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Lecture 9

# Lecture 9 - Lecture 9 Enzyme Kinetics Consider reaction S->...

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Lecture 9 Enzyme Kinetics Consider reaction S --> P [P] time Enzyme-catalyzed reaction E + S ES E+P, k1, k-1, etc are rate constants Assume initial conditions: ES is formed quickly and [ES] is constant; [P] is small, so (k-2)[E][P] ~ 0 Because [ES] is constant: d[ES]/dt = 0 so formation of ES is balanced by use of ES: k1([E total ] – [ES])[S] – (k2 + k-1)[ES] = 0 Rearranging, k1([E total ] – [ES])[S] = (k2 + k-1)[ES]; and ([E total ] – [ES])[S] = [(k2 + k-1)/k1][ES] If we define K m = (k2 + k-1)/k1, then ([E total ] – [ES])[S] = K m [ES] and [ES] = [E total ][S]/(K m +[S]) Define the rate of reaction V = k2[ES] = k2[E total ][S]/(K m +[S]) Note that k2[E total ] = V max (reaction is fastest when all enzyme is complexed with S) V = V max [S]/(K m +[S]) the Michaelis-Menten equation! k1 k-1 k2 k-2 catalyzed uncatalyzed

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Plotting enzyme data Michaelis-Menten V = V max [S]/(K m +[S]) Vmax as [S]<<Km, V = V max [S]/K m V (straight line with respect to [S]) ½ Vmax V = V max /(K m /[S] + 1) as [S] increases, V --> V max when [S] = K m , V = ½ V max K m [S] Lineweaver-Burke 1/V = (K m +[S])/(V max [S]) 1/V = (K m /V max )(1/[S]) + 1/ V max 1/V max slope = K m /V max
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