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Michaelis-Menten Calculator

Result

71.4286µmol/min

Result: 71.4286 µmol/min
How the result movesmM → µmol/min

The reaction velocity is v = Vmax × [S] ÷ (Km + [S]). At [S] = Km the enzyme runs at exactly half of Vmax; far above Km it approaches Vmax and more substrate stops helping. Km and [S] must share one concentration unit — the velocity comes out in the unit of Vmax.

Worked examples

Case 1
Maximum velocity Vmax
100µmol/min
Michaelis constant Km
2mM
Substrate concentration [S]
5mM

71.4286µmol/min

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Case 2
Maximum velocity Vmax
100µmol/min
Michaelis constant Km
2mM
Substrate concentration [S]
2mM

50.0000µmol/min

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Case 3
Maximum velocity Vmax
200µmol/min
Michaelis constant Km
5mM
Substrate concentration [S]
15mM

150.0000µmol/min

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How it's calculated

v = (Vmax × [S]) ÷ (Km + [S])

  1. StepEnter Vmax, the rate the enzyme reaches at full saturation.
  2. StepEnter Km and the substrate concentration in the same unit.
  3. ResultRead the velocity — at [S] = Km it is exactly half of Vmax.

What this number means

An enzyme has no single speed. It runs at whatever rate the substrate on offer allows, and this page gives that rate for one particular offer. The equation splits cleanly in two: Vmax sets the scale, and the ratio of [S] to Km decides what fraction of that scale you get. The defaults show it. With Km = 2 mM and [S] = 5 mM the substrate sits at 2.5 times Km, so the enzyme runs at 2.5 ÷ 3.5 = 0.7143 of full speed — against a Vmax of 100 µmol/min, 71.4286 µmol/min. Any other pair with the same ratio yields the same fraction, which is why Km and [S] only ever have to match each other and never Vmax. What the result does not contain is the two numbers it was built from. Km and Vmax are measured, and measured afresh for every enzyme, substrate, temperature and pH; the page stores neither, because there is no single correct value to store. You bring them, usually from a fitted curve of your own, and the velocity is only as good as that fit. The equation then assumes a steady state: one substrate, a complex forming and breaking down at a constant level, nothing inhibiting. Two substrates, an allosteric enzyme or a competitive inhibitor all bend the same curve, and the arithmetic here is then the wrong tool rather than an imprecise one.

Km is a concentration, not a speed

Km is the substrate concentration at which the reaction runs at exactly half of Vmax, so it carries the unit of [S]. A low Km means half speed is reached at little substrate, usually read as tight binding.

Km and [S] must share one unit

Only the ratio of Km and [S] enters the formula, so both have to use the same concentration unit. The velocity then comes out in whatever unit Vmax was given in.

Far above Km, more substrate stops helping

At 500 times Km the velocity is 99.8004 against a Vmax of 100 — the curve only approaches Vmax. Adding substrate there changes almost nothing.

The equation assumes a steady state

It holds for a single substrate, an enzyme-substrate complex forming and breaking down at a constant level, and no inhibitors. Multiple substrates, allosteric regulation or inhibition need extended models.

Commonly misread

A low Km means the enzyme works fast.

Km is a concentration, not a rate: it says at what substrate level half of Vmax is reached. The speed itself is set by Vmax.

Doubling the substrate doubles the velocity.

At [S] = Km the enzyme is already at half of Vmax, and at 500 times Km it reaches 99.8004 out of 100. The curve flattens as it approaches Vmax.

Vmax in µmol/min forces Km and [S] into mM.

Only Km and [S] have to match each other. Enter both in µM and the velocity still comes out in the unit of Vmax.

Reference table

Vmax, Km, [S]Substrate against KmVelocity (µmol/min)
100, 2, 0no substrate0.0000
100, 2, 2equal to Km50.0000
50, 1, 1equal to Km25.0000
100, 2, 52.5 times Km71.4286
200, 5, 153 times Km150.0000
10, 0.5, 1020 times Km9.5238
100, 2, 1000500 times Km99.8004

Questions

How do I calculate the Michaelis-Menten reaction velocity?

Multiply Vmax by the substrate concentration and divide by the sum of Km and that concentration: v = (Vmax × [S]) ÷ (Km + [S]). With Vmax = 100 µmol/min, Km = 2 mM and [S] = 5 mM the velocity is 500 ÷ 7 = 71.4286 µmol/min.

What is the Michaelis constant Km?

Km is the substrate concentration at which the reaction runs at exactly half of Vmax. A low Km means the enzyme reaches half speed at little substrate, usually read as tight binding; a high Km means more substrate is needed.

What does Vmax represent?

Vmax is the maximum rate the enzyme reaches when every active site is occupied. Beyond that point more substrate no longer speeds the reaction up, and the velocity only approaches Vmax.

Which units do Vmax, Km and [S] need?

Km and the substrate concentration must use the same concentration unit, because only their ratio enters the formula. The velocity then comes out in whatever unit Vmax was given in — here µmol/min against mM.

What assumptions does the equation make?

A single substrate, a steady state in which the enzyme-substrate complex forms and breaks down at a constant level, and no inhibitors. Multiple substrates, allosteric regulation or inhibition need extended models.

Sources and last check

  1. chem.libretexts.org

Information, not professional advice.