# Michaelis-Menten Equation Calculator URL: https://speedycalculator.com/calculators/chemistry/michaelis-menten-equation/ Last updated: 2026-10-30 > The Michaelis–Menten calculator finds the initial reaction rate of an enzyme from Vmax, Km and the substrate concentration – or solves for Km, Vmax or [S] instead – and draws the saturation curve, with kcat and catalytic efficiency when you know the enzyme concentration. ## Inputs - Solve for - Michaelis constant (Km) (µM, mM, M, nM) - Maximum reaction rate (Vmax) (µM/min, µM/s, mM/min, mM/s, M/s, nM/min, nM/s) - Substrate concentration [S] (µM, mM, M, nM) - Measured reaction rate (v) (µM/min, µM/s, mM/min, mM/s, M/s, nM/min, nM/s) - Total enzyme concentration [E]ₜ (optional, 0 = skip) (µM, mM, M, nM) — Concentration of active sites; used for kcat and kcat/Km. ## Results - Reaction rate v - Michaelis constant Km - Maximum rate Vmax - Substrate concentration [S] - Saturation curve - Rate as a share of Vmax (saturation) - [S] ÷ Km - Rate at [S] = Km (Vmax ÷ 2) - Substrate needed for 90% of Vmax (9 × Km) - Turnover number kcat - Catalytic efficiency kcat/Km - Kinetic regime ## 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 ``` ## Explanation 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] ÷ Kmv ÷ VmaxWhat it means0.19.1%Nearly first order – rate ∝ [S]0.533.3%Rate still very sensitive to [S]150%Definition of Km375%Mixed region990%Close to saturation9999%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. ## 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. ## FAQ **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. ## Sources - [OpenStax – Biology 2e](https://openstax.org/details/books/biology-2e) - [LibreTexts Chemistry](https://chem.libretexts.org/) - [IUPAC Compendium of Chemical Terminology (Gold Book)](https://goldbook.iupac.org/)