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Team Based Learning Exercise For

Medical biochemistry and Molecular Biology

Protein Structure

Authors: Edward E. McKee, Ph. D. Department of Biochemistry and Molecular Biology, Indiana University School of Medicine – South Bend

Selma Omer, Ph.D. Global Health Sciences, University of California – San Francisco

Instructors Guide to TBL 2-3 IRAT/GRAT 4-6 IRAT/GRAT Instructor’s Guide 7-9 Application 10 -13

Application Instructor’s Guide 14-17

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Instructors Guide Introduction: TBL is designed to be a highly interactive small group exercise that promotes the application of knowledge, higher order thinking, peer to peer teaching, communication, and professionalism. This TBL has generated good intra-team and inter-team discussions and students have consistently reported that they enjoy the exercise. This TBL also gives students hands-on practice with rapid self-assessment in their understanding of amino acid and protein structure and its relevance to medicine. TBL Format: A standard TBL format is used. Our class is divided into 5-6 person teams that are maintained throughout the year. The Individual Readiness Assurance Test (IRAT) and Group Readiness Assurance Test (GRAT) are given as closed book exams. The IRAT is taken online (20-25 min) in class via a course server (ANGEL). The GRAT portion is taken with Immediate Feedback Assessment Technique (IF-AT) cards (http://www.epsteineducation.com/home/about/default.aspx) that are collected for scoring (20 min). A brief discussion period follows to allow teams to defend and challenge answers. Following a short break, the application part of the exercise is begun. This is “open book”, and discussed one question at a time. Teams are allowed one computer connected to the web, and searches need to be conducted by a single team member under the direction of the team. The “scenario” and laboratory data are shown on a classroom screen in power point slides. The application questions are shown one at a time on a second classroom screen in power point slides. (These are included together as one Word document on MedEdPortal for ease of reviewing.) Teams respond and submit their answers one question at a time online (ANGEL). Team answers are provided simultaneously to the class after each question by Audience response system polling (http://www.turningtechnologies.com/), or by traditional flash cards. Team discussion follows each question. Students readily supply this discussion to support and defend their team’s answer, or to challenge an answer by another team. The facilitator should resist taking part in the discussion and should not indicate the preferred answer until discussion is completed. In this TBL, in our experience, while teams may occasionally submit an incorrect answer, the students usually arrive at a consensus that is in fact the preferred answer without any input from the facilitator. In the absence of computer resources, the IRAT/GRAT can be taken as a hard copy collected exam for later grading, manually or by scantron. It is also possible to use hard copies of the application questions, but they should be distributed and collected one question at a time. Very often the next question builds on, and provides the answers to, the previous question. Student Preparation – Amino Acid and Protein Structure: This material is well-covered in most Biochemistry Texts. At our center, this material is not given as formal lectures. The students may choose from a variety of resources in addition to a text book to learn this material, including: instructor prepared notes, instructor prepared Power Point presentation; and, Camtasia recordings (Camtasia records power point presentations with annotations and audio that can be edited) of previous lectures delivered on this material. All material is available on a Course server (Angel). The material covers basic amino acid and protein structure, peptide bond formation, protein folding, basic methods of protein purification, and clinical examples of misfolded proteins associated with diseases. Additional reading may include:

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1. Byers, P H, Graham, J M, Weiss, L, et al. (1989). Osteogenesis imperfecta. The position of

substitution for glycine by cysteine in the triple helical domain of the pro alpha 1(I) chains of type I collagen determines the clinical phenotype. Journal of Clinical Investigation, 84(4), 1206-14.

2. Chernoff, Y. O. Do amyloids remember their origin? Insights into the prion species barrier. (2004) Molec. Cell 14: 147-152.

3. Wilkinson B., Gilbert H.F., Protein disulfide isomerase, 2004, Biochem. Biophys. Acta 35–44.

4. Eskenazi D, Acosta D, Introduction to Biochemistry: Protein Structure and Enzyme Basics. MedEdPORTAL; Available from: www.mededportal.org ID 8122 (Very basic review tutorial.)

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IRAT/GRAT Protein Structure

1. Which structure represents the primary form of histidine at a pH of 7? The pK of the imidazole ring hydrogen of histidine in solution is 6.0.

2. Which of the following amino acids is involved in forming disulfide bonds?

a. Cysteine b. Histidine c. Proline d. Tryptophan e. Valine

3. Bonding between which amino acid grouping results in a peptide bond?

a. The amino group of both residues b. The carboxyl groups of both amino acids c. The carboxyl group of one residue and the amino group of the second d. The primary carbon of the first residue and the amino group of the second e. The primary carbon of one residue and the carboxyl group of the second

4. Which of the following is the BEST example for a helical wheel?

a. Amino acid side chain placement when looking down an α-helix axis. b. A motif of super secondary structure like the β-barrel. c. Construction of coiled coils such as collagen. d. Cross-section found in certain fibrous proteins viewed by electron microscopy. e. Interaction of side chains in the formation of the α-helix.

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5. Hemoglobin A is a protein which is composed of two α subunits and two β subunits. The arrangement of these four subunits constitutes which type of structure?

a. Primary b. Secondary c. Tertiary d. Quaternary

6. Which one of the choices below represents a non-covalent interaction in protein that

stabilizes both secondary and tertiary interactions?

7. Which one of the statements below best describes protein folding into correct secondary

and tertiary protein structures?

a. The initial formation of secondary structure in a protein typically requires a chaperonin.

b. The final folded form of a protein will be the one that offers the maximum stability.

c. Protein surfaces are generally composed of hydrophobic residues. d. Protein interiors are generally composed of polar residues. e. A peptide synthesized form a sequence of 100 amino acids chosen at random will

typically fold into a single stable structure.

8. What is a true functional characteristic of α-helix or β-sheet structures?

a. An α-helix is typically composed of two or more polypeptide chains. b. β-sheets are quite readily stretchable. c. β-turns often contain proline. d. Secondary structure is significantly stabilized by interactions of the amino acid

side chains. e. The α-helix relies on ionic bonds between the α-carbon and the amide hydrogen.

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9. Which one of the procedures below will separate a large number of proteins by both

charge and molecular weight (size)?

a. Gel exclusion chromatography b. Isoelectric focusing c. Affinity chromatography d. Ion exchange chromatography e. Two Dimensional Electrophoresis

10. A mixture of three proteins (A, B, and C) was eluted in the order A-B-C when the

mixture was passed through a column of Sephadex G-100 in a buffer at pH 8.0. When they were passed through a DEAE-cellulose column at the same pH, the proteins were not retained. What do these finding suggest?

a. Protein A is more acidic than proteins B and C at pH 8.0. b. Protein A is more basic than proteins B and C at pH 8.0. c. The three proteins have the same molecular weight. d. Protein B is larger than Protein C. e. The three proteins are negatively charged at pH 8.0.

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Instructors Guide IRAT/GRAT: 1. The pK of a free carboxyl group (COOH) is ~ 2 and will be completely deprotonated (COO-)

at pH 7. The pKs of amino groups (NH3+) are ~ 8.5-10 (pK = 9.2 for histidine) and will be

completely protonated (NH3+) at pH 7. Histidine has an imidazole group with a pKa of 6.00

i.e. near neutrality; a significant fraction of the side chains will become uncharged at neutral pH. This can be calculated from the Henderson-Hasselback equation:

7 = 6 + log N/NH+ thus, 10 = N/NH+

or about 91% of the imidazole group will be present in the uncharged form. (Answer B). In many enzyme catalyzed reactions His residues facilitate the reaction by serving as a proton donor or acceptor.

2. Cysteine (A) is the only amino acid with a sulfhydryl side chain that under the right

conditions of folding and oxidation can form di-sulfide bonds with other cysteine sulfhydryl groups within the same peptide chain or between peptide chains. This is a common motif in extracellular and vascular proteins, but is mush less common in the cytoplasm.

3. Two amino acids are covalently joined through a substituted amide linkage forming a covalent bond called the peptide bond. It is formed by the removals of the elements of water (dehydration) from the alpha carboxyl group of one amino acid and the amino group of the second (Figure on right) (Answer C). The alpha amino group acts as a nucleophile to displace the hydroxyl group of the other amino acid. This is an example of a condensation reaction. Under physiological conditions the equilibrium for the reaction favors amino acids over dipeptide.

4. A helical wheel is a type of plot or visual representation used to illustrate the properties of alpha helices in proteins. The sequence of amino acids that make up a helical region of the protein's secondary structure are plotted in a rotating manner where the angle of rotation between consecutive amino acids is 100°, so that the final representation looks down the helical axis (answer A). The plot can reveal whether hydrophobic amino acids are concentrated on one side of the helix, usually with polar or hydrophilic amino acids on the other. This arrangement is common in alpha helices within globular proteins, where one face of the helix is oriented toward the hydrophobic core and one face is oriented toward the solvent-exposed surface. Specific patterns

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characteristic of protein folds and protein docking motifs are also revealed, as in the identification of leucine zipper dimerization regions and coiled coils.

5. (See Figure on right.) Primary structure is the amino acid sequence of a peptide linked together by peptide bonds. (Not answer A.) Secondary structure refers to a particularly stable arrangement of amino acid residues giving rise to recurring structural patterns, such as an alpha helix of β sheet (not answer B). Tertiary structure describes the three dimensional folding of a single polypeptide (not answer C). When the protein has more than one subunit as in the case of hemoglobin, their arrangement together is the quaternary structure (answer D)

6. While tertiary structure is stabilized

by many types of non-covalent interactions, including electrostatic (answer A), hydrophobic (answer C), hydrogen bonds (answer B) and the covalent disulfide bond (answer D), secondary structure is stabilized only by hydrogen-bonds (answer B).

7. Secondary structures like alpha helices form readily since they make optimal use of stabilizing hydrogen bonding between the hydrogen atom attached to the electronegative nitrogen atom of a peptide linkage and the electronegative oxygen carbonyl oxygen atom of the fourth amino acid on the amino-terminal side of the peptide bond. They typically occur rapidly in the right environment without the aid of a chaperonin (not answer A). Chaperonins do function to increase the speed at which the correctly folded tertiary structure is formed. The primary structure of a protein determines its ultimate folding, which will be the form that offers maximal stability (answer B is true). Hydrophobic interactions are important for stabilizing protein conformations. The interior of the protein is a densely packed core of hydrophobic amino acids tucked away form the water (not answer D), while protein surfaces are composed of polar hydrophilic residues facing the aqueous environment (not answer C). When water surrounds a hydrophobic molecule the arrangement of hydrogen bonds results in a solvent layer which correlates with a decrease in the entropy of water, however when non polar groups are clustered together , there is a decrease in the extent of the solvent layer because each group no longer presents its entire surface to the solution. Biological proteins

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are highly evolved because they fold consistently and perform a useful function. Random sequence proteins have nearly infinite ways to fold and are unlikely to form a consistent structure (not answer E).

8. Beta turns are 1800 structures typically involving 4 amino acids with the carbonyl oxygen on

the first residue forming a hydrogen bond with the amino-group hydrogen on the fourth. Beta turns often contain Gly and Pro (answer C). The former because it is small and flexible, the latter because peptide bonds involving the amino nitrogen of proline can assume the cis configuration useful in tight turns. The α-helix is a secondary structure domain within a single peptide subunit (not answer A) that is independent of electrostatic interactions (not answer E) and amino acid side chain interactions so long as the side-chains do not hinder formation (not answer D). β-sheets are already fully extended and cannot be stretched (not answer B). Note: secondary structures like alpha helices can be prevented by certain arrangements of side-chains. For example, long blocks of Glu at pH 7 prevent helical formation because the negatively charged carboxyl groups repel each other and similarly with positively charged residues. The bulk and shape of Asn, Ser, Thr and Cys can also destabilize the alpha helix if they are close together.

9. Isoelectric focusing and ion exchange chromatography separate proteins based on charge only (not answer B or D). Gel exclusion chromatography relies on molecular weight (not answer A). In affinity chromatography separation depends on the interaction of a highly specific binding site on a protein to its ligand (not answer C). In two dimensional electrophoresis protein samples are first separated based on their charge by isoelectric focusing, then further fractionated by SDS electrophoresis based on their molecular weight (answer E)

10. Proteins A, B and C were separated by gel exclusion chromatography on a Sephadex column.

Molecules that are too large to pass through the Sephadex beads take the shortest route to the bottom of the column and are eluted first. Smaller molecules enter the beads and take a longer path along the column. Protein A is therefore larger than protein B which is larger than protein C (answer D, not answer C). Since all three proteins were not retained in the DEAE-cellulose column (positively charged column or anion exchanger) at pH8 indicates that all three proteins were either neutral or positively charges at that pH and could not be distinguished from each other (not answers A, B, or E).

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1. Application - Protein Structure

1. Under experimental conditions ribonuclease can be completely denatured by exposure to a concentrated urea solution and strong reducing agent. It will gradually refold into an active ribonuclease once these compounds are removed. An enzyme called disulfide isomerase that catalyzes the making and breaking of disulfide bonds has been isolated. In the presence of this enzyme inactive scrambled ribonuclease is rapidly converted into its enzymatically active form. What might these observations imply relative to the folding and active structure of ribonuclease?

a. For ribonuclease to fold correctly the cysteine residues must become fully reduced to the

sulfhydryl. b. Ribonuclease must be a homo-oligomer (at least a homo-dimer) joined by disulfide

bonds. c. Ribonuclease must contain cysteine residues that yield one stable structure among a

variety of possible disulfide bond formations. d. The folding of ribonuclease does not require disulfide bonds. e. The three dimensional structure of active ribonuclease must consist of more than one

peptide chain connected only by disulfide bonds.

2. In contrast, mature active insulin is rapidly inactivated by exposure to disulfide isomerase and will not recover even if the enzyme is removed. What might these observations imply relative to the structures of active ribonuclease and insulin?

a. For insulin to fold correctly the cysteine residues must become fully reduced to the

sulfhydryl. b. Insulin must be a homo-oligomer (at least a homo-dimer) joined by disulfide bonds. c. Insulin must contain only two cysteine residues with only one disulfide bond possible. d. The folding of insulin does not require disulfide bonds. e. The active 3 D structure of insulin must consist of more than one peptide chain

connected only by disulfide bonds.

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3. Osteogenesis imperfecta (OI) is an autosomal dominant disease of varying degrees of bone weakness. The disease is caused by a large number of different mutations in Type I collagen. Type I collagen is composed of 3 subunits that form a coiled coil. Many of the mutations that cause OI involve the substitution of another amino acid for glycine. In general glycine is a highly conserved amino acid in evolution. Why should this amino acid be so important in a protein sequence?

a. Glycine participates in disulfide bond formation. b. Glycine is a non-polar amino acid, allowing it to create an inner hydrophobic region

important in forming secondary structure. c. Glycine is the smallest amino acid, allowing it to fit into places where no other

amino acid will fit. d. The charged carboxyl group on glycine often forms an important salt bridge that

stabilizes protein structure. e. The hydrogen side-chain on glycine plays an

important role in forming the hydrogen bonds that stabilize α-helical structures.

4. Scrapie is a progressive neurodegenerative condition first described in sheep, but occurs in many other mammals. It was the first member of a class of disorders called the transmissible spongiform encephalopathies (TSE), which also includes mad-cow disease and the human conditions called kuru and Crutzfeldt-Jacob disease. The disease is transmitted by a single species of protein now called the “Prion Protein”. The protein exists in two forms, the normal PrPc (Figure, right panel) and the abnormal PrPSc (Figure, left panel). Exposure to the abnormal form leads to the disease, but only after a long latent period. Knockout mice that lack the gene for PrP do not get this disease, even if infected with the abnormal protein. Which one of the statements below is MOST correct concerning the PrPc and PrPSc structures?

a. An important stabilizing motif of PrPc is the formation of regular hydrogen bonds

between adjacent amino acid side chains over large stretches of secondary structure. b. An important stabilizing motif of PrPSc is the formation of regular hydrogen bonds

between adjacent amino acid side chains over large stretches of secondary structure. c. An important stabilizing motif of PrPc is the formation of regular hydrogen

bonds between each carbonyl carbon and the amide hydrogen 4 amino acids away over large stretches of secondary structure.

d. An important stabilizing motif of PrPSc is the formation of regular hydrogen bonds between each carbonyl carbon and the amide hydrogen 4 amino acids away over large stretches of secondary structure.

e. The lack of secondary structure in the N-terminal random coil region of PrPc suggests that this region contains a large number of bulky aromatic amino acids.

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Teams may do 5A and 5B together before showing answers. 5A. When PrPSc from a hamster is inoculated into a mouse, symptoms develop in 500 days. When PrPSc from those mice is inoculated into mice, symptoms are developed in only 140 days. Conversely, when PrPSc from mice is inoculated into hamsters, symptoms develop in 400 days. When PrPSc from those hamsters is inoculated to hamsters, symptoms develop in only 75 days. From these observations, which one of the statements below provides the best explanation of how exogenous PrPSc causes neurodegeneration?

a. Exogenous PrPSc blocks the synthesis of normal PrP leading to loss of PrP function. b. Exogenous PrPSc interacts with endogenous PrP to encourage the mis-folding of

PrP into PrPSc slowly increasing the endogenous PrPSc concentration. c. Exogenous PrPSc is a toxic molecule that directly kills certain nerve cells. d. Exogenous PrPSc must bind to endogenous PrP and the complex then acts as a toxin

to kill nerve cells. e. Exogenous PrPSc stimulates the over expression of endogenous PrP.

5B. Which of the following conclusions is BEST supported by these data?

a. Once same species PrPSc is formed, the progression to disease is similar in both hamsters and mice.

b. PrPc genes must be significantly divergent between species. c. Foreign species PrPSc is better at converting PrPc to PrPSc than same species PrPSc. d. The onset of disease symptoms correlate closely with the initial formation of same-

species PrPSc e. The slowest part of the pathological process is the initial formation of PrPSc from

endogenous PrPc.

6. A student is interested in characterizing and separating two proteins identified as A and B. He has loaded the mixture into a carboxy-methyl-cellulose column in low salt buffer at pH 8.0. Both of the proteins eluted in the void fraction of the column. From the list below, which one has the best chance of improving the separation results in the next experiment? Load the protein mixture onto the CM-cellulose column:

a. in high salt at pH 5.0. b. in high salt at pH 8.0. c. in high salt at pH 9.5. d. in low salt at pH 5.0 e. in low salt at pH 9.5.

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7. The two proteins from the question above (A and B) were retained on a CM-cellulose column in low salt at pH 5.0. The proteins were eluted from the column by increasing the salt in the elution buffer in the order A-B. However, if separated on a Sephadex-G100 column they eluted in the order B-A. These results suggest that: a. Protein A has a lower molecular weight and is more acidic than protein B. b. Protein A has a higher molecular weight and is more acidic than protein B. c. Protein A has a lower molecular weight and is less acidic than protein B. d. Protein A has a higher molecular weight and is less acidic then protein B.

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Instructors Guide – Application

1. A strong reducing agent will reduce disulphide bonds if present to SH, and urea destabilizes tertiary structure denaturing ribonuclease and resulting in its inactivation. When the urea and reducing agent are removed the protein spontaneously refolds into its tertiary structure. Since scrambled ribonuclease will refold without any help, all of the information needed must be present in the primary sequence of the scrambled molecule. Because disulfide isomerase speeds up the process, the implication is that the folding process must involve a number of disulfide bonds (not answer d) that have to be made correctly to form the lowest energy active molecule (answer c). There is no necessity that the molecule be oligomeric, although it could be (not answers b or e). The oxidation environment is important for disulfide bonds and if too reduced, (SH), they cannot take part in disulfide bond formation (not answer a). This is why there are few disulfide bonded proteins in the cytosol, which is a fairly reduced environment compared to the ER lumen and the extracellular space, which is less reducing and conducive to disulfide bond formation. 2. This is a more difficult question and teams take a little more time answering this because we have not yet covered the structure and processing of insulin. In fact, that is the point, if they know the structure then this question does not elicit as much discussion. However, because they don’t the intra-team discussions are good and they usually get the correct answer. The data indicates that 1) insulin must contain disulfide bonds that are broken and reformed by the isomerase, (not answers a or d) and 2) that mature insulin does not have the all the information needed in its remaining primary structure to reform correctly once scrambled, even in the presence or absence of isomerase. Thus, not answer (b) as this type of structure should be able to form de novo and not answer (c) as this would also seemingly reform. 2) Also implies that the structure is not a single peptide and that there are multiple ways for information to reform once scrambled, and suggests that the lowest energy form is no longer the active form. The three dimensional structure of insulin is two separate protein chains joined by S-S bonds. The protein chain is originally folded as pre-proinsulin, a single primary sequence containing the A, B, and C domains (chains) (See figure below.) The folding of the protein directs the disulfide bonds to form on the cysteine residues as shown between the A and B domain in the ER (in the presence of PDI) shortly after the protein is translated. This is presumably the most stable and lowest energy configuration of the protein at this point. The C chain is cleaved prior to secretion to yield insulin in its final form as two peptides connected only by disulfide bonds. If the disulfide bonds are broken in mature insulin, the folding information that directed their correct formation (the interaction of the A, B and C domains) is gone and the active structure is no longer the most stable form.

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3. Many different types of mutations in the COL1A1 gene cause osteogenesis imperfecta of differing degrees of severity, (types I-IV). In Type I, the mutant allele is either deleted or not expressed and normal collagen is made at a 50% level. In Types II-IV, a variety of different types of mutations lead to abnormal collagen fibers. One commonly occurring mutation is the substitution of a single glycine in a Gly-X Pro repetitive region with another amino acid. Since glycine is the smallest amino acid, it can fit into places where no other amino acid residues residue can fit (answer c), therefore substitutions with other amino acid residues can interfere with formation of the mature triple-stranded collagen molecule and prevent the production of mature type I collagen. (Byers, 1989 and Mottes 1999). Answer (a) is incorrect, glycine contains no sulfur. Answer (d) is incorrect because the carboxyl group is occupied in the peptide bond and will not form a salt bridge unless it is the carboxyl terminal amino acid. Answer (e) is incorrect as the glycine side-chain hydrogen plays no role in maintaining α-helical structures.

4. This question requires the student to recognize that PrPc is predominantly alpha helix while PrPsc has shifted to strongly beta sheet and to know the stabilizing features that define the alpha helix and beta sheet. Answers (a) and (b) are incorrect; neither structure is stabilized by H-bonds between adjacent amino acid side-chains. Answer (c) provides the correct mechanism of alpha helix stabilization for an alpha helical structure. Answer (d) is incorrect as it describes a correct mechanism of alpha helix stabilization, but for a beta sheet structure. Answer (e) is incorrect as aromatic amino acids can be observed in secondary structure and do not necessarily promote random coil. 5A and B. There have been many cross species and same species experiments with purified prions. While different forms of prions sometimes generate different results, what is illustrated here is typical of abnormal prions (PrPsc) that can “jump” to another species. That is, certain forms of one species PrPsc when inoculated into another species (mouse PrPsc into hamster, or hamster PrPsc into mouse) can cause “wasting” disease in the new species, but only after a very long latent period. The latent period, while still lengthy, is always much shorter when a same species PrPsc is used as the inoculate. The notion here is that the “foreign” PrPsc takes longer to encourage the endogenous PrPc to convert to the PrPsc form than endogenous PrPsc takes to encourage conversion of endogenous PrPc to additional PrPsc. 5A. The goal of this question is to get the teams to think about what is happening during the latent period and answer (b) is the best description of the correct answer. Answer (a) implies the disease is caused by loss of function of PrPc. This is clearly incorrect, since prion knockout mice do not have, and can not get the disease even if inoculated. Answer (c) is incorrect because toxicity would be highest immediately after inoculation, not after a lengthy (~year) latent period. Answer (d) is more tempting than answer (c), since one could argue that the binding of PrPsc to PrPc could be slow, but since there is a fixed amount of PrPsc in the inoculate, which gets diluted in the body, there would be only a small fixed amount of the PrPSc-PrPc complex in nervous tissue. This mechanism provides no way to amplify the signal, and since one would expect some PrPsc to be lost from the animal with time (entropy is hard to block completely) it is hard to imagine that a very slowly forming protein-protein complex would increase as the level of PrPsc decreases. Answer (e) is incorrect since one would expect that the maximum overexpression of PrPc should occur soon after injection of the inoculate, when PrPsc levels are highest.

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5B. The goal of this question is to get the teams to think about conclusions that can be reached from these data. Answer (a) is false as disease progression is much faster in hamsters than in mice. Answer (b) is false and requires teams to realize that cross species infectivity would not occur unless the proteins were highly similar rather than divergent. The proteins cannot be identical however, or there would be no difference between cross species and same species. Answer (c) is false because same species PrPsc is better at converting endogenous PrPc to PrPsc than the “foreign” PrPsc. Answer (d) is false since the answer implies that the latency period after inoculation with same species PrPsc should be short, since we are injecting the form of PrPsc directly that would be associated with symptoms. Answer (e) is true and reflects that the slowest part of the process is the very slow initial conversion of endogenous PrPc to PrPsc. Teams will sometimes struggle with question 5. This does not seem to be based on any particular misconception, so much as just a weak understanding of the pathophysiology of prion disease, which is greatly improved by end of team discussions. (Questions 4-5 can also be used to get the teams to discuss other important forms of cross species prion infectivity, such as sheep to cow, and cow to human as in the Mad cow disease scare in England in the 1990s. What is the likelihood of future human disease from this English cohort? Additional questions could also be added examining the relationship between the human forms of kuru and Crutchfield-Jacob syndrome.)

6. The Carboxy-methyl-cellulose column into which protein A and B were loaded is an anion exchanger with a negatively charged ligand. The affinity of each protein for the charged group on the column is affected by the pH, which determines the ionization state of the molecule and the concentration of competing free salt ions in the surrounding solution. Separation can therefore be optimized by gradually changing the pH or salt concentration. Since both proteins in this scenario eluted at a void fraction of the column it implies that at pH 8.0 the proteins are neutral or negatively charged and were not retarded by the negative charges on the column. Lowering the pH might improve the separation technique since it will result in altering the charge on the protein from neutral or negative at pH 8.0 to being more positive at pH 5.0 which will allow the proteins to interact with the charges on the column. High salt competes with protein column interactions and is used to drive proteins off of these kinds of columns.

7. The proteins were eluted from the column by increasing the salt concentration. The sodium ions in the salt will compete with the positively charged proteins for the resins. The protein that binds the strongest is the most positive and will compete better with the sodium ion than those that bind more weakly and will emerge from the column last. Since the order of elution was A-B, protein A is less positive and more acidic than protein B. In Sephadex-G100 columns proteins are separated based on size, larger proteins taking the most direct path between the beads. Since the order of elution was B-A, this indicates that A has a lower molecular weight. (The most common error that is made here is to select answer C, that A is less acidic than B because students sometimes get confused about the definition of acidic.)

References:

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1. Byers, P H, Graham, J M, Weiss, L, et al. (1989). Osteogenesis imperfecta. The position of

substitution for glycine by cysteine in the triple helical domain of the pro alpha 1(I) chains of type I collagen determines the clinical phenotype. Journal of Clinical Investigation, 84(4), 1206-14.

2. Chernoff, Y. O. Do amyloids remember their origin? Insights into the prion species barrier. (2004) Molec. Cell 14: 147-152.

3. Freedman R.B., Protein disulfide isomerase: multiple roles in the modification of nascent secretory proteins. 1989, Cell 57: 1069–1072.

4. Kocisko, D. A. et al., Species specificity in the cell-free conversion of prion protein to protease-resistant forms: A model for the scrapie species barrier. (1995) PNAS 92: 3923-3927

5. Maeda, R. (2007). Promotion of insulin aggregation by protein disulfide isomerase. Biochimica et biophysica acta. Proteins and proteomics, 1774(12), 1619-1627

6. Mottes, M. Four new cases of lethal osteogenesis imperfecta due to glycine substitutions in COL1A1 and genes. (1998). Human Mutation, 12, 71-.72

7. Vanik, D. L. Surewicz, K.A. and Surewicz, W.K. Molecular basics of barriers for interspecies transmissibility of mammalian prions. (2004) Molec. Cell 14: 139-145

8. Wilkinson B., Gilbert H.F., Protein disulfide isomerase, 2004, Biochem. Biophys. Acta 35–44.