Hong Life CalculatorsUnderstanding the Michaelis-Menten enzyme kinetics calculator — concept and principle
The Michaelis-Menten equation describes how the initial rate of an enzyme-catalysed reaction rises with substrate concentration and levels off at a maximum. This calculator returns the velocity v for a chosen [S], Vmax and Km, and lets you add a reversible inhibitor of competitive, non-competitive or uncompetitive type.
How it is calculated
- No inhibitor: v = Vmax × [S] ÷ (Km + [S]).
- Inhibition factor: α = 1 + [I] ÷ Ki (α = 1 when Ki is 0 or [I] is 0).
- Competitive: v = Vmax[S] ÷ (αKm + [S]) — apparent Km rises, Vmax unchanged.
- Non-competitive (pure): v = (Vmax/α)[S] ÷ (Km + [S]) — Vmax falls, Km unchanged.
- Uncompetitive: v = Vmax[S] ÷ (Km + α[S]) — both apparent Vmax and Km fall by α.
v takes the units of Vmax (for example µmol/min); Km, [S], [I] and Ki must share one concentration unit.
The principle
The model assumes enzyme E binds substrate S to form a complex ES, which breaks down to product. If the substrate is in large excess and ES reaches a steady state, the fraction of enzyme in the ES form is [S] ÷ (Km + [S]), and the rate is that fraction times Vmax. Km is therefore the substrate concentration at which v is exactly half of Vmax. Inhibitors either compete with S for the active site, bind ES only, or bind both forms equally, which changes the equation in the three ways above.
Worked example
- Inputs: Vmax = 100, Km = 2, [S] = 5, no inhibitor (Ki = 0, [I] = 0), so α = 1.
- Denominator: Km + [S] = 2 + 5 = 7.
- v = 100 × 5 ÷ 7 = 71.43 (in Vmax units).
- Fraction of Vmax: 71.43 ÷ 100 = 71.43%.
| Vmax | Velocity v |
|---|---|
| 25 | 17.85714286 |
| 50 | 35.71428571 |
| 75 | 53.57142857 |
| 100 | 71.42857143 |
| 125 | 89.28571429 |
| 150 | 107.1428571 |
| 200 | 142.8571429 |
How to read the result
The ratio [S]/Km tells you where you are on the curve: at [S] = Km, v = 50% of Vmax; at [S] = 4 Km, 80%; at [S] = 9 Km, 90%. Below Km the rate is nearly proportional to [S]; far above Km the enzyme is saturated and adding substrate barely helps. A low Km means the enzyme reaches half-maximal rate at low substrate concentration.
Common mistakes
- Mixing concentration units, such as Km in mM and [S] in µM.
- Applying the equation to rates measured after substrate is depleted rather than initial rates.
- Confusing pure non-competitive with mixed inhibition, which changes both Km and Vmax by different factors.
Frequently asked questions
How do I find Vmax and Km from data?
Measure initial rates at several [S] and fit the equation by non-linear regression. Lineweaver-Burk double-reciprocal plots are useful for visualising inhibition type but distort errors.
Why does a competitive inhibitor not change Vmax?
At very high [S] the substrate outcompetes the inhibitor for the active site, so the maximum rate is still reached.
What is kcat?
kcat = Vmax ÷ [E]total, the number of substrate molecules each active site converts per unit time.
Sources: L. Michaelis and M. L. Menten, Biochemische Zeitschrift 49, 333–369 (1913) · G. E. Briggs and J. B. S. Haldane, Biochemical Journal 19, 338–339 (1925) · D. L. Nelson and M. M. Cox, Lehninger Principles of Biochemistry, chapter on enzymes