Michaelis–Menten Calculator
Enter maximum velocity Vmax, Michaelis constant Km, and substrate concentration [S] to compute the enzyme-catalyzed reaction rate v and its fraction of Vmax, for enzyme-kinetics analysis.
Input Data
Results
At a glance:The Michaelis–Menten equation describes the initial rate of a single-substrate enzyme-catalyzed reaction vs substrate concentration: v = Vmax·[S] ÷ (Km + [S]). Vmax is the maximum rate at saturation; Km (Michaelis constant) is the substrate concentration at half Vmax, reflecting substrate affinity (smaller Km = higher affinity). At [S] ≪ Km the reaction is near first-order; at [S] ≫ Km the rate approaches Vmax (zero-order).
Formula
Reaction rate: v = Vmax × [S] ÷ (Km + [S]).
At [S] = Km, v = ½Vmax.
Fraction: v ÷ Vmax = [S] ÷ (Km + [S]).
$$v = \dfrac{V_{max}\,[S]}{K_m + [S]}$$$$\dfrac{v}{V_{max}} = \dfrac{[S]}{K_m + [S]}$$How to Use
- Enter the saturated maximum velocity Vmax (custom unit).
- Enter Km and substrate [S] (same unit for both).
- The right panel shows rate v and its fraction of Vmax instantly.
Rate vs substrate at Vmax=100, Km=5 (same units)
| Substrate [S] | Rate v | Fraction of Vmax |
|---|---|---|
| 1 | 16.67 | 16.7% |
| 5 (= Km) | 50.00 | 50% |
| 10 | 66.67 | 66.7% |
| 25 | 83.33 | 83.3% |
| 50 | 90.91 | 90.9% |
At [S] = Km the rate is exactly half-saturated (½Vmax); even at [S] = 10×Km the rate is only ~91% Vmax — reaching saturation needs very high substrate.
Case Studies
Half-saturation substrate
An enzyme Vmax = 100 µmol/min, Km = 5 mM. At [S] = 5 mM (= Km):
v = 100 × 5 ÷ (5 + 5) = 50 µmol/min, exactly ½Vmax.
This verifies the definition of Km: the substrate concentration at half-maximal rate.
Comparing two enzymes' affinity
Enzyme A: Vmax = 200, Km = 2; Enzyme B: Vmax = 200, Km = 20, compare at [S] = 8.
A: v = 200 × 8 ÷ (2 + 8) = 160; B: v = 200 × 8 ÷ (20 + 8) ≈ 57.1.
At the same substrate, the smaller-Km enzyme A is much faster, showing higher substrate affinity.
FAQ
What does Km represent, and is smaller better?
Km is the substrate concentration at half Vmax, reflecting enzyme–substrate affinity. A smaller Km means the enzyme reaches half-saturation at lower substrate, i.e. higher affinity. But 'better' depends on context: high affinity helps at low substrate, while some regulatory enzymes need a higher Km to be sensitive to substrate changes.
How do Vmax and kcat (turnover number) differ?
Vmax is the maximum rate at a given enzyme amount, proportional to enzyme concentration. kcat = Vmax ÷ [E]total is the max catalytic events per enzyme molecule per second, independent of enzyme amount, reflecting intrinsic efficiency. kcat/Km is the overall catalytic efficiency (specificity constant).
What assumptions underlie Michaelis–Menten?
Main assumptions: measuring initial rate (substrate consumed < 10%, product negligible, no reverse reaction), enzyme and substrate reach steady state quickly, single binding site with no cooperativity. Allosteric or multi-substrate systems need other models (e.g. Hill equation).
Why can't more substrate raise the rate further?
When [S] ≫ Km, nearly all enzyme is occupied (saturated); the rate is then limited by the enzyme's intrinsic turnover, approaching Vmax (zero-order). Adding more substrate provides no more free enzyme, so the rate plateaus.
Must [S] and Km use the same unit?
Yes. They appear together in (Km + [S]), so they must share a unit (e.g. both mM) to be meaningful. Vmax and v share units (e.g. µmol/min). This calculator does not fix units; just keep inputs consistent.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.