BCH 3023 Fall 2008 Exam 2, Form C Name: ANSWER KEY

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1 Name: ANSWER KEY In class, we discussed one method to linearize the Michaelis-Menton equation. There are other methods to do this, one being an Eadie-Hofstee plot. Given the Eadie-Hofstee plot below, answer questions 1 and 2. This plot might also help with question Eadie-Hofstee Plot of Enzyme Kinetic Data velocily vf[s], sec -1 Rearrangement of Michaelis-Menten Equation: v = VmAx[S]/(Km + [S]) v (Km + [S]) = VmAx[S] (v 4S1)(Km + [S]) VmAx t[s1)(km) + V = VmAx V = /[S])(Km) + VmAx Therefore: Slope = - Km = 2.0 mm intercept = VmAx = 100 mm/sec "X"- intercept = VmAx/Km = 50 sec-1 I. What is the value for the Km? a. 20 mm b. 20 p,m c. 2.0 mm d M e. 0.2 mm 1

2 2. What is the VmAx value? a. 1.0 secimm b. 1.0 mm/sec c. 2.0 mm/sec d. 100 InM/sec e mm/sec 3. What is the pattern for a pure noncompetitive inhibitor on an Eadie-Hofstee plot? a. A series of parallel lines. b. A series of lines that intersect on the v/is1-axis (the "X"-axis). c. A series of lines that intersect on the v-axis (the "Y"-axis). d. A series of lines with different slopes that do not intersect on either axis. e. A series of lines that yield different values of Km,app Pure non-competitve inhibition has no effect on the Km and VmAx decreases. Thus, the slope does not change while the intercepts on both axes decrease as II] increases. 4. In the binding of 02 to myoglobin, the relationship between the concentration of 02 (or the partial pressure of 02) and the fraction of binding sites occupied, 0, can best be described as: a. sigmoidal b. linear with a negative slope c. linear with a positive slope d. hyperbolic e. concave down 5. A metabolic pathway proceeds according to the scheme R > S T U V W. A regulatory enzyme, X. catalyzes the first reaction in the pathway. Which of the following is most likely correct for this pathway? a. Either metabolite U or V is likely to be a positive modulator, increasing the activity of X. b. The first product. S. is probably the primary negative modulator of X, leading to feedback inhibition. c. The last product. W. is likely to be a negative modulator of X. leading to feedback inhibition. d. The last product, W. is likely to be a positive modulator, increasing the activity of X. e. The last reaction will be catalyzed by a second regulatory enzyme. 2

3 6. You discover a mutant form of hemoglobin. Hbmut and characterize how Hbmut binds 02. Your studies yield a Hill number, nh, of 1.8 for the binding of 02 to Hbmut. Which of the following is true about Hbmut, based on this result? a. Hbmut is composed of only a single subunit b. Hbutut is composed of only two subunit c. Hbmut binds exactly 1 mol of 02 per mole of protein d. Hbmut binds exactly 2 mol of 02 per mole of protein e. Hbmut binds > 2 mol of 02 per mole of protein 7. Which of the following is an anomeric pair? a. ct-d-glucose and p-d-glucose b. oc-d-glucose and p-l-glucose c. L-glucose and L-fructose d. L-glucose and L-galactose e. D-glucosamine and D-glucose 8. When the linear form of D-glucouronic acid cyclizes, the product is a(n): a. hemiacetal b. hemiketal c. lactone d. disaccharide e. glycoside 9. Which of the following is NOT a common method of controlling an enzymatic reaction inside the cell? a. reversible covalent modification b. changes intracellular enzyme concentration c. reversible allosteric control d. adjusting the intracellular ph e. changes in the intracellular substrate concentration 10. How does a decrease in blood ph from 7.4 to 7.2, effect the affinity of hemoglobin for 02? a. Affinity increases, as shown by an increase in P50 b. Affinity increases, as shown by a decrease in P50 c. Affinity decreases, as shown by an increase in P50 d. Affinity decreases, as shown by a decrease in P50 e. No change, affinity is the same 3

4 11. Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as: kcat E+S E S > E+P Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: a. 1(1 GE] total [ES]) b. k] [SMEltotal [ES]) c. kcat[es] d. (k-i + kcat)[es1 e. lc] [ES] 12. You discover an enzyme that catalyzes the reaction: A B. Based on your experiments, you determine that the Km for substrate A is 3.0 0,4 and the kcat = 0.5 sec-1. At 8.0 mm [A], you determine that the velocity of the reaction is M/min. What was the concentration of total enzyme, [Etr, used in this experiment? a. 6 nm b. 20 nm c mm d. 16 mm e. Cannot calculate without knowing the VmAx At 8.0 mm substrate, the enzyme is operating at VmAx. Why, because the substrate is in great excess of KM, ,113.0!AM = 2,667. MAX 0.61_11klimin kat = 0.5 sec-1= 30 min-1 "MAX heat leltotat teltotal = VMAX/kcat (0.61_1M1mht)/(30 min') = M = 20 nm 13. A mutant of chymotrypsin has been identified that replaces His-57 with an Amine. This mutant retains a low level of enzymatic activity. Based on your knowledge of the chymotrypsin mechanism, how would the plot of kcat vs. ph be affected for the mutant chymotrpsin? a. No effect, the plot of kcat vs. ph for the mutant would be identical to wildtype b. The kcat vs. ph plot for the mutant would be fiat c. The pk, from the kcat vs. ph plot for the mutant would be 4.2 d. The pk, from the kcat vs. ph plot for the mutant would be 7.0 e. The pk, from the kcat vs. ph plot for the mutant would be

5 14. Which of the following is NOT a reducing sugar. a. Amylose b. Sucrose c. L-fructose d. Lactose e. D-galactosamine 15. The following data were obtained in a study of an enzyme known to follow Michaelis- Menten kinetics: Vo Substrate added (timolimin) (namon) , The Kin for this enzyme is approximately: a. 1 nikl. b. 1,000 mivi. c. 2 m.m. d. 4 mm. e. 6 mm. 16. What fraction of ligand binding sites are occupied, t9, when rligand] =Kd? a. 100'1/0 b. 80% c. 50% d. 25% e. 10% 0 = 111,]/(11_1 + K,I) see equation 5.8 on p. 156 When 111 = Kd = 11_1/(11_1 + = 11_1/211,1 = = 50% 5

6 17. The steady-state assumption states that if: lci = the rate constant for ES formation the rate constant for ES dissocation kcal the constant for product formation a. Ic.1 is negligible compared to kcat b. kcat is negligible compared to k.-1 c. The product concentration at the beginning of the reaction is low. d. The rate of formation of ES exceeds its rate consumption over the course of the reaction. e. The rate of formation of ES is equal to the rate of its consumption over the course of the reaction. 18. Radioactive carbon, 14C, has a half-life of 5700 years. Calculate the fraction of 14C atoms that decay per century. a % b % c. 1.2% d. 3.5% e. 18% k = In(2)/tv, In(2)/5700 years = 1.22 x i0 yr-1 [Ali[Alo= cc' where [A] is the amount remaining at time = t and [A]o is the initial concentration [ A ] / % = e 2 2 x 10-4 yr-1)(100 yr) [Al = (100/0)(0.988) = 98.8% (AMOUNT REMAINING) Amount that decayed = 100% %= 1.2% 19. A freshly prepared solution of pure a-d-glucose has a specific rotation of +112'. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value of +52.5'. Why? Note that the specific rotation of pure p-d-glucose is +19'. a. Oxidation of a-d-glucose produces D-gluconic acid b. Oxidation of a-d-glucose produces D-glucouronic acid c. Epimerization of a-d-glucose to a-d-malmose d. In solution, glucose exits in equilibrium between the both anomers of the pyranose form and the straight-chain aldehyde. The equilibrium mixture must contain both a-d-glucose and p-d-glucose. e. In solution, the straight-chain aldehyde form of glucose must predominate. 6

7 20. Compound, I. is an inhibitor of an enzymatic reaction. Based on the experimental data shown in the table below, calculate the inhibition constant (Ki or [I] KM,app VMAX,app mm 80 m1vi/min 3!AM 2.0 m1v1 58 m1v1/min 5 pa4 2.2 m1v1 49 m1vi/min 9!AIV1 2.1 m1v1 38 m1vi/min 16 pm 2.0 mm 27 mmilmin a. 0.5 pm b c. 8 JAI d. 16 p,m e. 25 p1v1 Km, pp does not change as III increases and VmAx,app decreases as III increases. This means that the inhibitor is a pure non-competitive inhibitor. Therefore, VmAx,app = VMAXIa where a = 1 + Now plug in values from table above all Ow the same value of Ki'. We designed the problem this way! 49 mni/min = (80 mm/min)/(1 + (5 f.tm/ Ki')) Invert equation! 1/49 mini/min = (1 + (5 AM/ K1'))/(80 mini/min) 5 4M/ K,' = (80 mni/min)/(49 mm/min) 1 = Ki' = 5 0,1/0.632 = 7.9 p.a/i 21. Which one of the following statements is true of enzyme catalysts? a. They increase the equilibrium constant for a reaction, thus favoring product formation. b. They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates. c. They lower the activation energy for the conversion of substrate to product. d. To be effective they must be present at the same concentration as their substrates. e. They decrease the reverse rate of product going to substate. 7

8 22. A transition state analog: a. binds covalently to the enzyme. b. binds to the enzyme more tightly than the substrate. c. binds very weakly to the enzyme. d. is too unstable to isolate. e. must be almost identical to the substrate. 23. Penicillin and related drugs inhibit the enzyme t h i s enzyme is produced by a. P-lacamase; bacteria b. transpeptidase; human cells c. transpeptidase; bacteria d. lysozyme human cells e. aldolase; bacteria 24. What is a zymogen (proenzyme)? a. It is in the inactive form of an enzyme that is activated by proteolytic cleavage. b. It is a molecule that can bind to a site distinct from the active site of an enzyme and activate it. c. It is a molecule that can reversibly activate an enzyme. d. It is the active form of an enzyme fonned through phosphorylation of key residues e. It is the suicide inhibitor of an enzyme. 25. In competitive inhibition, an inhibitor: a. binds at several different sites on an enzyme. b. binds covalently to the enzyme. c. binds only to the ES complex. d. binds irreversibly at the active site. e. lowers the characteristic Vrna of the enzyme. 8

9 26. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the KAI for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 I f, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 1.tmol) of product to be formed? a. 1.5 min b. 3.0 min c. 6.0 min d min e min When IS] is 1000 x KM, then the enzyme is saturated with S and v = VmAx. Therefore, the experiment is being carried out under VmAx conditions and any increase in 1S1 will not change the initial rate. A decrease in [Eltot al by a factor of 3, means that the initial rate decreases by a factor of 3 and it will take 3 times longer to produce 12 1.imol. of product. 9