Concept explainers
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
To calculate: The equilibrium constant for fructose-1-P hydrolysis reaction and the free energy for the reaction.
Introduction:
The mathematical expression for Keq and
Where Keq is equilibrium constant,
Explanation of Solution
Initial [Fructose-1-P] =
[Fructose-1-P] after reaching equilibrium =
So [Fructose] produced =
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Chapter 3 Solutions
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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. 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 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. 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_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. Draw the Titration Curve for a Weak Acid and Determine its pKa from the Titration Curve When a 0.1 M solution of a weak acid was titrated with base, the following results were obtained: Plot the results of this titration and determine the pK a of the weak acid from your graph.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 the Composition of Anomeric Sugar Mixtures -D-Glucose has a specific notation, []220, of + 112.20. whereas -D-glucose has a specific notation of +18.70. What is the composition of a mixture of - and -D-glucose, which has a specific notation of 83 .U0?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 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. 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. 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_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. 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. 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
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning