Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 14, Problem 18P
Interpretation Introduction
To propose:
To propose the environmental condition in which enzyme Y is unable to bring substrate to the transition state.
Introduction:
There are certain pH values under which enzyme Y can bring substrate to transition state. However, temperature is not the attribute for the enzyme Y. It is only dependent on the pH value.
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Chapter 14 Solutions
Biochemistry
Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 3PCh. 14 - Prob. 4PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 6PCh. 14 - Prob. 7PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οf this...Ch. 14 - Answers to all problems are at the end οf this...
Ch. 14 - Answers to all problems are at the end of this...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Prob. 26P
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- 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 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_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-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, Î’, X, and Y. Problems 12 through IS refer to these graphs.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. How Varying the Amount of Enzyme or the Addition of Inhibitors Affects v Versus [S] Plots Using Figure 13.7 as a model, draw curves that would be obtained in v versus [S] plots when a. twice as much enzyme is used. b. half as much enzyme is used. c. a competitive inhibitor is added. d. a pure noncompetitive inhibitor is added. e. an uncompetitive inhibitor is added. For each example, indicate how Vmax and Km change.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. Graphical Analysis of Negative Gooperativity in KNF Allosteric Enzyme Kinetics The KNF model for allosteric transitions includes the possibility of negative cooperativity Draw Lineweaver-Burk and Hanes-Woolf plots for the case of negative cooperatively m substrate binding. (As a point of reference, include a line showing the classic Michaelis-Menten response of v to [S].)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_forward
- Answers to all problems are at (he end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Understanding State Functions Define a slate function. Name three thermodynamic quantities that are state functions and three thatarrow_forwardAnswers 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. Interpreting Kinetics Experiments from Graphical Patterns The following graphical patterns obtained from kinetic experiments have several possible interpretations depending on the nature of the experiment and the variables being plotted. Give at least two possibilities for each.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 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 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 - VI The enzyme catalase catalyzes the decomposition of hydrogen peroxide: 2H2O22H2O+O2The turnover number (kcat) for catalase is 40,000,000 sec-1. The Km of catalase for its substrate H2O2 is 0.11 M. a. In an experiment using 3 nanomole/L of catalase, what is Vmax? b. What is v when [H2O2] = 0.75 M? c. What is the catalytic efficiency of catalase? d. Does catalase approach catalytic perfection?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
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