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Michaelis–Menten kinetics calculator

Enzyme reaction rate at a given substrate concentration.

Published 9 August 2026 · Updated 24 September 2026

What this calculator does

The Michaelis-Menten equation gives reaction rate as v = Vmax[S] ÷ (Km + [S]). It produces a hyperbolic curve: rate rises steeply with substrate at low concentrations, then flattens as the enzyme becomes saturated and approaches Vmax without ever quite reaching it.

Km is the substrate concentration at which the rate is exactly half of Vmax, and that is its definition rather than a consequence. Substituting [S] = Km into the equation gives Vmax·Km ÷ 2Km, which is Vmax over 2 regardless of what the numbers are. A low Km means the enzyme reaches half speed at low substrate, which is usually described as high affinity.

The formula

Formulav = Vmax[S] / (Km + [S])

Substrate concentration is multiplied by Vmax and divided by the sum of Km and the substrate concentration. The percentage of Vmax follows from the same expression, and depends only on the ratio of substrate to Km rather than on their absolute values. The rate at [S] = Km is reported separately as a check, and always comes to exactly half of Vmax.

TermMeaning
VmaxMaximum rate, approached when the enzyme is fully saturated with substrate.
KmThe Michaelis constant: substrate concentration giving half of Vmax. Lower means higher apparent affinity.
[S]Substrate concentration.
SaturationThe flattening of the curve at high substrate, where adding more makes almost no difference.

The inputs explained

FieldWhat to enter
Vmax (µmol/min)Maximum reaction velocity, in whatever rate units your assay uses.
Km (mM)The Michaelis constant, in the same concentration units as the substrate.
Substrate concentration [S] (mM)Substrate concentration, in the same units as Km.

When to use it

Predicting rate at a given substrate level

The direct use of the equation: given the two constants from a characterisation experiment, find the rate at any concentration.

Choosing assay conditions

Assays measuring Vmax need saturating substrate, and the table shows how much is actually required: even ten times Km only reaches about 91% of maximum.

Comparing two enzymes

Km comparisons describe affinity. An enzyme with Km of 1 mM reaches half speed at a tenth the substrate of one with Km of 10 mM.

Worked examples

Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.

How does reaction rate change with substrate concentration?

A range of substrate concentrations against a fixed Vmax and Km.

Vmax 100, Km 10
Substrate [S]Reaction rate v% of VmaxRate at [S]=Km
1 mM9.091 µmol/min9.09%50.000 µmol/min
5 mM33.333 µmol/min33.3%50.000 µmol/min
10 mM50.000 µmol/min50.0%50.000 µmol/min
20 mM66.667 µmol/min66.7%50.000 µmol/min
50 mM83.333 µmol/min83.3%50.000 µmol/min
100 mM90.909 µmol/min90.9%50.000 µmol/min
At [S] = Km = 10 the rate is exactly 50.000, half of Vmax, which is the definition of Km rather than a coincidence. The saturation is slow: ten times Km reaches only 90.9% of Vmax, so genuinely saturating an enzyme needs far more substrate than intuition suggests.

Questions

What does Km actually mean?

The substrate concentration at which the reaction runs at half its maximum rate. It is often described as an inverse measure of affinity: a low Km means half speed is reached at low substrate, which implies the enzyme binds it readily. Strictly it combines binding and catalytic steps, so it is an apparent affinity rather than a binding constant.

Why does the rate never reach Vmax?

Because the equation approaches it asymptotically. Vmax would require infinite substrate concentration, since there is always some fraction of enzyme not currently bound. In practice anything above about 90% of Vmax is treated as saturating, which needs roughly ten times Km.

What are the assumptions behind the equation?

A single substrate, a steady state in which the enzyme-substrate complex concentration is constant, substrate in large excess over enzyme, and no product inhibition or cooperativity. Enzymes with multiple substrates or allosteric behaviour need different models, and cooperative enzymes give a sigmoid curve rather than a hyperbola.

How are Vmax and Km measured?

By measuring initial rates across a range of substrate concentrations and fitting the curve. Historically this was done with a Lineweaver-Burk double reciprocal plot, which linearises the data but distorts the error structure badly. Non-linear regression on the original curve is the modern and better method.

For temperature effects on reaction rate, see the Q10 coefficient calculator. For microbial population growth, see the bacterial growth calculator.