Michaelis-Menten Equation Calculator

Solve the Michaelis-Menten equation for rate, Km, Vmax or substrate concentration.

Concentrations in M, mM, µM or nM; rates per second or per minute.

Updated October 19, 2026

CreatorFrank Zhao

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Introduction

This page is the Michaelis-Menten equation with a box for each of its four quantities. Leave one box empty, give it the other three, and it solves the missing one — most often the Michaelis constant $\mathrm{K_m}$ from a measured rate, or the rate you should expect from a known $\mathrm{K_m}$ .

It is written for the bench moment when you have an initial-rate measurement in front of you and one number — the plateau, or the constant — is missing. It is built around a single curve rather than a fit, so with three values it gives the fourth exactly, and with a rate measured at one substrate concentration it gives the constant that your numbers imply.

Any three of four values

The box left empty is the one the page solves

M, mM, µM, nM

Concentrations on one scale; rates per second or per minute

Half speed at the constant

Where the curve is halfway up, and the easiest check

Quick start

Say which of the four values you are missing

The four boxes are the reaction rate $\mathrm{V}$ , the maximum speed $\mathrm{V_{max}}$ , the substrate concentration $[\mathrm{S}]$ and the Michaelis constant $\mathrm{K_m}$ . Fill in three, leave the fourth empty, and that empty box is the one the page solves.

Put the two concentrations on one scale

$\mathrm{K_m}$ and $[\mathrm{S}]$ are both concentrations, and each box carries its own unit menu: nanomolars, micromolars, millimolars or molars. It does not matter which scale you read them on as long as both boxes describe your experiment on the same one.

Put the two rates on one clock

Measurements are often recorded per minute and the plateau quoted per second, or the other way around. Give each rate box the number in the unit its menu shows; the two are brought onto the same clock before anything is solved.

Enter the measured rate and read the answer

A rate assay run at $[\mathrm{S}] = 0.25\ \mathrm{mM}$ gave $\mathrm{V} = 0.2\ \mathrm{s^{-1}}$ , and the same amount of enzyme reached $\mathrm{V_{max}} = 0.8\ \mathrm{s^{-1}}$ in a saturated run. With $\mathrm{K_m}$ left empty:

\mathrm{K_m}

=

\frac{0.25\ \mathrm{mM} \times (0.8 - 0.2)}{0.2}

= 0.75\ \mathrm{mM}

The constant comes back in the same unit as the substrate concentration, and the box that was left empty is the one that fills in. Type a value into it instead and the page solves a different box.

Check the rate against the ceiling

The ratio $\mathrm{V}/\mathrm{V_{max}}$ is the fraction of the ceiling this assay is running at — here 0.2 divided by 0.8, so a quarter. That fraction is also $[\mathrm{S}]/([\mathrm{S}] + \mathrm{K_m})$ , which is a quick way to see whether the two concentrations are consistent: a quarter of the ceiling with 0.25 mM of substrate implies a constant of about 0.75 mM, exactly what the page returned.

Reading the curve from both ends

Everything on the page comes from one statement: the rate rises with substrate along a rectangular hyperbola that starts at zero and flattens towards $\mathrm{V_{max}}$ .

\mathrm{V} = \frac{\mathrm{V_{max}} \times [\mathrm{S}]}{[\mathrm{S}] + \mathrm{K_m}}

\mathrm{V}

The reaction rate: product formed per unit of time at that substrate concentration

\mathrm{V_{max}}

The plateau the curve approaches — the rate the enzyme would reach if it were saturated

\mathrm{K_m}

The Michaelis constant: the substrate concentration that gives exactly half of Vmax

[\mathrm{S}]

Substrate concentration, on the same scale as Km

f = \mathrm{V}/\mathrm{V_{max}}

The fraction of the ceiling in use, a pure number between zero and one

Half of Vmax defines the constant

Set the substrate concentration equal to the constant and the fraction collapses to one half. That is the whole meaning of $\mathrm{K_m}$ : it is a concentration, it is where the curve is halfway up, and it is the single easiest number to check an answer against.

[\mathrm{S}] = \mathrm{K_m} \quad \Rightarrow \quad \mathrm{V} = \frac{\mathrm{V_{max}}}{2}

How much substrate a target rate needs

Turn the equation around and the substrate concentration falls out of the fraction of the ceiling you are aiming for. Ninety per cent of $\mathrm{V_{max}}$ needs nine times the constant; ninety-nine per cent needs ninety-nine times it. This is the reason a saturated-rate measurement is expensive, and the reason the last few per cent of the curve are rarely worth chasing.

[\mathrm{S}] = \frac{\mathrm{K_m} \times f}{1 - f}, \qquad f = \frac{\mathrm{V}}{\mathrm{V_{max}}}

The same statement solved four ways

Because it is one equation, each quantity can be the unknown, which is why three boxes are always enough:

\mathrm{K_m}

=

\frac{[\mathrm{S}]\,(\mathrm{V_{max}} - \mathrm{V})}{\mathrm{V}}

\mathrm{V_{max}}

=

\frac{\mathrm{V}\,(\mathrm{K_m} + [\mathrm{S}])}{[\mathrm{S}]}

[\mathrm{S}]

=

\frac{\mathrm{K_m} \times \mathrm{V}}{\mathrm{V_{max}} - \mathrm{V}}

Where a single reading is thin

Solving for $\mathrm{K_m}$ leans on the difference $\mathrm{V_{max}} - \mathrm{V}$ , so the further the assay sits from the plateau, the smaller that difference and the more a rate error is magnified. The factor is $\mathrm{V_{max}}/(\mathrm{V_{max}} - \mathrm{V})$ : two at half saturation, about ten at nine tenths of the ceiling.

One per cent of rate error, eleven per cent of constant

Take a preparation whose true constant is 1 mM with a plateau of 1 per second. At 10 mM substrate the curve predicts about 0.909 per second — but 10 mM is nine tenths of the way up, so the derived constant is hypersensitive there. If the assay reads 0.9 per second instead, barely one per cent low, the page returns about 1.11 mM:

\mathrm{K_m}

=

\frac{10\ \mathrm{mM} \times (1 - 0.9)}{0.9}

\approx 1.11\ \mathrm{mM}

The same one per cent error at 1 mM substrate moves the constant by about two per cent. The measurement is not wrong in either case; it is simply being asked for information that the shape of the curve does not carry well far from $\mathrm{K_m}$ .

Rates and concentrations keep their own scales

The concentration terms form the dimensionless fraction $[\mathrm{S}]/([\mathrm{S}] + \mathrm{K_m})$ , so the two concentration boxes only have to agree with each other, and the two rate boxes only have to agree with each other. That is also why the rate boxes carry a per-time unit and no concentration: in this equation the rate unit cancels against the plateau.

Equal to Vmax has no finite answer

The plateau is an asymptote, not a reachable point

Ask the page for the substrate concentration that produces exactly $\mathrm{V_{max}}$ and the rearrangement divides by zero: only an infinite substrate concentration would saturate the enzyme completely. Rather than print an impossibility, the page leaves the substrate box empty with a short explanation, and the rate box is flagged because reaching the plateau is the conflict.

Aim slightly below the plateau instead — 90 or 95 per cent of $\mathrm{V_{max}}$ is a well-posed question and a realistic target. A rate measured above the plateau is flagged for the same reason: the two rates, or the two concentrations, are almost always on different scales.

None of the four boxes takes a zero either. A rate of zero, or no substrate at all, is the absence of a reaction rather than a point on this curve, so the box says so instead of returning a number.

Worked examples

How much substrate for ninety per cent of the plateau?

Situation: An assay is being set up around an enzyme with a known constant of 100 µM. The plateau for this amount of enzyme was measured at 2 per second, and the aim is an initial rate of 1.8 per second — ninety per cent of it, fast enough to measure comfortably and still short of the flat part of the curve. The substrate concentration is the unknown.

Entered

Michaelis constant100 µM

Max reaction speed2 per second

Reaction rate1.8 per second

The page returns

[\mathrm{S}]

=

\frac{100\ \mathrm{\mu M} \times 1.8}{2 - 1.8}

= 900\ \mathrm{\mu M} = 0.9\ \mathrm{mM}

In practice: 900 µM is nine times the constant, which is exactly what ninety per cent of the plateau predicts. From a 100 mM substrate stock that is roughly a 1 in 111 dilution, so a serial dilution from the stock is the practical next step rather than weighing substrate per tube.

The plateau needs the same enzyme as the rate

Situation: Two numbers are being compared across a dilution step: a plateau of 1.2 per second measured on undiluted enzyme, and an initial rate of 0.15 per second measured after the preparation was diluted tenfold. Substrate was 0.4 mM in both runs, and the constant is the unknown.

Entered

Max reaction speed1.2 per second (undiluted)

Reaction rate0.15 per second (tenfold diluted)

Substrate concentration0.4 mM

The page returns

\mathrm{K_m}

=

\frac{0.4\ \mathrm{mM} \times (1.2 - 0.15)}{0.15}

= 2.8\ \mathrm{mM}

The arithmetic is correct, and that is the point: nothing on the page can tell that the two rates came from different amounts of enzyme, so keeping the dataset on one enzyme amount is yours to do.

In practice: Ten times less enzyme should give roughly a tenth of the plateau, so a rate of 0.15 against a plateau of 1.2 is a mismatch: the 2.8 mM that comes back is mostly describing the dilution, not the enzyme. Measure the plateau and the rate with the same amount of enzyme in the cuvette, or scale the plateau to the dilution before entering it.

Getting a Km you can trust

Measure where the curve bends

The constant is best constrained by rates taken near half saturation, where the curve is steepest and the derived value is least sensitive to a rate error. A substrate well below $\mathrm{K_m}$ puts you on the straight opening of the curve, where the rate is roughly proportional to substrate and the plateau is barely felt.

One amount of enzyme behind both rates

$\mathrm{V_{max}}$ is a property of how much enzyme is in the cuvette, so it moves when the preparation is diluted and the rate moves with it. Doubling the enzyme doubles both, leaves the ratio unchanged and therefore leaves the constant unchanged: $\mathrm{K_m}$ is the number that travels between preparations. If the amount of enzyme is what you need to pin down, the enzyme activity calculator balances activity, mass and volume the other way round — the mass that delivers a target activity, or the activity a weighed amount produces.

Keep the concentrations on one scale

A concentration constant entered in micromolars next to a substrate concentration in millimolars is out by a factor of a thousand, and the derived constant lands a thousand times away from the truth. The ratio of the two boxes is what the equation uses, so read both in the unit you actually pipetted.

A concentration above 1 M — the size a micromolar and millimolar mix-up produces — is shown so you can see how far off it is, and flagged as past what an assay can reach.

Build the substrate series deliberately

Rates that are intended to describe one curve should come from one dilution series of the same substrate stock, made up in the same buffer on the same day. The solution dilution calculator turns a concentrated stock into the working concentrations, and each concentration is then a single number to enter here.

One more habit pays off later: record the enzyme amount, the temperature and the buffer alongside the constant. A $\mathrm{K_m}$ without those conditions cannot be compared with another one, and comparing constants is usually what the number is for.

Questions

Why does the substrate box stay blank when the rate equals Vmax?

Because no finite substrate concentration produces it. Solving for the concentration divides by $\mathrm{V_{max}} - \mathrm{V}$ , and that difference is zero at the plateau, so the equation is asking for the concentration at which an enzyme is completely saturated. The page leaves the box empty with a note naming the conflict, keeps the numbers you typed, and flags the rate box. Choose a target just below the plateau instead.

I measured at one substrate concentration. Is the constant I get trustworthy?

The algebra is exact: with the rate, the plateau and the concentration, there is one constant that fits, and the page returns it. Whether that constant survives a repeat measurement is a different question, because the calculation amplifies whatever error the rate carries — about twofold near $\mathrm{K_m}$ , roughly tenfold at nine tenths of the plateau. A reading near half saturation gives a far more stable value than one taken close to the plateau, and several concentrations with a proper fit are better still.

My rates are in micromoles per litre per second. Can I enter those numbers?

Yes. The equation uses the ratio of the two rates and the sum of the two concentrations, so as long as both rate boxes describe the same clock — both per second, or both per minute — the numbers can be in any consistent rate unit and the answers will be numerically right. Give each box the number in the unit its menu shows; the page brings the two onto one clock before it solves anything.

Does the constant change if I read the concentrations in micromolars?

No. Switching a concentration box between molars, millimolars, micromolars and nanomolars only changes how the number is displayed; the underlying concentration is the same solution, and the derived constant stays the same concentration expressed on the scale you chose. The same holds for the two rate boxes and the per-second or per-minute choice.

Where it stops

The curve describes initial rates at steady state. It treats the enzyme as a trace component and the substrate concentration you type as the free concentration. A rate measured after appreciable substrate has been consumed, or with the enzyme approaching the substrate concentration, does not belong on this curve.
One substrate, no regulation. Cooperative enzymes give a sigmoidal curve, inhibitors and activators bend it, product inhibition and the reverse reaction flatten it late in the reaction, and two-substrate mechanisms need their own treatment. None of those shapes is this hyperbola.
It solves an equation; it does not fit a dataset. There is no averaging and no uncertainty, so several measurements still have to be analysed elsewhere before a constant can be quoted with a confidence interval. The page is the arithmetic on one set of numbers.
The constants belong to the conditions. pH, temperature, buffer, ionic strength and the enzyme preparation itself all move $\mathrm{K_m}$ and $\mathrm{V_{max}}$ . Treat the result as the value implied by the numbers you entered, confirm it with your own replicate measurements, and keep the conditions with it when you quote it.

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