The MOLECULES of LIFE

The MOLECULES of LIFE Physical and Chemical Principles Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 16 Principles of Enzyme Catalysis Problems True/False and Multiple Choice 1. The initial reaction velocity for an enzyme reaction reaches a maximum at high substrate concentration because the free enzyme can no longer regenerate at the end of each reaction cycle. True/False 2. The turnover number for an enzyme obeying Michaelis Menten kinetics is: a. k 2. b. k cat /K M. c. k 1 /k 1. d. (k 1 + k 2 ). e. ΔG. 3. Catalytic antibodies are generally less efficient than natural enzymes that catalyze the same reactions. True/False 4. A metabolic enzyme generates the amino acid methionine. For a given substrate concentration, an experiment conducted in the presence of high initial concentrations of methionine generates less new methionine than an experiment conducted with no initial methionine present. This is likely an example of: a. A ping-pong mechanism of substrate binding. b. A proximity effect. c. Substrate strain. d. Product inhibition. e. A reaction intermediate. 5. Which of the following is not a commonly observed feature of proteases? a. The catalytic triad in the active site. b. Exclusively hydrophobic residues in the active site. c. A cysteine residue in the active site. d. Metal ions coordinated in the active site. e. A pair of acidic residues in the active site. 6. An enzyme inhibitor is observed to alter the K M but not the V max of a reaction. This inhibitor is most likely: a. A noncompetitive inhibitor. b. A competitive inhibitor. c. An allosteric inhibitor. d. A substrate-dependent noncompetitive inhibitor. e. A covalent inhibitor. 7. Due to its extremely slow dissociation kinetics, the protein bovine pancreatic trypsin inhibitor (BPTI) has broad specificity and inhibits more proteases than protease inhibitors that are small molecules. True/False Fill in the Blank 8. In the schemes for the catalyzed reactions considered in this chapter, S, E, and P refer to,, and, respectively. substrate, enyzme, product 9. The specificity constant or catalytic efficiency is the ratio between and. k cat, K M 10. In a plot of initial velocity versus substrate concentration, an allosteric enzyme displays a curve, whereas a non-allosteric enzyme that obeys Michaelis Menten kinetics displays a curve. sigmoidal, hyperbolic 11. The geometry of competitive inhibitors commonly mimics the of the reaction that the enzyme normally catalyzes. transition state

2 Chapter 16: Principles of Enzyme Catalysis 12. Proteins are not the only polymers that act as catalysts. Catalytic molecules are also essential for cells, including playing an essential role in protein synthesis. RNA/ribozyme a. 2.5 2.0 Quantitative/Essay 13. At 25 C, an enzyme accelerates a reaction by a factor of 10 5 over the uncatalyzed reaction in water. If the effect of the enzyme is solely to reduce the energy of the transition state, by what amount does it reduce the energy of the transition state (E A )? k cat /k uncat = e ( (ΔGo cat ΔG o uncat )/RT) 105 = e ( (ΔGo cat ΔG o uncat )/8.314 298) ln(10 5 ) = ( (ΔG o cat ΔG o uncat )/2.5) ΔG o cat ΔG o uncat = 28.5 kj mol 1 14. Show that the equations plotted in Lineweaver Burk and Eadie Hofstee plots are equivalent. Start with the Lineweaver Burk equation: 1/v 0 = 1/V max + K M /V max [S] Multiply all by V max, V max /v 0 = 1 + K M /[S] Multiply all by v 0, V max = v 0 + v 0 K M /[S] Rearrange for v 0, v 0 = v 0 K M /[S] + V max (which is the Eadie Hofstee equation). k1 15. In a reaction, + k2 E S E S E + P, calculate k 1 the value of K M if the forward rate constant (k 1 ) for E S formation is 4.3 10 6 sec 1 M 1, the reverse rate constant (k 1 ) for E S dissociation is 2.4 10 2 sec 1, and the turnover number (k 2 ) is 10 3 sec 1. K M = (k 1 + k 2 )/k 1 = (2.4 10 2 sec 1 + 10 3 sec 1 )/4.3 10 6 sec 1 M 1 = 3.3 10 4 M 16. Presented below are Lineweaver Burk plots for enzymatic reactions with (red) and without (blue) inhibitor. What type of inhibition is occurring in each case? 1/v (M 1 sec) 1.5 0.5 0 0 0.2 0.4 0.6 0.8 1/[S] (M 1 ) b. 0.7 0.6 1/v (M 1 sec) 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1/[S] (M 1 ) c. 4.5 4.0 3.5 1/v (M 1 sec) 3.0 2.5 2.0 1.5 0.5 0 0 0.2 0.4 0.6 0.8 1/[S] (M 1 )

PROBLEMS and solutions 3 a. Competitive inhibition. b. Substrate-dependent noncompetitive inhibition. c. Noncompetitive inhibition. 17. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. [S] (mm) v, no inhibitor 1 2.50 1.11 2 4.00 2.00 5 6.25 3.85 10 7.69 5.56 20 8.70 7.14 a. Graph a Lineweaver Burk plot. b. What are the apparent values of V max and K M for each experiment? c. What is the inhibition mechanism? d. If the concentration of inhibitor is 0.5 mm, what is the value of K I? a. Red line is with inhibitor, blue line is with no inhibitor. 1/v (mm 1 sec) 0.8 0.6 0.4 0.2 0.0 no inhibitor with inhibitor v, with inhibitor 0.0 0.2 0.4 0.6 0.8 1/[S] (mm 1 ) b. First calculate the slope of the lines the data are nearly perfectly linear, so we will just use the first two data points to calculate the relevant parameters. For no inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (0.4 0.25)/ (1 0.5) = 0.3 0.25 = 0.3 0.5 + 1/V max 0.1 = 1/V max V max = 10 mm sec 1 K M /V max = 0.3, K M = 3 mm For with inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (0.9 0.5)/(1 0.5) = 0.8 0.5 = 0.8 0.5 + 1/V max 0.1 = 1/V max V max = 10 mm sec 1 K M /V max = 0.8, K* M = 8 mm c. The slope changes but the y intercept does not. This is reflected in the apparent K M increase upon addition of inhibitor and the constant V max between the two reactions. These observations are indicative of a competitive inhibitor. d. For a competitive inhibitor, the apparent K M with inhibitor is related to the K M without inhibitor: K* M = K M (1 + [I]/K I ) 8 mm = 3 mm (1 + 0.5 mm/k I ) 8 mm = 3 mm + 1.5 mm/k I 5 mm = 1.5 mm/k I K I = 1.5 mm/5 mm K I = 0.3 mm 18. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. [S] (mm) v, no inhibitor 1 00 0.923 5 1.154 53 10 1.176 71 50 1.195 87 100 1.198 89 v, with inhibitor a. Graph a Lineweaver Burk plot for each set of data. b. What are the apparent values of V max and K M for each experiment? c. What is the inhibition mechanism? d. If the concentration of inhibitor is 10 nm, what is the value of K I? a. Red line is with inhibitor, blue line is with no inhibitor. 1/v (mm 1 sec) 0.8 0.6 0.4 0.2 0.0 no inhibitor with inhibitor 0.0 0.2 0.4 0.6 0.8 1/[S] (mm 1 )

4 Chapter 16: Principles of Enzyme Catalysis b. For no inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (1 0.85)/(1 0.1) = 0.167 1 = 0.167 1 + 1/V max V max = mm sec 1 K M /V max = 0.167, K M = 0.167 = 0.2 mm With inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (0.95 0.92)/(0.2 0.02) = 0.167 0.95 = 0.167 0.2 + 1/V max V max = 9 mm sec 1 K M /V max = 0.167, K M * = 0.167 9 = 0.182 mm c. The y intercept changes, but the slope does not. This is reflected in the apparent K M decrease upon addition of inhibitorand the apparent V max decrease, but the maintenance of a constant ratio between the two constants. These observations are indicative of a substrate-dependent noncompetitive inhibitor. d. 10 nm is 0.00001 mm, K* M = K M /(1 + [I]/K I ) K* M = K M /(1 + [I]/K I ) 0.182 mm = 0.2 mm/(1 + 0.00001 mm/k I ) 0.00001 mm/k I = 0.2 mm /0.182 mm 1 K I = 0.1 μm or 100 nm 19. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. [S] (µm) v, no inhibitor (µm sec 1 ) 10 8.93 6.94 20 10.42 8.93 30 13 9.87 100 12.02 11.57 200 12.25 12.02 v, with inhibitor (µm sec 1 ) a. Graph a Lineweaver Burk plot for each set of data. b. What are the values of V max and K M for each experiment? c. What is the inhibition mechanism? d. If the concentration of inhibitor is 100 nm, what is the value of K I? a. 0.16 1/v (µm 1 sec) 0.14 0.12 0.10 0.08 0.06 0.04 0.02 000 no inhibitor with inhibitor 0.0 0.02 0.04 0.06 0.08 0.10 0.12 1/[S] (µm 1 ) b. First calculate the slope of the lines the data are nearly perfectly linear, so we will just use the first two data points to calculate the relevant parameters. For no inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (0.112 0.096)/ (0.1 0.05) = 0.32 0.112 = 0.32 0.1 + 1/V max 0.08 = 1/V max V max = 12.5 μm sec 1 K M /V max = 0.32, K M = 4 μm For no inhibitor: Slope = K M /V max = (y 1 y 2 )/(x 1 x 2 ) = (0.144 0.112)/ (0.1 0.05) = 0.64 0.112 = 0.64 0.05 + 1/V max 0.08 = 1/V max V max = 12.5 K M /V max = 0.64, K* M = 8 μm c. The slope changes but the y intercept does not. This is reflected in the apparent K M increase upon addition of inhibitor and the constant V max between the two reactions. These observations are indicative of a competitive inhibitor. d. For a competitive inhibitor, the apparent K M with inhibitor is related to the K M without inhibitor: K* M = K M (1 + [I]/K I ) 8 = 4(1 + 0.1/K I ) 8 = 4 + 0.4/K I 4 = 0.4/K I K I = 0.4/4 K I = 0.1 μm or 100 nm 20. An experiment with 10 nm of an enzyme obeying Michaelis-Menten kinetics yields a V max of 7 10 3 M sec 1. a. What is the turnover number (k 2 )? b. The experiment is repeated in the presence of a noncompetitive inhibitor and the V max is reduced to

PROBLEMS and solutions 5 5 10 4 M sec 1. What fraction of the enzyme is bound to the inhibitor? a. V max = k 2 [E] 7 10 3 M sec 1 = (k 2 )(1 10 8 M) k 2 = 7 10 5 sec 1 b. V max = k 2 (1 f)[e] (1 f) = 5 10 4 M 1 sec 1 /(7 10 5 sec 1 1 10 8 M) 1 f = 0.07 f = 0.93 Therefore 93% of the enzymes are bound to the inhibitor. 21. Given the following three data tables of substrate concentrations and initial velocities for enzymes that obey Michaelis Menten kinetics, estimate K M for each enzyme in molar units. a. b. c. [S] (mm) v 0 1 266.7 3 553.8 5 705.9 50 1121.5 500 1191.7 5000 1199.2 [S] (nm) v 0 (mm min 1 ) 4 123.5 5 137.4 6 148.5 10 177.3 100 240.2 1000 249.0 [S] (mm) v 0 (M hour 1 ) 1 0.00 10 0.01 100 0.07 200 0.10 1000 0.17 5000 0.19 Recognize v0 = V max /(1 + K M /[S]), so a concentration near K M will show half the velocity of the highest velocity observed, assuming it is close to V max. a. ~3 5 10 3 M b. ~4 10 9 M c. ~200 10 3 M 22. When the bi-substrate analog PALA is added to the enzyme ATCase at low concentration it increases the rate of reaction of aspartate and carbamylphosphate. However, at higher concentrations it decreases the reaction rate. How can PALA act as both an activator and inhibitor of ATCase? PALA mimics the two substrates of ATCase. Its affinity is greater than the substrate, and therefore outcompetes substrate for the active site. At moderate concentrations (where not every site on ATCase is occupied by PALA), the equilibrium of ATCase is shifted to favor the active conformation. Since the active conformation has a higher affinity for substrate, binding sites that are not occupied by PALA are able to increase the reaction velocity. At high concentrations of PALA all binding sites become occupied by inhibitor and turnover decreases. 23. In the search for the catalytic mechanism of an enzyme, three mutations of charged residues to alanine (which is uncharged) are made and compared with the wild type (WT) enzyme. At otherwise identical conditions and concentrations of enzymes, the following initial velocities (µm sec 1 ) are measured as a function of ph. ph WT Arg55Ala Glu63Ala Lys113Ala 4 1.3 10 3 1.3 10 3 1.3 10 2 1.5 10 3 5 8.7 10 5 8.7 10 5 1.3 10 2 9.5 10 5 6 8.1 10 4 8.1 10 4 1.3 10 2 8.1 10 4 7 5.3 10 3 5.3 10 3 1.3 10 2 5.2 10 3 a. Explain which residue likely acts as a general acid/ base during catalysis? b. What is a possible mechanism for the slightly increased reaction velocities observed in the Lys113Ala mutants at lower ph? a. Glu63 likely participates directly in catalysis. All other mutations have much higher catalytic efficiency (comparable to wild type). Additionally, Glu63Ala no longer displays any ph dependency when mutated to an uncharged residue. b. Having a Lys near the catalytic Glu will shift the pka down and make it less likely to donate a hydrogen to the reaction. Removing this charge with the Lys113Ala mutant shifts the pka slightly and leads to increased reactivity and increased reaction velocities.

6 Chapter 16: Principles of Enzyme Catalysis 24. Why is triose phosphate isomerase considered to be an example of a perfect enzyme? Triose phosphate isomerase is considered a perfect enzyme because it catalyzes its reaction so fast that it is diffusion controlled.