Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 15, Problem 6P
Interpretation Introduction
To explain:
How consumption of caffeine can affect on glycogen phosphorylase activity.
Introduction:
Formation of 5’AMP from cAMP by 5’ phosphodiesterase can be inhibited by caffeine.
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Biochemistry
Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...
Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Answers to all problems are at the end of this...Ch. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24P
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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. Graphical Analysis of MWC Allosteric Enzyme Kinetics (Integrates with Chapter 1.1) Draw both Line weaver-Burk plots and Hanes-Woolf plots for an MWC allosteric enzyme system, showing separate curves for the kinetic response in (a) the absence of any effectors, (b) the presence of allosteric activator Λ, and (c) the presence of allosteric inhibitor I.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. Describe the Effects on cAMP and Glycogen Levels in Cells Exposed to Cholera Toxin Cholera toxin is an enzyme that covalently modifies the G-subunit of G proteins. (Cholera toxin catalyzes the transfer of ADP-ribose from NAD+ to an arginine residue in Gan ADP-ribosylation reaction.) Covalent modification of G� inactivates its GTPase activity. Predict the consequences of cholera to.vin on cellular cAMP and glycogen levels.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 Rale Constants to Calculate Km, Kinetic Efficiency (kcat/Km) and Vmax - II Triose phosphate isomerase catalyzes the conversion of glyceraldehyde-3-phosphate to dihydroxy-acetone phosphate. Glyceraldehyde3PdihydroxyacetonePThe Km of this enzyme tor its substrate glyceraldehyde-3-phosphate is 1.8 10-5 M. When [glyceraldehydes-3-phosphate] = 30 M, the rate of the reaction, v, was 82.5 mol mL-1 sec-1. a. What is Vmax for this enzyme? b. Assuming 3 nanomoles per mL of enzyme was used in this experiment ([Etotal]) = 3 nanomol/mL), what is kcat for this enzyme? c. What is the catalytic efficiency (kcat/Km) for triose phosphate isomerase? d. Does the value of kcat/Km reveal whether triose phosphate isomerase approaches catalytic perfection? e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that imposes the physical limit on kinetic perfection?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. 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 οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Comparison of Emzymatic and Nonenzymatic Rate Constants The for alkaline phosphatase—catalyzed hydrolysis of melhylphoiphate is approximately 14/sec at pH 8 and 25ºC. The rate constant for the uncatalyzed hydrolysis of methyl phosphate under the same conditions is approximately I0-15/sec. What is the difference in the free energies of activation of these two reactions?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. Calculating C°' and from KeqCellular Concentrations Hexokinase catalyzes the phosphorylation of glucose from ATP. yielding glucose- 6-P and ADP. The standard-stale free energy change for hydrolysis of glucose-6-Ρ is — 13.9 kJ/mol. Calculate the standard-state free energy change and equilibrium constant for the hexokinase reaction.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 - 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. Consult the following reference (Samanta U. and Bahnson. B. J., 2008. Crystal structure of human plasma platelet-activating factor acetylhydrolase. Journal of Biological Chemistry 283:31617-3U624). consider the active-site structure of this enzyme in the Light of material in this chapter, and write a detailed mechanism for the PAF acetylhydrolase.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. Quantitative Relationships Between Rate Constants to Calculate Km, Kinetic Efficiency (kcat/Km) and Vmax - III The citric acid cycle enzyme fumarase catalyzes the conversion of fumarate to form malate. Fumarate+H2OmalateThe turnover number, kcat, for fumarase is 800/sec. The Km of fumarase for its substrate fumarate is 5M. a. In an experiment using 2 nanomole/mL of fumarase, what is Vmax? b. The cellular concentration of fumarate is 47.5 M. What is v when [fumarate] = 47.5 M? c. What is the catalytic efficiency of fumarase? d. Does fumarase 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. 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_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. Use examples from the ActiveModel for Human GaleLtin-1 to describe the hydrophobic effect.arrow_forward
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