Enzyme Kinetics Calculator

What the Michaelis-Menten equation tells you

The Michaelis-Menten equation describes how fast an enzyme-catalysed reaction runs as a function of substrate concentration. At low substrate the enzyme has plenty of free active sites, so the rate climbs almost in proportion to how much substrate you add. As substrate rises, the active sites fill up and the rate levels off toward a ceiling called Vmax. This calculator evaluates that curve at one substrate concentration, so you can predict a reaction rate or check a value read from a plot.

Use it when you already have estimates of Vmax and Km, for example from a Lineweaver-Burk or nonlinear fit, and want to know the expected velocity at a new concentration. It is also handy for planning an assay: choosing a substrate concentration near Km keeps the rate sensitive to inhibitors, while a concentration several times higher than Km gives a rate close to Vmax and makes the reading less sensitive to small pipetting errors. The tool does not fit Vmax or Km from raw data, and it assumes a single substrate, steady-state conditions and no inhibitor.

Formula and variables

v = Vmax × [S] / (Km + [S])

v is the initial reaction velocity, in the same rate units you used for Vmax (for example µmol per minute).
Vmax is the maximum velocity reached when the enzyme is saturated with substrate.
Km is the Michaelis constant, the substrate concentration at which v equals exactly half of Vmax. It is expressed in concentration units.
[S] is the substrate concentration, which must use the same concentration units as Km.

Two useful limits follow from the equation. When [S] is much smaller than Km, v is approximately (Vmax/Km) × [S], a straight line. When [S] is much larger than Km, v approaches Vmax. The calculator also reports v as a percentage of Vmax, which is simply [S] / (Km + [S]).

Worked example

Suppose an enzyme has Vmax = 100 µmol/min and Km = 2 mM, and you run the assay at [S] = 10 mM.

Numerator: 100 × 10 = 1000.
Denominator: 2 + 10 = 12.
v = 1000 / 12 = 83.3333 µmol/min.

Entering 100, 2 and 10 gives “Reaction Rate: 83.3333” and 83.3% of Vmax. As a sanity check, setting [S] equal to Km (2 mM) gives 100 × 2 / 4 = 50, exactly half of Vmax, which confirms the meaning of Km.

Common mistakes and how to interpret the result

Mixing concentration units. Km in mM with [S] in µM shifts the answer by a factor of 1000. Convert first.
Reading the output as a unit-bearing number. The result carries whatever rate unit Vmax had; the calculator does not label it.
Expecting a full-saturation reading at [S] = 5 Km. Even then the rate is only 83% of Vmax; you need about 20 Km to reach 95%.
Ignoring inhibitors. Competitive inhibitors raise the apparent Km and non-competitive inhibitors lower the apparent Vmax, so use the apparent values measured under those conditions.

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Frequently Asked Questions

What is Km in plain terms?

Km is the substrate concentration at which the reaction runs at half of Vmax. A small Km means the enzyme reaches high rates at low substrate, which is often described as high apparent affinity, although Km also depends on the catalytic rate.

Can I use this to calculate Vmax or Km?

Not directly. The calculator solves for v given Vmax, Km and [S]. To estimate Vmax and Km you would fit several (substrate, rate) measurements with a nonlinear regression or a linearised plot, then enter the fitted values here.

Why does the rate never exceed Vmax?

Vmax is reached when every enzyme molecule is occupied with substrate, so adding more substrate cannot speed things up further. The equation approaches Vmax asymptotically but never passes it.

Do the units of Vmax and Km have to match?

The rate unit is free, but Km and [S] must share the same concentration unit. The ratio [S]/(Km + [S]) is dimensionless only when both are in the same unit.