Final exam 2022 answer key

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Name N number 1 Protein Biochemistry (BIOL-GA.1045) Final exam December 14, 2022 Be brief. All questions can be answered in the space provided. Please write your name on all pages. Name:_________________________________________________________ Signature_______________________________________________ ID# (N number)________________________

Name N number 2 1. (10 points) Consider the folding of proteins A and B . A can fold by itself, whereas B requires a chaperone. Data from protein folding studies below show how much unfolded protein ( A un and B un ) remains over time starting with 100% unfolded protein (assuming no back reaction). The reaction with B was performed in the presence of a small amount of the chaperone enzyme and in the excess of B . a) (5 points) What is the rate law of the folding reaction of A un into A fold ? b) (5 points) Why is the reaction for B a straight line? Explain briefly. Continued… d[A un ]/dt= -k fold [A un ] or -d[A un ]/dt = k fold [A un ] or d[A fold ]/dt = k fold [A un ] (i.e this is a first order reaction that only depends on the concentration of A un , leading to an exponential decrease in the concentration of A un ) Because it is an enzymatic reaction. It is showing constant B un turnover by the enzyme under saturating substrate conditions.

Name N number 3 2. (10 points) The chelate effect allows two tethered interaction domains with weak affinity to bind each other with high affinity and specificity. a) (5 points) Does the chelate effect primarily improve k on or k off ? Explain. b) (5 points) Does the length of the linker between the two domains impact the strength of the chelate effect? Explain briefly. Continued… k on because the tethering prevents the interaction domain from diffusing too far away after dissociation, allowing for easy recapture (i.e. it increases the effective concentration of the domain in the vicinity of the binding partner) Yes, longer linkers will reduce the chelate effect because the domains will be able to explore a larger volume away from the binding partner (this reduces the effective concentration leads to a lower k on )

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Name N number 4 3. (10 points) Your friend studies a dimeric protease that consists of an enzymatic subunit and an essential activating subunit. This protease is highly active. To slow the reaction, she dilutes the protease concentration in her samples. She observes the following results: a) (5 points) Your friend speculates that the loss activity is due to the dissociation of the activating subunit from the protease. Briefly explain why this is a plausible model and why the loss of activity occurs at a protease concentration of ~10 -8 M. b) (5 points) When she incubates the protease with a crosslinker before diluting the protease, the loss of activity is not observed. Explain how the crosslinker prevents a loss in activity at low protease concentrations. Continued… The crosslinker connects the two subunits, thereby preventing their full dissociation and keeping the protease active. The data suggest that the activating subunit binds the protease with a K D ≈ 10 -8 M. Below the K D , protease and activating subunit are largely dissociated, leading to a loss of protease activity

Name N number 5 4. (10 points) Consider the following reaction: A + B ⇆ C. At equilibrium, only 3% of the substrates (A, B) are converted to product (C). To improve product yield to >95%, your friend intends to add an enzyme that catalyzes this reaction. a) (5 points) Briefly explain why adding enzyme would not improve product yield. b) (5 points) In your research you identified a compound that allows you to remove C from solution. Would this compound be useful in improving product yield? Briefly explain why or why not. Continued… Yes, that compound would be helpful. By removing C from solution, you can remove it from the equilibrium. This prevents the back reaction and thus allows the reaction to be driven forward (in other words your compound is a product sink). Enzymes do not shift the equilibrium; they can only speed up the reaction. Adding enzyme will not increase the yield.

Name N number 6 5. (10 points) When studying the mechanism of action of chymotrypsin, a partitioning experiment with hydroxylamine and water was performed using substrates with different leaving groups. a) (5 points) There are two potential outcomes of this experiment: i. the hydroxylamine:water product ratio changes as function of the leaving group; ii. the ratio is unaffected by the nature of the leaving group. What would each of these outcomes tell you about the mechanism of the enzymatic reaction? b) (5 points) Why does a group with a higher pK a usually constitute a worse leaving group than one with a lower pK a ? Continued… If the hydroxylamine:H 2 O product ratio changes as the leaving group becomes worse, it means that this nucleophilic attack is happening directly to the substrate - this could either be a concerted reaction or a direct nucleophilic attack of the water/hydroxylamin on the substrate. If the hydroxylamine:H 2 O ratio does not change, it means that it is not acting on the substrate, pointing to a reaction with an acyl-enzyme intermediate A lower pKa means the group will be better at being an acid. An acid releases its proton which usually leaves a negatively charged group, so being a good acid depends on the stability of holding a negative charge, which is what happens when a leaving group goes through a reaction involving a nucleophilic attack.

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Name N number 7 6. (10 points) The enzyme, OMPD, catalyzes a decarboxylation reaction during pyrimidine biosynthesis. The proposed reaction mechanism is: (the curved dashed lines on the carboxylic groups indicate electron orbitals). a) (5 points) Propose a plausible role for Asp 91 in promoting the decarboxylation reaction. b) (5 points) OMPD is inhibited by 6-azauridine (a nucleoside analog). Addition of extra substrate does not improve the rate of product formation in the presence of 6- azauridine. However, product formation is restored to normal rates after dialysis is performed with the enzyme against an excess of reaction buffer without 6- azauridine. What type of competitor is 6-azauridine? Briefly explain your reasoning. Continued… Loss of the carboxyl group of the substrate is favored within in the enzyme because it avoids the charge-charge repulsion between the negative charge of Asp 91 and the carboxyl group of the substrate (this is an example of the enzyme stressing the substrate). A non-competitive inhibitor. It is not a competitive inhibitor because it cannot be competed out with excess substrate, and it is not a covalent inhibitor because it dissociates during dialysis.

Name N number 8 7. (10 points) The histone H3/H4 tetramer moonlights as a copper reductase, converting Cu 2+ to Cu + , using NADH as a cofactor. a) (5 points) Based on your knowledge of NADH, what is the likely role of NADH in this reaction. b) (5 points) Why do enzymes usually need cofactors, such as metals or NADH, to catalyze redox reactions? Continued… Because most amino acids are not redox active and thus cannot catalyze this type of reaction (cysteine is the only amino acid that is redox active). NADH is an electron donor, which is necessary to reduce Cu 2+ to Cu +

Name N number 9 8. (10 points) You prepare inside-out vesicles (IOVs) of the inner mitochondrial membrane in the presence of a pH sensitive dye. You add a large amount of ATP to these IOVs and observe that the pH drops inside the IOVs. a) (5 points) Why is there a change in pH inside the IOVs upon ATP addition? b) (5 points) Would you expect all ATP to be used up in this reaction? Briefly explain your reasoning. Continued… Because the F1-ATPase hydrolyzes ATP and pumps protons into the vesicle. (The enzyme performs the reverse reaction it would normally perform) No, because the protons would build up until the gradient is so steep that it is no longer energetically favorable to accumulate more protons.

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Name N number 10 9. (10 points) Actin filaments are self-assembling polymers with a (+) end and a (-) end. These ends differ in their K D for binding G-actin. a) (5 points) Briefly explain why the K D is different at the two ends. Mention the relevant co-factors. b) (5 points) You mix actin filaments with a large amount of G-actin. At the beginning of the reaction, the concentration of G-actin lies above the K D at both ends and you observe elongation of the filaments at both ends. What do you expect to happen to filament growth as this reaction proceeds? Explain briefly. Continued… The two ends differ in their nucleotide binding. The (+) end is made up of ATP actin, the (-) end of ADP actin. The nucleotides affect the structure of the binding interface, thereby affecting the K D . Growth of the filaments will deplete G-actin from solution. When the concentration of free G actin drops below the K D of the (-) end the filaments will start to treadmill.

Name N number 11 10. (10 points) After isolating and sequencing a fungus from a woodland population, you identify two mutations in the gene GAP1 when compared to the same fungal species from a desert population. The mutations are A42F and E160D. These mutations increase fitness by increasing growth. You generate these mutations in your lab fungal strain, individually and in combination, then measure the growth rate. Your results are as follows: A42F à slow growth E160D à normal growth A42F + E160D à fast growth a) (5 points) Which of the two mutations likely happened first? Explain your reasoning. b) (5 points) After sequencing other related fungal species, you find numerous amino acid changes in GAP1 across the different species. Do you expect these amino acid changes to result in large structural changes (e.g., a change from an alpha-helix to a beta-sheet)? Briefly explain your answer. Continued… (the following two questions are optional) No. Protein structure is much more strongly conserved than amino acid identity. Many amino acids can be exchanged without using secondary structure element, whereas large scale structural changes will in most cases destroy protein function and lead to a strong fitness defect. E160D likely happened first followed by A42F. In this order, there was no fitness cost because the E160D mutation allowed the A42F mutation to have a positive effect. If A42F had happened first, the cells would have been at a fitness disadvantage.

Name N number 12 RESURRECTION QUESTIONS (OPTIONAL) R1. (5 points) Affinity purification of a protein of interest from a cell extract also purifies another protein. To test whether the two proteins form a complex, you repeat the purification after first treating the cell extract with a crosslinker. SDS-PAGE analysis of both samples yields the following result: What is the nature of the new band that appeared in the crosslinked sample? R2. (5 points) The graph below indicates that the second peak (fractions 4-6) contains the pure enzyme we were assaying for. Which fraction(s) would you keep to maximize the yield from this purification step? Briefly explain. THE END. The result suggests that crosslinking stabilized the two proteins in a heterodimeric complex that contains one copy of each protein Fractions 4, 5 and 6. To recover the maximal amount of enzyme (yield) we need to collect all the fraction that contain enzymatic activity (note that combining those fractions would obviously dilute the enzyme compared to fraction 5 because it is present at much lower concentrations in fractions 4 and 6 (as indicated by A280)

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Final exam 2022 answer key

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