Midterm FA08Answ-1 scheffler

Midterm FA08Answ-1 scheffler - METABOLIC BIOCHEMISTRY Immo...

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METABOLIC BIOCHEMISTRY Fall 2008 Immo E. Scheffler MIDTERM EXAM All answers are to be written into the Blue Book. Leave the first inside page blank for scoring. There are 12 questions. Make sure that each answer is clearly identified with the question number at the top or left side of the page. Useful Information: Avogadro's number: 6.02 x 10 23 molecules / mole 1 Faraday = 96,494 Coulomb / mole = 96,494 Joules / Volt / mole Gas constant (R) = 8.31 Joules K -1 mol -1 = 1.987 cal K -1 mol -1 = 0.082 liter atm K -1 mol -1 1 calorie = 4.184 Joules QUESTION 1 (10 min) Write the capital letter corresponding to your answers to (1a), (1b), (1c), (1d) into the blue book a) Enzymes are potent catalysts because they: A) are consumed in the reactions they catalyze. B) are very specific and can prevent the conversion of products back to substrates. C) drive reactions to completion while other catalysts drive reactions to equilibrium. D) increase the equilibrium constants for the reactions they catalyze. E) lower the activation energy for the reactions they catalyze. b) The benefit of measuring the initial rate of a reaction V 0 is that at the beginning of a reaction: A) [ES] can be measured accurately. B) changes in [S] are negligible, so [S] can be treated as a constant. C) changes in K m are negligible, so K m can be treated as a constant. D) V 0 = V max . E) varying [S] has no effect on V 0 . c) Which of the following statements about a plot of V 0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false ? A) As [S] increases, the initial velocity of reaction V 0 also increases. B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at K m . C) K m is the [S] at which V 0 = 1/2 V max . D) The shape of the curve is a hyperbola. E) The y-axis is a rate term with units of μ m/min.
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d) An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the K m for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 μ mol. If, in a separate experiment, one- third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 μ mol) of product to be formed? A)
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This note was uploaded on 09/11/2009 for the course BIBC BIBC 102 taught by Professor Price during the Spring '08 term at UCSD.

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Midterm FA08Answ-1 scheffler - METABOLIC BIOCHEMISTRY Immo...

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