Concept explainers
Answers to all problems are at the end οf this book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book.
Understanding the Implications of Transition State Stabilization
As noted in Section 14.4. a true transition state can bind to an enzyme active site with a KT as Low as 7 X 10-26 M. This is a remarkable number, with interesting consequences. Consider a hypothetical solution of an enzyme in equilibrium with a ligand that binds with a KD of 10-27 M. If the concentration of free enzyme. [EL]. is equal to the concentration of the enzyme-ligand complex. [EL], what would [L]. the concentration of free ligand, be? Calculate the volume of solution that would hold one molecule of free ligand at this concentration.
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Biochemistry
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Quantitative Relationships Between Rate Constants to Calculate Km, Kinetic Efficiency (kcat/Km) and Vmax - I Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results: k1=2108M1sec1k1=1103sec1k2=5103sec1a. What is Ks, the dissociation constant for the enzyme-substrate complex? b. What is Km, the Michaelis constant for this enzyme? c. What is kcat (the turnover number) for this enzyme? d. What is the catalytic efficiency (kcat/Km) for this enzyme? e. Does this enzyme approach kinetic perfection? (That is, does kcat/Km approach the diffusion-controlled rate of enzyme association with substrate?) f. If a kinetic measurement was made using 2 nanomoles of enzyme per mL and saturating amounts of substrate, what would Vmax equal? g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give v = 0.75 Vmax? h. If a kinetic measurement was made using 4 nanomoles of enzyme per mL and saturating amounts of substrate, what would Vmax equal? What would Km equal under these conditions?arrow_forwardAnswers to all problems are at the end οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Using Site-Direcled Muta.nts to Understand an Enzyme Mechanism In this chapter, the exponent in which Craik and Rutter replaced Asp102 with Asn in trypsin (reducing activity 10,000 -fold) was discussed. On the basis of your knowledge of the catalytic triad structure in trypsin, suggest a structure for the “uncatalytic triad of Asn-His-Ser in this mutant enzyme. Explain why the structure you have proposed explains the reduced activity of the mutant trypsin. See the original journal articles (Sprang, et al., 1987. Science 237:905-913) to Craik, et al., 1987. Scieence 237:909-913) to see Craik and Rutter's answer to this question.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Graphing the Results from Kinetics Experiments with Enzyme Inhibitors The following kinetic data were obtained for an enzyme in the absence of any inhibitor (1), and in the presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [ET] is the same in each experiment. Graph these data as Lineweaver-Burk plots and use your graph to find answers to a. and b. a. Determine Vmax and Km for the enzyme. b. Determine the type of inhibition and the K1 for each inhibitor.arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Assessing the-Metabolic Consequences of Life Without Enzymes The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without just one of the thousands of enzymes in the human body. For example, consider life without fnnctose-1,6-btsphosphatase, an enzyme in the gluconeogenesis pathway in Liver and kidneys (see Chapter 22). which helps product new glucose from the food we eat: Fructose-1.6-blsphosphate + H2O Fmrlose-6-P + Pi The human brain requires glucose as its only energy source, and the typical brain consumes about 120 g (or 480 kilocalories) of glucose dally. Ordinarily, two pieces of sausage pizza could provide more than enough potential glucose to feed the brain for a day. According to a national fast-food chain, two pieces of sausage pizza provide 1340 kilocalories. 48% of which is from fat. Fats cannot be converted to glucose in gluconeogenesis, so that leaves 697 kilocalories potentially available for glucose synthesis. The first-order rate constant for the hydrolysis of fructose-l.6-bispliosphate in the absence of enzyme is 2 10-20 /sec. Calculate how long it would take to provide enough glucose for one day of brain activity from two pieces of sausage pizza without the enzyme. The following graphs show the temperature and pH dependencies of four enzymes, A, Β, X, and Y. Problems 12 through IS refer to these graphs.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Determining the Branch Points and Reducing Ends of Amylopectin A 0.2-g sample of amylopectin was analyzed to determine the fraction of the total glucose residues, that are branch points in the structure. The sample was exhaustively methylated and then digested, yielding 50-mol of 2,3-dimethylgluetose and 0.4 mol of 1,2,3,6- letramethylglucose. What fraction of the total residues are branch points? I low many reducing ends does this sample of amylopectin have?arrow_forwardAnswers to all problems are at the end οf this book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Understanding the Very Tight Binding of Transition States Another consequence of tight binding (problem 9) is the free energy change for the binding process. Calculate Gfor an equilibrium with a KD of 10-27 M. Compare this value to the free energies of the noncovalent and covalent bonds with which you are familiar. What are the implications of this number, in terms of the binding of a transition state to an enzyme active site?arrow_forward
- Answers to all problems are at the end οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Calculation of Rate Enhancement from Energies of Activation The relationships between the free energy terms defined in the solution to Problem 4 earlier are shown in the following figure. If the energy of the ES complex is 10 kJ/mol lower than the energy of E + S, the value of Ge:is 20 kJ/mol, and the value of Ge:is 90 kJ/mol what is the rate enhancement achieved by an enzyme in this case?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. CalculatingGandSfromH The equilibrium constant for some process AB 0.5 at 20°C and 10 at 30°C. Assuming that G is independent of temperature, calculate H for this reaction. GandSat20Candat30C Why- is it important in this problem to assume that H is independent of temperature?arrow_forwardAnswers to all problems are at the end of this book Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Solving the Sequence of an Oligopeptide From Sequence Analysis Data Analysis of the blood of a catatonic football fan revealed large concentrations of a. psychologic octapeptide. Amino acid analysis of this oclapeplide gave the following results: 2 Ala lArg 1 Asp 1 Mel 2 Tyr I Val 1NH/ The following facts were observed: Partial acid hydrolysis of the octapeptide yielded a dipeptide of the structure Chymolrypsin treatment of the octapeplide yielded two tetrapeptides, each containing an alanine residue. Trypsin treatment of one of the tetrapeptides yielded two dipeptides. Cyanogen bromide treatment of another sample of the same tetrapeplide yielded a tripeplideand free Tyr. N-lerminal analysis of the other tetrapeptide gave Asn. What is the amino acid sequence of this oclapeplide?arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Solving the Sequence of an Oligopeptide From Sequence Analysis Data Amino acid analysis of ail oligopeptide seven residues long gave The following fads were observed: a. Trypsin treatment had no apparent effect. b. The phenylthiohydantoin released by Lid mini degradation was c. Brief chymotrypsin treatment yielded several products, including a dipeptide and a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Lyi. and Met. d. Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and free Lys. What is the amino acid sequence of this heptapeptide?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Assessing the Formation and Composition of Limit Dextrins Prolonged exposure of amylopectin to starch phosphorylase yields a substance called a limit dextrin. Describe the chemical composition of limit dextrins. and draw a mechanism for the enzyme-catalyzed rcactioa that can begin the breakdown of a limit dextrin.arrow_forwardAnswers to all problems are at the end οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Characterizing a Covalent Enzyme Inhibitor Tosyl-L-phenylalanme cfaloromethyl ketone (TPCK) specifically inhibits chymotrypsin by covalently labeling His57 Propose a mechanism for the inactivation reaction, indicating the structure of the produce(s). State why this inhibitor is specific tor cJiymotrypsin. Propose a reagent based on the structure of TPCK that might be an effective inhibitor of trypsin.arrow_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning