What does Km actually represent?

Km is the substrate concentration at which v = Vmax/2 — mathematically it just falls out of the equation, but physically it captures how tightly the enzyme binds its substrate. Under the rapid-equilibrium limit Km ≈ Kd (the ES dissociation constant); under the Briggs-Haldane steady-state (the general case) Km = (k₋₁ + kcat)/k₁, which includes the catalytic step. So Km is not exactly a binding constant — a fast catalytic step pushes Km above Kd. A low Km (µM or lower) means high apparent affinity; a high Km (mM) means the enzyme is nowhere near saturated under typical cellular substrate levels.

What is kcat/Km and why is it called the specificity constant?

kcat/Km is a bimolecular rate constant with units M⁻¹·s⁻¹ that governs the apparent second-order reaction between free enzyme and free substrate. In the limit [S] ≪ Km, v = (kcat/Km)·[E][S], which is the same form as a simple bimolecular rate law. When an enzyme has two competing substrates, the ratio of their kcat/Km values equals the ratio of the rates the enzyme processes them at low [S] — hence 'specificity constant.' Diffusion sets a hard upper bound of ~10⁸–10⁹ M⁻¹s⁻¹; enzymes that reach it (carbonic anhydrase, catalase, triose-phosphate isomerase) are called 'catalytically perfect' (Lehninger 7e Box 6-1).

When does the Michaelis-Menten model fail?

Three big classes. (1) Allosteric enzymes: cooperative binding gives sigmoidal v vs [S] curves that fit the Hill equation, not MM. Hemoglobin (not an enzyme but the classic case) and phosphofructokinase both do this. (2) Multi-substrate enzymes at physiological substrate ratios — ternary-complex mechanisms give apparent Km values that depend on the co-substrate concentration. (3) Anything past ~5% substrate consumption without regeneration — the model assumes [S] stays approximately constant during the initial-rate measurement, which is only true early. Also fails for uncompetitive inhibition regimes at fixed [I] because both Vmax and Km change together.

How do I extract Vmax and Km experimentally?

Direct nonlinear least-squares fit of v vs [S] to the MM equation — this is what any modern kinetics software does and it gives the least biased error bars. Linear transformations (Lineweaver-Burk 1/v vs 1/[S], Eadie-Hofstee v vs v/[S], Hanes-Woolf [S]/v vs [S]) were popular before computers because they turn the fit into a straight line, but they all distort the error weighting — Lineweaver-Burk is the worst because it puts the most emphasis on the least-reliable low-[S] points. Use them for pedagogy or to eyeball a data set; use nonlinear regression for anything you'll publish (Cornish-Bowden 2012 §4).

How much substrate do I need to hit near-Vmax?

The fraction of Vmax is f = [S]/(Km + [S]), so [S]/Km = f/(1−f). To hit 90% of Vmax you need [S] = 9·Km; for 99% you need 99·Km. If your Km is 100 µM and you want to saturate, work at ~10 mM substrate; if your substrate is only soluble to 1 mM you'll top out at 91% Vmax and need to extrapolate. This ratio is what makes 'low Km enzymes' operationally useful — they saturate at physiological concentrations.

Michaelis-Menten Calculator — v, Vmax, Km, [S] Solver

The Michaelis-Menten equation is the workhorse of enzyme kinetics: one hyperbolic rate law that connects three physical quantities you can measure — initial rate, substrate concentration, and enzyme concentration — into two intuitive parameters, Vmax and Km. Nearly every quantitative statement in biochemistry about enzyme performance, drug binding, or metabolic flux traces back to fits of experimental data to this one equation.

Michaelis-Menten Calculator

Solve v, Vmax, Km, or [S] for a Michaelis-Menten enzyme. Keep units consistent across v/Vmax and Km/[S]. Adding [E]_total unlocks kcat and the specificity constant kcat/Km.

Equations used

v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]}

k_{c a t} = \frac{V _{ma x}}{[ E ] _{t o t a l}}, \frac{k _{c a t}}{K _{m}}

Steady-state Michaelis-Menten model — Briggs & Haldane (1925), Biochem. J. 19, 338. Assumes reversibly-formed ES complex and no substrate depletion. See Lehninger 7e §6.3 for the derivation.

The equation

Under the steady-state approximation of Briggs & Haldane (1925), applied to the two-step mechanism E + S ⇌ ES → E + P,

$v = \frac{V_{max}\,[S]}{K_m + [S]}$

v is the initial rate of product formation (before substrate depletion matters).
Vmax is the asymptotic rate as [S] → ∞ (every enzyme molecule is bound and turning over).
Km is the substrate concentration at which v = Vmax/2.
[S] is the free-substrate concentration; under conditions where [S] ≫ [E], [S] ≈ [S]_total.

Two limits worth memorising

Low [S] ([S] ≪ Km): $v \approx (V_{max}/K_m)\,[S]$ . The reaction looks first-order in substrate; the effective rate constant is Vmax/Km, a proxy for kcat/Km.
High [S] ([S] ≫ Km): $v \approx V_{max}$ . The enzyme is saturated; adding more substrate does nothing.

The transition between the two happens over roughly one order of magnitude in [S] around Km.

Worked example — hexokinase

Hexokinase phosphorylates glucose with Km(glucose) = 1.5 × 10⁻⁴ M and Vmax = 6.5 µmol·min⁻¹ under a given set of conditions. What is v at [glucose] = 5.0 × 10⁻⁵ M?

Worked example — hexokinase at sub-Km [glucose]

Vmax: 6.5 µmol·min⁻¹
Km: 1.5 × 10⁻⁴ M
[S]: 5.0 × 10⁻⁵ M

Formula
$v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]}$
Substitute
$v = \frac{6.5 \cdot 5.0 \times 1 0 ^{- 5}}{1.5 \times 1 0 ^{- 4} + 5.0 \times 1 0 ^{- 5}} = \frac{3.25 \times 1 0 ^{- 4}}{2.0 \times 1 0 ^{- 4}}$
Note
[S] is one-third of Km, so v is 25 % of Vmax — check: f = [S]/(Km + [S]) = 0.25, and 0.25 · 6.5 = 1.625 µmol/min. Notice how sensitive v is to [S] in this regime: doubling [glucose] would take v to 40 % of Vmax, not 50 %, because the curve is hyperbolic, not linear.
Result
$v \approx 1.63 μ mol/min$

kcat and the specificity constant

If you know the enzyme’s molar concentration [E]_total, you can convert Vmax into a catalytic-constant (turnover number):

$k_{cat} = \frac{V_{max}}{[E]_{total}}$

kcat is the number of substrate molecules each active site processes per second at saturation. Reasonable values range from 1 s⁻¹ (many regulatory enzymes) to 10⁷ s⁻¹ (catalase, which decomposes H₂O₂ at ~10⁷ per second per site).

The composite constant kcat/Km is the enzyme’s specificity constant — a bimolecular rate constant with units M⁻¹·s⁻¹ that governs the reaction under sub-saturating substrate. It has a fundamental physical upper bound: ~10⁸–10⁹ M⁻¹s⁻¹ from the diffusion-controlled encounter rate of a small molecule with a protein active site. Enzymes at that limit are called catalytically perfect.

Enzyme	kcat (s⁻¹)	Km (M)	kcat/Km (M⁻¹s⁻¹)
Carbonic anhydrase (H₂O + CO₂)	4 × 10⁵	8 × 10⁻³	5 × 10⁷
Catalase (H₂O₂)	1 × 10⁷	2.5 × 10⁻²	4 × 10⁸
Triose-phosphate isomerase	4 × 10³	5 × 10⁻⁴	8 × 10⁶
Chymotrypsin (Ac-Phe-Ala)	5 × 10¹	5 × 10⁻³	1 × 10⁴

Values above are order-of-magnitude and taken from Lehninger 7e Box 6-1.

Common pitfalls

Confusing Km with Kd. Km ≥ Kd (with equality only when kcat ≪ k₋₁). A fast enzyme has Km above its true binding constant.
Using non-initial rates. The equation is for the initial rate before [S] has changed noticeably. Fit points collected while [S] falls and you’ll get systematically wrong Km.
Assuming [S] = [S]_total. True only when [S]_total ≫ [E]. In tight-binding or high-affinity systems, the free-substrate concentration differs and the MM analysis needs a modification (Morrison equation for tight-binding inhibitors).
Extrapolating past ~99% Vmax. You need infinite substrate to reach Vmax exactly. Curve-fit codes handle this fine; back-of-envelope tables of “the enzyme is saturated at 10 mM” are informal shorthand, not equations.

Frequently asked questions

What does Km actually represent?: Km is the substrate concentration at which v = Vmax/2 — mathematically it just falls out of the equation, but physically it captures how tightly the enzyme binds its substrate. Under the rapid-equilibrium limit Km ≈ Kd (the ES dissociation constant); under the Briggs-Haldane steady-state (the general case) Km = (k₋₁ + kcat)/k₁, which includes the catalytic step. So Km is not exactly a binding constant — a fast catalytic step pushes Km above Kd. A low Km (µM or lower) means high apparent affinity; a high Km (mM) means the enzyme is nowhere near saturated under typical cellular substrate levels.
What is kcat/Km and why is it called the specificity constant?: kcat/Km is a bimolecular rate constant with units M⁻¹·s⁻¹ that governs the apparent second-order reaction between free enzyme and free substrate. In the limit [S] ≪ Km, v = (kcat/Km)·[E][S], which is the same form as a simple bimolecular rate law. When an enzyme has two competing substrates, the ratio of their kcat/Km values equals the ratio of the rates the enzyme processes them at low [S] — hence 'specificity constant.' Diffusion sets a hard upper bound of ~10⁸–10⁹ M⁻¹s⁻¹; enzymes that reach it (carbonic anhydrase, catalase, triose-phosphate isomerase) are called 'catalytically perfect' (Lehninger 7e Box 6-1).
When does the Michaelis-Menten model fail?: Three big classes. (1) Allosteric enzymes: cooperative binding gives sigmoidal v vs [S] curves that fit the Hill equation, not MM. Hemoglobin (not an enzyme but the classic case) and phosphofructokinase both do this. (2) Multi-substrate enzymes at physiological substrate ratios — ternary-complex mechanisms give apparent Km values that depend on the co-substrate concentration. (3) Anything past ~5% substrate consumption without regeneration — the model assumes [S] stays approximately constant during the initial-rate measurement, which is only true early. Also fails for uncompetitive inhibition regimes at fixed [I] because both Vmax and Km change together.
How do I extract Vmax and Km experimentally?: Direct nonlinear least-squares fit of v vs [S] to the MM equation — this is what any modern kinetics software does and it gives the least biased error bars. Linear transformations (Lineweaver-Burk 1/v vs 1/[S], Eadie-Hofstee v vs v/[S], Hanes-Woolf [S]/v vs [S]) were popular before computers because they turn the fit into a straight line, but they all distort the error weighting — Lineweaver-Burk is the worst because it puts the most emphasis on the least-reliable low-[S] points. Use them for pedagogy or to eyeball a data set; use nonlinear regression for anything you'll publish (Cornish-Bowden 2012 §4).
How much substrate do I need to hit near-Vmax?: The fraction of Vmax is f = [S]/(Km + [S]), so [S]/Km = f/(1−f). To hit 90% of Vmax you need [S] = 9·Km; for 99% you need 99·Km. If your Km is 100 µM and you want to saturate, work at ~10 mM substrate; if your substrate is only soluble to 1 mM you'll top out at 91% Vmax and need to extrapolate. This ratio is what makes 'low Km enzymes' operationally useful — they saturate at physiological concentrations.

Sources

Nelson & Cox, Lehninger Principles of Biochemistry, 7th ed. (2017), §6.3 — Enzyme Kinetics as an Approach to Understanding Mechanism
Berg, Tymoczko, Gatto & Stryer, Biochemistry, 9th ed. (2019), §8.4 — The Michaelis-Menten Model Accounts for the Kinetic Properties of Many Enzymes
Cornish-Bowden, Fundamentals of Enzyme Kinetics, 4th ed. (2012), Ch. 2
Briggs & Haldane (1925), Biochem. J. 19, 338 — A note on the kinetics of enzyme action
IUBMB Enzyme Nomenclature

Last updated: 2026-09-18.