Michaelis–Menten Calculator (Km, Vmax, Fit)
Work with the Michaelis–Menten equation. Solve v = Vmax[S] ÷ (Km + [S]) for the rate, the substrate, Km or Vmax, see how inhibitors shift the curve, get kcat and kcat/Km, or fit Km and Vmax with standard errors to your own data.
How to Use
- Choose what to solve for: the rate v, the substrate [S], Km or Vmax; Inhibition for a rate with an inhibitor; or Fit data to find Km and Vmax from your measurements.
- Enter the known values with their units: concentrations in M, mM, µM or nM and rates in µM/min, mM/min, nM/s and so on.
- Optionally enter the enzyme concentration [E] to get the turnover number kcat and the specificity constant kcat/Km.
- For a fit, paste one “[S], v” pair per line (a header line is fine), pick their units, and the tool fits the curve by non-linear least squares.
- Switch the plot between the saturation curve, a Lineweaver–Burk plot and an Eadie–Hofstee plot; Show Work lists every step and the fit statistics.
Worked Example
Rate and turnover. An enzyme has Vmax = 100 µM/min and Km = 40 µM, and the assay uses [S] = 120 µM with 0.05 µM enzyme. v = 100 × 120 ÷ (40 + 120) = 75 µM/min. kcat = 100 ÷ 0.05 = 2,000 per minute = 33.33 s⁻¹, and kcat/Km = 33.33 ÷ (40 × 10⁻⁶) = 8.33 × 10⁵ M⁻¹s⁻¹.
Competitive inhibition. Add 10 µM of an inhibitor with Ki = 5 µM. The factor α = 1 + 10 ÷ 5 = 3, so the apparent Km is 3 × 40 = 120 µM and v = 100 × 120 ÷ (120 + 120) = 50 µM/min, 33.33% inhibition.
The common mistake: reading Km and Vmax off a Lineweaver–Burk line. For the example data set (7 points from 5 to 500 µM), the straight line through 1/v against 1/[S] gives Km = 38.04 µM and Vmax = 97.31 µM/min. Fitting the curve itself by non-linear least squares gives Km = 40.13 ± 0.96 µM and Vmax = 100.1 ± 0.7 µM/min. The reciprocals give the slowest, least certain points the most weight, so the line is pulled away from the best answer.
Show Work
Formulas
Michaelis, Menten and the Straight Lines
In 1913 Leonor Michaelis and Maud Menten, working in Berlin, published careful measurements of how the enzyme invertase splits sucrose and showed that the rate follows a hyperbola in the substrate concentration. Victor Henri had proposed the same form about a decade earlier; Michaelis and Menten’s contribution was to measure initial rates under controlled pH so the equation could be tested properly. In 1925 George Briggs and J. B. S. Haldane derived it again using the steady-state assumption, which is the derivation taught today.
Before computers, fitting a curve was hard, so biochemists turned the hyperbola into straight lines. Hans Lineweaver and Dean Burk published their double-reciprocal plot in 1934, and it became one of the most cited papers in biochemistry; George Eadie (1942) and B. H. J. Hofstee (1952) proposed the v against v/[S] form. These plots remain useful for spotting the type of inhibition, but non-linear least squares, as used here, gives better estimates of Km and Vmax.
About This Tool
This calculator solves the Michaelis–Menten equation for any one of its four quantities, works out kcat and kcat/Km when you give the enzyme concentration, and shows how competitive, uncompetitive and mixed inhibitors change the apparent Km and Vmax. Its fitting mode uses the Levenberg–Marquardt method, started from the Lineweaver–Burk line, and reports standard errors from the covariance matrix and R², along with the Lineweaver–Burk and Eadie–Hofstee estimates for comparison. Every mode can be drawn as a saturation curve or either straight-line plot.
Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: Data Regression Tool, Molarity Calculator, and Bacterial Growth Calculator.
Frequently Asked Questions
What is the Michaelis–Menten equation?
It gives the initial rate of an enzyme reaction as v = Vmax[S] ÷ (Km + [S]). With Vmax = 100 µM/min, Km = 40 µM and [S] = 120 µM, v = 100 × 120 ÷ (40 + 120) = 75 µM/min, which is 75% of Vmax.
What does Km mean?
Km is the substrate concentration at which the rate is half of Vmax. A lower Km means the enzyme reaches half speed at a lower substrate level. To reach 90% of Vmax you need [S] = 9 × Km, so an enzyme with Km = 40 µM needs 360 µM substrate.
How do I calculate kcat and kcat/Km?
kcat = Vmax ÷ [E]. With Vmax = 100 µM/min and 0.05 µM enzyme, kcat = 2,000 per minute = 33.33 s⁻¹. Dividing by Km = 40 µM gives kcat/Km = 33.33 ÷ (40 × 10⁻⁶ M) = 8.33 × 10⁵ M⁻¹s⁻¹, the specificity constant.
How do inhibitors change Km and Vmax?
With [I] = 10 µM and Ki = 5 µM (a factor of 1 + 10 ÷ 5 = 3) on an enzyme with Km = 40 µM and Vmax = 100 µM/min: a competitive inhibitor raises Km to 120 µM and leaves Vmax alone; an uncompetitive one divides both by 3 (13.33 µM, 33.33 µM/min); a pure non-competitive one leaves Km at 40 µM and cuts Vmax to 33.33 µM/min. At [S] = 120 µM the rates are 50, 30 and 25 µM/min instead of 75.
Why fit the curve directly instead of using a Lineweaver–Burk plot?
Taking reciprocals magnifies the error of the slowest points. For the example data on this page the Lineweaver–Burk line gives Km = 38.04 µM and Vmax = 97.31 µM/min, while non-linear least squares on the same points gives Km = 40.13 ± 0.96 µM and Vmax = 100.1 ± 0.7 µM/min. The double-reciprocal plot is still good for seeing the type of inhibition.
How do I use the Michaelis–Menten Calculator (Km, Vmax, Fit)?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
Designing an assay
To run an enzyme at 90% of Vmax, use [S] = 9 × Km: 360 µM for an enzyme with Km = 40 µM.
Fitting lab data
Seven rates measured from 5 to 500 µM substrate fit to Km = 40.13 ± 0.96 µM and Vmax = 100.1 ± 0.7 µM/min with R² = 0.9996.
Comparing enzymes
kcat = 33.33 s⁻¹ and Km = 40 µM give kcat/Km = 8.33 × 10⁵ M⁻¹s⁻¹, the number used to rank how well enzymes use a substrate.
Drug inhibition
A competitive inhibitor at twice its Ki triples the apparent Km, cutting the rate at 120 µM substrate from 75 to 50 µM/min (33.33% inhibition).
Biochemistry homework
Find Km from one point: v = 25 µM/min at [S] = 10 µM with Vmax = 100 µM/min gives Km = 10 × 75 ÷ 25 = 30 µM.
Last updated: