There are 28 questions in all. Questions 1-4 are essays worth a total of 40 points; the point value for each question is in parentheses. (5) = 5 points. Questions 5-28 are multiple choice; 5-16 are each worth 3 points, and 17-28 are each worth 2 points.

1. Hexokinase is an example of an important type of transferase. (8)
a. Write (do not draw) the reaction catalyzed by hexokinase.
b. Hexokinase uses three methods of catalysis, one of which is acid-base. Name and
     describe one of the other methods used by hexokinase.
c. What distinguishes acid-base catalysis from other methods of catalysis?

2. Protein kinase A (PKA) covalently modifies either S or T (S/T) in this sequence of glycogen synthase:   – R – R/K – x – S/T –. Modified glycogen synthase is less active. (10)
a. Write the reaction catalyzed by PKA when glycogen synthase is a substrate.
      Use GS-OH for the substrate.
b. Name one way this reaction resembles the reaction catalyzed by hexokinase.
c. Name one way the active site of PKA must be different from the active site of
      hexokinase.
d. Name the class of enzyme required to restore glycogen synthase activity and the two
     products of the reaction it catalyzes.

3. An imaginary enzyme Examase catalyzes the reaction below with Michaelis-Menton kinetics. I is an inhibitor (either competitive or mixed) of Examase. (12)

The V_maxfor the enzyme is 1,000 mM s^-1 and the K_Mfor O is 200 mM.
Answer either a or b but not both. Answer both c and d.

      a. Use the chart below to draw a Michaelis-Menton graph for Examase. Label the axes,
     the V_max, and the K_M. (5)

      b. Explain the difference between the two types of inhibition (competitive and mixed)
      in terms of interactions of I with O, S, P, H₂O, and/or Examase. (5)

      c. Fill in the table for the two types of inhibition. You may write I (increase),
     D (decrease), or N (no change) to show the effect.

      d. List two ways a Lineweaver-Burk graph would be different from the graph in (a).
      What does a Lineweaver-Burk graph show for competitive inhibition?

4. Protein kinase A (PKA) is allosterically activated by cyclic AMP (cAMP). Both the active and allosteric sites of PKA bind nucleotides. (10)
a. How are the nucleotides different, and what change, if any, occurs to each as a
      result of binding to PKA?
b. How is the activation of PKA reversed?

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

           5. The most favorable reaction is the one with
(A) K_eq = 9.4 x 10²; (B) K_eq = 0.192; (C) ΔG°′ = 5.1 kJ/mol; (D) ΔG°′ = − 12.3 kJ/mol.

6-8 Use this reaction, in which succinate is converted to succinyl-CoA. An intermediate is shown in brackets.

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

           7. When succinyl-CoA is a product of the reaction, GTP is a              , and the first product of the reaction is                                                      .
(A) prosthetic coenzyme / P_i ; (B) co-substrate coenzyme / GDP;
(C) prosthetic co-factor / succinyl-CoA; (D) co-substrate coenzyme / P_i .

           8. Derivation of the Michaelis-Menton equation wouldn't have worked with this reaction because (A) there's an intermediate; (B) there's more than one substrate;
(C) the reaction is reversible; (D) product formation is favored.

          9. The enzyme that is most efficient under cellular conditions is the one with the
(A) highest V_max; (B) lowest K_M; (C) lowest k_cat/K_M; (D) highest k_cat/K_M.

          10. If inhibitor and enzyme concentrations are constant, the inhibitor that has the greatest effect is the one with (A) α = 2.5; (B) α = 0.025; (C) α′ = 1; (D) α′ = 0.001.

          11. The                        is calculated by dividing v₀ at high substrate concentration by the concentration of the enzyme. (A) k_cat; (B) k_cat/K_M; (C) V_max; (D) K_M.

          12. The active site of enolase is unusual because it includes
(A) His⁰ (His with no charge); (B) His⁺; (C) Lys⁰; (D) Ser⁻.

     ß      14. What occurs first after substrate 1 (S₁) binds in the active site of chymotrypsin?
(A) the peptide bond –N– accepts H⁺;
(B) a histidine side chain becomes protonated;
(C) a serine oxygen attacks the peptide bond C=O;
(D) the oxyanion is stabilized.

          15. The diagram shows                  , and the next thing that happens is                      .
(A) S₁/ formation of an oxyanion; (B) substrate 2 (S₂) / H⁺ donation by S₂;
(C) a stable enzyme intermediate / S₂ binds; (D) product 1 (P₁) / P₁ leaves.

          16. Enolase, but not chymotrypsin, uses (A) acid-base catalysis; (B) covalent catalysis; (C) metal ion catalysis; (D) induced fit.

17-20. Use molecules A-F.

          17. Which molecule is fructose-2-phosphate?

          18. Which fructose-phosphate has the α configuration? If more than one does, answer with both letters.

          19. Which disaccharide contains two glucose residues? If more than one does, answer with both letters.

          20. Disaccharide E has a/n                                       glycoside bond and its reducing end has the                                         configuration.
(A) α–(1→4) / α ; (B) α–(1→4) / β ; (C) α–(1→6) / α ; (D) α–(1→6) / β .

          21. Molecule 21-22 is a          with the N that forms a glycoside bond to ribose inside a
          . (A) purine / circle; (B) pyrimidine / square; (C) purine / square; (D) pyrimidine / triangle.

          22. Molecule 21-22 is (A) adenine; (C) cytosine; (G) guanine; (T) thymine; (U) uracil.

          23. Molecule 23 is (A) ATP; (B) deoxyAMP; (C) deoxyCMP; (D) TMP.

          24. Molecule 24-26 is a/n
(A) triacylglycerol; (B) glycerophospholipid; (C) sphingolipid; (D) eicosanoid.

          25. How many carbons are in the mono-unsaturated fatty acid residue in Molecule 24-26? Write the number in the blank.

          26. The circled atom in Molecule 24-26 is part of a                      residue.
(A) fatty acid; (B) glucose; (C) sugar alcohol that is not glycerol; (D) glycerol.

          27. The circled atom in Molecule 27-28 is part of a/n                  residue.
(A) fatty acid; (B) glycerol; (C) sphingosine; (D) alcohol that is not glycerol.

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