Transcript Document

ENZYME KINETICS
Consider reaction S (substrate) --> P (product)
amount of enzyme
amount of substrate
catalyzed
[P]
4x
8x
2x
4x
2x
1x
1x
uncatalyzed
time
time
time
•Formation of product is faster in the catalyzed reaction than in
the uncatalyzed reaction and initially is linear with time.
•We measure the enzyme activity by the rate of formation of product in the
catalyzed reaction.
•Enzyme activity is proportional to the amount of enzyme and increases with
an increase in the amount of substrate added.
What more can we say about the enzyme and the reaction from the
dependence of the reaction rate on the conditions?
Analys is of an enz yme-catalyzed reaction
E+S
k1
ES
k-1
k2
E + P,
k-2
where k1, k-1, k2, k-2 are rate constants:
the ra te of E + S
ES is k1[E][S], etc.
Assume initial conditions: ES is formed quickly and
[ES] is constant; [P] is small, so (k-2)[E][P] ~ 0 and
the equations simplify to:
E+S
k1
k-1
ES
k2
E+P,
Analys is of an enz yme-catalyzed reaction
E+S
k1
ES
k-1
k2
E+P,
k-2
where k1, k-1, k2, k-2 are rate constants
Assume initial conditions: ES is formed quickly and
[ES] is constant; [P] is small, so (k-2)[E][P] ~ 0 and
the equations simplify to:
E+S
k1
ES
k2
E+P,
k-1
Because [ES] is constant: the formation of ES must be
balanced by the removal of ES:
k1([E total] Ğ[ES])[S] = (k2 + k-1)[ES];
and ([E total] Ğ[ES])[S] = [(k2 + k-1)/k1][ES]
Analys is of an enz yme-catalyzed reaction (continued)
E+S
k1
ES
k2
E+P,
k-1
Here is where we are so far:
([E total] Ğ [ES])[S] = [(k2 + k-1)/k1][ES]
If we define Km ("Michaelis constant"), = (k2 + k-1)/k1
then ([ Etotal] Ğ [ES])[S] = Km[ES];
solving for [ES], we get [ES] = [Etotal][S]/(K m+[S])
Define the rate of reaction: V = k2[ES], so
V = k2[Etotal][S]/(K m+[S])
Note that k2[Etotal] = Vmax (reaction is fastest when
all enzyme is complexed with S)
V = V max [S]/(K m+[S])
t he Michaelis-Menten equation!
This allows us to measure important c haracteristics
of the enzyme, Km and Vmax, by varying substrate
and measuring the reaction rate.
Vmax
V
1/2 Vmax
Km
[S]
Lineweaver-Burke
Take the inverse of Michaelis-Menten eqn:
1/V = (K m+[S])/(V max [S])
= (K m/Vmax)(1/[S]) + 1/ Vmax
Plot 1/[V] vs 1/[S]
Slope = Km/Vmax
(allows l inear interpolation)
X-intercept = -1/Km
Y-intercept = 1/Vmax
1/V
1/Vmax
-1/Km
1/[S]
Vmax
Eadie-Hofstee
V(Km + [S]) = V max [S]
VKm + V[S] = V max [S]
VKm/[S] + V = V max
V = V max - Km(V/[S])
Plot V vs V/[S]
V
V/[S]
Slope = - Km
(allows a linear interpolation with a more even
distribution of points.)
Y intercept = Vmax
X intercept = Vmax/Km
Vmax/Km
An example of the three plotting methods using some hypothetical data
and Excel to make the graphs:
Michaelis-Menten
45
40
V
12
20
29
35
40
35
30
25
V
S
1
2
4
8
12
20
15
10
5
0
0
2
4
6
8
s
Vmax = 45 - 50 (?)
Km = 2.5 - 3 (depending on Vmax)
10
12
14
Lineweaver-Burke
1/[S]
1
0.5
0.25
0.125
0.083
1/V
0.083
0.05
0.034
0.029
0.025
Slope
ÔK
m/VmaxÕ 0.063
y
intercept
Ô1/V
maxÕ 0.0195
Km
Slope/y
intercept
3.24
Vmax
1/y
intercept
51.3
(I used Excel’s functions to get the slope
and the y intercept and then calculated
Km and Vmax from those values.)
Eadie-Hofstee
V
12
20
29
35
40
45
40
35
30
25
v
V/[S]
12
10
7.25
4.37
3.33
20
15
"-Km"
Slope
3.07
10
5
0
0
Vmax
y
intercept
49.9
5
10
15
v/s
(Again, I used Excel’s functions to get the slope
and the y intercept.)
Some implications of the enzy me measurements:
1. If k2 << k-1, then Km = (k2 + k-1)/k1 reduces to
K m = k-1/k1
In th is case, Km is close to a Òdissociation constantÓ,
which describes the tendency of a ligand to come off
the p rotein (or other molecule) that bin ds to it.
We get the sa me thing if [ES] << Etotal:
([E total] Ğ [ES])[S] = Km[ES] reduces to Km =
[Etotal][S]/[ES]
Note that dissociation constant is the inverse of
Òaffinity.Ó
Kms for vari ous enzymes range from 0.0004 to 122
mM. The low value shows very tight binding of
substrate to enzyme; the high value, very loose
binding.
Some more implications
2. The ÒTurnover numberÓ(kcat) describes the rate of
product formation per uni t enzyme at saturating [S]:
kcat = V max/[Etotal] (maximum catalytic activity).
Turnover number is mols P formed per mol E per
second: units in s -1
Turnover numbers for various enz ymes range fro m
0.5 s -1 for lys ozyme to 40,000,000 s -1 for catalase.
3. Catalytic efficiency is kcat/Km
Efficiency is higher as turnove r number rises and/or
Km falls (affinity rises)
For the most efficient enzymes, catalyti c efficiency is
limited by the diffusion of substrate to the enzyme.
The maximum is 108 to 109 M-1s-1
Summary
•The reaction of an enzyme with its substrate to form product
can be described quantitatively: the reaction rate (V) depends
on the amount of enzyme (Vmax), the concentration of substrate
([S]), and the affinity of the enzyme for substrate (Km).
•Vmax and Km can be determined by measuring V as a function
of [S]. There are several methods for plotting the results to get
Vmax and Km.
•Under certain conditions, Km provides an indication of the affinity
of an enzyme for its substrate. “Turnover number” and “Catalytic
efficiency” provide even better measures of enzyme activity.