Chemistry 340 Exam 2 Fall 2006
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There are 24 questions in all. Questions 1–20 are multiple choice and are each worth 3 points. Questions 21–24 are essays and problems worth a total of 40 points.

These are constants you may find helpful: R = 8.314 J/mole–K; T = 298 K

1-5. Use this reaction:                     
The enzyme for this reaction is allosterically inhibited by ATP and acetyl-CoA.
ΔG°′ = – 31.0 kJ/mol

              1. The enzyme that catalyzes the reaction is a/n
(A) isomerase; (B) transferase; (C) oxidoreductase; (D) ligase.

              2. The enzyme requires the co-factor ADP, which is a/n (A) a prosthetic group;
(B) co-substrate; (C) allosteric activator; (D) uncompetitive inhibitor.

              3. The enzyme catalyzes an ordered sequential reaction, which means that
(A) both phosphoenolpyruvate and ADP are in the active site at the same time;
(B) both ADP and pyruvate are in the active site at the same time;
(C) both phosphoenolpyruvate and ATP are in the active site at the same time;
(D) ATP is released before phosphoenolpyruvate binds.

              4. Keq for the reaction is (A) a large positive number (> 1,000); (B) a small positive number (< 0.001); (C) approximately = 1; (D) a large negative number.

              5. When a molecule of ATP binds the enzyme's                    site, it acts as a               .
(A) active, negative modulator; (B) active, positive modulator;
(C) allosteric, competitive inhibitor; (D) allosteric, negative modulator.

              6. This is a reaction coordinate diagram for an imaginary reaction in which
A ↔ X, via a transition state M.
ΔG for the noncatalyzed reaction in which
A → X is about
(A) +100; (B) +150 ; (C) + 50 ; (D) – 50 .


              7. What type of enzyme is a dehydrogenase? (A) oxidoreductase; (B) hydrolase;
(C) isomerase; (D) ligase; (E) lyase; (F) transferase.

8-13. Use these definitions: [E] = enzyme concentration; [S] = substrate concentration;
[P] = product concentration; ES = enzyme-substrate complex

              8. KMfor an enzyme is (A) = Vmax/[S]; (B) the slope of a Michaelis-Menton curve; (C) an indication of its affinity for the substrate when the rate-limiting step is ES → P; (D) = [E]/Vmax.

              9. kcat /KMfor an enzyme is (A) its specificity constant; (B) a measure of the enzyme's efficiency at high [P]; (C) the y-axis for a Lineweaver-Burk chart;
(D) all of the above.

10-11. Use this chart for enzymes F, G, and H. The Vmaxfor all enzymes is 200, but they have different KM's.

              10. Which enzyme has a KM= 20?

              11. When a Lineweaver-Burk chart is prepared for these enzymes, the one with the greatest slope is (F), (G), (H), or (D) all of them have the same slope.


              12. Derivation of the Michaelis-Menton equation was simplified by considering only (A) competitive inhibitors; (B) enzymes with quaternary structure;
(C) reactions with one substrate; (D) reactions in which [P] is large.

              13. An important assumption that was made in deriving the Michaelis-Menton equation was that (A) the reaction is reversible; (B) [ES] is constant;
(C) v0 = Vmax; (D) E + S → ES irreversibly.

              14. For a competitive inhibitor, α = 1 + ([I]/KI), and α modifies KMin the Michaelis-Menton equation. As a result, inhibition is greater when [I] is                     and KI is                 .
(A) large, large; (B) large, small; (C) small, large; (D), small, small.

              15. Hexokinase catalyzes a reaction that involves                       after the substrate binds, and that requires the co-factors               and              .
(A) proteolysis, NAD+, Mg2+; (B) induced fit, NAD+, Zn2+;
(C) proteolysis, ATP, Mg2+; (D) induced fit, ATP, Mg2++ .

              16. The reaction catalyzed by enolase requires                      in the active site to stabilize negative charges and a/an           to function as H+ acceptor.
(A) a lysine side chain, aspartate side chain; (B) Mg2+, lysine side chain;
(C) Zn2+, histidine side chain; (D) an oxyanion hole, glutamate side chain.

              17. The coagulation cascade involves the conversion of prothrombin to thrombin by Factor X, which is a serine protease. Which choice is not correct?
(A) Prothrombin is larger than thrombin; (B) Factor X catalyzes a ping pong reaction; (C) Factor X is a hydrolase; (D) Factor X requires Zn2+ in its active site.

18-20. Identify the molecules below using one of these terms:
(A) adenine; (B) β-fructose; (C) cytosine; (D) dihydroxyacetone; (E) eicosanoid;
(F) fatty acid; (G) guanine; (H) glyceraldehyde; (I) α-fructose; (K) α-glucose;
(L) cholesterol

           18.                       19.                 20.

21. Analysis of an enzyme and its reversible inhibitor resulted in the Michaelis-Menton and Lineweaver-Burk charts below. Based upon the charts,
a. What type of inhibitor is this (competitive, uncompetitive, or mixed)? (2)
b. Explain how you know. (2)
c. Describe the interaction of the inhibitor (I) with the enzyme (E) and define the
     term(s) KI and/or KI′ if they are relevant. (6)




22. The diagrams below represent two stages in the reaction catalyzed by chymotrypsin.
a. For one diagram, label the parts of the catalytic triad. (3)
b. Label the diagrams 1 and 2 to indicate which occurs first during the reaction. (1)
c. Write the reaction catalyzed by chymotrypsin as a two-step reaction. (4)
d. Choose one diagram and describe what will happen next during the reaction. (2)

23. The molecule below is a membrane component that can form weak interactions with membrane proteins. (10)
a. Label each circled part of the molecule with the type of residue indicated.
     A residue may not be completely circled, but you should imagine that it is.
     Be as specific as possible (for example, write adenine rather than purine).
b. Label one hydrophilic part of the molecule with L and one hydrophobic part with B.
c. Name an amino acid whose side chain would form a weak interaction with the
    hydrophilic part of the molecule you labeled and name the type of interaction.


d. Name an amino acid whose side chain would form a hydrophobic interaction
    with the hydrophobic part of the molecule.

24. Glycogen synthase a is converted to glycogen synthase b by protein kinase A (PKA).
The kinetics of glycogen synthase are described using the term K0.5, rather than KM.

a. PKA causes the K0.5 for glycogen synthase to                                                      

     and its activity to                                                .
     Write increase, decrease, or stay the same in each blank. (2)

b. What type of regulation–covalent or allosteric–is involved; how do you know? (4)
c. What is K0.5 and why is it more appropriate for some enzymes than KM? (4)

 

 

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