Chromatographic Separation of Proteins - Modern Biomodernbio.com/sites/default/files/modern_biology_basic_program_3... · Chromatographic Separation of Proteins. Preface This laboratory

BASIC LABORATORY COURSE 3Chromatographic Separation of Proteins

Preface This laboratory manual forms an integral part of a complete teaching system. In addition to the manual, the system consists of the chemicals and materials required to perform a series of experiments in modem biology.

This teaching system was designed principally for students taking a laboratory course in general biology, physiology, or basic biochemistry. I have assumed that most readers have had an introductory lecture course in the biological sciences. However, I have attempted to write the manual such that even a stranger to biology could follow it and perform the exercises.

The laboratory manual is divided into two major sections. The first section provides basic information on the biology and chemistry of proteins and is also intended to acquaint the student with the principles and techniques of chromatography . Students should be familiar with this material before advancing to the second section of the manual. The second part of the manual shows students how to apply what they have learned to perform five exercises in modem biology. The exercises deal with the structure, function, and isolation of proteins, and stress the relevance of molecular biology to animal physiology. Each experiment provides state-of-the-art information and each experiment can be completed within a 2 hour laboratory session. A background information section is given for each exercise and students should be required to read this material prior to performing the exercise in the laboratory. Study questions are found at the end of each exercise. Students derive answers to these questions by integrating the background material found in the text with the results of the experiments.

John N. Anderson, Ph.D. Professor of BiologyDepartment of Biological SciencesPurdue UniversityWest Lafayette, Indiana 47907

© 2013 by John N. Anderson, All rights reserved.

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Table Of cOnTenTsPart A. Background Information

I. Protein Composition and Structure: A Review of the Basics Amino Acids-Building Blocks of Proteins .................................................... 1 The Peptide Bond .......................................................................................... 2 Hierarchies of Protein Structure .................................................................... 3

II. Chromatography Chromatographic Separation of Small Biomolecules .................................... 6 Chromatographic Separation of Proteins ....................................................... 7

Part B. Laboratory ExercisesIntroductory Remarks ......................................................................................... 10

Experiment 1. Separation of Molecules by Gel Filtration Chromatography ..................................................... 12

Experiment 2. Determination of the Molecular Weight of Human Hemoglobin ........................................................... 19

Experiment 3. Dye Binding of Serum Albumin ............................................ 24

Experiment 4. A Comparison of the Properties of a-Amylase from Human Saliva and Pancreas ................................................... 27

Experiment 5. Isolation and Immunological Analysis of a Protein from Egg White ...................................................................... 37

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BACKGROUND INFORMATIONI. Protein Composition and Structure:

A Review of Tile Basics

Proteins occupy a central position in the structure and function of all living organisms. Some proteins serve as structural components while others function in communication, defense,and cell regulation. The enzyme proteins act as biological catalysts which control the pace and nature of essentially all biochemical events. Indeed, although DNA serves as the genetic blueprint of a cell, none of the life processes would be possible without the proteins.

Amino Acids - Building Blocks of ProteinsThe fundamental unit of proteins is the amino acid. The common amino acids have the general structure shown in Figure I. Each amino acid has an amino group (NH2 and a carboxylic acid group (COOH) attached to a central carbon atom called the alpha carbon. Also attached to the alpha carbon are a hydrogen atom and an R- group or side chain.

Figure 1. General Structure of Alpha-Amino Acids.

The C stands for a carbon atom; C* is the alpha carbon; His hydrogen; N is nitrogen, O is oxygen, -NH2 an amino group and -COOH is the carboxylic acid group. R is a general term for any one of several different side chains that determine the nature of different amino acids.

There are 20 amino acids commonly found in proteins and these differ from each other in the nature of the R-groups attached to the alpha carbon. A convenient classification of amino acids depends on the number of acidic and basic groups that are present. Thus, the neutral amino acids contain one amino and one carboxyl group. The acidic amino acids have an excess of acidic carboxyl over amino groups. The basic amino acids possess an excess of basic amino groups. Table I lists the major amino acids found in proteins.


Table 1. Amino Acids Found in Proteins.

Neutral Amino AcidsGlycine Alanine Valine Leucine Isoleucine Serine Threonine Cysteine Methionine Phenylalanine Tyrosine Tryptophan Proline

Acidic Amino Acids Basic Amino Acids Aspartic Acid Arginine Glutamic Acid Lysine Histidine

The Peptide Bond and the Primary Structure of ProteinsProteins are composed of amino acids linked into chains by peptide bonds as shown in Figure 2. Two amino acids joined by a single peptide bond form a di -peptide; three amino acids form a tri-peptide, and a large number of amino acids joined together constitute a polypeptide. A protein is a polypeptide chain that contains more than 50-100 amino acids. The monomer units in the chain are known as amino acid residues. The average protein contains about 350 amino acid residues although proteins with as many as 1000 residues and those with as few as 100 are not uncommon.

Figure 2. Formation of a Peptide Bond.

Peptide bonds are enclosed in the dotted boxes. The dotted circle shows how a peptide bond is formed with the production of H2O.


The sequence or order of amino acids along a polypeptide chain is referred to as the primary structure of the protein. The primary structure of the protein myoglobin is given in Figure 3. This protein serves to bind and store oxygen in muscle. The primary structure of over 500 different proteins is now known.

Figure 3. The Primary Structure of Whale Myoglobin.

[Amino or N-terminus]Val-Leu-Ser-Giu-gly-Giu-Trp-Gin-Leu-Val-Leu-His-Val-Tyr-Aia-Lys-Val- Giu-Aia-Asp-Val-Ala-Giy-His-Gly-Gin-Asp-lle-Leu-lle-Arg-Leu-Phe-Lys - Ser-His-Pro-Glu-Thr-Leu-Glu-Lys-Phe-Asp-Arg-Phe-Lys-His-Leu-Lys-Thr- Giu-Ala-Glu-Met-Lys-Aia-Ser-Giu-Asp-Leu-Lys-Giy-His-His-Giu-Ala-Giu Leu-Thr-Ala-Leu-Giy-Aia-lle-Leu-Lys-Lys-Lys-Giy-His-Giu-Ala-Giu-Lys Leu-Lys-Pro-Leu-Ala-Gln-Ser-His-Ala-Thr-Lys-His-Lys-lle-Pro-lle-Lys -Tyr-Leu-Giu-Phe-lle-Ser-Giu-Aia-lle-lle-His-Val-Leu-His-Ser-Arg-His -Pro-Giy-Asn-Phe-Giy-Aia-Asp-Aia-Gin-Giy-Aia-Met-Asn-Lys-Aia-Leu-Giu -Leu-Phe-Arg-Lys-Asp-lie-Ala-Ala-Lys-Tyr-Lys-G lu-Leu-G ly-Tyr-G ln-Gly

This protein consists of 153 amino acid residues. Each residue in the protein is given a 3-letter abbreviation (thus: lysine = Lys); valine = Val). Proteins are always written with the free amino or N-terminus toward the left.

Three-Dimensional Protein StructureIn the cell,the polypeptide chain is folded into a highly ordered shape or con formation. Most proteins are globular in shape and these proteins are usually soluble in water or in aqueous media containing salts. This group includes the enzymes, antibodies, and a variety of other proteins. Less frequently, proteins are long and fibrous and most of these elongated molecules are insoluble in water and serve a role in the maintenance of cell structure.

The three-dimensional structure of a protein is due to the type and sequence of its constituent amino acids. Since the amino acid sequence of each protein is unique, it follows that different proteins assume different shapes. Thus, there is a remarkable diversity of three-dimensional protein forms. The conformation of a protein is usually of critical importance in the protein’s function. For example, a protein can be unfolded into a polypeptide chain that has lost its original shape. In general, proteins such as enzymes are rendered nonfunctional upon unfolding because functional activity is dependent on the proteins native shape. This process is called denaturation. Most proteins can be denatured by heating, by certain detergents, and by extremes of pH. The ionic detergent, sodium dodecyl sulfate (SDS), is often used to denature proteins. The denaturing treatment can frequently be reversed, for example by removing the detergent or by neutralizing the pH.


During this renaturing process, the polypeptide chain spontaneously refolds into its original conformation and the protein regains its biological activity. A similar folding process occurs in the cell for when a polypeptide is constructed on the ribosomes, it folds into a biologically active conformation. Thus, the three-dimen sional folding of a protein and its biological properties are directed by the sequence of amino acid residues along the polypeptide chain.

Biochemists have identified three structural levels that define the three-di mensional shape of a protein. These levels of organization are secondary structure, tertiary structure,and quaternary structure. Figure 4 shows examples of these levels of organization. The major force involved in the formation and maintenance of these structures are various types of weak, noncovalent bonds that are formed between the amino acid residues and between the amino acid residues and water. Although a noncovalent bond typically has less than 1/20 the strength of a covalent bond, a large number of noncovalent bonds participate in the folding of a single protein into its native conformation.

Figure 4. Levels of Structural Organization.


Secondary StructureThe spatial arrangement of the protein backbone that is generated from the folding of the polypeptide chain is called the secondary structure of the protein. The secondary structures of proteins are stabilized by hydrogen bonds in which a hydrogen serves as a bridge between oxygen and nitrogen atoms (-C=O••••••HN-). A common secondary structure is the a-helix which consists of a single polypeptide chain coiled into a rigid cylinder. In the a-helix, each peptide bond along the polypeptide is itself hydrogen bonded to other peptide bonds. Many enzymes contain small regions of a-helices, while long sections of the a-helix are often found in proteins involved in cell structure. Another type of secondary structure of proteins is the b-sheet, which is a central organizing feature of enzymes, antibodies, and most other proteins that perform nonstructural functions. Here, a single polypeptide chain folds back and forth upon itself to produce a rather rigid sheet. Hydrogen bonds between neighboring polypeptide chains are a major stabilizing force for the b-sheet conformation.

Tertiary StructureThe tertiary structure of a protein describes the detailed features of the three dimensional conformation of the polypeptide chain. It is brought about by the interactions between the amino acid side chains which cause the folding and bending of a-helix and b-sheet segments of the protein. One very important interaction at this level of organization involves the hydrophobic and hydrophilic side chains of the amino acid residues. Hydrophobic amino acids, such as phenylalanine and leucine, show limited solubility in water. Thus, these hydropho bic residues in a protein tend to cluster on the inside of the protein in order to avoid contact with the aqueous environment. Hydrophilic amino acids such as glutamic acid and lysine are readily soluble in water, and thus these amino acids arrange themselves on the surface of the protein molecule, where they can interact with water and with other hydrophilic side chains. The consequence of these interactions is that a polypeptide chain typically folds spontaneously into a stable, usually globular structure, with the hydrophobic side chains packed into the central core of the protein and the hydrophilic side chains forming the irregular, external surface.

Quaternary StructureSome proteins contain more than one polypeptide chain. For example, each molecule of human hemoglobin consists of four polypeptide chains which are held together by a variety of noncovalent bonds. The arrangement of the polypeptides in such proteins is called the quaternary structure.


II. ChromatographyA. Chromatographic Separation of Small BiomoleculesIn 1906, the Russian botanist Tswell introduced the term chromatography (“to write with color”) as a result of his studies on the separation of plant pigments using a column of calcium carbonate. Chromatography, like many other important discoveries, remained largely unnoticed for almost one-half century but in the past 30 years has become one of the most important tools for the rapid separation of biological molecules.

Paper chromatography is used to separate small molecules such as amino acids, sugars, and dyes. In a common form of this technique, the sample is applied as a spot to a sheet of adsorbent paper and a solvent is allowed to flow through the sheet from one edge. As the solvent travels across the paper by capillary action, it picks up those sample molecules that are soluble in it and not bound to the paper. Consequently, sample molecules that move the fastest are those that are most soluble in the solvent and have the lowest affinity for the adsorbent paper. A simple optional experiment, described in Table 2, illustrates the separation of the compo nents of ink by paper chromatography.

Table 2. Separation of the Components of Writing Ink by Paper Chromatography

(An Optional Experiment)1. Place samples of different inks as small spots along the shorter edge of a

piece of Whatman No. I filter paper (20x30cm), 2cm from the edge.

2. After the samples have dried, staple the two longer edges together to form a cylinder.

3. Place the end of the cylinder containing the samples in a jar containing I em of water. As the water rises, the different colored components of the inks will migrate at different speeds, which are related to their solubility in water and adsorption to the paper.


B. Chromatographic Separation of ProteinsA major goal in the modern biology laboratory is the isolation of specific proteins from complex mixtures. Such isolation procedures frequently involve chromatography. Proteins are most often separated by column chromatography, in which a mixture of different proteins is passed through a column or tube that contains a matrix material. The separation occurs because proteins interact with various matrices in different ways. In ion-exchange chromatography, small beads that carry either a positive or negative charge form the column matrix,and proteins are separated according to their charge. An acidic protein that carries a negative charge will bind to a positively charged matrix. Thus, such a matrix can be used to separate negatively charged acidic proteins (which will interact with the matrix and be retained) from positively charged proteins (which will flow through the matrix). The acidic proteins can then be removed or eluted from the column by disrupting their interactions with the matrix. In hydrophobic chromatography,the columns are packed with beads that contain hydrophobic side chains. Proteins with exposed hydrophobic regions bind to the matrix, but proteins without such regions do not.

Gel filtration, another type of chromatographic method, will be used in the laboratory exercises described in this manual. Gel filtration is the technique of separating molecules of different size by passing them through a gel column. In the modem biology laboratory, this method is most commonly used for the separation of proteins, although the technique can be used for the separation of other types of biomolecules as well.

The steps in the separation of two proteins of different size by gel filtration are shown in Figure 5 and described below.

1. Preparation of the Gel ColumnThe two most common polymers used to form the gel are polyacrylamide and dextran. The polymers are cross-linked to themselves to form small beads which swell upon addition of water to form structures that resemble microscopic porous sponges. The hydrated beads,which are typically about 0.1mm in diameter,are then packed into a column and washed with buffer.

2. Sample ApplicationA sample containing a mixture of proteins of different sizes is carefully introduced to the top of the gel bed. Buffer is added to the top of the column and the proteins enter the gel.


3. Sample SeparationThe porous (sponge-like) nature of the hydrated gel beads forms the basis of this separation method. Molecules larger than the largest pores of the gel cannot enter the pores (they are excluded) and therefore pass rapidly down the column between the beads. The path of smaller molecules through the gel bed is much longer than that of the large molecules because they move in and out of the pores of the gel (they are included). Thus, the ability of the small molecules to penetrate the pores in the gel results in the retardation in the rate of their migration down the column. The porosity of the beads is an important feature in gel filtration, since it determines the size range of proteins that can be separated by the method. Different types of beads are commercially available which differ in the extent of cross-linking and thus pore size. Beads with large pores are used for separating large protein molecules while small proteins and peptides are separated on gels with small pores. Sephadex G-100 will be used in the experiments described in this manual and this material permits separation of proteins in the molecular weight range of 5,000-100,000.

4. Samnle CollectionThe column eluate is allowed to run dropwise into a series of test tubes and the amounts and types of the proteins in each tube (fraction) are determined.


Figure 5. Stages of a Separation of Two Proteins of Different Size on a Gel Filtration Column

1. PREPARATION OF TilE GEL COLUMN- Hydrated beads (large open circles) are packed into a cylindrical tube or column to form the gel bed.

2. SAMPLE APPLICATION - A sample containing a mixture of a large protein (large black dots) and a small protein (small black dots) is added to the top of the gel bed.

3. SAMPLE SEPARATION - Buffer is applied to the top of the column and the proteins enter the gel bed. The smaller proteins penetrate the gel beads and their movement down the column is retarded. The larger proteins are excluded from the gel beads and thus move rapidly down the column.

4. SAMPLE COLLECTION - The two proteins are now separated and can be collected in the column eluate.


PARTB LABORATORY EXERCISES Introductory Remarks

A. FEATURES OF THIS LABORATORY1. This laboratory course was designed for 8 groups of students working

in teams of 14. Each group will be responsible for the preparation and maintenance of one gel filtration column. We encourage full cooperation between members of each group and an equitable distribution of work and responsibility.

2. Each exercise consists of a background information section, an experimental procedure, and study questions. Before each laboratory, the student should read the background information section preceding the description of the exercise and then study the directions for doing the experiment. During this study, the student should understand the reason for each step in the procedure.

3. After completing the experiment, the student should clean the materials he/she has used in the laboratory.

B. QUANTITIES AND THEIR MEASUREMENTTo successfully perform the experiments in this manual, the student must be familiar with the following metric units of measurements.

Length English Equivalent1 meter(m) = 100 centimeters1 centimeter (em) = 10 millimeters (mm) = 0.3937 inches

Volume1 liter (l) = 1000 milliliters = 1.06 quartsmilliliter (ml) = 1000 microliters1 microliter (µ1) = 10-3ml

Mass1 Kilogram (Kg) = 1000 grams = 2.2046 pounds1 gram (g) = 10-3Kg1 milligram (mg) = 10-3g1 microgram (mg) = 10-6g

A solution is a homogenous mixture whose composition may be varied. The solute is the component present in lesser quantity and the solvent is the constituent


present in greater quantity. Because the composition of a solution may vary, it is necessary to specify the concentration of the solutes to describe the solution. The concentration of a solute is often expressed as the amount of solute per amount of solution. In this manual,concentration is frequently expressed as the mass of solute per unit volume of solution. For example, 10µg protein/ml means that 10µg of protein is dissolved in I ml of solution.

The basic tools you will have available for measuring liquids are the macropipetors and two types of transfer pipets. The macropipetors are calibrated in mls and should be used for measuring volumes greater than 2 mls. A diagram of the transfer pipets is given in Figure 6 showing the approximate volumes of liquid that are dispensed if the pipets are filled to the indicated positions. The large transfer pipets should be used for measuring volumes between 0.5-1.ml and the small transfer pipets for volumes less than 0.5 mi. Thus, to transfer 0.1ml (100µl) from tube A to tube B, place the tip of a small transfer pipet in tube A, depress the bulb slightly with your thumb and index finger, slowly relax your index finger until the solution has filled the pipet to the 100µl level,transfer the pipet to tube Band depress the bulb. A separate small pipet should be used by the entire class for dispensing each solution provided with this Experiment Package. The small transfer pipets will also be used for the application of samples to the chromatographic columns. Students should practice using these devices prior to beginning the experimental exercises.

Figure 6. Transfer Pipets.

Large Type Small Type

ApproximateVolume Approximate

Volume1.0ml (1,000µl)0.4ml (400µl)

0.1ml (100µl)

0.05ml (50µl)

0.01ml (10µl)

0.75ml (750µl)

0.5ml (500µl)

0.25ml (250µl)


Experiment 1. Separation of Of Molecules by Gel Filtration Chromatography

Background InformationIn order to characterize the behavior of a molecule in gel filtration, its. elution volume (Ve) is measured. The elution volume of a molecule is defined as the volume of buffer required to elute the molecule from the column. The elution volume of a molecule is,of course,dependent on its size. As we shall see in Exercise 2, this value can be used to estimate the molecular weight of an unknown protein. In order to standardize the elution volume, the void volume (Vo) of the column is determined. The void volume is the elution volume for a substance that is completely excluded from the gel. A common material for determining the void volume is blue dextran,a very large colored polysaccharide with a molecular weight of about 2,000,000. This substance does not enter the pores in the gel beads because of its large size and therefore flows only between the beads during its journey down the column. Thus, Vo is equal to the volume of liquid between the beads in the column bed. The ratio Ve/Vo is a common expression used in gel filtration chromatography and, in general, this value is dependent solely on the size of the molecules under analysis. Figure 1-1 shows the measurements of the elution volumes of three molecules: blue dextran; the protein myoglobin; and the small molecular weight dye, phenol red. The molecular weights of these molecules are 2,000,000,17,000, and 354, respectively. In this example, the Ve values for these molecules are 5ml, 15ml and 20ml respectively; VeNo for myoglobin is 3 and for phenol red is 4.


Figure 1-1. Elution Volumes of Blue Dextran, Myoglobin and Phenol Red


Objective To separate a mixture of blue dextran, myoglobin and phenol red by gel filtration chromatography and to determine their elution volumes. Note that this exercise requires about two hours to complete. About one hour is required to pour the column and one hour to perform the separation. If necessary, these two sections can be performed during different lab periods.

Materials Provided7 grams of Sephadex G-100 (fine grade). This material must be allowed to

swell in excess column buffer before the columns are poured. For this purpose, add about 260ml of column buffer to the 7 grams of dry beads, stir briefly and let stand at room temperature for at least three hours. Seven grams of the dry gel will form about 90ml of hydrated column bed volume. The 90ml of hydrated beads is more than enough to prepare 8 columns.

Column Buffer - The buffer is provided as a 50X concentrate and must be diluted before use as described in the instructor manual. The diluted buffer contains

0.lM NaCL,10mm Tris-HCl, pH 7.4. The buffer should be made up at leastl day

before the laboratory and stored at room temperature.8 columns. Each column (8mm by 200mm) consists of a 20ml reservoir, a

9ml column tube, a lower fitting with cap and a bed support, which is a porous polyethylene disc.

8- 24 well fraction collection plates.8 small transfer pipets8 large transfer pipets2 macropipets (12ml).Blue Dextran - Myoglobin mixture - Let this tube warm to room temperature

and then shake the tube to ensure that the material is in solution.Phenol red - The phenol red is in a 20ml bottle and will be used in all

subsequent exercises.8 small tubes.8 tube racks.

Materials Not Provided8 ring stands and clamps such as burette clamps to hold the columns.<optional-

see below).Metric rulers8 small (50ml) beakers or other similar containers.


Procedure1. Assembling the chromatographic

columns

A. As shown in Figure 1-2, the column consists of a 20 ml reservoir, a 9 ml column, a lower column fitting with blue cap and a porous disc that is used as a gel bed support. Assemble the column as indicated in the Figure.

B. Secure the column in a vertical position with blue cap end down, by clamping the column to a ring stand or by taping the column

2. Packing the Column BedProper packing of the column bed is very important for good performance. In addition, the gel bed that is prepared today will be used for all subsequent experiments in this manual so it is imperative that it be prepared correctly.

A. Place a beaker or the collection plate under the column to catch the buffer that will flow through, remove the column reservoir and place the blue bottom cap on the column.

B. Place 25ml of the mixed gel slurry into a small beaker.

C. Fill the column to about 20% of its length with column buffer. Thoroughly mix the gel slurry to make sure that it is uniformly suspended,and add part of the slurry to fill the column.

D. Allow a portion of the gel to settle.

E. As thegel is settling you will observe two layers forming: a bottom opaque layer of gel and a top layer of buffer. A 2-5 em gel bed will form in 5-10 minutes. At this time, place the reservoir on the column, fill it with the mixed gel slurry and then remove the blue cap. The buffer will now flow out of the column and the gel will slowly settle.

F. Add more gel slurry to the reservoir in small portions until the gel bed is 3.5 em from the top of the column tube.

Figure 1-2. CHROMATOGRAPHIC COLUMN


G. Place the blue cap on the column outlet to halt the flow of buffer and fill the column reservoir with buffer. When not in use, the column should be stored in this fashion to prevent the gel from drying. If the gel bed does dry out, it should be removed from the column, rehydrated and repacked into the column as described above. The same procedure should be performed if the column flow rate slows substantially during the course of the experiments described in this manual.

3. Determining the Column Flow RateThe rate at which buffer will flow through your column should be roughly 8-l0ml/hour. However, this value is subject to column variation and must be determined experimentally for your column.

A. Fill the column reservoir with column buffer and place the dry collection plate under the column to catch the buffer that will flow through.

B. Remove the blue cap from the column and allow the buffer to collect in one well of the plate for exactly 10 minutes.

C. Measure the volume of the buffer in the well by drawing the solution into a small transfer pipet to the 0.4ml level (see Figure 6).

D. In the space provided below, record the flow rate of your column.

FLOW RATE

_____________________ml/10 minutes

_____________________ml/hour

* ____________________ml/3 minutes

*This value will be used as the volume of each column fraction.

4. Separating Phenol Red, Myoglobin and Blue Dextran.

A. Using a small transfer pipet (see Figure 6), one student from each of the 8 groups should place 0.lml of the Myoglobin-Blue dextran mixture and 0.05ml (about 4 drops) of phenol red into a small tube and place the tube in a tube rack.

B. Remove the column reservoir and bottom blue cap and let the buffer drain down to the level of the gel bed.


C. Transfer the 0.15ml sample to the column. In this manipulation, place the tip of the pipet about l-2mm above the gel bed and carefully layer the sample onto the upper bed surface.

D. Let the sample drain into the bed and then add about 0.2ml of column buffer to wash the sample into the bed.

E. Wash out the pipet with buffer, add buffer to the column until it is filled, and then attach and carefully fill the reservoir with buffer.

F. Place the collection plate under the column and begin to collect the first fraction.

G. After 3 minutes, move the collection plate so that the second well is under the column and collect the second fraction for 3 minutes.

H. Repeat the above process until the 3 colored substrates have been eluted from the column. Approximately 15-20 fractions should be collected. During collection, watch the separation of the three substances as they progress down the column. White paper held behind the column will facilitate visualization of the colored bands.

I. Examine the column fractions in the collection plate and identify those fractions that contain the highest concentrations of blue dextran, myog lobin and phenol red. These can be seen most clearly when the collection plate is placed on a sheet of white paper. In the space provided below, record this information.

FRACTION NUMBER

Blue Dextran _________________

Myoglobin _________________

Phenol Red _________________

Since you have already determined the volume of the column fractions (see section 3D. on column flow rates) you can now calculate the elution volumes of the 3 molecules.


Study Questions and Analysis

1. Record, in the spaces provided below, the elution volumes (Ve) for blue dextran, myoglobin and phenol red and the void volume of the column. Also, calculate the Ve/Vo values for myoglobin and phenol red. These values will be used in the next exercise.

Blue Dextran Ve = ______________

Myoglobin Ve = ______________

Phenol red Ve = ______________

Void volume Vo = ______________

Myoglobin Ve/Vo = ______________

Phenol red Ve/Vo = ______________

2. From the results of your experiment, what can you conclude about the size of myoglobin in relation to the sizes of phenol red and blue dextran?

3. Gel filtration chromatography is frequently used to remove salts such as NaCl from a protein preparation. Why?


Experiment 2. Determination of the Molecular Weight of Human Hemoglobin

Background InformationTable 2-1 lists the molecular weights of the colored proteins that you will study in this exercise and the molecular weights of phenol red and blue dextran.

Table 2. Molecular Weights of Molecule Used in This ExerciseMOLECULE CHEMICAL NATURE COLOR MOLECULAR* WEIGHT

HORSE MYOGLOBIN PROTEIN BROWN 17,000

COW (BOVINE) HEMOGLOBIN PROTEIN BROWN 64,000

RABBIT HEMOGLOBIN PROTEIN RED TO BE DETERMINED IN THIS EXERCISE

PHENOL RED DYE RED 354

BLUE DEXTRAN POLYSACCHARIDE BLUE 2,000,000** (100,000)**

*The terms “molecular weight” and “dalton” are used interchangeably in this manual. For example, a 20,000-dalton protein has a molecular weight of 20,000. A dalton is a unit equal to 1.0000 on the atomic mass scale; this unit is very nearly equal to that of a hydrogen atom. The average amino acid residue in a protein is 120 daltons. Thus, a protein with a molecular weight of 20,000 contains 167 amino acid residues (20,000/120=167).

**The column packing material used in this experiment is Gl00 which means that all macromolecules that have molecular weights that are greater than 100,000 are excluded from the beads and have the same elution volumes (The void volume). Consequently, the molecular weight used for blue dextran should be 100,000.

A brief description of the functions and properties of the proteins listed in the table is given below.

Myoglobin - Myoglobin and hemoglobin also have an iron containing heme group and the iron is involved in oxygen binding. Myoglobin binds and stores oxygen in muscle and hemoglobin is involved in the transport of oxygen in blood.

Hemoglobin - Red blood cells, or erythrocytes, carry the protein hemoglobin in the circulation. This protein serves to transport oxygen


from the lungs to the tissue. Hemoglobin is a globular protein made up of 4 subunits, each of which contains a polypeptide chain attached to heme. Each heme group contains one iron atom, which can bind one 02 molecule. The polypeptide chains of hemoglobin are referred to as the globin portion of the molecule. Normal adult hemoglobin has •fow sub-units made up of two alpha (a) and two beta (b) polypeptide chains. When oxygen is bound to hemoglobin, the protein is bright red in color. However, when oxygen is lost from the protein, or when the protein is exposed to certain chemical agents, it becomes a dark blue or brown. The cow hemoglobin that will be analyzed in today’s experiment is brown while the hemoglobin from the blood of the rabbit should be oxygenated and thus red.

The determination of the molecular weight of a protein is often of critical importance for with this information, one can compare proteins and calculate the number of their amino acids residues. Gel filtration chromatography has proven to be a useful tool for protein molecular weight determinations as well as for the separation of proteins of different sizes. In this exercise, we will determine the molecular weight of human or rabbit hemoglobin.

The molecular weight of a protein such as hemoglobin can be estimated by comparing its elution profile with the elution patterns of standard proteins of known molecular weights. As shown in Figure 3, a linear relationship is obtained if the logarithms of the molecular weights of proteins are plotted against their respective elution volumes. To determine the molecular weight of an unknown protein, its elution volume is determined and compared to the elution volumes of the standard proteins as shown in Figure 3. In practice, the void volume of the column is also measured and the ratios of Ve/Vo are plotted against log molecular weights for the series of reference proteins.


Figure 2-1. Determination of the Molecular Weight of an Unknown Protein

To determine the molecular weight of an unknown, protein standards of known molecular weight (A= 80,000; B = 55,000; C = 14,000) are separated on a column and their elution volumes (Ve) are plotted against the logarithms of their molecular weight The elution volume of the unknown protein is used to determine its molecular weight by extrapolation from the standard graph. In !he example shown above, the unknown has a molecular weight of about 28,000.


Objective To determine the molecular weight of hemoglobin from your blood or from the blood of the rabbit.

Materials Provided8 gel filtration columns packed with gel, collection plates, column buffer,

small transfer pipets and test tubesErythrocyte lysis bufferCow hemoglobin Rabbit blood Phenol red

Materials Not Provided (Optional)Sterile finger lancets70% alcoholSterile cotton

Procedure

I. Separation of cow hemoglobin and phenol red. A. Using a small transfer pipet, one student from each of eight groups

should place 0.1ml of cow hemoglobin and 0.05 ml (about 4 drops) of phenol red into a small tube.

B. Remove the column reservoir and bottom blue cap and let the buffer drain down to the level of the gel bed.

C. Layer the sample onto the column gel bed as described in Exercise I and collect fractions every 3 minutes until the three colored substances have been eluted from the column.

D. In the space provided below, record the fractions that contain the highest concentration of cow hemoglobin, cytochrome C and phenol red, and calculate their elution volumes.

FRACTION ELUTION NUMBER VOLUME

Bovine Hemoglobin __________ __________

Phenol Red __________ __________

II. Human Hemoglobin -(optional)- If this section is not performed proceed to Section III (below). A. Place about 601-11 (5-6 drops) of erythrocyte lysis buffer into a

small test tube.

B. Disinfect your finger with 70% alcohol, allow it to dry and puncture it with a sterile finger lancet. CAUTION: Do not exchange lancets


with other students. Allow 1-3 drops of blood to fall into the test tube and mix the contents of the tube thoroughly. The erythrocyte lysis buffer contains the detergent Nonidet P-40 which serves to lyse (break open) the eryth rocytes which in turn liberate their hemoglobin.

C. Using a small transfer pipet, add 0.05ml (about 4 drops) of phenol red and layer the entire sample onto your chromatographic column. Collect fractions every 3 minutes and identify the fractions that contain the hemoglobin and phenol red.

FRACTION ELUTION NUMBER VOLUME

Human Hemoglobin __________ __________

Phenol Red __________ __________

III. Rabbit Hemoglobin

If human hemoglobin was not analyzed in the preceding section, perform the experiment described above using the rabbit blood that is provided with the Chemical Package. Erythrocyte lysis buffer has been added to the rabbit blood so you should add phenol red and then subject the mixture to chromatography as described above.

Study Questions aud Analysis1. On semilog paper, plot the elution volumes of cytochrome C, myoglobin,

and bovine hemoglobin as a function of their molecular weights (see Table 2-1) and determine the size of human or rabbit hemoglobin. The elution volume of myoglobin should be found on page 17.

2. Interferon is a protein with a molecular weight of 25,000. What elution volume would it have if it had been included in this experiment?

3. Give the approximate number of amino acids found in human (or rabbit)hemoglobin (see footnote -Table 2-1).


Experiment 3. Dye Binding of Serum Albumin

Background Information

A. Interaction Between Molecules in Biological SystemsOne of the most important principles of modem biology is that two molecules with complementary surfaces tend to bind or stick together, whereas molecules without such surfaces do not. This principle can be illustrated by considering the specificity of enzyme action. Enzymes accelerate the velocity of virtually all reactions that occur in biological systems, including those involved in breakdown, synthesis and chemical transfers. In so doing, they are responsible for performing essentially all the changes associated with life processes.

The general expression frequently used to describe an enzyme reaction is:

Certain important features of the nature of enzyme reactions are evident from this diagram:

l. The term substrate refers to the compound that is acted upon by the enzyme. In general, enzymes exhibit a high degree of substrate specificity in that they usually catalyze only a single chemical reaction.

2. The enzyme binds to the substrate to form an enzyme-substrate complex. This interaction is responsible for the specificity of enzyme action, since only those compounds that “fit” into the substrate binding site can be acted upon by the enzyme.

3. The enzyme is not destroyed during the reaction but rather is set free after the formation of the end product. Thus, the liberated enzyme is available to combine with more substrate to produce more product.


The binding of an enzyme to its substrate is only one example of the many specific molecular interactions that occur in biological systems. An analogous binding process occurs when an antibody binds antigen and this interaction will be described in more detail in Exercise 5. In fact, the selective interaction of biomolecules with each other often permits the self-assembly of these molecules into geometrically regular aggregates such as structural fibers, tubes, spheres and even functional cellular organelles. In today’s laboratory, you will study the specificity of the binding of a protein from blood for small dye molecules and determine whether the native structure of the protein is required for its binding activity.

B. Serum AlbuminBlood is a remarkable tissue containing cellular elements (erythrocytes, leu kocytes and platelets) suspended in a liquid medium called plasma. Whole blood, or plasma, clots upon standing and if the clot is removed, the remaining straw colored fluid is called serum. Serum has basically the same composition as plasma except that it lacks certain proteins that are involved in the clotting process.

Serum contains a variety of small molecular weight components as well as hundreds of different serum proteins. The major protein in serum is albumin, which functions as a carrier molecule for the transport of certain small molecular weight compounds in blood. Molecules that bind to serum albumin include bilirubin, fatly acids, hormones, and some synthetic dyes.

Objectives To determine the molecular weight of serum albumin and to study the binding specificity of this protein for small synthetic dye molecules using a gel filtration assay. In addition, you will examine the effects of the detergent sodium dodecyl sulfate (SDS) on the binding activity of serum albumin. This detergent is a strong protein denaturant (see page 3).

Materials ProvidedBovine (cow) serum albumin (5%) Phenol redBromophenol blue24 small test tubes and 8 tube racks8 gel filtration columns, collection plates, column buffer and small transferpipets.SDS (Sodium Dodecyl Sulfate)

ProcedureThe small transfer pipets should be used in this exercise.

1. One member from each group should place 0.1 ml of bovine serum albumin into two tubes and place the tubes in a rack.

2. Add 3 drops of phenol red and 2 drops of bromophenol blue to each tube.


3. Add I drop of SDS to one tube and mark the tube +SDS.

4. Apply the mixture from the tube lacking SDS to your gel filtration column and collect fractions every 3 minutes as described in Exercise I. Record the position of the two dyes in the elution profile.

5. Apply the protein-dye mixture from the tube containing SDS to the column and record the position of the two dyes in the elution profile as described above.

Study Questions and Analysis1. Which of the dye molecules binds to serum albumin? Which does not?

2. Estimate the molecular weight of bovine serum albumin by comparing its elution volume to the elution volumes of hemoglobin, myoglobin and cyto chrome (Exercises I and 2).

3. How would you determine the molecular weight of serum albumin in your blood?


Experiment 4. A Comparison of the Properties of α-Amylase from Human Saliva and Pancreas

Background InformationGel filtration chromatography has the advantage of being able to yield the molecular weight of a specific protein in a complex mixture, provided the protein bas a biological activity that can be measured. For example, a crude cell exttact contains hundreds of different enzymes yet it is possible to determine the molecular weight of a single enzyme protein in this extract by passing the extract through a column and determining the elution volume of the enzyme activity in the column eluates. In today’s laboratory, this procedure will be used to estimate the molecular weight of an enzyme that is involved in the breakdown of complex carbohydrates.

A. Basic Carbohydrate ChemistryCarbohydrates, together with lipids and proteins, play a fundamental role in the life of aU organisms. The carbohydrates include such substances as sugars, starch, and cellulose. These compounds are composed of carbon, hydrogen, and oxygen. The simplest type of carbohydrates are the monosaccharides, simple sugars that contain 3-7 carbon atoms. There are about 20 different monosaccharides that are found in living organisms. One of the most common monosaccharides is glucose, whose structure is shown in Figure 4-1.

Figure 4-1. The Structure of Glucose.

In nature, glucose occurs in a ring form like that shown in the figure.


Polysaccharides are macromolecules (giant molecules) that contain many monosaccharide units linked together by glucosidic bonds. The major polysaccharides digested in the human gastrointestinal tract are glucose polymers that comprise the plant starches and animal glycogens. The starches consist of two components called amylose and amylopectin (Figure 4-2). Amylose, which turns blue when stained with iodine, is made up of glucose molecules arranged in long straight chains. Amylopectin, which turns purple to red when stained with iodine, is a branched polymer of glucose. Animal glyocogen, which is a storage form of glucose in liver and muscle, is similar in structure to amylopectin in that it is a complex branched carbohydrate.

B. AmylaseDigestion of carbohydrate is initiated in the mouth by the action of a hydrolytic enzyme in saliva called a-amylase. This enyzme is a major protein found in saliva. The optimal pH for this enzyme is about 7, so its action is inhibited when food enters the stomach by the acidic stomach juice. In the small intestine, polysaccharide digestion is reinitiated by pancreatic a-amylase, an enzyme that is similar but not identical to salivary a-amylase. Both enzymes cleave the glucosidic bonds that link glucose molecules into starches and glycogen (Figure 4-2). The a amylase first hydrolyzes the polysaccharides to products called dextrins, which contain several to several hundred simple sugar molecules. The dextrins, in turn, are broken down under the influence of α-amylase to di, tri- and oligosaccharides. These sugars are subsequently broken down to glucose by intestinal oligosaccharidases and the glucose molecules cross the intestinal wall and enter the blood. Figure 4-3 shows the breakdown of starch under the influence of a-amylase.

Figure 4-2. Structure of the Components in Starch Amylose.

Amylose and amylopectin are glucose polymers. The portion of amylose shown above contains 5 glucose residues.


Figure 4-3. Hydrolysis of Starch by α-Amylase.

C. Detection and Measurement of Enzyme ActivityIn this exercise, you will study the hydrolysis of starch by the enzyme, α -amylase. The enzyme reaction can be represented in a simplified form as follows:

α-AMYLASEH2O

STARCH OLIGOSACCHARIDES (SUBSTRATE)

The reaction can be detected either by following the decrease in the amount of starch present in an enzyme reaction mixture or by measuring the appearance of oligosaccharides. Iodine staining is frequently used to follow the course of starch disappearance because starch turns blue-black when stained with iodine, while lhe oligosaccharides do not

Two parameters that influence the amount of starch breakdown are time and amylase concentration. As shown in Figure 4-4, if starch is in excess, !hen there is an increase in the amount of starch breakdown with increasing reaction time and with increasing amylase concentration. If time is held constant, then the amount of starch breakdown is directly related to the amylase concentration. This feature of the reaction enables one to determine the amount of amylase activity in a sample.


Figure 4-4. Effects of Time and Amylase Concentration on Starch Breakdown

In this experiment α-amylase activity will be quantified by the method outlined below:

1. An agar gel containing starCh is cast in a petri dish.

2. Amylase or cellular extracts containing amylase are placed in holes made in the gel.

3. The gel is incubated at room temperature for one day. Amylase diffusion into the gel is accompanied by digestion of the starch.

4. The starCh in the gel is detected by iodine staining, and enzyme activity is noted by transparent rings around the sample wells in a blue-black gel.

Figure 4-5 illustrates the relationship between ring diameter and a-amylase concentration applied to the gel. To determine the amount of amylase in an unknown sample, the sample is applied to the gel and the diameter of the resulting ring around the sample well is compared to the diameters of the rings around the wells containing known concentrations of α-amylase. For example, a ring diameter of 13mm means that the sample contains 10μg of amylase activity (see dotted lines on the graph in Figure 4-5).


Figure 4-5. Determination of Amylase Activity.

Amylase solutions (50μl) containing (A) 0, (B) 0.8, (C) 4, (D) 20 (E) 100 and (F) (500)μg amylase per ml were incubated for 24 hours in an agar gel which was then stained with iodine solution.

The diameters of the rings from the above experiment were measured and plotted as a function of amylase concentration. Note that the data are presented as a semi logarithmic plot


Objective To determine the amounts and molecular weights of amylase in saliva and in pancreatic extracts.

Materials Provided8 columns containing Sephadex G-1008 collection plates16 small test tubes and 16 large test tubes8 tube racks*Column bufferMacropipetAgarStarch solution16 Petri dishes8 Glass pasteur pipetes14 small transfer pipets: one pipet per student group, one for the starch

solutionand one for each of the 5 amylase solutions•Amylase (500μg/ml)•Amylase (100μg/ml)•Amylase (20μg/ml)•Amylase (4μg/ml)•Amylase (0.8μg/ml)Pancreatic extract*I2Kl solutionPhenol red

*Prepared as described in Appendix 2 of the instructor manual.

Materials Not ProvidedBalance- Desirable but not absolutely necessary (see below)

Rulers, marking pensA container such as a 600ml beaker to boil waterBunsen burner with wire gauze on a tripod

Procedure:

I. Preparation of the Sample from SalivaAdd one drop of your saliva to 0.4ml of column buffer in a small tube and mix thoroughly.

II. ChromatographyIn this exercise, 4 groups of students will study a-amylase in saliva and the other 4 groups will study a-amylase in the pancreatic extract provided in the Chemical Package. At the end of the experiment, students should compare their results.


I. One member from each group should place 0.1ml of the saliva sample (4 groups) or 0.1 ml of the pancreatic extract (4 groups) into a small test tube.

2. Add 50μl (about 4 drops) of phenol red to the tube and apply the sample to the column.

3. Collect fractions every 3 minutes until the phenol red has been eluted from the column and record the fraction number that contains the phenol red.

III. Determination of Amylase Activity in Column FractionsA. Preparation of the Agar Gels

1. Each group of students should obtain two large glass test tubes. To each tube, add 20ml of column buffer and 0.32 grams of agar. The agar can be weighed out directly on an appropriate balance. If a balance is not available, 0.32 grams of agar can be estimated by filling one of the small 0.5ml tubes provided with the package with agar until full.

2. Add 0.4ml of the mixed starch solution to each tube and swirl the tubes until the agar forms a suspension.

3. Place the test tubes into a boiling water bath and allow the agar suspension to come to a vigorous boil. The level of the water in the bath should be about equal to the level of the solution in the tubes.

4. Pour the contents of each tube into the bottom section of a petri dish.

5. Let lhe agar cool for at least 15 minutes. The agar gels can be used immediately or they can be stored covered in the refrigerator in a sealed plastic bag for up to one week.

6. Place lhe pattern shown in Figure 4-6 beneath the dishes and use the Pasteur pipets provided in the Chemical Package to cut the wells in the agar according to the pattern. Push the wide end of the Pasteur pipet into the agar to cut the wells. Withdraw the pipet and remove the agar segments from the wells by piercing them with the narrow end of the pipet Repeat the process using the pattern until all sample wells on both plates are formed.

7. With a marking pen, place the letter A on one plate and the letter B on the other. Turn the plates over and number the wells #1-#9 with a marking pen as shown in Figure 4-6.


Figure 4-6. Pattern for Arrangement of Wells in the Petri Dish.

B. Sample ApplicationUsing the small transfer pipets, carefully fill the wells as indicated below. Draw a small amount of sample (about 01ml = 100μl) into the pipet and slowly eject it into the well until the well is almost full. At this time,the well should contain about 0.05ml (50μl). Five pipets should be used by the entire class to dispense the amylase solutions. Each group should then use a single pipet to add the saliva, pancreatic extracts and portions of the column fractions to the sample wells. Rinse this pipet between samples by drawing up and expelling water three times.


PLATE A PLATE B

WELL NUMBER SAMPLE WELL NUMBER SAMPLE (Column Fraction Number)

1 Column buffer 1 22 Amylase 0.8µg/ml 2 43 Amylase 4µcg/ml 3 64 Amylase 201µg/ml 4 85 Amylase IOOµg/ml 5 106 Amylase 500µg/ml 6 127 Saliva- sample 7 148 Saliva- sample 8 169 Pancreatic Extract 9 18

After all samples have been loaded into the wells, place the lid on the dish. The dish should not be moved at this time. The plates should remain at room temperature for 12-24 hours before iodine staining. If longer times are desired, the plate should be placed in the freezer after one day. C. Detection of Amylase Activity

1. Place about 5ml of the diluted iodine solution onto the agar in the dishes.

2. After 10-20 minutes, discard the iodine and fill the petri dishes with water.

3. After 10-20 minutes, discard the water, measure the diameter (in mm) of the clear rings around each well and record your results on the following page. The rings can be seen most clearly by holding the dish over a light source such as a desk lamp.

4. The plastic plates and pipets should be washed with a mild detergent and rinse with water.


PLATE A PLATE B

WELL NUMBER RING DIAMETER (mm) WELL NUMBER RING DIAMETER (mm)1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9

Study Questions and Analysis1. On semilog paper (provided below) plot the diameters of the rings from

wells #2-#6 (PLATE A) as a function of the concentration of a-amylase.

2. Determine the amount of amylase activity in the column fractions and estimate the molecular weight of amylase by comparing its elution volume with the elution volumes of serum albumin, hemoglobin, myoglobin and cytochrome C (Exercises 1-3).

3. Explain the basis for the formation of the rings in the agar-starch gels.


Experiment 5. Isolation and Immunological Analysis of a Protein from Egg White

Background Information

A. Isolation of the Protein Ovalbumin from Egg WhiteA knowledge of the solubility of proteins is frequently useful in the isolation of these macromolecules. For example, the major protein found in chicken egg white is ovalbumin (also called egg albumin) and this protein is soluble in distilled water. Egg globulins, in contrast, are proteins that are not soluble in water but are soluble in dilute salt solutions. Thus, if egg white is added to water, the globulins precipitate and can readily be removed from the soluble albumin preparation by centrifugation or filtration. This differential solubility of ovalbumin forms the basis of the first step in the isolation of ovalbumin that will be performed in today’s laboratory.

In most cases, the purification of a single protein involves a series of steps with each step exploiting differences in protein solubility, size or charge. Gel filtration chromatography is frequently the method of choice for the size fractiona tion step in a protein purification scheme and this procedure will be used as the second step in today’s ovalbumin purification procedure.

In order to determine the elution volume of ovalbumin, and thus its molecular weight, a method must be used to detect this specific protein in fractions from the gel filtration column. However, ovalbumin is a colorless protein that does not possess enzymatic activity. For these reasons,an immunological assay will be used to identify the position of ovalbumin in the elution profile of the gel filtration column in today’s experiment.

B. Antibodies as Tools in Modern BiologyThe immune system consists of a variety of cells, tissues and organs and about 1012 antibody molecules. The major function of the immune system is to protect the organisms from viruses,bacteria, protozoans and larger parasites. When these organisms enter the body, macromolecules on their surfaces induce the production of specific antibodies that appear in the serum of the infected animal. The antibodies, in turn, combine with these foreign macromolecules, thereby rendering the invading organisms inactive and noninfective.

The macromolecules that elicit antibody production are called antigens and are most often proteinaceous in nature. Although antigens are frequently compo nents of foreign organisms, purified foreign proteins will serve as antigens in that they will stimulate the formation of antibodies when injected into a suitable test animal such as the rabbit. The antibodies are recovered in the serum of the immunized animal and this serum is referred to as an antiserum.


Antibodies against specific proteins are extremely important tools used in the modem biology laboratory for they permit one to identify and quantify specific protein antigens. In the experiment described in this section, you will perform the double-diffusion technique to identify the egg white protein ovalbumin. The technique is based on the observation that specific antibody-antigen complexes undergo precipitation reactions that are readily visualized in an agar gel. The method was developed by Organ Ouchterlony more than 40 years ago and is still in wide use today. In this method, an agar gel is prepared in a petri dish and a small amount of antigen and antibody solution are placed separately in small wells cut into the agar. The antigen and antibody diffuse outward toward each other at rates that are related to their concentration, size and shape. A milky line of precipitate forms where one antigen encounters its antibody. This technique is illustrated in Figure 4-1.

Figure 4-1. Double-Diffusion Technique for Analysis of Antigen-Antibody Complexes.

1. An agar gel is cast in a dish and six small wells are produced by cutting circular holes in the agar.

2. Antiserum prepared against the egg white protein ovalbumin is placed in the center well and ovalbumin (A), serum albumin (B) a-amylase (C), total egg white (D) and serum (E) are placed in wells on the periphery of the plate. Following time to permit the diffusion of antibodies and antigens, the plate is viewed for the presence of milky white precipitation lines. Note that a single line is formed when purified ovalbumin or total egg white is placed in the wells.

Objective: To isolate ovalbumin from the chicken eggs.

Materials Provided8 columns containing BIO-GEL P1008 collection platesSmall test tubes, tube racks, and large test tubes


Column bufferAgar16 Petri dishes8 glass Pasteur pipetsMacropipetsCheese cloth14 small transfer pipetsx Ovalbumin 20 mg/ml x Ovalbumin 4 mg/ml Ovalbumin 0.8 mg/mlx Ovalbumin 0.16 mg/mlxx Anti-ovalbumin antibodies Phenol redx Prepared as described in Appendix 2 of the instructor manual.xx This antibody preparation was produced by rabbits that were injected

with purified ovalbumin.

Materials Not ProvidedOne fresh eggOne 500ml beaker

Procedure

I. Extraction of Ovalbumin from the Chicken EggThis section should be performed by the instructor or by the entire class as a group.

1. Carefully break the egg and separate the yolk from the white by passing the yolk between the two egg shell halves while allowing the white to flow into a 500ml beaker or another suitable container.

2. Slowly add l00ml of distilled or deionized water to the egg white and stir. A white precipitate composed of the egg globulins will form.

3. To remove the precipitate either:

A. Centrifuge a portion (about 20%) of the extract and pour off and save the supernatant ovalbumin fraction. Label this fraction “ovalbumin extract”.

B. Filter a portion (about 20%) of the extract through 8 layers of cheesecloth provided with the Chemical Package and label the filtrate “ovalbumin ex tract”.


II. Chromatography1. One member from each group should place 0.15ml of the ovalbumin

solution into a small test tube containing 50μ1 (about 4 drops) of phenol red.

2. Apply the sample to the column and collect fractions every 3 minutes until the phenol red has been eluted. Record the fraction number that contains the highest concentration of phenol red.

Ill. Identification or Ovalbumin in tbe Column FractionsA. Preparation of the Agar Gels

1. Prepare agar gels in two petri dishes as described in Experiment 4 except omit the starch from the gel.

2. Label one plate A and one plate B.

3. Place the pattern shown in Figure 5-2 beneath the dishes and use the Pasteur pipets to cut the wells in the agar according to the pattern.

4. Tum the plates over and number the wells #l-#9 with a marking pen as shown in Figure 5-2.

Figure 5-2. Pattern for the Arrangement of Wells in the Petri Dish.


B. Sample Application and Ovalbumin Detection1. Using the small transfer pipets, carefully fill the wells as indicated below

and as detailed in Experiment 4.

PLATE A WELL NUMBER SAMPLE 1 Anti-ovalbumin antibodies

2 Column buffer

3 Ovalbumin 20 mg/ml

4 Ovalbumin 4 mg/ml

5 Ovalumin 0.8 mg/ml

6 Ovalbumin 0.16 mg/ml

7 “Ovalbumin Extract”



PLATE B WELL NUMBER SAMPLE 1 Anti-ovalbumin Antibodies

2 Column Fraction 4

3 Column Fraction 6

4 Column Fraction 8

5 Column Fraction 10





2. Place the lids on the dishes.

3. Observe the dishes daily for about 3 days for the development of precipitation lines.

4. After 3-6 days, examine the gel carefully for precipitation lines and describe your results. The intensity of the precipitation lines is proportional to the concentration of ovalbumin in the sample.


Study Questions and Analysis1. Estimate the elution volume, the molecular weight, and the number of

amino acid residues for ovalbumin.

2. Estimate the amount of ovalbumin present in one chicken egg.

3. Lysozyme is an egg white protein. This protein is also an enzyme that hydrolyzes the carbohydrates in the capsules of bacteria, which results in the lysis (disintegration) of the bacteria. Describe two methods or procedures that you would use to determine the amount of lysozyme that is present in one egg.

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