Fall 2005: CH395G - Exam 2 - Multiple Choice (2 pts each)

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1 Fall 2005: CH395G - Exam 2 - Multiple Choice (2 pts each) These constants may be helpful in some of your calculations: Avogadro s number = 6.02 x molecules/mole; Gas constant (R) = x 10-3 kj/k mol 1. The melting point of a lipid a. increases with chain length, and increases with the amount of unsaturation b. increases with chain length, and decreases with the amount of unsaturation c. decreases with chain length, and decreases with the amount of unsaturation d. all of the above e. none of the above 2. In a simple enzyme-catalyzed reaction, as substrate concentration increases the reaction velocity a. continues to increase linearly at higher substrate concentrations b. goes from 1 st order to 2 nd order with respect to substrate concentration c. increases at a lower rate at higher substrate concentrations d. none of the above e. all of the above 3. Competitive inhibitors have the following effects on K M and V max a. They alter the value of K M but not V max b. They alter the value of V max but not K M c. They alter the value of K M and V max d. none of the above e. all of the above 4. The height of the activation barrier determines a. the free energy change of the reaction b. the rate of the reaction c. the probability that the reaction will be spontaneous d. a and c e. a, b, and c 5. A hydropathy plot is used to a. determine the water-solubility of a protein. b. deduce the quaternary structure of a membrane protein. c. determine the water content of a native protein. d. extrapolate for the true molecular weight of a membrane protein. e. predict whether a given protein sequence contains membrane-spanning segments.

2 6. Which of these is a general feature of the lipid bilayer in all biological membranes? a. Individual lipid molecules are free to diffuse laterally in the surface of the bilayer. b. Individual lipid molecules in one face (monolayer) of the bilayer readily diffuse (flip-flop) to the other monolayer. c. Polar, but uncharged, compounds readily diffuse across the bilayer. d. The bilayer is stabilized by covalent bonds between neighboring phospholipid molecules. e. The polar head groups face inward toward the inside of the bilayer. 7. Which of the following statements is true of enzyme catalysts? a. They bind to substrates but are never covalently attached to substrate or product. b. They increase the equilibrium constant for a reaction, thus favoring product formation. c. They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates. d. They lower the activation energy for the conversion of substrate to product. e. To be effective, they must be present at the same concentration as their substrates. 8. The benefit of measuring the initial rate of a reaction V 0 is that at the beginning of a reaction a. [ES] can be measured accurately. b. changes in [S] are negligible, so [S] can be treated as a constant. c. changes in K m are negligible, so K m can be treated as a constant. d. V 0 = V max. e. varying [S] has no effect on V Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as k 1 k 2 E + S ES E + P k -1 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression a. k 1 ([E t ] [ES]). b. k 1 ([E t ] [ES])[S]. c. k 2 [ES]. d. k -1 [ES] + k 2 [ES]. e. k -1 [ES]. 10. The Michaelis-Menton constant has all of the following characteristics, except a. It is similar to the affinity constant between the enzyme and substrate. b. The dimension for the Michaelis-Menton constant is concentration, such as molarity. c. The Michaelis-Menton constant determines the Vmax. d. It is the substrate concentration necessary to reach 1/2 Vmax. e. The apparent Michaelis-Menton constant increases with a competitive inhibitor.

3 11. M N O P Q R Referring to the diagram above: Which two enzymes would be the most likely ones to regulate if this pathway is freely reversible and can go both ways? a. 1 and 2 b. 1 and 3 c. 1 and 5 d. 2 and 4 e. 4 and In the concerted model for subunit behavior (the choices are) - 1. Each subunit can exist in a relaxed (R) and taut (T) conformation. 2. All subunits of each oligomer will be in either the R or in the T conformation. 3. Some subunits of each oligomer can be in the R state while others are in the T state. a. 1 b. 2 c. 3 d. both 1 and 2 e. both 1 and Glycolipids are characterized by containing the following non-lipid component a. sugars b. glycerol c. phosphate d. sphingosine e. more than one of these characterize glycolipids 14. In the original mechanism for lysozyme, David Philips proposed in 1965 that strain on the substrate caused the D ring to adopt a half chair conformation which aided the reaction. A number of later experiments suggested that this strained D ring might not be part of the reaction pathway. However very recent evidence suggests that indeed the D ring adopts a half chair conformation. This recent evidence comes from a. kinetic studies with lactone-like substrates. b. NMR studies of the enzyme in solution. c. 1.5 angstrom resolution X-ray studies. d. comparison of enzyme binding affinities of several substrate analogs. e. theoretical studies on enzyme flexibility. 15. Citrate is isomerized to isocitrate in the citric acid cycle. The reaction is catalyzed by the enzyme aconitase. The G of the reaction is 5 kj/mol. The kinetics of the reaction are studied in vitro where 1 M citrate and 1 M isocitrate are added to an aqueous solution of the enzyme at 25 C. What is the equilibrium constant for the reaction? a b c x 10-7 d x e x 10-78

4 16. What are the equilibrium concentrations of citrate and isocitrate, respectively, from Question 15? a M and 0.23 M b M and 1.77 M c M and 1.12 M d M and 0.88 M e. none of the above are correct 17. The G for the formation of UDP-glucose from glucose-1-phosphate and UTP is about zero. Yet the production of UDP-glucose is highly favorable. What specifically is the driving force for this reaction? a. the rapid uptake of UDP-glucose b. the overabundance of UTP c. the hydrolysis of UTP d. the hydrolysis of pyrophosphate e. none of these 18. Which of the following statements is not true of bioinformatics? a. BLOSUM 62 is a matrix calculated from comparisons of sequences with no less than 62% divergence. b. PAM matrices are based on an explicit evolutionary model. c. PAM matrices are based on local alignments of closely related blocks of sequence. d. BLOSUM matrices are based only on highly conserved regions in series of alignments. e. PAM 250 scoring matrix means that about 250 mutations per 100 amino acids may have happened. 19. The Smith-Waterman algorithm refers to: a. a way to give generate scoring matrices. b. a way to align proteins to obtain the best possible score. c. a way to use scramblings to decide when a score is good enough. d. a way to calculate the Expect Values using the extreme value distribution.. e. a weighting function to account for gaps in doing multiple sequence alignments. 20. Glycogen is degraded via a phosphorolysis process that produces a. glucose b. glucose-1-phosphate c. glucose-6-phosphate d. a small, 3-sugar oligosaccharide that undergoes further degradation e. none of these

5 Fall 2005 CH395G Exam 2 Written Answer Questions 1. A. Draw pairs of free energy diagrams with correctly labeled axes that show the differences between the following reactions: a. A fast reaction versus a slow reaction (2) Fast Slow G G The fast reaction would have a lower activation energy b. A one-step reaction versus a two-step reaction (2) G One-step G Two-step c. An endergonic reaction versus an exergonic reaction (2) G Exergonic G Endergonic B. Substrates and reactive groups in an enzyme s active site must be precisely aligned in order for a productive reaction to occur. Why then is some conformational flexibility also a requirement for catalysis? (2) The enzyme s conformation must be flexible enough to allow substrates access to the active site, to stabilize the changing electronic structure of the transition state, and to accommodate the reaction products. C. In a chemical labeling study of an enzyme, a reagent that covalently modifies Lys residues also abolishes enzyme activity. Why doesn t this observation prove that the active site includes a Lys residue? (2) The experiment indicates that a Lys residue may be part of the active site. It is also possible that a modified Lys elsewhere on the enzyme affects catalytic activity by altering the conformation of the enzyme.

6 2. A. On the graph below, draw the changes in concentration of S (substrate), P (product), E (enzyme), ES (enzyme-substrate complex) for a simple enzyme-catalyzed reaction. (4) Concentration [S] [P] [ES] [E] Time TiTimeme B. What are three assumptions used in deriving the Michaelis-Menton equation? 1. k -1 >> k 2 therefore the first step of the reaction achieves equilibrium (3) 2. [ES] maintains a steady state, i.e. the rate of synthesis of ES is equal to the rate of consumption over most of the course of the reaction. 3. The reaction of ES P + E is not reversible. (3) C. When is K M almost equal to the dissociation constant for the enzyme-substrate complex and when is it not? When k 2 is extremely small with respect to k 1 or k -1, then K M is essentially the same as the dissociation constant.

7 3. A detailed understanding of the mechanism of viral replication of the human immunodeficiency virus-1 (HIV-1) will allow investigators to design more effective therapeutic treatment regimens for persons with HIV-1 and AIDS. One study has shown that a protein produced by the virus, called p6*, is an inhibitor of the HIV-1 protease, an enzyme important in the viral life cycle. The investigators measured the activity of the HIV-1 protease in the presence (10µM) and absence of p6*. The data are shown in the table below: [S] (µm) v o without p6* (nmol min -1 ) v o with p6* (nmol min -1 ) (4) A. On the attached graph paper, construct a Lineweaver-Burk plot showing both the noninhibited and the inhibited reactions. See the graph at the back. You should have two lines that intersect (within experimental error) on the y-axis of the graph. (4) B. Determine the K M and V max in the presence and in the absence of inhibitor. For the uninhibited data, they should get around K M = 54 µm and V max = 28.5 nmol/min. For the inhibited data, they should get around K M = 89 µm and V max = 26.0 nmol/min. Its ok if you don t get these exact numbers as long as your math from the graphed values is correct and your graph is decently drawn. x-intercept = -1/K M y-intercept = 1/V max (2) C. What type of inhibitor is p6*? Briefly explain. Competitive inhibitor - The V max is the same in the presence and absence of inhibitor (within experimental error), but the K M has increased nearly two fold, indicating that p6* is competing with the substrate for binding to the active site of the enzyme.

8 4. The standard free-energy of hydrolysis for Glc-6-P is 13.8 kj/mol and that for ATP is 30.5 kj/mol. (2) A. When the phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP, what is the standard free-energy change for this reaction? Glucose to Glc-6-P ATP to ADP Total kj/mol kj/mol kj/mol (6) B. What is the free-energy if in the cell it has been determined that the [ATP] = 0.5mM, [ADP] =4mM, [Glc] = 1mM and [Glc-6-P] = 5mM at 25 o C? G = G o + RT ln ([Glc-6-P][ADP])/([Glc}{ATP]) G = G o + RT ln (40) = kj/mol kj/mol = -7.6 kj/mol (2) C. Is the conversion of glucose to glucose-6-p still spontaneous under these conditions? Why? YES - G is negative so process is spontaneous as written.

9 5. (1) (1) (1) (1) (1) A. Name the key regulatory enzyme for the pathway PFK = phosphofructokinase B. What is the name of the missing metabolite X : 3PG = 3-phosphoglycerate C. Three steps in this pathway are thermodynamically irreversible. Identify them by number from the above scheme D. Which of the 10 reactions of glycolysis is a carbon-carbon bond cleavage? a. enolase b. phosphoglycerate mutase c. phosphoglucose isomerase d. aldolase e. glyceraldehydes-3-phosphate dehydrogenase E. Which of the 10 reactions of glycolysis is a dehydration reaction? a. enolase b. phosphoglycerate mutase c. phosphoglucose isomerase d. aldolase e. glyceraldehydes-3-phosphate dehydrogenase F. Fructose-2,6-bisphosphate is an important effector of this pathway. Describe its action and what controls its levels in the cell. (2) F2,6BP is potent activator of PFK-1 and inhibitor of F-2,6-Bpase. The levels of F-2,6-BP is controlled by PFK-2 and FBPase-2 (3) a bifunctional enzyme whose two activities are controlled by phosphorylation/dephosphorylation which in turn is controlled by hormones or neurotransmitters: phosphorylation by camp-dependent protein kinase inhibits PFK-2 activity and activates FBPase-2 activity glycolysis inhibited dephosphorylation by phosphatase activates PFK-2, inhibits FBPase-2 [F2,6P] incr., PFK activity incr.. glycolysis increases

10 6. Draw the structures of the following compounds: (2 points each) A. ATP B. Fructose-2,6-BP C. Pyruvate D. 3-Phosphoglycerate. E. A typical membrane glycerophospholipid

11 Graphs for 3a should look very similar to this.