Biology · Introductory biology · Worked example
Find Km and Vmax from enzyme rate data
An enzyme assay gives initial rates of 4.1, 6.5, 10.2, 13.1, 16.2 and 17.6 µmol/min at 1, 2, 4, 8, 16 and 32 mM substrate. Estimate Vmax and Km, and predict the rate at 10 mM.
Look at the shape
Doubling [S] from 1 to 2 mM raises the rate by more than half, but doubling it from 16 to 32 mM adds under 10%. The rate is leveling off: the enzyme is approaching saturation.
| [S] (mM) | v (µmol/min) |
|---|---|
| 1 | 4.1 |
| 2 | 6.5 |
| 4 | 10.2 |
| 8 | 13.1 |
| 16 | 16.2 |
| 32 | 17.6 |
Look past the largest rate
The largest rate, 17.6 µmol/min, is not Vmax. The curve is still rising at 32 mM, so Vmax is somewhat higher.
Fit the Michaelis–Menten curve
A least-squares fit of v = Vmax[S]/(Km + [S]) to all six points gives Vmax ≈ 19.9 µmol/min and Km ≈ 3.96 mM, with R² ≈ 0.999. Every residual is smaller than 0.25 µmol/min.
Check what Km means
At [S] = Km, the fitted rate should be half of Vmax.
Predict the rate at 10 mM
Substitute [S] = 10 mM into the fitted equation.
Result
Vmax ≈ 19.9 µmol/min and Km ≈ 3.96 mM; the predicted rate at 10 mM is about 14.3 µmol/min.
Your turn
For the same enzyme, at what substrate concentration is the rate three quarters of Vmax?
Show the answer and explanation
At [S] = 3Km ≈ 11.9 mM.
Set Vmax[S]/(Km + [S]) = ¾Vmax. Then 4[S] = 3Km + 3[S], so [S] = 3Km = 3(3.96) ≈ 11.9 mM.
Keep exploring
The enzyme kinetics tool opens with these six points and fits the curve. Move the exploration sliders to see how Vmax and Km each change its shape.
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