1. Basic enzyme kinetics Concepts building:
S: substrate used at high concentration
P: product (appearing at low concentrations)
v0 = DP/Dt, initial velocity
E: enzyme (added at low concentrations)
Zero order in S
First order in E
[E] (“ferment”) v0
[S] v0 E1 E2 E3
[E] = 0
First order in S
First order in E
In addition at low [S],
At high [S] (>> [E]),
v0 proportional to [E]
v0 independent of [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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2. At low [S] (>> [E]), S is all consumed, but E remains present
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. t P
Low S+E
Low S
At low [S] (>> [E]),
S is all consumed, but
E remains present
Thus, E must be recycled.
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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3. 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.
E + S E·S P k1 k2
k-1
Phenomenological equations
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4. E + S E·S P k1 k2
k-1
We need to find an expression for the rate of reaction v = dP/dt, as a function of S and E concentrations.
t = 0 => [S] = [S0]; [P] = 0; [E] = [ET]; [E·S] = 0 Initial conditions
[S] + [P] + [E·S] = Constant = [S0] (reactants) Mass conservation
[E] + [E·S] = Constant = [ET] (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
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:
Rapid and sustained equilibrium for the first step.
Rapid and sustained steady state for the whole reaction.
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5. Rapid equilibrium assumption: k-1 >> k2
E + S E·S P k1 k2
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:
Using our simplifying assumptions (S ~ S0), mass conservation of the enzyme (E = ET - ES) and isolating ES, we get:
and for initial velocity
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
Relative velocity
Molar fraction of productive enzyme
Saturation ratio of the enzyme
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6. Steady-state assumption: d[ES]/dt = 0
E + S E·S P k1 k2
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
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:
and for steady-state velocity,
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:
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7. 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.
Physically relevant part
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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8. The Michaelis-Menten equation
Hyperbolic form: Michaelis-Menten equation
v = Vmax/Km * S0
Vmax
S0 >> Km => v → Vmax
Vmax= k2.ET
S0 << Km => v → Vmax/Km * S0
Km= (k-1+k2)/k1
S0 = Km, v = Vmax/2 Km
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9. Linear forms of the Michaelis-Menten equation
Double reciprocal plot
Linear form: Lineweather-Burk equation
Slope = Km/Vmax
y intercept = 1/Vmax
x intercept = - 1/Km
Km/Vmax
1/Vmax
Vmax
- Km
Linear form: Eadie-Hofstee equation
Slope = - Km
y intercept = Vmax
x intercept = Vmax/Km
Vmax/Km
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