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Michaelis–Menten Equation Calculator

Calculate an enzyme’s initial reaction rate or solve for its Michaelis constant from compatible kinetic inputs.

By ChemistryCalc•Updated October 28, 2026•References
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Results update as you type. Choose units beside a measurement; changing units converts the current value.

On this pageCalculatorHow to useFormula & methodWorked exampleReading the saturation curveRespect the kinetic assumptionsFrequently asked questionsRelated calculatorsSources

01 Set up your calculation

  1. Choose whether to calculate an initial rate or infer Km.
  2. Enter Vmax and substrate concentration for the same assay conditions.
  3. Supply Km or the measured initial rate and inspect the calculated value.

02 The relationship

v = Vmax[S]/(Km + [S]); Km = [S](Vmax/v − 1).

For the assumptions behind these results, read the explanation below. You can also explore Enzyme Activity and Molarity.

WORKED EXAMPLE

Follow the numbers

With Vmax = 10 µmol/min and [S] = Km = 0.001 mol/L, v = 10 × 0.001/(0.001 + 0.001) = 5 µmol/min.

03 Reading the saturation curve

The Michaelis–Menten relationship describes a hyperbolic approach to a maximum initial rate. At [S] = Km, the rate is exactly half of Vmax. At much lower substrate concentration, rate is approximately proportional to substrate concentration.

The rate and Vmax share the same units, while Km and substrate concentration share concentration units. Vmax depends on the amount of active enzyme in the assay; it is not an intrinsic constant independent of enzyme loading.

04 Respect the kinetic assumptions

The simple model is intended for suitable initial-rate measurements where substrate depletion and product effects are small. Cooperative behavior, substrate inhibition, multiple interacting substrates, and enzyme instability can require other rate equations.

Solving for Km from one point assumes Vmax is already known reliably. A fitted series of initial rates over a suitable substrate range is usually a stronger basis for parameter estimation. Km is not universally identical to a binding dissociation constant.

Questions from the notebook

Why can a finite positive Km not give v above Vmax?

The fraction [S]/(Km + [S]) stays below one for positive Km.

Does zero substrate give zero rate?

Yes, in this simple model.

Does doubling enzyme amount double the prediction?

If that doubles Vmax while Km and conditions stay the same, the modeled rate doubles.

References & reading

Our explanations are written for ChemistryCalc. Read how we build and check our tools. These calculations are educational estimates where the underlying model requires assumptions.