Biochemistry 1 Exam 3 Questions and Answers

Page 1 Exam III Biochemistry I I. _________________/36 II. _________________/10 III. _________________/30 IV. _________________/ 9 V. _________________/15 TOTAL ________________/100 I. MULTIPLE CHOICE. (36 points; 3 points each) Choose the BEST answer to the question by CIRCLING the appropriate letter. No other answer will be accepted. 1. Which of the following statements about allosteric control of enzymatic activity is false ? A. B. C. D. E. 2. One benefit of measuring the initial rate of a reaction, v 0 , is that at the beginning of a reaction: A. B. C. D. E. 3. Allosteric proteins are generally composed of several subunits. Heterotropic allosteric effectors compete with substrate for binding sites. Binding of the effector changes the conformation of the enzyme. An effector may either inhibit or activate an enzyme. One explanation of allosteric control is the sequential model. changes in [S] are negligible, so [S] can be treated as a constant. [ES] can be measured accurately. v 0 = V max. changes in K m are negligible, so K m can be treated as a constant. varying [S] has no effect on v 0. Below is a plot of V 0 vs. [S] for a specific allosteric enzyme under different conditions. Line #4 is Vmax for this [enzyme]. 4 VO 1 2 3 [S] Which of the following is true based on the graph? A. B. C. D. E. Curve #3 will eventually cross curve #4 at very high [S]. Extrapolation of line #4 to the y-axis gives the value for K m. Adding a negative effector to #2 would result in curve 3. Adding a positive effector to #1 would result in curve 2. Curve 1 represents maximum inhibition. 4. Page 2 One of the assumptions stated by Michaelis & Menten is called the steadystate assumption. This assumption implies: A. B. C. D. E. 5. A transition-state analog: A. B. C. D. 6. F. increase decrease increase by 10-3 decrease by 10-3 increase by 106 2 increase by 10 Which of the following statements about heme structure in Hb & Mb is true ? A. B. C. D. E. 9. roles in DNA metabolism. amino acid sequences. secondary structures. roles in the metabolism of E. coli . If an enzyme were able to stabilize the transition state (DDG‡) by -8 kcal/mole the rate of the catalyzed reaction would A. B. C. D. E. 8. most often binds covalently to the enzyme is too unstable to isolate have K d values lower than those for the substrate have K a values lower than those for the substrate The functional differences, as well as differences in three-dimensional structure, between two different enzymes from E. coli result directly from their different: A. B. C. D. 7. K m = K eq. the K m is equivalent to the cellular substrate concentration. the enzyme cannot change the reaction equilibrium the [ES] does not change. there are two ways to decrease [ES] and one way to increase [ES]. Heme contains a tetrapyrrole ring with four methyl, four vinyl, and four propionate side chains. The iron atom in the heme may be present in the ferrous or the ferric state. The iron in free heme (no Hb or Mb) has six ligands. In deoxyhemoglobin, the iron atom is NOT in the plane of the tetrapyrrole ring (porphorin) The axial coordination positions of heme are occupied by tyrosine residues in hemoglobin. Which of the following is true of the binding energy derived from enzymesubstrate interactions? A. B. C. D. Most of it is used up simply binding the substrate to the enzyme. Most of it is derived from covalent bonds between enzyme and substrate. It is sometimes used to hold two substrates in the optimal orientation for reaction. It cannot provide enough energy to explain the large rate accelerations brought about by enzymes. Page 3 10. The Lineweaver-Burk plot is used to: A. B. C. D. E. 11. determine the equilibrium constant for an enzymatic reaction. illustrate the effect of temperature on an enzymatic reaction. solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. solve, graphically, for the ratio of products to reactants for any starting substrate concentration. extrapolate for the value of reaction rate at infinite enzyme concentration. In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, general base catalysis is identified by the numbers ________ and covalent catalysis is the process identified by the numbers _________. Chymotrypsin 195 Ser 195 Ser 5 B: 1 HO A. B. C. D. E. 12. 2 1 1 1 2 and and and and and N C H O 4; 2; 2; 5; 5; 3 3 3 2 3 C 2 4 R2 O B: R1 and and and and and 3 O + R 1 2 R -NH 2 5 4 5 3 4 Which one of the following might explain how a phosphoryl group can change the conformation of a protein to which it is attached. A. B. C. D. It introduces two positive charges that can affect eletrostatic interactions. It can form several highly directional hydrogen bonds with adjacent H-bond donor groups. It adds a large bulky hydrophobic group to the protein that can cause collaspe of this region into the interior. It adds a modification that can be used by a phosphatase to cleave the adjacent peptide bond. Page 4 II. TRUE-FALSE. (10 points; 1 pt. each) CIRCLE the appropriate answer. No other answer will be accepted. T F 13. Proteases can be used as esterases, which is an important property that was used for studies of their mechanism. T F 14. DIPF reacts with a "special" serine at the active site of one class of proteases. T F 15. The molecular chaperonins, GroEL/GroES, use a considerable amount of ATP to affect the folding of proteins in the cell that would otherwise be too slow . T F 16. The hydrophobic interior of a protein is an important driving force for protein folding and stability. T F 17. The ability of the fetus to "breath" depends on fetal hemoglobin that is composed of a and g subunits, which have a lower affinity for BPG than the adult b subunits. T F 18. A log V max versus pH curve that shows a pK a value of 6.5 implicates a His residue at the active site that is important for binding . T F 19. TLCK (tosyl-lysylchloromethylketone) is an affinity label for the active site of trypsin. T F 20. Ammonium sulfate is used in protein purification due to the position of +NH4 and -2SO4 in the Hofmeister series. T F 21. The specificity, or efficiency, constant for an enzyme-catalyzed reaction, k cat/K m, is sometimes called a pseudo-second order rate constant and has units of M-1s-1. T F 22. A "ping-pong bi bi" enzyme can be distinguished from a "sequential bi bi" enzyme using steady-state kinetics. 23. III. SHORT ANSWER. (30 points) Give a brief answer or fill in the blanks as directed to each problem or question below. Explain the physiological reason that chymotrypsinogen is inactive until it is outside the cell. What regulatory mechanism/strategy is used for regulating its activity, i.e., how is it activated? (4 pts) (use back if necessary) Page 5 24. Complete the following table (9 pts): Inhibitor: Enzyme Species Which to which it binds kinetic (E, ES, EP, etc.) constant is affected? L-B plot 1/v no I Competitive 1/[S] 1/v no I Uncompetitive 1/[S] 1/v no I Noncompetitive 1/[S] 25. List 2 catalytic strategies and 2 mechanistic strategies used by enzymes. Label clearly. (4 pts) 26. For the following reaction: E+S k1 k-1 (4 pts) ES k2 E+P a) Which of the above species refers to the "enzyme-substrate complex"? b) Write a rate equation for the production of ES. c) What is the rate of product formation from ES? d) If all the enzyme is bound with substrate, what is the effect on the rate of the forward reaction by adding more substrate? 27. 28. 29. Page 6 For protein stability, name one way to perturb the equilibrium, N <--> D, for a protein and name one way to observe the changes. Make a plot of your chosen observable versus the perturbation and show the shape of the curve. (6 pts) Describe TWO ways that serine proteases support the hypothesis that stabilization of the transition state rather than substrate binding, in this case the tetrahedral intermeditate, increases the rate of ctalysis. (3 pts) IV. MECHANISMS. (9 points) In the active site of LDH, the binding of the NADH is not in the same plane as the flat pyruvate substrate (sp2 carbon). Given what you know about transition state theory, why would this help increase the rate of the reaction in the stereospecific production of D-Lactate. (3 pts) 30. Page 7 Draw the tetrahedral intermediate in chymotrypsin that is produced just prior to peptide bond cleavage and formation of the acyl-enzyme intermediate. Include only the tetrahedral carbon, those atoms bonded to it, and labels for the substituents bonded to those atoms. Subsequently, draw arrows indicating the electron movements that would occur in the next step, the formation of the acyl-enzyme intermediate. DO NOT DRAW THE ENTIRE MECHANISM (6 pts) V. 31. FILL IN. (15 points) For the following paragraph, chose answers from the following list: (15 pts) hyperbolic sequential myoglobin heme His hemoglobin iron H-bonds salt-bridges Val 2 1 sigmoidal greater decreased Hill Bohr Menten BPG concerted [NOTE: Some items may be used more than once and some items may not be used at all.] Hemoglobin is a tetrameric protein consisting of a)_______________ a and a)_______________ b subunits. The structure of each subunit is remarkably similar to myoglobin, including the prosthetic group called b)_______________. One of the roles of the protein is to coordinate the iron with a c)_______________ (F8) residue leaving an unliganded position open to bind oxygen. The binding curves differ drastically between myoglobin and hemoglobin, oxygen binding better to d)_______________ than to e)________________ at all partial pressures of oxygen. The shapes of the curves are also different, myoglobin has a f)__________________ curve and hemoglobin has a g)___________________ curve. The h)_______________ coefficient for hemoglobin is 2.8 compared to 1.0 for myoglobin. Much of the oxygen binding behavior of hemoglobin can be explained by the i)_______________ model for allosteric enzymes originally proposed by Monod, Wyman and Changeaux. When oxygen binds to deoxyHb the j)_______________ moves 0.39 Å into the plane of the b)_______________ pulling a-helix F. This breaks several k)___________, in particular one between the carbonyl of a conserved Val residue and a Tyrhydroxyl near the carboxyl terminus, which is then bound less tightly enabling the breakage of several l)_______________. When the tetramer switches to the R-state there is a destabilization of m)_______________ binding. The R-state tetramer enables other oxygen molecules to bind easier at the empty sites remaining in the tetramer. In the reverse reaction the transition of hemoglobin from the oxy form to the deoxy form is helped by lowering the pH because the deoxyHb has n)___________ affinity for H+. This effect of pH on oxygen binding is called the o)_______________ effect. Page 1 No. on Test Correct Answer 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B A C D C B E D C C D B T T F The kinetics of the folding is not affected by GroEL/GroES, it is the thermodynamics. It does use ATP but the proteins would otherwise aggregate. 16 17 T T The lower affinity for BPG, which binds to the T-state, means that the T<-->R equilibrium lies further to the right than adult Hb, thus giving fetal Hb a higher affinity for oxygen. 18 F While the pK a is consistent with a His, the effect it has on V max implicates it in catalysis not binding. If it affected V max/K m, then it would be implicated in binding. 19 T The "L" in TLCK stands for lysyl and thus it has affinity for trypsin, not chymotrypsin. 20 21 22 T T T Lineweaver-Burke plots of different amounts of a second substrate at varying concentrations of the first substrate yield parallel lines for a "ping-pong bi bi" enzyme versus lines intersecting near the negative x-axis for a "sequential bi bi" enzyme. 23 Chymotrypsinogen is the inactive form of chymotrypsin. It is produced in cells and kept in this inactive state to protect the other proteins in those cells. It is activated by Proteolysis, which is in the mechanistic class of regulation called covalent modification. The Arg-15/Ile-16 peptide bond leaves a new N-terminal group at Ile-16, which becomes positively charged and draws it out of position that blocks the active site. Page 2 No. on Test Correct Answer 24 Inhibitor: Enzyme Species to which it binds (E, ES, EP,) Which kinetic constant is affected? L-B plot 1/v Competitive E +I no I Km 1/[S] +I 1/v Uncompetitive ES Km & Vmax no I 1/[S] 1/v Noncompetitive E & ES +I Vmax no I 1/[S] 25 Catalytic strategies: positioning/proximity strain polarization desolvation Mechanistic strategies: acid/base catalysis metal ion catalysis covalent catalysis 26 a) The "enzyme-substrate complex" refers to the ES species; the interaction between E and S. b) The rate equation for the production of ES is the rate of formation of ES minus the rate of its degradation. The degradation has two pathways, either back to E + S or on to product E + P: d [ES]/dt = rate = k1[E][S] - k-1[ E S ] - k2[ES]. c) The rate of product formation, d [P]/dt = rate = v = k2[ES]. d) No effect; increasing the substrate concentration when all the enzyme is in the ES complex does not increase the rate of the reaction. At this point the rate, v = V max = k2[E]T, where [E]T is the total enzyme concentration. It is working at its maximal rate and can work no faster. Page 3 No. on Test Correct Answer 27 ___ 28 There were two stablizing aspects discovered for serine proteases: 1) The oxy-anion hole from nitrogen residues of the backbone that are out of the plane of the sessile peptide bond, which make transient bonds to the negatively charged carbonyl oxygen in the tetrahedral intemediate. These new bonds contribute to the lowering of the activation energy by the enzyme catalyzed reaction. 2) The His-Asp residues become charged as His accepts a proton in the tetrahedral intermediate. This charge is stabliziaed by the negative charge on the Asp as a LOW-BARRIER H-BOND (LBHB) is formed. In other words, the Asp & His get closer than normal making a stronger bond. These LDHB are very high in bonding energy and contribute to the lowering of the activation energy by the enzyme catalyzed reaction. 29 30 D-lactate is tetrahedral at the former sp2 carbon of pyruvate. Thus, the position of NADH puts the hydride in the correct geometry corresponding to the transition state of the reaction. The substrates are bound to look like the transition state, and therfore lower the activation energy and increase the rate of the reaction. -O Ser-O C - - HN Peptide CH -Peptide R Page 4 No. on Test Correct Answer 31 Hemoglobin is a tetrameric protein consisting of a) 2 a and a) 2 b subunits. The structure of each subunit is remarkably similar to myoglobin, including the prosthetic group called b)heme . One of the roles of the protein is to coordinate the iron with a c)His (F8) residue leaving an unliganded position open to bind oxygen. The binding curves differ drastically between myoglobin and hemoglobin, oxygen binding better to d)myoglobin than to e)hemoglobin at all partial pressures of oxygen. The shapes of the curves are also different, myoglobin has a f)hyperbolic curve and hemoglobin has a g)sigmoidal curve. The h)Hill coefficient for hemoglobin is 2.8 compared to 1.0 for myoglobin. Much of the oxygen binding behavior of hemoglobin can be explained by the i)concerted model for allosteric enzymes originally proposed by Monod, Wyman and Changeaux. When oxygen binds to deoxyHb the j)iron moves 0.39 Å into the plane of the b)heme pulling a-helix F. This breaks several k)H-bonds , in particular one between the carbonyl of a conserved Val residue and a Tyrhydroxyl near the carboxyl terminus, which is then bound less tightly enabling the breakage of several l)salt-bridges . When the tetramer switches to the R-state there is a destabilization of m)BPG binding. The R-state tetramer enables other oxygen molecules to bind easier at the empty sites remaining in the tetramer. In the reverse reaction the transition of hemoglobin from the oxy form to the deoxy form is helped by lowering the pH because the deoxyHb has n)greater affinity for H+. This effect of pH on oxygen binding is called the o)Bohr effect.