ChemBench

Michaelis-Menten Calculator

The central equation of enzyme kinetics — how fast an enzyme works depends on how much substrate is available.

Vmax

[S] (substrate)

Reaction velocity

66.667

% of Vmax: 66.667

What you entered

Vmax: 100Km: 5[S] (substrate): 10

1
Km + [S]
5 + 10= 15
2
v = Vmax × [S] ÷ (Km + [S])
100 × 10 ÷ 15= 66.6667 mol/L/s
3
Fraction of Vmax
66.6667 ÷ 100 × 100= 66.6667%

Result

Reaction velocity: 66.667

At [S] = 10 mM (Km = 5 mM), the enzyme operates at 66.7% of its maximum velocity (v = 66.667 μmol/min).

Enzyme velocity across substrate concentrations

Reaction velocity is Vmax times substrate concentration over Km plus substrate concentration. At [S] equal to Km the enzyme runs at exactly half its maximum rate.

[S]	[S] relative to Km	Velocity	Share of Vmax
1	0.1 x Km	9.091	9.1%
5	0.5 x Km	33.333	33.3%
10	1 x Km	50.000	50.0%
20	2 x Km	66.667	66.7%
50	5 x Km	83.333	83.3%
100	10 x Km	90.909	90.9%

Rows use Vmax 100 and Km 10. The curve saturates rather than plateauing sharply: ten times Km still only reaches 91% of Vmax, which is why Vmax cannot be measured directly and is extrapolated instead.

Saturation kinetics explained

At low substrate concentrations, velocity increases nearly linearly — the enzyme has plenty of free active sites. As substrate concentration rises, the enzyme saturates and velocity plateaus at Vmax. The Km value is the substrate concentration at which velocity is exactly half of Vmax, and it measures the enzyme's affinity for its substrate.

Why Km matters in pharmacology

A low Km means the enzyme binds substrate tightly (high affinity) and reaches half-max velocity at low concentrations. Drug designers target enzymes by comparing their drug's binding affinity to the natural substrate's Km — competitive inhibitors effectively increase the apparent Km.

Frequently asked questions

My enzyme has Vmax = 100 μmol/min and Km = 5 mM. At [S] = 10 mM, what is the reaction velocity?

v = Vmax × [S] / (Km + [S]) = 100 × 10 / (5 + 10) = 66.7 μmol/min. The enzyme is at two-thirds of its maximum speed.

What happens at [S] = Km?

v = Vmax × Km / (Km + Km) = Vmax/2. By definition, Km is the substrate concentration at which the enzyme runs at exactly half its maximum velocity.

How does a competitive inhibitor affect Km and Vmax?

A competitive inhibitor increases the apparent Km (the enzyme needs more substrate to reach half-max speed) but does not change Vmax — with enough substrate, you can always overcome a competitive inhibitor.

How is Michaelis-Menten related to the rate law calculator?

Rate law describes simple chemical kinetics (zeroth, first, second order). Michaelis-Menten is specifically for enzyme-catalyzed reactions, which follow saturation kinetics — at low [S] the reaction looks first-order, at high [S] it looks zeroth-order.