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Unformatted text preview: urve appears “shallower” with a less steep climb to the asymptotic limit. The red data points correspond to the new KM’. You can see it takes a significantly higher [ATP]0 for the data to eventually reach Vmax in the presence of an inhibitor. Direct plot 2.5 v0 (uM/s) 2 1.5 1 0.5 0 0 10 20 30 40 50 60 [ATP]0 (uM) Chemistry 160 problem set 11 solutions Lineweaver Burk plot (with competitive inhibitor present) • Vmax is unchanged so the y intercept stays the same. • KM’ is larger, which results in two things: o x intercept is smaller, as x intercept =  1/KM’ o slope is larger, as slope = KM’/Vmax, so the line is steeper. The red data points correspond to values reflecting the presence of inhibitor. The blue data points correspond to the original kinetic data in Problem 1. Lineweaver Burk plot 2.5 2 1/v0 1.5 1 0.5 0  1  0.5 0 0.5 1 1.5 2 1/[ATP]0 Chemistry 160 problem set 11 solutio...
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This document was uploaded on 03/22/2014 for the course CHEM 160 at San Jose State.
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 Physical chemistry, pH

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