processing, theirsmall size (between 100 and 200 nucleotides) con-tributes a very small percentage to the total cellu-lar RNA. The role of snRNAs and the mechanismof intron splicing are currently a topic of intensiveresearch. In addition to snRNAs, a number of othersmall nuclear and cytoplasmic RNAs have been dis-covered and shown to serve a variety of biochemi-cal functions.

Chemical Properties of Nucleic Acids

Action of Alkali

The N-glycoside linkages of DNA and RNA arestable in mildly alkaline conditions. However, thephosphodiester linkages of DNA and RNA demon-strate markedly different chemical properties inmild alkali. For example, 0.3 M KOH (37°C)rapidly (,1 hr) catalyzes the cleavage of all phos-

phodiester bonds in RNAs and yields intermediatecyclic-29,39-phosphonucleosides that hydrolyze to nucleoside-29-phosphates and nucleoside-39-phosphates on further (12–18 hr) incubation (Fig.V-7). The phosphodiester bonds of DNA are sta-ble under these conditions because they lack the ad-jacent 29 hydroxyl group on D-ribose required toform the cyclic-29,39-phosphonucleoside interme-diates.

Action of Acid

Very mild acid, or dilute acid at or below room tem-perature, has little effect on nucleic acids. Indeed,dilute acids (e.g., trichloroacetic, HClO4, HCl) at 0to 25°C are routinely used to precipitate (isolate)nucleic acids and other polar macromolecules.Somewhat stronger acidic conditions (prolonged ex-posure to dilute acid or dilute acid at increased tem-perature or acid strength) cause depurination, i.e.,acid-catalyzed hydrolysis of purine N-glycosides.Thus, 15 min of treatment with 1 N HCl at 100°Cremoves most of the purines from DNA and RNA.Much harsher acidic conditions (several hours at100°C or more concentrated acid) are required toremove pyrimidine N-glycosides. Some limited

CH2

H HHH

O

OH

O

O

O

O2

OH2

CH2

H2O

H HHH

Base

Base

O

OHO

CH2

H HH

n

H

O

O

O

OH

OP

Base

Cyclic-29,39-phosphonucleosides

Nucleoside-39-phosphates

Nucleoside-29-phosphates

O

CH2

H HH

1

H

OOH Base

P O2

O2

O

P O2

O2

O

P O2O

P O2

O

O

CH2

H HHH

O

O

OH

OH

OH

Base

Figure V-7 Alkaline hydrolysis of RNA.


phosphodiester bond cleavage occurs during acid-catalyzed depurination. Even more phosphodiestercleavage occurs during acid-catalyzed depyrimidi-nation.

Physical Properties of Nucleic Acids

Several of the physical properties of nucleic acidshave already been mentioned in the discussion ofnucleic acid structure. The following propertiesplay important roles in the isolation, detection, andcharacterization of nucleic acids.

UV Absorption of Nucleic Acids

All nucleic acids, nucleosides, and nucleotidesstrongly absorb ultraviolet (UV) light. This UV ab-sorption arises from the nitrogenous bases, each ofwhich possesses its own unique absorption spec-trum (Fig. V-8). The sum of these individual ni-trogenous base spectra generates the overall UVspectrum of DNA and RNA. Several features ofthese spectra deserve further comment. First, thenucleoside spectra are a function of pH. This prop-erty reflects the titration of specific functionalgroups within the nitrogenous bases. Such pH-de-pendent spectral shifts provide an additional toolfor characterizing unknown solutions containingthese nucleosides. Second, the ratios of the ab-sorption at two different wavelengths (e.g.,A280/A260) at any pH provide criteria for identifi-cation of individual nitrogenous bases or nucleo-sides, and for the estimation of the relative pro-portions of individual nucleosides in anoligonucleotide. Third, a nucleic acid spectrum isthe sum of the individual nitrogenous base spectrapresent in the nucleic acid. Such summation yieldsa spectrum with a lmax of 260 nm. It follows, then,that the absorption of a solution containing a nu-cleic acid is proportional to the nucleic acid con-centration. In general, a 1-mg/ml solution of single-stranded DNA (lacking any degree of sec-ondary structure) has an A260 of 25 at neutral pHin a 1-cm light path (ε260 5 25 (mg/ml)21 cm21).Extensive secondary structure, as found in nativetRNA, rRNA, and DNA, reduces the UV absorp-tion of the stacked, hydrogen bonded bases so that

a 1-mg/ml solution of these types of nucleic acidshas an A260 of 20 at neutral pH in a 1-cm light path.It follows, then, that the disruption of secondarystructure within DNA, rRNA, and tRNA increasesthe A260 of a solution containing these molecules.

1.0

0

0

1.0

1.0A

bsor

banc

e of

1 3

10–4

M s

olut

ion

in a

1-c

m c

uvet

te

0

0

0220 260 300

1.0

1.0

GuanosinepH 7.0 E1cm260 = 11,700

pH 7.0 280/260 = 0.66

1M

AdenosinepH 7.0 E1cm260 = 14,900

pH 7.0 280/260 = 0.15

1M

CytidinepH 7.0 E1cm260 = 7500

pH 7.0 280/260 = 0.94

1M

UridinepH 7.0 E1cm260 = 9900

pH 7.0 280/260 = 0.39

1M

Thymidine (withdeoxyribose)

pH 7.0 E1cm260 = 8700

pH 7.0 280/260 = 0.74

Nanometers

1M

Figure V-8 UV spectral characteristics of nucleo-

sides. (——) 5 pH 7.0, (— — —) 5pH 1.0, and (- - -) 5 pH 13.0.


This increase in the absorption spectrum followingdenaturation (destruction of secondary structure) istermed the “hyperchromic effect” (Fig. V-9). Con-versely, the decrease in the absorption spectrum onrenaturation of these types of nucleic acids (restora-tion of secondary structure) is termed the“hypochromic effect.” These effects are observed inExperiment 19.

Acid–Base Titration of Nitrogenous

Bases, Nucleotides, Nucleosides, and

Nucleic Acids

The nitrogenous bases and phosphates of nucleicacids contain groups that titrate in aqueous media(Table V-1). Thus, nucleic acids have titrationcurves and act as buffers in solution and in biolog-ical extracts. Extremes of acid or alkali will intro-duce charges on nitrogenous bases and yield struc-tures that no longer participate in base stacking andhydrogen bonding. Extremes of acid or alkali de-stroy (denature) secondary structure in nucleicacids. Using Table V-1 and the Henderson–Has-selbach equation, you can calculate the net chargeon any nucleoside or nucleotide at any pH. Thesecharges, and their effect on the overall polarity ofthe molecule, serve as the basis for resolution of nu-

cleotides and nucleosides in chromatographic andelectrophoretic systems (see Experiments 2 and 4).

Hybridization

Secondary structure in nucleic acids can be dis-rupted by extremes of pH or by increasing the tem-perature of the nucleic acids residing in a solutionof low ionic strength. Disruption of double-stranded nucleic acids, which contain stacked basesand hydrogen-bonded base pairs (e.g., DNA),yields separate, single-stranded, complementarystrands of polynucleotides. At favorable tempera-tures, pHs, and ionic strength conditions, these sin-gle-stranded molecules will reassociate with them-selves or hybridize with other polynucleotidescontaining a complementary sequence. Biochemistsmake great use of this phenomenon. Specifically, ifhybridization is allowed to take place in the pres-ence of an isotopically labeled, single-stranded DNAmolecule with a known sequence, then a fragmentof DNA containing the complementary sequencecan be identified in a complex DNA sample. Thisprocess of carrying out DNA:DNA hybridizationon a solid support (Fig. V-10) was developed by Ed-ward Southern. Because of this, the technique iscalled Southern blotting. The technique has proven

20

10

DNA

220

Nanometers

300

Abs

orba

nce

of a

1-m

g/m

l sol

utio

n in

a 1

-cm

cuv

ette

260

20

10

RNA

220

Nanometers

300

Abs

orba

nce

of a

1-m

g/m

l sol

utio

n in

a 1

-cm

cuv

ette

260

Figure V-9 UV characteristics, right, of single-stranded RNA and, left, of DNA. (———) 5 pH 7.0,

(— — —) 5 pH 1.0, (– – –) 5 pH 13.0. The denaturation of double-stranded DNA at pH 1

and 13 causes an increase in UV absorbance, an illustration of the hyperchronic effect.

312Table V-1 Titration Properties of Nitrogenous Bases and Phosphates of Nucleic Acids and Their

Subunits

Name of Ionizing Group Primary Ionization Secondary Ionization

Phosphodiesters of

internucleotide

linkages of RNA

and DNA

Phosphomonoesters

of nucleotides or

terminal phosphates

Adenine in adenosine,

adenylic acids, AMP, etc.,

and RNA and DNA

Guanine in guanosine,

guanylic acids, GMP, etc.,

and RNA and DNA

Cytosine in cytidine,

cytidylic acids, GMP, etc.,

and RNA and DNA

Uracil in uridine,

uridylic acids, UMP, etc.,

and RNA

Thymine in thymidine,

thymidylic acids, TMP, etc.,

and DNA NCH3CH3

O2

N

R

O

2N

O

N

R

O

O

N

NH

R

pKa1 5 9.8

6 H1

O

CH3

R

O

O N

2N

R

O

O N

NH

pKa1 5 9.2

6 H1

R

O2

O N

N

NH2

R

O N

N

NH2

R

O N

1NH

pKa1 5 4.3

6 H1

H2N H2N

O

R

N

N N

2N

H2N

H

R

OO

N1

N N

NH

R

pKa1 5 2.3

6 H1

pKa2 5 9.2

6 H1

N

N N

NH

NH3

N

N NR

N

R

NH2

N

N N

N

R

pKa1 5 3.6

NH2

N

N N

1N

1

H

6 H1

O2

O2

O

R O PpKa2

5 6.0

OH

O2

O

R O P

OH

OH

O

R O PpKa1

5 0.9

6 H1 6 H1

OCH2

O

O2

O

O

OCH2

P

OCH2

O

OH

O

O

OCH2

PpKa1

5 1.5

6 H1


Chromosomal DNA Chromosomal DNA fragments

Molecular weightsize standards Sample

Restrictionendonuclease

Southern blotting

Northern blotting

Agarose-gelelectrophoresis

Incubate withsingle-stranded,radiolabeled DNAmolecule (probe) ( )

Exposeto film

Hybridization

Hybridization

Transfer DNA from gelto nitrocellulose

Molecular weightsize standards Sample

Transfer RNA from gelto nitrocellulose

Can identify the DNA fragment containing the sequence complementary to the radiolabeled probe

Exposeto film

Identify the RNA fragment containing the sequence complementary to that of the radiolabeled probe

Incubate withsingle-stranded,radiolabeled RNA orDNA molecule (probe) ( )

Isolate RNAAgarose-gelelectrophoresis

Film

Cell

Figure V-10 Southern blotting and Northern blotting.

314

covery of plasmids has revolutionized the study ofbiochemistry. Although the recombinant plasmidsused in modern molecular biology have been engi-neered for a number of different applications, thereare several features that are common to most plas-mids (Fig. V-11).

Recall that DNA replication initiates at a spe-cific DNA sequence termed the origin of replication.It is the specificity of this sequence on a plasmidthat will determine the host organism in which aparticular plasmid will be able to be propagated(replicated). An Escherichia coli plasmid contains anE. coli origin of replication, a Bacillus subtilis plas-mid contains a B. subtilis origin of replication, andso forth. Some recombinant plasmids contain theorigin of replication for more than one organism.These “shuttle” plasmids can therefore be propa-gated in more than one organism. Specific ele-ments within the origin of replication will deter-mine the frequency of replication of the plasmidresiding in the host cell. “Low-copy” plasmids haveorigins of replication that are under the same strin-gent control as the chromosome, which is ulti-mately regulated by the cell cycle. Hence, one ortwo molecules of a “low-copy” plasmid will bepresent in an individual host cell. “High-copy”plasmids contain a less stringent or “relaxed” ori-gin of replication. Since a high-copy plasmid is notunder the same replication control as the chromo-some during cell division, hundreds of moleculesof a high copy plasmid can be present in a singlecell.

All recombinant plasmids contain a genetic el-ement that provides a means of selecting for thosecells that harbor it. Most commonly, the plasmidwill contain a gene coding for an enzyme that willallow the cell to metabolize or inactivate a specificantibiotic. For instance, the b-lactamase gene (bla)on many E. coli plasmids will allow the cell tocleave the lactam ring structure common to thefamily of b-lactam antibiotics, thereby inactivat-ing them. A list of antibiotics, their mode of ac-tion, and the genes involved in their inactivationare listed in Table V-2. Antibiotic resistance genesare not the only genetic selection markers that canbe present in plasmids. Genes encoding enzymesinvolved in essential biosynthetic pathways canalso provide a means of selection, provided thatthe host strain into which the plasmid is intro-duced contains a null mutation (inactivation) of

to be a very powerful tool in molecular biology. Youcan identify a DNA fragment of interest in anycomplex mixture of nucleic acids (such as from achromosomal digest), and attempt to isolate andclone it. Southern blotting can also be used to iden-tify the position of various restriction endonucle-ase cleavage sites within the nucleotide sequence ofinterest, a technique that has been put to good usein forensic biochemistry and genetic screening formutations associated with various diseases and con-ditions (see Experiment 24, Fig. 24-5).

A similar technique, in which RNA is bound toa solid support and hydridized to an isotopically la-beled single-stranded RNA or DNA molecule, iscalled Northern blotting (Fig. V-10). As with thetechnique of Southern blotting, Northern blottingcan be used to determine if an RNA molecule witha particular sequence is present in a complex mix-ture of nucleic acids. You can imagine how this tech-nique may be useful in studies involving the regu-lation of gene expression: the mRNA transcribedfrom a particular gene can be identified and quan-tified at any time during the growth cycle, andchanges in the level of transcription of particulargenes can be monitored as specific alterations aremade to the growth media. With this technique, ithas been possible to study the effects of specific bi-ological componds (nutrients, drugs) on gene ex-pression, as well as the roles of various protein andnucleic acid structural elements that control theprocess of transcription. Hybridization is also a crit-ical element in the powerful technique known asthe polymerase chain reaction (PCR), described indetail in Experiment 24.

Plasmids and Other Cloning Vectors

Plasmids

Plasmids are naturally occurring, extrachromoso-mal, circular DNA elements that endow the hostorganism with potentially advantageous capabili-ties. Genes carried on these plasmids may conferresistance to specific antibiotics, allow the cell tometabolize harmful organic compounds, and/or al-low the cell to ward off infection by bacteriophage.Some plasmids also carry genes involved in bacte-rial conjugation (mating) and virulence. The dis-



Selectable genetic marker that allows the determination of which host cells took up the plasmid following transformation

Multiple cloning site contains a number of unique, closely spaced restriction endonuclease recognition sites where foreign DNA of interest may be inserted= Promoters (often inducible)

that allow the expression of the inserted gene

= Universal primer binding sites that allow the sequence determination of the insert DNA

Origin of replication (determines in what host the plasmid may be replicated); also determines whether the plasmid is “high copy” or “low copy”

Figure V-11 Features found on many popular plasmids used for different molecular biology applications.

that same gene on the chromosome. This idea ofselection for transformed cells is discussed later inthis section.

Recombinant plasmids have been engineered to“accept” fragments of foreign DNA. Most plasmidscontain a multiple cloning site (MCS) into whichforeign DNA can be inserted. The MCS is a shortsegment of DNA on the plasmid that contains anumber of unique restriction enzyme recognitionsequences found nowhere else on the plasmid. Theplasmid and the DNA insert can be digested withone or more of these restriction endonucleases andligated together to produce a new recombinantplasmid that carries the DNA insert of interest (Fig.V-12). Once this recombinant plasmid has been iso-

lated, it can be introduced into and faithfully prop-agated (cloned) by the host organism.

This process of inserting genes into plasmids isthe cornerstone of molecular biology. Once a geneof interest has been separated from the chromo-some and inserted into a plasmid, the gene can besubjected to a number of different manipulationsthat provide insight into its biological function. Inbiochemical studies, it is often necessary to expressthe gene carried on the plasmid, isolate the proteinthat it encodes, and study its function. To aid in thisprocess, many recombinant plasmids have been en-gineered to include powerful promoter sequences59 to the MCS. Genes cloned into the MCS down-stream (39) to the promoter can be expressed at high


Table V-2 Some Commonly Used Antibiotics, Their Modes of Action, and

Their Inactivation by Resistant Host Cells

Antibiotic Mode of Action Inactivation

Chloramphenicol Binds to 23S rRNA in the Chloramphenicol acetyl

50S ribosomal subunit and transferase acetylates the

inhibits translation antibiotic

Ampicillin Inhibits cross-linking of b-lactamase cleaves the ring

peptidoglycan chains in in the antibiotic

bacterial membrane

synthesis

Erythromycin Binds to 23S rRNA in the An enzyme methylates a

50S ribosomal subunit and specific adenine residue in the

inhibits translation rRNA so that it is unable to bind

the antibiotic

Tetracycline Binds to the 30S ribosomal A membrane protein actively

subunit and prevents the tRNA expels or transports the

from interacting with the antibiotic out of the cell

ribosome

Kanamycin and Bind to the 30S ribosomal Aminoglycoside phospho-

Neomycin subunit and inhibits transferase phosphorylates

translation the antibiotic

Streptomycin Binds to the 30S ribosomal A deletion of the S12

subunit and inhibits ribosomal protein prevents

translation binding of the antibiotic

levels in the host cell, and the proteins that they en-code can be purified. Often, you would like to con-trol the expression of genes cloned into plasmids.This becomes critical if the high level expression ofa particular gene product is found to be lethal tothe host cell. Many of the promoter elements thatflank the MCS, therefore, are able to be inducedthrough the addition of a small molecule or someother change in the condition of the growthmedium. In this way, the host strain can be grownin culture, and the gene of interest can be expressedat any particular time during the growth cycle. Alist of inducible promoters found in recombinantplasmids and their means of induction are listed inTable V-3.

Expression of a gene of interest is by no meansthe only manipulation that can be carried out oncethe gene of interest has been inserted into a plas-mid. Current techniques in molecular biology al-low you to make virtually an unlimited number ofmutations in the gene-carrying recombinant plas-mid: 1 specific base pair can be changed, 2 specificbase pairs can be changed, 50 base pairs can bedeleted, 20 base pairs can be added, and so forth.

The information obtained from the analysis of thesemutant genes can provide a great deal of insightinto the various domains, structure, and function ofthe proteins that they encode.

Bacteriophage l

Although plasmids are the most popular cloningvectors used in modern molecular biology, they aregenerally limited to use with DNA inserts less than10 kb in size. If the DNA fragment of interest isbetween 10 kb and 20 kb, you may wish to use a vi-ral cloning vector. One commonly used viral DNAvector that can be propagated in E. coli is the dou-ble-stranded, linear chromosome of bacteriophagel. This bacteriophage chromosome, about 48 kb insize, encodes a number of proteins required forreplication of the viral chromosome, head and tailassembly of the phage particle, and lysis of the hostcell. The genes that encode these proteins areknown to lie on either end of the linear chromo-some, and constitute about 65% of the total phagegenome. The remaining 35% of the bacteriophagegenome contains nonessential genes that can be

317

Plasmid

Plasmid 2Plasmid 1 Plasmid 3

Transform host cell and select for genetic marker

Digest plasmid within the multiplecloning site with restriction endonuclease

Insert DNA containing the geneor DNA fragment of interest

Petri plate containingselective media

Clone 1Clone 3

Clone 2

Ligate

Figure V-12 General cloning strategy utilizing plasmids.


Table V-3 Some Inducible Promoters Found in

Commercially Available Vectors

Promoter* Inducer

lac Isopropylthiogalactoside (IPTG)

tac Isopropylthiogalactoside (IPTG)

l Increased temperature (42°C)

BAD Arabinose

MMTV Dexamethasone

(mouse mammary

tumor virus)

GAL1 Galactose

AOX1 Methanol

MT CuSO4

(metallothionein)

DHSP Muristerone A

*Lower-case letters denote prokaryotic origin, upper-case letters

denote eukaryotic origin.

excised and replaced with any foreign DNA frag-ment that is 10 to 20 kb in size (Fig. V-13). Oncethe DNA fragment of interest has been inserted intothe bacteriophage l chromosome, the chimericchromosome can be packaged in vitro into phageparticles that will deliver the DNA into the E. colihost. Since the chimeric genome still contains thegenes required for replication, phage assembly, andcell lysis, the DNA fragment that is inserted intothe chromosome will be replicated (cloned), pack-aged, and used to infect neighboring E. coli cells.The plaque produced when cells infected with therecombinant phage are plated on a lawn of unin-fected E. coli cells contains progeny (clones) of a sin-gle phage. Since bacteriophage l is able to packageonly DNA between 40 and 50 kb into the phagehead, DNA inserts smaller than 10 kb and largerthan 20 kb in size cannot be cloned using this vec-tor. Other E. coli bacteriophages, such as the single-stranded M13 phage, have smaller chromosomesthat can accept DNA fragments smaller than 10 kbin size.

Cosmids

The cosmid is another variation of the plasmid andviral DNA vectors that has been developed to al-low the cloning of DNA fragments that range from40 to 50 kb. As the linear bacteriophage l DNA is

introduced into E. coli, DNA ligase will convert itto a closed, circular form by joining together thecohesive ends (12 bases in length) at either end ofthe molecule (Fig. V-14). The region at which thesetwo complementary strands are joined is the cos site.Following replication and packaging of this nowcircular chromosome into the phage head, the chro-mosome is again cleaved at the cos site to producethe linear form of the chromosome that will be usedto infect the next host cell. By placing the cos sitefrom bacteriophage l into a plasmid, the phage willbe able to package and process the vector in vitro,provided that the DNA fragment inserted in the mul-tiple cloning site is 40 to 50 kb in size. After the cos-mid is cleaved (linearized) and introduced into theE. coli host, it will be closed or circularized to re-form the cos site. Unlike the viral DNA vector, thiscircular DNA molecule contains none of the phagegenes required for replication, packaging, or cell ly-sis. Therefore, the cosmid will continue to be repli-cated from the bacterial origin of replication, justas with a plasmid. As stated earlier, the in vitro pack-aging system requires that the total size of the DNAmolecule containing the cos site falls within a spec-ified range. Therefore, DNA fragments smallerthan 40 kb and larger than 50 kb cannot be clonedwith cosmid vectors.

Yeast Artificial Chromosomes (YACs)

This type of DNA cloning vector, designed to ac-cept DNA fragments up to several hundred kilo-bases in size, contains the three elements that willallow propagation in yeast: an origin of replica-tion—the autonomously replicating sequence—a cen-tromere, and a telomere at either end of the lin-ear DNA molecule. The centromere is the regionon the chromosome that will attach to the spindleduring mitosis. The telomeres are the small (100-base-pair) repetitive sequences found at the end oflinear eukaryotic chromosomes that are crucial inthe DNA replication process.

Restriction Endonucleases andOther Modifying Enzymes

The discipline of molecular biology requires theability to manipulate the gene or DNA sequence un-der study. In the early 1970s, an enzyme was isolated


Plaques containingprogeny (clones) ofa single recombinantbacteriophage


Ligate


Nonessential genes

Nonessential genes

Chromosomal DNA

Recombinant (chimeric) chromosome

Infect E. coliand plate on a lawnof uninfected cells

In vitro packaging is

successful if insert D

NA

is 10–20 kb in size

Bacteriophage l chromosome

Genes involved in replication,packaging, and cell lysis

Figure V-13 Cloning genes using bacteriophage l.

that catalyzes the hydrolysis of the phosphodiesterbackbone in both strands of a double-stranded DNAmolecule at a highly specific sequence. The enzymewas termed a restriction endonuclease. Currently, over700 restriction endonucleases have been isolatedfrom a variety of different prokaryotic organisms.Why would an organism produce enzymes thatcleave DNA molecules? It is thought that restrictionendonucleases act as a defense mechanism to blockor restrict infection by bacteriophage. If the host cell

that the phage is invading is able to destroy the phagechromosome before the genes carried on it are ex-pressed, the phage will be unable to replicate andlyse the host cell. In a sense, you might consider re-striction endonucleases to be a crude immune sys-tem for organisms that produce them.

How does the cell prevent these same restric-tion endonucleases from destroying its own chro-mosomal and plasmid DNA? It has been deter-mined that restriction endonucleases are one


Ligate

Digest with restrictionendonuclease

Digest with restrictionendonuclease

Chromosomal DNA

Cosmid

Selectable genetic marker

Selectablegeneticmarker

Selectablegeneticmarker

cos site

cos site

cos site

Multiplecloningsite

Infect E. coli andplate on selective media

In vitro packaging is

successful if insert D

NA

is large enough

Colonies containing the cosmid will grow since the phage does not contain any genes for replication or cell lysis

Figure V-14 Cloning genes with the use of cosmids.


component of a two-component system collectivelyreferred to as the restriction-modification system. Therestriction endonuclease component will recognizea specific base-pair sequence in the DNA and cat-alyze the hydrolysis of the phosphodiester back-bone in both strands of the DNA duplex. The sec-ond component of this system, the DNA methylase,catalyzes the transfer of a methyl group from S-adenosylmethionine to a specific adenine or cyto-sine base within the same DNA sequence recog-nized by the endonuclease (Fig. V-15). Once therecognition sequence has been methylated, theDNA is resistant to the action of the restriction en-donuclease. As the bacterial chromosome is repli-cated, the DNA methylase methylates the appro-priate sequences and protects the DNA fromhydrolysis. Since the invading phage DNA is notprotected by methylation, it will be destroyed bybacterial restriction endonucleases and exonucle-ases (see below) before it can cause damage.

There are three types of restriction-modifica-tion systems. Type I and type III systems consist ofa single, multisubunit enzyme that conducts boththe endonuclease and methylase activities. In anATP-dependent mechanism, type I restriction en-donucleases scan the DNA, bind to a specific recog-nition sequence, and cleave both strands of theDNA duplex at a site 1 to 10 kb from the recogni-tion sequence. Type III restriction endonucleasesdo not require ATP and usually cleave both strandsof the DNA duplex within 100 base pairs of therecognition sequence. For reasons that are not en-tirely clear, S-adenosylmethionine is required forthe activity of both the type I and type III en-donucleases, despite the fact that it is not consumedduring the hydrolysis of the phosphodiester back-bone. While the type III DNA methylases modify(methylate) a single strand of the DNA duplexwithin the recognition sequence, type I DNAmethylases methylate both strands of the duplexwithin the recognition sequence.

Type II restriction-modification systems differfrom their type I and type III counterparts in thatthe endonuclease and DNA methylase activities areconducted by two separate enzymes (not a singlemultisubunit complex). The restriction endonucle-ase cleaves both strands of the DNA duplex withina defined recognition sequence, while the compan-ion DNA methylase methylates a specific basewithin the same recognition sequence. In contrast

to the type III endonucleases, the hydrolysis of thephosphodiester backbone does not require the pres-ence of ATP or S-adenosylmethionine. All of therestriction endonucleases discussed above are sim-ilar in that they utilize a divalent cation as a cofac-tor, most commonly Mg21 or Mn21.

Because type II restriction endonucleases cleavethe DNA at a specified position within (rather thanadjacent to) a defined recognition sequence, they arethe enzymes of choice in most practical applicationsin molecular biology. Some commercially availabletype II restriction endonucleases, their recognitionsequences, and their cleavage sites are listed in TableV-4. Note that some of these enzymes make stag-gered cuts across the often palindromic recognitionsequence, producing 59 single-stranded DNA over-hangs, while others cleave straight across the DNAduplex, producing blunt-ended DNA fragments.

In addition to the very useful type II restrictionendonucleases discussed above, a number of othercommercially available enzymes can be utilized invarious applications in molecular biology (TableV-5). Exonucleases have been discovered that hy-drolyze DNA and/or RNA from either the 39 or 59terminus. Some of these enzymes act on single-stranded nucleic acids, others on both strands of theDNA duplex. The majority of these modifying en-zymes are used to remove nucleic acids from cellextracts during the early phases of protein purifi-cation. Other applications include the preparationof single-stranded DNA templates for sequenceanalysis, preparation of blunt-ended DNA frag-ments for cloning, DNA footprinting (see Experi-ment 22), removal of phosphoryl groups from the59 terminus of cleaved vector DNA to improvecloning efficiency, and radiolabeling of the 59 hy-droxyl group of chemically synthesized oligonu-cleotides for DNA sequencing and Southern blot-ting.

Ligation of DNA Fragments

The ability of a cell to join (ligate) DNA fragmentsis essential for its survival. This reaction is per-formed by an enzyme called DNA ligase. Recall thatDNA ligase plays a critical role in DNA replica-tion, linking together the Okazaki fragments syn-thesized by DNA polymerase on the lagging strandof the replication fork. DNA ligase is also a key


DNA methylase

DNA methylase

EcoRI recognition site

H3N

S-adenosylhomocysteine

H

NH2

CH2

CH2

CH2

CH

HHH

OH1

S

OH

O

O N

N N

N

COO2

H3N

S-adenosylmethionine

Adenine

H

NH2

CH2

CH2

CH3

CH2

CH

HHH

OH1

1

CH3

CH3

S

OH

O N

N N

N

COO2

NH2

N

N N

N

G A A T T C

N 6-methyladenine

HN CH3

N

N N

N

Ribose

G A A T T C

SmaI recognition site

Cytosine

NH2

N

N

H

H

C C C G G G

O

N 4-methylcytosine

HN

H

CH3

N

N H

C C C G G G

Figure V-15 The DNA methylase component of the restriction modification system methylates a specific

cytosine or adenine residue, making it incapable of being acted on by the companion restric-

tion endonuclease.


component in four different DNA repair mecha-nisms: base-excision repair, base mismatch repair,nucleotide excision repair, and recombinational(SOS) repair. Homologous recombination, or ex-change of DNA fragments within the genome thathave similar sequences, also requires DNA ligase.This latter type of recombination, which occurs fre-quently during meiosis, is responsible for the ge-netic and phenotypic diversity seen within popula-tions of higher eukaryotes.

Although DNA ligase is essential for numerousin vivo processes, it has also proven to be a verypowerful tool in the growing field of recombinantDNA technology. As discussed earlier in this sec-tion, a gene of interest can be excised from the

chromosome with restriction endonucleases, iso-lated, and inserted into a vector. The ligation of theDNA fragment of interest into the vector is cat-alyzed in vitro by DNA ligase (Fig. V-16). The re-action catalyzed by DNA ligase is energy-requiring.The enzyme first attaches an adenyl group (fromATP or NAD1) to the 59-phosphoryl group pres-ent on one of the two DNA strands. The phos-phoamide adduct produced in this reaction is thensubject to nucleophilic attack at the 59 phosphorylgroup by the 39 hydroxyl group present on the other

Table V-5 Some Commercially Available

Modifying Enzymes

Enzyme Activity

Deoxyribonuclease I Cleaves both single- and double-

(DNaseI) stranded DNA

Exonuclease III Removes nucleotides from the

39 ends of double-stranded DNA

Lambda Degrades the 59 phosphorylated

exonuclease strand of double-stranded DNA

Nuclease P1 Completely hydrolyzes DNA

and RNA to form nucleotide

monophosphates

Nuclease S1 Hydrolyzes single-stranded

nucleic acids to form nucleotide

monophosphates or 59-phosphoryl

oligonucleotides

Mung bean Hydrolyzes single-stranded

nuclease nucleic acids to form

nucleotide monophosphates or

59-phosphoryl oligonucleotides

Snake venom Removes nucleotides from the

phosphodiesterase I 39-hydroxyl group terminus of

DNA and RNA

Ribonuclease H An endonuclease that degrades

the RNA strand in a DNA–RNA

duplex

Ribonuclease I Hydrolyzes the phosphodiester

bonds in RNA to form

oligonucleotides

Polynucleotide Catalyzes the phosphorylation

kinase of the 59-hydroxyl group terminus

of DNA or RNA. Also catalyzes an

exchange reaction between the

g-phosphate group of ATP and

the 59-phosphoryl group on DNA

and RNA

Alkaline Removes 59-phosphoryl groups

phosphatase from nucleic acid fragments

Table V-4 Some Commercially Available Type II

Restriction Endonucleases

Restriction Endonuclease Recognition Sequence

BamHI 59-GQGATTC-39

39-CCTAAqG-59

BglII 59-AQGATCT-39

39-TCTAGqA-59

ClaI 59-ATQCGAT-39

39-TAGCqTA-59

EcoRI 59-GQAATTC-39

39-CTTAAqG-59

EcoRV 59-GATQATC-39

39-CTAqTAG-59

HindIII 59-AQAGCTT-39

39-TTCGAqA-59

KpnI 59-GGTACQC-39

39-CqCATGG-59

NdeI 59-CAQTATG-39

39-GTATqAC-59

NotI 59-GCQGGCCGC-39

39-CGCCGGqCG-59

SacI 59-GAGCTQC-39

39-CqTCGAG-59

SalI 59-GQTCGAC-39

39-CAGCTqG-59

Sau3AI 59-QGATC-39

39-CTAGq-59

SmaI 59-CCCQGGG-39

39-GGGqCCC-59

SspI 59-AATQATT-39

39-TTAqTAA-59

XhoI 59-CQTCGAG-39

39-GAGCTqC-59


PyrophosphateH

NH2

NH2 CH2

HHH

OH

2O1

OH

P

O

O N

N N

N O2P

O

OPLysDNAligase

O

O

2O

H

NH2

CH2

HHH

OH OH

O N

N N

NP

O

O

2O

2O

H

NH 2

CH 2 HHH

OHOH

O

N

N

N

N

P

OO

O2P

OO

OO2

2O 2O

1

1 NH2LysDNAligase

Adenosine triphosphate (ATP)

H

NH2

NH2 CH2

HHH

OH

O

OH

P

O

O N

N N

NP

O

O PLysDNAligase

O

OO

2O2O 2O

1

HO

P

OO

2

OO2

HO

Adenylation

1

Adenosine monophospate (AMP)

Strands linked via aphosphodiester bond

T4 DNA Ligase

Figure V-16 The reaction catalyzed by DNA ligase (T4).


Figure V-16 (continued) The reaction catalyzed by DNA ligase (E. coli ).

H

NH2

CH2

HHH

OH OH

O N

N N

NP

O

O

2O

2O

H

NH 2

CH 2 HHH

OHOH

O

N

N

N

N

P

OO

2OP

OO

OO2

NH2LysDNAligase

Nicotinamide adenine dinucleotide (NAD1)

Nicotinamide mononucleotide (NMN)

H

NH2

NH2

HHH

OH OH

O N

N N

N

LysDNAligase

O

P

O

P

O

OH H

HH

OH OH

O N

2O1

1

1

O2O

1

HO

P

OO

2

O2O

HO

Adenylation

1

1

Adenosine monophospate (AMP)

Strands linked via aphosphodiester bond

E. coli DNA Ligase

NH2

CH2

CH2

CH2

C

O

H

NH2

NH2

HHH

OH OH

O N

N N

N

LysDNAligase

O

O

P

O2

P

O

OH H

HH

OH OH

O N

2O

O2

NH2

CH2

C

O

1

326

ability to take up extracellular DNA) can be inducedas the bacteria are starved for particular nutrients.If the bacteria are exposed to a recombinant vectorafter induction of the competence system, the hostcell will take up the DNA and become transformed(eukaryotic cells that have taken up a recombinantvector are referred to as being transfected ). It is be-lieved that the competence system has evolved incertain bacterial species as a means of obtaining po-tentially advantageous genes from other organismsduring times of environmental stress. Transforma-tion by this process is illustrated in Experiment 20.

Other commonly used prokaryotic and eukary-otic host cells do not posess a genetically controlledcompetence system. Despite this fact, these typesof cells can become competent after a series ofchemical and/or physical treatments. For instance,if E. coli cells are first treated with a solution ofCaCl2 or RbCl2 (see Table V-6) and heat shockedin the presence of a recombinant vector, a certainpopulation or percentage of the cells (,106 per mgof DNA) will become transformed. The hypotoniccation solution used in these procedures is believedto swell the outer membrane, allow expulsion ofperiplasmic proteins into the surrounding solution,and allow the negatively charged DNA to bind tothe membrane. On heat shock, transient pores arecreated in the membrane, which allows the DNAto enter the cell. Although the host cells are neardeath following this heat shock treatment, many ofthem will soon recover and faithfully propagate theintroduced vector. Transformation of yeast cells in-volves a similar protocol utilizing lithium acetate,polyethylene glycol (PEG), and heat shock (seeTable V-6). Although the detailed mechanism ofyeast-cell transformation is different from that ofE. coli cells described above, the method is com-monly used to produce stable yeast transformants.

Nearly any type of cell (prokaryotic or eukary-otic) can be transformed by the technique of elec-troporation. Protoplasts are first prepared by enzy-matic or chemical disruption of the host-cellmembrane polysaccharides. Next, the recombinantvector is introduced to the protoplast suspensionresiding in a very low ionic strength buffer (or dis-tilled water). This DNA–protoplast suspension isthen subjected to one or several 250-V pulses de-livered from a cathode and anode placed directlyinto the solution. This applied voltage gradient willcause a certain population of the cells (,1010 per

DNA strand. At the expense of one molecule ofATP, two DNA strands are joined via a phosphodi-ester bond. Although eukaryotic and T4 phageDNA ligase use ATP as the adenyl group donor inthis reaction, E. coli DNA ligase utilizes NAD1 (Fig.V-16).

Since the type II restriction endonucleases pro-duce DNA fragments with 59 phosphoryl groupsand 39 hydroxyl groups, DNA fragments isolatedafter digestion can be easily joined together in vitroin the presence of T4 DNA ligase and ATP. Sincemany of these same restriction endonucleases cleavethe DNA at staggered positions within a specificrecognition sequence, single-stranded DNA over-hangs are present on the ends of these fragments.As long as there is complementarity between thesesingle-stranded DNA overhangs on two DNA frag-ments, DNA ligase will be able to connect them. Ifthe vector DNA and the insert DNA are both di-gested with the same restriction endonuclease, thiswill be the case (Fig. V-17). You may even be ableto join two DNA fragments produced from diges-tion with two different restriction endonucleases,provided that both enzymes produce complemen-tary (compatible) single-stranded DNA overhangs.Although the reaction usually requires higher con-centrations of DNA ligase and longer incubationtimes, the enzyme can even join together blunt-ended (no single stranded overhangs) DNA frag-ments produced from digestion with some type IIrestriction endonucleases.

Transformation of Host Cells

Once the desired recombinant vector has beenconstructed in vitro, it is then necessary to intro-duce the vector into the appropriate host cell. If aviral cloning vector such as bacteriophage l is used,this process becomes trivial. After the recombinantbacteriophage chromosome containing the gene ofinterest is packaged in vitro into an infectiousphage particle, the genetic material packaged in thephage head will be delivered into the host cell viathe normal, phage-directed, transfection mecha-nism.

Some bacterial species, such as Bacillus subtilis,have a genetically specified system that will allowthem to take up extracellular DNA. Expression ofthe genes that make up this competence system (the



LigateLigate

Digest withBamHI (G GATCC)

Chromosomal DNA

Digest withBglII (A GATCT)

Chromosomal DNA

Plasmid digested with BamHI

The single-stranded DNA overhangs produced by BglII and BamHI are complementary.The BamHI sitesand BglII sites are lost following ligation.

∗∗∃7

&7

∃∗∃7&&

&&7∃

∗∃

7&7∃∗∗BamHI sites

still intact

∗∗∃7

&&

∗∗∃7&&

&&7∃

∗∗

&&7∃∗∗

∗

∗∃7&&

∗∗∃7&&

∗∃7&7 ∃

∃ 7&7∃∗

Produces fragments with 5´-GATCsingle-stranded overhangs

∗∃7&& ∗

∗ &&7∃∗

Produces fragments with 5´-GATCsingle-stranded overhangs

Figure V-17 Cloning genes using a single or compatible enzymes (BamHI/BglII).

microgram of DNA) to take up the vector. It is ex-tremely important when using this technique to removeall of the salt from the solution. If this is not done, thesolution will arc (a spark will appear), heat, and killall of the host cells. Even with a very-low-ionic-strength solution, a mortality rate of up to 80% isnot uncommon with electroporation. Its major ad-

vantage is the nearly 10,000-fold increase in trans-formation efficiency (number of transformed cellsproduced per microgram of vector DNA) comparedto the traditional E. coli transformation method de-scribed earlier.

Evidence indicates that transformation effi-ciency can be improved with yeast transformation

328Table V-6 Preparation of E. coli and Yeast Competent Cells

E. coli (CaCl2 method)

1. Grow 100 ml of E. coli cells to an OD600 of 0.35.

2. Harvest cells by centrifugation.

3. Resuspend cell pellet in 20 ml of 0.1 M CaCl2. Incubate on ice for 10 min.


5. Resuspend cell pellet in 4 ml of ice cold 0.1 M CaCl2.

6. Use 200–300 ml of these cells immediately in the transformation reaction.

7. Incubate the cells on ice for 30 min and heat shock at 42°C for 1–2 min.

8. Add 800 ml of rich media and incubate at 37°C for 1 hr.

9. Plate on selective media.

E. coli (RbCl2 method)

1. Grow 50 ml of E. coli cells to an OD600 of 0.35.


3. Resuspend the cell pellet in 16 ml of transformation buffer 1:

12 g of RbCl21.2 g of MnCl2?2H2O

1 ml of 1 M potassium acetate, pH 7.5

1.5 g of CaCl2?2H2O

150 g of glycerol

Bring total volume of solution to 100 ml with distilled water and filter sterilize

4. Incubate the cells on ice for 15 min.


6. Resuspend cell pellet in 4 ml of transformation buffer 2:

1.2 g of RbCl220 ml of 0.5 M MOPS buffer, pH 6.8

11 g of CaCl2?2H2O

150 g of glycerol

Bring total volume of solution to 100 ml with distilled water and filter sterilize

7. Use 200–300 ml of these cells immediately in the transformation reaction. Cells prepared by this method can also be

flash frozen in a dry ice ethanol bath and stored at 270°C for later use.

8. Incubate the cells on ice for 30 min and heat shock at 42°C for 1–2 min.

9. Add 800 ml of rich media and incubate at 37°C for 1 hr.

10. Plate on selective media.

Yeast

1. Grow 300 ml of yeast cells to an OD600 of 0.5.

2. Harvest the cells by centrifugation and resuspend the cell pellet in 10 ml of distilled water.

3. Harvest the cells by centrifugation and resuspend the cell pellet in 1.5 ml of Tris/EDTA/lithium acetate:

10 mM Tris, pH 7.5

1 mM EDTA

100 mM lithium acetate

4. Add 1–5 mg of plasmid DNA plus 100 mg of single-stranded carrier (salmon sperm) DNA to a sterile 1.5-ml

microcentrifuge tube.

5. Add 200 ml of the cells prepared in step 3 to the same tube.

6. Add 600 ml of the following solution to the same tube:

40% wt/vol polyethylene glycol (PEG-8000)

10 mM Tris, pH 7.5

1 mM EDTA

100 mM lithium acetate

7. Incubate the cells at 30°C for 30 min.

8. Add 90 ml of dimethyl sulfoxide (DMSO), and heat shock the cells at 42°C for 15 min.

9. Harvest the cells by centrifugation and resuspend the cell pellet in 1 ml of Tris/EDTA buffer (10 mM Tris, pH 7.5, 1

mM EDTA).

10. Plate cells on selective media.


and electroporation if a defined concentration ofsingle-stranded carrier DNA is included in the host-cell suspension during transformation. This carrierDNA is usually from a nonhomologous source (suchas salmon sperm) to ensure that the genetic integrityof the host genome is maintained. Carrier DNAcontaining homologous gene sequences may havethe potential to recombine with sequences on thehost cell genome and alter the genotype of the hostcell. For reasons that are not entirely understood,a larger percentage of host cells will take up the de-sired vector in the presence of this carrier DNA.

Finally, a new transformation technique hasbeen developed that literally “fires” or “shoots” thevector DNA into the desired host cell. This “genegun” technique works particularly well on plantcells, which contain a rather tough and imperme-able cell wall. The technique involves coating mi-croscopic heavy metal particles (such as tungsten orgold) with vector DNA and projecting them at highspeed toward the host cell. These particles actuallypierce the cell wall and physically carry the DNAinto the cell. Another advantage of this techniqueis that different cell types can be transformed in situ,or directly in the tissue where that cell type is nor-mally found. By doing this, you can determine theeffect of the gene in the developing organ or tissue.

Selection and Screening for Transformed Cells

Once a recombinant DNA vector has been intro-duced into the appropriate host cell, it is then timeto select for those cells that received the vector. Ingeneral, you do this by growing the transformedcell population in conditions under which onlythose that received the plasmid will be able to sur-vive. For instance, suppose that an E. coli strain istransformed with a plasmid carrying the bla (b-lactamase) gene and plated on rich media con-taining ampicillin. Since E. coli does not carry thebla gene on the chromosome, only those cells thattook up the plasmid will express b-lactamase, me-tabolize (inactivate) the antibiotic, and formcolonies on the plate. This type of drug selectionis feasible for any cell type provided that the se-lectable marker (gene) is not normally expressedby the host cell.

Antibiotic resistance genes are not the only ge-netic markers that can be carried on a vector that

will allow a powerful selection for transformed cells.Selection can also be made with the use of variousnutritional genetic markers. Suppose that a popula-tion of yeast cells with a null mutation (deletion) ofthe leu2 gene on the chromosome is transformedwith a vector carrying the leu2 gene. The leu2 geneencodes b-isopropylmalate dehydrogenase. Thisenzyme oxidizes b-isopropylmalate to a-ketoiso-caproate, the last intermediate in the leucinebiosynthetic pathway. Because of the null mutation inthe leu2 gene, this strain will not be able to grow in me-dia that does not contain leucine. If the yeast cells aretransformed with a vector carrying the leu2 geneand plated on media that does not contain leucine,only the cells that took up the vector will be ableto synthesize leucine and form colonies on the plate.Obviously, this same vector could not be selectedfor in a yeast strain that did not contain a deletionof the leu2 gene on the chromosome. Similar se-lection techniques have been devised for differenttypes of host cells that contain chromosomal dele-tions of genes involved in the biosynthesis of histi-dine, methionine, tryptophan, lysine, isoleucine,phenylalanine, threonine, and valine. The only re-quirement for selection is that the vector must endow thehost cell with a phenotype that will allow you to distin-guish it from those cells that did not take up the vector.

Although it is essential that one be able to selectfor cells that took up the vector following trans-formation, it is also desireable to screen for cells thatharbor the desired recombinant vector. Suppose, forinstance, that you have isolated a gene of interestby cleaving the parent DNA with BamHI (see Fig.V-18). You digest the pBR322 plasmid with thesame restriction endonuclease, mix it with the gene-carrying DNA fragment, and perform an in vitroligation reaction. The ligation mixture is used totransform E. coli cells, which are then plated on richmedia containing ampicillin. As can be seen in Fig-ure V-18, two plasmids (or more) may result fromthis procedure: the pBR322 plasmid may re-ligatewithout the gene of interest, or the gene of inter-est may ligate into pBR322 at the BamHI site. Sinceboth plasmids will give rise to ampicillin resistantcolonies, how would you go about determiningwhich of these colonies harbor the pBR322 plasmidcontaining the gene of interest?

The pBR322 plasmid carries genes that conferresistance to both ampicillin and tetracycline. TheBamHI site lies within the tetracycline resistancegene. If the gene of interest has been inserted into


Figure V-18 Insertional inactivation of the tetracycline resistance gene following cloning in pBR322.

ampR 5 ampicillin-resistant; tetR 5 tetracycline-resistant; ori 5 origin of plasmid replication

Digest with BamHI

Digest with BamHI

Chromosomal DNA

pBR322(4361 bp)

AmpR

AmpR

AmpR

TetR

TetR

TetR

TetR

TetR

TetR

AmpR

ori

ori

ori

ori

BamHIsite

Replica plate on tetracyclinecontaining media

Two possible plasmids

Transform E. coliand plate on ampicillin-containing media.

Cells that do not grow on this plate have an insertion in the tetracyclineresistance gene.

Cells that grow on ampicillin took up either of two plasmids.

Ligate

1


the plasmid at the BamHI site, the tetracycline re-sistance gene will be disrupted (insertional inactiva-tion), and cells harboring the plasmid will be unableto grow in the presence of tetracycline. Therefore,if ampicillin-resistant colonies are replica plated ona tetracycline containing plate and do not grow,then they contain the pBR322 plasmid with an in-sertion (most likely of the gene of interest) at theBamHI site. Replica plating is the process of liftingindividual colonies from one plate and transferringthem to the exact same position on a new plate. Fora plate that contains a large number of colonies, thiscan be performed by placing a sterilized felt patchacross the surface of the plate and pressing firmlyto allow the felt to come into contact with the bac-teria. This felt patch can then be placed over thesurface of a fresh plate and pressed firmly to obtainan exact replica of the original plate.

A similar type of insertional inactivation screen-ing process for recombinant vectors is described inthe introduction to Experiment 21. Many modernE. coli cloning vectors contain multiple cloning siteswithin the lacZ a-peptide coding sequence. TheLacZ a-peptide consists of the first 150 amino acidsof b-galactosidase. It is known that the LacZ a-peptide can be removed (cleaved) from the enzymeand added back to the C-terminal domain of b-galactosidase to produce the fully active enzyme. Ifthis plasmid is introduced into an E. coli strain inwhich the chromosomal copy of the lacZ gene hasthe region of the a-peptide deleted, the host strainwill be complemented (restored) for b-galactosidaseactivity. If this same plasmid has a gene inserted inthe multiple cloning site within the lacZ a-peptidesequence, the lacZ a-peptide reading frame will bedisrupted, and the host cell will not display any b-galactosidase activity. Any cell that is not comple-mented for b-galactosidase activity (contains a plas-mid with an insertion at the multiple cloning site)will give rise to white colonies on an IPTG/Xgalplate, while cells that received the unmodified plas-mid will give rise to blue colonies. This screeningprocess involving a complementation is described ingreater detail in Experiment 21.

REFERENCES

Balbas, P., et al. (1986). Plasmid Vector pBR322 and ItsSpecial-Purpose Derivatives—A Review. Gene 50:3.

Chu, G., Hayakawa, H., and Berg, P. (1987). Electro-poration for the Efficient Transfection of Mam-malian Cells with DNA. Nucleic Acids Res 15:1311.

Collins, J., and Hohn, B. (1978). Cosmids: A Type ofPlasmid Gene-Cloning Vector That Is Packageablein Vitro in Bacteriophage l Heads. Proc Natl AcadSci 75:4242.

Dagert, M., and Ehrlich, S. D. (1979). Prolonged Incu-bation in Calcium Chloride Improves the Compe-tence of Escherichia coli Cells. Gene 6:23.

Davidson, J. N. (1972). The Biochemistry of Nucleic Acids,7th ed. New York: Academic Press.

Davies, J., and Smith, D. I. (1978). Plasmid Deter-mined Resistance to Antimicrobial Agents. AnnuRev Microbiol 32:469.

Doerfler, W. (1983). DNA Methylation and Gene Ac-tivity. Annu Rev Biochem 52:93.

Dugaiczyk, A., Boyer, H. W., and Goodman, H. M.(1975). Ligation of EcoRI Endonuclease-GeneratedFragments into Linear and Circular Structures. JMol Biol 96:171.

Friedberg, E. C. (1985). DNA Repair. New York: Free-man.

Gerard, G. F. (1983). Reverse Transcriptase. In: Jacob,S. T., ed. Enzymes of Nucleic Acid Synthesis and Modi-fication: Volume I. DNA Enzymes. Boca Raton, FL:CRC Press.

Hohn, B. (1979). In Vitro Packaging of l and CosmidDNA. Methods Enzymol 68:299.

Hung, M. C., and Wensink, P. C. (1984). Different Re-striction Enzyme-Generated Sticky DNA Ends CanBe Joined in Vitro. Nucleic Acids Res 12:1863.

Ish-Horowicz, D., and Burke, J. F. (1981). Rapid andEfficient Cosmid Cloning. Nucleic Acids Res 9:2989.

Kornberg, T., and Kornberg, A. (1974). Bacterial DNAPolymerases: In: Boyer, P. D., ed. The Enzymes,3rd ed. Vol. 10, pp. 119–144. New York: AcademicPress.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Nucleotides and Nucleic Acids. In: Princi-ples of Biochemistry, 2nd ed. New York: Worth.

Mathews, C. K., and van Holde, K. E. (1990). NucleicAcids. In: Biochemistry. Redwood City, CA:Benjamin/Cummings.

McClelland, M., and Nelson, M. (1987). The Effect ofSite-Specific DNA Methylation on Restriction En-donucleases and DNA Modification Methyltrans-ferases—A Review. Gene 74:291.

Messing, J. (1983). New M13 Vectors for Cloning.Methods Enzymol 101:20.

Minton, N. P., Chambers, S. P., Prior, S. E., Cole, S.T., and Garnier, T. (1988). Copy Number and Mo-bilization Properties pf pUC Plasmids. Bethesda ResLab Focus 10(3):56.

332

Sykes, R. B., and Mathews, M. (1976). The Beta-Lactamases of Gram-Negative Bacteria and TheirRole in Resistance to Beta-Lactam Antibiotics.J Antimicrob Chemother 2:115.

Ullmann, A., Jacob, F., and Monod, J. (1967). Charac-terization by in Vitro Complementation of a Pep-tide Corresponding to an Operator-Proximal Seg-ment of the b-Galactosidase Structural Gene ofEscherichia coli. J Mol Biol 24:339.

Watson, J. D., Hopkins, N. H., Roberts, J. W., Steitz,J. A., and Weiner, A. M. (1987). Molecular Biology ofthe Gene, 4th ed. Menlo Park, CA: Benjamin/Cum-mings.

Oishi, M., and Cosloy, S. D. (1972). The Genetic andBiochemical Basis of the Transformability of Es-cherichia coli K12. Biochem Biophys Res Comm49:1568.

Patterson, T. A., and Dean, M. (1987). Preparation ofHigh Titer Lambda Phage Lysates. Nucleic Acids Res15:6298.

Roberts, R. J. (1988). Restriction Enzymes and TheirIsoschizomers. Nucleic Acids Res (suppl) 16:r271.

Saenger, W. (1984). Principles of Nucleic Acid Structure.New York: Springer-Verlag.

Sambrook, J., Fritsch, E. F., and Maniatis, T. (1989).Molecular Cloning: A Laboratory Manual, 2nd ed.Plainview, NY: Cold Spring Harbor LaboratoryPress.

Stoker, N. G., Fairweather, N. F., and Spratt, B. G.(1982). Versatile Low-Copy-Number Plasmid Vec-tors for Cloning in Escherichia coli. Gene 18:335.


N (

333

EXPERIMENT 19

Isolation of Bacterial DNA

Theory

To isolate a functional macromolecular componentfrom bacterial cells, you must accomplish threethings. First, you must efficiently disrupt the bac-terial cell wall and cell-membrane system to facili-tate extraction of desired components. Second, youmust work under conditions that either inhibit ordestroy the many degradative enzymes (nucleases,proteases) released during cell disruption. Finally,you must employ a fractionation procedure thatseparates the desired macromolecule from othercellular components in satisfactory yield and purity.

The isolation of bacterial DNA described in thisexperiment, patterned after the work of Marmur(1961), accomplishes these objectives. Bacterialcells are disrupted by initial treatment with the en-zyme, egg-white lysozyme, which hydrolyzes thepeptidoglycan that makes up the structural skele-ton of the bacterial cell wall. The resultant cell wallsare unable to withstand osmotic shock. Thus, thebacteria lyse in the hypotonic environment. The de-tergent, sodium dodecyl sulfate, (SDS, sodium do-decyl sulfate) then completes lysis by disruptingresidual bacterial membranes. SDS also reducesharmful enzymatic activities (nucleases) by its abil-ity to denature proteins. The chelating agents, cit-rate and EDTA (ethylenediamine tetraacetic acid),also inhibit nucleases by removing divalent cationsrequired for nuclease activity.

This experiment employs a variety of fraction-ation methods to purify the bacterial DNA. Per-chlorate ion is used to dissociate proteins fromDNA. Chloroform–isoamyl alcohol is used to de-nature and precipitate proteins by lowering the di-

electric constant of the aqueous medium. The pre-cipitated material is removed by centrifugation.DNA is then isolated from the solution by ethanolprecipitation, which allows spooling of the long, fi-brous DNA strands onto a glass rod. Although theresidual RNA and proteins in the solution also co-agulate, they do not form fibers and are left in thesolution as a granular precipitate. The resultantcrude DNA is dissolved and precipitated again withisopropanol under conditions that favor specificDNA precipitation. These steps form a general pro-cedure for the isolation of chromosomal DNA fromvirtually any bacterium. If you intend subsequentlyto assay the biological activity of the isolated DNAin a transformation assay (as in Experiment 20), itis necessary to start this isolation with the appro-priate species and strain of bacteria possessing thegenotype of interest.

This experiment will also provide you with ex-perience in characterizing DNA samples by deter-mining their ultraviolet absorption spectra and ob-serving their thermal denaturation by means of thehyperchromic effect. These topics were discussed inthe introduction to Section V.


For growth of wild-type (Trp�) Bacillus subtilis (if this ex-periment is to be done together with Experiment20. Other kinds of frozen bacterial cells can be usedif Experiment 20 will not be performed.)

Stock culture of wild-type B. subtilis (Bacillus GeneticStock Center strain 1A2)

Inoculating needle


Luria broth (see Appendix)

37°C Incubator-shaker

Fermenter

Centrifuges, continuous and laboratory

Sterile saline solution (8.5 g NaCl/liter)

0.15 M NaCl, 0.l M Na2EDTA

Lysozyme solution, 10 mg/ml

25% (wt/vol) SDS solution

5.0 M NaClO4

Chloroform–isoamyl alcohol (24:1, vol/vol)

95% Ethanol

Dilute (1/10X) saline-citrate (0.015 M NaCl, 0.0015 MNa3-citrate)

Standard (1X) saline-citrate (0.15 M NaCl, 0.015 M Na3-citrate)

Concentrated (10X) saline-citrate (1.5 M NaCl, 0.15 MNa3-citrate)

3.0 M Sodium acetate, 1 mM EDTA, pH 7.0

Isopropanol

Chloroform

Spectrophotometer and quartz cuvettes

Protocol

Growth of Wild-Type Bacillus subtilis for

DNA Isolation

This portion of the experiment may be performedby an instructor in advance.

1. Inoculate (sterile technique) 100 ml of sterileLuria broth (200- to 400-ml flask) with a sam-ple of wild-type Bacillus subtilis obtained froma stock culture slant. Incubate this solution ona shaker at 37°C for 15 to 24 hr (until the cul-ture is densely turbid and is entering the sta-tionary phase of growth).

2. Transfer the starter culture (sterile technique)into a fermenter containing 10 to 12 liters ofsterile Luria broth and continue the bacterialgrowth at 37°C while stirring (�400 rpm) andaerating at 1 liter of air per liter of fluid perminute. Monitor the growth periodically by de-termining the turbidity of the culture (ab-sorbance at 660 nm) during the 3- to 5-hrgrowth until this culture just enters the sta-tionary phase of growth.

3. Stop the growth by cutting off both air and ag-itation. Use a continuous centrifuge (e.g.,Sharples) to harvest the cells as quickly as pos-sible.

4. Suspend the cell paste in an equal volume ofcold (1 to 4°C) sterile saline solution. Completethe cell-washing procedure by harvesting theB. subtilis cells (10 min, 20,000 � g, 1 to 3°C)with a high-speed centrifuge.

5. Decant the supernatant fluid from the cell pel-let after centrifugation and store the washedcells in Parafilm or waxed-paper-wrapped lotsat �20°C until use (2-g lots are convenient).The expected yield is 35 to 45 g of cells per 10liters.

Isolation of Bacterial DNA

1. Suspend 2 g of bacterial cell paste in 25 ml of0.15 M NaCl, 0.1 M Na2EDTA solution. Thebacterial paste is most easily suspended by firstadding a small amount of the NaCl–EDTA so-lution, making a slurry, and then slowly addingthe rest of the NaCl–EDTA solution. Aftersuspension, add 1 ml of a 10-mg/ml lysozymesolution and incubate the suspension at 37°Cfor 30 min, gently agitating occasionally.

2. After the 30-min incubation, complete the ly-sis of the bacteria by adding 2 ml of a 25% SDSsolution and heating this preparation for 10min in a 60°C water bath. Cool the solution toroom temperature in a bath of tap water.

3. Add 7.5 ml of 5.0 M NaClO4 to the 25 ml oflysed bacteria cells and mix gently. Then add32.5 ml of CHCl3–isoamyl alcohol (24:1) (i.e.,a volume equal to the volume of lysed cellpreparation containing 1.0 M perchlorate).Slowly shake the solution (30–60 oscillationsper minute, by hand or on a low-speed shaker)in a tightly stoppered flask for 30 min at roomtemperature.

4. Transfer the suspension to a 250-ml polypropy-lene centrifuge bottle, and separate the result-ing emulsion by centrifuging for 5 min at10,000 � g at room temperature.

5. During the centrifugation step, remove anyresidual enzymes from a glass rod by heatingone end of the rod in a flame and allowing thesterile end to air cool in a thoroughly washed250-ml beaker.

6. After the centrifugation step, carefully pipettethe clear aqueous phase (top layer) away fromthe coagulated protein emulsion at the inter-face between the aqueous and organic phasesof the centrifuged solution. Do not attempt to

EXPERIMENT 19 Isolation of Bacterial DNA 335

remove all of the aqueous phase; rather, avoidcontaminating the aqueous phase with mater-ial from the other layers. Place the aqueousphase, which contains the extracted nucleicacids, in the 250-ml beaker. Dispose of theCHCl3-containing waste in the special containerprovided by your instructor.

7. Gently stir the nucleic acid solution with thesterilized rod while slowly and gently adding 2volumes (about 60 ml) of 95% ethanol. Pourthe ethanol gently down the side of the beaker sothat it is layered over the viscous aqueous phase.Continue to stir this preparation gently with thesterilized glass rod so that the ethanol is verygradually mixed throughout the entire aqueousphase. Avoid stirring vigorously, which wouldshear the DNA strands and make the DNAmore difficult to isolate. The DNA will forma white fibrous film at the water–ethanol in-terface, and you should be able to spool all ofthe gelatinous, threadlike, DNA-rich precipi-tate onto the glass rod. Continue gentle stir-ring until the aqueous and ethanol phases arecompletely mixed and no more DNA precipi-tate can be collected on the glass rod. Drain offexcess fluid from the spooled crude DNA by press-ing the rod against the walls of the beaker until nofurther fluid can be squeezed from the spooledpreparation.

8. Dissolve the crude DNA by stirring the glassrod with its spool of material in 9 ml of dilute(1/10X) saline-citrate in a test tube. Difficultyin dissolving the sample indicates failure to re-move sufficient alcohol from the sample dur-ing step 7. If such difficulty is encountered,continue working the sample within the solu-tion until you obtain as uniform a suspensionas possible.

9. To the DNA solution add 1 ml of 3.0 M sodiumacetate, 1 mM EDTA, pH 7.0 solution andtransfer the preparation to a 100-ml beaker.Gently swirl the sample while slowly drippingin 6.0 ml of isopropanol. If fibrous DNA isreadily apparent, collect the DNA threads bystirring and spooling with a sterilized glass rodas described in step 7. If a gel-like preparationdevelops, add an extra 1.0 ml of isopropanoland stir to spool the DNA onto the sterilizedglass rod. Remove excess fluid from the spooledDNA by pressing the sample against the wallsof the beaker.

10. Wash the sample (by submersion and briefswirling of the DNA on the rod) in test tubescontaining, in order, 10 ml of 70% ethanol(made by adding 2.6 ml H2O to 7.4 ml of 95%ethanol), and then 10 ml of 95% ethanol.

11. If you plan to isolate the DNA only for use inthe genetic transformation outlined in Experi-ment 20, or if a day or longer will elapse be-fore you will characterize the DNA as describedbelow, store the DNA in a stoppered tube (2°C)as a spool submerged on the rod in 95%ethanol. Dissolve the DNA as described in step1 of the following section when you are readyto use it.

Characterization of DNA

1. Remove the rod and the spooled DNA fromthe 95% ethanol, blot away all obvious resid-ual fluid with a clean piece of filter paper, andthen dissolve the DNA by stirring the glass rod,with its spooled DNA, in a test tube contain-ing 10 ml of dilute (1/10X) saline-citrate (ster-ile solution if the DNA is to be used in Experiment20). This solution can be stored at 2°C.

Spectral Characterization of DNA

2. Dilute a 0.5 ml sample of the DNA solutionwith 4.5 ml of standard (1X) saline-citrate anddetermine the absorbance of this diluted sam-ple at 260 nm against a dilute saline-citrateblank containing no DNA. If the absorbance at260 nm is greater than 1.0, dilute the samplewith a precisely known volume of standard (1X)saline-citrate until you obtain an absorbance ofbetween 0.5 and 1.0.

3. Determine the absorbance of this solution at 5-nm intervals between 240 and 300 nm, so as toobtain an absorption spectrum for the isolatedDNA. (Remember that quartz cuvettes must beused in this wavelength range.) Blank the spec-trophotometer to read zero absorbance with 1Xsaline-citrate at each wavelength. (If equipmentfor the automated collection of this absorbancespectrum is available, follow the instructionsfor that instrument.) Include this absorptionspectrum in your report.

4. Calculate the A260�A280 ratio for your sample,and compare it with the value expected for pureDNA (2.0).


5. Assuming that a 1-mg/ml solution of nativeDNA has an absorbance at 260 nm of 20 andthat all of your absorbance at 260 nm is due tonative DNA, calculate the concentration (inmilligrams per milliliter) of the DNA in yourundiluted DNA solution and the total yield (inmicrograms or milligrams) of DNA you ob-tained.

6. Assuming 100% recovery of the cellular DNAand assuming the wet bacterial cell paste was80% water, calculate the percent of the dryweight in the bacterial cells that is DNA. Howdo your results agree with the DNA content ofa typical bacterium, which is about 3% of thecell’s dry weight? If your results do not agreewith this value, discuss possible reasons for thediscrepancy.

Melting of DNA—The “Hyperchromic Effect”

1. If your laboratory is equipped with a spec-trophotometer with a thermoprogrammer de-signed for determination of melting curves ofnucleic acid samples (Gilford, Beckman Instru-ments), determine a melting curve of a sampleof your DNA dissolved in the required volume(usually 1 ml or less) of standard (1X) saline-citrate so as to give an initial absorbance ofabout 1 at 260 nm. Use the results from thepreceding section to decide how much youneed to dilute the stock DNA solution. Followthe instructions provided by the instrument’smanufacturer in programming the rate of tem-perature increase, protecting the sample againstevaporation, etc.

2. If your laboratory does not have access to aspectrophotometer with a thermoprogrammer,use the “quick cool” method described in steps3 through 6 below to obtain a melting curvefor your DNA. This method does not observea true equilibrium between native and dena-tured (“melted”) DNA and is appreciably lessaccurate than the method described in step 1,but it will suffice to illustrate the principles un-derlying the thermal separation of DNAstrands.

3. Using your stock DNA solution from step 11,prepare 10 ml of a diluted DNA solution in thedilute saline-citrate (1/10X) that has a final ab-sorbance at 260 nm of about 0.5. Use the re-

sults from the preceding section to decide howmuch you need to dilute the stock DNA solu-tion. Determine the absorbance at 260 nm ofthis dilute solution at room temperature (about20°C).

4. Transfer 1.0 ml of the same dilute DNA solu-tion into each of six microcentrifuge tubes andcap them tightly. Incubate one tube at each ofthe following temperatures for 20 min: 50°C,65°C, 75°C, and 85°C. In addition, incubatetwo of the tubes at 100°C (i.e., a boiling waterbath).

5. After 20 min of incubating the tubes at theindicated temperatures, quickly cool all ofthem by placing them in an ice-water bath andgently agitating them. Allow one of the tubesheated at 100°C to cool slowly to room tempera-ture by placing it in a rack at your bench (30 to40 min will be required).

6. As quickly as possible after they have cooled,determine the absorbance at 260 nm of the so-lution in each tube. For a blank, use a sampleof dilute saline-citrate (1/10X) that contains noDNA.

7. During heating, the double-stranded structureof DNA “melts”; that is, the strands separate.The nucleotide bases in the separated strandsare no longer stacked in a regular fashion,which results in an increase in their absorbanceof UV light. This is called the hyperchromic ef-fect or hyperchromicity. When melted DNA isquickly cooled in solutions with very low saltconcentrations, the strands are very slow to reassociate into their original base-paired, double-stranded form. Thus, you can estimatethe degree of melting at each temperature eventhough you are measuring the absorbance at260 nm at room temperature. The DNA sam-ple that was heated to 100°C and slowly cooledto room temperature, on the other hand, is ex-pected to form normal double-stranded DNAagain (reanneal) and to exhibit little hyper-chromic effect. Do your data support this ex-pectation?

8. Prepare a plot of the absorbance at 260 nm ateach temperature for the unheated sample andeach of the rapidly cooled samples. Estimatethe melting temperature (Tm) for your DNA.(Tm is the temperature at which the DNA sam-ple is half melted. Assume that the DNA is

EXPERIMENT 19 Isolation of Bacterial DNA 337

completely melted at 100°C. Calculate the hy-perchromic effect for your DNA as the percentincrease in absorbance at 260 nm on completemelting. Calculate the percent guanine plus cy-tosine (% GC) base pairs in your DNA usingthe following formula

% GC � 2.44 (Tm � 81.5°C � 16.6 log10[Na�])

For dilute (1/10X) saline-citrate, this becomes% GC � 2.44 (Tm � 53.9°C ). B. subtilis DNAis known to contain about 43% GC. How doesthis agree with your results? Discuss the resultsof this experiment.

Exercises

1. The DNA of cells containing nuclei (eukary-otic cells) is tightly associated with proteinscalled histones, most of which are rather basicproteins (i.e., they have high isoelectric points).Considering the basic character of most his-tones and the amino acids that contribute tothis basic character, what enzymes would prob-ably be most effective in disrupting histonesfrom eukaryotic cellular DNA? Would you ex-pect the procedure used in this experiment forthe isolation of bacterial DNA to be useful forthe isolation of DNA from eukaryotic cells?

2. This isolation of bacterial DNA utilizesreagents in the neutral pH range. What effects,if any, would you expect extremes in acid or al-kali to have on the fibrous character of theDNA isolated in this procedure? Why?

3. Examine the procedure described in Experi-ment 21 for the isolation of plasmid DNA frombacteria. It is very different from the procedureused to isolate chromosomal DNA in this ex-periment. Could it be used to isolate chromo-somal DNA from bacteria? Why or why not?Explain why such different procedures are usedfor isolating these two types of DNA.

4. What other physical and chemical treatmentsbesides heating would be expected to allow youto demonstrate the hyperchromicity of single-stranded DNA? Explain how these treatmentsbring about strand separation.

5. Most naturally occurring RNA molecules aresingled-, not double-stranded duplexes, yetRNA also shows a hyperchromic effect onmelting. Explain.

6. The equation used for calculating % GC basespairs in this experiment predicts that increas-ing % GC and increasing Na� concentrationincreases the Tm of DNA. Explain the physicalbasis for these effects.

REFERENCES

Mandel, M., and Marmur, J. (1968). Use of ultravioletabsorbance-temperature profiles for determiningthe guanine plus cytosine content of DNA. MethodsEnzymol 12B:195.

Marmur, J. (1961). A procedure for the isolation ofdesoxyribonucleic acid from microorganisms. J MolBiol 3:208.

Puglisi, J. D., and Tinoco, I. (1989). Absorbance melt-ing curves of RNA. Methods Enzymol 180:304.

D w

339

as motility, the ability to degrade and utilize alter-native metabolites, and endospore formation). Thedevelopment of competence involves expression ofa large number of com genes and is subject to a com-plex regulatory system. Many details have yet to belearned, but a number of steps in transformationhave been identified. First, double-stranded DNAis bound by specific proteins found on the cell sur-face of competent, but not noncompetent, cells.There appears to be little sequence specificity in thebinding of DNA, but only DNA molecules with ahigh degree of sequence homology to the DNA inthe B. subtilis chromosome will eventually recom-bine and lead to genetic changes in the recipientcells. Very large DNA molecules undergo one ormore cleavages to yield smaller double-strandedDNAs. The DNA is then transported into the cellin a process that requires metabolic energy. Then,the DNA is converted to the single-stranded formin a process that probably involves the randomdegradation of one of the two strands of the im-ported DNA. The single-stranded DNA forms aheteroduplex with homologous sequences of thechromosomal DNA. Subsequent integration, re-combination, and repair of mismatches in the DNAproceed by steps that are not fully understood, butwhich involve a number of the genes required forother forms of genetic recombination.Transformation of B. subtilis cells is easily stud-

ied in the laboratory if the genes under observa-tion lead to a readily selectable phenotype. For ex-ample, consider competent recipient cells that areauxotrophic for a required metabolite, such as theamino acid tryptophan; that is, they have a geneticdefect in a gene (trp gene) that specifies one of the

EXPERIMENT 20

Transformation of a Genetic Characterwith Bacterial DNA

Theory

Transformation is the process by which exogenousDNA is taken up by bacterial cells and permanentlyalters the cells’ inherited characteristics. Distinctionis made between “natural” transformation and “ar-tificial” transformation. Numerous bacterial speciesare naturally transformable; that is, they have thegenetic capacity to develop competence for transfor-mation naturally under certain growth conditions asone of their normal physiological responses. Exam-ples of naturally competent bacteria include mem-bers of the genera Bacillus, Streptococcus, Haemophilus,and Neisseria. Many other bacterial species and eu-karyotic cells do not develop competence naturally,but they can be made artificially competent fortransformation by laboratory procedures that ren-der them capable of taking up exogenous DNA,such as a combination of osmotic shock and treat-ment with CaCl2, electroporation, or removal of cellwalls to form protoplasts. The development of suchtechniques has played an extremely important rolein the development of modern molecular biology.(See the introduction to Section V and Experiment21 for a further discussion of artificial competence.)Artificial transformation of Escherichia coli cells

will be performed as part of Experiment 21. In thisexperiment, you will study natural transformationusing strains of the bacterium Bacillus subtilis. Nat-ural transformation has been subjected to extensivestudy by genetic and biochemical analysis. B. sub-tilis cells are not able to take up DNA and undergotransformation under all growth conditions, butgenerally develop competence during periods ofnutrient limitation (along with other responses such


*Additional details in the Appendix.

enzymes required for the biosynthesis of trypto-phan (such cells are called Trp�). Trp� cells can-not grow on a medium lacking tryptophan. If, how-ever, Trp� cells are grown under conditions leadingto the development of competence, and they aretreated with DNA isolated from a B. subtilis strainthat had a normal copy of the trp gene that was de-fective in the recipient cells, some of the recipientcells will take up the normal gene, recombine itinto their chromosome, and inherit a normal trpgene. Such cells will now be able to grow on amedium lacking tryptophan. Cells that did not re-ceive the trp gene will remain unable to grow on amedium that lacks tryptophan. The Trp� cells willgrow as individual colonies on Petri dishes con-taining agar medium without tryptophan, but theTrp� cells that were not transformed will not grow.In this experiment you will use DNA isolated

from a wild-type (Trp�) strain of B. subtilis as de-scribed in Experiment 19 to transform a competentTrp� mutant strain of B. subtilis (strain 168) fromtryptophan auxotrophy (Trp�) to tryptophan pro-totrophy (Trp�). The results will also enable youto determine the transformation efficiency obtainedwith this protocol.


For growth of competent recipient Bacillus subtilis strain168 (Trp�):

Stock culture of B. subtilis strain 168 (Bacillus GeneticStock Center strain 1A1)

Inoculating needle

0.8% Nutrient broth, 0.5% glucose, 50 �g/ml trypto-phan

37°C incubator-shaker

Competence growth medium I (SPI)*

Competence growth medium II (SPII)*

Minimal medium agar Petri plates*

Minimal medium agar Petri plates supplemented with50 �g/ml tryptophan*

Centrifuge

Sterile centrifuge bottles or tubes with caps

Sterile glass rod

Sterile synthetic medium*

Sterile 10-ml pipettes

Sterile saline (8.5 g NaCl/liter)

Sterile microcentrifuge tubes

Sterile 100% glycerol

For transformation:

Sterile test tubes with caps (preferably 13-�-100-mmtubes)

Sterile 0.1-ml and 1.0-ml pipette tips

Sterile minimal medium*

DNA from wild-type B. subtilis (from Experiment 19)

Standard (1X) saline-citrate (0.15 M NaCl, 0.015 M Na3-citrate) (sterile)

Pancreatic DNase (1 mg/ml in sterile 1X saline-citrate)

Recipient B. subtilis strain 168 cells in sterile saline

Sterile 0.1 M EGTA, pH 7.9

40 to 45°C water bath

37°C incubator-shaker

95% Ethanol

Bent glass rods

Minimal medium agar Petri plates*

Minimal medium agar Petri plates supplemented with50 �g/ml tryptophan*

Tubes containing 9.9 ml of sterile saline

Tubes containing 0.9 ml of sterile saline

Masking tape

Protocol

Growth of Competent Recipient Bacillus

subtilis Strain 168 (Trp�)

This portion of the experiment may be performedby an instructor in advance.

NOTE: Sterile technique is essential in the fol-lowing transfers, incubations, and centrifugationsto avoid contamination with other bacteria.

1. Inoculate 5 ml of 0.8% nutrient broth, 0.5%glucose, and 50 �g/ml tryptophan with a sam-ple of Bacillus subtilis strain 168 (Trp�) obtainedfrom a stock culture slant.

2. Incubate this culture on a shaker at 37°C for12 hr overnight.

3. After this incubation, use a sterile inoculatingneedle to streak out a few droplets of thegrowth medium on one Petri plate containingminimal medium agar and another plate con-taining minimal medium agar supplementedwith 50 �g/ml tryptophan. The results of theseplatings are necessary to confirm the Trp� phe-notype of the cells. Incubate the plates at 37°Cfor 24 hr and examine the plates. Many colonies

EXPERIMENT 20 Transformation of a Genetic Character with Bacteria DNA 341

should appear on the plates containing trypto-phan, and no colonies should be found on theplates lacking tryptophan.

4. After removing samples for plating to confirmthe Trp� phenotype, use the remaining cultureto inoculate 100 ml of SPI medium in a 500-ml flask. Incubate the culture at 37°C overnightwith shaking. The solution should be turbid inthe morning.

5. Transfer 5 to 10 ml of this culture to 100 ml offresh SPI medium in a 500-ml sidearm flask sothat the final cell suspension has an absorbanceat 500 nm of about 0.1 when viewed against anSPI medium blank. (NOTE: If you do not usea Klett colorimeter and a sidearm flask, remove1-ml samples for absorbance readings in a spec-trophotometer. Use sterile technique for allsamplings. Do not return samples from the cu-vettes to the cell suspension.) Incubate this cul-ture with shaking at 37°C.

6. Follow the growth of the cells in the culture bytaking readings from the sidearm with a Klettcolorimeter or using the sampling techniquedescribed in step 5. Plot the absorbance as afunction of time of growth on semilog paper.

7. When the turbidity of the culture (cell growth)stops increasing exponentially (usually 4 to 5 hrafter inoculation), transfer 50 ml of the cultureto a 2-liter flask containing 450 ml of fresh SPIImedium, which is the same as SPI medium ex-cept that it contains 0.5 mM CaCl2 and 2.5 mMMgCl2. Incubate this culture with shaking at37°C for 90 min.

8. Cells from the culture in step 7 should have de-veloped maximal competence and can be used

directly for transformation. However, it is usu-ally much more convenient to store the cells asfrozen cell cultures and to thaw and use themwhen needed. Frozen competent cells are pre-pared as follows:Centrifuge the cell culture in a high-speed

centrifuge (10 min, 20,000� g, 1 to 4°C) us-ing sterilized centrifuge bottles. Working asep-tically and quickly, decant all but about 1/10 ofthe original volume (about 50 ml) of the SPIIgrowth medium. Resuspend (sterile glass rod)the cell pellet in the remaining growth medium.Add 5 ml of sterile glycerol and mix thoroughly.The resultant cell suspension should have anabsorbance at 660 nm of roughly 0.5 (i.e., �108

cells per milliliter). Store these cells in 0.6-mlaliquots in sterile 1.5-ml microcentrifuge tubesat �70°C until use in the transformation ex-periments below. This procedure yields suffi-cient cells for about 80 pairs of students.

Transformation

The following procedures require sterile technique; that is,use of sterilized glassware, pipettes, Petri plates, and otherequipment. Do not use equipment that has lost a cover-ing cap and consequently has been exposed to the air forlong periods. If you are unfamiliar with sterile techniquesfor transferring bacterial cultures, consult your instructorfor details of the techniques required in this experiment.

1. Start by carrying out the special treatments ofDNA required for tubes 4 and 6 in Table 20-1.

2. If frozen competent cells are to be used, thawthe entire 0.6-ml sample by incubation for

Table 20-1 Protocol for Transformation of Competent B. subtilis Cells with DNA

Tube Number

Component (volumes in ml) 1 2 3 4 5 6

Minimal growth medium 0.8 0.8 0.8 0.8 0.8 0.8

B. subtilis DNA from wild type* — 0.1 0.2 0.1 0.1 0.1†

Standard (1X) saline-citrate 0.2 0.1 — — 0.2 0.1

DNase solution — — — 0.1‡ — —

Competent B. subtilis cells (Trp�) 0.1 0.1 0.1 0.1 — 0.1

*DNA dissolved in standard (1X) saline-citrate as described in Experiment 19.

†Denature this DNA sample before use in transformation by heating in a boiling water bath for 5 min and cooling rapidly in an ice bath.

‡Incubate the growth medium � DNA � DNase at 37°C for 15 min before adding the competent cells.


5 min at 40 to 45°C. Add 6 �l of sterile 0.1 MEGTA, pH 7.9, and mix gently. Use these cellsfor step 3.

3. Using sterile technique, pipette the followingingredients into six empty sterile test tubes inthe order shown in Table 20-1. Note that tubes4 and 6 require special treatment of the DNAbefore initiating the incubations. Label thetubes clearly with their numbers and your ini-tials.

4. Cover the tubes with caps and incubate themfor 2 hr at 37°C, preferably in a slowly shaking

water bath, but alternatively in a 37°C incuba-tor with occasional gentle hand agitation. Dur-ing the 2-hr incubation, set up and familiarizeyourself with the details of the following dilu-tion and plating operations.You will determine the number of cells in

each bacterial suspension by use of the agarplating method. In this method, you spread asmall sample of known volume (or a known di-lution of such a sample) over the surface of asterile Petri plate containing an agar-stabilizedgrowth medium (Fig. 20-1). The sample dif-

Sterilepipette

Bacterialsolution ora dilutionof same

Lift lid slowlyand add solutionaliquot

Use flamedglass rod tospread aliquot

Bent glass rodin 95% Ethanol

Flame away95% Ethanol

Incubateinvertedplate at 37°C

Close lid,wait 5 min,invert

Figure 20-1 Steps of the procedure for plating bacteria.

EXPERIMENT 20 Transformation of a Genetic Character with Bacteria DNA 343

fuses into the growth medium, embedding sin-gle bacterial cells in the agar. On incubation,the bacterial cells take up nutrients and divide.Each original bacterial cell capable of growthon the growth medium eventually yields a vis-ible colony of cells growing in the same spot onthe agar where the original single bacterial cellwas embedded. Thus, the number of coloniesobtained from the sample placed on the plateequals the number of viable cells originallypresent in the sample. This assumes that all ofthe cells can grow on the growth medium pro-vided in the Petri plate. If the growth mediumis altered such that only cells of a certain phe-notype can grow on the plate, the number ofcolonies will reflect only the cells of that sub-population.Analysis of the transformation performed in

this experiment requires agar plating to deter-mine the number of transformed cells (Trp�

cells) in each of the six tubes of the experiment.The Petri plates used in this determination ob-viously do not contain tryptophan in thegrowth medium. In a separate experiment, youwill determine the total number of cells (Trp�

plus Trp� cells) in the transformation mixtureby plating a diluted sample of known volumeand known dilution on a Petri plate in whichthe agar growth medium does contain trypto-phan. The number of transformants (Trp�)cells divided by the total number of cells (Trp�

plus Trp�) is the efficiency of that transforma-tion experiment.The number of transformed cells in each

tube will be small compared with the totalnumber of cells (perhaps 0.1 to 1%). You cantherefore determine the number of trans-formed cells in each tube by plating a 0.1-mlsample (or a limited dilution of a 0.1-ml sam-ple) on minimal medium agar (i.e., growthmedium containing sugars and salts but lack-ing tryptophan). In contrast, analysis of the to-tal number of cells in each tube requires ex-tensive dilution of the contents of each tubebefore plating on minimal medium agar platessupplemented with tryptophan.

5. After the 2-hr incubation at 37°C, use steriletechnique to perform the following steps fordilution and plating, labeling all tubes andplates with the dilutions and medium used.

(NOTE: Once you obtain a specified bacterialdilution, conserve pipettes by using the samepipette both for plating and for subsequent di-lutions.) Use the technique illustrated in Fig-ure 20-1 to plate the samples and spread themevenly over the agar surface.

6. Determine the number of transformed (Trp�) cellsin each of the six assay tubes by plating a 0.1-ml sample from each tube on labeled platescontaining minimal medium agar (withouttryptophan).

7. The Trp� colonies obtained from tubes 2 and3 may be too numerous to count. For this rea-son you must confirm the number of trans-formed (Trp�) cells in tubes 2 and 3 by trans-ferring a 0.1-ml sample from each of thesetubes into separate marked tubes containing9.9 ml of sterile saline. Mix these solutionsthoroughly and plate a 0.1-ml sample of the100-fold diluted samples on separate markedplates containing minimal medium agar. Savethe diluted culture samples for later use.

8. Determine the total number of cells (Trp� plusTrp�) in tubes 1 and 2 by preparing a 100-folddilution of tube 1 by adding 0.1 ml of culturefrom tube 1 to 9.9 ml of sterile saline. The samedilution was already performed for tube 2 instep 5. Take a 0.1-ml sample from the dilutionsof tubes 1 and 2 and transfer each to separatemarked tubes containing 9.9 ml of sterile saline.Plate a 0.1-ml aliquot of each of these latter10,000-fold dilutions on minimal medium plustryptophan agar plates. Finally, transfer sepa-rate 0.1-ml samples of both 10,000-fold dilutedsamples to marked tubes containing 0.9 ml ofsterile saline, mix the contents, and plate sep-arate 0.1 ml samples of these 100,000-fold di-lutions on minimal medium plus tryptophanagar plates. Two dilutions of the cultures intubes 1 and 2 are prepared in case the numberof colonies on the plate containing the 10,000-fold dilution are too numerous to count. In thiscase, the 100,000-fold dilution of the same cul-ture should produce a number of colonies thatyou can count readily.

9. Immediately after placing each 0.1-ml sampleof culture on a Petri plate, spread the sampleuniformly over the surface of the plate with asterile bent glass rod (see Fig. 20-1). Be sure thatyou sterilize the glass rod between spreadings


of each sample by dipping it in ethanol andflaming it.

10. Allow all the plated samples to diffuse into theagar for 5 to 10 min. Then, invert each Petriplate (agar side up), write your name or initialson each, combine the plates in stacks held bymasking tape, and incubate all plates for 48 to72 hr at 37°C. Count the colonies on eachplate.

11. Use the dilution factors and the number ofcolonies detected on each plate to determinethe number of Trp� cells per milliliter of undi-luted growth medium in each of the six trans-formation tubes. How do the results obtainedfor tubes 2 and 3 with the 100-fold diluted sam-ples agree with those obtained when 0.1 ml ofundiluted cells was plated? Explain your find-ings.

12. Use the dilution factors and colony counts todetermine the total number of viable cells(Trp� plus Trp�) present per milliliter of undi-luted culture in tubes 1 and 2. Assuming thatall of the Trp� cells are transformants and as-suming that the total number of viable cells intubes 2 through 6 is the same, calculate the ef-ficiency of transformation in tubes 2 to 6. In-terpret all your results in terms of the contentsof each tube and your knowledge of the processof transformation. Which tubes served as con-trols for the experiment? Explain.

Exercises

1. What biological advantage do bacteria derivefrom having the capacity to develop naturalcompetence? Suggest why competence devel-ops optimally in nutrient-limited cultures.

2. A student performing this experiment detects2 � 107 total viable cells (Trp� plus Trp�) permilliliter in tube 1 of this experiment, yet thestudent cannot detect a significant number ofviable cells in tube 2 of this experiment. Sug-gest an explanation for this result.

3. Assume that indoleglycerolphosphate syn-thetase, an enzyme essential for tryptophanbiosynthesis in B. subtilis, has a molecularweight of 50,000, that amino acids in proteinshave an average molecular weight of 100, andthat nucleotides in nucleic acids have an aver-age molecular weight of 333. Calculate the min-imal molecular weight of a transforming frag-ment of double-stranded DNA that could carrythe complete genetic information necessary forsynthesis of indoleglycerolphosphate syn-thetase.

REFERENCES

Anagnostopoulos, C., and Spizizen, J. (1961). Require-ments for transformation in Bacillus subtilis. J Bacte-riol 81:741.

Dubnau, D. (1993). Genetic exchange and homologousrecombination. In: Sonenshein, A. L., Hoch, J. A.,and Losick, R., eds. Bacillus subtilis and Other GramPositive Bacteria: Biochemistry, Physiology and Molecu-lar Genetics.Washington, DC: American Society forMicrobiology, pp. 555.

Mahler, I. (1968). Procedures for Bacillus subtilis trans-formation. Methods Enzymol 12B:846.

Sadaie, Y., and Kada, T. (1983). Formation of compe-tent Bacillus subtilis cells. J Bacteriol 153:813.

Wilson, G. A., and Bott, K. F. (1968). Nutritional fac-tors influencing the development of competence inthe Bacillus subtilis transformation system. J Bacteriol95:1439.

345

EXPERIMENT 21

Constructing and Characterizing a Recombinant DNA Plasmid

Theory

The discovery of plasmids and restriction endonu-cleases is largely responsible for the enormous tech-nological advances made in the field of molecularbiology over the past 20 years. A desired DNA frag-ment or gene sequence can be identified, excisedfrom the chromosome, isolated, and cloned into anumber of different plasmids or vectors. This genecan then be sequenced, mutated, and/or expressedto gain insight into the function of the protein thatit encodes. The plasmid carrying the gene of in-terest can be introduced into a host cell and faith-fully propagated during the cell’s normal cycle ofDNA replication and cell division. In this experi-ment, you will subclone the gene for aminoglyco-side-39-phosphotransferase from plasmid pUC4Kinto pUC19. This gene will confer resistance to theantibiotic, kanamycin, a phenotype that can be ex-ploited in the selection process for cells that ob-tained the desired plasmid. The resultingpUC19/4K recombinant plasmid will be selectedfor, and the construction of the plasmid will be ver-ified using restriction endonuclease digestion andagarose-gel electrophoresis.

Plasmid pUC4K is a commercially available Es-cherichia coli vector that contains the kanamycin re-sistance (kanR) gene from transposon (Tn) 903 (Fig.21-1). The gene is flanked on the 59 and 39 sidesby four different restriction endonuclease recogni-tion sites. Thus, the kanR gene can be excised frompUC4K and inserted into virtually any other exist-ing plasmid to create a new vector that confers kanR

to its host. The kanR gene can also be inserted intothe open reading frame of a gene of interest, al-

lowing you to study the effect of deletion of thegene (null mutation) on the organism. This type ofgene disruption experiment has proven to be veryuseful in efforts to determine the function of par-ticular gene products in various biologicalprocesses.

Plasmid pUC19 (and its derivative, pUC18) isone of the most popular and widely used E. colicloning vectors in molecular biology (see Fig. 21-1).Like pUC4K, pUC19 contains the gene that con-fers ampicillin resistance (ampR) to its host (bla, b-lactamase). This vector also contains a multiplecloning site (MCS) within the sequence of theLacZa peptide. The LacZa peptide, which encodesthe N-terminal 150 amino acids of b-galactosidase,will function in trans to complement (restore) b-galactosidase activity in a host strain deleted for theLacZa peptide. The position of the MCS on pUC19will play an important role in the selection processto identify cells that harbor the desired pUC19/4Kplasmid. If cloned into pUC19 in the same readingframe as the LacZa peptide sequence, a gene of in-terest can be expressed as a fusion protein with theN-terminus of the LacZa peptide. Like the LacZapeptide, the expression of the fusion protein willnow be under the control of the lac promoter, whichis inducible with the addition of isopropylthio-b-galactoside (IPIG) to the growth medium. Recallthat the expression of genes under the control of thelac promoter is normally low in the presence of thelac repressor protein (LacI), the gene for which isalso contained in pUC19 (see Fig. 21-1) (See intro-duction to Experiment 7).

On Day 1 of the experiment, you will digestpUC4K with EcoRI and XhoI. You will also digest


ampR ampR

lacZ'

kanRla

cz'

ori

ori

TAAPstI (417)Sal I (411)BamHI (405)EcoRI (396)

Pst I (1657)Sal I (1663)

BamHI (1669)EcoRI (1678)

ATG

HindIII (993)

XhoI (1503)kanR gene translationalstart codon

kanR genepromoter

pUC4K(3914 base pairs)

pUC19(2686 base pairs)

kanR gene translationalstop codon

lacI

lacZ'

Multiple cloning site (MCS)(see below)

lac promoter

Multiple cloning site in pUC19:

G A A T T C G A G C T C G G T A C C C G G G G A T C C T C T A G A G T C G A C C T G C A G G C A T G C A A G C T T

396 452

EcoRI SacI KpnI SmaIXmaI

BamHI XbaI SalIAccI

HincII

PstI SphI HindIII

Figure 21-1 The pUC19 and pUC4K plasmids.

pUC19 with EcoRI, BamHI, and SalI. Followingdigestion, the DNA fragments resulting fromthese reactions (Fig. 21-2) will be mixed togetherand precipitated with ethanol. Next, T4 DNA lig-ase and ATP will be added, and the ligation reac-tion will be allowed to proceed overnight. Asshown in Fig. 21-2, the 59 single-stranded DNAoverhang produced in the XhoI digest on pUC4Kwill be compatible with the 59 single-strandedDNA overhang produced in the SalI digest onpUC19. Although DNA ligase will be able to jointhese two fragments together, the resulting se-quence will no longer be recognized by XhoI orSalI. The 59 single-stranded DNA overhangs pro-duced in the EcoRI digests on pUC4K and pUC19will also be compatible, and will be joined by DNAligase to regenerate the EcoRI site. This type of di-rectional, or forced, cloning ensures that the kanR

gene will ligate into pUC19 in a specific orienta-tion (see Fig. 21-2). If both the kanR gene andpUC19 were digested with EcoRI only, this wouldnot be the case (see Fig. 21-3).

Also note that the XhoI digest on pUC4K cleavesoff the promoter and start codon (ATG) for the kanR

gene, rendering it incapable of being transcribed ortranslated by the host cell (see Fig. 21-1). However,the XhoI/SalI ligation of the kanR gene into pUC19will put it into the same reading frame as the LacZapeptide sequence. What will be produced is alacZ/kanR gene fusion protein that will confer kanR

to the host cell and that is inducible in the presenceof IPTG. This differs from the case in pUC4K,where the kanR gene is constitutively expressed un-der the control of a different promoter.

Since the ligation reaction is performed in thepresence of a number of different DNA fragments

347

EcoRI (396)XhoI (1503)

EcoRI BamHI SalI

EcoRI (1678)

pUC4K pUC19

CTTA

AG

GA

AT

TC

G

AG

CT

G

C

TC

GA

C

G

AGCTG

CTTAA

G

CT

CG

AC

GA

ATT

C

Digest with XhoI and EcoRI Digest with SalI, EcoRI, and BamHI

5'–TCGAG

3'–CG–3'CTTAA–5'

C–3'GAGCT–5'

5'–AATTC

3'–G

5'–AATTC

3'–C

G–3'CTTAA–5'

5'–AATTC

3'–G

5'–AATTC

3'–G

5'–GATCC

3'–G

G–3'CAGCT–5'

G–3'CAGCT–5'

G–3'CCTAG–5'

ampR

ampR

kanR

(XhoI)

(XhoI)

(EcoRI at 396)

(EcoRI at 396)

(EcoRI)

(EcoRI)(EcoRI at 1678)

(EcoRI at 1678)

(SalI)

(SalI)

(BamHI)

(BamHI)

1

1

1

1

T4 DNAligase 1ATP

ampR

ampR

lacZ'

lacIori

kanR

kanR

lacZ'kanR

The EcoRI end at 396 in pUC4K is compatible (complementary) with the EcoRI end produced by the same digestion on pUC19. Likewise, the single-stranded DNA overhang produced by XhoI in pUC4K is complementary to the single-stranded DNA overhang produced by SalI in pUC19. DNA ligase will therefore be able to join these two DNA strands.

kanR gene is now in the same reading frame as lacZ α peptide. Since the expression of the lacZ α peptide is inducible with IPTG the kanR gene is now IPTG–inducible as well (expressed as a fusion protein with the N-terminus of lacZ α peptide).

1107 bp ~2600 bp

175 bp ~25 bp

2632 bp

pUC19

XhoI/SalI hybrid site no longer

recognized by either enzyme

lac promoter

EcoRIsite

ampR

Figure 21-2 DNA fragments produced from pUC4K and pUC19 following restriction enzyme digestion.


kanR

AA

TT

C

AAT

TC

CTTA

A

CT

TA

A

ampR

pUC19 digestedwith EcoRI

G

G G

G

lacZ'

lacZ

'

kanRG

AA

TT

C

GA

AT

T

C

CT

TA

AG

CT

TA

AG

ampR

kanR

GA

AT

TC

GA

AT

TC

CT

TA

AG

CT

TA

AG

ampR

lacZ

'

lacZ'

lacZ

'

lacZ'

kanR gene lies in an orientation opposite that of lacZ'

kanR gene lies in the same orientationas lacZ'

lac

prom

oter

lac

prom

oter

Figure 21-3 Two recombinant pUC19/4K plasmids are possible if both plasmids are digested with a sin-

gle enzyme. The two resulting plasmids differ in the orientation of the kanR gene in pUC19.

The arrow (→) indicates the direction of transcription of the kanR gene.

with compatible single-stranded DNA overhangs,the desired pUC19/4K recombinant plasmid is byno means the only plasmid that could be producedduring the ligation reaction. For instance, it is pos-sible that the EcoRI/XhoI fragment carrying the kanR

gene from pUC4K will ligate back into pUC4K. Itis also possible that the EcoRI/SalI fragment will lig-

ate back into pUC19. The possibility of reformingeither of the two parent plasmids is minimized, how-ever, because both events would require a success-ful three-point ligation (see Fig. 21-4). The two-point ligation that is required to produce the desiredpUC19/4K plasmid is statistically much more likelyto occur.

EXPERIMENT 2I Constructing and Characterizing a Recombinant DNA Plasmid 349

G

TC

GA

C

GG AT C C

G

AATT

C

CA

GC

T

G

C C TA G G

CT

TA

A

G

TC

GA

G

C

G

AAT

TC

G

AAT

TCC

GA

GC

T

C

TTA

AG

CT

TA

A

G

ampR

kanR

kanR

pUC4K

ampR

pUC19

C

TC

GA

C

G

AATT

C

GA

GC

TG

CT

TA

A

G

ampR

pUC19/4K

Figure 21-4 On a statistical basis, the two-point ligation required to form the recombinant pUC19/4K

plasmid is much more likely than the three-point ligations required to form pUC19 or

pUC4K.

How do you distinguish between cells carryingthe pUC19 plasmid and the pUC19/4K recombi-nant plasmid? Recall that pUC19 carries the genefor ampR, while pUC19/4K carries the genes forampR and kanR. Any colony that will grow in thepresence of ampicillin but not kanamycin mostlikely harbors pUC19. In addition to the antibi-otic selection just described, pUC19 will also con-fer another unique phenotype to the host cell thatwill distinguish it from those carrying pUC19/4K.

The host cell used in this experiment is Epicuriancoli XL1-Blue. This strain contains a deletion onthe chromosome of the first 450 base pairs of itslacZ coding sequence (the LacZa peptide). ThepUC19 plasmid will be able to restore b-galac-tosidase activity in the host strain, since this plas-mid contains an uninterrupted LacZa peptide cod-ing sequence (the strain will be complemented forb-galactosidase activity). The LacZa peptide ex-pressed from the pUC19 plasmid will be able to


Table 21-1 Basis of Selection for Clones

Carrying the Desired pUC19/4K

Recombinant Plasmid

KanR Color on

IPTG- IPTG/Xgal

Plasmid KanR? AmpR? Inducible? Plate

pUC19 No Yes — Blue

pUC4K Yes Yes No White

pUC19/4K Yes Yes Yes White

noncovalently interact with the C-terminal frag-ment of LacZ produced by the host cell, creatinga functional b-galactosidase enzyme. This tech-nique of selection is therefore referred to as a-com-plementation. If cells containing pUC19 are placedon an agar plate containing IPTG and 5-bromo-4-chloro-3-indolyl-b,D-galactopyranoside (Xgal),LacZa peptide expression will be induced, and theXgal substrate will be converted to a blue-coloredprecipitate. Blue colonies on an IPTG/Xgal platewill therefore be indicative of the presence of anintact LacZa peptide sequence in the plasmid(pUC19). Since pUC19/4K will have the kanR

gene inserted into the LacZa peptide sequence, itwill not be able to complement the host strain forb-galactosidase activity, and it will produce whitecolonies on an IPTG/Xgal plate.

How do you distinguish between cells carryingthe pUC4K plasmid and the pUC19/4K recombi-nant plasmid? This is not as easy as the situationdescribed above, since both plasmids contain an in-sertion in the LacZa peptide sequence. Both plas-mids will confer kanR and ampR to the host strainand both will produce white colonies in the pres-ence of IPTG and Xgal. Still, there is one differ-ence between these two plasmids that you will beable to exploit during the selection process. Recallthat the kanR gene is constitutively expressed fromthe pUC4K plasmid, while the kanR gene is underthe control of the lac promoter in the pUC19/4Krecombinant plasmid. Therefore, any white colonythat grows well on a plate containing onlykanamycin (no IPTG) most likely carries pUC4K.In contrast, any colony that grows well on akanamycin plate containing IPTG, but poorly on aplate containing only kanamycin, most likely car-ries pUC19/4K. The basis for the different types

of selection that will be used in this experiment areoutlined in Table 21-1.


pUC19 (0.25 mg/ml)—Pharmacia catalog #27–4951-01

pUC4K (0.5 mg/ml)—Pharmacia catalog #27–4958-01

Distilled water (sterile)

PstI (10 units/ml) with 10X buffer—Gibco BRL catalog#15215-023

EcoRI (10 units/ml) with 10X buffer—Gibco BRL cata-log #15202-021

XhoI (10 units/ml) with 10X buffer—Gibco BRL catalog#15231-020

SalI (10 units/ml) with 10X buffer—Gibco BRL catalog#15217-029

BamHI (10 units/ml) with 10X buffer—Gibco BRL cat-alog #15201-031

HindIII (10 units/ml) with 10X buffer—Gibco BRL cat-alog #15207-020

Water baths at 37°C, 15°C, and 42°C

Phenol (Tris-saturated)

Chloroform

Isoamyl alcohol

3 M sodium acetate (pH 5.2)

Absolute ethanol

Dry ice/ethanol bath (270°C)

Microcentrifuge


70% vol/vol ethanol in distilled water


P-20, P-200, and P-1000 Pipetmen with disposable tips(sterile)

T4 DNA ligase (1 unit/ml) with 5X buffer—Gibco BRLcatalog #15224-025

Epicurian coli XL1-Blue competent cells—Stratagene,catalog #200130

Yeast-tryptone (YT) broth (1% Bactotryptone, 0.5%yeast extract, 0.5% NaCl)

100 mM IPTG solution in sterile distilled water (filter-sterilized)

Ampicillin (100 mg/ml) prepared in sterile distilled wa-ter—Sigma catalog #A-9518

Kanamycin (100 mg/ml) prepared in sterile distilled wa-ter—Sigma catalog #K-4000

Agar

Culture shaker with test tube racks at 37°C

Petri plates (disposable, plastic)

Sterile glass rod (for spreading bacteria over the plate)

Sterile toothpicks

37°C incubator


Large (16 3 125 mm) glass test tubes (sterile)

1-kb DNA ladder size standard—Gibco BRL catalog#15615-016

Ethidium bromide solution in 0.5X TBE buffer (0.5mg/ml)—Toxic!

Polaroid camera with film and a 256-nm light box

GTE lysis buffer (25 mM Tris, pH 8.0, 50 mM glucose,10 mM EDTA, 10 mg/ml lysozyme)

Fresh solution of 0.2 N NaOH and 1% SDS in distilledwater

Potassium acetate solution (60 ml of 5 M potassium ac-etate, 11.5 ml of glacial acetic acid, 28.5 ml of dis-tilled water)

RNaseA (10 mg/ml in water)—Boil this solution for 10 minto inactivate any DNase

6X DNA loading buffer (0.25% wt/vol bromophenolblue, 0.25% wt/vol xylene cyanole, 30% wt/volglycerol in distilled water)

Agarose (electrophoresis grade)

5X TBE buffer (54 g/liter Trizma base, 27.5 g/liter boricacid, 20 ml/liter of 0.5 M EDTA, pH 8.0)

Agarose-gel electrophoresis apparatus with casting boxand 10-well comb

5-bromo-4-chloro-3-indolyl-b,D-galactopyranoside (Xgal),40 mg/ml solution in dimethylformamide—Toxic!(Sigma catalog #B-4252)

Protocol

Day 1: Restriction Endonuclease Digestion

of pUC19 and pUC4K, Ethanol

Precipitation of DNA Fragments, and DNA

Ligation

1. Set up the following reactions in two labeled1.5-ml sterile microcentrifuge tubes:

Digestion of pUC19 Digestion of pUC4K

14 ml of distilled 14 ml of distilled

water (sterile) water (sterile)

2 ml of 10X React 2 ml of 10X React

Buffer 2 Buffer 2

1 ml of pUC19 2 ml of pUC4K

(0.25 mg/ml) (0.50 mg/ml)

1 ml of EcoRI 1 ml of EcoRI

(10 units/ml) (10 units/ml)

1 ml of BamHI 1 ml of XhoI

(10 units/ml) (10 units/ml)

1 ml of SalI

(10 units/ml)

2. Incubate both reactions at 37°C for 1.5 hr.3. Transfer both reactions to a single tube and add

160 ml of sterile distilled water.4. Add 200 ml of Tris-saturated phenol;chloro-

form;isoamyl alcohol (25;24;1). Mix vigor-ously in a Vortex mixer for 1 min and centrifugefor 3 min to separate the aqueous and organicphases. This step is done to denature and re-move the restriction endonucleases from theDNA solution.

5. Remove the aqueous (upper) phase and place itin a clean microcentrifuge tube. Repeat step 4(extraction of aqueous phase) with 200 ml ofchloroform:isoamyl alcohol (24;1). This step isdone to remove all traces of phenol from thesolution, which may denature the T4 DNA lig-ase during the ligation reaction. Dispose of theorganic phase from this extraction in a wastebeaker designated by the instructor.

6. Transfer the upper (aqueous) phase to a freshmicrocentrifuge tube. Do not transfer any of theorganic phase during this process. Dispose of theorganic phase from this extraction in a wastebeaker designated by the instructor.

7. Add 10 ml of 3 M sodium acetate (pH 5.2) tothe aqueous phase, as well as 800 ml of absoluteethanol. Cap and invert the tube several timesto mix. Submerge the closed tube in a dryice/ethanol bath (270°C) for 10 min. This isdone to precipitate plasmid DNA.

8. Centrifuge the sample at 4°C in a microcen-trifuge for 30 min. Carefully remove the su-pernatant from the precipitated DNA pellet.Depending on how pure the DNA is, this pellet maybe translucent and difficult to see. It may help youto place the outside hinge of the microcentrifuge tubetoward the outside of the rotor during the centrifu-gation so that you will know where the DNA pel-let is at the bottom of the tube. You may also findthe use of Pellet Paint to be helpful during this pro-cedure. This product, sold by Novagen (catalog#69049-3), is a pink chromophore that will bindthe DNA and give the pellet a color that will beeasier to identify at this point.

9. Add 1 ml of ice cold 70% ethanol to the DNApellet, centrifuge at 4°C for 5 min, and care-fully remove the supernatant from the precip-itated DNA pellet. This step is done to removethe salt that was added to the DNA to aid inthe precipitation. This salt, if not removed, couldinterfere with the DNA ligation reaction.


10. Allow the DNA pellet to air dry (,10 min) ordry it down briefly (,2 min) in the vacuum mi-crocentrifuge.

11. Resuspend the DNA pellet in 20 ml of steriledistilled water. Store the DNA on ice untilready to proceed with the ligation reaction.

12. Carefully mix the following together in a ster-ile, 1.5-ml microcentrifuge tube:

15 ml of DNA sample (from step 11)

4 ml of T4 DNA ligase buffer (250 mM Tris, pH 7.6,50 mM MgCl2, 5 mM ATP, 5 mM dithiothre-itol, 25% wt/vol polyethylene glycol-8000)

1 ml of T4 DNA ligase (1 unit/ml)

13. Incubate the reaction at room temperature for1 hr and then at 15°C overnight. Store the re-action at 4°C for use on Day 2. Also, store theremainder of your DNA sample from step 11(5 ml) at 4°C for use on Day 2.

Day 2: Transformation of E. coli Host Cells

1. Add 30 ml of sterile distilled water to your 20-ml ligation reaction. Add 2.5 ml of 3 M sodiumacetate (pH 5.2) and 200 ml of absolute ethanol.Cap the tube and invert several times to mix.Submerge the capped tube in a dry ice/ethanolbath (270°C) for 10 min.

2. Centrifuge the sample at 4°C for 30 min. Re-move the supernatant from the precipitatedDNA pellet and add 1 ml of ice cold 70%ethanol. Centrifuge the sample again at 4°C for5 min. Remove the supernatant from the pre-cipitated DNA pellet.

3. Air dry the DNA pellet as before or dry brieflyin the vacuum microcentrifuge.

4. Resuspend the dried DNA pellet in 5 ml of ster-ile distilled water.

5. Obtain a microcentrifuge tube from the in-structor containing 160 ml of freshly thawedEpicurian coli XL1-Blue competent cells (withb-mercaptoethanol added as per the manufac-turer’s instructions). Keep these cells on ice at alltimes!

6. Add 40 ml of these cells to four prechilled (onice) round-bottomed, 15-ml polypropyleneculture tubes. Again, keep the cells on ice at alltimes and label the tubes 1 to 4.

7. To tube 1, add 5 ml of the ligation mixtureprepared in step 4. To tube 2, add 5 ml of a

1-ng/ml solution of untreated pUC19. Totube 3, add 5 ml of the 1X ligation buffer (noDNA). To tube 4, add 5 ml of unligated DNAsample left over from Day 1 (see Table 21-2and step 13 of the Day 1 protocol).

8. Incubate the cells with these samples on ice for15 min, swirling gently every 2 min. Quicklyremove the tubes from the ice and place themin a 42°C water bath. Incubate the tubes at42°C for exactly 45 sec. Quickly remove thetubes from the water bath and incubate themagain on ice for 2 min. On heat shock, the com-petent E. coli cells will take up the intact plas-mid DNA produced during the ligation reac-tion. The competent cells are very compromised atthis point, and they will die if the incubation is notperformed at exactly 42°C for exactly 45 sec. Theincubation on ice will prevent the cells from dyingafter the heat shock.

9. Add 1 ml of YT broth (use sterile technique)and 30 ml of sterile 100 mM IPTG solution toeach tube. Incubate the cells at 37°C with gen-tle shaking for 1 hr. During this time, the cellswill recover from the heat shock and begin toexpress genes (such as kanR in the pUC19/4Kplasmid) under the control of the lac promoter.

10. Obtain five agar plates containing IPTG (40mg/ml), Xgal (40 mg/ml), and ampicillin (100mg/ml). Obtain two agar plates containingIPTG (40 mg/ml), Xgal (40 mg/ml), andkanamycin (50 mg/ml).

11. Remove 100- and 200-ml aliquots from cul-ture tube 1 and place on two separateIPTG/Xgal/amp plates (see Table 21-3). Platethe same volumes of cells from culture tube1 on the two separate IPTG/Xgal/kan plates.

Table 21-2 Procedure for E. coli

Transformation

Volume of

E. coli

XL1-Blue

Tube DNA Sample Volume (ml) cells (ml)

1 Ligation mixture 5 40

2 pUC19 (untreated) 5 40

3 Ligation buffer (no DNA) 5 40

4 Unligated DNA 5 40


Table 21-3 Procedure for Plating and

Selection of Transformed Bacteria

Transformation Volume to Be Components in

Tube Plated (ml) Agar Plate

1 100 IPTG/Xgal/amp

1 100 IPTG/Xgal/kan

1 200 IPTG/Xgal/amp

1 200 IPTG/Xgal/kan

2 200 IPTG/Xgal/amp

3 200 IPTG/Xgal/amp

4 200 IPTG/Xgal/amp

12. Remove 200 ml from culture tube 2 and place onan IPTG/Xgal/amp plate. Remove 200 ml fromculture tube 3 and place on an IPTG/Xgal/ampplate. Remove 200 ml from culture tube 4 andplace on an IPTG/Xgal/amp plate. Be sure thatyou label each plate so that you know what antibioticthe plate contains and what transformation mixturewas placed on each plate.

13. Using sterile technique (to be demonstrated bythe instructor), use a sterile bent glass rod tospread the bacteria evenly over the surface ofeach plate. Allow the surface of the plates todry for 5 min, invert the plates (agar side up),and place them in the 37°C incubatorovernight. The colonies that appear on theseplates will be grown in culture the next day, andplasmid DNA will be isolated from them onDay 3. The plates must be incubated at 37°Cfor at least 24 hr. After this incubation, theplates may be stored at 4°C for several days. Wehave found that the blue color will develop muchbetter in colonies that contain a plasmid with an in-tact LacZa peptide sequence if this is done. Incu-bating the plates at 4°C will make it much easierto distinguish between blue and white colonies whenthe plates are counted (see below).

Off day: Growth of Transformed Cells

1. Remove your seven agar plates from the 37°Cincubator. Obtain four large glass culture tubesfrom the instructor that each contain 3 ml ofsterile YT broth supplemented with 100 mg/mlampicillin. Label the tubes 1 to 4.

2. Analyze the two IPTG/Xgal/kan plates con-taining the transformation mixture from tube 1 tofind four colonies that are white in color. Obtaintwo fresh agar plates: one containingkanamycin and one containing kanamycin andIPTG. On the bottom surface of each plate,use a marker to divide the plate into four quad-rants, labeled 1 to 4.

3. Using a sterile toothpick, stab into the middleof a white colony that is present on thisIPTG/Xgal/kan plate. Remove the toothpickand stab the end that came into contact withthe bacteria first on quadrant 1 of the kan plateand then into quadrant 1 of the IPTG/kan plate.Finally, drop the toothpick (bacteria end down)into tube 1 containing ampicillin supplementedYT broth. Repeat step 3 for each of the otherquadrants on the two plates, picking a newwhite colony on the IPTG/Xgal/kan plate eachtime.

4. Incubate the cultures at 37°C with shakingovernight. These strains will be used on Day 3to isolate plasmid DNA. Place the two replicaplates produced in step 3, agar side up, in the37°C incubator overnight.

5. Count the number of colonies from transfor-mation culture tube 1 present on the twoIPTG/Xgal/kan plates. How many of thesecolonies are blue and how many are white? Ex-plain these results in terms of what you knowabout the properties of each of the plasmidsthat these bacterial colonies may contain.

6. Count the number of colonies from transfor-mation culture tube 1 present on the twoIPTG/Xgal/amp plates. How many of thesecolonies are blue and how many are white? Ex-plain these results in terms of what you knowabout the properties of each of the plasmidsthat these bacterial colonies may contain.

7. Count the number of colonies present on theIPTG/Xgal/amp plate containing bacteriafrom transformation culture tube 2. Why wasthis control experiment performed? Whatcolor are these colonies? Explain. Describewhat a low number of colonies on this platewould indicate, as well as possible causes forthis result.

8. Count the number of colonies present on theIPTG/Xgal/amp plate containing bacteriafrom transformation culture tube 3. Why was


this control experiment performed? Describewhat a large number of colonies on this platewould indicate, as well as possible causes forthis result.

9. Count the number of colonies present on theIPTG/Xgal/amp plate containing bacteriafrom transformation culture tube 4. Why wasthis control experiment performed? Whatcolor are these colonies? Explain. Describewhat a large number of blue or white colonieswould indicate, as well as possible causes forthis result.

Day 3: Isolation of Plasmid DNA

1. Remove the four culture tubes from the 37°Cshaker and add 1.5 ml of each culture to fourseparate microcentrifuge tubes labeled 1 to 4.Cap each tube and centrifuge for 2 min at roomtemperature to harvest the cells. Remove thesupernatant from the bacterial pellet.

2. Add the remaining 1.5 ml of each culture to theappropriate microcentrifuge tubes and repeatstep 1.

3. Add 100 ml of GTE (lysis) buffer to each bac-terial cell pellet. Resuspend each cell pelletthoroughly by repeated pipetting or mixingwith a Vortex mixer. Incubate the tubes at roomtemperature for 5 min.

4. Add 200 ml of freshly prepared 0.2 N NaOH–1% SDS solution to each tube. Cap and invertseveral times to mix. The solution will turnfrom turbid to more translucent as the cells arecompletely lysed and the DNA and proteins aredenatured. Incubate the tubes on ice for 5 min.

5. Add 150 ml of 3 M potassium acetate solutionto each tube. Cap and invert several times tomix. The precipitate that forms contains muchof the denatured proteins produced in step 4,as well as chromosomal DNA that was not ableto renature correctly when the pH was quicklylowered. Incubate the tubes on ice for 10 min.

6. Centrifuge the tubes for 15 min at 4°C. Re-move the supernatant from each tube and placein four fresh microcentrifuge tubes labeled 1 to4. The supernatant contains the plasmid DNA,while the pellet contains denatured proteins,chromosomal DNA, membrane components,and other cell debris. The tubes containing thispellet may be discarded.

7. Add 500 ml of Tris-saturated phenol:chloro-form:isoamyl alcohol (25;24;1) to each tubecontaining the supernatant fraction. Mix eachtube with a Vortex mixer for 1 min, centrifugefor 5 min at room temperature, and transfer theaqueous (upper) phases to four fresh micro-centrifuge tubes labeled 1 to 4. This step isdone to remove any remaining proteins andlipid components from the plasmid DNA. Dis-pose of the organic phase from this extractionin a waste beaker designated by the instructor.

8. Repeat the aqueous phase extraction proceduredescribed in step 7 with 500 ml of chloro-form;isoamyl alcohol (24;1). Transfer the aque-ous (upper) phases to four fresh microcentrifugetubes. This step is done to remove all traces ofphenol from the solution. Dispose of the or-ganic phase from this extraction in a wastebeaker designated by the instructor.

9. Add 1 ml of absolute ethanol to each of the fouraqueous fractions. Cap the tubes and invertseveral times to mix. Incubate the tubes on icefor 10 min.

10. Centrifuge the tubes at 4°C for 15 min. Re-move the supernatant from the precipitatedDNA pellet (which may also contain significantamounts of RNA, making the pellet appearwhite in color). Add 1 ml of ice cold 70%ethanol to each DNA pellet, centrifuge for 5min at 4°C, and remove the supernatant fromthe DNA pellet.

11. Allow the pellets to air dry for 10 min or drybriefly in the vacuum microcentrifuge. Storethe four, dried DNA pellets at 220°C for useon Day 4.

Day 4: Restriction Endonuclease Digestion

and Agarose-Gel Electrophoresis of

Plasmid DNA

1. In this experiment, you will perform restrictiondigests on two of the four plasmid DNA sam-ples that you have isolated. If it is possible, se-lect the plasmid DNA isolated from two bac-terial colonies that were able to grow well onthe IPTG/kan plates, but not as well on thekanamycin plates without IPTG. This is de-termined by analyzing the growth of the fourcolonies (clones) on the kan plate and thekan/IPTG plate produced in step 3 from the


previous day. Remember that our method of se-lecting between cells carrying the pUC4K andpUC19/4K plasmids is that the kanR gene onpUC19/4K is IPTG-inducible, while the kanR geneis expressed constitutively in cells carrying thepUC4K plasmid. The lac promoter is known tobe somewhat “leaky,” showing some expressionof genes under its control (such as the kanR

gene) even in the absence of IPTG. Therefore,you may find that cells carrying pUC19/4Kmay show some growth on the kan plate with-out IPTG. Ideally, you will want to analyzecolonies (clones) that grew well only on the kanplate with IPTG. If you only have colonies thatgrew equally well on both the kan plate and theIPTG/kan plate, choose any two of the fourplasmids for analysis.

2. Resuspend the two plasmid DNA sample pel-lets in 32 ml of sterile distilled water.

3. Set up the following reactions for each of the twoplasmid samples in 1.5 ml microcentrifuge tubes:

(see step 2). At this point you should have eightsamples for agarose-gel analysis.

5. Prepare a 1% TBE agarose gel by adding 0.5 gof electrophoresis-grade agarose to a 250-mlErlenmeyer flask containing 50 ml of 0.5XTBE buffer (dilute the 5X TBE buffer stock1:10 with distilled water to prepare the 0.5Xworking solution). Preweigh the flask before heat-ing and record this value.

6. Microwave the flask for 2 to 3 min, or untilthe agarose has been fully dissolved (no smallpieces of undissolved agarose remain). Placethe flask on the balance and add distilled wa-ter to the flask until its mass is the same as thatbefore the flask was heated. Water will evapo-rate from the flask as it is heated. This watermust be replaced to ensure that a 1% agarosesolution is maintained. Depending on theamount of water that has evaporated, the 1%agarose gel prepared in step 5 may now begreater than 1%.

PstI Digests HindIII Digests SalI Digests

8 ml of plasmid DNA 8 ml of plasmid DNA 8 ml of plasmid DNA

1.5 ml of 10X React 2 1.5 ml of 10X React 2 1.5 ml of 10X React 10

1 ml of PstI (10 units/ml) 1 ml of HindIII (10 units/ml) 1 ml of SalI (10 units/ml)

1 ml of RNaseA 1 ml of RNAseA 1 ml of RNAseA

3.5 ml of water 3.5 ml of water 3.5 ml of water

The RNAseA is at a concentration of10 mg/ml, and is added to completely digestall of the RNA remaining in the sample. If thisis not done, a large RNA “spot” will be pres-ent on the gel following ethidium bromidestaining, which will prevent you from seeinglow molecular weight DNA fragments on thegel.

4. After a 1-hr incubation at 37°C, add 3 ml of 6XDNA sample buffer to each of the reactiontubes, as well as to the 8 ml of undigested plas-mid remaining from each of the two samples

7. Allow the solution to cool for 5 min, swirlingthe flask gently every so often to prevent thegel from hardening. When the outside of theflask is cool enough to touch, pour the solutioninto an agarose-gel cast fitted with a 10-wellcomb at one end (consult the instructor). Al-low the gel to set (,40 min). Remove the comb,and place the gel in an agarose-gel elec-trophoresis chamber. Add 0.5X TBE buffer tothe chamber until it completely covers the gel andfills the wells.

8. Load the samples prepared in step 4 as follows:

Lane 1 Lane 2 Lane 3 Lane 4 Lane 5 Lane 6 Lane 7 Lane 8 Lane 9

Restriction enzyme * PstI HindIII SalI None * PstI HindIII SalI

Plasmid 1 1 1 1 † 2 2 2 2

*This lane contains your undigested plasmid samples.

†This lane should be loaded with a solution containing 1 mg of 1-kb DNA ladder size standard, water, and DNA sample buffer (prepared

by the instructor).


12,21611,13510,150 9,152 8,144 7,126 6,108 5,020 4,072 3,054

2,036 1,636

1,015

506

396 344 295

Often don’t resolve or separate well on a 1% TBE gel

Size standards easiest to identify on a 1% TBE agarose gel and of the greatest use in producing a standard curve

Figure 21-5 Number of base pairs in each band

of the 1-kb DNA ladder.

9. Attach the negative electrode (cathode) to thewell side of the chamber and the positive elec-trode (anode) to the other side of the chamber.Remember that the DNA is negatively charged,and will migrate through the gel toward thepositive electrode.

10. Perform the electrophoresis at 50 mA, constantcurrent. You will see two dye fronts develop:one from the bromophenol blue dye (dark blue)and one from the xylene cyanole dye (lightblue). The latter of the two dyes will migratethe same as a DNA fragment of approximately4 kb, while the former dye will migrate thesame as a DNA fragment of about 0.5 kb.

11. Continue the electrophoresis until the fastermoving of the two dye fronts (bromophenolblue) migrates about three-fourths of the waythrough the gel.

12. Turn off the power supply, disassemble theelectrophoresis apparatus, carefully remove theagarose gel, and submerge it in a small tray con-taining a 0.5-mg/ml solution of ethidium bro-mide in 0.5X TBE (enough to completely coverthe gel). Wear gloves at all times when workingwith ethidium bromide! Incubate at room tem-perature for about 20 min. The ethidium bro-mide will enter the gel and intercalate betweenthe base pairs in the DNA strands. When ex-posed to ultraviolet light, the ethidium bromidewill fluoresce, and the DNA will appear as or-ange or pink bands on the gel.

13. Destain the gel for 10 min in a solution of 0.5XTBE buffer without ethidium bromide. This isdone to remove ethidium bromide from all por-tions of the gel that do not contain DNA.

14. Place the gel in a dark box fitted with a 254 nmlight source. Close the doors of the box andturn on the light source to visualize the DNAbands on the gel. Take a photograph of the gelfor later analysis.

16. Prepare a plot of the number of base pairs ver-sus the distance traveled (in centimeters) foreach DNA fragment present in the 1-kb DNAladder lane. If the 1-kb DNA size standards re-solved well, you should be able to differentiatebetween the relative mobilities of the 0.5-, 1.0-, 1.6-, 2.0-, and 3.0-kb DNA fragments (seeFig. 21-5) for use in preparing the standardcurve. Do not attempt to determine the relative mo-bilities of the larger-molecular-weight DNA size

standards on the gel if they did not resolve or sepa-rate well. This will only add error to the standardcurve over the region where the DNA size stan-dards did display good resolution.

17. Have you successfully constructed the desiredpUC19/4K recombinant plasmid? Explain youranswer in terms of what size DNA fragmentsresulted from each digest and your knowledgeof the composition of the parent plasmids usedin this experiment (pUC19 and pUC4K). Pre-pare a restriction map of the newly constructedplasmid. The map should include the totalnumber of base pairs in the plasmid, the iden-tity and position of all restriction sites in theplasmid, and the position of all of the differentgenes present in the plasmid.

18. If you do not think that you have isolated thedesired pUC19/4K plasmid, prepare a map ofthe plasmid that you think you have isolated.Support your drawing with an explanation ofhow the digestion of this plasmid with the var-ious restriction enzymes would produce the re-sults that you obtained.


Exercises

1. Why is it important to perform a transforma-tion control experiment using undigested plas-mid? Describe two possible results that couldbe obtained in this control experiment, alongwith an explanation of what each result wouldindicate.

2. Why is it important to perform a transforma-tion control experiment using no plasmidDNA? Describe two possible results that couldbe obtained in this control experiment, alongwith an explanation of what each result wouldindicate.

3. Why is it important to perform a transforma-tion control experiment using unligated plas-mid? Describe all the possible results that couldbe obtained in this control experiment, alongwith an explanation of what each result wouldindicate.

4. The pUC19 and pUC4K plasmids both conferantibiotic resistance to the E. coli host strainsthat they are transformed into. Other than drugresistance, can you think of any other gene(s)that could be carried on a plasmid that wouldallow you to select for cells that obtained it ontransformation? For each of the genetic ele-ments that you propose, describe the relevantfeatures of the genotype of the host strain thatyou would use in the experiment, as well as howyou would select for cells that obtained theplasmid following transformation.

5. Can you think of any advantages of having thelacI gene carried on the pUC19 plasmid?

6. You have obtained a plasmid that is able to bepropagated in Escherichia coli but not in Bacillussubtilis. Why is this plasmid able to be propa-gated in one bacterium but not another? Howwould you propose to alter the plasmid so thatit is able to be propagated both in E. coli andB. subtilis?

7. Where were plasmids originally isolated from?What function(s) do naturally occurring plas-mids serve?

REFERENCES

Berger, S. L., and Kimmel, A. R. (1987). Guide toMolecular Cloning Techniques. Methods EnzymolVol. 152.

Garrett, R. H., and Grisham, C. M. (1995). Recombi-nant DNA: Cloning and Creation of ChimericGenes. In: Biochemistry. Orlando, FL: Saunders.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Recombinant DNA Technology. In: Princi-ples of Biochemistry, 2nd ed. New York: Worth.

Sambrook, J., Fritsch, E. F., and Maniatis, T. (1989).Molecular Cloning, 2nd ed. Plainview, NY: ColdSpring Harbor Laboratory Press.

Wilson, K., and Walker, J. M. (1995). Molecular Biol-ogy Techniques. In: Principles and Techniques of Prac-tical Biochemistry, 4th ed. Hatfield, UK: CambridgeUniversity Press.

) g

359

EXPERIMENT 22

In Vitro Transcription from a Plasmid Carrying a T7 RNA Polymerase–Specific Promoter

Theory

Transcription is one of two processes that allow acell to synthesize the proteins encoded in its DNA.In the first process, the genes contained on thechromosome are transcribed into molecules of RNA.In the second process, these RNA messages aretranslated by the ribosomes and transfer RNAs (tRNAs) to produce proteins with specific aminoacid sequences. RNA polymerase is the enzyme thatsynthesizes RNA. This enzyme will read a single-stranded DNA template in the 3� to 5� directionand construct an RNA molecule in the 5� to 3� di-rection that is complementary to it (Fig. 22-1). Insome respects, RNA synthesis is similar to DNAsynthesis: both molecules are synthesized in the 5�to 3� direction, both of the enzymes involved re-quire a template DNA strand to direct the synthe-sis, and both of the enzymes involved use nucleo-side triphosphates as the basic monomeric units inthe synthesis reaction.

Despite these similarities, there are some veryimportant differences between the reactions car-ried out by DNA polymerase and RNA poly-merase. Remember that DNA is synthesized withthe use of four deoxyribonucleoside triphosphates(dATP, dGTP, dTTP, and dCTP). In the double-stranded DNA helix, adenine bases hydrogen-bondwith thymine bases, and guanine bases hydrogen-bond with cytosine bases. RNA, however, is syn-thesized with the use of four ribonucleosidetriphosphates (ATP, GTP, CTP, and UTP). Ade-nine and guanine bases in the DNA template willdirect the addition of uracil and cytosine bases tothe growing RNA molecule, respectively, while

thymine and cytosine bases in the DNA templatewill direct the addition of adenine and guaninebases, respectively, to the growing RNA molecule.The net result of this is that RNA polymerase willproduce an RNA molecule identical to that of thenontemplate (coding) DNA strand, with uracil inplace of thymine bases.

Another difference between DNA polymeraseand RNA polymerase is that the latter enzyme doesnot require the presence of a single-stranded nu-cleic acid primer to initiate the polymerization re-action. Given the correct sequence of DNA (a pro-moter, see below), RNA polymerase will bind theDNA template and begin transcription. This is incontrast to DNA polymerase (see Experiment 24),which requires a short nucleic acid primer to initi-ate synthesis of DNA. Unlike DNA polymerase,RNA polymerase does not have a 3� to 5� exonu-clease or “proofreading” activity. As a result, an er-ror (a single base change) will be introduced intoan RNA molecule for every 104 to 106 nucleotidesadded to the growing macromolecule during syn-thesis. This error rate is acceptable, since manymolecules of RNA are produced from a single DNAtemplate, and because the population of a particu-lar RNA molecule is constantly being regeneratedor turned over with time (the half-life of mRNA isrelatively short). The 3� to 5� proofreading activityof DNA polymerase decreases the error rate inDNA synthesis to one in 109 to 1010 bases. Becausethe DNA in a cell provides the permanent geneticinformation, you can understand why the accuracyof DNA synthesis during replication is more criti-cal than that of RNA synthesis in the life cycle ofa cell.

360

sequence and spacing of these two elements are crit-ical for allowing the bacterial RNA polymerase tobind the DNA template and initiate transcription.Promoters for eukaryotic RNA polymerases arequite variable. Still, there are some recurring ele-ments found about 25, 40, and 100 base pairs to the5� side of the transcriptional start site. These se-quences, as well as others that are often thousandsof base pairs from the transcriptional start site, arebelieved to be the binding sites for transcription fac-tors (proteins) that regulate the activity of the eu-karyotic RNA polymerases. A list of different RNApolymerases, and the promoter sequences that theyrecognize, are listed in Table 22-1.

Where does transcription of a DNA templateend? The process of transcription termination isbest understood in bacteria. Rho-independent ter-minators are characterized by a self-complementary

Where does RNA polymerase begin the processof transcription on the DNA template? The DNAelement on which RNA polymerase binds and ini-tiates transcription is termed the promoter. The sim-plest promoters are those recognized by viral RNApolymerases. Usually, these are a contiguous se-quence of 15 to 30 base pairs that direct the viralRNA polymerase to the transcriptional start site(Fig. 22-2). The viral promoter sequences possess a5� to 3� polarity, allowing you to determine whichof the two DNA strands will act as the template fortranscription. Bacterial promoters consist of two dif-ferent but conserved DNA sequences. One of theseelements is located about 10 base pairs on the 5� sideof the transcriptional start site, while the other ele-ment is located about 35 base pairs to the 5� side ofthe transcriptional start site. Although these two el-ements consist of as little as six base pairs each, the


Coding strand 5' Promoter DNA

Movement of RNApolymerase

RNA polymerase

Template strand 3' GACTCAGGGATCCGCATTCATG

CTGAGTCCCTAGGCGTAAGTAC

ATP

GTP

UTP

CTP

3'5'

5' 3'

GACTCAGGGATCCGCATTCATG

CTGAGTCCCTAGGCGTAAGTAC

GACUCAGGGAUCCGCAUUCAUG5' –O P

O–Transcript

O

O P O P + 21 PPiβ γ

O

3'5'

RNA polymerase

O–

O

O–

Oγ β α

Transcription start site

Figure 22-1 Schematic diagram of the process of transcription. RNA polymerase will read the template

DNA strand 3� to 5� and produce a transcript in the 5� to 3� direction that is identical to the

sequence of the coding DNA strand, with uracil (U) in place of thymine (T). A radiolabeled

transcript can be produced by including an �-[32P] nucleotide, which will become part of the

product, or by including a �-[32P] nucleoside triphosphate that is known to be the first

nucleotide in the transcript.

a particular sequence of bases in the DNA, then thatstretch of bases will be protected from chemicalsand nucleases that cleave DNA. A single strand ofDNA in a double helix is first labeled with a ra-dioactive group at either the 3� or 5� end. The DNAduplex is then incubated with the RNA polymeraseand treated with a chemical or DNase enzyme thatproduces a number of different-sized DNA frag-ments. The DNA duplex is then denatured, and theDNA fragments are resolved by polyacrylamide-gelelectrophoresis. Any region of the DNA duplex pro-tected by the polymerase during the DNase treat-ment will be absent on the autoradiogram of the gel,compared to the same radiolabeled DNA duplex notprotected by the polymerase. The “footprint” lefton the DNA by the polymerase provides insight intowhat sequences and what regions of the DNA arebound by the RNA polymerase (Fig. 22-3).

DNA sequence located roughly 20 bases on the 5�side of the termination site. The RNA produced ontranscription of this region is able to form a hair-pin structure that causes the RNA polymerase to“stall” and eventually dissociate from the DNAtemplate. A stretch of uridyl nucleotides at the 3�end of the RNA hairpin is part of the rho-inde-pendent terminator sequence that is also believedto help destabilize the enzyme–DNA complex. Arho-dependent terminator also contains a self-com-plementary sequence capable of forming a hairpinstructure. In an unknown mechanism, the rho pro-tein hydrolyzes ATP and destabilizes the RNApolymerase–DNA complex as it stalls at the hairpin, eventually leading to the terminationof transcription.

How were the promoter sites for the variousRNA polymerases determined? If the enzyme binds

EXPERIMENT 22 In Vitro Transcription from a Plasmid Carrying a T7 RNA Polymerase–Specific Promoter 361

Table 22-1 RNA Polymerases and the Promoters That They Recognize

RNA Polymerase Molecular Weight Subunits Promoter

SP6 (from SP6 phage) �100 kDa 1 5�-ATTTAGGTGACACTATAGAACTC-3�

T7 (from T7 phage) �100 kDa 1 5�-TAATACGACTCACTATAGGGAGA-3�

Bacterial (E. coli )* �324 kDa 5 5�-TTGACA-3� (�35 region)

5�-TATAAT-3� (�10 region)

Eukaryotic† 500–700 kDa �12 5�-GGCCAATCT-3� (�110 region)

5�-GGGCGG-3� (�40 region)

5�-TATAAAA-3� (�25 region)

*Promoter sequence recognition by bacterial RNA polymerase is governed by its sigma subunits. The sequence shown is recognized by

RNA polymerase with the major subunit found in rapidly growing, well-nourished cells. Other sequences are recognized under other physio-

logical conditions that lead to formation or activation of alternate sigma subunits.

†The promoter sequences for eukaryotic RNA polymerases I, II, and III are numerous and, in some cases, not well defined. The promoter

sequence shown is the consensus sequence recognized by RNA polymerase II.

Top strand 5'

SP6 promoterSP6 start site

T7 start site

Bottom strand 3' ATTTAGGTGACACTATA

TAAATCCACTGTGATAT

TATAGTGAGTCGTATTA

ATATCACTCAGCATAAT

3'5'

SP6 polymerase will read the bottom (template) strand 3' 5' and produce a transcript in the 5' 3' direction that is identical to the top (coding) strand, with uracil (U) in place of thymine (T).

T7 polymerase will read the top (template) strand 3' 5' and produce a transcript in the 5' 3' direction that is identical to the bottom (coding) strand, with uracil (U) in place of thymine (T).

T7 promoter

Figure 22-2 Promoters specify where RNA polymerase will bind the DNA and initiate transcription. The

polarity of the promoter sequence will specify the coding and template strands of the DNA.


Separate both samples, along with a radiolabeled nucleic acid size standard, on a high-percentage polyacrylamide gel and expose to film. The radioactive bands present on the gel will indicate the region on the DNA that was protected by the polymerase.

Region of DNA (promoter) protected from digestion by

bound RNA polymerase (RNA polymerase leaves

its “footprint” on the DNA)

Sizestandards

5' 3' DNA fragment believedto contain a promoter3' 5'

5' 3'3' 5'

Incubate with T4 polynucleotide kinase and γ–[32P]ATP to label one of the DNA strands at the 5' end ( 5' )

Add DNase I or methidium–EDTA–Fe at a concentration that will allow each strand to be cleaved at a single, random site in the DNA

5' 3'3' 5'

5' 3'3' 5'

5'5'5'5'5'5'

5'5'5'5'5'5'

Incubate without RNApolymerase

Incubate with RNApolymerase

Figure 22-3 DNA “footprinting” to identify the promoter region bound by RNA polymerase.


pSP72(2462 bp)

ampR

AT TA G G T G A C A C TATA G A A C T C G A G C A G C T G A A G C T T G C AT G C C T G C A G G T C G A C T C TA G A

TA AT C C A C T G T G ATAT C T T G A G C T C G T C G A C T T C G A A C G TA C G G A C G T C C A G C T G A G AT C T

G G AT C C C C G G G TA C C G A G C T C G A AT T C AT C G AT G ATAT C A G AT C T G C C G G T C T

C C TA G G G G C C C AT G G C T C G A G C T TA A G TA G C TA C TATA G T C TA G A C G G C C A G A

C C C TATA G T G A G T C G TAT TA

G G G ATAT C A C T C A G C ATA AT

5 '3 '

EcoRI EcoRV Bgl IISacI ClaIKpnI

XhoI PvuII

BamHI

XbaIPstISphI Hind IIISal IAccI

SmaI

SP6 promoter

SP6 transcriptionstart site

T7 promoterT7 transcriptionstart site

3'5'

MCS

Figure 22-4 The pSP72 plasmid map (simplified).

In this experiment, you will carry out an in vitrotranscription reaction from a plasmid (pSP72, Fig.22-4) containing a T7 RNA polymerase specificpromoter. A 32P-labeled nucleoside triphosphate(labeled at the � position) will be included in thereaction to produce a radioactive RNA molecule.The size and sequence of the various transcripts thatyou will produce can be predicted by the restric-tion enzymes that the plasmid will be digested withprior to the beginning of the transcription reaction.Since the plasmid template for the transcription re-action will be digested, you will be preparing “run-off ” transcripts from the T7 promoter. As a result,

the termination of the transcription reaction willnot rely on the presence of any transcriptional ter-minator sequence (rho-dependent or -independent)on the plasmid. The exact molecular weight or sizeof the various transcripts (the exact number of nu-cleotides that they contain) will be verified by poly-acrylamide-gel electrophoresis at the end of the ex-periment.

Although this may appear to be a simple exper-iment, the same techniques may be modified to an-alyze the effects of different mutations in the RNApolymerase and/or promoter sequence on theprocess of transcription. Suppose that you wanted


to determine which of the bases in the T7 promoterwere absolutely critical for transcription. You couldproduce a number of different point mutationswithin the T7 promoter sequence and use these astemplates in the same kind of run-off transcriptionexperiments. By comparing the results of these ex-periments with those that you will obtain in yourexperiment (using the consensus T7 promoter inthe template DNA), you could make quantitativedeterminations of how each point mutation in thepromoter affects transcription. In the same fashion,T7 RNA polymerase mutants could be testedagainst the wild-type enzyme to determine whichamino acid residues are important to carry out tran-scription from the consensus T7 promoter.

Caution: This experiment will require the use of

significant amounts of 32P-labeled nucleotide.

Wear safety goggles and gloves at all times.

Use Plexiglas shields when working directly with

the radioisotope. Cover your work surface with

absorbent paper to localize any spills that may

occur. Consult the laboratory instructor for the

proper disposal of the solid and liquid radioac-

tive waste that will be produced in this experi-

ment.


P-20, P-200, and P-1000 Pipetmen with sterile dispos-able tips

1.5-ml microcentrifuge tubes

Microcentrifuge

HindIII (10 units/�l) with the supplied 10X reactionbuffer—Gibco BRL catalog #15207-020

BamHI (10 units/�l) with the supplied 10X reactionbuffer—Gibco BRL catalog #15201-031

EcoRI (10 units/�l) with the supplied 10X reactionbuffer—Gibco BRL catalog #15202-021

pSP72 plasmid (�0.1 mg/ml in sterile distilled water)—Promega catalog #P-2191

37°C water bath

Dry ice/ethanol bath (�70°C)

Sterile distilled water

Phenol:chloroform (1:1, water saturated)

Vortex mixer

Chloroform:isoamyl alcohol (24:1)

3 M sodium acetate (pH 5.2)

Absolute ethanol


TE buffer (10 mM Tris, pH 7.5, 1 mM EDTA)

10X Transcription mix

0.4 M Tris, pH 8.0

0.15 M MgCl2Bovine serum albumin (1 mg/ml)

0.5 mM ATP

10 mM CTP

10 mM GTP

10 mM UTP

0.1 mM dithiothreitol

�-[32P]ATP (50,000 cpm/�l)—ICN Biomedicals cat-alog #32007x

T7 RNA polymerase (20 units/�l)—Gibco BRL catalog#18033-100

Apparatus for polyacrylamide gel electrophoresis

Reagents for 15% polyacrylamide gel containing 7 Murea

40% acrylamide solution (380 g/liter acrylamide, 20g/liter N,N�-methylenebisacrylamide)

Urea

5X TBE buffer (see below)

TEMED (N,N,N�,N�-tetramethylethylenediamine)

10% wt/vol ammonium persulfate in water—preparedfresh

5X TBE buffer (54 g/liter Tris base, 27.5 g/liter boricacid, 20 ml/liter 0.5 M EDTA, pH 8.0)

Loading buffer (7 M urea in water with 0.1% wt/vol xy-lene cyanole and bromophenol blue)

RNA size standards (�20, 61, and 97 nucleotides)—pre-pared by the instructor

Power supply

Film cassette

Plastic wrap

Razor blade

Kodak X-omat film

Film developer or Kodak developing solutions

Dark room

Protocol

Day 1: Preparation of DNA Template and

in Vitro Transcription Reaction

1. Set up the following reactions in three labeled1.5-ml microcentrifuge tubes:


Tube 1 Tube 2 Tube 3

8 �l of pSP72 plasmid 8 �l of pSP72 plasmid 8 �l of pSP72 plasmid

8 �l of 10X React 3 8 �l of 10X React 3 8 �l of 10X React 2

2 �l of EcoRI 2 �l of BamHI 2 �l of HindIII

62 �l of water 62 �l of water 62 �l of water

Mix all the components thoroughly (do not usea Vortex mixer) and incubate in a 37°C waterbath for 1.5 hr.

2. To remove the restriction endonucleases fromthe reaction, add 80 �l of phenol:chloroform(1:1) to each tube and mix with a Vortex mixerfor 1 min. Centrifuge the samples at room tem-perature for 1 min, remove the aqueous phase(top layer) from each tube, and place these inthree fresh microcentrifuge tubes. Add 80 �l ofchloroform:isoamyl alcohol (24:1) to each ofthe three aqueous samples, cap, and mix with aVortex mixer for 1 min. Centrifuge the samplesat room temperature for 1 min and transfer thethree aqueous phases (top layers) to three freshmicrocentrifuge tubes. This last extraction isdone to remove all traces of phenol from thesolution. It is critical that all of the phenol andchloroform be removed from the aqueous phases inthis final extraction, since they may denature theRNA polymerase during the transcription reaction.Dispose of the organic phases produced inthese extractions in a waste container desig-nated by the instructor.

3. Add 10 �l of 3 M sodium acetate (pH 5.2) tothe aqueous phase in each of the three tubes,along with 200 �l of absolute ethanol. Cap thetubes and invert them several times to mix.

4. Incubate the tubes in a dry ice/ethanol bath(�70°C) for 10 min to precipitate the plasmidDNA. Pellet the precipitated DNA by cen-trifugation for 30 min at 4°C. Place the micro-centrifuge tubes in the rotor with the cap hinge fac-ing the outside during this step. The DNA pelletwill be translucent and very difficult to see, but youwill know that the pellet will be on the same side ofthe tube as the cap hinge. Carefully remove thesupernatant from the DNA pellet. You may alsofind the use of Pellet Paint to be helpful dur-ing this procedure. This product, sold by No-vagen (catalog #69049-3), is a pink chro-mophore that will bind the DNA and give thepellet a color that will be easier to identify.

5. Add 1 ml of ice cold 70% ethanol to each tube,centrifuge for 5 min at 4°C, and carefully re-move the supernatant from the precipitatedDNA pellet. This wash is done to remove thesalts that were added to the DNA solution toaid in its precipitation, which may affect the ac-tivity of the T7 RNA polymerase.

6. Dry the DNA pellets in each of the three tubesin the vacuum microcentrifuge (�5 min), or al-low the pellets to air dry for about 20 min.

7. Resuspend the DNA pellet in each of thesethree tubes with 8 �l of TE buffer. Label thetubes with your name and either H (HindIII),E (EcoRI), or B (BamHI), depending on whichrestriction enzyme was used to treat each of theplasmid samples.

8. Add 1 �l of 10X transcription mix and 1 �l ofT7 RNA polymerase to each of the three re-action tubes. Mix thoroughly but do not use aVortex mixer. Remember that the transcriptionmix is radioactive. Wear safety goggles and gloveswhen working with radioisotopes. Anything thattouches the solution at this point will be radioactive.Be careful with disposal of radioactive pipette tips,microcentrifuge tubes, etc. Your lab instructor willprovide you with radioactive waste storage contain-ers. Flash spin the samples to the bottom of thetubes for 1 sec and incubate all three reactionsat 37°C for 1 hr.

9. After the 1-hr incubation, store the three reac-tion tubes at �20°C for use on Day 2.

Day 2: Polyacrylamide-Gel Electrophoresis

of RNA Transcripts

1. Obtain a 10-well, 15% polyacrylamide gel con-taining 7 M urea from the instructor. Place thegel in the electrophoresis apparatus.

2. Add 0.5X TBE buffer to the upper and lowerbuffer chambers. Remove the comb from thegel. Connect the negative electrode (cathode)to the top buffer chamber and the positive


electrode (anode) to the lower buffer chamber.The RNA that is to be loaded on this gel isnegatively charged and will migrate throughthe gel toward the positive electrode. Apply200 V (constant voltage) to the gel for 20 minbefore the samples are loaded. This pretreatmentof the gel will remove ions left over from thepolymerization reaction that may affect themigration of the RNA on the gel.

3. Add 5 �l of loading buffer to each of the threesamples prepared on Day 1 that have beenthawed at room temperature. Remember thatthese samples are radioactive. Wear safety gogglesand gloves at all times, and be careful in the dis-posal of radioactive waste. Also, add 5 �l of load-ing buffer to a 10-�l sample of RNA size mark-ers that have been prepared by the instructor.The urea present in the loading buffer and inthe gel works to disrupt any hydrogen bonding(intramolecular or intermolecular) within theRNA molecules that may cause them to run atan apparent higher or lower molecular weightthan they really are.

4. Turn off the power supply to the gel and washthe wells out thoroughly with 0.5X TBE bufferto remove any precipitated urea and unpoly-merized acrylamide that may be clogging thewells. Load the three samples in three consec-utive wells on the gel. Next to your three sam-ples, load the RNA size standard sample. Sinceyou will be using only 4 of the 10 wells, anothergroup can skip a lane and load their four samples.Be sure that you know which of the four samplesare in which lane.

5. Reconnect the power supply and apply 200 V(constant voltage). Continue the electrophore-sis until the bromophenol blue (dark blue) dyejust migrates off of the bottom of the gel intothe lower buffer chamber. Any unincorporatedradiolabeled nucleotide that was present in the sam-ples will run off of the bottom of the gel with thedye. Remember that the buffer in the lower cham-ber is now radioactive.

6. Turn off the power supply and remove theplates containing the gel. Separate the plates toexpose the gel and put a notch in one cornerof the gel with a razor blade to allow you toorient the gel. Place the gel on a sheet ofWhatman 3 MM filter paper. Wrap the gel inplastic wrap and place it in the film cassette.

7. Expose the gel to film overnight (�18 hr) at�70°C. Before closing the cassette, use a felt-tipped marking pen to mark the position on thefilm that corresponds to the notch made on thecorner of the gel. Exposing the film at low tem-peratures will reduce the radioactive scatteringand produce more defined RNA bands on theautoradiogram. Develop the film the followingday.

8. Align the autoradiogram with the gel and markthe position of the wells on the film. Identify theRNA size marker bands on the autoradiogram(three of them in one lane). The instructor willprovide you with the information that you needto determine the number of nucleotides thateach standard RNA marker contains.

9. Prepare a plot of number of nucleotides versusdistance migrated (in centimeters) for eachRNA size standard band on the autoradiogram.Using this standard curve, and the distances mi-grated by each of the transcripts produced inthe three reactions (the other three lanes on thegel), determine the molecular weight and thenumber of nucleotides present in the tran-scription reactions containing pSP72 digestedwith EcoRI, BamHI, and HindIII.

10. Using the map of the pSP72 plasmid, deter-mine the exact sequence of the transcript thatyou would expect following run-off transcrip-tion from the T7 promoter using the pSP72plasmid digested with the three different re-striction enzymes (assume that the plasmid wasdigested to completion with each restrictionenzyme). Do the sizes of the RNA moleculespresent in the three sample lanes on the au-toradiogram agree well with your predicted re-sults? Explain your answer.

11. Is there more than one radiolabeled RNA frag-ment present in each of the three sample lanes?If so, determine whether they are lower orhigher in molecular weight than that of the pre-dicted RNA molecule. What may have oc-curred to produce these RNA fragments of un-expected size?

12. Look at the autoradiogram and compare the in-tensities (the dark color) of each of the RNAmolecules in the three sample lanes. Are alltranscripts of equal radioactive intensity? Whyor why not? Based on the predicted sequenceof your transcripts (determined in step 10),


which of the three transcripts should be themost and the least radioactive? Be as quantita-tive as you can in answering this question.

Exercises

1. Determine the exact sequence of the transcriptsthat you would expect to be produced if thesame templates used in this experiment (pSP72digested with BamHI, EcoRI, and HindIII) weresubjected to an in vitro transcription reactionusing the SP6 RNA polymerase rather than theT7 RNA polymerase.

2. In this experiment, you used �-[32P]ATP as theradiolabeled nucleotide. Would your resultshave been any different if you had used �-[32P]GTP as the radiolabeled nucleotide? Ex-plain your answer in terms of the mechanismof the reaction catalyzed by RNA polymerase.Could you have used another �-labeled nu-cleotide and still obtained a radiolabeled RNAtranscript? Explain your answer.

3. You digest 10 �l of the pSP72 plasmid (0.1mg/ml) to completion with BamHI. This di-gested plasmid is used as a template for in vitrotranscription reactions (50 �l total reaction vol-ume) using �-[32P]CTP (1 �Ci/�mol) and SP6RNA polymerase. After the transcription reac-tion is complete, you isolate the RNA transcriptfrom the unincorporated nucleotides and de-termine that 1 �l of the 50-�l reaction contains20,000 dpm of 32P. Calculate the number of mi-crograms of RNA produced in this reaction.What specific activity of �-[32P]GTP wouldyou have had to use in this reaction to obtainthe same mass of RNA transcript containingthe same amount of radioactivity (the same spe-cific activity of the RNA transcript)?

4. You are conducting a study of the mechanismof action of T7 RNA polymerase. You have pu-rified a total of 10 point mutants in the enzyme

that are unable to carry out transcription fromthe T7 promoter. You are not sure if these mu-tations affect the ability of the enzyme to rec-ognize the promoter or if they render the en-zyme unable to perform the polymerization reaction. Design a set of experiments thatwould allow you to distinguish between thesetwo possibilities.

5. Often, phage genomes contain genes under thecontrol of both viral and bacterial promoters.One gene that is commonly under the controlof the bacterial promoter is that which encodesthe phage RNA polymerase. In terms of the lifecycle of a phage, why has the phage evolved tohave genes under the control of both types ofpromoters? What will ultimately determinewhat species of bacteria a certain phage will beable to infect? (Why can a particular phage in-fect one species of bacteria but not another?)

REFERENCES

Chamberlin, M., McGrath, J., and Waskell, L. (1970).New RNA Polymerase Form Escherichia coli In-fected with Bacteriophage T7. Nature 228:227.

Chamberlin, M., and Ryan, T. (1982). BacteriophageDNA-Dependent RNA Polymerases. In: Boyer,P.D., ed. The Enzymes, Vol. 15: Nucleic Acids,Part B, p. 87. Academic Press, New York.

Dunn, J. J., and Studier, F. W. (1983). The CompleteNucleotide Sequence of Bacteriophage T7 DNAand the Locations of T7 Genetic Elements. J MolBiol 135:917.

Lehninger, A. L., Nelson, D.L., and Cox, M. M.(1993). RNA Metabolism. In: Principles of Biochem-istry, 2nd ed. New York: Worth.

Rosa, M. D. (1979). Four T7 RNA Polymerase Pro-moters Contain an Identical 23 Base Pair Sequence.Cell 16:815.

Stahl, S., and Zinn, K. (1981). Nucleotide Sequence of the Cloned Gene for Bacteriophage T7 RNAPolymerase. J Mol Biol 148:481.

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369

EXPERIMENT 23

In Vitro Translation: mRNA, tRNA, and Ribosomes

Theory

As discussed in Experiment 22, the flow of geneticinformation from the genes encoded by the DNAinto proteins is a two-step process. During transcrip-tion, RNA polymerase binds at the promoter regionson the DNA and transcribes a single-stranded mes-senger RNA (mRNA). The RNA polymerase en-zyme is instructed as to the sequence of each mRNAmolecule that it produces by reading one of the twoDNA strands (template strand) in the 3� to 5� direc-tion and synthesizing an RNA in the 5� to 3� direc-tion that is complementary to it (or identical to thesequence of the coding DNA strand, with U in placeof T). Once an mRNA is constructed, it will takepart in a process known as translation. Beginning ata defined positionwithin the mRNA, the ribosome willread the base sequence of the mRNA in sets of three(a codon) and produce a protein or peptide specifiedby that particular messenger molecule. Translationis carried out in a four-step mechanism, which is dis-cussed in detail below. The mechanism of transla-tion is best understood in prokaryotes. The detailsof the mechanism of translation in eukaryotes re-mains an area of intensive research. Because of this,the four-step mechanism described below applies tothe prokaryotic system. Immediately following this,the mechanism of translation in eukaryotes is de-scribed, and differences between eukaryotic andprokaryotic translation are highlighted.

Step 1: Synthesis of Aminoacyl tRNA

The first step involves activation or “charging” ofthe various transfer RNAs (tRNAs) in the cell with

their appropriate amino acids. The general struc-ture of the tRNA is shown in Figure 23-1. The two-dimensional “cloverleaf ” structure of the moleculeimparted to it by the ordered array of intramolec-ular hydrogen bonds between neighboring baseshas two important features. First, each tRNA con-tains an anticodon loop that will allow it to base-pairwith a complementary sequence (codon) on themRNA and deliver the appropriate amino acid tothe growing peptide chain. The second is the con-served 3� end of the molecule (5�-GCCA-3� or

Conserved 3' end of tRNA that forms an ester bond with the amino acid5'

3'ACC

Anticodon

(G/A)

Figure 23-1 Two-dimensional structure of tRNA.


5�-ACCA-3�) that will form a covalent (ester) bondwith the appropriate amino acid and deliver it tothe site of translation. The sequence of the accep-tor stem, which is made up of the 5� and 3� ends ofthe tRNA, is important in the recognition of a giventRNA by the correct “charging enzyme.”

The enzymes that link the amino acids to the3� ends of the various tRNAs are called aminoacyltRNA synthetases. There is at least one aminoacyltRNA synthetase that can recognize each of the 20amino acids. Due to the degeneracy of the geneticcode (more than one codon that specifies a partic-ular amino acid), a single aminoacyl tRNA syn-thetase may be able to charge a number of differ-ent tRNAs with the same amino acid. For example,phenylalanine is specified by the codons UUU andUUC. The aminoacyl tRNA synthetase specific forphenylalanine will therefore be able to charge atRNA with AAA or AAG in the anticodon loop withthis amino acid. The mechanism by which an aminoacid is linked to the 3� end of the tRNA is shownin Figure 23-2. The process is energetically costly,requiring the hydrolysis of two high-energy phos-phate bonds.

Step 2: Initiation

Translation in prokaryotes begins with the forma-tion of the ribosome complex at a defined positionon the mRNA, termed the Shine–Delgarno se-quence, or the ribosome binding site (RBS). Inprokaryotic mRNA, this is a relatively small (4–7nucleotides) region rich in purine nucleotides lo-cated less than 10 nucleotides to the 5� side of the translational start site (Fig. 23-3). This Shine–Delgarno sequence is complementary to the 16Sribosomal RNA (rRNA) associated with the 30Sribosomal subunit, and directs it to bind the mRNAat that position. The 30S ribosomal subunit will notbind this region on the mRNA without the aid of an associated protein called initiation factor 3(IF-3). The 30S ribosomal subunit will bind themRNA in such a manner that the peptidyl (P) site ofthe complex is occupied by a specialized codon withthe sequence, AUG. It is at this AUG “start” codonwhere translation will eventually begin.

The next step of the initiation process inprokaryotes involves the delivery of a specializedaminoacyl tRNA to the P site of the 30S ribosomalsubunit. This specialized tRNA has a UAC anti-

codon loop sequence capable of interacting with theAUG start codon, and is charged with an unusualamino acid called N-formylmethionine (fMet, Fig.23-4). The fMet-tRNA is brought to the P sitethrough its association with a GTP binding proteincalled initiation factor 2 (IF-2, Fig. 23-3).

To complete the initiation process, the 50S ri-bosomal subunit binds to the 30S–mRNA complex,with the subsequent hydrolysis of GTP and the re-lease of IF-2 and IF-3 from the 70S ribosomal com-plex (Fig. 23-3). At this point, the aminoacyl site (Asite) of the ribosome is ready to accept the incom-ing aminoacyl tRNA specified by the codon thatoccupies this site.

Step 3: Elongation

As with initiation, elongation is a multistep process.In the first step, another GTP-binding protein,elongation factor Tu (EF-Tu), delivers the nextaminoacyl tRNA to the A site on the ribosome. Asthis occurs, EF-Tu hydrolyzes GTP and leaves theribosome (Fig. 23-5).

Next, fMet-tRNA will deliver the carboxyl ter-minus of fMet to the amino terminus of the aminoacid linked to the tRNA at the A site (a peptidyltransfer reaction), with the subsequent formation ofa peptide bond between the two amino acids. Atthis point, the tRNA at the A site is covalentlylinked to a dipeptide (Fig. 23-5).

To remove the now uncharged fMet-tRNAfrom the P site, the ribosomal complex will shift to-ward the 3� end of the mRNA by three bases, po-sitioning a new codon in the A site and ejecting the“spent” fMet-tRNA. At this point, the dipeptidebound by the tRNA that formerly occupied the Asite now occupies the P site. This final step in trans-lation elongation is fueled by the hydrolysis of an-other GTP molecule by a translocase enzyme calledelongation factor G (EF-G, Fig. 23-5).

The process of elongation will continue in the5� to 3� direction down the mRNA. Each time theribosome shifts position by three bases to accept an-other aminoacyl tRNA, the growing peptide chainwill increase in length by one amino acid. Becausethe carboxyl group of the peptide linked to thetRNA at the P site always forms a covalent bondwith the amino group of the amino acid on thetRNA at the A site, peptides are always synthesizedfrom the amino terminus to the carboxyl terminus.

371

NN

PO

�OO

OH

OH

PO

�

�OO

2�O

PO

�

�OO

�O

PO

�OO

PO

�OO

OP

O

O�

O

O CC

NH

3

RH�

O

CC

H3N

RH�

O

CC

H3N

R

O�

HN

N

H

CH

2

NH

2

H

O

NN

O

PO

�OO

OH

OH

O

HH

H H

HH

HH

HH

NN

H

CH

2

NH

2

H

O

NN

PO

�

OO

OH

O

NN

H

CH

2

NH2

H

O Am

inoa

cyl t

RN

ANN

PO

�OO

OH

OH

O

NN

H

CH

2

NH

2

H

O

Ade

nosi

ne-5

�-m

onop

hosp

hate

tRN

A

�

Am

ino

acid

�

Ade

nosi

ne-5

�-tr

ipho

spat

e

NN

PO

�OO

OH

OH

O

NN

H

CH

2

NH

2

H

O

5�-A

min

oacy

lade

nyla

te

Pyro

phos

phat

ase

�

Figure 23-2 Reaction catalyzed by aminoacyl tRNA synthetase.


5' AStart codon

Shine–Delgarno sequence

mRNAU

P A

G

30S Ribosome IF-3

3'

5' 3'

A U G

IF-2

GTP

IF-3

IF-2 PiIF-3

P A

30S Ribosome

50S Ribosome

A U GU A C

P A

30S Ribosome

70S Ribosomal Complex

50S Ribosome

+

++

+

fMet tRNA

fMet

GDP

fMet

Figure 23-3 Schematic diagram of translational initiation in prokaryotes.

Step 4: Termination

Unlike translational initiation, elongation, andaminoacyl tRNA synthesis, translational termina-tion is a spontaneous process that does not requirethe input of energy (GTP hydrolysis). There arethree codons on the mRNA that will trigger theend of translation when they appear at the A siteon the ribosome: UAA, UGA, and UAG. As the ri-bosome encounters these codons, one of two re-lease factor proteins will bind at the A site and hy-

drolyze the ester bond linking the peptide chain tothe tRNA at the P site. Following this event, the50S and 30S ribosomal subunits will dissociate fromthe mRNA (Fig. 23-6).

Eukaryotic Translation

One major difference between the prokaryotic andeukaryotic translational machinery is the composi-tion of the two ribosomal subunits. Unlike the small

EXPERIMENT 23 In Vitro Translation: mRNA, tRNA, and Ribosomes 373

C

O O

C CNH

CH2

CH2

S

CH3

H O�

H

Figure 23-4 The structure of N-formylmethio-

nine, which is the amino terminal

residue of all newly synthesized

prokaryotic proteins. In many cases,

the N-formyl group or both the

N-formyl group and methionine are

posttranslationally cleaved from the

protein.

prokaryotic ribosomal subunit (30S, with 21 pro-teins and 16S rRNA), the small eukaryotic ribo-somal subunit (40S) consists of 33 proteins and an18S rRNA. The large ribosomal subunit also dif-fers between prokaryotic and eukaryotic organisms:the prokaryotic subunit (50S) is made up of 34 pro-teins, a 23S rRNA, and a 5S rRNA, while the eu-karyotic subunit (60S) consists of about 50 proteinsand three separate rRNAs (5S, 28S, and 5.8S). Thetwo prokaryotic subunits combine to form the func-tional 70S ribosome, while the eukaryotic subunitscombine to form the functional 80S ribosome.

Another difference between prokaryotic and eu-karyotic translation is the nucleic acid sequencesthat recruit the small ribosomal subunit to themRNA. As stated earlier, the Shine–Delgarno se-quence interacts with the 16S rRNA of the 30S rib-osomal subunit in the prokaryotic system, the crit-ical step in translation initiation. In the eukaryotic

5' 3'P A1 2 3

AA1 AA2

5' 3'P A

1 2 3

AA2AA1

AA3

P A1 2 3

AA1AA2

A second aminoacyl tRNA enters the A site on the ribosome. This process is aided by EF-Tu, which hydrolyzes one molecule of GTP to GDP + Pi

Translocation of the ribosome. This process is catalyzed by EF-G, with the hydrolysis of one molecule of GTP to GDP + Pi.

This process frees the A site on the ribosome to accept the next aminoacyl tRNA. Again, the next aminoacyl tRNA is aided by EF-Tu, which hydrolyzes one molecule of GTP to GDP + Pi. The unchanged fMet tRNA is ejected from the ribosome.

Peptidyl

transfer

fMet tRNA

Figure 23-5 Schematic diagram of translational elongation in prokaryotes.


3'

P A125 A AU

AA125

AA124

AA125

5'

P A125 A AU

P A

AA125

AA124

Releasefactor

Releasefactor

Releasefactor

Completed peptide

50S Ribosome

COO–

AA1H3N+

+ + +

5'

mRNA

3'

30S ribosome

Figure 23-6 Schematic diagram of translational termination in prokaryotes.

system, a 7-methyl-guanosine “cap” is added to the5� end of the mRNA (Fig. 23-7). It is this 5� capthat facilitates the binding of the 40S ribosomalsubunit to the mRNA (see below).

Eukaryotic translation initiation is far morecomplicated than the prokaryotic system describedearlier. To begin the process, two initiation factor

proteins (eIF3 and eIF4C) bind to the 40S ribosomalsubunit and prepare it to bind to the mRNA. Next,a cap-binding protein (CBPI) binds the 7-methyl-guanosine cap at the 5� end of the mRNA, alongwith three other initiation factor proteins (eIF4A,eIF4B, and eIF4F). In an ATP-requiring process,this protein complex “scans” the mRNA until the


O

H2N

CH2

CH3

HN

N

H HH H

N

N

O

OH OH

CH2

H HHH

O

O OCH3

O

O

O

P

�O

�O

O

O

P

�O O

�O O

�O O

O

P

PO

Base

Methylated incaps 1 and 2

CH2

H HHH

O

O OCH3

P

Base

Methylatedin cap 2

Figure 23-7 Structure of 7-methyl-guanosine

capped eukaryotic mRNA. (From

Stryer, L. (1988) Biochemistry,

3rd ed. New York: Freeman, Figure

29-27. Reprinted with permission.)

AUG start codon is identified. At this point, the 40Sribosome/eIF3/eIF4C complex binds the mRNA.

The eIF2 protein (a GTPase) delivers the firstaminoacyl tRNA to the P site on the 40S ribosome.Unlike the prokaryotic system, in which translationbegins with f-Met, eukaryotic peptides begin withL-methionine. Although this initiating methionineis unmodified, the tRNA that delivers it to the P

site of the ribosome is different from the tRNA thatdelivers methionine residues (internal) to the grow-ing polypeptide. eIF2 hydrolyzes GTP to GDP andPi during this process. The eIF2 protein releasesthe GDP and recycles to the GTP form with theaid of another protein—eIF2B. Finally, the proteineIF5 binds the mRNA/Met-tRNAmet/40S ribo-somal complex, causing eIF3 and eIF4C to dissoci-ate. At this point, the 60S ribosomal subunit bindsto form the functional 80S ribosome. The processof eukaryotic translation initiation is depicted inFigure 23-8.

The elongation reaction described earlier forthe prokaryotic system is essentially the same as thatfound in the eukaryotic system. The final differencebetween the two systems lies in the mechanism oftranslational termination. Recall that the prokary-otic system utilizes two different release factor pro-teins. The eukaryotic system relies on a single re-lease factor protein, eRF.

In this experiment, you will perform a numberof in vitro translation reactions. The ribosomesused in this experiment were obtained from wheatgerm, a eukaryotic organism. After the material isground into a fine paste, the mixture is diluted withbuffer to extract most of the proteins and othersmall molecules from the cells. This cellular extractis then subjected to centrifugation at 30,000 � g.The insoluble material harvested following this stepcontains unlysed cells, cellular debris, and intact mi-tochondria. The supernatant, or S-30 fraction, con-tains all of the components needed to perform invitro translation (ribosomes, tRNA, initiation fac-tor, elongation factor, etc.).

The mRNA that you will use in these reactionsis poly(U), which will direct the ribosome to syn-thesize a poly-phenylalanine (poly-Phe) peptide. Byincluding a small concentration of radiolabeled (3H)phenylalanine (Phe) in the reaction, the relativeamounts of the peptide produced in the differentreactions can be determined through isolation ofthe peptide and scintillation counting (see below).

In vitro, the fidelity of translation is strongly in-fluenced by the concentration of Mg2� ions in thereaction. In the range from 1 to 4 mM Mg2�, theribosome will require a Shine–Delgarno sequence(prokaryotic) or a 7-methyl-guanosine cap (eukary-otic) on the mRNA before translation will initiate.For this reason, in vitro translation experiments per-formed with naturally occurring mRNAs are most


Met

60Sribosome

P A

40S ribosome3'

5' Cap

CBPI

eIF4F

ATP

ADP + PieIF4AeIF4BeIF4FCBPI

eIF4CeIF3

eIF4BeIF4A

mRNA

3'

P A

40S ribosomeeIF4C

eIF2GTP

eIF3

eIF5

P A

40S ribosome

P A

40S ribosome

80S ribosomal complex

60S ribosome

Met

Met

5'

5'

5' 3'

3'

+

eIF3eIF4CeIF5GDP + Pi

Figure 23-8 Translation initiation in eukaryotes.


often carried out at low Mg2� concentrations. If theMg2� concentration is raised to the range from 8to 12 mM, the ribosome will initiate transcriptionat random positions throughout the mRNA. Thus,by conducting your in vitro translation reactions under high Mg2� concentrations, the ribosomeswill efficiently translate the poly(U) mRNA into apoly-Phe peptide in the absence of any formal trans-lational start site.

The 3H-poly-Phe peptides produced in theseexperiments will be isolated by exploiting their abil-ity to precipitate in a solution containing 5%trichloroacetic acid (TCA). Strong acids such asTCA are effective in inducing the precipitation ofall ionic macromolecules. Thus, the proteins, tRNA,and mRNA present in the translation reactions willall become insoluble on the addition of 5% TCA.Remember that there are three individual pools of3H-Phe present in this ongoing translation reac-tion: the free 3H-Phe, the 3H-Phe covalently linkedto the tRNA, and the 3H-Phe that has been incor-porated into the growing peptide. The free 3H-Phewill remain soluble in 5% TCA due to its small size,and therefore will not be present in the precipitate.The 3H-Phe covalently linked to the 3� end of thetRNA, however, will be insoluble in 5% TCA andwill be present in the precipitate (Fig. 23-9). If this3H aminoacyl tRNA pool is not removed from the pre-cipitate, you will overestimate the amount of 3H-Phethat was incorporated into the peptide when the reac-tions are subjected to scintillation counting.

How can you remove the 3H-Phe tRNA fromthe precipitate to obtain an accurate measure ofthe amount of 3H-poly-Phe peptide produced?There is one difference in the properties of thesetwo molecules that you can exploit. Unlike the pep-tide bonds linking the 3H-Phe monomers in thepeptide, the ester bond linking the 3H-Phe to the3� end of the tRNA is labile in 5% TCA at 90°C.Therefore, if the precipitate is heated at 90°C for10 min, all of the 3H-Phe associated with the tRNAwill be hydrolyzed and converted to soluble (free)3H-Phe that will diffuse back into the bulk solu-tion. At this point, the only 3H-Phe present in the so-lution will be that which incorporated into the peptide(Fig. 23-9).

To quantify the relative amount of 3H-poly-Pheproduced from one reaction to the next, the pre-cipitate will be filtered through a 0.45-�m mem-brane that will trap the insoluble material and al-

low the soluble 3H-Phe in the reaction to passthrough. When this filter is dried and subjected to scin-tillation counting, the amount of 3H present (in countsper minute) will be directly proportional to the amountof 3H-poly-Phe produced in the reaction (Fig. 23-9).

There are four variables in the general in vitrotranslation protocol that you will investigate. First,you will vary the Mg2� concentration to study itseffect on the fidelity of translation initiation. Sec-ond, you will investigate the specificity of the 3H-Phe-tRNA and the mRNA codon with which it isable to interact. Third, you will study the effects ofdifferent ribonuclease enzymes on the various RNAcomponents present in the reaction (see below). Fi-nally, you will investigate the ability of the ribo-some to carry out translation in the presence ofthree different inhibitors with different modes ofaction (see below).

The two RNase enzymes that will be used inthis experiment are RNase A and RNase T1. RNaseA is a heat-stable nuclease that will hydrolyze thephosphodiester bonds linking the pyrimidine bases(U or C) in the molecule. RNase A will cleave boththe mRNA and the tRNA present in the reaction.RNase T1 differs from RNase A in that it is a base-specific nuclease. RNase T1 will cleave the phos-phodiester bonds on the 3� side of guanine. Becauseof its specificity, RNase T1 will act only on the tRNApresent in the reaction (the poly(U) mRNA doesnot contain guanine).

The three antibiotic inhibitors of translation thatwill be used in this experiment are chlorampheni-col, cycloheximide, and puromycin (Fig. 23-10).Chloramphenicol is specific for prokaryotic ribosomes,blocking the transfer of the peptide on the tRNA atthe P site to the amino acid linked to the tRNA atthe A site (the peptidyl transfer reaction). Since thesource of the ribosomes used in this experiment iswheat germ (eukaryotic), we would predict thatchloramphenicol would not have a great effect ontranslation. The mechanism of cycloheximide-mediated inhibition is the same as that describedabove for chloramphenicol, except for the fact thatit is specific for the 80S eukaryotic ribosome.Puromycin is a more broad translational inhibitor,effective on both eukaryotic and prokaryotic ribo-somes. It acts as a substrate analog of aminoacyltRNA. When it binds at the A site of the ribosome,it induces premature termination of translation (Fig.23-10).


To vacuum

1 3H-Phe

2 3H-Phe-tRNA

3 3H-poly-Phe

1 3H-phe 1 3H-Phe

3H-Phe

2 3H-Phe-tRNA

3 3H-poly-Phe2 tRNA

3 3H-poly-Phe

Add 5% TCA 90°C, 10 min

Isolate filter and analyze

by scintillation counting

Ongoing in vitro translation reaction (all radioactive isotopes are soluble)

Soluble

Insoluble

Soluble

Insoluble

Pass solution throughHAWP filter

Insoluble material (3H-poly-Phe) is trapped by filter

The number of 3H cpm present on thefilter is directly proportional to the amount of

3H-poly-Phe produced in the reaction.

Figure 23-9 Analysis of poly-Phe peptide produced following in vitro translation.


C

O

NH CH2

CH3

NO2

CH

OH

CH3

CH2

O

NCH CH

CH

OH

OH

Cl

Cl

Chloramphenicol

O

O

Cycloheximide

HC

NH2

R

C O

O OH

N

N

N

N

NH2

H HH H

CH2

O

P

O

P A3�5�

Peptidyltransferase

P A3�5�

P sitepeptidyl-tRNA

A sitepuromycin

puromycin

HC

CH2

H2N

C

O

OCH3

NH OH

N

N

N

N

CH3

O

H HH H

CH2

OH

CH3

..

mRNA

O

HC

NH

R

CO HC

CH2

N

C

O

O

OCH3

NHOH

N

N

N

N

CH3

O

H HH H

CH2

OH

CH3

H

AA

Structureof peptidyl-puromycin

Figure 23-10 Structure of translation inhibitors,

and the mechanism of action of

puromycin. (From Lehninger, A.L.,

Nelson, D.L., and Cox, M.M. Prin-

ciples of Biochemistry, 2nd ed.

New York: Worth, Figure 26-34.

Reprinted with permission.)





S-30 fraction from wheat germ—commercial (Fisher cat-alog #L-4440) or freshly prepared (see Appendix)supplemented with 125 mM ATP (potassium salt),2.5 mM GTP (potassium salt), 100 mM creatinephosphate, and 0.5 mg/ml creatine

Reaction buffer (300 mM HEPES, 30 mM dithiothreitol[pH 7.1])

Salt solution I (1.3 M KCl, 225 mM magnesium acetate)

Salt solution II (1.3 M KCl, 25 mM magnesium acetate)3H-Phenylalanine solution (0.1 mM total concentration

at an activity of 0.005 �Ci/ml)

Poly(U) solution (1 mg/ml) in RNase free distilled wa-ter—Sigma catalog #P-9528

Poly(A) solution (1 mg/ml) in RNase free distilled wa-ter—Sigma catalog #P-9403

Trichloroacetic acid solution (5% wt/vol) in distilled water

Fine-tipped forcepts

Millipore filters (HAWP, 0.45-�m pore size)

Filtration apparatus with vacuum pump

30°C water bath

90°C waterbath

Scintillation vials with caps

Scintillation fluid (0.5% wt/vol diphenyloxazole, 10%wt/vol Triton X-100 in toluene)


100°C oven

RNase A solution (0.5 mg/ml) in distilled water—Sigmacatalog #R-5125

RNase T1 solution (0.5 mg/ml) in distilled water—Sigmacatalog #R-1003

Chloramphenicol solution (0.5 mM) in distilled water—Sigma catalog #C-0378

Cycloheximide solution (0.5 mM) in distilled water—Sigma catalog #C-7698

Puromycin solution (0.1 mM) in distilled water—Sigmacatalog #P-7255

Plastic squirt bottle

Caution: This experiment requires the use of

low levels of tritium (3H). Wear gloves and

safety goggles at all times when working with

the radioisotope. Cover your work surface with

absorbent paper to localize any spills that may

occur. Consult the instructor for the proper dis-

posal of all solid and liquid radioactive waste

that will be produced in the experiment.

Protocol

Day 1: Effect of Mg2� and Various mRNAs

on Translation

1. Set up the following reactions in five labeledmicrocentrifuge tubes:

Volumes (�l)

Component 1 2 3 4 5

Reaction buffer 12 12 12 12 12

Salt solution I 12 12 12 — 12

Salt solution II — — — 12 —

1 mg/ml Poly(U) 10 10 — 10 —

1 mg/ml Poly(A) — — — — 10

S-30 fraction 20 20 20 20 20

Distilled water 16 16 16 16 16

2. Add 10 �l of 3H-Phe solution to tube 1 andpipette up and down rapidly several times tomix. Immediately add 750 �l of 5% TCA to stopthe reaction. Place tube 1 on ice for later analy-sis. This reaction will serve as a time zero blankfor the remaining reactions that will indicatethe amount of soluble 3H-Phe in the reactionthat will unavoidably contaminate the precipi-tated proteins.

3. Add 10 �l of 3H-Phe solution to each of tubes2 to 5. Pipette up and down rapidly severaltimes to mix. Incubate all four tubes in a 30°Cwater bath for 30 min.

4. Following the 30-min incubation, add 750 �lof 5% TCA solution to each of tubes 2 to 5.Cap the tubes and invert them several times tomix. Place tubes 1 to 5 in a 90°C water bathfor 10 min. This high-temperature incubationwill hydrolyze the ester bond linking the 3H-Phe to the tRNA. The 3H-Phe that has beenincorporated into the peptide is stable at thistemperature.

5. Remove all five tubes from the 90°C water bathand allow them to cool to room temperature.

6. Using a fine-tipped forceps, place a 0.45-�mHAWP filter in the filtration apparatus (to bedemonstrated by the instructor). Fill a squirtbottle with 5% TCA and apply a small volumeto the filter to moisten it.

7. Apply the entire contents of tube 1 to the fil-ter and turn on the vacuum line to the unit.


Immediately fill the now empty tube 1 with 5%TCA and pour the contents on the filter. Re-peat this wash procedure on tube 1 a total offive times to harvest all of the precipitated ma-terial and transfer it to the filter.

8. Apply one last aliquot (�5 ml) of 5% TCA tothe filter and allow the filter to dry under vac-uum for 20 sec (allow all of the solvent to flowthrough the filter and continue to apply thevacuum for an additional 20 sec). This finalwash step is done to remove any trace amountsof soluble 3H-Phe that may be bound to thefilter or to the precipitate adsorbed to the fil-ter.

9. Place the filter in a scintillation vial and placethe vial (uncapped) in a 100°C oven for exactly10 min. This heat treatment will dehydrate thefilter and cause it to curl up in the vial. Avoidexcessive heating, which will cause the filter toturn yellow and possibly quench the 3H � par-ticles in the sample as they are being counted(color quenching, see introduction to Experi-ment 3).

10. Perform steps 6 to 9 on the contents of each oftubes 2 to 5. You will require a clean, unused fil-ter for each of the remaining four reaction tubes.When you are finished, you will have five scintil-lation vials (labeled 1 to 5), each corresponding tothe appropriately numbered reaction tube.

Reaction Time Course

11. Obtain four fresh 1.5 ml microcentrifuge tubes.Label these tubes 6 to 9, add 750 �l of 5% TCAto each, and place all four of the tubes on ice.

12. Obtain a fifth fresh microcentrifuge tube andadd the following: 10 �l of reaction buffer; 10�l of salt solution I; 50 �l of 1 mg/ml Poly(U);100 �l of S-30 fraction; and 30 �l of distilledwater.

13. Add 50 �l of 3H-Phe solution to this tube, andpipette up and down rapidly several times tomix. Incubate this tube in a 30°C water bath.

14. After a total of 15 min of incubation at 30°C,remove a 50-�l aliquot of the reaction and addit to tube 6. Cap tube 6 and invert several timesto mix. Place tube 6 back on ice and return thereaction tube to the 30°C water bath.

15. After a total of 30 min of incubation at 30°C,remove a 50-�l aliquot of the reaction and addit to tube 7. Cap tube 7 and invert several times

to mix. Place tube 7 back on ice and return thereaction tube to the 30°C water bath.



18. Perform steps 4 to 9 (heating at 90°C and fil-tration steps) on each of the reaction timepointsin tubes 6 to 9. At the end of the procedure, youwill have a total of nine scintillation vials (labeled1 to 9), each containing a dried filter coresspondingto the appropriately numbered reaction or reactiontimepoint tube.

19. Add 5 or 10 ml of scintillation fluid to each ofthe nine scintillation vials and place them in ascintillation counter rack marked with yourname.

20. Place the rack in the scintillation counter andcount each vial for 2 min in the 3H channel(consult the instructor) to obtain a 3H cpmvalue for each filter.

21. Subtract the counts per minute (cpm) value forvial 1 (time � zero, reaction control) from allof the other cpm values obtained for vials 2 to9. This is done to account for the fact that somesoluble 3H may not have been removed from theinsoluble material (precipitate) during the filtra-tion step.

22. Compare the relative level of 3H-poly-Phepresent in vial 2 compared to that in vial 3. Ex-plain your results in terms of the componentsthat were present in these two reactions.

23. Compare the relative level of 3H-poly-Phepresent in vial 2 compared to that in vial 4. Ex-plain your results in terms of the componentsthat were present in these two reactions, as wellas your knowledge of the dependence of trans-lational initiation on Mg2� concentration.

24. Compare the relative level of 3H-poly-Phepresent in vial 2 compared to that in vial 5. Ex-plain your results in terms of the componentsthat were present in these two reactions, as wellas your knowledge of the genetic code.

25. Using the cpm data obtained from tubes 6 to9, prepare a plot of 3H-poly-Phe cpm versus


reaction time (15, 30, 45, and 60 min). Arethe kinetics of this in vitro translation reac-tion linear over 60 min (does this plot yield astraight line through the data points)? If not,for what length of time is this reaction linearwith time?

26. Determine the slope of the line through the lin-ear portion of the curve, and calculate an ini-tial velocity (v0) for the reaction in units of cpmof 3H incorporated per min.

Day 2: Effect of Nucleases and Inhibitors

on Translation

1. Obtain eight fresh 1.5-ml microcentrifugetubes and label them 10 to 17. Set up the fol-lowing reactions in these tubes:

5. Place tubes 10 to 16 (but not 17, which should bekept on ice) in a 90°C water bath for 10 min.

6. After the 10-min incubation, remove tubes 10to 16 from the 90°C water bath and allow themto cool at room temperature.

7. Perform steps 6 to 9 from the Day 1 protocol(filtration and TCA wash steps) on the contentsof tubes 10 to 17. When you are finished, youshould have eight scintillation vials (labeled 10 to17) corresponding to each of the appropriately num-bered reaction tubes.

8. Add 5 or 10 ml of scintillation fluid to each ofthe nine scintillation vials and place them in ascintillation counter rack marked with yourname.

9. Place the rack in the scintillation counter andcount each vial for 2 min in the 3H channel

Volumes (�l)

Component 10 11 12 13 14 15 16 17

Reaction buffer 2 2 2 2 2 2 2 2

Salt solution I 2 2 2 2 2 2 2 2

1 mg/ml Poly(U) 10 10 10 10 10 10 10 10

0.5 mg/ml RNase A — — 2 — — — — —

0.5 mg/ml RNase T1 — — — 2 — — — —

0.5 mM chloramphenicol — — — — 6 — — —

0.5 mM cycloheximide — — — — — 6 — —

0.1 mM puromycin — — — — — — 6 —

S-30 Fraction 20 20 20 20 20 20 20 20

Distilled water 6 6 4 4 — — — 6

2. Add 10 �l of 3H-Phe solution to tube 10 andpipette up and down rapidly several times tomix. Immediately add 750 �l of 5% TCA to stopthe reaction. Place tube 10 on ice for lateranalysis. This reaction will serve as a time zeroblank for the remaining reactions that will in-dicate the amount of 3H-Phe in the reactionthat will unavoidably contaminate the precipi-tated proteins.

3. Add 10 �l of 3H-Phe solution to each of tubes11 to 17. Pipette up and down rapidly severaltimes to mix. Incubate all seven tubes in a 30°Cwater bath for 30 min.

4. After the 30-min incubation, add 750 �l of 5%TCA to each of tubes 11 to 17. Cap the tubesand invert several times to mix.

(consult the instructor) to obtain a 3H cpmvalue for each filter.

10. Subtract the cpm value for vial 10 (time � zeroreaction control) from all the other cpm valuesobtained for vials 11 to 17. This is done to ac-count for the fact that some soluble 3H may nothave been removed from the insoluble material(precipitate) during the filtration step.

11. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 12.Explain your results in terms of the compo-nents that were present in these two reactions,and your knowledge of the mechanism of ac-tion of RNase A.

12. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 13.


Explain your results in terms of the compo-nents that were present in these two reactions,and your knowledge of the mechanism of ac-tion of RNase T1.

13. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 14.Explain your results in terms of the compo-nents that were present in these two reactions,and your knowledge of the mechanism of ac-tion of chloramphenicol.

14. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 15.Explain your results in terms of the compo-nents that were present in these two reactions,and your knowledge of the mechanism of ac-tion of cycloheximide.

15. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 16.Explain your results in terms of the compo-nents that were present in these two reactions,and your knowledge of the mechanism of ac-tion of puromycin.

16. Compare the relative level of 3H-poly-Phepresent in vial 11 compared to that in vial 17.Explain your results in terms how these two re-actions were treated following the addition of5% TCA.

Exercises

1. For each of the synthetic mRNAs described be-low, predict all the possible peptides that wouldbe produced in an in vitro translation reactionperformed under conditions of high Mg2� con-centration:

a. Poly(G)b. Poly(A)c. Poly(UC)d. Poly(UAC)

2. You have purified protein X. You believe thatit may interact with (bind) protein Y. You havenot purified protein Y, but you have the genethat encodes it cloned into the pSP72 plasmid

(see Experiment 22) in an orientation thatplaces it under the control of the T7 promoter.In addition, you have a monoclonal antibodythat has been raised against protein X. Describean experiment that you could perform, involv-ing both in vitro translation and one of the im-munochemical techniques described in SectionIV, that would allow you to determine quicklywhether protein X interacts with protein Y. As-sume that you also have a number of radiola-beled amino acids at your disposal.

3. Why is the fidelity of translation initiation invitro dependent on the concentration of Mg2�?To answer this, you must think about the in-teraction between the mRNA and the 30S ri-bosomal subunit.

4. Using your knowledge of the multiple steps in-volved in translation that require an input ofenergy, how many total molecules of GTP mustbe hydrolyzed to produce a peptide that is 10amino acids in length?

REFERENCES

Conway, T. W., and Lipmann, F. (1964). Characteriza-tion of Ribosome-linked GTPase in E. coli Extracts.Proc Natl Acad Sci USA 52:1462.

Gualerzi, C. O., and Pon, C. L. (1990). Initiation ofmRNA Translation in Prokaryotes. Biochemistry29:5881.

Laemmli, U. K. (1970). Cleavage of Structural Proteinsduring the Assembly of the Head of BacteriophageT4. Nature 227:680.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Protein Metabolism. In: Principles of Bio-chemistry, 2nd ed. New York: Worth.

Moldave, K. (1985). Eukaryotic Protein Synthesis.Annu Rev Biochem 54:1109.

Prescott, L. M., Harley, J. P., and Klein, D. A. (1993).Synthesis of Nucleic Acids and Proteins. In: Micro-biology, 2nd ed. Dubuque, IA: Wm. C. Brown.

Roberts, P. E., and Paterson, B. M. (1973). EfficientTranslation of TMV-RNA and Globin 9S RNA ina Cell-Free System from Commercial WheatGerm. Proc Natl Acad Sci USA 70:2330.

385

EXPERIMENT 24

Amplification of a DNA Fragment Using Polymerase Chain Reaction

Theory

Polymerase chain reaction (PCR) is a techniquethat allows the amplification of a specific fragmentof double-stranded DNA in a matter of hours. Thistechnique has revolutionized the use of molecularbiology in basic research, as well as in a clinical set-ting. PCR is carried out in a three-step process(Fig. 24-1). First, the template DNA that containsthe target DNA to be amplified is heated to dena-ture or “melt” the double-stranded DNA duplex.Second, the solution is cooled in the presence ofan excess of two single-stranded oligonucleotides(primers) that are complementary to the DNA se-quences flanking the target DNA. Since DNA syn-thesis always occurs in the 59 to 39 direction (read-ing the template strand 39 to 59), you must ensurethat the two primers are complementary to (willanneal to) opposite strands of the DNA duplex thatflank the region of target DNA that is to be am-plified. Third, a heat-stable DNA polymerase isadded, along with the four deoxyribonucleotidetriphosphates (dNTPs), so that two new DNAstrands that are identical to the template DNAstrands can be synthesized. If this melting, anneal-ing, and polymerization cycle is repeated, thefragment of double-stranded DNA located be-tween the primer sequences can be amplified overa millionfold in a matter of hours.

The heat-stable DNA polymerase (Taq) com-monly used in PCR reactions was isolated from athermophilic bacterium, Thermus aquaticus. Sincethis enzyme is heat-stable, it can withstand the hightemperatures required to denature the DNA tem-plate after each successive round of polymerization

and retain its activity. Since the development of thetechnique, biotechnology companies have devel-oped a number of improved and specialized poly-merase enzymes for use in PCR (Table 24-1). Manyof these polymerases are marketed as being moreprocessive and/or “accurate” than the traditionalTaq enzyme, since they display 39 to 59 exonuclease(proofreading) activity. The Taq DNA polymerasehas no proofreading activity, increasing the possi-bility of introducing point mutations (single basepair changes) in the amplified DNA product.

In this experiment, you will amplify a fragmentof pBluescript II (a plasmid), which includes the mul-tiple cloning site (MCS) of the vector (Fig. 24-2).The pBluescript II plasmid comes in the S/K formand the K/S form. These two plasmids are identicalexcept for the orientation of the MCS (see Fig. 24-2).Using restriction enzymes and agarose-gel elec-trophoresis, you will determine which of these twoplasmids was used as a template in the PCR reaction.The sequences of the two primers that will be usedin the PCR reaction are shown under “Supplies andReagents.” Primer 1 will anneal to positions 188 to211 (59 to 39) on one strand of the plasmid, whilePrimer 2 will anneal to positions 1730 to 1707 (59 to39) on the opposite strand of the plasmid (Fig. 24-3).On amplification, a 1543-base-pair fragment of DNAwill be produced that includes the multiple cloningsite of the plasmid. SstI (an isoschizomer of SacI) andKpnI will then be used to determine whether the S/Kor K/S form of the pBluescript II plasmid was usedas a template in the amplification reaction.

Although this experiment is designed to intro-duce you to the basic technique of PCR, youshould be aware that PCR can be used in a variety


5' 3' Template DNA

5' 3'

3' 5'

Step 1: DenaturationThe solution is heated to denature the double-stranded template DNA.

5' 3'3' 5' primer 1

3' 5'

Step 2: Primer annealingThe solution is cooled in the presence of a high concentration of single-stranded oligonucleotides (primers) that will bind to opposite strands of the template DNA.

3'primer 2 5'

5' 3'

3' 5'

5'

5' 3' 5'

Step 3: PolymerizationIn the presence of dNTPs, MgCl2, and the appropriate buffer, Taq polymerase will polymerize two new DNA strands that are identical to the template strands.

The cycle is repeated numerous times to achieve exponential amplification of the DNA sequence located between the two primers.

Figure 24-1 The basic principle underlying the technique of polymerase chain reaction (PCR).

Table 24-1 Some Commercially Available Polymerase Enzymes for Use

with Polymerase Chain Reaction

Polymerase Enzyme Relevant Features Company

Pfu Low error rate (proofreading) Stratagene

Produces blunt-ended PCR products

Taq 2000 High processivity (for long PCR products) Stratagene

Exo2 Pfu Specially designed for use with PCR sequencing Stratagene

AmpliTaq High polymerase temperature (minimizes false-priming) Perkin Elmer

UlTma Excellent proofreading activity Perkin Elmer

rTth High processivity (for PCR products 5–40 kb in length) Perkin Elmer

Platinum Taq Temperature activation of polymerase (minimizes false priming) Life Technologies

Vent Excellent proofreading activity New England Biolabs

easy cloning into a desired vector following PCR(Fig. 24-4a). In addition to introducing restrictionrecognition sequences, PCR can be used to add(Fig. 24-4b) or delete (Fig. 24-4c) small sequencesfrom a gene of interest. Provided that the DNAprimers are long enough to allow for sufficient basepairing on either side of the desired mutation,nearly any sequence can be added to, or deletedfrom, a gene of interest. PCR can also be used to

of different applications. One popular applicationfor PCR is its use in the introduction of specific mu-tations in the product DNA that is amplified fromthe template DNA. For example, suppose youwanted to produce a PCR product that had restriction-enzyme recognition sequences at eitherend. If these recognition sequences are present inthe single-stranded DNA primers, the target DNAwill be amplified to include these sites to allow for

387

Prim

er2

(1730–1707)

Primer 1 (188–211)

col E1 origin

pBluescript II S/K(2961 base pairs)

SC

MZc

alam

pR

KpnI (657)

SacI (759)

Prim

er2

(1730–1707)

Primer 1 (188–211)

col E1 origin

pBluescript II K/S(2961 base pairs)

SC

MZc

alam

pR

KpnI (759)

SacI (657)

Figure 24-2 Plasmid maps of pBluescriptII (S/K) and pBluescriptII (K/S).

388

gene and “walk” down the chromosome to obtainnew sequence information and identify new openreading frames (genes). It is even possible to obtainmRNA from a cell and use PCR to reverse transcribethe message to obtain the cDNA for a number ofdifferent genes (RT-PCR).

Since polymerase chain reaction can be used toamplify a specific DNA fragment in the presenceof countless other sequences, it has proven to be apowerful tool in a number of different fields ofstudy. One area that has benefited tremendouslyfrom this technique is the field of forensic science.Any biological sample recovered from a crime scenethat contains DNA (a single hair, a drop of blood,skin cell, saliva, etc.) can be subjected to PCR.Primers complementary to repetitive DNA se-quences found throughout the human genome are

change a single base pair in the gene of interest(Fig. 24-4d ). Studies of protein mutants producedby these methods have proven useful in studies of protein–protein interaction, protein function, andprotein structure. The thing to keep in mind whendesigning mutagenic primers is that any base pairchanges, deletions, or additions present in sequence ofthe primer, in contrast to that of the DNA template,will be present in the amplified DNA product.

Mutation analysis is one of many applications ofthe technique of PCR. Variations on the basic tech-nique described in this experiment will allow youto obtain the sequence of a DNA fragment from aslittle as 50 fmol of sample (PCR cycle sequencing).Once a gene has been cloned and localized to a par-ticular region on a chromosome, PCR can be usedto amplify DNA fragments on either side of the


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Sac I/Sst I

KpnI

primer 15´ 3´

primer 2 5´3´

pBluescript II (S/K)

Figure 24-3 Annealing of primers to pBluescript plasmids. Note that only the portion of the plasmids

between base pairs 1 and 1920 is shown.

analyses can provide a basis for a detailed study onthe process of evolution. Considering the enor-mous impact that the technique of PCR has hadover a wide range of fields in the past 10 years, itis not surprising that the Nobel Prize was awardedfor this discovery.


0.5-ml thin-walled PCR tubes



pBluescript II (S/K) plasmid in TE buffer (1 mg/ml)—Stratagene catalog #212205

used to produce a set of amplified DNA products.If these PCR fragments are subjected to Southernblotting (see introduction to Section V) after beingdigested with a number of different restriction en-zymes, the size and pattern of restriction fragmentsproduced provide a DNA “fingerprint” (Fig. 24-5).Traditionally, forensic science has relied only onblood chemistry and restriction mapping of wholechromosomal DNA, which often require larger bi-ological samples and are less discriminating.

Polymerase chain reaction has also proven tobe useful in the fields of archaeology and evolu-tion. Ancient biological samples recovered fromdigs and expeditions can be subjected to PCR togain insight into the genetic composition of ex-tinct species, including some of our earliest an-cestors. The information obtained from these

EXPERIMENT 24 Amplification of a DNA Fragment Using Polymerase Chain Reaction 389

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1

121

241

361

481

601

721

841

961

1081

1201

1321

1441

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Sac I/Sst I

KpnI

primer 15´ 3´

primer 2 5´3´

pBluescript II (S/K)

Figure 24-3 (continued )

390

pBluescript II (K/S) plasmid in TE buffer (1 mg/ml)—Stratagene catalog #212207

Sterile distilled water

TE buffer (10 mM Tris, pH 7.5, 1 mM EDTA)

12.5 mM MgCl2 in distilled water

Deoxyribonucleotide (dNTP) solution—1.25 mM eachof dATP, dCTP, dGTP, and dTTP in water

Primer 1—10 mM in distilled water (59-AAAGGGC-GAAAAACCGTCTATCAG-39)

Primer 2—10 mM in distilled water (59-TTTGCCG-GATCAAGAGCTACCAAC-39)

Taq polymerase (1 unit/ml)—Gibco BRL catalog #18038-042

10X Taq Polymerase buffer (supplied with enzyme)—Gibco BRL

PCR thermocycler

Agarose (electrophoresis grade)

5X TBE buffer (54 g/liter Tris base, 27.5 g/liter boricacid, 20 ml/liter of 0.5 M EDTA, pH 8.0)

6X agarose gel DNA sample buffer (0.25% (wt/vol)bromophenol blue, 0.25% (wt/vol) xylene cyanole,30% (wt/vol) glycerol in water)

1 kb-DNA ladder size marker—Gibco BRL catalog#15615-016

Casting box and apparatus for agarose-gel electrophore-sis

5' CTTAAACCGTC 3'CACGGCTAAC

3'500 base pairs

GAATTTGGCAG 5'GTGCCGATTG

CTTAAACCGTC 3'CACGGCTAAC

3'500 base pairs



3'500 base pairs

GAATTTGGCAG GTGCCGATTG

5'EcoRI recognition

site

Hind III recognition site

500 basesCTTAAACCGTC

CTTA

AGGGG–5'

CTTA

AGGGG–5'

5'–GACAAGCTT

5'–GACAAGCTT

3'CACGGCTAAC

3'–GTGCCGATTG

3'500 basesCTTAAACCGTC″3'


Denature template DNA and anneal primers that contain a specific restriction-enzyme recognition sequence at the 5' end

Polymerization

a

3

1

1

Figure 24-4 (a) Using PCR to introduce restriction sites on either end of the amplified DNA product. The

amplified DNA in subsequent cycles will include the EcoRI site and HindIII site at either end

for easy cloning into a desired vector. (b) Using PCR to add base pairs within a DNA frag-

ment of interest. The amplified DNA in subsequent cycles will include the five additional

base pairs specified by the mutagenic primer. (c) Using PCR to delete base pairs within a

DNA fragment of interest. The amplified DNA in subsequent cycles will be missing the five

base pairs not present in the mutagenic primer. (d ) Using PCR to change a single base pair

in the template DNA. The amplified DNA in subsequent cycles will include the single-base-

pair change encoded by the mutagenic primer.


3'500 base pairs


3'500 bases

GAATTTGGCAG GTGCCGATTG

5'500 bases


3'– GTGCCGATTG 5'

3'500 basesCTTAAACCGTC″3'

GAATTTGGCAG

G

GG

GG

5'5'

5'

GTGCCGATTG

Denature template DNA and anneal primers that contain additional bases. The primer will “loop out” as it anneals to the template.

Polymerization results in the following product DNA strands.

5'500 bases

1Added bases specifiedby the primer

CTTAAACGGGGGCGTC GACGGCTAAC 3'

Mutagenic primer


3'500 base pairs


3'500 bases

1GAATTTGGCAG GTGCCGATTG

5'500 bases


3'– GTGCCGATTG 5'

3'500 bases

CTTGTC

GAACAG

TC

TG

T

5'

5'

5'

GTGCCGATTG

Denature DNA template and anneal a primer that does not contain some of the bases in the template. The template will “loop out” as the primer anneals.


5'

b

c (TTTGC bases are deleted by the mutagenic primer)

<500 basesCTTGTC CACGGCTAAC 3'

3'

Mutagenic primer

1

1


392 5' CTTAAACCGTC 3'CACGGCTAAC

3'500 bases


3' GAATTTGGCAG GTGCCGATTG

5'500 bases


3'– GTGCCGATTG 5'

3'500 basesCTTA ACCGTC″3'

GAATTTGGCAG

C

5'5'

5'

GTGCCGATTG

Denature DNA template and anneal a primer that has a single mismatch in the template DNA sequence.


5'500 bases

1Base change encodedby the mutagenic primer

CTTACACCGTC CACGGCTAAC 3'

Mutagenic primer

d

1


Power supply

KpnI (10 units/ml) with 10X reaction buffer—Gibco BRLcatalog #15232-036

SstI (10 units/ml) with 10X reaction buffer—Gibco BRLcatalog #15222-037

0.5 mg/ml solution of ethidium bromide in 0.5X TBEbuffer

Light box with a 256-nm light source

Polaroid camera with film

Protocol

Day 1: Standard PCR Amplification Reaction

1. Obtain four thin-walled PCR tubes containingthe following:

Volume of Concentration

pBluescript II of Vol. of

Tube DNA pBluescript II Water

A — — 10 ml

B 10 ml 1 fg/ml —

C 10 ml 1 pg/ml —

D 10 ml 1 ng/ml —

Remember the metric units: f 5 femto 5 10215,p 5 pico 5 10212, n 5 nano 5 1029. Store thesetubes on ice while you prepare the Master Mixfor the PCR reaction.

2. Prepare the following Master Mix in a sterile,1.5-ml microcentrifuge tube (mix all compo-nents thoroughly by repeated pipetting up anddown):

Volume of

Component Stock Solution Stock Solution

MgCl2 68 ml 12.5 mM

dNTPs 68 ml 1.25 mM

(of each dNTP)

Primer 1 42.5 ml 10 mM

Primer 2 42.5 ml 10 mM

Taq buffer 42.5 ml 10X

Sterile water 110.5 ml —

Taq polymerase 8.5 ml 1 unit/ml

3. Add 90 ml of this Master Mix to each of thefour PCR reaction tubes prepared in step 1 (A–D) and mix thoroughly.


Biological samplecontaining DNA

(hair, blood, etc.)

Extract chromosomalDNA from sample.

Perform PCR with primers that are complementary with

highly repetitive sequences found throughout the genome.

Amplified DNA products that vary in size and sequence from individual to individual.

Southern blot1) Transfer to

nitrocellulose2) Probe with a

radiolabeled DNA fragment

3) Expose to film

Film—radiolabeled DNA fragment will hybridize with complementary sequences in the sample DNA. The number of repetitive sequences in the chromosomal DNA and the position of restriction sites will not be exactly the same in two individuals.

It is apparent that the DNA sample found at the crime scene is from Subject 2. The different pattern of DNA fragments producted after digestion (restriction-fragment–length polymorphisms, RFLPs) ensures that innocent subjects will not be convicted of crimes that they did not commit.

Digest productDNA with restriction endonucleases and

separate by agarose-gel electrophoresis.

Do the same with DNA from a subject.

Sizestandards Sample

DNA fromSubject 1

DNA fromSubject 2

Sizestandards Sample

DNA fromSubject 1

DNA fromSubject 2

Figure 24-5 Use of PCR in forensic science.

4. Overlay each sample in tubes A to D with 40 mlof mineral oil. Some thermocyclers are equippedwith heated lids that make the mineral oil unnec-essary. The mineral oil will keep the reaction vol-ume at the bottom of the tube as the solution is re-

peatedly heated and cooled during the reaction. Ifthis were not done, the solution would begin to con-dense on the walls of the tube and possibly alter theconcentrations of the different components in the reaction.

394

3. Add 6 ml of 6X DNA sample buffer to each ofthese two reactions following the 1-hr incuba-tion. Label these tubes 7 (KpnI digest) and 8(SstI digest).

4. Set up six additional samples for agarose-gelelectrophoresis, each containing the following:

Volume of DNA Volume of 6X DNA

Tube Sample (Undigested) Sample Buffer

1 5 ml (PCR reaction A) 1 ml

2 5 ml (PCR reaction B) 1 ml

3 20 ml (PCR reaction B) 4 ml

4 5 ml (PCR reaction C) 1 ml

5 5 ml (PCR reaction D) 1 ml

6 5 ml of 0.2 mg/ml 1 ml

1-kb DNA ladder

5. Prepare a 10-well, 1% TBE agarose gel byadding 0.5 g of agarose to 50 ml of 0.5X TBEbuffer in a 250 ml Erlenmeyer flask (the 0.5XTBE buffer is prepared by diluting the 5X TBEbuffer stock 1:10 with distilled water). Weighthe flask and record its mass. Microwave theflask until all of the agarose is completely dis-solved (,2 min). Weigh the flask again and adddistilled water until the flask reaches itspreweighed mass. This is done to account forthe fact that some water is lost to evaporationduring the heating process. If this water is notadded back to the flask, the gel may not actu-ally be 1% agarose.

6. Allow the flask to cool at room temperature(swirling occasionally) for ,5 min. Pour the so-lution into a gel cast fitted with a 10-well combat one end. Allow the agarose to set (,40 min).

7. Carefully remove the comb, remove the gelfrom the cast, and place it in the agarose-gelelectrophoresis unit. Add 0.5X TBE buffer tothe chamber until the buffer covers the top ofthe gel and fills all of the wells.

8. Load the DNA samples in order (1–8) into sep-arate wells of the gel. Connect the negativeelectrode (cathode) to the well side of the unitand the positive electrode (anode) to the otherside of the unit (the side opposite the wells inthe gel). Remember that the DNA is negativelycharged and will migrate toward the positiveelectrode.

5. Label the four tubes with your name and placethem in the thermocycler set to room temper-ature.

6. Perform 30 PCR cycles using the following pa-rameters:

a. Ramp from room temperature to 94°C in50 sec and hold at 94°C for 1 min (denatu-ration).

b. Ramp down to 55°C in 40 sec and hold at55°C for 1 min (primer annealing).

c. Ramp up to 72°C in 10 sec and hold at 72°Cfor 2 min (polymerization).

d. Ramp from 72 to 94°C in 15 sec (denatura-tion).

The 30-cycle program will take approximately2.5 hr. After completion of the 30 cycles, a“soak file” should be included to hold the sam-ples at 4°C until they are ready to be removed.

7. Remove the samples from the thermocyclerand store them at 220°C for use on Day 2.

Day 2: Restriction-Enzyme Analysis of

Amplified DNA Product

1. Place the samples obtained at the end of Day1 at room temperature and allow them to thaw.Remove the mineral oil from the four PCR re-actions by poking the tip of the P-200 Pipet-man down through the mineral oil to the bot-tom of each tube. Draw up the bottom(aqueous) phase into the tip up to the level ofthe mineral oil. Remove the tip from the tubeand wipe the mineral oil off of the outside ofthe tip using laboratory tissue paper. Expel eachaqueous PCR reaction (A–D) into a new, ster-ile 1.5-ml microcentrifuge tube.

2. Prepare a sample of the PCR reaction in tubeD for restriction digest in two separate 1.5-mlmicrocentrifuge tubes as follows:

KpnI Digest SstI Digest

5 ml of PCR reaction 5 ml of PCR reactionfrom tube D from tube D

16.5 ml of sterile water 16.5 ml of sterile water

2.5 ml of 10X KpnI buffer 2.5 ml of 10X SstI buffer

1 ml of KpnI (10 units/ml) 1 ml of SstI (10 units/ml)

Incubate both reactions at 37°C for 1 hr.


actions B to D, estimate the total amount ofproduct produced in each PCR reaction. Hint:the 1636 bp DNA band in the 1-kb DNA ladderlane contains approximately 100 ng of DNA.Therefore, if the PCR product band is of the sameintensity as this marker, then the PCR product bandalso contains about 100 ng of DNA.

16. Based on the size of the DNA fragments pro-duced from the KpnI and SstI digests, deter-mine whether you amplified a portion of thepBluescript II (S/K) plasmid or the pBluescriptII (K/S) plasmid. Justify your answer in termsof what size DNA fragments were produced ineach digest.

17. Determine whether there were any “unex-pected” DNA bands in the lanes containingsamples of PCR reactions B to D. If there were,explain what you think may have caused these,as well as how you might alter the conditionsof the PCR reaction to eliminate them.

18. Determine the number of molecules (N0) of the2961-base-pair pBluescript II plasmid that werepresent in each of PCR reactions B to D. Basedon your estimation of the mass of DNA prod-uct present in each of these PCR reactions(step 15) and the molecular weight of the prod-uct (320 Da/nucleotide, or 640 Da per basepair), calculate the number of molecules ofDNA product (N) that were produced in eachPCR reaction. From these two pieces of infor-mation, estimate the amplification efficiency(E) for each PCR reaction using the followingequation:

N 5 N0 (1 1 E)n

where n 5 the number of amplification cycles(30). The amplification efficiency (E) can havea value of between 0 and 1.0. Zero representsno amplification and 1.0 represents 100% am-plification efficiency. Which of the three PCRreactions (B, C, or D) showed the highest andthe lowest amplification efficiency? Explainwhy you think the amplification efficiencymight vary with respect to template DNA con-centration.

19. Calculate the mass of DNA product that wouldbe present in PCR reaction tubes B to D if theamplification efficiency were 100% (E 5 1.0).

9. Apply 50 mA (constant current) and continuethe electrophoresis until the bromophenol blue(dark blue) tracking dye has migrated aboutthree-fourths of the way to the end of the gel.

10. Turn off the power supply, disconnect the elec-trodes, and place the gel in a solution of 0.5XTBE containing 0.5 mg/ml ethidium bromide.Wear gloves at all times when handling ethidiumbromide. Incubate the gel at room temperaturefor 20 min. During this time, the ethidium bro-mide will enter the gel and intercalate betweenthe base pairs in the DNA. The DNA will ap-pear as pink or orange bands on the gel whenexposed to ultraviolet light, as the ethidiumbromide–DNA complex fluoresces.

11. Destain the gel by placing it in a solution of0.5X TBE buffer without ethidium bromide.Incubate at room temperature for 10 min. Thisis done to remove any ethidium bromide fromareas of the gel that do not contain DNA.

12. Place the gel in a light box fitted with a cam-era, appropriate filter, and a 254-nm lightsource. Take a photograph of your gel for lateranalysis.

13. Prepare a plot of number of base pairs versusdistance traveled (in centimeters) for as manyDNA fragments present in the 1-kb DNA lad-der lane as possible. If the 1-kb DNA laddersize standards resolved well, you should be ableto differentiate accurately between the relativemobilities of the 0.5-, 1.0-, 1.6-, 2.0-, and 3.0-kb fragments for use in preparing the standardcurve (do not try to calculate the relative mobilitiesof the higher-molecular-weight DNA size standardfragments if they did not resolve well [see Exper-iment 21, Day 4, step 16]). Also, refer to Fig-ure 21-5 for the sizes of the DNA fragmentspresent in the 1-kb DNA ladder.

14. Using the standard curve prepared in step 13,calculate the number of base pairs present inall of the other sample lanes on the gel. Whatis the size of the DNA fragment produced inPCR reactions A to D? Is there any DNA pres-ent in the lane containing a sample of PCR re-action A? Explain. Which of the PCR reactions(B, C, or D) produced the most and the leastamount of product? Explain.

15. Based on the intensities of the DNA bands pres-ent in the lanes containing samples of PCR re-


396

Design two degenerate primers (21 nucleotidesin length) that could be used to try to amplifythe homologous gene from the bacterial chro-mosome. You will need a chart of the geneticcode to design this primer pair. The degener-acy of the primer arises due to the fact thatthere is “wobble” in the genetic code at thethird position for many amino acids. Calculatethe degeneracy of each primer that you havedesigned. This can be done simply by multi-plying the number of possible codons for eachamino acid over the length of the amino acidsequence that the primer spans. For example,glycine has four possible codons and phenyl-alanine has two possible codons. Therefore, thedegeneracy of a primer spanning these twoamino acids is 8. Also, calculate the molecularweight and the number of base pairs that youmay expect if the PCR reaction is successful.

REFERENCES

Arnheim, M., and Erlich, H. (1992). Polymerase ChainReaction Strategy. Annu Rev Biochem 61:131.

Erlich, H. A., Gelfand, D., and Sninsky, J. J. (1991).Recent Advances in Polymerase Chain Reaction.Science 252:1643.

Lehninger, A. L., Nelson, D. L., and Cox, M. M.(1993). Recombinant DNA Technology. In: Princi-ples of Biochemistry, 2nd ed. New York: Worth.

Saiki, R. K., Gelfand, D. H., Stoeffel, B., Scharf, S. J.,Higuchi, R., Horn, G. T., Mullis, K. B., and Erlich,H. A. (1988). Primer Directed Amplification ofDNA with a Thermostable DNA Polymerase. Science 239:487.

Sambrook, J., Fritsch, E. F., and Maniatis, T. (1989).Molecular Cloning, Plainview, NY: Cold Spring Harbor Laboratory Press.

Exercises

1. The maximum number of molecules of prod-uct produced (assuming E 5 1.0) is often notpossible due to the fact that the amount ofprimers and/or dNTPs may be limiting. Basedon the number of primers present in reactionsB to D, calculate the maximum number of mi-crograms of product that could be produced.Based on the number of dNTPs present in re-actions B to D, calculate the maximum num-ber of micrograms of product that could beproduced. Which of the two components,dNTPs or primers, would ultimately limit theamount of product that can be produced inthese reactions?

2. Calculate the melting temperature (Tm) of the1543-base-pair product and the 2961-base-pairpBluescript II plasmid used in this experimentwith the following formula:

Tm 5 81.5°C 1 16.6 log [Na1] 1 (0.41)(% GC) 2(675/number of base pairs) 2 (% formamide) 2

(number of mismatched base pairs)

The [Na1] in these reactions is 0.05 M, the %GC of the product is 56.2, and the % GC ofthe plasmid is 50.4.

3. You wish to amplify a portion of a bacterialgene that has homologs both in yeast and mice.This gene has already been cloned from bothof these organisms, and it has been found thattwo amino acid sequences at the amino and car-boxyl termini are conserved in both proteins:

mouse: NH2 O LKVAPWYVDGSE O (105 amino

acids) OLFGLCTANDHKVQ O COOH

yeast: NH2 O PRYAPWYVDGTC O (105 amino

acids) O GRILCTANDHGRN O COOH


T R

399

In the past several decades, research scientists havecloned and sequenced hundreds of thousands ofgenes from a variety of organisms. More recently,the scientific community has undertaken the mas-sive project of sequencing entire genomes from aselected set of model organisms (including humans)that are most often used in studies of biochemistry,molecular biology, and genetics. The overwhelm-ing amount of DNA and protein sequence infor-mation generated by these projects necessitated thedevelopment of biological databases that could or-ganize, store, and make the information accessibleto research scientists around the world. There hasalso been a striking increase in the numbers of high-resolution structures of biological macromoleculesdetermined by x-ray diffraction studies or by nu-clear magnetic resonance (NMR) spectroscopy.This rich database of important biological infor-mation must also be made readily available to theresearch community. This need has led to thegrowth of a new area of research commonly re-ferred to as information science or bioinformatics. In acooperative effort, research scientists and computerscientists have established a large number of bio-logical databases, most of which are accessible viathe World Wide Web.How do you find the different biological data-

bases on the Internet? In an effort to publicize thenumerous sites that are being maintained, severalInternet sites have been created to act as directo-ries. Two of these database directories that we havefound to be quite useful are the Biology Workbench(http://biology.ncsa.uiuc.edu) and Pedro’s BioMolec-ular Research Tools (http://www1.iastate.edu/�pe-dro/research_tools.html).

The Biology Workbench is a site that has beendeveloped by the Computational Biology Groupat the National Center for Supercomputing Ap-plications (NCSA) at the University of Illinois. Pedro’s BioMolecular Research Tools is a similardirectory maintained by a group at Iowa State.Both of these sites serve as search engines for anumber of general and specialized DNA and pro-tein databases, some of which are described below.Many of these databases are directly accessiblethrough the Biology Workbench. A more recentdirectory site has been developed by ChristopherM. Smith of the San Diego Supercomputer Cen-ter. The CMS Molecular Biology Resources site(http://www.sdsc.edu/ResTools/cmshp.html) listsnearly 2000 biological Web sites. Unlike the othertwo directories, this one lists sites according to thedesired application. For instance, if you wish toanalyze the coding region(s) within a fragment ofDNA, it will list sites that will aid you. If you wishto perform a phylogenetic analysis on a set of re-lated genes, it will list sites that will aid you.

GenBank

GenBank is the National Institutes of Health se-quence database. It is the oldest, largest, and mostcomplete general database. Together, GenBankand the EMBL database (see below) comprise theInternational Nucleotide Sequence Database Collabo-ration. Newly discovered genes must be submittedto either GenBank or EMBL before they can bepublished in a research journal. Gene sequencescan be located on GenBank either by submittingthe name of the gene or the GenBank accession

SECTION VI

Information Science

Introduction

400

cular Biology Laboratory, which specializes in thearea of bioinformatics. Since EMBL is constantlyexchanging information with GenBank, it too isquite large and current. The EMBL database hasnumerous specialized databases that draw informa-tion from it (see Table VI-1).

ExPASy and ISREC

ExPASy is a database maintained by the Swiss Insti-tute of Bioinformatics (SIB) that is dedicated to theorganization of protein sequences. ISREC is a sim-ilar database maintained by the Bioinformatics Group

number that is assigned to the gene. Often, asearch performed on the basis of the gene namewill identify several homologs of the gene thathave been identified in different organisms. Gen-Bank is most easily accessed from an Internet sitemaintained by the National Center for BiotechnologyInformation (http://www.ncbi.nlm.nih.gov/).

European Molecular Biology

Laboratory (EMBL)

The EMBL database is maintained by the HinxtonOutstation (Great Britain) of the European Mole-

SECTION VI Information Science

Table VI-1 An Incomplete Listing of Biological Databases and Other Useful Sites on the World Wide

Web

Database Source Area of Specialization

3D-ALI EMBL Protein structure based on amino acid sequence

AA Analysis EMBL Protein identification based on amino acid sequence

PIR

AA Compldent SwissProt Protein identification based on amino acid sequence

ExPASy

AbCheck Kabat Antibody sequences

ALIGN EERIE Sequence alignments

MIPS

AIIAII SwissProt Protein sequence alignments

ASC EMBL Protein analytical surface calculations

ATLAS MIPS General DNA and protein searches

BERLIN CAOS/ RNA databank (5S rRNA sequences)

CAMM

BLAST GenBank Alignment of sequences

EERIE

BLOCKS EMBL Conserved sequences in proteins

BioMagResBank BIMAS Protein structure determined by NMR

BMCD CARB Crystallization of macromolecules

Coils ISREC Protein structure (coiled coil regions)

DrugBank NIH Structure of different drugs

DSSP EMBL Protein secondary structure

EMP Pathways NIH Metabolic pathways

ENZYME ExPASy Enzymes

EPD EMBL Eukaryotic promoters

FlyBase Harvard Sequences identified in Drosophila

FSSP EMBL Groups proteins into families based on structure

Gene Finder Baylor Introns, exons, and RNA splice sites

GuessProt SwissProt Identify proteins by isoelectric point and molecular weight

ExPASy

HOVERGEN CAOS/ Homologous genes identified in different vertebrates

CAMM

Kabat Kabat Proteins related to immunology

Specialized Biological Databases

As science has become increasingly specialized overthe past two decades, so too has the area of bioin-formatics. There are numerous specialized data-bases that deal with different aspects of biology.Some of these specialize in the area of proteins,while others specialize in the area of DNA. In gen-eral, these sites differ in two respects: the generaldatabase(s) from which they draw information, andthe parameters that they require the user to defineto conduct the search. For instance, the AA Analy-sis database searches the more general SwissProtdatabase to identify or group proteins on the basis

at the Swiss Institute for Experimental Cancer Research.As shown in Table VI-1, a number of specializeddatabases draw information from these two data-bases.

SwissProt

SwissProt is a computational biology database spe-cializing in protein sequence analysis maintained bythe Swiss Federal Institute of Technology (ETH) inZurich, Switzerland. Like the other general data-bases described above, a number of more special-ized biological databases draw information fromthis source.

SECTION VI Introduction 401

LIGAND GenomeNet Ligands (chemicals) for different enzymes

MassSearch EMBL Identifies proteins on the basis of mass

SwissProt

MGD Jackson Mouse genome sequences

Laboratory

MHCPEP WEHI Binding peptides for major histocompatibility complex

MPBD EMBL Molecular probes

NRSub National Institute of Sequences from the genome of Bacillus subtilis

Genetics

(Japan)

NuclPepSearch SwissProt General DNA searches

NYC-MASS Rockefeller Protein mass spectrometry

PHD EMBL Protein secondary structure

Phospepsort EERIE Phosphorylated peptides

pI/Mw ExPASy Calculation of isoelectric point and molecular weight

PMD EMBL Mutant proteins

GenomeNet

ProDom EMBL Protein domains

ProfileScan ExPASy Searches based on sequence profile

PROPSEARCH EMBL Protein homolgy based on amino acid composition

PROSITE ExPASy Protein sites and patterns

PSORT GenomeNet Prediction of protein sorting signals based on amino acid composition

RDP University of Illinois Ribosomal proteins

ReBase EMBL Restriction enzymes

REPRO EMBL Repeated sequences in proteins

SAPS ISREC Protein sequence statistical analysis

SSPRED EMBL Predict protein secondary structure

Swiss-2DPage ExPASy Two-dimensional PAGE

TFD NIH Transcription factors

TFSITE EMBL Transcription factors

TMAP EMBL Transmembrane regions in proteins

TMpred ISREC Transmembrane regions in proteins

VecBase CAOS/ Sequence of cloning vectors

CAMM

402

EDITSEQ

The EDITSEQ application allows you to manuallyenter DNA or protein sequence information intoyour computer. This application has several featuresthat make it useful to the research scientist. First, itcan identify open reading frames (possible gene se-quences) within a DNA sequence. Second, it canprovide the percent base composition (A,G,C,T),the percent GC, the percent AT, and the meltingtemperature of the entire sequence or a small sub-set of that sequence. Third, EDITSEQ can trans-late a nucleotide sequence into a protein sequence.Finally, the application is capable of translating orreverse translating a nucleotide sequence of interestusing codes other than the standard genetic code.

GENEMAN

The GENEMAN application is a tool that allowsyou to access and search for DNA and protein se-quences located in six different biological databases.The search for a sequence of interest can be madeas broad or restrictive as desired, since there are 12different “fields” (definition, reference, source, ac-cession number, etc.) to choose from when the searchis performed. In addition to performing databasesearches to find sequences of interest, GENEMANallows you to search the database for sequences thatshare homology with the sequence of interest, or forentries that contain a particular conserved sequence.Any number of different DNA or protein sequencesfound in these databases can be isolated and storedas a sequence file for later analysis.

MAPDRAW

The MAPDRAW application provides a detailed re-striction map of the DNA sequence of interest,whether it has been entered manually or importedfrom a database. Since MAPDRAW is able to iden-tify 478 different restriction endonuclease recogni-tion sequences, you may wish to simplify your re-striction map by applying selective filters. Thesefilters will identify restriction endonuclease sitesspecifically on the basis of name, the 5� or 3� single-stranded overhangs that they produce, the frequency(number) of sites contained within the sequence,and/or the restriction endonuclease class (type I ortype II, see Section V ). In addition, a detailed de-

of their amino acid composition. If you were to iso-late a protein and determine its amino acid com-position, you could enter the data into the AAAnalysis database in an attempt to identify it. Keepin mind that new databases are constantly being de-veloped. Table VI-I presents an entensive, but in-complete, list of specialized databases.

Computer Software Programs for Analyzing Sequences

Although biological databases contain a great dealof sequence information, they are very limited intheir ability to analyze DNA and protein sequences.Fortunately, computer software is available that willallow a researcher to import a sequence of interestfrom a database and subject it to a variety of dif-ferent analytical applications. Of the many softwareprograms currently on the market, we find Laser-gene by DNASTAR, Inc. (Madison, WI) to be themost comprehensive and easy to use. This softwareis available in both PC and Macintosh formats.Lasergene is under copyright by DNASTAR, Inc.It is illegal to install the software on a new computerwithout purchasing it. If you would like to use Laser-gene strictly for teaching purposes at no charge to yourinstitution, you may contact DNASTAR (1–608–258–7420) to establish an educational license agree-ment. DNASTAR is very helpful in the installationof the software, and they provide a wealth of liter-ature for students to help them understand the op-eration of the various applications contained in thesoftware. A brief description of the various appli-cations in Lasergene is found below. These de-scriptions are intended to inform you of the capa-bilities of the Lasergene software. A more detaileddescription of the software is available in the “User’sGuide” supplied by DNASTAR. A less detailed butadequate description is provided in the “Installing,Updating, and Getting Started” manual also pro-vided by DNASTAR.DNASTAR is one of many companies that sup-

plies computer software useful in the field of bioin-formatics. Other popular computer software pack-ages are offered by DNAStrider, the GeneticComputer Group, and GeneJockey II. If you haveexperience with these software packages, you maycontact the company to establish an educational li-cense agreement.


stringency of eight different criteria to which thesearch will adhere (primer length, number of basesin the primers capable of forming inter- or in-tramolecular hydrogen bonds with neighboringbases, etc.). As the sets of primer pairs are selected,they are presented along with thermodynamic andstatistical data that you may find useful. In additionto altering the parameters for primer selection,PRIMERSELECT also allows you to alter variousconditions for the proposed PCR reaction, includ-ing primer concentration and ionic strength.

PROTEAN

The PROTEAN application provides a great dealof general information about protein sequences, en-tered manually or imported into the program froma database. A single report provides the protein’smolecular weight, amino acid composition, extinc-tion coefficient, isoelectric point, and theoreticaltitration curve. The application also provides alarge number of different secondary structure pre-dictions for the protein. The Garnier–Robson andChou–Fasman algorithms predict alpha helices, betasheets, and turn regions based on the linear aminoacid sequence of the protein. The algorithm of Kyteand Doolittle identifies hydrophobic stretches ofamino acids, indicating possible transmembrane re-gions in the protein. An antigenic index of the pro-tein calculated by the Jameson–Wolf method indi-cates possible antigenic peptides that could be usedto raise antibodies specific for that protein. TheEmini algorithm predicts amino acid residues thatare likely to reside on the surface of the native pro-tein, and so forth.Two Lasergene applications that are not ex-

plored in Experiment 25 are SEQMAN II andGENEQUEST. SEQMAN II is capable of align-ing over 64,000 individual DNA sequences into asingle, continuous sequence (a contig). Since it iscapable of accepting information generated inEDITSEQ or derived from selected automatedDNA sequencers, it is suited to both large- andsmall-scale sequencing projects. GENEQUEST isdesigned to greatly simplify the identification ofspecific genetic elements within a DNA sequence,such as open reading frames, transcription startsites, transcription stop sites, translation start sites,translation stop sites, binding sites for transcrip-tion factors, and so forth.

scription of any restriction endonuclease can be ob-tained with this application simply by “clicking” onthe enzyme of interest. MAPDRAW will display theamino acid sequence of the double-stranded DNAin all six potential open reading frames (three forthe top strand, and three for the bottom strand).This feature enhances the researcher’s ability toidentify the correct reading frame within a largeDNA sequence.

MEGALIGN

The MEGALIGN application is a comprehensivetool for establishing relationships between differentDNA and protein sequences. If multiple sequencesare to be compared, the Clustal and the Jotun Heinalgorithms are at your disposal. If only two se-quences are to be compared, the Wilbur–Lipman,Martinez–Needleman–Wunsch, and Dotplot algo-rithms can be applied. The results of these align-ments can be viewed or displayed in a variety of for-mats with MEGALIGN. First, the alignment canbe displayed as a consensus sequence that the “ma-jority” (a parameter that can be defined by the op-erator) of the sequences in the alignment contain.Second, you may wish to view the alignment in atabular format showing the percent similarity andthe percent divergence among the sequences in theset. Finally, you have the option of viewing thealignment in the form of a phylogenetic tree, indi-cating how closely related two sequences are. Youcan also examine DNA or protein sequences thatmay have evolved from a common ancestor.

PRIMERSELECT

The PRIMERSELECT module is an extremelyvaluable tool for designing oligonucleotides formolecular biology applications, including poly-merase chain reaction (PCR), DNA sequencing,and Southern blotting. The application assists inidentifying suitable regions within a DNA templatesequence to which an oligonucleotide with a com-plementary sequence will hybridize with a high de-gree of specificity. The PRIMERSELECT appli-cation searches for primer pairs on the templatebased on the PCR product length, the upper primerrange, the lower primer range, or combinations ofthe three. In addition to these parameters, you canrestrict the search for primers by increasing the

SECTION VI Introduction 403

404

REFERENCES

An, J., Nakama, T., Kubota, Y., and Sarai, A. (1998).3DinSight: An Integrated Relational Database andSearch Tool for the Structure, Function and Prop-erties of Biomolecules. Bioinformatics 14:188.

Barlow, D. J., and Perkins, T. D. (1990). Applicationsof Interactive Computer Graphics in Analysis ofBiomolecular Structures. Nat Prod Rep 7:311.

Brazma, A., Jonassen, I., Eidhammer, I., and Gilbert,D. (1998). Approaches to the Automatic Discoveryof Patterns in Biosequences. J Comput Biol 5:279.

Clewley, J. P. (1995). Macintosh Sequence AnalysisSoftware. DNAStar’s LaserGene. Mol Biotechnol3:221.

Doolittle, R. F. (1996). Computer Methods for Macro-molecular Sequence Analysis. Methods Enzymology,266.

Froimowitz, M. (1993). HyperChem: A Software Pack-age for Computational Chemistry and MolecularModeling. Biotechniques 14:1010.

Gasterland, T. (1998). Structural Genomics: Bioinfor-matics in the Driver’s Seat. Nat Biotechnol 16:625.

Hellinga, H. W. (1998). Computational Protein Engi-neering. Nat Struct Biol 5:525.

Kanehisa, M. (1998). Grand Challenges in Bioinfor-matics. Bioinformatics 14:309.

Moszer, I. (1998). The Complete Genome of Bacillussubtilis: From Sequence Annotation to DataArrangement and Analysis. FEBS Lett 430:28.

Plasterer, T. N. (1997). MAPDRAW: Restriction Map-ping and Analysis. Methods Mol Biol 70:241.

Plasterer, T. N. (1997). PRIMERSELECT: Primer andProbe Design. Methods Mol Biol 70:291.

Plasterer, T. N. (1997). PROTEAN: Protein SequenceAnalysis and Prediction. Methods Mol Biol 70:227.

Sanchez-Ferrer, A., Nunez-Delicado, E., and Bru, R.(1995). Software for Retrieving Biomolecules inThree Dimensions on the Internet. Trends BiochemSci 20:286.

Sayle, R. A., and Milner-White, E. J. (1995). RAS-MOL. Biomolecular Graphics for All. TrendsBiochem Sci 20:374.

Smith, T. F. (1998). Functional Genomic—Bioinfor-matics Is Ready for the Challenge. Trends Genet7:291.

The Protein Data Bank (PDB)and RasMol

The Protein Data Bank is a unique database inthat it specializes in the three-dimensional struc-tures of proteins and other biomolecules. It ismaintained by the Brookhaven National Laboratory(http://www.pdb.bnl.gov/ ). (Note that this data-base will move to Rutgers University in the fu-ture.) This database allows you to visualize a pro-tein in three dimensions, provided that atomiccoordinate information for it is available fromcrystallography or NMR studies. Once you haveaccessed PDB at the above address, you will selectSoftware and Related Information on the home page.Here, you will find a free molecular visualizationprogram called RasMol (http://www.umass.edu/microbio/rasmol/rasquick.htm). After you haveaccessed the RasMol home page, you will find adetailed description of how to install the RasMolsoftware on your computer (select Getting and In-stalling RasMol on the home page). At this point,you are ready to search the PBD for a three-dimensional view of a protein or other moleculeof interest.Once the coordinates of a molecule have been

imported into RasMol, there are a number of ana-lytical tools that can be used to visualize the manydifferent interactions taking place within it. For in-stance, RasMol allows you to determine how manypeptide chains there are in the protein, the posi-tions of particular atoms within the protein, the po-sitions of different ions and hydrophobic aminoacids, alpha helices and beta sheets, the distance be-tween two particular atoms or residues in the pro-tein, points of contact between the protein and itsligands, the amino terminal and carboxy terminalamino acids, inter- and intramolecular disulfidebonds within the protein, and hydrogen bonds pres-ent between different atoms in the protein.In addition, RasMol has the capability to rotate

the molecule on its X, Y, and Z axes, giving a truethree-dimensional view of the molecule. If you areinterested in a quick tutorial session, you can go to http://www.umass.edu/microbio/rasmol/rastut.htm. If you are interested in how to investigate the different interactions listed above, you can go to http://www.umass.edu/microbio/rasmol/raswhat.htm.


. N

405

1. Using the Biology Workbench (http://biol-ogy.ncsa.uiuc.edu) or the GENEMAN applica-tion of Lasergene, search GenBank for a geneof interest (enter the name of the gene or pro-tein for the query).

2. What are some of the “fields” that you can useto restrict your search of the GenBank data-base? How many entries did you obtain in yoursearch? Explain why you may have obtainedseveral entries in response to your search, aswell as what they represent.

3. Select one of these entries and save it as a Se-quence file in your computer for later analysis.

4. Using the EDITSEQ application of Lasergene,obtain some statistical information about theDNA sequence that you have selected, searchfor open reading frames (ORFs) within the se-quence, and translate the ORFs into protein se-quences.

a. Open your sequence file (saved in step 3) byselecting Open from the File menu.

b. Use Command-A to select the entire DNAsequence and select Find ORF from theSearch menu. If there is more than oneORF present in your DNA sequence, theFind ORF function can be repeated severaltimes to identify all of them.

c. Click on the largest ORF in your DNA se-quence to highlight it, then select Trans-

late DNA from the Goodies menu. At thispoint, you should see a new window con-taining the amino acid sequence encoded bythe DNA.

d. To obtain some statistical information aboutyour DNA sequence, use Command-A to

EXPERIMENT 25

Obtaining and Analyzing Genetic andProtein Sequence Information via theWorld Wide Web, Lasergene, and RasMol

This computer exercise is designed to introduce youto the wealth of DNA and protein sequence infor-mation available on the World Wide Web. Usingthe Biology Workbench and Lasergene software(DNASTAR, Madison, WI), you will search Gen-Bank for a particular sequence of interest and studyit in depth. We use glutathione-S-transferase, MAPkinase, EcoRI, and CheY for this exercise. We en-courage you to experiment with different genes forthis exercise, perhaps one that is of interest to youin your own research. The only requirement is thatyou must use a gene that encodes a protein forwhich the atomic coordinates are known (consultthe instructor before the beginning of the experi-ment if you are not sure).


Lasergene software (contact DNASTAR, Inc. in Madi-son, WI)

200-Mhz computer with access to the Internet

Computer printer

Protocol

The following protocol explains the application

of the Lasergene and RasMol software in the

Macintosh format. If you are using IBM comput-

ers, this protocol must be modified for use in

the IBM format. A Lasergene “User’s Guide”

can be obtained from DNASTAR, Inc. RasMol

software compatible for IBM computers can be

installed after accessing the RasMol home

page.

406 SECTION VI Information Science

highlight the entire sequence and selectDNA statistics from the Goodies menu.What is the %A, %G, %T, and %C in yoursequence? What is the %A-T and %G-Ccomposition of your DNA sequence? Whatis the melting temperature of this DNA sequence as determined by the Davis–Botstein–Roth method?

e. Print out the report of the DNA statisticsand select Quit from the EDITSEQ File

menu.

5. Using the MAPDRAW application of Laser-gene, generate a detailed restriction map ofyour DNA sequence.

a. To open your DNA sequence, select Open

from the File menu. At this point, yourDNA sequence (double-stranded) will beshowing with the positions of all of the po-tential 478 restriction endonuclease recog-nition sequences that it may contain. Belowthis, you will see the amino acid sequencespecified by your DNA sequence in all sixreading frames (three for the top strand andthree for the bottom strand).

b. Select Unique sites from the Map menuto show the position of all of the restric-tion endonuclease sites that appear onlyonce within your DNA sequence. Print outthis report, which you will need for a laterexercise.

c. Select Absent sites from the Map menu toobtain a list of restriction endonucleases thatwill not cleave within your DNA sequence(the recognition sequence for these enzymesare not present within your DNA sequence).Print out this report, which you will needfor a later exercise.

d. Experiment with some of the different re-striction enzyme “filters” available in theMAPDRAW application (apply the New

Filter option under the Enzyme menu).When you feel that your restriction map iscomplete and easy to read, print out the re-port displayed in the window for use in alater exercise.

e. To view all of the potential ORFs in yourDNA sequence, select ORF map from theMap menu. You will see the ORFs encodedby the top strand displayed in red and the

ORFs encoded by the bottom strand dis-played in green. You will likely see a “clus-ter” of ORFs specified by either the top orthe bottom strand that all end in the sameplace, but have different start sites. These“staggered” start sites represent all potentialATG (met) start codons in the interior ofthe full-length gene. Single-click with themouse on the largest ORF specified by thetop strand. In the upper left portion of thescreen, you will see the range of base pairsspecified by this ORF. Record the range of thisORF (where it begins and ends) in your note-book. You will need this information laterwhen you design a set of primers that couldbe used to amplify the gene by polymerasechain reaction (PCR).

f. Select Quit from the MAPDRAW File

menu.

6. Using the information obtained in step 5 andyour knowledge of molecular biology tech-niques (see Experiment 21 and Section V), de-sign a protocol that will allow you to clone yourgene into the multiple cloning site of a vectorof your choice (pUC18/19, pBluescript S/K orK/S, etc.). Your instructor will provide youwith a list of restriction-enzyme recognition se-quences, where these enzymes cleave withinthis sequence, and restriction maps of a num-ber of different cloning vectors for this exer-cise. If you cannot find restriction sites withinyour sequence that will allow you to easily clonethe entire gene, you may be forced to design aprotocol that will allow you to clone a portionof the gene. Alternatively, your instructor mayhave previously introduced “phantom” restric-tion endonuclease recognition sites at the 5�and/or 3� end of the gene to make this gene se-quence easier to work with. Your detailed pro-tocol should include the following:

a. The restriction endonucleases with whichyou will digest your DNA sequence.

b. The restriction endonucleases with whichyou will digest your cloning vector.

c. A complete map of the recombinant plas-mid that would result if your ligation reac-tion were successful.

d. The host strain into which you would trans-form your ligation mixture.

EXPERIMENT 25 Obtaining and Analyzing Genetic and Protein Sequence Information via the World WideWeb, Lasergene, and RasMol

407

e. The method that you would use to selectboth for transformed cells and for those cellsthat may contain the desired recombinantplasmid.

f. A set of restriction endonuclease digestionsand agarose-gel electrophoresis experimentsthat would allow you to verify the structure(sequence) of the recombinant plasmid afteryou have re-isolated it from the host cell.(What enzymes would you digest the plas-mid with and what size DNA fragmentswould you expect to be produced in thesedigests?)

7. Using the PRIMERSELECT application ofLasergene, determine a set of primers thatcould be used to amplify your DNA sequenceusing PCR.

a. Select Initial conditions from the Condi-

tions menu to view the salt concentrationand primer concentration at which the pro-posed PCR experiment will be performed.These are fairly standard conditions, so youwill not need to change them for purposesof this exercise. Record these values in yournotebook and close this window.

b. Select Primer characteristics from theConditions menu. This window will allowyou to increase or decrease the stringencyof your search for suitable primer pairs. Wewill experiment later with the primer lengthminimum, primer length maximum, themaximum dimer duplexing, and the maxi-mum hairpin values, so make a note of theirlocations in this window.

c. Open your DNA sequence by selecting En-

ter sequence from the Filemenu. Next, se-lect Primer locations from the Conditions

menu and select upper and lower primer

ranges from the pull-down menu. Using theupper and lower ranges for your gene ob-tained with the MAPDRAW application ofLasergene (step 5E), enter the appropriateranges that will encompass the entire coding re-gion of the gene.

d. Select PCR primer pairs from the Locate

menu to obtain a set of primers that fit thecriteria specified by the default values (step7B). If your parameters were too stringent,you may not have received any entries. Ex-

periment with the different parameters de-scribed in step 7B to obtain 5 to 10 primerpairs that are 17 to 24 bases in length. The lo-cated primer pairs will be listed in order ofdecreasing “score.”

e. Double-click on the primer pair with thehighest score to obtain a schematic view of theDNA product that would be produced in aPCR reaction performed with this set ofprimers.

f. Select Amplification summary from theReport menu and print a copy of this re-port. What is the 5� to 3� sequence of yourtwo primers? To which strands of the DNAtemplate (top or bottom) will each primeranneal? What are the predicted meltingtemperatures (Tm) of these two primers?What is the predicted length of the DNAproduct that would result in a PCR reactionperformed with this set of primers? Whatannealing temperature would you use in thisPCR reaction?

g. Select Composition summary from theReport menu and print a copy of this re-port. What is the molecular weight of yourtwo primers? What is the conversion factor(nM/A260 and �g/A260) for each primer?What is the %G, %C, %A, and %T com-position of your two primers?

h. Select Quit from the PRIMERSELECTFile menu.

8. Using the PROTEAN application of Laser-gene, predict some of the secondary structuralfeatures of your protein based on its amino acid(primary) sequence.

a. Open your protein sequence file by select-ing Open from the File menu.

b. Select Composition from the Analysis

menu and print a copy of this report. Whatis the molecular weight of your protein?How many amino acids does it contain?What is the molar extinction coefficient (at280 nm) for your protein? What is the iso-electric point for your protein? What is thepercent (by weight and by frequency) of thecharged, acidic, basic, hydrophobic, and po-lar amino acids in your protein?

c. Select Titration curve from the Analysis

menu to obtain a theoretical titration curve

408 SECTION VI Information Science

for your protein and print out a copy of thisreport. What is the net charge on your pro-tein at pH 5.0, 7.0, 8.0, and 9.0?

d. Return to the original PROTEAN windowto display a list of secondary structure pre-dictions for your protein. Print out a copyof this report. Using the Chou–Fasman andGarnier–Robson algorithms, locate the pre-dicted alpha helices, beta sheets, and turnregions in your protein. What stretches ofyour protein are predicted to be quite hy-drophobic? What residues in your proteinare likely to be found on the surface? Whatpeptide contained within your protein couldyou use to raise an antibody against yourprotein (the sequence of the peptide)?

9. Using the RasMol program, visualize your pro-tein in three dimensions and compare its struc-ture to the secondary structural predictions foryour protein generated by the PROTEAN ap-plication of Lasergene.

a. Open your sequence file and the RasMacprogram by selecting Open from the re-spective File Menus. Be sure that the view-ing window and command line window arevisible.

b. You can rotate your protein by clicking anddragging within the viewing window. Ex-periment with the RasMol viewing tools un-til you have obtained a good view of everysurface of your protein.

c. Experiment with different options in boththe Display and Colours menu to obtainseveral different views or models of your pro-tein (ribbon diagram, space-filling model,atomic backbone, etc.). Which of these mod-els do you find to be most useful and why?If your protein is an enzyme, can you locatesome of the residues that may comprise theactive site?

d. Experiment with the different Select andLabel commands to highlight some differ-ent residues in your protein.

e. Using the Cartoons command in the Dis-

play menu and the Structure command inthe Colours menu, “select” and “label”both the alpha helices and beta sheets inyour protein. Compare these experimentalresults with the secondary structural pre-dictions generated by the Chou–Fasmaand Garnier–Robson algorithms in thePROTEAN application of Lasergene.Which of the two algorithms performedbetter in predicting the structural charac-teristics of your protein? Which algorithmdid a better job at predicting the �-helicalregions within your protein sequence?Which algorithm did a better job at pre-dicting the �-sheet regions within yourprotein sequence? Is the same true forother proteins that members of your classhave analyzed?

c h

411

Appendix: Reagents and MaterialsSuggestions for Preparation andTeaching Staff

The following appendix lists details for the prepa-ration of reagents and equipment for 100 studentsworking in pairs (50 groups). It allows for studentwaste of reagents. Smaller classes will need a frac-tion of these quantities. Lesser quantities of reagentswill often suffice if reagent usage is carefully con-trolled by an instructor. At times, specific commer-cial sources of reagents are listed. These choices aremostly for convenience; alternative sources of thesereagents may be equally satisfactory.

Experiment 1: Photometry

Spectrophotometer with cuvettes—At least 1 required

Examination of an Absorption Spectrum

Adenosine solution (5 � 10�5 M)—Dissolve 2.67 mg ofadenosine (free base) in 200 ml of water. Store at4°C.

Riboflavin solution (5 � 10�5 M)—Dissolve 3.76 mg ofriboflavin in 200 ml of water. Store at 4°C.

Folin–Ciocalteau Assay

Lysozyme solution (1 mg/ml)—Dissolve 200 mg oflysozyme in 200 ml of water. Store at 4°C.

200 ml of lysozyme at “unknown” concentrations (storeat 4°C). Each group requires 2.5 ml of one of these“unknown” solutions:

2 mg/ml lysozyme—Dissolve 150 mg of lysozyme to75 ml of water.

1.5 mg/ml lysozyme—Add 18.7 ml of 2 mg/mllysozyme to 6.3 ml of water.

1 mg/ml lysozyme—Add 12.5 ml of 2 mg/ml lysozymeto 12.5 ml of water.

0.5 mg/ml lysozyme—Add 12.5 ml of 1 mg/mllysozyme to 12.5 ml of water.

0.25 mg/ml lysozyme—Add 12.5 ml of 0.5 mg/mllysozyme to 12.5 ml of water.

1% (wt/vol) CuSO4 � 5H2O—Dissolve 2 g of CuSO4 �5H2O in 200 ml of distilled water.

2% (wt/vol) Sodium tartrate—Dissolve 4 g of sodiumtartrate � 2H2O in 200 ml of distilled water.

2% (wt/vol) Na2CO3 in 0.1 N NaOH—Dissolve 20.0 gof NaOH and 100 g of Na2CO3 in 5 liters of dis-tilled water.

Folin–Ciocalteau Phenol Reagent (2 N)—200 ml re-quired. (Sigma catalog #F-9252).

Large (16 � 125 mm) glass test tubes—500 required.

Experiment 2: Chromatography

Chromatography columns—25 pieces; 5- to 8-mm insidediameter clean glass tubing approximately 15 in.long, or 50 equivalent chromatography columns.

CM-Sephadex (G-50) in l.0 M potassium acetate,pH 6.0—Stir 150 g of fresh or used CM-Sephadex(G-50, medium grain size) into 2.5 liters of dis-tilled water and vacuum filter (Büchner funnel) offthe excess fluid. Rinse the accumulated filter cakeon the filter with 2 volumes of 95% ethanol, fol-lowed by 2 volumes of distilled water. Remove andresuspend the filter cake in 4 volumes of 0.1 N HCl(415 ml concentrated HCl into 4.6 liters of dis-tilled water), then vacuum filter and rinse with2 volumes of distilled water as before. Remove andresuspend the filter cake in 2 volumes of 1.0 Mpotassium acetate, pH 6.0 (i.e., dissolve 490 g of

412

Volumetric dispensing bottle—Provide one bottle or ap-paratus capable of dispensing 5 or 10 ml of scin-tillation fluid, depending on the size of the scintil-lation vials to be used.

Scintillation fluid—Dissolve 25 g of 2,5-diphenyloxazolein 5 liters of reagent-grade toluene.

Chloroform—50 ml required.

Quenched 14C unknown—In a hood, mix 37.5 to 47.5 mlof the 14C-toluene standard with 2.5 to 12.5 ml ofchloroform to obtain 50 ml.

Scintillation vials—400 required.

Scintillation counter—1 required.

Marking pen—preferably 1 per group.

Experiment 4: Electrophoresis

NOTE: The quantities of materials to be used anddetails of preparation of the PAGE gels for elec-trophoresis will depend on the number of studentswho share a single electrophoresis gel and the char-acteristics of the electrophoresis apparatus to beused. Appropriate adjustments must be made to fitthe specific circumstances of your laboratory. Thevolumes given below are convenient volumes forpreparation, but are generally well in excess of theamount required for 50 student pairs.

1.5 M Tris chloride, pH 8.8—Dissolve 91 g of Tris (tris-hydroxymethyl-aminomethane) free base in about375 ml of distilled water. Titrate the solution topH 8.8 with concentrated HCl and dilute to 500 mlwith distilled water.

1.0 M Tris chloride, pH 6.8—Dissolve 30.5 g Tris (tris-hydroxymethyl-aminomethane) free base in about175 ml of distilled water. Titrate the solution topH 6.8 with concentrated HCl and dilute to 250 mlwith distilled water.

30% acrylamide (292 g/liter acrylamide, 8 g/liter N,N�-methylenebisacrylamide)—Caution: Acrylamide istoxic!Dissolve 146 g of acrylamide and 4 g of N,N�-methylene-bis-acrylamide in 500 ml of distilledwater. Filter through a 0.45-�m-pore-size mem-brane (to remove undissolved particulate matter)and store in a dark bottle labeled with the abovewarning at 4°C.

10% (wt/vol) SDS—Dissolve (gently to avoid foaming) 5 gof sodium dodecyl sulfate in 50 ml of distilled wa-ter.

TEMED (N,N,N�,N�-tetraethylethylenediamine)—Usethe commercial liquid neat. (Caution: unpleasantodor.) Approximately 500 �l will be required.

potassium acetate in 4.75 liters of distilled waterand titrate to pH 6.0 with 15 to 25 ml of glacialacetic acid), and vacuum filter again. Repeat thissuspension and filtration step in 2 volumes of 1.0Mpotassium acetate, pH 6.0, a second time. Finally,resuspend the filter cake in 2 volumes of 1.0 Mpotassium acetate, pH 6.0. At this point, you willhave approximately 1 liter of the swollen resin.

CM-Sephadex (G-50) in 0.1 M potassium acetate,pH 6.0—Perform the ethanol, acid, and 1.0 Mpotassium acetate, pH 6.0 washings of CM-Sephadex (G-50) as described above. After the sec-ond wash, resuspend the filter cake in 4 volumesof 0.1 M potassium acetate, pH 6.0 (i.e., 1 part1.0M potassium acetate, pH 6.0, � 9 parts distilledwater). Remove the excess fluid with vacuum fil-tration and resuspend the filter cake in 2 volumesof 0.1 M potassium acetate, pH 6.0. At this point,you will have approximately 1 liter of swollen resin.

1.0 M potassium acetate, pH 6.0—Dissolve 490 g potas-sium acetate in 4.75 liters of distilled water andtitrate to pH 6.0 with 15 to 25 ml of glacial aceticacid. Dilute to 5.0 liters.

0.1 M potassium acetate, pH 6.0—Mix 0.5 liters of 1.0 Mpotassium acetate, pH 6.0 with 4.5 liters of distilledwater.

Glass wool—10 in3 required.

Chromatography sample—dissolve 25 mg of Blue Dex-tran, 50 mg of cytochrome c, and 25 mg of DNP-glycine together in 25 ml of distilled water.

Tape—1 roll required.

Pasteur pipettes with dropper bulbs—50 required.

Experiment 3: RadioisotopeTechniques

3H and 14C unknowns—In a fume hood, mix sufficientquantities of 3H-toluene, 14C-toluene and unla-beled toluene to obtain 2 to 10 nCi of 3H-tolueneand 1 to 2 nCi of 14C-toluene per 0.1 ml of toluenesolution. You will need 10 ml of this solution for50 groups to perform the experiment.

14C-toluene standard—In a fume hood, mix sufficient 14C-toluene and unlabeled toluene to yield a knownquantity of 14C-toluene (e.g., 1 to 2 nCi of 14C-toluene) per 0.1 ml of toluene solution. Label thesample with the 14C-toluene/ml with known dis-integrations per minute (dpm). You will need250 ml of this solution. Save 50 ml of this reagentfor the quenched 14C unknown. Dispense the other200 ml to the students.

APPENDIX Reagents and Materials Suggestions for Preparation and Teaching Staff

isoleucine, arginine, and cystine are less suitablefor titrations than the others); also suitable aresalts of weak acids (e.g., sodium formate, sodiumacetate) and bases (e.g., tris-hydroxymethylamino-methane). Take care to record the precise saltform (sodium salt, hydrochloride) and hydrationof the salt used. Each group will require 800 mgof their unknown.

pH meters—50 required.

2 N HCl—Slowly add 415 ml of concentrated HCl to2.085 liters of distilled water. Keep tightly stop-pered. The precise concentration should be deter-mined by titration with commercial standard base.

2 N NaOH—Dissolve 200 g of NaOH in 500 ml of dis-tilled water and dilute to 2.5 liters. Protect fromCO2 with a soda-lime tube. The precise concen-tration should be determined by titration withstandard acid.

Buffers for standardizing pH meters—Use commercialpreparations; 4.0, 7.0, and 10.0 are suitable. Weperform a 2-point calibration for the acid titrationwith pH 4.0 and 7.0. For the base titration, we re-calibrate the pH meter with solutions at 7.0 and10.0. Approximately 500 ml of each standard willbe required.

10-ml Burettes and holders—50 required.

Magnetic stirring motors with stirring bars—50 required.

NOTE: Instruction in the operation and principles of op-eration of pH meters may be necessary. Provideburette brushes and solvent (CHCl3) for cleaningburettes and removing stopcock grease, unless bu-rettes with Teflon stopcocks are used.

Experiment 6: SequenceDetermination of a Dipeptide

1-fluoro-2,4-dinitrobenzene (FDNB) solution—Add1.25 ml of FDNB to 23.75 ml of absolute ethanol.

Ninhydrin spray—Dissolve 5 g of ninhydrin in 2.5 litersof n-butanol. Add 2.5 ml of 2,4-collidine immedi-ately before use.

Dipeptides (gly-leu, ala-phe, leu-gly, ala-val)—Prepare25 glass test tubes (13 � 100 mm) and 25 micro-centrifuge tubes each containing 2 mg of one ofthe dipeptides. Number each of the tubes in match-ing sets. Keep a record of the students’ names andtheir unknown numbers. Any dipeptide can beused, provided that cysteine, tryptophan, arginine,glutamine, and asparagine are not present.

DNP-amino acid standards (for thin-layer chromatogra-phy)—Dissolve 1 mg of each standard in 1.5 ml of

10% (wt/vol) ammonium persulfate in water—This solu-tion must be freshly prepared by the students. Provideammonium persulfate as a dry powder. Studentsshould prepare the solution needed by dissolving20 mg in 200 �l of distilled water immediately be-fore use.

SDS–PAGE electrophoresis buffer—Dissolve 30 g ofTris free base, 188 g of glycine, and 10 g of SDS(gently to avoid foaming ) in 9 liters of distilled wa-ter. Adjust the pH to 8.3 with concentrated HCl,and dilute to 10 liters with distilled water.

4X SDS–PAGE sample buffer—Mix together 17.5 ml ofdistilled water, 12.5 ml of 1.0 M Tris � HCl (pH 7.0),15 ml of glycerol, and 5 ml of 2-mercaptoethanol.(Caution: unpleasant odor.) Dissolve 4 g of SDS and2.0 mg of bromophenol blue in this mixture carefully,to avoid foaming. The stock 1.0 M Tris chloride, pH7.0, is prepared by dissolving 60.5 g of Tris free basein about 400 ml of distilled water, titrating to pH 7.0with concentrated HCl, and diluting to 500 ml withdistilled water.

Coomassie Blue staining solution—Dissolve 1.25 g ofCoomassie Brilliant Blue R-250 in 500 ml ofdestaining solution (see next entry).

Destaining solution—Mix 2 liters of methanol and 2.5liters of distilled water, then carefully add 500 mlof glacial acetic acid.

Power supplies and electrophoresis apparatus—8 re-quired.

Protein molecular weight standards—Suggested stan-dards are phosphorylase (97,400 Da), bovine serumalbumin (66,200 Da), ovalbumin (45,000 Da), car-bonic anhydrase (31,000 Da), soybean trypsin in-hibitor (21,500 Da) and lysozyme (14,400 Da) at1 mg/ml each in a single mixture in 0.01 M Trischloride buffer, pH 7.0 (prepared by dilution of thestock 1 M buffer above). Approximately 2 ml ofthis solution will be required.

Unknown protein solutions (approximately 1 mg/ml in0.01 M Tris chloride buffer, pH 7.0)—It is conve-nient to prepare the unknowns in larger volumes(100 to 500 �l) and to dispense them to studentsas 20-�l samples in microcentrifuge tubes. Any ofthe proteins used as standards or other readilyavailable proteins with molecular weights withinthe range of the standards may be used.

Experiment 5: Acid–BaseProperties of Amino Acids

Amino acid unknowns—5 g each; any of the commonamino acids can be used (aspartic acid, leucine,

APPENDIX Reagents and Materials Suggestions for Preparation and Teaching Staff 413

414

column (250 � 4.6 mm)—1 required (Alltech cat-alog #28935).

PicoTag Eluant A (Waters catalog #88108)—2 liters re-quired.

140 mM potassium acetate, pH 5.3 (optional)—Dissolve27.48 g of potassium acetate in 1.9 liters of distilledwater. Adjust to pH 5.3 with 2 M glacial acetic acid.Bring the final volume of the solution to 2 literswith distilled water. Filter sterilize and de-gas thesolution before use in HPLC.

60% (vol/vol) Acetonitrile—Mix 1.2 liters of HPLC-grade acetonitrile with 800 ml of HPLC-grade water.

Amino Acid Standard H (Sigma catalog #AA-S-18)—Approximately 1 ml required.

Pressurized N2 tank—1 required.

Chromatography chambers for thin layer chromatogra-phy—for 25 plates.

Chromatography jars for paper chromatography—50 re-quired.

Experiment 7: Study of theProperties of �-Galactosidase

0.08 M sodium phosphate buffer, pH 7.7—Dissolve 191 gof Na2HPO4 (anhydrous) and 18 g of NaH2PO4 �H2O in 14 liters of distilled water. Bring the pHto 7.7 with 10 M HCl or 10 M NaOH. Bring thefinal volume to 15 liters with distilled water. Storeat room temperature.

�-galactosidase stock solution—Dissolve 5000 units oflyophilized �-galactosidase (Sigma catalog #G-6008) in 10 ml of 0.08 M sodium phosphate buffer,pH 7.7. Make dilutions of this stock solution in thesame buffer containing 1 mg/ml bovine serum albu-min (1:10 to 1:500). Using these dilute solutions, de-termine what dilution of the stock enzyme solutionwill give a A420/min of about 0.10, using the sameprotocol as described in Day 1 of the protocol. Forthe class, prepare a stock �-galactosidase solutionthat is five times as concentrated as the dilution thatgave a A420/min of about 0.10. Prepare 250 ml ofthe appropriate dilution and divide this into 1.0-mlaliquots. Each group will require about 4 ml of thisenzyme solution to perform the experiment: 1 ml onDay 1, 2 ml on Day 2, and 1 ml on Day 3.

0.08 M sodium phosphate buffers, pH 6.4, 6.8, 7.2, and8.0—Dissolve 11.04 g of NaH2PO4 � H2O in900 ml of distilled water. Bring the solutions to theappropriate pH with 10 M NaOH. Bring the finalvolume of each solution to 1.0 liter with distilledwater. Store at room temperature.

acetone. DNP-gly, DNP-ala, DNP-leu, DNP-ser,and DNP-phe are useful for the four dipeptideslisted above. If the dipeptides that you choose con-tain other N-terminal amino acids, DNP-aminoacid standards of them must be prepared.

Amino acid standards (for paper chromatography)—Dis-solve 2 mg of each amino acid in 2 ml of distilledwater. Gly, leu, ala, phe, val, gln, lys, and ser areuseful. Store at 4°C.

Peroxide-free diethyl ether—Place two 500-ml canistersof this reagent in the fume hood.

Whatman 3 MM chromatography paper—Each grouprequires an 8.5-in.-by-8.5-in. piece for paper chro-matography.

4-ml glass (hydrolysis) vials with Teflon-lined screwcaps—100 required.

6 N HCl—Add 50 ml of concentrated HCl to 50 ml ofdistilled water.

4.2% NaHCO3—Dissolve 10.5 g of NaHCO3 in 250 mlof water.

Silica-gel thin-layer chromatography plates (Fisher cata-log #05-713-317)—Heat at 100°C for 10 min justbefore use. (Store dessicated at room temperature.)Twenty-five will be required; we try to fit twogroups on one plate.

pH paper (1–11)—One roll for every 2 to 4 groups is re-quired.

Heating block—Each group will require two compart-ments in the heating block that will hold the 4-mlglass vials.

Pasteur pipettes—500 to 1000 required.

Large (16 � 125 mm) and small (13 � 100 mm) glass testtubes—Approximately 500 of each required.

Vacuum centrifuge—1 required.

Mobile phase for thin-layer chromatography—Mix 1.5liters of chloroform, 600 ml of t-amyl alcohol, and60 ml of glacial acetic acid. Store tightly coveredin a dark bottle at room temperature.

Mobile phase for paper chromatography—Mix 1.8 litersof n-butanol, 450 ml of glacial acetic acid, and750 ml of water. Store tightly covered in a darkbottle at room temperature.

Wash reagent for PTC-coupling of amino acids—Mix6 ml of absolute ethanol, 6 ml of triethylamine, and3 ml of water. Store tightly covered at 4°C.

PTC coupling reagent—Mix 21 ml of absolute ethanol,3 ml of triethylamine, 3 ml of phenylisothiocyanate(Sigma catalog #P-1034), and 3 ml of water. Storetightly covered at 4°C.

HPLC unit—1 required.

UV/Vis Spectrophotometer and chart recorder forHPLC—1 of each required.

Alltech Adsorbosphere HS C18 reverse-phase HPLC


Experiment 8: Purification ofGlutamate–OxaloacetateTransaminase from Pig Heart

0.05 M potassium maleate buffer (pH 6.0)—Dissolve 29 gof maleic acid and 9.3 g of disodium EDTA �2 H2O in 4 liters of distilled water. Adjust pH to6.0 with KOH. Bring the total volume of the so-lution to 5 liters with distilled water. Store at 4°C.If the buffer is to be stored for more than a week,one drop of toluene or chloroform may be addedto prevent bacterial growth.

0.04 M -ketoglutarate—Prepare this reagent just before useand store at 4°C. Dissolve 17.5 g of -ketoglutaricacid in 2.6 liters of distilled water. Adjust the pH to6.0 with NaOH. Bring the total volume of the solu-tion to 3.0 liters with distilled water.

0.03 M sodium acetate buffer—Dilute 17.2 ml of glacialacetic acid to 9 liters with distilled water. Adjustthe pH to 5.4 with NaOH. Bring the total volumeof the solution to 10 liters with distilled water.Store at room temperature. If the buffer is to bestored for more than a week, one drop of tolueneor chloroform may be added to prevent bacterialgrowth.

Saturated ammonium sulfate solution—Dissolve 7 kg ofammonium sulfate (reagent grade) in 10 liters ofdistilled water. Store at room temperature.

1% CuSO4 � 5H2O—Dissolve 5 g of CuSO4 � 5H2O in500 ml of water. Store at room temperature.

2% sodium tartrate solution—Dissolve 10 g of sodiumtartrate � 2H2O in 500 ml of distilled water. Storeat room temperature.

2% Na2CO3 in 0.1 N NaOH—Dissolve 20 g of NaOHand 100 g of Na2CO3 in 5 liters of distilled water.Store at room temperature.

Folin–Ciocalteau Phenol Reagent (2 N)—Approximately1 liter will be required. (Sigma catalog #F-9252).

1 mg/ml bovine serum albumin (or lysozyme) standard—Dissolve 600 mg of bovine serum albumin (frac-tion V) or lysozyme in 600 ml of distilled water.Store at 4°C.

0.10 M potassium phosphate buffer (pH 7.5)—Dissolve14.1 g of anhydrous K2HPO4 and 2.18 g of anhy-drous KH2PO4 in 800 ml of distilled water. Adjustto pH 7.5. Bring the total volume of the solutionto 1.0 liter with distilled water. Store at 4°C. If thebuffer is to be stored for more than a week, onedrop of toluene or chloroform may be added toprevent bacterial growth.

0.10 M potassium phosphate/0.12 M aspartate buffer—Dissolve 261 g of anhydrous K2HPO4 and 240 gof L-aspartic acid in 12 liters of distilled water. Ad-

2.5 mM o-nitrophenyl-�,D-galactopyranoside (ONPG)—Dissolve 1.507 g of ONPG in 2.0 liters of distilledwater. Store at 4°C in small aliquots.

10 mM o-nitrophenyl-�,D-galactopyranoside (ONPG)—Dissolve 0.6 g of ONPG in 200 ml of distilled wa-ter. Store at 4°C.

750 mM methyl-�,D-galactopyranoside (MGP)—Dis-solve 5.826 g of MGP in 40 ml of distilled water.Store at 4°C.

150 mM methyl-�,D-thiogalactoside (MTG)—Dissolve1.261 g of MTG in 40 ml of distilled water. Storeat 4°C.

200 mM isopropylthio-�,D-galactopyranoside (IPTG)—Dissolve 1.430 g of IPTG in 30 ml of dis-tilled water. Filter sterilize the solution and store at4°C.

100 mM adenosine 3�,5� cyclic monophosphate (cAMP)—Dissolve 527 mg of cAMP (sodium salt) in 15 ml ofdistilled water. Filter sterilize the solution and storeat 4°C.

20% (wt/vol) glucose—Dissolve 10 g of D-glucose in50 ml of distilled water. Filter sterilize the solutionand store at 4°C.

1.0 M Na2CO3—Dissolve 132.5 g of Na2CO3 in 1 literof distilled water. Bring the final volume to 1.25liters with distilled water, and store at room tem-perature.

0.1% sodium dodecyl sulfate (SDS)—Dissolve 50 mg ofSDS in 50 ml of distilled water. Store at room tem-perature.

YT broth—Dissolve 50 g of bactotryptone, 25 g of yeastextract, and 25 g of NaCl in 4 liters of distilled wa-ter. Bring the final volume to 5 liters with distilledwater, and autoclave in 1-liter aliquots. Store at4°C; warm to room temperature before use.

Chloroform—About 50 ml required.

Small (13 � 100 mm) glass test tubes—3000 required.

Large glass test tubes (16 � 125 mm, sterile) with caps—1200 required.

Colorimeter tubes or cuvettes for spectrophotometer—50 required.

Heated water baths—5 required.

1.5-ml Microcentrifuge tubes—Approximately 500 re-quired.

Spectrophotometer—1 required.

E. coli strains EM1, EM1327, EM1328, EM1329—Theequivalent strains can be obtained from the E. coli Genetic Stock Center, Department of Biol-ogy, Yale University, New Haven, CN (website address: http://cgsc.biology.yale.edu). These areEM1 � CA8000, EM1327 � CA8445-1 (crp�),EM1328 � CA8306 (cya�), and EM1329 � 3.300(lacl�).


416

more than a week, one drop of toluene or chloro-form may be added to prevent bacterial growth.

Carboxymethylcellulose slurry (equivalent of 75 g of drymaterial)—Slowly wet 75 g of dry resin in distilledwater. Draw off the supernatant (after the resin hassettled) with vacuum filtration. Resuspend the fil-tered resin cake in 1.5 L of 0.5 M NaOH (30 g ofNaOH dissolved in 1.5 L of distilled water). Allowthe resin to settle, draw off the supernatant, andwash the resin cake twice, as before, in 2.5 L vol-umes of distilled water. Resuspend the resin cakein 2.5 L of 0.5 M HCl (100 ml of concentratedHCl in 2.4 L of distilled water). Allow the resin tosettle, draw off the supernatant, and wash the resincake twice, as before, in 2.5 L volumes of distilledwater. Repeat the wash procedure described abovewith 1.5 L volumes of 0.03 M sodium acetatebuffer, pH 5.0, until the pH and ionic strength ofthe drawn-off supernatant is the same as that of thesodium acetate buffer (pH and conductivity me-ter). Resuspend the resin cake in 1.5 L 0.03 Msodium acetate buffer, pH 5.0. Add 2 g/litersodium azide for storage to prevent bacterialgrowth. Remove the azide by washing again in 0.03M sodium acetate buffer, pH 5.0 when it is readyto be used by the class.

Fresh pig hearts—10 required.

Dialysis tubing (8000–12000 molecular weight cutoff)—Heat tubing at 100°C for 15 to 20 min in a solu-tion containing 1 g of NaHCO3 and 0.1 g of disodium EDTA per 100 ml. Rinse thoroughly indistilled water after the tubing has cooled and storewet and covered at 4°C. Pierce Chemical Co. of-fers SnakeSkin dialysis tubing in molecular weightcutoffs from 3500 to 10,000 Da. It is supplied in adry roll ready to use (no boiling or hydration re-quired as with conventional dialysis tubing). It isvery convenient for use with large laboratorygroups.

Supplies and reagents for SDS–PAGE—See Experiment4. We normally use a 12% polyacrylamide gel toanalyze the purity of GOT in this experiment.

Experiment 9: Kinetic andRegulatory Properties ofAspartate Transcarbamylase

Escherichia coli strain EK1104 containing plasmid pEK2.Obtain strain No. 700695 from American TypeCulture Collection, 10801 University Blvd., Man-assas, VA 20110.

just to pH 7.5 with dilute KOH. Bring the totalvolume of the solution to 15 liters with distilledwater. Store at 4°C. If the buffer is to be stored formore than a week, one drop of toluene or chloro-form may be added to prevent bacterial growth.

0.10 M -ketoglutarate (pH 7.5)—Prepare this reagent justbefore use and store at 4°C. Dissolve 8.76 g of -ketoglutaric acid in 480 ml of distilled water. Ad-just the pH to 7.5 with NaOH. Bring the total vol-ume of the solution to 600 ml.

250-ml plastic centrifuge bottles—50 required.

50-ml plastic centrifuge tubes—50 required.

Bunsen burners—50 required.

Pasteur pipettes—200 required.

Spectrophotometer with cuvettes—At least 1 required.



Cheesecloth—1 large roll required.

Glass wool

Chromatography columns (20 ml capacity) with tubing—50 required.

Meat grinder—1 required.

Blender—1 required.

3 mM NADH—Prepare this reagent just before use and storeat 4°C. Dissolve 68.7 mg of Na2 � NADH � 3H2O(if other salt and hydration forms are supplied, ad-just weight accordingly) in 30 ml of 0.10 M potas-sium phosphate buffer, pH 7.5. Because this reagentis expensive and unstable, only the quantity needed foreach day should be prepared at the start of each labora-tory period. This reagent is stable for 2 to 4 hr.

Malic dehydrogenase—Prepare this reagent just before useand store at 4°C. Dilute commercially available en-zyme in 0.10 M potassium phosphate buffer,pH 7.5, in which 1 mg/ml bovine serum albuminhas been dissolved. The final concentration ofmalic dehydrogenase should be 200 units/ml. Pre-pare a total of 20 ml of this reagent.

0.03 M sodium acetate buffer, pH 5.0—Dilute 34.4 mlof glacial acetic acid to 19 liters with distilled wa-ter. Adjust the pH to 5.0 with NaOH. Do not over-titrate or back titrate with acid. It is essential that thebuffer does not contain excess salt. Bring the total vol-ume of the solution to 20 liters with distilled wa-ter. Store at 4°C. If the buffer is to be stored formore than a week, one drop of toluene or chloro-form may be added to prevent bacterial growth.

0.08 M sodium acetate buffer, pH 5.0—Dilute 4.6 ml ofglacial acetic acid to 900 ml with distilled water.Adjust the pH to 5.0 with NaOH. Bring the totalvolume of the solution to 1 liter with distilled wa-ter. Store at 4°C. If the buffer is to be stored for


concentrated HCl and dilute to 500 ml with dis-tilled water. Store at 1 to 4°C.

Sonicator—1 required.30 and 60°C Water baths—One of each required.0.08 M Sodium phosphate buffer, pH 7.0—Dissolve

21.25 g of NaH2PO4 � H2O and 35.0 g ofNa2HPO4 (anhydrous) in distilled water to a finalvolume of 5 liters. The pH should be 7.0.

0.125 M Sodium L-aspartate—Dissolve 8.3 g of L-asparticacid in 350 ml of distilled water, bring the pH to7.0 with NaOH (2 N or more concentrated; addi-tion of NaOH may be necessary to dissolve all ofthe aspartic acid), and dilute to 500 ml with dis-tilled water.

0.036 M Carbamyl phosphate—Dissolve 690 mg ofdilithium carbamyl phosphate in sufficient distilledwater (use weight appropriate to hydration state in-dicated on label) to yield 125 ml of solution. If thesolution is cloudy, warm it briefly to clarify. Pre-pare just before the experiment and store at 1 to 3°C.

l.0 mM Carbamyl aspartate standard—Dissolve 17.5 mgof carbamyl aspartate in distilled water, titratingwith dilute NaOH to bring the solid into solution(do not go above pH 7.0). Dilute to 100 ml withdistilled water. Prepare just before experiment andstore at 1 to 3°C.

Antipyrine-H2SO4 reagent—Carefully add 1.25 liters ofconcentrated H2SO4 to 1.25 liters of distilled waterand allow to cool. Dissolve 12.5 g of antipyrine inthe H2SO4 solution. Label with warning: strong acid.

Diacetylmonoxime reagent—Prepare in advance 1.5 litersof 5% acetic acid (75 ml glacial acetic acid dilutedto 1.5 liters with distilled water). Just before the ex-periment, dissolve 12 g of diacetylmonoxime (bu-tanedione monoxime) in the 1.5 liters of 5% aceticacid. Keep refrigerated in a foil-covered bottle.

2% Perchloric acid—Add 30 ml 70% HClO4 to 1 literof distilled water.

40 mM ATP—Dissolve 585 mg of Na2ATP � 2H2O (useweight appropriate to salt and hydration form in-dicated on the label) in 20 ml of distilled water,titrate to pH 6.0 to 8.0 with 1 N NaOH, and di-lute to 25 ml with distilled water. Store frozen; thawjust before use and keep at 1 to 3°C.

10.0 mM CTP—Dissolve 138 mg of Na2CTP � H2O(use weight appropriate to salt and hydration formindicated on the label) in 20 ml of distilled water,titrate to pH 6.0 to 8.0 with 1 N NaOH, and di-lute to 25.0 ml with distilled water. Store frozen;thaw just before use and keep at 1 to 3°C.

5 mM p-Mercuribenzoate—Dissolve 95 mg of sodium p-chloromercuribenzoate in 50 ml of 5 mM KOH.Prepare on the day of use. Store in a foil-wrapped con-

Materials for Growth Medium and

Culturing Bacteria

The growth and harvesting of the bacterial cellsfor this experiment is usually performed by a teach-ing assistant or a technician in advance of the ex-periment. The cells and cell extracts can be storedfrozen at �20 or �70°C before the class will usethem. The amount of growth medium requiredwill depend on the amount of cells required. Gen-erally, 1 liter of bacterial culture will provide suf-ficient enzyme for a class of 50 students workingin pairs.

Prepare the growth medium as follows (theamounts given are for 1 liter of growth medium):Dissolve the following in 870 ml of distilled water:6 g of Na2HPO4, 3 g of KH2PO4, 1 g of NH4Cl,5 g of Casamino acids, and 0.5 g of NaCl. Auto-clave in a capped 2- or 3-liter culture flask. Sepa-rately autoclave (in separate flasks or bottles): 4 gof glucose in 50 ml of distilled water and 12 mg ofuracil in 25 ml of distilled water. Prepare (by filtersterilization) solutions of 40 mg of L-tryptophan in50 ml of distilled water and 50 mg of ampicillin in2.0 ml of distilled water. When the sterilized solutionshave cooled, add the small-volume components to thelarge flask (using sterile technique), holding back 0.5 mlof the uracil solution to add to the inoculation medium(see next paragraph). Separately, prepare autoclavedstock solutions of the following components: 1 MMgSO4 � 7H2O (24.65 g/100 ml H2O); 0.1 MCaCl2 � 2H2O (1.47 g/100 ml H2O), 0.5% thiamine(50 mg/10 ml), 1 mM FeSO4 � 7H2O (27.8mg/100 ml), and 1 mM ZnSO4 � 7H2O (28.7mg/100 ml). Using sterile technique, add 1 ml ofeach of these solutions to the large flask.

Prepare the inoculation medium as follows: Us-ing sterile technique, withdraw 10 ml from thegrowth medium described above and place it in asterile growth tube. Add 0.5 ml of sterile uracil fromthe previous paragraph to bring the inoculationmedium to 30 �g/ml. Add 150 �g ampicillin per mlof the medium by adding 30 �l of a 50-mg/ml solutionof ampicillin in H2O, which has been filter-sterilized.Grow and harvest the bacteria as described in theprotocol for Experiment 9.

Shaking incubator at 37°C.

0.1 M Tris chloride buffer, pH 9.2—Dissolve 6.05 g Trisfree base in distilled water, titrate to pH 9.2 with


418

10 mM reduced glutathione solution—Dissolve 0.3 g ofTris free base in 40 ml of distilled water. Adjust thepH to 8.0 with HCl. Add 0.15 g of reduced glu-tathione to the solution, adjust the pH to 8.0 again(if necessary), and bring the final volume of the so-lution to 50 ml with distilled water. Prepare thisreagent just before use.

IPTG solution for strain induction—Dissolve 0.24 g ofisopropylthio-�,D-galactopyranoside in 2 ml ofdistilled water. Filter-sterilize this solution. Addthis entire solution to a growing 2-liter culture ofE. coli strain TG1 harboring plasmid pGEX-2T toinduce expression of glutathione-S-transferase.

Preparative centrifuge with a rotor for 250-ml centrifugebottles—You will need one centrifuge and eight250-ml polypropylene centrifuge bottles rated forup to 17,000 to 20,000 � g.

Reagents and supplies to perform SDS–PAGE—See Ex-periment 4. We suggest using a 12 or 15% poly-acrylamide gel for this experiment.

Purified GST standard (prepared by instructor)—Pre-pare a 1-liter culture of the GST expression strain,harvest the cells, and purify the protein accordingto the protocol. Use 1 ml of the immobilized glu-tathione Sepharose 4B per 1 liter of culture. Ad-just the concentration of the purified protein to1 mg/ml, and store at �20°C in 1-ml aliquots. Onepreparation of GST will give enough protein to actas a standard for many semesters.

Experiment 11: Microanalysis of Carbohydrate Mixtures byIsotopic, Enzymatic, andColorimetric Methods

Unknown Carbohydrate Solutions

Prepare separate 30 mg/ml stock solutions of D-glucose, D-ribose, and D-glucosamine by dissolving300 mg of each carbohydrate in 10 ml of distilledwater. Prepare “unknown” solutions containing be-tween 5 and 10 mg/ml of each carbohydrate by mix-ing different proportions of these stock solutionswith distilled water. Each group will require about1 ml of the “unknown” carbohydrate solution.

Reduction of Carbohydrates

[14C]glucose—Dilute 2 �l of a 200- to 300-mCi/mmol so-lution of [14C]glucose in 1 ml of distilled water. Eachgroup requires 5 �l of this solution. Dilute com-

tainer to protect from light. The 5 mM KOH is pre-pared by dissolving 14 mg of KOH in a total of 50 ml of distilled water.

Experiment 10: AffinityPurification of Glutathione-S-Transferase

pGEX-2T/E. coli culture stock—We purchase pGEX-2Tfrom Pharmacia (catalog #27-4801-01). This 25-�g package size of DNA can be resuspended in 25�l of TE buffer (see Experiment 24). To prepare aculture stock, transform E. coli strain TG-1 (Strat-agene catalog #200123) with 10 �l of a 1 ng/ml so-lution of the plasmid prepared in sterile distilledwater. If you prefer, E. coli strain DH5 (Gibco-BRL catalog #18265-017) or BL21 (Pharmacia cat-alog #27-1542-01) would perform well. After ob-taining colonies on an agar plate supplementedwith 100 �g/ml ampicillin, grow a 2-ml culture ofthis stock (from a single colony) in YT broth con-taining 100 �g/ml ampicillin at 37°C for severalhours. When the 2-ml solution achieves an OD600

of about 0.3, add 140 �l of dimethylsulfoxide(DMSO). Allow the culture to incubate at 37°C forabout 30 min with gentle agitation, and freeze theculture in a sterile vial at �70°C. Each semester, asterile inoculation loop can be used to streak outthe culture on a fresh agar YT-agar plate contain-ing 100 �g/ml ampicillin. NOTE: If you are unfa-miliar with bacterial transformation and prepara-tion of competent E. coli cells, refer to Experiment21 and Table V-6.

YT broth—Dissolve 20 g of bactotryptone, 10 g of yeastextract, and 10 g of NaCl in 1 liter of distilled wa-ter. Bring the final volume to 2 liters with distilledwater, and autoclave in a single, 4-liter culture flask.

Sonicator—1 required.

PBS buffer—Dissolve 13.15 g of NaCl, 3.4 g ofNa2HPO4, and 0.72 g of NaH2PO4 in 1.2 liters ofdistilled water. Adjust pH to 7.3 with HCl orNaOH if required. Bring the total volume of thesolution to 1.5 liters with distilled water.

Glutathione Sepharose 4B—We obtain this from Phar-macia (catalog #17-0756-01). Prepare a 50% slurryof this resin in PBS buffer according to the man-ufacturer’s instructions. NOTE: The resin can beregenerated and used in following semesters bywashing the resin in 3 M NaCl. The Sepharose4B matrix is chemically stable in the pH range of4 to 9.



Heat blocks—5 to 10 required.

Descending chromatography tank—1 to 5 required, de-pending on its size.

0.5-ml microcentrifuge tubes—50 required.


Fume hood—1 required (it is very important for the stu-dents’ safety that this is provided).

Whatman 3 MM chromatography paper—50 strips (1 � 22 in. each).

4-ml Glass (hydrolysis) vials with Teflon-lined screwcaps—400 required.

Enzymatic Determination of Glucose

Glucose hexokinase reagent—Dissolve the contents of twoGlucose (HK) reagent vials (Sigma catalog #16-50)each in 50 ml of distilled water. Store at 4°C. Eachgroup requires 2 ml of this reagent.

Elson–Morgan Determination of

Glucosamine

1 M Na2CO3—Dissolve 21.2 g of Na2CO3 in a total vol-ume of 200 ml of distilled water.

Glucosamine standard solution (0.5 �mol/ml)—Add98.8 �l of the 30-mg/ml D-glucosamine stock so-lution to 27.4 ml of distilled water.

Acetylacetone reagent—Add 6 ml of acetylacetone to294 ml of 1 M Na2CO3. This reagent must be pre-pared on the day that the experiment is performed.

95% ethanol—Add 50 ml of distilled water to 950 ml ofabsolute ethanol. Store tightly capped at 4°C.

p-Dimethylaminobenzaldehyde reagent—Dissolve 7.83 gof p-dimethylaminobenzaldehyde in 300 ml of a 1:1mixture of 95% ethanol and concentrated HCl.Store tightly covered at 4°C. Prepare on the day ofthe experiment.


Experiment 12: Glucose-1-Phosphate: Enzymatic Formationfrom Starch and ChemicalCharacterization

NOTE:The measurements in Experiment 12 will beseriously interfered with if residues of phosphate-containing laboratory detergents are present in theglassware. Students should be made aware of thepotential problem, and a supply of phosphate-free

mercial 14C-glucose (sp act � 200 mCi/mmol) toyield a solution with about 2,000 cpm/�l in the14C/3H window. Store cold or frozen, tightly stop-pered. From this solution, accurately withdraw a 5 or10 �l sample and transfer quantitatively to a 1 � 1/2inch segment of Whatman No. 1 paper, and deter-mine, using the same counting conditions as theclass, (1) the cpm in the 14C/3H channel per �l glu-cose solution, and (2) the cpm in a full 3H channelper �l glucose solution. (1) plus (2) should equal thetotal 14C cpm (which you can also determine bycounting the full 14C channel). From these data, cal-culate and post for class use in analyzing data:

� ,

and

�

NaB3H4—Dissolve 5 mg of 15- to 150-mCi/mmolNaB3H4 (15–25 mCi/mmol is most convenientand economical; New England Nuclear Corp. orAmersham/Searle can supply 3H-NaBH4) in250 �l 0.1 N NaOH. Store frozen; stable about 1month. Determine the specific activity accuratelyas follows: Mix 5 �l of a glucose solution of knownconcentration (�20 �moles/ml is convenient) with5 �l 14C-glucose (same as used by class) and 5 �l3H-NaBH4 solution just prepared. Proceed as de-scribed in Experiment 14. From the 3H-cpm in the3H-glucitol on the chromatogram, determine thespecific activity of the 3H-NaBH4 (the value to begiven to the class), expressing it as 3H-cpm permole of monosaccharide reduced. This value willdepend on the efficiency of counting, so it shouldbe determined under conditions identical to thoseused by the class. Each group requires 5 �l of thissolution.

0.75 N H2SO4—Add 21 �l of concentrated HCl to979 ml of distilled water. Each group requires 5 �lof this solution.

Mobile-phase solvent for paper chromatography—Mixtogether 45.4 ml of 88% formic acid, 154.6 ml ofdistilled water, 120 ml of glacial acetic acid, and720 ml of ethyl acetate. Store tightly stoppered ina dark glass bottle at room temperature.

Scintillation fluid—Dissolve 30 g of diphenyloxazole in6 liters of toluene. Store tightly stoppered in a darkglass bottle at room temperature.


(2)(1) � (2)

Fraction of total 14C

in 3H channel

(1)(1) � (2)

Fraction of total 14C

in 14C/3H channel


420

decant. Wash with 10 liters of 4% NH4OH anddecant. Wash with 10 liters of 4% NaOH and de-cant. Rinse with deionized water with stirring anddecanting until the pH of wash water is 9 or lower.Store under water. Each pair of students will needabout 250 ml of moist resin. The resin should becollected at the end of the experiment (Do not al-low accidental mixing with Dowex 50.) and promptlyregenerated as follows: Wash the resin with 5%KOH (50 g/liter, about 10 liters for the above 2.5-liter quantity of resin), decant, and wash thor-oughly with distilled water until the pH is 9.5 to10.5. Store tightly covered under water. A few daysbefore use of the regenerated IR-45 resin, removeyellow organic material that bleeds from the beadsby washing thoroughly with methanol, and thenagain with distilled water. The same procedure canbe used with Dowex-50 if yellow material accu-mulates in the water in which it is stored.

Absolute methanol—Freshly opened reagent-grade sol-vent should be used. Approximately 30 liters willbe required.

Glass tubes—Approximately 1 in. in diameter and 2 ft inlength, or equivalent columns for chromatography.The group will require 50 of these tubes.

Glass wool.

Reducing agent—Dissolve 75 g of NaHSO3 and 25 g ofp-methylaminophenol in 2.5 liters of distilled wa-ter. Store in a brown bottle. Stable for 10 days.

Acid molybdate reagent—Cautiously add 680 ml concen-trated H2SO4 to 1.8 liters of distilled water and al-low to cool. Dissolve 125 g (NH4)6Mo7O24 �4 H2O in 2.5 liters of distilled water and add tothe above sulfuric acid solution. Make up to 5 literswith distilled water.

1.0 mM Phosphate standard—Dissolve 680 mg ofKH2PO4 in 5 liters of distilled water.

5% KOH—Dissolve 500 g KOH in 10 liters of distilledwater.

0.01 M KI, 0.01 M I2—Dissolve 450 mg of KI in 250 mlof distilled water. Add 325 mg of I2 and mix untildissolved. Stable about 1 month.

Charcoal (finely powdered, e.g., Norit)—100 g required.

Vacuum desiccator containing fresh P2O5—1 required.

High-speed centrifuges—1 or 2 required.

300-ml Centrifuge bottles or cans—50 to 100 required.

Nelson’s A reagent—Dissolve 62.5 g of Na2CO3 (anhy-drous), 62.5 g of potassium sodium tartrate, 50 gNaHCO3, and 500 g Na2SO4 (anhydrous) in 1.75liters of distilled water and dilute to 2.5 liters withdistilled water.

Nelson’s B reagent—Dissolve 37.5 g CuSO4 � 5 H2O in

detergent, such as sodium dodecyl sulfate, shouldbe made available for washing glassware during thisexperiment.

Phenylmercuric nitrate—2.5 g required.

0.8 M Potassium phosphate buffer, pH 6.7—Dissolve950 g of K2HPO4 � 3 H2O and 545 g of KH2PO4

in approximately 9 liters of distilled water and ad-just to pH 6.7 with HCl or KOH, as necessary, be-fore making up to 10 liters.

Soluble starch—500 g required.

Filter aid (Celite 535 or similar material)—125 g re-quired.

14% NH4OH—Mix 500 ml of concentrated NH4OHwith 500 ml of distilled water.

2.0 N HCl—Mix 165 ml of concentrated HCl with850 ml of distilled water. (It is important that boththis reagent and the 2.0 N NaOH below be exactly2.0 N, or that equal volumes of each reagent yielda neutral [pH 6 to 8] solution. A teaching assistantshould check this by titration.)

2.0 N NaOH—Dissolve 80 g of NaOH in 1 liter of dis-tilled water. (See note under 2.0 N HCl above.)

l.0 N HCl—Mix 41.5 ml of concentrated HCl with460 ml of distilled water. (It is important that equalvolumes of this reagent and the 1.0 N NaOH de-scribed in the next entry neutralize each other[pH 6 to 8]. This should be checked by titration.)

l.0 N NaOH—Dissolve 20 g of NaOH in a total volumeof 500 ml of distilled water.

Potatoes—Stored 24 hr at l to 5°C and tested for phos-phorylase activity. Do not freeze. Approximately 30potatoes will be required.

Blenders—1 or 2 required.

Cheesecloth—1 or 2 rolls required.

Mg(Acetate)2 � 4 H2O—1.5 kg required

Dowex 50 (H� form)—Mix 1 kg of Dowex 50, X-8, 20to 50 mesh, with 20 liters of dilute HCl (3 liters ofconcentrated HCl � 17 liters of distilled water),stir for 5 min, fill with additional H2O, and decant.Wash by decantation or suspension and vacuum fil-tration until the eluate no longer yields any titrat-able acidity (pH 3 to 4). Drain the resin until amoist paste is obtained. Each pair of students willneed about 350 ml of moist Dowex 50. The resinshould be collected at the end of the experiment(Do not allow accidental mixing with IR-45) andpromptly regenerated by the above procedure.Store tightly covered under a water layer. (See notefollowing IR-45 preparation.)

IR-45 (OH� form)—Add 5 liters of wet IR-45 to a bat-tery jar. Wash with 10 liters of 4% Na2CO3 and


Experiment 13: Isolation andCharacterization of ErythrocyteMembranes

Fresh pig blood—Approximately 4 liters required.

Heparin sulfate/EDTA solution—Dissolve 80 mg of heparin sulfate in 40 ml of 0.5 M EDTA. Stir thissolution into 4 liters of pig blood immediately after it iscollected to prevent clotting.

Isotonic phosphate buffer (310 mosM, pH 6.8)—Dis-solve 85.6 g of NaH2PO4 (monobasic monohy-drate) and 88.0 g of Na2HPO4 (dibasic anhydrous)in 9.6 liters of distilled water. Chill the solution,adjust the pH to 6.8 with HCl, and bring the finalvolume to 10 liters with distilled water.

Hypotonic phosphate buffer (20 mosM, pH 6.8)—Add600 ml of isotonic phosphate buffer to 8.7 liters ofdistilled water. Adjust the pH to 6.8 if necessary.

Plastic centrifuge bottles (250–300 ml)—At least six re-quired.

Plastic centrifuge tubes (30–50 ml)—50 required.

15-ml Plastic conical tubes with screw caps—100 re-quired.

1.5-ml Microcentrifuge tubes—500 required.

Disposable Pasteur pipettes—Approximately 500 re-quired.

4-ml Glass vials with Teflon-lined screw caps—Approx-imately 400 required.

100°C Oven—One required.

Silica-gel thin-layer chromatography plates—We pur-chase these from Fisher (catalog #05-713-317).The plates are prepared (activated) by heating in a100°C oven for 5 to 10 min to remove any mois-ture from the matrix. Remove from the oven andstore desiccated at room temperature until they areready to be used. Approximately 25 plates will berequired (1 plate/2 groups).


Thin-layer chromatography tanks—5 to 10 required, de-pending on their size.

I2 Chamber—Fill the bottom of a rectangular glasschamber with I2 (solid) and cover tightly. Allow 6to 8 hr for the chamber to saturate with I2 vaporsbefore the thin-layer chromatography plates aredeveloped.

Pressurized N2 tank—1 required.

Perchloric acid (70%)—Aldrich catalog #31, 142-1 orequivalent. 200 ml required.

Cholesterol assay reagent, pH 6.5 (See Supplies andReagents list for the formulation)—We purchasethis reagent in dry, mixed form from Sigma (cata-

250 ml H2O, and add 5 drops of concentratedH2SO4.

Arsenomolybdate reagent—Dissolve 125 g (NH4)6Mo7

O24 � 4 H2O in 2.25 liters of distilled water, andadd 105 ml of concentrated H2SO4. Dissolve 15.0g of Na2HAsO4 � 7 H2O in 125 ml of distilled wa-ter, and add to the above acid molybdate solution.Store in a brown bottle for 24 hr at 37°C. Thisreagent should be yellow with no green tint.

Glucose standard (1.0 �mol/ml in 1.0 M NaCl)—Dis-solve 36 mg (weighed to 0.1 mg) and 11.2 g NaClin distilled water and dilute to 200 ml with distilledwater.

Large sheets Whatman No. 1 chromatography paper—20 sheets required.

1% Glucose-1-phosphate standard—Dissolve 100 mgglucose-1-phosphate dipotassium salt in 10 ml ofdistilled water; store at 0 to 5°C.

1% Glucose-6-phosphate standard—Dissolve 100 mgglucose-6-phosphate Na or K salt in 10 ml of dis-tilled water; store at 0 to 5°C.

1% Glucose standard in 1 N HCl—Dissolve 100 mg glu-cose in 10 ml of 1 N HCl (concentrated HCl:H2O;1:11); store at 0 to 5°C.

1% Inorganic phosphate standard—Dissolve 100 mg ofKH2PO4 in 10 ml of distilled water.

10% Mg(NO3)2 in ethanol—Dissolve 5 g of Mg(NO3)2� 6 H2O in 10 ml of absolute ethanol.

Methanol—2.5 liters required.

88% Formic acid—500 ml required.

Aniline–acid–oxalate spray—Dissolve 4.5 g oxalic acid in1 liter of distilled water, add 9 ml of aniline. Storein a brown bottle.

Modified Hanes–Ischerwood spray reagent—Mix thefollowing: 250 ml of 4% ammonium molybdate(10 g (NH4)6Mo7O24 � 4 H2O dissolved in H2Oto yield 250 ml of solution), 100 ml of 1.0 N HCl(see instructions earlier in Experiment 12 list),50 ml of 70% perchloric acid, and 600 ml of dis-tilled water. Prepare this reagent on the day of use.Use phosphate (detergent)-free glassware for mix-ing, storing, and spraying.

Acid-FeCl3 in acetone—Mix 750 mg of FeCl3 � 6H2O,15 ml of 0.3 M HCl (12.5 ml concentrated HCl �487.5 ml of H2O), and 485 ml of acetone until dis-solved. Store in a tightly stoppered bottle.

1.25% Sulfosalicylic acid in acetone—Dissolve 6.25 g ofsulfosalicylic acid in 500 ml of acetone. Prepare onthe day of the experiment. Store in a tightly stop-pered bottle.

UV lamp—1 or 2 required.

100°C oven—1 required.


422

Folin–Ciocalteau reagent (2 N)—200 ml required.

Lysozyme solution (1 mg/ml)—Dissolve 100 mg oflysozyme in 100 ml of distilled water. Store at 4°C.

Phosphate standard solution—Dissolve 13.8 mg ofNaH2PO4 � H2O in 100 ml of distilled water.

5% Ammonium molybdate solution—Dissolve 5 g ofammonium molybdate in 100 ml of distilled water.

Amidol reagent—Dissolve, in order, 25 g of sodium bisul-fite and 1 g of 2,4-diaminophenol in 100 ml of dis-tilled water. Make this reagent fresh on the day thatthe assay is performed.

SDS cleaning solution—Dissolve 10 g of sodium dode-cyl sulfate in 2 liters of distilled water.

Experiment 14: ElectronTransport

Oxygen electrode system with recorder—For example,Yellow Spring Instruments Model 53 oxygen elec-trode (Yellow Springs, OH), or equivalent, at-tached to a linear 10-mV recorder. Five of theseunits will be required.

Oxygen electrode electrolyte—Fluid should be that spec-ified by the manufacturer of the oxygen electrode.Approximately 250 ml of this solution will be re-quired.

Oxygen electrode membranes—As specified by the man-ufacturer of the oxygen electrode—5 required.

Magnetic stirrers and stirring bars—Standard magneticstirrers with stirring bars (or cut up paper clips)small enough to fit in sample chamber of the oxy-gen electrodes—5 required.

Sample injection syringe with tubing end (5 required)—Equip a calibrated 1-ml syringe with a short nee-dle and slip a piece of polyethylene tubing 1 to 2in. in length over the needle. The tubing diame-ter and length should be the minimum necessaryto reach to the bottom of the sample chamber ofthe oxygen electrode.

Fresh pig heart—5 required.

Meat grinder—1 or 2 required.

1 mM Ethylenediaminetetraacetate (EDTA), pH 7.4—Dissolve 7.44 g of disodium EDTA in 20 liters ofdistilled water and titrate to pH 7.4 with NaOH.(You will use 5 to 10 liters of this to prepare otherreagents.)

Blender—1 or 2 required.

0.02 M Potassium phosphate, 1 mM EDTA (pH 7.4)—Dissolve 13.95 g of K2HPO4 and 2.7 g of KH2PO4

in 5 liters of 1 mM EDTA, pH 7.4.

High-speed centrifuge—This centrifuge must be capa-

log #352-50) and reconstitute each vial accordingto the manufacturer’s instructions. Each group willrequire 6 ml.

Cholesterol standard solutions (1 mg/ml, 2 mg/ml, and4 mg/ml)—We purchase prepared sets of thesestandards from Sigma (catalog #C-0534). Thesestandards are supplied as 100, 200, and 400 mg/dlsolutions, which are equivalent to 1, 2, and 4-mg/ml solutions, respectively, in an aqueous/isopropanol solvent. They are ready to use in thecholesterol quantitation assay.

Phospholipid standard solutions in CHCl3—We purchasesphingomyelin from Sigma (catalog #S-1131). Dis-solve 25 mg of sphingomyelin in 0.5 ml of CHCl3.We purchase L--phosphatidylcholine from Sigma(catalog #P-6638). It is supplied as a 10 mg/ml so-lution in CHCl3 and is ready to use as a standard.We purchase L--phosphatidylethanolamine fromSigma (catalog #P-9137). It is supplied as a 10mg/ml solution in CHCl3 and is ready to use as astandard. We purchase L--phosphatidyl-L-serinefrom Sigma (catalog #P-7769). Dissolve 25 mg ofL--phosphatidyl-L-serine in 0.5 ml of CHCl3. Wepurchase L--phosphatidylinositol from Sigma(catalog #P-2517). It is supplied as a 10 mg/ml so-lution in CHCl3 and is ready to use as a standard.We purchase cholesterol from Sigma (catalog#8667). Dissolve 500 mg of cholesterol in 10 ml ofCHCl3. Store all of these solutions tightly capped at�20°C.

Heating block—5 to 10 required, depending on the ca-pacity of the blocks for the 4-ml vials.

Preparative centrifuge—At least one required.

Clinical tabletop centrifuge (low speed)—At least one re-quired.

Spectrophotometer to read absorbance at 500 and660 nm with cuvettes—At least one required.

Chloroform:methanol (1:2)—Mix 150 ml of chloroformwith 300 ml of methanol. Store tightly covered ina dark bottle at room temperature.

0.1 M HCl—Add 500 �l of concentrated HCl to 60 mlof distilled water.

Mobile phase for thin-layer chromatography—Mix 1.25liters of chloroform, 750 ml of methanol, 200 ml ofglacial acetic acid, and 50 ml of distilled water. Storetightly covered in a dark bottle at room temperature.

1% Copper sulfate solution—Dissolve 1 g of CuSO4 �5H2O in 100 ml of distilled water.

2% Sodium tartrate solution—Dissolve 2 g of sodiumtartrate in 100 ml of distilled water.

2% Sodium carbonate solution—Dissolve 40 g ofNa2CO3 and 8 g of NaOH in 2 liters of distilledwater.


0.01 M Methylene blue—Dissolve 935 mg of methyleneblue in 250 ml of distilled water.

Antimycin A (180 �g/ml 95% ethanol)—Dissolve 9.0 mgof Antimycin A in 50 ml 95% (vol/vol) ethanol.

95% (vol/vol) Ethanol—Mix 47.5 ml of absolute ethanolwith 2.5 ml of distilled water.

6 M KOH—Dissolve 17 g of KOH in 50 ml of distilledwater.

DPIP solution—Dissolve 67.5 mg of 2,6-dichlorophenol-indophenol (DPIP) solution in 250 ml of distilledwater.

Experiment 15: Study of thePhosphoryl Group TransferCascade Governing GlucoseMetabolism: Allosteric andCovalent Regulation of Enzyme Activity

1 M Tris, pH 8.0—Dissolve 88.8 g of Tris � HCl and53 g of Tris free base in about 700 ml of distilledwater. The pH of this solution will be approxi-mately 8.0. Adjust the pH if necessary with diluteNaOH or HCl. Bring the final volume of the so-lution to 1 liter with distilled water.

1 M Glycerophosphate, pH 8.0—Dissolve 108 g of glyc-erophosphate (disodium hydrate, Sigma catalog#G-6501) in about 350 ml of distilled water. Ad-just the pH to 8.0 with HCl. Bring the final vol-ume of the solution to 500 ml with distilled water.Filter-sterilize the solution and store at 4°C.

60 mM Mg(acetate)2, pH 7.3—Dissolve 0.64 g of mag-nesium acetate tetrahydrate in 40 ml of distilledwater. Adjust the pH to 7.3 with dilute NaOH.Bring the final volume of the solution to 50 ml withdistilled water. Filter-sterilize the solution andstore at 4°C.

Concentrated enzyme storage buffer—Add 0.5 ml of 1 MTris, pH 8.0 (see above), 0.5 ml of 1 M glyc-erophosphate, pH 8.0 (see above), 10 �l of �-mer-captoethanol, and 40 �l of 0.5 M EDTA, pH 8.0,to 6 ml of distilled water. The 0.5 M EDTA, pH8.0, is prepared as follows: dissolve 186 g of di-sodium EDTA dihydrate in 800 ml of distilled wa-ter. Adjust the pH to 8.0 with concentrated NaOH.Bring the final volume of the solution to 1 literwith distilled water. Add 4 ml of glycerol to thissolution and mix thoroughly.

Reaction buffer—Dilute 2.5 ml of 1 M Tris, pH 8.0, and2.5 ml of 1 M glycerophosphate, pH 8.0, in 45 mlof distilled water.

ble of centrifuging 300 to 600 ml of fluid at at least5000 � g. A Sorvall RC2B centrifuge is satisfac-tory.

0.25 M Potassium phosphate, 1 mM EDTA (pH 7.4)—Dissolve 35.0 g of K2HPO4 and 6.7 g of KH2PO4

in 1 liter of 1 mM EDTA, and adjust to pH 7.4.

l.0 M Acetic acid—Dilute 29.5 ml of glacial acetic acidwith 470 ml of distilled water.

1.0 mM NAD�—Dissolve 40 mg of Na2NAD� � 4 H2Oin 50 ml of distilled water. Refrigerate until use.Stable for 1 week.

0.3 M Potassium phosphate (pH 7.4)—Dissolve 210 g ofK2HPO4 and 40.5 g of KH2PO4 in 5 liters of dis-tilled water; adjust pH to 7.4 if necessary with l.0 MKOH or HCl.

0.05 M Sodium succinate—Dilute 0.5 M sodium succi-nate, prepared as below (5 ml of 0.5 M sodium suc-cinate � 45 ml of H2O). Refrigerate. Stable for 1week.

0.5 M Sodium succinate—Dissolve 33.9 g of sodium suc-cinate in 250 ml of distilled water. Refrigerate. Sta-ble for 1 week.

0.25 M Malate (Na salt)—Dissolve 1.675 g of malic acidand 500 mg of NaOH in approximately 40 ml ofdistilled water; adjust pH to 6 to 8 with 1 N NaOH,and dilute to 50 ml with distilled water. Refriger-ate. Stable for 1 week.

0.25 M �-Hydroxybutyrate—Dissolve 1.3 g of �-hydroxybutyric acid in 25 ml of distilled water bygradual addition of 12.5 ml of 1 N NaOH (finalpH should be in the range from 6 to 8); dilute toa final volume of 50 ml with distilled water. Alter-natively, use the sodium salt of �-hydroxybutyrate(1.575 g) and adjust the pH with HCl. Refriger-ate. Stable for 1 week.

6 � 10�4 M Cytochrome c—Quantities used depend onpurity of starting material. The molecular weightis 13,000. Prepare 50 ml of this solution. Refrig-erate. Stable for 1 week.

Ascorbic acid or sodium ascorbate—Solutions are pre-pared by the students just before use (see Experi-ment 14). Approximately 25 g will be required.

1 N NaOH—Dissolve 2.0 g of NaOH in 50 ml of dis-tilled water. (Not needed if sodium ascorbate isprovided.)

0.5 M Malonate—Suspend 2.6 g of malonic acid in 6 mlof distilled water. Adjust to pH 6 to 8 with 4.0 NNaOH (80 g NaOH/500 ml) and dilute to 50 mlwith distilled water.

0.1 M KCN—Dissolve 325 mg of KCN in 50 ml of dis-tilled water on the day of use. Keep in a tightlystoppered bottle marked Poison. Provide apropipette for dispensing.


424

Film cassette for autoradiography—2 to 5 required, de-pending on the size of the gels and cassettes.

Dark room with solutions for film development.

Kodak X-omat film—2 to 5 large pieces required.


Glass test tubes (13 � 100 mm)—1000 required.

10-ml Pipettes with bulbs—50 required.

30 mM cysteine, pH 7.0—Dissolve 527 mg of L-cysteinehydrochloride monohydrate in 80 ml of distilledwater. Adjust the pH to 7.0 by slowly adding solidsodium bicarbonate. Bring the final volume of thesolution to 100 ml with distilled water. Prepare thisreagent on the day that it is to be used.

80 mM cysteine, pH 6.8—Dissolve 1.4 g of L-cysteinehydrochloride monohydrate in 80 ml of distilledwater. Adjust the pH to 6.8 by slowly adding solidsodium bicarbonate. Bring the final volume of thesolution to 100 ml with distilled water. Prepare thisreagent on the day that it is to be used.

250 mM Tris, pH 7.7—Dissolve 28.6 g of Tris � HCl and8.3 g of Tris free base in 800 ml of distilled water.The pH of the solution should be at or near pH7.7. Adjust the pH to 7.7 with dilute HCl or NaOHif necessary. Bring the final volume of the solutionto 1 liter with distilled water.

250 mM glycerophosphate in 250 mM Tris, pH 7.7—Dissolve 54 g of glycerophosphate (disodium salt)in 700 ml of 250 mM Tris, pH 7.7. Adjust the pHto 7.7 with dilute NaOH or HCl, if necessary, andbring the final volume of the solution to 1 liter with250 mM Tris, pH 7.7.

40 mM glycerophosphate in 80 mM cysteine, pH 6.8—Dissolve 864 mg of glycerophosphate (disodiumsalt) in 70 ml of 80 mM cysteine, pH 6.8. Bringthe final volume of the solution to 100 ml with80 mM cysteine, pH 6.8.

Glycogen phosphorylase kinase solution—Dilute 40 �l ofthe 2500-unit/ml glycogen phosphorylase kinasestock solution in 4.96 ml of 30 mM cysteine, pH 7.0.

30 mM ATP solution—Dissolve 165 mg of adenosine 5�-triphosphate (disodium salt) in 8 ml of distilled wa-ter. Adjust the pH to 7.7 with very dilute NaOHand bring the final volume of the solution to 10 mlwith distilled water. Divide the solution into 20- to50-�l aliquots and store at �70°C. Thaw on icejust before use.

100 mM Mg(acetate)2, pH 7.7—Dissolve 21.45 g of mag-nesium acetate terahydrate in 900 ml of distilledwater. Adjust the pH to 7.7 with NaOH, and bringthe final volume of the solution to 1 liter with dis-tilled water.

500 mM potassium phosphate, pH 6.8—Dissolve 19.05 gof monobasic potassium phosphate (anhydrous)

2 mM ATP solution—Dissolve 11 mg of adenosine 5�-triphosphate (disodium salt) in 10 ml of 60 mMMg(acetate)2, pH 7.3. Divide into 20- to 50-�laliquots and store at �70°C. Thaw on ice just be-fore use.

Spectrophotometer with cuvettes or colorimeter tubes—At least 1 required.

Glycogen phosphorylase b stock solution—Dissolve25 mg (�750 units) of glycogen phosphorylase b(Sigma catalog #P-6635) in approximately 750 �lof concentrated enzyme storage buffer (resuspendthe lyophilized protein gently, do not use a vortexmixer). Divide into 20 �l aliquots and store at�20°C. Dilute this �1000-unit/ml stock solutionto the required concentration (see protocol) justbefore use. Store the dilute enzyme on ice until itis ready to be used.

Glycogen phosphorylase kinase stock solution—Dissolvea 2500-unit package size of glycogen phosphory-lase kinase (Sigma catalog #P-2014) in 1 ml of con-centrated enzyme storage buffer (gently, do not usea vortex mixer). Divide into 20-�l aliquots andstore at �20°C. Dilute this 2500-unit/ml stock so-lution to the appropriate concentration (see pro-tocol) just before use. Store the dilute enzyme onice until it is ready to be used.

4X SDS/EDTA buffer—Dissolve 3.51 g of Tris � HCland 0.335 g of Tris free base in 50 ml of 0.5 MEDTA, pH 8.0. Adjust the pH to 7.0 with diluteNaOH or HCl if necessary. Add 30 ml of glycerol,10 ml of �-mercaptoethanol, and 10 ml of distilledwater. Dissolve (avoid foaming) 8 g of SDS intothis solution, as well as 4 mg of bromophenol blue.Store the solution tightly capped at room temper-ature.

-[32P]ATP (4000 Ci/mmol)—We purchase this fromICN Biochemicals (catalog #35001x). We dilutethis solution 1:80 with distilled water for use inthe in vitro phosphorylation assay. At this specificactivity, a 12- to 18-hr exposure time is usuallyrequired. If you choose to do a greater dilution of this reagent, longer exposure times will be re-quired. For safety purposes, we keep this reagentshielded and in the hands of the teaching assis-tant. The students bring their prepared reactionsto the teaching assistant, who adds the reagent forthem.

Prestained high-molecular-weight protein markers—Wepurchase these from Gibco BRL (catalog #26041-020) and prepare them according to the manufac-turer’s instructions.

Plastic wrap—1 large roll required.

Whatman 3 MM filter paper—Approximately 8 to 10sheets required.


buffer. Divide into 20-�l aliquots and store at�20°C. Dilute this 1000-units/ml stock enzymesolution 1:500 in 40 mM glycerophosphate, pH6.8, shortly before use. Store on ice.

Glycogen phosphorylase b control solution (2 units/ml)—Dilute the 1000-units/ml enzyme stock so-lution (see above) 1:500 in 40 mM glycerophos-phate, pH 6.8.

Glycogen phosphorylase b control solution (2 units/ml in60 mM AMP)—Dilute the 1000-units/ml enzymestock solution (see above) 1:500 in 40 mM glyc-erophosphate containing 60 mM AMP. This solu-tion is prepared by dissolving 2.08 g of adenosine5�-monophosphate (monosodium salt) in 100 ml of40 mM glycerophosphate, pH 6.8.

Reagents and supplies for SDS–PAGE—See Experi-ment 4.

Experiment 16: Experiments inClinical Biochemistry andMetabolism

Fresh pig blood—Approximately 100 to 200 ml required.

Heparin sulfate/EDTA solution—Dissolve 4 mg of he-parin sulfate in 2 ml of 0.5 M EDTA. Stir this so-lution into 200 ml of pig blood immediately after it iscollected to prevent clotting.

5-ml and 10-ml Pipettes—50 of each required.



Glucose standard stock solution (44.48 mM)*—Dissolve80.1 mg of D-glucose in 10 ml of distilled water.

3.00 mM glucose standard solution*—Add 67 �l of44.48 mM glucose standard stock solution to 933�l of distilled water.

11.10 mM glucose standard solution*—Add 250 �l of44.48 mM glucose standard stock solution to750 �l of distilled water.

16.65 mM glucose standard solution*—Add 374 �l of44.48 mM glucose standard stock solution to626 �l of distilled water.

4.44 mM lactate standard—Dissolve 4.0 mg of D-lacticacid (free acid, crystalline) in 10 ml of distilled water.

and 62.71 g of dibasic potassium phosphate (anhy-drous) in 800 ml of distilled water. The pH of thesolution should be near 7.2. Adjust the pH of thesolution to 6.8 with HCl, and bring the final vol-ume of the solution to 1 liter with distilled water.

4% (wt/vol) glycogen solution—Dissolve 10 g of rabbitliver glycogen (Sigma catalog #G-8876) in distilledwater to a total volume of 250 ml. Filter sterilizethe solution and store at 4°C.

300 mM MgCl2—Dissolve 30.5 g of MgCl2 � 6 H2O indistilled water to a final volume of 500 ml.

100 mM EDTA, pH 8.0—Dissolve 18.61 g of disodiumEDTA dihydrate in 400 ml of distilled water. Ad-just the pH to 8.0 with NaOH and bring the finalvolume of the solution to 500 ml with distilled wa-ter.

6.5 mM nicotinamide adenine dinucleotide phosphate so-lution (NADP)—We purchase this reagent inpreweighed vials from Sigma (catalog #240-310).To prepare the 6.5 mM reagent, dissolve 10-mgvial of NADP in 2.0 ml of distilled water. Preparethis reagent immediately before use.

0.1% (wt/vol) -D-glucose-1,6-diphosphate solution—Dissolve 0.1 g of -D-glucose-1,6-diphosphate(potassium salt, hydrate) in 100 ml of distilled wa-ter. Filter-sterilize the solution and store in smallaliquots at �20°C.

Phosphoglucomutase solution (10 units/ml)—We pur-chase this enzyme from Sigma (catalog #P-3397)as a crystalline suspension in 2.5 M ammonium sul-fate. Consult the specification sheet for the prod-uct (lot-specific, usually 100–300 units/mg of pro-tein) to determine a dilution that will yield a10-unit/ml solution. The dilution of this enzymeis done in distilled water immediately before use.Store the remainder of the crystalline suspensionat 4°C for use in future semesters.

Glucose-6-phosphate dehydrogenase solution (10 units/ml)—We purchase this enzyme from Sigma (cata-log #G-6378), which is derived from baker’s yeast.Reconstitute the enzyme in distilled water as perthe manufacturer’s instructions. Consult the spec-ification sheet (lot-specific, usually 200–400units/mg of protein) to determine a dilution of theenzyme stock that will yield a 10-unit/ml solution(dilution is performed in distilled water). Store theremainder of the reconstituted stock enzyme solu-tion in small aliquots at �20°C for use in futuresemesters.

Glycogen phosphorylase a enzyme solution (2 units/ml)—We purchase this enzyme from Sigma (catalog #P-1261). Dissolve 25 mg ( �625 units) of glyco-gen phosphorylase a (gently, do not use a Vortexmixer) in 625 �l of concentrated enzyme storage


*Sigma offers a combined glucose/urea standard set (catalog #16-11)

that can be used for these assays. These standards contain 800 mg/dL

(44.48 mM glucose, 133.3 mM urea), 300 mg/dl (16.65 mM glucose,

50 mM urea), and 100 mg/dl (5.55 mM glucose, 16.66 mM urea) of

each compound. These standards can be diluted appropriately to

prepare the standard solutions required for this assay.

426

Anti-�-galactosidase rabbit serum*—Each group will re-quire 180 �l. Store at �20°C

Polyoxyethylenesorbitan monolaurate (Tween 20, Sigmacatalog #P-5927)—about 10 ml required.

Immulon-2 96-well flat-bottomed microtiter (ELISA)plate (50 required)—We purchase these fromFisher (catalog #1424561), but any standard ELISAplate would suffice.

Blocking solution—Dissolve 50 g of bovine serum albu-min (BSA) and 2.5 ml of polyoxyethylenesorbitanmonolaurate (Tween 20) in 5 liters of 1X PBS.

Wash solution—Dissolve 5 ml of Tween 20 in 10 litersof 1X PBS.

�-Galactosidase stock solution—Dissolve 5 mg of �-galactosidase (Sigma catalog #G-6008) in 50 ml of1X PBS. Store at �20°C.

�-Galactosidase solution (10 �g/ml)—Add 20 ml of the�-galactosidase stock solution to 180 ml of 1X PBS.Store at �20°C.

�-Galactosidase solution (40 �g/ml)—Add 8 ml of �-galactosidase stock solution to 12 ml of 1X PBS.Store at �20°C.

�-Galactosidase solutions (“unknown concentrations”)—Prepare about 5 ml of each unknown concentra-tion by making the appropriate dilution of the �-galactosidase stock solution in 1X PBS. We havefound that unknown concentrations in the rangefrom 10 �g/ml to 100 �g/ml work well in the com-petitive ELISA. Store each unknown solution at�20°C.

4 N H2SO4—Dilute 666 ml of 12 N H2SO4 to 2 literswith distilled water. Store covered at room tem-perature. Label with the warning Caution: StrongAcid.

HRP substrate—Mix equal volumes of each of the3,3�,5,5�-tetramethylbenzidene and H2O2 solu-tions (Kirkegaard & Perry catalog #50-7600) toprepare the substrate. This should be done right be-fore the students are ready to develop their microtiterplates. About 750 ml of this substrate is requiredfor the full three-day experiment.

Urea standard stock solution (35.70 mM)*—Dissolve21.4 mg of urea in 10 ml of distilled water.

3.57 mM urea standard*—Add 100 �l of 35.70 mM ureastandard stock solution to 900 �l of distilled water.



Serum lactate reagent—Sigma catalog #735-10. Prepareby adding distilled water to the dry mixture as perthe instructions. Each group will require 6 ml.

Serum urea nitrogen reagent—Sigma catalog #66-20.Prepare by adding distilled water to the dry mix-ture as per the instructions. Each group will re-quire 12 ml.

Serum glucose reagent—Sigma catalog #16-50. Prepareby adding distilled water to the dry mixture as perthe instructions. Each group will require 10 ml.

Serum creatine kinase reagent—Sigma catalog #DG-147-K. Prepare Reagents A and B by adding dis-tilled water to the dry mixtures as per the supplier’sinstructions. Mix Reagents A and B at the recom-mended ratios. Each group will require 5 ml of thisworking solution.

Experiment 17: PartialPurification of a PolyclonalAntibody, Determination of Titer, and Quantitation of anAntigen Using the ELISA

10X Phosphate-buffered saline (10X PBS)—Dissolve5.12 g of NaH2PO4 � H2O, 44.96 g of Na2HPO4 �7H2O, and 87.6 g NaCl in 1.5 liter of distilled water. Adjust the pH to 7.3 with HCl or NaOH asneeded. Bring the final volume of the solution to2 liters with distilled water. Store short term atroom temperature or long term at 4°C. Dilute thissolution 1:10 with distilled water to prepare thePBS solution (1X) used in the blocking and washsolutions described below.

Normal or pre-immune rabbit serum—Each group willrequire 180 �l. Store at �20°C. We have also pur-chased normal rabbit serum from Sigma (catalog#R-4505).


*Sigma offers a combined glucose/urea standard set (catalog #16-11)

that can be used for these assays. These standards contain 800 mg/dL

(44.48 mM glucose, 133.3 mM urea), 300 mg/dl (16.65 mM glucose,

50 mM urea), and 100 mg/dl (5.55 mM glucose, 16.66 mM urea) of

each compound. These standards can be diluted appropriately to

prepare the standard solutions required for this assay.

* As stated in the text, the dilutions for the primary antibodies used in

this experiment are approximate. We have found that the titer of the J1

monoclonal antibody varies somewhat from lot to lot. Obviously, the

titer of the rabbit polyclonal antibody will also vary from animal to

animal. Because of this, we suggest that the experiment be performed in

advance of the class in case some small adjustments of the dilutions stated in

the text are necessary. The secondary antibody dilutions described in the

text are much less variable from semester to semester. We raise the

rabbit-anti-�-galactosidase polyclonal antibody by shipping the antigen

(Sigma catalog #G-6008) to CoCalico Biologicals, Inc. in Reamstown,

PA. This company will raise the antibody and send you batches of the

serum. The company will test the serum antibodies in an ELISA to

determine if they are adequate for your application.

020) and prepare them according to the manufac-turer’s instructions.

Coomassie Blue staining solution—Dissolve 400 ml ofmethanol, 100 ml of glacial acetic acid, and 2.5 gof Coomassie Brilliant Blue R-250 in 500 ml of dis-tilled water. Store tightly covered at room tem-perature.

Coomassie Blue destaining solution—Dissolve 400 ml ofmethanol and 100 ml of glacial acetic acid in 500 ml of distilled water. Store tightly covered atroom temperature.

Methanol—200 ml required to prepare 1 liter of trans-fer buffer.

Transfer buffer—We suggest that this solution be preparedfresh on the day of the experiment. Dissolve 11.6 g ofTris base, 5.86 g of glycine, and 0.038 g of sodiumdodecyl sulfate in 700 ml of distilled water. AdjustpH to 9.2 with HCl or NaOH as needed. Add200 ml of methanol. Bring the final volume of thesolution to 1 liter with distilled water. Store thissolution tightly covered at room temperature.

Immobilon-P PVDF membrane (0.45-�m pore size)—We purchase this from Millipore (catalog #IPVH000 10). One large roll required.

Whatman 3 MM filter paper—Approximately 50 sheetsrequired, depending on the size of the gels and thetype of transfer apparatus used.

Western blot blocking solution—We suggest that this so-lution be prepared fresh on the day of the experiment.Dissolve 2.42 g of Tris base, 14.62 g of NaCl, and86.22 g of nonfat dry (powdered) milk to 700 mlof distilled water. Bring the final volume of the so-lution to 1 liter with distilled water.

Tween 20 (polyoxyethylenesorbitan monolaurate)—Wepurchase this from Sigma (catalog #P-5927). Ap-proximately 5 ml will be required.

Western blot wash solution—We suggest that this solutionbe prepared fresh on the day of the experiment. Theamount given is sufficient for one group of students.Dissolve 2.42 g of Tris base, 14.62 g of NaCl, and86.22 g of nonfat dry (powdered) milk to 700 ml ofdistilled water. Add 0.5 ml of polyoxyethylenesor-bitan monolaurate (Tween 20). Bring the final vol-ume of the solution to 1 liter with distilled water.

Mouse-antiglutathione-S-transferase monoclonal anti-body—We purchase this from Sigma (catalog #G-1160). This antibody is supplied as a sterile liquid.Dilute as specified in the protocol in blocking so-lution. Prepare about 5 liters ( �100 ml per group).

HRP-conjugated goat-antimouse IgG—We purchasethis reagent from Pierce Chemical Co. (catalog#31430). This reagent is supplied as a lyophilizedpowder. Reconstitute and store as specified by the

Mouse-anti-�-galactosidase monoclonal antibody( J1)*—Keep the stock antibody (Sigma catalog#G-8021) frozen at �80°C. We have found thatthe titer drops dramatically after short-term stor-age at �20°C. Dilute the antibody immediately be-fore use in blocking solution (about a 1:1000 dilu-tion). Each group will require about 1.2 ml of thediluted antibody to perform the three-day experi-ment.

HRP-conjugated-goat-antirabbit IgG*—Each group willprepare 2.5 ml of this solution by making serial di-lutions of the stock antibody (Kirkegaard & Perrycatalog #074-1506) in blocking solution. The finaldilution should be about 1:9000. Prepare immedi-ately before use.

HRP-conjugated-goat-antimouse IgG*—Each groupwill prepare 0.8 ml of this solution by making se-rial dilutions of the stock antibody (Kirkegaard &Perry catalog #074-1802) in blocking solution. Thefinal dilution should be about 1:10,000. Prepareimmediately before use.

Saturated ammonium sulfate solution—Dissolve 350 g ofammonium sulfate (reagent grade) in 0.5 liter ofdistilled water. Store at room temperature.


10-ml Pasteur pipettes—50 required.


96-Well microplate spectrophotometer—1 required,with a printer.

Experiment 18: Western Blot toIdentify an Antigen

Reagents and apparatus for SDS–PAGE—See Experi-ment 4.

Power supplies—Enough to power all available SDS–PAGE and Western blot transfer units.

Prestained high-molecular weight protein markers—Wepurchase these from Gibco-BRL (catalog #26041-


* As stated in the text, the dilutions for the primary antibodies used in

this experiment are approximate. We have found that the titer of the J1

monoclonal antibody varies somewhat from lot to lot. Obviously, the

titer of the rabbit polyclonal antibody will also vary from animal to

animal. Because of this, we suggest that the experiment be performed in

advance of the class in case some small adjustments of the dilutions stated in

the text are necessary. The secondary antibody dilutions described in the

text are much less variable from semester to semester. We raise the

rabbit-anti-�-galactosidase polyclonal antibody by shipping the antigen

(Sigma catalog #G-6008) to CoCalico Biologicals, Inc. in Reamstown,

PA. For a very fair price, this company will raise the antibody and send

you batches of the serum. The company will even test the serum

antibodies in an ELISA to determine if they are adequate for your

application.

428

Stock culture of wild-type B. subtilis (Bacillus GeneticStock Center strain 1A2)—1 required.

Inoculating needle—1 required.

Luria broth—Luria broth: Prepare five 100-ml and five10-liter solutions containing 10 g/liter Bacto-Tryptone, 5 g/liter Bacto-Yeast extract, 5 g/literNaCl, 0.5 g/liter antifoam agent (e.g., PolyglycolP-2000), and 1 g/liter glucose (autoclave sepa-rately). Autoclave the main solution and separatesolution of concentrated glucose for 20 min at 105to 110°C and cool; use sterile conditions to add theglucose solution to the main solution and usewithin 24 hr.

37°C Incubator-shaker—1 or 2 required.

Fermenter capable of handling 50 liters of growthmedium.

Centrifuges—For harvesting the 10-liter culture, a continuous-flow centrifuge such as a Sharples ismost convenient. For washing the cells, a refriger-ated high-speed laboratory centrifuge (e.g., SorvallRC-2B) is necessary.

Sterile saline solution—Dissolve 4.25 g of NaCl in500 ml of distilled water and autoclave (20 min at105–110°C).

Parafilm or wax paper—1 or 2 rolls required.

Bacterial cell paste—If performing Experiment 20 also,grow wild-type Bacillus subtilis (Bacillus GeneticStock Center strain 1A2, Columbus, OH) as de-scribed for Experiment 20 or buy wild-type B. sub-tilis from Grain Processing Corp. Muscatine, IA,or General Biochemicals, Chagrin Falls, OH. Ifnot performing Experiment 20, grow or buy anybacterial species (late log phase). Provide frozencells for students. Approximately 50 g will be re-quired.

0.15 M NaCl, 0.l M Na2EDTA—Dissolve 21.9 g of NaCland 93 g of disodium EDTA � 2H2O in 2.5 litersof distilled water.

Lysozyme solution, 10 mg/ml—Dissolve 1.25 g of crys-talline egg white lysozyme in 125 ml of distilledwater. Refrigerate until use. Stable for 24 to 48 hr.

25% sodium dodecyl sulfate solution—Dissolve 125 g ofsodium dodecyl sulfate in 400 ml of distilled waterand dilute to 500 ml with distilled water.

5.0 M NaClO4—Dissolve 1.53 kg of NaClO4 (anhy-drous) in 2 liters of distilled water and dilute to 2.5liters with distilled water.

Chloroform-isoamyl alcohol (24:1)—Mix 4.8 liters ofchloroform and 200 ml of isoamyl alcohol. Storetightly capped in a dark bottle at room tempera-ture.

95% Ethanol—Add 250 ml of distilled water to 4.75 litersof absolute ethanol. Store tightly capped.

Dilute (1/10 X) saline-citrate (0.015 M NaCl, 1.5 mM

manufacturer’s instructions. Dilute as specified inthe protocol in blocking solution. Prepare about 5 liters (�100 ml per group).

Razor blades—10 to 20 required (students can share these).

Aluminum foil—1 large roll required.

Small metal or plastic trays for membrane incubations—50 required.

SuperSignal West Pico Chemiluminescent Substrate—Pierce catalog #34080. Prepare the working solutionimmediately before use by mixing equal volumes ofthe SuperSignal luminol/enhancer solution and thestable peroxide solution. Each group will require 5to 10 ml, depending on the size of the membrane.

Plastic wrap—1 large roll required.

Kodak X-omat film—5 to 7 large pieces required, de-pending on the size of the membranes and film cas-settes.

Film cassettes—5 to 7 required, depending on the sizeof the membranes.

Darkroom—1 required, preferably fitted with a “safelight.”

Transfer apparatus—We use the BIORAD Transblot SDmodel. With mini-gels, we can fit at least threeblots per unit.

Reagents for Colorimetric Detection of

HRP (optional)

10 mM Tris-buffered saline (TBS), pH 7.4—Dissolve 1.8 g of Tris free base and 13 g of NaCl to a totalvolume of 1.4 liters with distilled water. Adjust thepH to 7.4 with 1 M HCl. Bring the final volumeof the solution to 1.5 liters with distilled water.

4-Chloro-1-naphthol—This reagent can be purchasedfrom Pierce Chemical Co. (catalog #34010). Seethe protocol for the preparation of this substratesolution.

Absolute ethanol—200 ml required.

3% (wt/wt) hydrogen peroxide in distilled water—Prepare this reagent immediately before use byadding 10 ml of a 30% (wt/wt) hydrogen peroxidestock solution (Sigma catalog #H-1009) to 90 mlof distilled water.

Experiment 19: Isolation ofBacterial DNA

For Growth of Wild-Type (Trp�) Bacillus

subtilis

Usually a teaching assistant or technician performsthis portion of the experiment. This section is de-signed to yield cells for 10 students.


of (NH4)2SO4, 750 mg of Na3citrate � 2H2O,375 mg of MgSO4 � 7H2O, and 750 mg of yeastextract in 689 ml of distilled water. Glucose solution:Dissolve 3.75 g of glucose in 10 ml of distilled wa-ter. Casamino acids solution: Dissolve 50 mg ofCasamino acids in 1 ml of distilled water. Trypto-phan solution: Dissolve 37.5 mg of L-tryptophan in50 ml of distilled water. Autoclave (20 min at 105to 110°C) all samples separately, cool, and asepti-cally mix in a sterile 1-liter flask. Transfer 250 mlto another sterile 1-liter flask, and retain the restfor preparation of SPII medium (next entry).

Competence growth medium II (SPII)—Prepare stocksolutions of 1.0 M CaCl2 (1.47 g of CaCl2 � 2H2Oin 10 ml of H2O) and 1.0 M MgCl2 (2.03 g ofMgCl2 � 6H2O in 10 ml of H2O). Sterilize thesesolutions by autoclaving (20 min at 105 to 110°C),allow them to cool, and store at room temperature.On the day of use for preparation of competentcells, prepare 500 ml of SPII medium by adding225 �l of sterile 1.0-M CaCl2 and 1.125 ml of ster-ile 1.0-M MgCl2 to 499 ml of sterile SPI mediumfrom the preceding entry.

Minimal medium agar Petri plate (see section on trans-formation below)—1 required.

Minimal medium agar Petri plate supplemented with50 �g/ml tryptophan (see section on transforma-tion below)—1 required.

Centrifuge—1 required.

Sterile centrifuge bottles—Autoclave a sufficient numberof bottles to centrifuge 500 ml of culture medium.

Sterile glass rod—Place a fire-polished glass rod in acapped test tube and autoclave (20 min at 105 to110°C).

Sterile synthetic medium—Prepare a 10-ml and two 200-ml solutions containing: 2 g/liter (NH4)2SO4,14 g/liter K2HPO4, 6 g/liter KH2PO4, 1.9 g/literNa3citrate � 2H2O, 0.72 g/liter MgSO4, and50 mg/liter of each of the following amino acids:L-histidine, L-tryptophan, L-arginine � HCl, L-valine, L-lysine � HCl, L-threonine, L-aspartic acid,and L-methionine. Glucose (5 g/liter final medium)should be prepared as a 50-fold concentrated so-lution and autoclaved separately. Autoclave themain solution and the concentrated glucose solu-tion for 20 min at 105 to 110°C and cool; use ster-ile conditions to add glucose to the main solution,and use the medium within 24 hr.

Sterile 10 ml pipettes—Autoclave in a pipette can as forsterile pipettes in the section on transformation. Atleast 2 required.

Sterile 1.5-ml microcentrifuge tubes (for aliquots of com-petent cells if these are to be frozen)—Approxi-mately 80 of these will be required.

Na3citrate)—Dilute 25 ml of concentrated (10X)saline-citrate solution (see below) with 2.475 litersof distilled water. Stable for 1 month.

Standard (1X) saline-citrate ( 0.15 M NaCl, 15 mMNa3citrate)—Dilute 250 ml of concentrated (10X)saline-citrate solution (see below) with 2.25 litersof distilled water. Stable for 1 month.

Concentrated (10X) saline-citrate (1.5 M NaCl, 0.15 MNa3citrate)—Dissolve 43.85 g of NaCl and 22 gof Na3citrate � 2H2O in 450 ml of distilled waterand dilute to 500 ml with distilled water. Use 275ml of this solution to make up the dilutions ofsaline-citrate above and dispense the rest to the stu-dents. Stable for 1 month.

3.0 M Sodium acetate, 1 mM EDTA, pH 7.0—Dissolve61.5 g of sodium acetate (anhydrous) and 95 mg ofdisodium EDTA � 2H2O in 175 ml of distilled wa-ter; titrate to pH 7.0 with a few drops of 1.0 MHCl (concentrated HCl:H2O � 1:11) and dilute to250 ml with distilled water.

Isopropanol—1 liter required.

Spectrophotometer with quartz cuvettes—At least 1 re-quired.

Chloroform—50 ml required.

Experiment 20: Transformationof a Genetic Character withBacterial DNA

For Growth of Competent Recipient Bacillus

subtilis Strain 168 (Trp�)

Usually a teaching assistant or technician performsthis portion of the experiment. This section is de-signed to yield cells for 30 to 40 pairs of students.

Stock culture of B. subtilis strain 168 (Bacillus GeneticStock Center strain 1A1, Columbus, OH).

Inoculating needle—1 required.

0.8% Nutrient broth, 0.5% glucose, 50 �g/ml trypto-phan—Dissolve 0.24 g of nutrient broth in 10 mlof distilled water, 0.15 g of glucose in a second10 ml of distilled water, and 1.5 mg of tryptophanin a third 10 ml of distilled water. Autoclave(20 min at 105 to 110°C) all samples separately,cool, and aseptically mix in a sterile 100-ml flask.

37°C incubator-shaker—1 required.

Competence growth medium I (SPI)—First prepare750 ml of SPI medium; 500 ml of this will be usedto prepare the SPII medium (see next entry). TheSPI medium is composed of the following four sep-arately autoclaved solutions—Salts solution: Dis-solve 10.5 g of K2HPO4, 4.5 g of KH2PO4, 1.5 g


430

37°C incubator-shaker—1 required.

95% Ethanol—Add 125 ml of distilled water to 2.375liters of absolute ethanol. Store in a tightly cappedbottle.

Bent glass rods—Bend Pyrex glass rods so they fit in a150-ml beaker. Fire polish (O2 torch) the ends. SeeFigure 20-1 for details. Prepare 25 of these.

Minimal medium agar Petri plates—Add 33.75 g ofagar to sufficient salts and yeast extract to pre-pare 2.25 liters of the minimal medium describedabove (medium lacking the monosodium gluta-mate and glucose samples, which are autoclavedseparately). Warm (�50°C) this mixture in suffi-cient water to dissolve (leaving room for later ad-ditions of sterile Na-glutamate and glucose solu-tions). Autoclave (20 min at 105–110°C) thissolution and separate Na-glutamate and glucosesolutions. Just after autoclaving (while still hot),aseptically mix the three solutions to create min-imal medium agar solution; pour 10-ml aliquotsinto each of 225 sterile Petri plates. Cover theplates, allow medium to solidify, invert the plates,and store (refrigerated) until use. Stable for 1week.

Minimal medium agar Petri plates supplemented with50 �g/ml tryptophan—Add 18.75 g of agar and62.5 mg of L-tryptophan to sufficient salts andyeast extract to prepare 1.25 liters of the minimalmedium described above (medium lacking themonosodium glutamate and glucose samples,which are autoclaved separately). Autoclave this so-lution, and separate solutions of Na-glutamate andglucose for 20 min at 105–110°C. Just after auto-claving (while still hot), aseptically mix the threesolutions and then pour 125 sterile agar medium-containing plates as above. Refrigerate; stable for1 week.

Tubes containing 9.9 ml of sterile saline—Dissolve12.75 g of NaCl in 1.5 liters of distilled water. Dis-pense 9.9-ml aliquots of this solution into each of135 test tubes (18 � 150 mm) with loose-fittingsterile caps. Autoclave (20 min at 105–110°C),cool, and store capped until use. Save remainingsaline solution for dilution tubes (see following en-try).

Tubes containing 0.9 ml of sterile saline—Dispense 0.9-ml aliquots of above saline solution (12.75 gNaCl/1.5 liters H2O) into each of 60 test tubes(18 � 150 mm) with loose-fitting caps. Autoclave(20 min at 105–110°C), cool, and store capped un-til use.

Masking tape (for taping stacks of Petri plates togetherduring incubation)—1 or 2 rolls required.

Sterile 100% glycerol (if competent cells are to be storedfrozen)—Autoclave for 20 min at 105 to 110°C.Approximately 5 ml will be required.

For Transformation

Sterile test tubes with caps—Autoclave (20 min at 105 to110°C) test tubes (preferably 13 � 100 mm tubes)with sterile caps loosely set over tops. Approxi-mately 150 of these will be required.

Sterile 0.1- and 1.0-ml pipette tips: Autoclave (20 min at105–110°C) in covered racks. You will requireabout 400 of the 0.1-ml tips and about 25 of the1.0 ml tips.

Sterile minimal medium—Prepare and autoclave (20 minat 105–110°C) 25 capped 18 � 150 mm test tubes,each containing 10 ml of the following medium:2 g/liter (NH4)2SO4, 14 g/liter K2HPO4, 6 g/literKH2PO4, 0.2 g/liter MgSO4, 1 g/liter Na3citrate �2H2O, 1 mg/liter yeast extract, 5 g/liter mono-sodium glutamate (autoclave separately), and5 g/liter glucose (autoclave separately). Prepare 5liters of minimal medium, because it is also needed forpreparation of the Petri plates (see below).

DNA from wild-type B. subtilis—As isolated and dis-solved in sterile 1X saline-citrate (a minimum of 0.6ml per pair of students) as described in Experiment19. Prepare about 15 ml of this reagent.

Standard (1X) saline-citrate (0.15 M NaCl, 0.015 M Na3-citrate) (sterile)—Prepare as in the Appendix toExperiment 19. Autoclave (20 min at 105–110°C).Prepare about 250 ml of this reagent.

Pancreatic DNase (1 mg/ml in sterile 1X saline-citrate)—Dissolve 1 mg of pancreatic DNase (WorthingtonBiochemical Corp., Freehold, N.J.) in 5 ml of ster-ile 1X saline-citrate. Refrigerate or freeze until use.If possible, pass through a sterile bacterial filter(e.g., Millipore, type HAW) into a sterile test tube.Stable for 48 hr.

Recipient B. subtilis strain 168 cells in sterile saline—These cells are conveniently provided as thawed0.6-ml aliquots of frozen competent cells preparedas described in Experiment 20. You will require ap-proximately 250 ml of these cells.

Sterile 0.1 M EGTA, pH 7.9—Dissolve 1.9 g of EGTA(ethylene glycol-bis(�-aminoethyl ether) N, N, N�,N�-tetraacetic acid) in 35 ml of distilled water,titrate to pH 7.9 with 1 N NaOH, and dilute to50 ml final volume with distilled water. Autoclavein a capped test tube (20 min at 105–110°C) andcool.

40 to 45°C water bath—1 required.


nal volume of the solution to 100 ml with distilledwater. Autoclave or filter sterilize the solution andstore at room temperature.

Phenol (Tris-saturated)—Add 500 ml of liquefied phe-nol to 500 ml of 0.5 M Tris � HCl, pH 8.0. Gen-tly but thoroughly mix the solution for 15 min. Al-low the two phases to separate, and aspirate off asmuch of the aqueous (upper) phase as possible. Re-peat this extraction two more times with (500 mlper extraction) with 100 mM Tris � HCl, pH 8.0.After the final extraction, leave a sufficient layer ofTris � HCl, pH 8.0, over the equilibrated phenol.Store this solution in a tightly capped dark bottleat 4°C (stable for several weeks).

Phenol:chloroform:isoamyl alcohol (25:24:1)—Mix to-gether 100 ml of Tris-saturated phenol, 96 ml ofchloroform, and 4 ml of isoamyl alcohol. Storetightly capped in a dark bottle at room temperature.

Chloroform:isoamyl alcohol (24:1)—Mix together 192 mlof chloroform and 8 ml of isoamyl alcohol. Storetightly capped in a dark bottle at room tempera-ture.

Absolute ethanol—Approximately 500 ml required.

Dry ice/ethanol bath—Grind or crush enough dry ice tofill a large tray about half full. Slowly add ethanolor acetone to the dry ice, stirring constantly witha spatula, until a bubbling gel or slurry is achieved.Allow this slurry to stand for 10 min before theprecipitations are performed.


70% (vol/vol ethanol)—Mix 300 ml of distilled waterwith 700 ml of absolute ethanol. Store tightlycapped at �20°C and place on ice for the studentsto use.

Vacuum microcentrifuge—1 required.

Incubating culture shaker with test tube racks—1 required.

Sterile glass rods for spreading cultures over petriplates—50 required.

Sterile toothpicks (autoclaved)—Wash wooden tooth-picks thoroughly with distilled water (to removeany preservatives that may inhibit bacterialgrowth). Place in a shallow container with a coverand autoclave for 20 min.

37°C Incubator—1 required.

T4 DNA ligase (1 unit/�l) with 5X reaction buffer—Weobtain this from Gibco BRL (catalog #15224-025),but any other supplier of quality T4 ligase will do.

100 mg/ml ampicillin solution—Dissolve 1 g of ampi-cillin (sodium salt, Sigma catalog #A-9518) in a to-tal of 10 ml of distilled water. Filter-sterilize thesolution and store at �20°C.

50 mg/ml kanamycin sulfate solution—Dissolve 0.5 g ofkanamycin sulfate (Sigma catalog #K-4000) in

Experiment 21: Constructing andCharacterizing a RecombinantDNA Plasmid

TE buffer—Prepare a 1 M stock solution of Tris by dis-solving 121.1 g of Tris free base in 700 ml of dis-tilled water. Adjust the pH to 7.5 with concentratedHCl. Prepare a 0.5 M EDTA stock solution by dis-solving 93 g of Na2EDTA � 2H2O in 400 ml ofdistilled water. Adjust the pH to 8.0 by slowly dis-solving in NaOH pellets. Bring the final volumeof the solution to 500 ml with distilled water. Toprepare the TE buffer, add 1 ml of the 0.5 MEDTA stock solution and 10 ml of the 1 M Trisstock solution to 989 ml of distilled water.

pUC19 (0.25 �g/�l)—Pharmacia catalog #27-4951-01.This plasmid is supplied in 25-�g quantities(0.5 �g/�l). Dilute the contents of the tube withan equal volume of TE buffer. Store in 10-�laliquots at �20°C.

pUC4K (0.5 �g/�l)—Pharmacia catalog #27-4958-01.Use as supplied.

HindIII (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15207-020), but any other supplier of quality restrictionendonucleases will do.

BamHI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15201-031), but any other supplier of quality restrictionendonucleases will do.

EcoRI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15202-021), but any other supplier of quality restrictionendonucleases will do.

XhoI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15231-020), but any other supplier of quality restrictionendonucleases will do.

SalI (10 units/�l) with 10X reaction buffer supplied—Weobtain this from Gibco BRL (catalog #15217-029),but any other supplier of quality restriction en-donucleases will do.

PstI (10 units/�l) with 10X reaction buffer supplied—Weobtain this from Gibco BRL (catalog #15215-023),but any other supplier of quality restriction en-donucleases will do.

Heated water baths—One at 37°C, one at 15°C (placein cold room and adjust the temperature), and oneat 42°C.

3 M sodium acetate, pH 5.2—Dissolve 40.8 g of sodiumacetate � 3H2O in 70 ml of distilled water. Adjustthe pH to 5.2 with glacial acetic acid. Bring the fi-


432

depending on the size of your electrophoresischambers.

GTE lysis buffer—Dissolve 1.8 g of D-glucose, 0.606 g ofTris free base, and 0.744 g of disodium EDTA � 2H2O in 150 ml of distilled water. Adjust the pHto 8.0 with diluteHCl. Bring the final volume of thesolution to 200 ml with distilled water. Autoclavethe solution and store at 4°C. If you wish, you mayadd lysozyme to this solution to a final concentra-tion of 10 �g/ml. We find that this is not necessary.

Solution of 0.2 N NaOH and 1% SDS—Prepare a 10 NNaOH stock solution by dissolving 200 g of NaOHin 300 ml of water. Bring the final volume of thesolution to 500 ml with distilled water and store ina plastic (not glass) bottle at room temperature.Prepare a separate 10% (wt/vol) SDS solution bydissolving 10 g of sodium dodecyl sulfate in 70 mlof water. You may have to heat the solution slightlyto allow all of the SDS to dissolve. Bring the finalvolume of the solution to 100 ml with distilled wa-ter and store at room temperature. To prepare the0.2 N NaOH/1% SDS solution immediately be-fore the experiment, mix 2 ml of the 10 N NaOHstock solution and 10 ml of the 10% (wt/vol) SDSstock solution with 88 ml of distilled water.

5 M potassium acetate solution—Dissolve 29.4 g ofpotassium acetate in distilled water to a final vol-ume of 60 ml. Add 11.5 ml of glacial acetic acidand 28.5 ml of distilled water. Autoclave the solu-tion and store at 4°C.

6X DNA sample buffer—Dissolve 0.25 g of xylenecyanole FF and 0.25 g of bromophenol blue in70 ml of distilled water. Add and thoroughly mixin 30 ml of glycerol. Store the solution at roomtemperature.

1-kb DNA ladder size standard—We purchase this fromGibco BRL (catalog #15615-016). Add 50�l of 1-kb DNA ladder to 775 �l of distilled water. Add175 �l of 6X DNA sample buffer. Mix thoroughly,but gently, and store at 4°C. Each group will load20 �l of this solution.

Ethidium bromide (0.5 �g/ml) in 0.5X TBE buffer—Dissolve 0.5 mg of ethidium bromide in 1 liter of0.5X TBE buffer. Place the entire 1-liter solutionin a shallow tray for students to incubate their gelsin. A more concentrated solution of ethidium bro-mide (1–2 �g/ml) can be prepared if you wish todecrease the required staining time. Caution:

ethidium bromide is a suspected carcinogen.

Polaroid camera fitted to a 256-nm light box—1 required.

Film for camera—50 exposures required.

RNaseA—We purchase this enzyme from Sigma (cata-log #R-5125). Dissolve 100 mg gently in 10 ml ofdistilled water. Boil this solution for 5 min to in-activate any DNase that may be present (RNaseA

10 ml of distilled water. Filter-sterilize the solutionand store at �20°C.

40 mg/ml Isopropylthio-�,D-galactopyranoside (IPTG)solution—Dissolve 0.4 g of IPTG (Sigma catalog#I-6758) in 10 ml of distilled water. Filter-sterilizethe solution and store at �20°C.

40 mg/ml 5-bromo-4-chloro-3-indolyl-�,D-galactopyra-noside (Xgal) solution—Dissolve 0.4 g of Xgal(Sigma catalog #B-4252) in 10 ml of dimethylfor-mamide (DMF). DMF is very toxic and can be ab-sorbed through the skin. Avoid contact with this reagentand do not breathe in its vapors. Work in the fume hood.Filter-sterilize the solution and store at �20°C.

100 mM IPTG solution—Dissolve 0.715 g of IPTG in30 ml of distilled water. Filter-sterilize the solutionand store at 4°C.

Epicurian coli XL1-Blue subcloning grade competentcells—Stratagene catalog #200130.

Disposable plastic petri plates—500 required.

Large (16 � 125 mm) glass culture tubes with caps—250required. Sterilize by autoclaving for 15 min priorto use.

YT broth—Dissolve 10 g of Bactotryptone (Fisher cat-alog #DFO 123), 5 g of yeast extract (Fisher cata-log #DFO 127), and 5 g of NaCl in 800 ml of dis-tilled water. Bring the final volume of the solutionto 1 liter with distilled water. Autoclave for 25 minto sterilize. You will need 600 ml of this media forthe growth of the students cultures on the “off day”of the experiment. To prepare the YT media agarplates, add 15 to 20 g/liter agar (Fisher catalog#DFO 140) to each 1 liter batch of YT media be-fore autoclaving. You will need to make �6.25 litersof this YT agar media for the IPTG/Xgal/ampplates. As each 1-liter batch cools following the au-toclave step, add 1 ml of 40-mg/ml IPTG solution,1 ml of 40-mg/ml Xgal solution, and 1 ml of 100-mg/ml ampicillin solution to the YT agar media.After the addition of these reagents and thoroughmixing, pour about 25 ml of the YT/agar mediaper plate. You will also need to make �2.5 liters ofthis YT agar media for the IPTG/Xgal/kan plates.As each 1-liter batch cools following the autoclavestep, add 1 ml of 40-mg/ml IPTG solution, 1 mlof 40-mg/ml Xgal solution, and 1 ml of 50-mg/mlkanamycin solution to the YT agar media. Afterthe addition of these reagents and thorough mix-ing, pour about 25 ml of the YT/agar media perplate. After the plates have cooled and hardened,store them at 4°C in plastic bags.

5X TBE agarose-gel electrophoresis buffer—Refer toExperiment 4 (“Electrophoresis”) for this recipe.The stock solution will be diluted 1:10 to producea 0.5X solution for use in this experiment. You willneed 1 to 2 liters of the 5X TBE stock solution,


20 mg/ml Bovine serum albumin—Dissolve 2 g of bovineserum albumin (RNase-free) in 100 ml of distilledwater. Filter-sterilize the solution and store at�20°C.

TE buffer—See Experiment 21.

0.1 M dithiothreitol solution—Dissolve 1.54 g of DL-dithiothreitol in 100 ml of distilled water. Filter-sterilize (do not autoclave) and store at �20°C in1.0-ml aliquots.

-[32P]ATP—Obtain 250 �Ci of a solution at a specificactivity of 3,000 Ci/mmol. We use ICN catalog#32007x.

T7 RNA polymerase—We obtain this in 2500-unit (50units/�l) aliquots from Gibco BRL (catalog#18033-100). The enzyme is diluted immediatelybefore use to 20 units/�l in sterile distilled water.

10X Transcription mix—Mix the following reagents to-gether to prepare 100 �l of this solution. Preparethis reagent immediately before use:

40 �l of 0.4 M Tris-HCl, pH 8.0

15 �l of 0.15 M MgCl25 �l of 20 mg/ml bovine serum albumin

0.5 �l of 100 mM ATP

10 �l of 100 mM CTP

10 �l of 100 mM GTP

10 �l of 100 mM UTP

1 �l of 0.1 M dithiothreitol (DTT)

6 �l of sterile distilled water

2.5 �l of a 1:100 dilution of the -[32P]ATP stock insterile distilled water

Loading buffer—Dissolve 42 g of urea, 0.25 g of xylenecyanole FF, and 0.25 g of bromophenol blue in dis-tilled water to a final volume of 100 ml. Store atroom temperature.

Absolute ethanol—Approximately 100 ml required.

15% Polyacrylamide gel—Refer to Experiment 4.

0.5X TBE buffer—See Experiment 21.

Film cassettes—5 to 10 required, depending on the sizeof the gels.

Plastic wrap—1 roll required.

Razor blades—5 to 10 required (students can share these).

Kodak X-omat film—5 to 10 pieces required, dependingon the size of the gels and film cassettes.

Darkroom and solutions for film development.

Electrophoresis chambers and power supplies—25 re-quired.

RNA size markers—Add 20 �l of the pSP72 plasmid tothree microcentrifuge tubes. One of these plasmidsamples will be digested with XhoI, the other withXbaI, and the third with EcoRV. Add 6 �l of theappropriate 10X restriction-enzyme buffers to eachof the three tubes. Add 29 �l of sterile distilled

is heat stable while DNase is not). Allow the solu-tion to cool to room temperature and store at 4°C.

Agarose (electrophoresis grade)—25 g required.

Agarose-gel electrophoresis apparatus with casting moldsand 10 well combs—50 required.

Power supplies—Enough to power all required elec-trophoresis chambers.

Experiment 22: In VitroTranscription from a PlasmidCarrying a T7 RNA PolymeraseSpecific Promoter

HindIII (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15207-020), but any other supplier of high-quality re-striction endonucleases will do.

BamHI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15201-031), but any other supplier of high-quality re-striction endonucleases will do.

EcoRI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco BRL (catalog #15202-021), but any other supplier of high-quality re-striction endonucleases will do.

pSP72 plasmid (100 �g/ml)—Promega catalog #P-2191.This plasmid is supplied as a 1 �g/�l solution inTE buffer. Dilute this solution 10-fold with TEbuffer. Each group will require 24 �l of the plas-mid to perform the experiment. Store at �20°C.

37°C Water bath—1 required.

Dry ice/ethanol bath—See Experiment 21.

Phenol:chloroform (1:1)—Mix 250 ml of chloroformwith 250 ml of Tris-saturated phenol (see Experi-ment 21). Add about 75 ml of water to the solu-tion and allow the phases to separate. Leave a layerof water over the organic phase and store in a darkbottle tightly capped at room temperature.

Chloroform:isoamyl alcohol (24:1)—See Experiment 21.

3 M sodium acetate, pH 5.2—See Experiment 21.

0.4 M Tris, pH 8.0—Dissolve 3.55 g of Tris � HCl and2.12 g of Tris free base in 100 ml of distilled wa-ter. The pH should be at or near 8.0. If necessary,adjust the pH to 8.0 with dilute HCl or NaOH.Store at room temperature.

100 mM nucleotide solutions (UTP, ATP, CTP, GTP)—We buy these as a set from Boehringer Mannheim(catalog #1277057). Each nucleotide is suppliedseparately in the set at a concentration of 100 mM.

0.15 M MgCl2—Dissolve 3.05 g of MgCl2 � 6H2O in100 ml of distilled water. Filter-sterilize or auto-clave the solution and store at room temperature.


434

of acid-washed sand in a cold mortar. Vigorouslygrind the dry wheat germ and sand with a cold pes-tle for 1 to 2 min. Add 5 ml of cold grinding buffer(20 mM HEPES, 100 mM KCl, 1 mM magnesiumacetate, 2 mM CaCl2, 6 mM dithiothreitol, pH 7.6)and continue grinding until you obtain an evenmush ( �1 min more). Continue grinding andadding 5-ml aliquots of fresh grinding buffer untila total of 25 ml have been added. Centrifuge thehomogenate (30,000 � g) for 10 min at 4°C andcollect the S-30 supernatant, being careful to avoidthe yellow fatty surface layer above the cell debrisand mitochondria. Pass 15 ml of the supernatantover a 50 � 2 cm Sepadex G-25 column pre-equi-librated with at least 600 ml of buffer (20 mMHEPES, 120 mM KCl, 5 mM magnesium acetate,6 mM dithiothreitol, pH 7.6). Collect the first 10fractions (1.5 ml each) as they elute from the col-umn. Using a Pasteur pipette, slowly drip the frac-tions into a pool of liquid nitrogen. Allow eachdrop to freeze and sink to the bottom of the ves-sel. When all of the material has been applied tothe liquid nitrogen, frozen, and sunk to the bot-tom of the vessel, pour off the N2 and place thefrozen S-30 fraction in small, airtight vials to befrozen at �80°C.

1.5-ml Microcentrifuge tubes—2,000 required.

Fine-tipped forceps—50 required.

Millipore filters (HAWP, 0.45-�m pore size)—2000 required.

Filtration apparatus with vacuum pump (or house vac-uum)—50 required.

Water baths at 30°C and 90°C—One of each required.

100°C Oven—1 required.



Plastic squirt bottles—50 required.

Reaction buffer—Dissolve 7.15 g of N-[2-hydrox-yethyl]piperazine-N�-[2-ethanesulfonic acid](Sigma catalog #H-7523) and 0.46 g of dithio-threitol (Sigma catalog #D-5545) in 80 ml of dis-tilled water. Adjust the pH of the solution to 7.1with dilute NaOH. Bring the final volume of thesolution to 100 ml with distilled water and store at4°C.

Salt solution I—Dissolve 9.7 g of KCl and 4.83 g of mag-nesium acetate tetrahydrate in 80 ml of distilledwater. Bring the final volume of the solution to100 ml with distilled water and store at 4°C.

Salt solution II—Dissolve 9.7 g of KCl and 0.483 g ofmagnesium acetate tetrahydrate in 80 ml of dis-tilled water. Bring the final volume of the solutionto 100 ml with distilled water and store at 4°C.

water to each tube, as well as 5 �l of the appro-priate restriction enzyme (10 units/�l). Incubatethese tubes at 37°C for 2 hr. Ethanol-precipitatethe digested plasmid in each of these tubes byadding 8 �l of 3 M sodium acetate, 200 �l of ab-solute ethanol, and incubating in a dry ice/ethanolbath for 10 min. Centrifuge at 4°C for 30 min toharvest the DNA and discard the supernatant. Af-ter a 70% ethanol wash, dissolve the DNA pelletsin 6 �l of sterile distilled water. Add the followingto each tube:

4 �l of 5X T3/T7 RNA polymerase buffer (suppliedwith enzyme)

2 �l of 0.1 M dithiothreitol

4 �l of a 2.5 mM solution of GTP, CTP, and UTP(see below)

1 �l of a 1 mM ATP solution (see below)

1 �l of RNasin (20–40 units/�l) (Promega catalog#N2511)

2 �l of the undiluted 10mCi/ml solution of -[32P]ATP

2 �l of T7 RNA polymerase (20 units/�l)

The 2.5 mM CTP, UTP, and GTP solution is preparedby adding 1 �l of each 100 mM nucleotide stocksolution to 47 �l of distilled water. The 1 mM ATPsolution is prepared by adding 1 �l of the 100 mMATP stock to 99 �l of distilled water. Incubatethese three transcription tubes at 37°C for 1 hr.Store at �20°C. To prepare the markers, the re-actions will be mixed together in a defined ratio.You will have to experiment with an exact dilutionand ratio of these transcripts to produce RNA sizemarkers that will be visible as defined bands on thefilm after an 18-hr exposure. The XhoI transcriptwill be 97 nucleotides in length, the XbaI transcriptwill be 61 nucleotides in length, and the EcoRVtranscript will be 20 nucleotides in length. Re-member that the EcoRV- and XbaI-digested tran-scripts will have less radioactivity per mole thanthe XhoI-digested transcript.

Experiment 23: In VitroTranslation: mRNA, tRNA, andRibosomes

S-30 Fraction from wheat germ—This can be obtainedcommercially (Fisher catalog #L-4440) in a formthat is ready to use in the in vitro reaction. If youwish to prepare it yourself, perform the followingprotocol. Place 6.75 g of wheat germ and 6.75 g


Supplies and Reagents list for this exercise. Weusually order 100 nmol of each primer. Consult thespecification sheet to determine exactly how muchof each primer is present when you receive it. Add0.5 to 1.5 ml of sterile distilled water to each tubeto bring the final concentration of each primer to100 pmol/�l. Each group will need 42.5 �l of eachprimer at a concentration of 10 pmol/�l (10 �M).Dilute each of the primer stock solutions 10-foldin sterile distilled water to achieve this concentra-tion. Store the remainder of the primer stocks at�20°C for use in future semesters (100 nmol ofprimers is enough for approximately two semes-ters).

Template DNA—The pBluescript II (S/K) and (K/S)plasmids can be purchased from Stratagene (cata-log #212205 and 212207, respectively). The 20-�gpackage size for each plasmid is enough to performthe experiment for approximately 40 semesters.Using serial dilutions, prepare 1 ng/�l, 1 pg/�l, and1 fg/�l stock solutions.

dNTPs—We purchase 100 mM stock solutions of dATP,dTTP, dGTP, and dCTP from Gibco-BRL (cata-log #10297-018). Add 62.5 �l of each 100 mMstock dNTP solution to 4.750 ml of sterile distilledwater to prepare the 1.25 mM dNTP stock solu-tion. Each group will need 68 �l of the dNTP so-lution to perform the experiment. Store the re-mainder of the 100 mM dNTP stock solutions (insmall aliquots) at �20°C for use in future semes-ters.

Taq polymerase—We purchase this from Gibco-BRL(catalog #18038-042). The enzyme is supplied with10X Taq polymerase dilution buffer and a vial of50 mM MgCl2. Each group will require 68 �l of12.5 mM MgCl2. To prepare, add 1 ml of 50 mMMgCl2 stock to 3 ml of sterile distilled water. Toadjust the 5-units/�l Taq polymerase stock to 1unit /�l for use in the experiment, add 100 �l of 5-units/�l Taq polymerase stock solution to 400 �lof 1X Taq polymerase buffer. This 1X Taq dilutionbuffer is prepared by adding 100 �l of the 10X Taqpolymerase buffer stock to 900 �l of distilled wa-ter. The dilution of the Taq polymerase to 1 unit/�lshould be performed immediately before the start of theexperiment to preserve the activity of the enzyme. Eachgroup will require 8.5 �l of the 1-unit /�l Taq poly-merase solution to perform the experiment.

1-kb DNA ladder—We purchase these from Gibco-BRL(catalog #15615-016). To prepare the 0.2 �g/�l so-lution, add 50 �l of the 1 �g/�l stock solution to200 �l of TE buffer.

5X TBE agarose-gel electrophoresis buffer—See Exper-iment 21.

3H-Phenylalanine solution—Dissolve 1.65 mg of L-phenylalanine (Sigma catalog #P-8324) in 100 mlof distilled water. Add an amount of high specificactivity (greater than 1000 Ci/mmol) [3H]-phe-nylalanine that contains a total of 500 �Ci. Storeat 4°C.

Poly(U) solution—Dissolve 10 mg of Sigma catalog #P-9528 in 10 ml of distilled water. Store at �20°C.

Poly(A) solution—Dissolve 25 mg of Sigma catalog #P-9403 in 25 ml of distilled water. Store at �20°C.

Trichloroacetic acid solution—Dissolve 250 g oftrichloroacetic acid (Sigma catalog #T-9159) in 5 liters of distilled water. Store at 4°C.

Scintillation fluid—Dissolve 50 g of diphenyloxazole and100 ml of Triton X-100 in 9 liters of toluene. Storetightly covered in a dark bottle at room tempera-ture.

RNase A solution—Dissolve 5 mg of Sigma catalog #R-5125 in 10 ml of distilled water. Store at 4°C.

RNase T1 solution—We use product #R-1003 fromSigma, which is supplied as a crystalline suspen-sion in 3.2 M ammonium sulfate (300,000–600,000units/mg of protein). Transfer a volume equivalentto 1 mg of protein (consult the specification sheet)to a microcentrifuge tube and centrifuge for 5 minto collect the crystals. Draw off the supernatantand resuspend the solid crystals in 2 ml of distilledwater.

Chloramphenicol solution—Dissolve 16.2 mg of Sigmacatalog #C-0378 in 100 ml of distilled water. Storeat �20°C in small aliquots.

Cycloheximide solution—Dissolve 14.1 mg of Sigma cat-alog #C-7698 in 100 ml of distilled water. Store at�20°C in small aliquots.

Puromycin solution—Dissolve 5.4 mg of Sigma catalog#P-7255 in 100 ml of distilled water. Store at�20°C in small aliquots.

Experiment 24: Amplification ofa DNA Fragment UsingPolymerase Chain Reaction

TE buffer—See Experiment 21.

0.5-ml Thin-walled PCR tubes—250 required.


Primers—We order these from Integrated DNA Tech-nologies, Inc., 1710 Commercial Park, Coralville,IA 52441. This company can be reached at 1-800-328-2661, or on the World Wide Web(http://www.idtdna.com). The sequence of theprimers needed for this experiment is shown in the


436

KpnI (10 units/�l) with 10X reaction buffer supplied—We obtain this from Gibco-BRL (catalog #15232-036), but any supplier of high-quality restrictionendonucleases will do.

SstI (10 units/�l) with 10X reaction buffer supplied—Weobtain this from Gibco-BRL (catalog #15222-037),but any supplier of high-quality restriction en-donucleases will do.

Agarose-gel electrophoresis chambers with casting moldsand 10 well combs—25 required.


Power supplies—Enough to power 25 agarose-gel elec-trophoresis chambers.

6X DNA sample buffer—See Experiment 21.

Agarose (electrophoresis grade)—Approximately 12.5 grequired.

Ethidium bromide solution (0.5 �g/ml) in 0.5X TBEbuffer—See Experiment 21.

Polaroid camera fitted to a 256-nm light box—1 re-quired.

Film for camera—At least 25 exposures will be required.

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Experimental Biochemistry (1)