Michaelis Menten Calculator
The saturation curve, both directions: velocity from substrate, or the substrate a target velocity demands — with the half-speed definition doing the teaching.
Michaelis Menten Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Single-substrate steady-state model
- No inhibitor or allosteric terms
In short: An enzyme with Vmax 100 µM/min and Km 2 µM running at [S] = 2 µM moves at 50.000000 µM/min — exactly half speed, because Km is the substrate concentration of half velocity by definition. At [S] = 18 µM the same enzyme reads 90.000000 (90% of the ceiling); the ladder card prices 0.1×Km → 9.090909% and 10×Km → 90.909091%. Ask the reverse question — what substrate buys 50 µM/min? — and the solver answers 2.000000 µM. Targets at or above Vmax get refused: enzymes approach the ceiling, never arrive.
Formula
v = Vmax·[S]/(Km + [S]) · solve [S] = Km·v/(Vmax − v) · half speed at [S] = Km
Briggs–Haldane steady-state kinetics gives the rectangular hyperbola: velocity rises with substrate but saturates, because the enzyme’s active sites — not the substrate — become the scarce resource. Km is the substrate concentration at exactly half Vmax (a grip measurement: lower Km, tighter binding); Vmax is the ceiling the turnover number sets on the enzyme concentration present. The algebra inverts cleanly as long as the target velocity stays below the ceiling.
Worked Example
- Enter Vmax and Km for the enzyme.
- Type a substrate level to read the velocity —
- …or blank [S] and enter a target velocity to solve it.
- Check the ladder card for where you sit on the curve.
Defaults: Vmax 100, Km 2, [S] 2 → 50.000000 µM/min (the half-speed point). [S] 18 → 90.000000. Solver: blank [S], target 50 → 2.000000 µM; target 99 → 198.000000 µM. Target 100 or above → refused at the door: no finite substrate buys the ceiling.
Strengths & Limits Of This Model
Where this engine is strong
- Solves both directions of the hyperbola
- The asymptote refused, not fudged
Where it stops
- No curve fitting from data
- No multi-substrate kinetics
Practical Use Cases
Assay design
pick [S] that saturates
Inhibitor work
where on the curve you sample
Teaching
the hyperbola, both directions
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Michaelis Menten Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why is v exactly half of Vmax when [S] = Km?
Because that is what Km means: the formula gives v = Vmax·Km/(Km+Km) = Vmax/2. The half-speed point is not a coincidence the equation produces — it is the definition the equation was built around, which is why Km is reportable as a single concentration. Any enzyme’s grip on its substrate can be quoted as the concentration that buys half the ceiling.
What does a low Km tell me about the enzyme?
A tight grip: half-speed at a whisper of substrate means the enzyme works well even when its substrate is scarce. Physiological enzymes often carry Km values near their substrate’s real concentrations — the range where the curve is steepest and control is most responsive. High-Km enzymes need abundance before they engage; the ladder card shows how slowly velocity climbs once you are past a few multiples of Km.
Why can’t I solve for a substrate that reaches Vmax?
Because the algebra and the physics agree for once: the formula v = Vmax·[S]/(Km+[S]) approaches Vmax asymptotically and reaches it only at infinite substrate. A target at or above the ceiling has no finite answer, and the solver says so instead of printing a negative or an absurdity. Enzymes approach; they never arrive — the door is the asymptote doing its job.
Is v in these units real units?
The units are whatever Vmax carries — µM/min here by convention of the input labels. The arithmetic is indifferent: if Vmax is quoted in µmol/min per mg (a specific activity), v comes out in the same currency. For absolute units per preparation, the enzyme activity page prices the assay that measures the slope this page then models — the two pages are the measurement and the curve.
How do inhibitors fit this picture?
They change the two constants: competitive inhibitors raise the apparent Km (the grip weakens) while Vmax survives at high substrate; noncompetitive inhibitors cut Vmax with Km untouched. Measure v across a substrate series under each condition and fit constants — this page then replays any single point on the curve. The classic Lineweaver–Burk straight-line trick was the old fit, at the price of amplifying exactly the data points you trust least (approximate).
What assumptions hide behind the curve?
Steady state (the ES complex holds constant while you measure), initial rates (product has not accumulated to reverse the reaction), and a single substrate. Real assays honor them by measuring early and diluting — the Briggs–Haldane derivation is a few lines on those assumptions and the page’s whole curve stands on them.
Why does 10×Km only buy 90.909091% of the ceiling?
Because saturation is asymptotic: each additional factor of substrate buys less headroom — half at Km, nine-tenths at ten times Km, ninety-nine percent at a hundred. That fading return is why assays run at saturating substrate (≥10×Km) but not absurdly above: the last percent of saturation costs more than it measures. The ladder card prices the whole ladder at a glance.
What is the relationship between Vmax and kcat?
Vmax = kcat·[E]: the ceiling is the turnover number — catalytic events per enzyme site per second — multiplied by how much enzyme is in the beam. Double the enzyme and you double Vmax while Km (a property of the site, not the quantity) stays put — which is why Km identifies an enzyme’s grip and Vmax reports its concentration. The activity page prices [E]’s contribution; this page shows its ceiling.