classes_winter08_153AHID41_midterm_2_key

classes_winter08_153AHID41_midterm_2_key - Name Midterm #2...

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Name Midterm #2 Chem 153AH February 29, 2008 Name (Last) (First) Student ID Problem I (20pts) 2 (1 Opts) 3 (25pts) 4 (1 Opts) 5 (20pts) 6 (1 5pts) TOTAL Make sure that you have 5 pages and write your name on each page. Score
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I. (Total of 20 points)The lineweaver-burk plot of an enzyme is shown on the left (a double reciprocal plot of substrate concentration ([S]) versus initial velocity (Vo)). The enzyme catalyzed reaction exhibits simple Michaelis-Menten kinetics. Two graphs are shown, one with and without an inhibitor. The total enzyme concentration in both plots ([Et]) is equal to 0.5 micromolar. (1A-3pts) Determine the maximum velocity of the enzyme in the absence of inhibitor. \ 4 ,/ = 0s ~5 a-~wlQ,dv -'-% (IB-2pts) Determine the turn-over number of the enzyme in the absence of -. inhibitor. VmA = cT k, kz= 2 h4W CTICL- - Y ctL4 I (1Cdpts) Determine the Michaelis constant (Km) of the enzyme in the absence of the inhibitor. fqD-Spts) When is Km a measure of the enzyme's affinity for its substrate?
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This note was uploaded on 07/09/2008 for the course BIO CHEM 153A taught by Professor Staff during the Summer '06 term at UCLA.

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classes_winter08_153AHID41_midterm_2_key - Name Midterm #2...

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