Lecture 9
Enzyme Kinetics
Consider reaction S > P
[P]
time
Enzymecatalyzed reaction
E + S ES E+P, k1, k1, etc are rate constants
Assume initial conditions: ES is formed quickly and [ES] is constant; [P] is small, so
(k2)[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 + k1)[ES] = 0
Rearranging, k1([E
total
] – [ES])[S] = (k2 + k1)[ES]; and
([E
total
] – [ES])[S] = [(k2 + k1)/k1][ES]
If we define K
m
= (k2 + k1)/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 MichaelisMenten equation!
k1
k1
k2
k2
catalyzed
uncatalyzed
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Plotting enzyme data
MichaelisMenten
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]
LineweaverBurke
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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 Spring '09
 DavidSmart
 Catalysis, vmax, k2, Vmax Km LineweaverBurke

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