MB Discussion, Spring 2012 Practice Problems 1. Effect of enzymes on reactions Which of the listed effects would be brought about by any enzyme catalyzing the following simple reaction? k 1 S P where K eq = [P] / [S] k 2 a) Decreased K eq b) Increased K eq c) Increased k 1 d) Increased k 2 e) Increased!G f) Decreased!G g) More negative!g 2. Applying Michaelis-Menten kinetics a) At what substrate concentration would an enzyme with a of 60!M operate at one quarter of its maximal rate? [ ] [ ] v 0 = 1 4 v! insert into M-M: 1 max 4 v = v S max max + S [ S] = 1 3 K = 20µM M! 1 ( 4 K + S M [ ]) = [ S] b) Determine the initial rates of the reaction (in terms of v max ) that would be obtained at the following substrate concentrations [S]: ", 2, 10 Plugging this into M-M: v 0 = v max 2 + 2 = 1 3 v max Likewise: v 0 = v 2K max M = 2 + 2 3 v maxand v 0 = v 10K max M = 10 +10 11 v max otice how at [S] = 10, v 0 is already very close to v max. c) Assume k cat = 20 min -1. In an experiment, [S] = 20 mm, and the intitial velocity, v 0 was 480 nm min -1. What was the total enzyme concentration [E t ] used in the experiment? First, you need to realize that we re at a substrate concentration [S], which is much higher than the. Thus, we re at the special case where v 0 = v max. v max = k cat [ E] t! [ E] t = v max 480nM min = = 24nM 20min "1 k cat d) In an experiment [E t ] = 0.5!M, and the measured v 0 = 5!M min -1. What was the substrate concentration [S] in the experiment? First, you need to calculate the v max for the enzyme:
MB Discussion, Spring 2012 [ ] t = 0.5µM " 20min #1 =10µM min v max = k cat E ow you can realize that v 0 =! v max. This the other special case, at which [S] =. Therefore the substrate concentration is 60 "M. (Alternatively, you could insert the values for v 0, v max and into the M-M equation. owever, the calculation will be much longer that way.) 3. Estimation of v max and by inspection (Lehninger h. 6, Qu. 11) Although graphical methods are available for accurate determination of the v max and of an enzyme-catalyzed reaction, sometimes these quantities can be readily estimated by inspecting values of v 0 at increasing [S]. Estimate the v max and of the enzyme-catalyzed reaction for which the following data were obtained: [S] (M) v 0 (!M/min) 2.5 # 10-6 28 4.0 # 10-6 40 1 # 10-5 70 2 # 10-5 95 4 # 10-5 112 1 # 10-4 128 2 # 10-3 139 1 # 10-2 140 Inspection of the v 0 values shows you that they don t change very much between 2 # 10-3 and 1 # 10-2 M substrate concentration. Therefore, you know that you re at the v max of the enzyme at these high substrate concentrations. V max = 140 "M/min. ow you can make use of the property [S] =, when v 0 = v max /2. V max /2 is 70 "M/min and you conveniently find a data point for this (at [S] = 1 # 10-5 ). Thus = 1 # 10-5. 4. Enzyme inhibition You are studying the enzyme enolase, which is the second to last step of glycolysis (more of this to come next week). It catalyzes the reaction of 2- phophoglycerate (2PG) to phophoenolpyruvate (PEP). You are interested in the inhibition of the enzyme by phosphonoacetohydroxamate (PhA shown below). P 2-3 - 2 P 2-3 - P 2-3 - 2-Phosphoglycerate (2PG) Phosphoenolpyrvate (PEP) Phosphonoacetohydroxamate (PhA) Measurements for the rate of formation of PEP are carried out. If increasing concentrations of PhA are added, the initial reaction reaction rate is lower. After data analysis, the following Lineweaver-Burk plot is obtained:
MB Discussion, Spring 2012 a) Based on the Lineweaver-Burk plot, what type of inhibitor is PhA? ow do the apparent kinetic parameters (, v max ) change in this type of inhibition when inhibitor is added? ompetitive Inhibition (K m!, no $V max ) Shown below is the mechanism for enolase (from Lehniger p. 213): - P 3 2- - - P 3 2- + - P 2-3 - + 2 Lys Glu Lys Glu Lys Glu b) In the above mechanism, identify the high-energy intermediate ( the transition state ) of the reaction and circle it. (See above) c) What cofactor is required for enzyme catalysis? What is its role in catalysis? Magnesium ions ( ) are required for the reaction. They are necessary for substrate binding because they neutralize the negative charge on the carboxyl group and on the phosphate group. In addition, the stabilizes high-energy intermediate by neutralizing the extra charge on the carbanion intermediate. d) Based on the chemical structure of PhA and the mechanism of enolase, propose an explanation for why PhA acts as the type of inhibitor you determined in a. Draw a possible way how PhA could bind to enolase consistent with this model. PhA might serve as a structural analogue of the high-energy intermediate (the carbanion from c). It has a similar shape and can form similar interactions with the Mg2+ ions. If PhA was bound to the active site, it would directly compete with the binding of the 2PG substrate. This would then explain why PhA was a competitive inhibitor.
MB Discussion, Spring 2012 5. Enzyme inhibitors (continued!) Identify the following Lineweaver-Burk plots as examples of either competitive, uncompetitive or mixed inhibition. Uncompetitive Inhibition ompetitive Inhibition Mixed Inhibition 6. Applying the Michaelis-Menten equation (part II!) For an enzyme-catalyzed reaction, the presence of 5 nm of a reversible inhibitor yields a Vmax value that is 80% of the value in the absence of the inhibitor. The Michaelis constant value remains unchanged. a. What type of inhibition is likely occurring? Inhibition is most likely mixed (noncompetitive) with! =!' since it is reversible and only V max is affected. b. What proportion of the enzyme molecules have enzyme bound? Since!'V max = 0.8 V max, 80% of the enzyme remains uninhibited. Therefore, 20% of the enzyme molecules have inhibitor bound. c. alculate the inhibition constant. "' = V max V = 1 I ' =1.25 " 1.25 =1+ " K app ' I max 0.8 K I = 5nM 1.25 "1 = 20nM 7. Understanding the atalytic Mechanism of hymotrypsin a. Describe the nature and purpose of the oxyanion hole in chymotrypsin. The oxyanion hole consists of the backbones of both Ser195 and Gly. These groups allow -bonds to form to the carbonyl oxygen of the tetrahedral intermediate. It helps to stabilize the transition state so cleaving can occur. b. If the following peptide were cleaved by chymotrypsin, which R side chain must fit into the specificity/hydrophobic pocket of the enzyme? (int: R(n-1), R(n), R(n+1) or R(n+2))
MB Discussion, Spring 2012 The scissile bond refers to the bond that is going to be cleaved by the enzyme. If the boxed peptide bond is the one to be cleaved then R(n- 1)is the side chain that will fit into the specificity pocket. c. If the residues in the specificity pocket of chymotrypsin were changed to Asp and Glu, what residues would this modified chymotrypsin cleave after? If the residues of the specificity pocket were changed to Asp and Glu (these are negatively charged amino acids) then you would expect the modified chymotrypsin to bind to side chains that were positively charged (ie. Lys and Arg) and hence it would cleave after these residues d. In the reaction mechanism diagrammed below, place a box around the residues that comprise the catalytic triad. e. omplete the diagram with arrows indicating the movement of electrons as the reaction proceeds. f. ircle the tetrahedral intermediate(s). (see next page)
MB Discussion, Spring 2012 E + S α 2 is-57 2 R 2 2 α 2 is-57 2 + arboxylate product (P 2 ) 2 E + P 2 (1) (6) E - S α 2 R2 is-57 2 2 α 2 is-57 2 2 (2) (5) E - P 2 oxyanion hole α 2 R 2! is-57 2 2 oxyanion hole α 2! is-57 2 2 E - TI 1 (3) (4) E - TI 2 Acyl E + P 1 α 2 is-57 2 R 2 Amine product (P 1 ) 2 α 2 is-57 2 2 Acyl E + 2