Michaelis–Menten Enzyme Kinetics: Derivation and Interpretation
A steady-state derivation applied to a biological catalyst, giving two constants whose meanings are constantly confused.
The mechanism
The enzyme binds the substrate reversibly, and the complex then converts to product, releasing the enzyme unchanged. The key observation the model must explain is saturation: at low substrate concentration the rate rises roughly linearly, but at high concentration it plateaus.
The derivation
Apply the steady-state approximation to the complex:
Total enzyme is conserved, so [E] = [E]0 − [ES]. Substituting and rearranging:
Since the rate of product formation is k2[ES]:
What the two constants mean
| Limit | Rate | Interpretation |
|---|---|---|
| [S] << KM | v ≈ (Vmax/KM)[S] | First order in substrate — most enzyme is free |
| [S] = KM | v = Vmax/2 | This is the operational definition of KM |
| [S] >> KM | v ≈ Vmax | Zero order — the enzyme is saturated |
A low KM means the enzyme reaches half its maximum rate at low substrate concentration, so it binds effectively. Vmax depends on total enzyme concentration, so it is not an intrinsic property; dividing by [E]0 gives the turnover number k2, which is.
Linearising the data
The Lineweaver–Burk form takes reciprocals:
Plotting 1/v against 1/[S] gives a straight line with intercept 1/Vmax and slope KM/Vmax. It is convenient, and it is what most exam questions use, but it has a known statistical weakness: taking reciprocals magnifies the error in measurements at low substrate concentration, which are exactly the least reliable points. Mentioning that limitation is worth a mark where the question invites comment.
Inhibition, and how the plot reveals the type
| Type | Binds to | KM | Vmax | Lineweaver–Burk signature |
|---|---|---|---|---|
| Competitive | Free enzyme, at the active site | Increases | Unchanged | Lines meet on the y-axis |
| Non-competitive | Enzyme and complex, elsewhere | Unchanged | Decreases | Lines meet on the x-axis |
| Uncompetitive | The complex only | Decreases | Decreases | Parallel lines |
The reasoning behind competitive inhibition is worth being able to state: the inhibitor competes for the active site, so adding enough substrate outcompetes it and the maximum rate is still reachable — hence Vmax is unchanged while more substrate is needed to reach half of it.
Frequently asked questions
Why does the rate saturate at high substrate concentration?
Because every enzyme molecule is bound, so the rate is limited by how fast the complex converts to product. Adding more substrate cannot help when there is no free enzyme.
Is the steady-state assumption always valid here?
It requires substrate to be in large excess over enzyme, which is usual in practice. Early in the reaction, before the complex builds up, it does not hold.
What does a very high turnover number indicate?
That the catalytic step is extremely fast. Some enzymes approach the limit set by how quickly substrate can diffuse to them, and are described as diffusion-controlled.
How is this connected to the steady-state approximation in ordinary kinetics?
It is the same approximation applied to a different intermediate. Recognising that makes the derivation reproducible rather than memorised.
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