Most enzymes speed up as you add substrate, but only up to a point. At low substrate concentration almost every extra substrate molecule finds a free active site, so the rate climbs in proportion to [S]. At high concentration the active sites are nearly always occupied and the rate levels off at the maximum velocity, Vmax. The Michaelis–Menten equation describes this hyperbolic curve with just two parameters: Vmax and the Michaelis constant Km.
Km is the substrate concentration that gives exactly half of Vmax. A small Km means the enzyme reaches high activity at low substrate levels; a large Km means it needs a lot of substrate. Km equals (k₋₁ + k_cat)/k₁ for the scheme E + S ⇌ ES → E + P, so it is only a true dissociation constant when product formation is much slower than ES breaking apart.
Dividing Vmax by the total enzyme concentration gives the turnover number kcat – how many substrate molecules one active site converts per second – and kcat/Km, the catalytic efficiency, which compares enzymes at low substrate concentration. Diffusion-limited enzymes such as catalase and triose phosphate isomerase approach 10⁸–10⁹ M⁻¹·s⁻¹.
How to use the Michaelis-Menten Equation Calculator
- Solve for
- Michaelis constant (Km)
- Maximum reaction rate (Vmax)
- Substrate concentration [S]
- Measured reaction rate (v)
- Total enzyme concentration [E]ₜ (optional, 0 = skip) — Concentration of active sites; used for kcat and kcat/Km.
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Choose what you want to solve for, enter the other three quantities in any concentration or rate unit, and optionally the total enzyme concentration. The calculator rearranges v = Vmax[S]/(Km + [S]), reports how saturated the enzyme is, and plots the curve with Km (half-Vmax point) and your operating point marked.
Worked example
An enzyme has Vmax = 50 µM/min and Km = 20 µM. At [S] = 30 µM the rate is v = 50 × 30 ÷ (20 + 30) = 30 µM/min, 60% of Vmax, and [S]/Km = 1.5. Raising the substrate to 9 × Km = 180 µM would give 45 µM/min (90%). With 0.05 µM enzyme, kcat = 50 µM/min ÷ 0.05 µM = 1,000 min⁻¹ = 16.67 s⁻¹ and kcat/Km = 16.67 s⁻¹ ÷ 20 µM ≈ 8.33 × 10⁵ M⁻¹·s⁻¹. Working backwards, a measured rate of 30 µM/min at 30 µM with Vmax = 50 µM/min gives Km = 30 × (50 − 30) ÷ 30 = 20 µM.
How saturation depends on [S]/Km
| [S] ÷ Km | v ÷ Vmax | What it means |
|---|---|---|
| 0.1 | 9.1% | Nearly first order – rate ∝ [S] |
| 0.5 | 33.3% | Rate still very sensitive to [S] |
| 1 | 50% | Definition of Km |
| 3 | 75% | Mixed region |
| 9 | 90% | Close to saturation |
| 99 | 99% | Practically zero order – rate ≈ Vmax |
Tips
- Keep enzyme and substrate concentrations in the same unit.
- Check that your data lie in the linear range of the standard curve.
- pH strongly affects enzyme activity and protein solubility.
- Everything is calculated in your browser.
Common mistakes to avoid
- Extrapolating a calibration curve far beyond the standards.
- Mixing units (U, kat, µmol/min).
- Ignoring dilution factors when reporting the original concentration.
Formula
v = Vmax·[S] ÷ (Km + [S]) · Km = [S]·(Vmax − v) ÷ v · Vmax = v·(Km + [S]) ÷ [S] · [S] = Km·v ÷ (Vmax − v) · kcat = Vmax ÷ [E]ₜ · efficiency = kcat ÷ Km
Worked example
Inputs — Solve for: Reaction rate v; Michaelis constant (Km): 20 µM; Maximum reaction rate (Vmax): 50 µM/min; Substrate concentration [S]: 30 µM; Total enzyme concentration [E]ₜ (optional, 0 = skip): 0.05 µM.
Results — Reaction rate v = 5.0000e-7; Michaelis constant Km = 2.0000e-5; Maximum rate Vmax = 8.3333e-7; Substrate concentration [S] = 3.0000e-5; Rate as a share of Vmax (saturation) = 60 %; [S] ÷ Km = 1.5.
Key terms
- Vmax
- The rate the enzyme would reach if every active site were occupied all the time.
- Michaelis constant (Km)
- The substrate concentration that gives half of Vmax.
- Turnover number (kcat)
- Substrate molecules converted per active site per second, Vmax ÷ [E]ₜ.
- Catalytic efficiency
- kcat/Km, the second-order rate constant for free enzyme meeting substrate.
- Initial rate
- The rate measured before more than a few per cent of substrate is used up.
Frequently asked questions
What does the Michaelis constant tell you?
Km is the substrate concentration at which the rate is half of Vmax. Values from about 1 µM to 10 mM are common; a lower Km means the enzyme works well at lower substrate levels.
Is a low Km the same as high affinity?
Only roughly. Km = (k₋₁ + kcat)/k₁, so it equals the ES dissociation constant only when kcat is much smaller than k₋₁. It is safer to call Km an apparent affinity.
How much substrate do I need to reach Vmax?
Strictly, Vmax is never reached. At [S] = 9 × Km the rate is 90% of Vmax and at 99 × Km it is 99%.
What units does Vmax have?
A concentration per time (for example µM/min) or an amount per time per mass of enzyme (µmol/min/mg). kcat = Vmax/[E]ₜ has units of s⁻¹.
What is a good kcat/Km?
Typical enzymes have 10³–10⁶ M⁻¹·s⁻¹; values near 10⁸–10⁹ M⁻¹·s⁻¹ mean the reaction is limited only by diffusion.
How do I get Km and Vmax from experimental data?
Measure initial rates at 6–10 substrate concentrations spanning roughly 0.2–5 × Km and fit the hyperbola by non-linear regression. Lineweaver–Burk (1/v vs 1/[S]) plots are useful for display but distort the errors.
Does the equation hold for every enzyme?
It assumes one substrate, steady state, initial rates, [E] much smaller than [S] and no product inhibition. Allosteric enzymes give sigmoidal curves and need the Hill equation instead.