The Michaelis-Menten Equation
The enzyme (E) and substrate (S) react to form the enzyme-substrate complex (ES).
E + S <=> ES <=> P + E
k1: the forward rate of E + S to ES
k2: the forward rate of ES to P + E
k-1: the reverse rate
k-2: the reverse rate
During the initial reaction where [P] is small (usually the first 60 seconds), k-2 will be negligible. During this time [S] will be essentially constant.
Formation of ES: V1 = k1[E][S]
Degradation of ES: V2 = k-1[ES] + k2[ES] = [ES] (k-1 + k2)
Steady state occurs when V1 = V2.
Setting the above equations equal yields the following:
Km = [E][S]/[ES] = (k-1 + k2)/ k1
Km is the Michaelis-Menten Constant
If Km is high => [ES] is low => low substrate affinity
If Km is low => [ES] is large => high substrate affinity
These correlations are only true if k2 << other rate constants. That is, when k2 is rate-limiting: k-1/k1 ≈ Km. i.e., when the conversion of ES into P + E is the rate-limiting step.
[E] and [ES] are very difficult to determine experimentally. Instead, use the total enzyme concentration ET defined as [ET] = [E] + [ES]
Substitute this into the equation of Km and rearrange:
Km = ([Et]/[ES] – 1)[S]
Next, eliminate [ES]. Recall that V0 = k2[ES]. When there’s a high concentration of substrate, the reaction achieves its maximum velocity at which point all the enzyme is associated to an ES complex. The total concentration of enzyme ET then becomes the same as the concentration of ES.
Thus, an alternate expression for Vmax is Vmax = k2[ET] : all enzyme is in ES complex.
Substituting these relations in to the Km equation above yields:
Km = (Vmax/V0-1)[S] => V0 = Vmax[S]/(Km + [S])
This is the Michaelis-Menten Equation. It’s graph is a hyperbolic curve, as seen before experimentally
What predictions are now possible?
- When [S] is low => V0 ≈ (Vmax * [S])/Km. This means V0 is proportional to [S]. This is not good for a cell. The reaction becomes very sensitive to changes in concentration of substrate. However, the Enzyme will never reach its maximum velocity, and thus will not work at its full potential.
- When [S] is high => V0 ≈ Vmax. This means V0 is proportional to Vmax , and thus the cell can not fine tune the activity of the enzyme because the E is always working at its maximum capacity, which is a wasteful degradation of substrate even when the product of the reaction is not needed.
- When [S] = Vmax => V0 ≈ Vmax/2. This is optimal for the cell because the E can both be sensitive to changes in [S], but it can also reach its maximum velocity.
This latter observation sets up an alternate definition of Km: the [S] needed to reach half of maximum velocity.
Note: as will be discussed in a future lecture, Km does not apply to allosteric enzymes. Those have a different property, which will be explained.