Deriving the Michaelis-Menten Equation from the Steady State
Balances formation and loss of the enzyme-substrate complex in a three-rate-constant model. Uses total enzyme concentration to derive the Michaelis-Menten equation, states the initial-rate assumptions, and explains when Km approximates a binding dissociation constant.
What a learner can do afterwards
- Derives the rate equation from the steady-state assumption and states where that assumption holds.
- Explains the Michaelis constant in terms of the rate constants that make it up.
- Explains what the maximum rate depends on and why adding more substrate cannot raise the rate above it.
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Use a one-site model: E + S forms ES (k1); ES returns to E + S (k-1) or forms E + P (k2). E is free enzyme, S substrate and P product. Brackets mean concentration. k-1 names the reverse constant, not subtraction. Total active enzyme is [E]T = [E] + [ES]. ES forms at rate k1[E][S] and is lost at rate (k-1 + k2)[ES]. At steady state, formation and total loss balance, so [ES] changes very little.
Measure initial rates after the brief buildup of ES, before much substrate is consumed or product accumulates. Substrate greatly exceeding total enzyme is a usual sufficient condition for the steady-state approximation. Steady state means [ES] is nearly constant during measurement. It need not be small relative to total enzyme. Immediately after mixing, ES may still be building up.
Substitute [E] = [E]T - [ES]. Expand the balance: k1[E]T[S] - k1[ES][S] = (k-1 + k2)[ES]. Add k1[ES][S] to both sides, then factor: k1[E]T[S] = (k1[S] + k-1 + k2)[ES]. With Km = (k-1 + k2)/k1, divide to obtain [ES] = [E]T[S]/(Km + [S]).
Divide the collected balance by k1: [E]T[S] = (Km + [S])[ES]. Divide by Km + [S] to isolate [ES]. Product forms at v = k2[ES], so multiplying gives v = k2[E]T[S]/(Km + [S]). With Vmax = k2[E]T, v = Vmax[S]/(Km + [S]). At [S] = Km, the substrate fraction is Km/(2Km) = 1/2.
Km = (k-1 + k2)/k1 includes dissociation and product formation. If only k2 rises, Km rises. The binding dissociation constant is Kd = k-1/k1. In the rapid-equilibrium approximation, k2 is much smaller than k-1, so Km is approximately Kd. A small Km alone does not prove tight binding.
Vmax = k2[E]T. With positive Km, [S]/(Km + [S]) stays below 1 at finite substrate concentration, so v approaches Vmax from below. Near saturation, most enzyme is in ES and turnover limits rate. Doubling active enzyme doubles Vmax if the other conditions stay fixed.
Balancing ES gives v = Vmax[S]/(Km + [S]), with Vmax = k2[E]T and Km = (k-1 + k2)/k1. Km approximates Kd when k2 is much smaller than k-1.
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Account for both exits from ES
The one-site model has three rate constants. E and S associate to form ES with constant k1. ES can dissociate to E and S with constant k-1, or produce E and P with constant k2. The name k-1 denotes the reverse constant, not subtraction.
Formation has rate k1[E][S]. Total loss has rate (k-1 + k2)[ES]. At steady state, these rates approximately balance. Individual complexes still form and disappear. Their nearly constant total concentration does not mean that binding or product formation has stopped.
Use total enzyme to solve the balance
Total active enzyme is free enzyme plus complex: [E]T = [E] + [ES]. Therefore [E] = [E]T - [ES]. Insert this expression into the formation term of the steady-state balance.
Expanding and collecting terms gives k1[E]T[S] = (k1[S] + k-1 + k2)[ES]. Divide by k1 and define Km = (k-1 + k2)/k1. Solving gives [ES] = [E]T[S]/(Km + [S]). Product forms at rate v = k2[ES], so v = Vmax[S]/(Km + [S]), where Vmax = k2[E]T.
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The ES concentration may rise briefly after enzyme and substrate are mixed. For this derivation, measure initial rate after that buildup but before much substrate has been consumed or much product has accumulated. The model then neglects reverse reaction from product.
Substrate greatly exceeding total enzyme is a usual sufficient condition for the steady-state approximation. ES must change little during measurement, but it need not be a small fraction of the enzyme. Near saturation, most enzyme can be in ES while the complex concentration remains nearly constant.
Distinguish binding from product formation
For reversible binding, the dissociation constant is Kd = k-1/k1. The Michaelis constant is Km = (k-1 + k2)/k1. Only Km contains the product-formation constant. Increasing k2 therefore raises Km when k1 and k-1 stay fixed, without changing Kd.
If k2 is much smaller than k-1, its contribution is small and Km is approximately Kd. This is the rapid-equilibrium approximation. Vmax describes another limit: Vmax = k2[E]T. As substrate concentration increases, the rate approaches that limit. Increasing active enzyme raises Vmax when the rate constants and other conditions stay fixed.
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