Fa07Midterm - METABOLIC BIOCHEMISTRY Immo E. Scheffler Fall...

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METABOLIC BIOCHEMISTRY Fall 2007 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 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 (2 min) The role of an enzyme in an enzyme-catalyzed reaction is to: A) bind a transition state intermediate, such that it cannot be converted back to substrate. B) ensure that all of the substrate is converted to product. C) ensure that the product is more stable than the substrate. D) increase the rate at which substrate is converted into product . E) make the free-energy change for the reaction more favorable. QUESTION 2 (4 min) The phosphorylation of a hexose requires an enzyme, hexokinase. Illustrate with an example (with structural formulae) the specificity of such an enzyme with regard to substrate(s) and products obtained.
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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 . QUESTION 4 (2 min) 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. QUESTION 5 (4 min) 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) 1.5 min B) 13.5 min C) 27 min D) 3 min E) 6 min QUESTION 6 (2 min) For enzymes in which the slowest (rate-limiting) step is the reaction k 2 ES P K m becomes equivalent to: A) k cat . B) the [S] where V 0 = V max . C)
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Fa07Midterm - METABOLIC BIOCHEMISTRY Immo E. Scheffler Fall...

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