BME+321+Homework+4+Solution+-st-dw-st

BME+321+Homework+4+Solution+-st-dw-st - BME 321 Homework 4...

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BME 321 Homework 4 (7 pts) Name: 1. For a standard Michaelis-Menten reaction, k 1 = 5 x 10 7 M -1 s -1 , k -1 = 2 x 10 4 s -1 , and k 2 = 4 x 10 2 s -1 . Calculate Ks (the dissociation constant of the first step in the enzymatic reaction) and Km (the Michaelis constant) for this reaction. (0.5 pts) Does substrate binding achieve rapid equilibrium or steady state or both? (0.5 pts) Note k1 indicates forward reaction while k-1 indicates backward reaction. Ks = = 4 x 10 -4 M Km = = 4 x 10 -4 M Since Km = Ks, the substrate binding will achieve both rapid equilibrium and steady state. 2. The Km of a Michaelis-Menten enzyme for a substrate (same reaction as above) is 1.0 x 10 -4 M. At a substrate concentration of 0.2M, v = 43 µM min-1 for a certain enzyme concentration. However, with a substrate concentration of 0.02M, v has the same value. First, show that this observation is accurate using numerical calculations. (0.5 pts) Then estimate what the best range of [S] is for measuring Km. (0.5 pts) First, calculate Vmax for the reaction: v = v = 43 x 10 -6 M min -1 ; [S] = 0.2M; Km = 1.0 x 10
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BME+321+Homework+4+Solution+-st-dw-st - BME 321 Homework 4...

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