⚗️ Michaelis, Menten Enzyme Kinetics Calculator

v = Vmax·[S] / (Km + [S]), where v is the reaction velocity, Vmax the maximum velocity at saturating substrate, [S] the substrate concentration, and Km the substrate concentration that gives half-maximal velocity.

Km is the substrate concentration at which the reaction runs at half of Vmax. A low Km means the enzyme reaches half-maximal rate at low substrate, often interpreted as higher apparent affinity for the substrate.

The velocity is exactly half of Vmax. You can see this by substituting [S] = Km into the equation: v = Vmax·Km/(2Km) = Vmax/2. The tool shows the fraction of Vmax so you can check this.

As [S] greatly exceeds Km, the velocity approaches Vmax and levels off, the enzyme is saturated and adding more substrate barely increases the rate.

Measure velocity at several substrate concentrations and fit the Michaelis, Menten curve (nonlinear regression), or linearise with a Lineweaver, Burk (double-reciprocal) plot. This calculator uses Vmax and Km you already have to predict velocity.

Any consistent units: [S] and Km in the same concentration unit, and v comes out in the same units as Vmax. The equation is unit-agnostic as long as [S] and Km match.

With Vmax = 100 µM/s, Km = 25 µM and [S] = 25 µM: v = 100 × 25 / (25 + 25) = 2500/50 = 50 µM/s, exactly half of Vmax, as expected when [S] equals Km.

kcat, the turnover number, is Vmax divided by the total enzyme concentration [E]t: kcat = Vmax/[E]t. It is the number of substrate molecules one active site converts per second at saturation. The ratio kcat/Km measures catalytic efficiency and is often used to compare enzymes.

It assumes a single substrate, an established steady state where the enzyme, substrate complex forms and breaks down at constant concentration, negligible product back-reaction (initial-rate conditions), and substrate in large excess over enzyme. Allosteric or cooperative enzymes follow a sigmoidal curve instead and need the Hill equation.

The equation describes the rate at the start of the reaction, before substrate is appreciably depleted and before product builds up enough to slow things down or run the reverse reaction. Measuring the initial slope (the first few percent of substrate conversion) keeps [S] essentially constant and the steady-state assumption valid, so the v you enter should be an initial rate.

A competitive inhibitor raises the apparent Km (more substrate is needed to reach half-maximal rate) but leaves Vmax unchanged, because high substrate outcompetes it. A pure non-competitive inhibitor lowers Vmax while leaving Km unchanged. Uncompetitive inhibition lowers both. Recognising which pattern your data show is how inhibitor mechanisms are classified.