Sample/practice exam 2017, questions and answers PDF

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Assignment 6

MBB 222 Fall 2017 (Craig)

Answer the questions to the best of your ability. Type out answers only single-spaced on a minimal number of pages. Hand in to your TA at the beginning of tutorial; keep a copy for yourself to go over in tutorial. Due in tutorial the week of Oct. 23-27. 1. Using the Michaelis-Menten rate equation, explain why (a) V0 = Vmax when [S] >> KM, (b) [S] = KM when V0 = Vmax/2, and (c) the rate (V0) is directly proportional to [S] when [S] > KM, KM is negligible in the ([S] + KM) denominator so V0 = Vmax x [S]/[S] = Vmax x 1; so V0 = Vmax (b) when Vo = Vmax/2, Vmax/2 = Vmax x [S]/([S] + KM); 2[S] = [S] + KM; [S] = KM (c) when [S] carbonic anhydrase > catalase (d) Enzymes interact with their substrates by random collisions. Thus, they cannot interact with their substrates any faster than the diffusion rate of 108 - 109 sec-1 M-1. Superoxide dismutase has a kcat/KM of 7 x 109 s-1 M-1 – is at the diffusion rate limit. 5. Why is it that you can overcome enzyme inhibition of a competitive inhibitor by adding an excess of substrate, but this will have no effect on a non-competitive inhibitor? A competitive inhibitor binds in the enzyme active site, so adding more substrate will outcompete the inhibitor. A non-competitive inhibitor binds at a site other than the active site and causes an allosteric change to the enzyme such that the enzyme can bind but can no longer process/turn over its substrate (or possibly can no longer bind). This effect will not be changed by adding more substrate.

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Assignment 6

MBB 222 Fall 2017 (Craig)

6. For the reaction coordinate shown of an enzyme catalyzed conversion of S to P, label the intermediate(s) and transition state(s). Is the reaction thermodynamically favorable? The peaks are the transition states and the valleys are the intermediates. It is not thermodynamically favorable because the G is positive.

7. Predict the effect of mutating (i.e. changing via genetic engineering) Asp102 to Asn in chymotrypsin on (a) substrate binding (KM) and (b) catalysis (kcat). (c) What about a mutation that changes one of the hydrophobic residues in the chymotrypsin binding pocket to a lysine? (d) What about a mutation that changes Gly193 in the oxyanion hole to Ala? To Pro? (a) Little or no effect on KM, because substrate binding depends on the presence of a deep hydrophobic pocket to bind to aromatic and large hydrophobic side chains – this hydrophobic pocket is close to but distinct from the active site catalytic triad of Asp102-His57-Ser195. (Note, the KM value is a reflection of the formation of the enzyme:substrate complex, not of formation of the acyl enzyme intermediate. All enzymes bind to their substrates with high specificity. In general, the better they bind the more likely catalysis will proceed. The formation of the acyl enzyme intermediate in chymotrypsin proteolysis is part of catalysis, not formation of the enzyme:substrate complex.) (b) Catalysis would be significantly reduced because the catalytic triad is no longer functional – Asp102 is needed to stabilize a positively-charged His57 after it has abstracted a proton away from Ser195; Asn cannot stabilize the positiviely charged His57 because it has a neutral rather than a negatively-charged side chain. Thus, for the Asp102Asn mutant, Ser195 will likely remain protonated and cannot perform nucleophilic attack of the carbonyl carbon of the peptide. (c) A charged residue in the specificity pocket would disrupt the binding pocket and prevent bulky hydrophobic side chains from binding – this would decrease the affinity of chymotrypsin for such side chains, thereby increasing the KM - if such a side chain did bind, the orientation of the peptide bond might not be optimal because of the poor fit, which would decrease the rate of catalysis (kcat). (d) Changing Gly193 to Ala would have no effect on the oxyanion hole, which stabilizes the negatively charged oxyanion of the tetrahedral intermediates via interactions between the electron-deficient backbone amide hydrogens of Gly193 and Ser195. Ala would still have an amide hydrogen. Changing this residue to proline, however, would remove one of two amide hydrogens in the oxyanion hole, thereby decreasing the stability of both tetrahedral intermediates. This would likely adversely affect kcat but would probably not affect KM. (Pro193 might also deform the backbone around this residue and shift the position of Ser195, part of the catalytic triad.) (Note, you do not need to know the residue names/numbers for the oxyanion hole.)

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Assignment 6

MBB 222 Fall 2017 (Craig)

8. For the pentapeptide Ala-Tyr-Gly-Ser-Val, draw the products of its complete chymotrypsin proteolysis. Use R for each side chain (eg. R1, R2 etc, beginning at the N-terminal residue of the amino acid, Ala). Label as P1 and P2. (The pH of the reaction is 7.0.)

The charges are shown for peptide products at pH 7. (FYI, chymotrypsin is produced in the pancreas and released into the duodenum between the stomach and the small intestine (pH 76.4).) 9. Which chymotrypsin amino acid is the base catalyst? Which is the covalent catalyst? base catalyst, His57; covalent catalyst, Ser195 10. Draw out the carbonyl carbon of peptide bond that will be cleaved by chymotrypsin in its 1st tetrahedral intermediate and the trigonal planar geometry of the acyl enzyme intermediate. Which intermediate is at a higher energy? Why?

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Assignment 6

MBB 222 Fall 2017 (Craig)

The tetrahedral intermediate is higher energy – it is a transition state – it is unstable because of the negative charge on the oxygen. The acyl enzyme intermediate is a more stable intermediate. 11. Provide an example of weak chemical interactions provided by chymotrypsin that stabilize (i) the substrate and (ii) its transition state. (i) The substrate binding pocket provides hydrophobic and van der Waals interactions for the R1 aromatic/bulky hydrophobic side chain. (ii) The amide nitrogens of Gly193 and Ser195 form the oxyanion hole that stabilizes the negatively-charged O in the tetrahedral transition states (one that forms after nucleophilic attack of the carbonyl carbon by the Ser195 alkoxide ion and one that forms after nucleophilic attack of the same carbon by water). 12. Briefly compare trypsin and subtilisin? (i.e. how are they similar, how are they different?) They are both serine proteases, using a catalytic triad (Ser-His-Asp) to hydrolyze peptide bonds, but they are not related in amino acid sequence or structure. They are distinct proteins that have converged to the same mechanism and function. 13. Compare and contrast myoglobin and hemoglobin. Both small proteins that are mostly -helical, 8 helices per subunit, have a hydrophobic cavity that binds heme and oxygen. Myoglobin: monomeric, 1 heme, in muscle cells, binds 1 O2 molecule tightly, binding not influenced by pH, CO2 concentration or BPG. Hemoglobin: tetrameric, 1 heme/monomer, in RBCs, binds 4 O2 molecules weakly, cooperatively, O2 binding is influenced by pH, CO2 concentration, BPG. 14. There are 2 histidines involved in coordinating the heme iron for both myoglobin and hemoglobin. How do their roles differ? His F8 directly coordinates with the heme Fe2+, whereas His E7 stabilizes O2 when it is bound.

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