Michaelis-Menten Calculator
Find an enzyme’s reaction rate from Vmax, Km, and the substrate concentration.
At this substrate level the enzyme runs at 50% of its maximum rate Vmax = 100. Half of Vmax is reached exactly when [S] = Km = 5.
The rate climbs steeply, then plateaus toward Vmax. The dashed line marks v = Vmax ÷ 2, the half-maximal rate reached when [S] = Km.
The Michaelis-Menten equation gives an enzyme’s rate as v = Vmax · [S] ÷ (Km + [S]). With Vmax = 100, Km = 5, and [S] = 5, you get v = 100 · 5 ÷ 10 = 50 — exactly half of Vmax. The rate always equals Vmax ÷ 2 when the substrate concentration equals Km.
What the Michaelis-Menten model describes
The Michaelis-Menten equation predicts the initial rate v of an enzyme-catalysed reaction from the substrate concentration [S]. Two constants define the enzyme: Vmax, the maximum rate reached when every active site is occupied, and Km, the Michaelis constant — the substrate concentration at which the enzyme runs at half of Vmax. At low [S] the rate rises almost linearly with substrate; at high [S] it plateaus toward Vmax as the enzyme becomes saturated.
Reaction rate v from Vmax, the Michaelis constant Km, and substrate concentration [S].
Worked example
An enzyme has Vmax = 100 µmol/min and Km = 5 mM. What is the rate when the substrate concentration is [S] = 15 mM?
- 1 Write the equation. v = Vmax · [S] ÷ (Km + [S]). You need Vmax, Km, and the current substrate concentration [S].
- 2 Substitute the values. v = 100 · 15 ÷ (5 + 15). Here Vmax = 100, Km = 5, and [S] = 15.
- 3 Simplify the denominator. Km + [S] = 5 + 15 = 20, and Vmax · [S] = 100 · 15 = 1500.
- 4 Divide to get the rate. v = 1500 ÷ 20 = 75. The enzyme runs at 75, which is 75% of Vmax at this substrate level.
Substrate level vs fraction of Vmax
How the fraction of Vmax rises with [S] measured relative to Km.
| Substrate [S] | Rate v | Fraction of Vmax |
|---|---|---|
| 0.5 · Km | Vmax ÷ 3 | ≈ 33% |
| Km | Vmax ÷ 2 | 50% |
| 2 · Km | 2·Vmax ÷ 3 | ≈ 67% |
| 10 · Km | 10·Vmax ÷ 11 | ≈ 91% |
How to read Km and the plateau
Km is the substrate concentration at half Vmax. It measures how much substrate the enzyme needs to work at half speed, so a low Km means high affinity — the enzyme reaches half its maximum rate with very little substrate. A high Km means the enzyme needs a lot of substrate to get going.
The rate plateaus because active sites saturate. As [S] climbs, more and more enzyme molecules are occupied at any instant; once nearly all sites are filled, adding substrate can barely raise the rate, so v approaches — but never quite reaches — Vmax.
Assumptions. The equation gives the initial rate under steady-state conditions with substrate in large excess over enzyme, no product inhibition, and no cooperativity between active sites. Enzymes that deviate — such as cooperative, multi-subunit ones — follow a sigmoidal curve instead and need a different model.