Basic enzyme kinetics
Concepts building:
• S: substrate used at
high concentration
• P: product (appearing at
low concentrations)
• v0 = DP/Dt, initial velocity
v0
• E: enzyme (added at
low concentrations)
Zero order in S
First order in E
v0
E3
E2
E1
[E] = 0
[E] (“ferment”)
At high [S] (>> [E]),
• v0 proportional to [E]
• v0 independent of [S]
First order in S
First order in E
[S]
In addition at low [S],
• v0 proportional to [E] and
• v0 proportional to [S]
• Thus, bimolecular reaction
in E and S => v0 = k [E] [S]
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If the reaction is bimolecular (in E and S), one can infer that at
least in these conditions, an intermediate is formed between
limiting S and E, so that:
E + S  [E·S]*  .?.  P
At high [S], the first step becomes saturated, and the reaction,
independent of S, remains limited by E (<< S) only.
P Low S+E Low S
At low [S] (>> [E]),
• S is all consumed, but
• E remains present
Thus, E must be recycled.
t
In addition, while E catalyses S  P, it often catalyses P  S
as well.
One may then (by symmetry) conclude that either S or P can
associate to E (forming [E·S] or [E·P]) and further reversibly
dissociate from it.
Thus,
E+S
[E·S]*
.?.
[E·P]*
E+P
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E+S
[E·S]*
.?.
[E·P]*
E+P
Having reached a working model for a reaction catalyzed by an
enzyme, we now set off to develop a kinetic model for it, which
eventually will generate experimentally testable predictions.
Using simplifying assumptions, one may reduce the complex
model above to a more manageable form.
Assumptions:
• S >> P
initial stage, P negligible and last step virtually irreversible.
• E << P, S
dilute enzyme solutions; reactants in relatively large excess.
The mechanism may then be reduced into a 2 steps reaction, for
simplicity.
k1
k
E+S
E·S 2 P
k-1
dP
Phenomenological equations
 k 2 ES
dt
dES
 k1E S  k1ES  k2 ES  k1E S  (k1  k2 )ES
dt
v
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E+S
k1
k-1
E·S
k2
P
We need to find an expression for the rate of reaction v = dP/dt, as a function of S and E
concentrations.
dES
 k1E S   (k 1  k 2 )ES
dt
t = 0 => [S] = [S0]; [P] = 0; [E] = [ET]; [E·S] = 0
Initial conditions
• [S] + [P] + [E·S] = Constant = [S0]
• [E] + [E·S] = Constant = [ET]
Mass conservation
(reactants)
(enzyme)
At short times, and using dilute enzyme solutions (assumptions), one may write S ~ S0, and
by substituting E, we get a differential equation homogeneous in ES
dES
 k1 ET  ES S0  (k 1  k 2 )ES  k1ET S0  (k 1  k 2  k1S0 )ES
dt
While it is possible to derive a complex analytical solution for this equation (see previous
lecture), there are two special cases for which the solution is easier, and even elegant!...
These are either:
1. Rapid and sustained equilibrium for the first step.
2. Rapid and sustained steady state for the whole reaction.
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Rapid equilibrium assumption: k-1 >> k2
k1
E+S
k2
E·S
P
k-1
Henri, shortly followed by Michaelis and Menten, postulated that if the rate constant k-1 for
unproductive dissociation of the complex ES (back to E and S) is much larger than k2, the
rate constant for the productive dissociation (yielding P and E), the reversible reaction in the
first step should rapidly reach an equilibrium, and remain in this state for a long time,
relatively to product accumulation.
For dissociation in the first step (reverse direction), we define KS as the equilibrium constant:
KS 
E S   k 1
ES  k1
Using our simplifying assumptions (S ~ S0), mass conservation of the enzyme (E = ET - ES)
and isolating ES, we get:
ES 
ET  S0
KS  S0
and for initial velocity
v 0  k 2  ES 
k 2  ET  S0
.
KS  S0
The expression k2ET, representing the rate of reaction at saturation with S (when S >> KS,
ES ~ ET) is referred to as maximal velocity or Vmax.
The relation v0 = k2 ES can be rearranged as v0 /k2 = ES; Dividing by ET = E + ES, we get
v0
v
ES
S0
 0 

k 2ET Vmax E  ES KS  S0
Relative
velocity
Molar fraction of
productive enzyme
Saturation ratio of
the enzyme
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Steady-state assumption: d[ES]/dt = 0
k1
E+S
k2
E·S
P
k-1
Haldane, in order to solve the central differential equation, postulated that if the reaction
reached a steady-state , in view of the large excess of S over E, the concentration of the
intermediate ES will remain constant for a long time. In this case, the rate of formation of the
complex, must equal that of its dissociation in both directions. Thus, at steady-state,
k1 [E] [S] = (k-1 + k2) [ES]
or
E S   k 1  k 2
ES 
k1
 Km
This relation defines Km as the Michaelis constant, and reveals its meaning in terms of
both concentrations and rate constants. Using our simplifying assumptions (S ~ S0), mass
conservation of the enzyme and isolating ES, we get:
ES 
ET  S0
K m  S0
and for steady-state velocity, v  k 2  ES 
k 2  ET  S0
K m  S0
The expression k2ET, represents again Vmax. The equation for v0 is of the same form as that
obtained in the previous case, with the difference that Ks, the equilibrium dissociation
constant is replaced by Km, a “dynamic equilibrium” constant, taking also into account the
productive dissociation of ES. Finally, The relation between both macroscopic constants is
given by:
Km 
k 1  k 2
k
 KS  2
k1
k1
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Dimensionless form of the Michaelis-Menten equation
In processing equations, it is always useful to put
them in dimensionless form (for easy mathematical
manipulation).
In order to do that, we divide the variables v and S0
by their respective ‘rulers’ parameters Vmax and Km.
v
Vmax  S0
v

K m  S0
Vmax
S0
Km

S
1 0
Km
of the form Y 
Physically
relevant part
X
1 X
The dimensionless form reveals a rectangular hyperbolic
form for the mathematical function, and allows for a
meaningful analysis of the limits (asymptotes).
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The Michaelis-Menten equation
Hyperbolic form: Michaelis-Menten equation
v  Vmax
S0
K m  S0
120
v = Vmax/Km * S0
100
Vmax
Vmax= k2.ET
S0 << Km => v → Vmax/Km * S0
v - mM/s
S0 >> Km => v → Vmax
80
60
40
Km= (k-1+k2)/k1
S0 = Km, v = Vmax/2
20
0
0
50
Km
100
150
200
S0 - mM
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Linear forms of the Michaelis-Menten equation
Double reciprocal plot
Linear form: Lineweather-Burk equation
0.06
0.05
y intercept = 1/Vmax
0.04
x intercept = - 1/Km
1/v - s/mM
1
1
K
1

 m 
v Vmax Vmax S0
Slope = Km/Vmax
0.03
Km/Vmax
0.02
1/Vmax 0.01
v - mM/s
120
Vmax
100
0
-0.1
0
0.1
0.2
0.3
0.4
0.5
1/S0 - 1/mM
80
60
Linear form: Eadie-Hofstee equation
- Km
40
v  Vmax
20
0
0
2
4
6
8
10
Vmax/Km
12
v
 Km 
S0
Slope = - Km
y intercept = Vmax
x intercept = Vmax/Km
v/S0 - 1/s
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