1. Rate enhancement and substrate specificity.

2. A cofactor.

3. Coenzymes and metals.

A9

4. Vitamins are converted into coenzymes.

5. Enzymes facilitate the formation of the transition state.

6. The intricate three-dimensional structure of proteins allows the construction of active sites that will recognize only specific substrates.

7. The energy required to reach the transition state (the activation energy) is returned when the transition state proceeds to product.

8. (a) . Using equation 5 in the text, ΔG°′ = − 11.42 kJ mol⁻¹ (−2.73 kcal mol⁻¹)

(b) k_F = 10⁻² s⁻¹ and k_R = 10⁻⁴ s⁻¹. The equilibrium constant and ΔG°′ values are the same for both the uncatalyzed and catalyzed reactions.

9. Protein hydrolysis has a large activation energy. Protein synthesis must require energy to proceed.

10. The enzymes help protect the fluid that surrounds eyes from bacterial infection.

11. Binding energy is the free energy released when two molecules bind together, such as when an enzyme and a substrate interact.

12. Binding energy is maximized when an enzyme interacts with the transition state, thereby facilitating the formation of the transition state and enhancing the rate of the reaction.

13. There would be no catalytic activity. If the enzyme–substrate complex is more stable than the enzyme–transition-state complex, the transition state would not form and catalysis would not take place.

14. Transition states are very unstable. Consequently, molecules that resemble transition states are themselves likely to be unstable and, hence, difficult to synthesize.

15. (a) 0; (b) +28.53; (c) −22.84; (d) −11.42; (e) +5.69.

16. (a)

Because [G6P]/[G1P] = 21, there is 1 molecule of G1P for every 21 molecules of G6P. Because we started with 0.1 M, the [G1P] is 1/22(0.1 M) = 0.0045 M and [G6P] must be 21/22(0.1 M) or 0.096 M. Consequently, the reaction does not proceed as written to a significant extent.

(b) Supply G6P at a high rate and remove G1P at a high rate by other reactions. In other words, make sure that the [G6P]/[G1P] is kept large.

17. K_eq = 19, ΔG°′ = −7.3 kJ mol⁻¹ (−1.77 kcal mol⁻¹)

18. The three-dimensional structure of an enzyme is stabilized by interactions with the substrate, reaction intermediates, and products. This stabilization minimizes thermal denaturation.

19. At substrate concentrations near the K_M, the enzyme displays significant catalysis yet is sensitive to changes in substrate concentration.

20. No, K_M is not equal to the dissociation constant, because the numerator also contains k₂, the rate constant for the conversion of the enzyme–substrate complex into enzyme and product. If, however, k₂ is much smaller than k₋₁, K_M ≈ K_d.

21. When [S] = 10 K_M, V₀ = 0.91 V_max. When [S] = 20 K_M, V₀ = 95 V_max.

So any Michaelis–Menten curves showing that the enzyme actually attains V_max are pernicious lies.

22. (a) 31.1 μmol; (b) 0.05 μmol; (c) 622 s⁻¹, a midrange value for enzymes (Table 8.5).

23. (a) Yes, K_M = 5.2 × 10⁻⁶ M; (b) V_max = 6.8 × 10⁻¹⁰ mol minute⁻¹; (c) 337 s⁻¹.

24. Penicillinase, like glycopeptide transpeptidase, forms an acyl-enzyme intermediate with its substrate but transfers the intermediate to water rather than to the terminal glycine residue of the pentaglycine bridge.

25. (a) V_max is 9.5 μmol minute⁻¹. K_M is 1.1 × 10⁻⁵ M, the same as without inhibitor.

(b) Noncompetitive.

(c) 2.5 × 10⁻⁵ M

(d) f_ES = 0.73, in the presence or absence of this noncompetitive inhibitor.

26. (a) V₀ = V_max − (V₀/[S]) K_M

(b) Slope = −K_M, y-intercept = V_max, x-intercept = V_max/K_M

(c) An Eadie–Hofstee plot.

27. Sequential reactions are characterized by the formation of a ternary complex consisting of the enzyme and both substrates. Double-displacement reactions always require the formation of a temporarily substituted enzyme intermediate.

28. The rates of utilization of substrates A and B are given by

and

Hence, the ratio of these rates is

Thus, an enzyme discriminates between competing substrates on the basis of their values of k_cat/K_M rather than of K_M alone.

29. The mutation slows the reaction by a factor of 100 because the activation free energy is increased by +11.42 kJ mol⁻¹ (⁺2.73 kcal mol⁻¹). Strong binding of the substrate relative to the transition state slows catalysis.

30. 11 μmol minute⁻¹

31. (a) This piece of information is necessary for determining the correct dosage of succinylcholine to administer.

(b) The duration of the paralysis depends on the ability of the serum cholinesterase to clear the drug. If there were one-eighth the amount of enzyme activity, paralysis could last eight times as long.

(c) K_M is the concentration needed by the enzyme to reach ½ V_max. Consequently, for a given concentration of substrate, the reaction catalyzed by the enzyme with the lower K_M will have the higher rate. The mutant patient with the higher K_M will clear the drug at a much lower rate.

32. (a) K_M is a measure of affinity only if k₂ is rate limiting, which is the case here. Therefore, the lower K_M means higher affinity. The mutant enzyme has a higher affinity.

A10

(b) 50 μmol minute⁻¹. 10 mM is K_M, and K_M yields ½ V_max. V_max is 100 μmol minute⁻¹, and so …

(c) Enzymes do not alter the equilibrium of the reaction.

33. Enzyme 2. Despite the fact that enzyme 1 has a higher V_max than enzyme 2, enzyme 2 shows greater activity at the concentration of the substrate in the environment because enzyme 2 has a lower K_M for the substrate.

34. (a) The most effective means of measuring the efficiency of any enzyme–substrate complex is to determine the k_cat/K_M values. For the three substrates in question, the respective values of k_cat/K_M are: 6, 15, and 36. Thus, the enzyme exhibits a strong preference for cleaving peptide bonds in which the second amino acid is a large hydrophobic amino acid.

(b) The k_cat/K_M for this substrate is 2. Not very effective. This value suggests that the enzyme prefers to cleave peptide bonds with the following specificity: small R group—large hydrophobic R group.

35. If the total amount of enzyme (E_T) is increased, V_max will increase, because V_max = k₂[E]_T. But K_M = (k₋₁ + k₂)/k₁; that is, it is independent of substrate concentration. The middle graph describes this situation.

36.

37. (a)

(b) This behavior is substrate inhibition: at high concentrations, the substrate forms unproductive complexes at the active site. The adjoining drawing shows what might happen. Substrate normally binds in a defined orientation, shown in the drawing as red to red and blue to blue. At high concentrations, the substrate may bind at the active site such that the proper orientation is met for each end of the molecule, but two different substrate molecules are binding.

38. The first step will be the rate-limiting step. Enzymes E_B and E_C are operating at ½ V_max, whereas the K_M for enzyme E_A is greater than the substrate concentration. E_A would be operating at approximately 10⁻² V_max.

39. The fluorescence spectroscopy reveals the existence of an enzyme–serine complex and of an enzyme–serine–indole complex.

40. (a) When [S⁺] is much greater than the value of K_M, pH will have a negligible effect on the enzyme because S⁺ will interact with E⁻ as soon as the enzyme becomes available.

(b) When [S⁺] is much less than the value of K_M, the plot of V₀ versus pH becomes essentially a titration curve for the ionizable groups, with enzyme activity being the titration marker. At low pH, the high concentration of H⁺ will keep the enzyme in the EH form and inactive. As the pH rises, more and more of the enzyme will be in the E⁻ form and active. At high pH (low H⁺), all of the enzyme is E⁻.

(c) The midpoint on this curve will be the pK_a of the ionizable group, which is stated to be pH 6.

41. (a) Incubating the enzyme at 37°C leads to a denaturation of enzyme structure and a loss of activity. For this reason, most enzymes must be kept cool if they are not actively catalyzing their reactions.

(b) The coenzyme apparently helps to stabilize enzyme structure, because enzyme from PLP-deficient cells denatures faster. Cofactors often help stabilize enzyme structure.

42. The slow second order step occurs when complementary sequences on two different molecules bind with one another. Once the initial sequence alignment occurs, the remainder of the molecule can quickly locate complementary sequences from this nucleation site and reanneal, a first order reaction.