Ans A Section 73 4 Chapter 7 Kinetics and Regulation 31 When reaction

Ans a section 73 4 chapter 7 kinetics and regulation

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Ans: A Section: 7.3 4
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Chapter 7 Kinetics and Regulation 31 When reaction conditions are such that the amount of substrate is far greater than the amount of enzyme present, then the following conditions are also met. A) The [substrate] is much less than K M . B) The V 0 is half V max . C) The enzyme is displaying second-order kinetics. D) The enzyme is displaying first-order kinetics. E) The enzyme is displaying zero-order kinetics. Ans: E Sections: 7.1 and 7.2 32 During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity versus substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. A) This is an enzyme that displays Michaelis–Menten kinetics, and you purify away a homotrophic inhibitor. B) This is an enzyme that displays Michaelis–Menten kinetics, but you must use a Lineweaver–Burk plot to determine K M and V max correctly. C) This is an allosteric enzyme, but you must use a Lineweaver–Burk plot to determine K M and V max correctly. D) This is an allosteric enzyme , and during purification you purify away a heterotrophic activator . E) This is an allosteric enzyme displaying a double-displacement mechanism, and during purification you purify away one of the substrates. Ans: D Sections: 7.2 and 7.3 33 After purifying the enzyme in the previous question, you determine the Mr to be 75,000. By assaying 5 μg of the enzyme under saturating [S] concentrations, you determine the V max to be 1.68 μmol/sec. Calculate the turnover number for this enzyme. A) 2.25 10 6 sec -1 B) 1.50 10 5 sec -1 C) 2.50 10 4 sec -1 D) 1.79 10 5 sec -1 E) You need to also know the K M for this enzyme to calculate turnover number. Ans: D Section: 7.2 5
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