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ExamIIFormAAnswer - Exam II: Form A 153L SPRING 2010 Name:_...

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Exam II : Form A Name:___________________ 153L SPRING 2010 TA:_________Section:______ June 1, 2010 TA Use Only: Be sure to: Write your name on the first and last page. Check that there are 5 pages including the cover page. Show all work – for math problems, if strategy is not shown , then points will not be awarded! Do not skip logical steps in explaining your reasoning. But keep the answers direct. Do not repeat the question in your answer. Run-on sentences will be grammatically corrected & then graded based new sentence count. “Shotgun” answers full of keywords that do not make sense will not result in full credit. If you don’t know the answer to a question, then write what you think – don’t try to guess. Multiple choice questions could have more than one right answer. Wrong answers results in deductions. If you finish during the last 10 minutes of the exam, then please remain seated. M-M stands for Michaelis-Menten. Useful information: v o = (Vmax[S])/(Km+[S]), alpha = 1 + [I]/K I Score: P1.______P2.______P3.______ P4________P5________= _______ / 80pts.
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1. Which of the following is true with regards to the concept of steady state in M-M kinetics? Circle all that apply. a. There is no net difference in thermal energy from the initial to the final state of the reaction. b. There is no net difference in the rate of breakdown and the rate of formation of the product. c. There is no net difference in the initial and final concentrations of substrate and product in the reaction. d. There is no net difference in the intermediate enzyme complex concentration although the complex is in a state of flux. e. This is only a short-lived state that occurs only when there is no reverse enzymatic reaction. Correct Answers: D & E (+4) Deduct +2 for every wrong answer. Not exceed worth of question. 2. Which of the following assumptions are related to the valid assumptions required to remove unknown variables from the M-M derivation or solve for M-M constants ? Circle all that apply. a. [ES] is significantly smaller than the [S] o . b. k 2 is significantly smaller than k -1 . c. We only detect v p when [ES] = [E] f d. k cat is considered a single reaction step. e. Velocities of infinite substrate concentrations can be predicted. f. Km = [S] where the v o = Vmax/2. Correct answer: A, D, E. (+6) Deduct +2 for every wrong answer. Not exceed worth of question. 3. (+8) Explain why Km is measuring binding affinity despite that the constant includes the turnover rate constant. k 2 <<< k -1 (+2) because conversion of covalent bonds that require more thermal energy than the break down of noncovalent bonds involved in the S binding. (+4) Km is thus k -1 /k 1 . (+2) 4. (+10 total) a. Explain the theory in how an uncompetitive inhibitor lowers the K m value? I binds to the ES. (+2) This stabilizes substrate binding since I must first be released in
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This note was uploaded on 10/07/2010 for the course POLI SCI 40 taught by Professor Sch during the Spring '10 term at UCLA.

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ExamIIFormAAnswer - Exam II: Form A 153L SPRING 2010 Name:_...

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