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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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G&G Ch. 13 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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G&G Ch. 13 (cont’d) 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 s and K m : k –1 k 1 = K s = 1x10 3 s -1 7x10 7 M -1 s -1 = 1.43x10 -5 M 1x10 3 s -1 + 2x10 4 s -1 7x10 7 M -1 s -1 = 3x10 -4 M
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