g&g_ch_13_p_1-4

g&g_ch_13_p_1-4 - G&G Ch. 13 Problem 1 (pg. 438) This...

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Problem 1 (pg. 438) This problem requires simple manipulation of the Michaelis-Menten equation in order to determine the ratio we’re being asked for: V max [S] K m + [S] = v 0 [S] K m + [S] = v 0 V max We are given that [S] = 4K m . Use this to substitute for [S]: 4K m K m + 4K m = v 0 V max 4K m 5K m = v 0 V max = 0.8 v 0 V max Problem 2 (pg. 438) Let’s use the Michaelis-Menten equation again. We’ll simply plug values in to the equation to get the v 0 . Remember to convert all your units to a common form!!! V max [S] K m + [S] = v 0 (0.1mmol/mL•s)(0.02mmol/mL) 0.002mmol/mL + 0.02mmol/mL = v 0 = 0.091mmol/mL•s = 91μmol/mL•s v 0
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Problem 3 (pg. 438) Remember that K s refers only to the rate constant for dissociation of the ES complex in to E and S, while K m refers to the rate constant for dissociation of the ES complex into both E and S and E and P: k –1 k 1 = K s k –1 + k 2 k 1 = K m ES E + S k –1 k 1 ES k –1 k 1 E + P E + S k 2 Use these equations to calculate K
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This note was uploaded on 07/16/2008 for the course BIS 102 taught by Professor Hilt during the Summer '08 term at UC Davis.

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g&g_ch_13_p_1-4 - G&G Ch. 13 Problem 1 (pg. 438) This...

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