Enzyme Kinetics Calculator

Calculate reaction velocity, kcat, and identify inhibition type from kinetic parameters.

Calculate

Vmax (any units) (umol/min)

Km (mM) (mM)

Substrate [S] (mM) (mM)

Enzyme concentration [E] (nM) (nM)

Apparent Km with inhibitor (mM) (mM)

0=no inhibitor

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What K_m Actually Means

The Michaelis constant is the substrate concentration at which reaction rate reaches half of V_max. Substituting [S] = K_m into the equation makes this fall out immediately.

v = V_max[S] / (K_m + [S])

[S] relative to K_m	Rate	Kinetic order
[S] « K_m	v ≈ (V_max/K_m)[S]	First order in substrate
[S] = K_m	v = V_max/2	Transitional
[S] = 10 K_m	v = 0.91 V_max	Approaching saturation
[S] » K_m	v ≈ V_max	Zero order — saturated

A low K_m means the enzyme reaches half-maximal rate at low substrate concentration, which is usually described as high apparent affinity. Strictly K_m equals the dissociation constant only when catalysis is slow relative to substrate release — otherwise it also contains kinetic terms.

Why k_cat/K_m Is the Meaningful Comparison

V_max depends on how much enzyme is present, so it cannot compare enzymes. Dividing by enzyme concentration gives k_cat, the turnover number — catalytic events per active site per second. Dividing that by K_m gives the specificity constant, which measures catalytic efficiency at low substrate concentration where enzymes usually operate.

The upper limit for k_cat/K_m is around 10⁸–10⁹ M⁻¹s⁻¹, set by how fast substrate can diffuse to the active site. Enzymes reaching it — catalase, carbonic anhydrase, triosephosphate isomerase — are described as catalytically perfect: they convert essentially every molecule they encounter.

Distinguishing Inhibitor Types

Inhibitor	K_m	V_max	Overcome by more substrate?
Competitive	Increases	Unchanged	Yes
Non-competitive	Unchanged	Decreases	No
Uncompetitive	Decreases	Decreases	No — worsens

Worked Examples

Example 1: [S]=1mM, Km=0.5mM, Vmax=100

V=100x1/1.5=66.7

Result: 67% of Vmax at 2x Km

[S]=10xKm gives 91% Vmax

Example 2: kcat: Vmax=100umol/min, [E]=10nM

kcat=100umol/min/(10nM)=167 s^-1 per nM

Result: Adjust units carefully

Carbonic anhydrase: 10^6 s^-1 among fastest

Example 3: Reaching 90% of V_max

K_m = 0.5 mM, want v = 0.9 V_max

Result: [S] = 4.5 mM, nine times K_m

Saturation approaches asymptotically. Ninefold K_m gives 90%; reaching 99% would need ninety-nine times K_m.

Example 4: Catalytic perfection

Carbonic anhydrase, k_cat/K_m near 10⁸ M⁻¹s⁻¹

Result: Diffusion limited

Essentially every encounter results in catalysis. No further improvement is possible without changing how fast substrate arrives.

Common Mistakes

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Treating K_m as a binding constant

K_m equals the dissociation constant only when catalysis is slow relative to substrate release. Otherwise it contains kinetic terms as well.

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Comparing enzymes by V_max

V_max scales with how much enzyme is present. Use k_cat for turnover or k_cat/K_m for catalytic efficiency.

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Assuming saturation at [S] = K_m

That is only half-maximal rate. Reaching 90% of V_max requires about ten times K_m.

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Using Lineweaver–Burk for parameter fitting

The double-reciprocal plot distorts error at low substrate concentrations. It remains useful for visualising inhibition type, but non-linear regression gives better parameters.

Frequently Asked Questions

Competitive vs non-competitive?

Competitive: Km increases, Vmax unchanged. Overcomes with more substrate. Non-competitive: Vmax decreases, Km unchanged. Cannot overcome with more substrate.

kcat/Km significance?

Specificity constant. Perfect enzyme approaches diffusion limit 10^8-10^9 M^-1s^-1. Drug target enzymes: design inhibitors with Ki << Km.

What does K_m tell you?

The substrate concentration giving half-maximal rate. A low K_m means the enzyme works efficiently at low substrate concentration.

Why use k_cat/K_m rather than V_max?

V_max depends on enzyme amount. The specificity constant measures intrinsic catalytic efficiency and allows comparison between enzymes and substrates.

What is a catalytically perfect enzyme?

One whose k_cat/K_m approaches the diffusion limit of about 10⁸–10⁹ M⁻¹s⁻¹, converting nearly every substrate molecule it encounters.

How do I identify inhibitor type?

Competitive inhibition raises K_m with V_max unchanged; non-competitive lowers V_max with K_m unchanged; uncompetitive lowers both.