Concept explainers
(a)
Interpretation :
The relative glycogen phosphorylase activity when neither AMP nor ATP is present should be estimated from figure 15.15.
Introduction:
Glycogen phosphorylase is an enzyme which catalyzes the rate limiting step in glucogenolysis. It converts glycogen in animals into glucose-1-phosphate. ATP and glucose-6-phosphate act as allosteric inhibitors for glycogen phosphorylase because they reduce the affinity of glycogen phosphorylase for its substrate, inorganic phosphate. ATP is the end product of the glycogen phosphorylase enzyme. And glucose-1-phosphate released is readily converted into glucose-6-phosphate and participate in ATP production pathways. AMP acts as an allosteric activator for glycogen phosphorylase.
(b)
Interpretation:
The relative glycogen phosphorylase activity when AMP is present should be estimated from the figure 15.15.
Introduction:
Glycogen phosphorylase is an enzyme which catalyzes the rate limiting step in glucogenolysis. It converts glycogen in animals into glucose-1-phosphate. ATP and glucose-6-phosphate act as allosteric inhibitors for glycogen phosphorylase because they reduce the affinity of glycogen phosphorylase for its substrate, inorganic phosphate. ATP is the end product of the glycogen phosphorylase enzyme. And glucose-1-phosphate released is readily converted into glucose-6-phosphate and participate in ATP production pathways. AMP acts as an allosteric activator for glycogen phosphorylase.
(c)
Interpretation :
The relative glycogen phosphorylase activity when ATP is present should be estimated from the figure 15.15.
Introduction:
Glycogen phosphorylase is an enzyme which catalyzes the rate limiting step in glucogenolysis. It converts glycogen in animals into glucose-1-phosphate. ATP and glucose-6-phosphate act as allosteric inhibitors for glycogen phosphorylase because they reduce the affinity of glycogen phosphorylase for its substrate, inorganic phosphate. ATP is the end product of the glycogen phosphorylase enzyme. And glucose-1-phosphate released is readily converted into glucose-6-phosphate and participate in ATP production pathways. AMP acts as an allosteric activator for glycogen phosphorylase.
Trending nowThis is a popular solution!
Chapter 15 Solutions
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 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. 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_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. 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. 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 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 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_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. Why zymogens Are Advantageous Why do you suppose proteolytic enzymes are- often synthesized as inactive zymogens?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. 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_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. 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_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. 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_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning