Turnover Number Calculators
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Turnover Number Formula
kcat = Vmax / [E]total
[E]total = total enzyme concentration (M). kcat units: s⁻¹ (substrate molecules per enzyme molecule per second at saturation). Example: Vmax = 5 × 10⁻⁸ M/s; [E] = 1 × 10⁻⁸ M: kcat = 5 × 10⁻⁸ / 1 × 10⁻⁸ = 5 s⁻¹.
Catalytic Efficiency (kcat/Km)
kcat/Km (M⁻¹s⁻¹) = catalytic efficiency; also called specificity constant or apparent second-order rate constant. Measures enzyme performance at low [S] (where most enzyme is unbound). Theoretical maximum: ~10⁸–10⁹ M⁻¹s⁻¹ (rate limited by diffusion). Enzymes with kcat/Km near diffusion limit are called 'perfect enzymes': acetylcholinesterase (~1.5 × 10⁸); carbonic anhydrase (~8 × 10⁷); catalase (~4 × 10⁷).
Notable kcat Values
- Carbonic anhydrase: ~600,000 s⁻¹ (fastest enzyme)
- Catalase: ~10,000,000 s⁻¹ (fastest overall)
- Lysozyme: ~0.5 s⁻¹ (very slow)
- DNA polymerase III: ~1,000 s⁻¹
- Ribonuclease A: ~1,000 s⁻¹
Glossary
Frequently Asked Questions
kcat (turnover number) = the maximum number of substrate molecules an enzyme molecule converts to product per second when fully saturated with substrate. Units: s⁻¹. Derivation: kcat = Vmax / [E]_total. At saturating [S]: all enzyme molecules are in ES complexes → maximum reaction rate. kcat reflects the rate-limiting step in the catalytic cycle — the chemical step or product release. High kcat: enzyme quickly processes each bound substrate molecule. Low kcat: enzyme is slow at the chemical conversion step (e.g., lysozyme 0.5 s⁻¹; RNA processing enzymes < 1 s⁻¹). Comparison: Vmax depends on both kcat and enzyme concentration; kcat is intrinsic to the enzyme.
kcat/Km (M⁻¹s⁻¹) = the apparent second-order rate constant for enzyme-substrate encounter; the best measure of enzyme efficiency under physiological conditions (where [S] << Km). High kcat/Km: enzyme converts substrate efficiently even at low [S] — the enzyme is productive on most encounters with substrate. Interpretation: kcat/Km = kcat / Km. A high kcat alone doesn't mean efficient if Km is also very high (substrate must be at very high concentration for saturation). Diffusion limit: ~10⁸–10⁹ M⁻¹s⁻¹ — the fastest physically possible rate; enzyme would need to process every substrate molecule it encounters. 'Catalytically perfect' enzymes (acetylcholinesterase, triose phosphate isomerase): kcat/Km approaches diffusion limit → evolution has maximized catalytic efficiency.
From Michaelis-Menten data: (1) Run an enzyme kinetics experiment: measure initial velocity (v₀) at multiple substrate concentrations [S]. (2) Fit data to v₀ = Vmax × [S] / (Km + [S]) using non-linear regression or Lineweaver-Burk plot. (3) Extract Vmax and Km. (4) Measure or know [E]_total (enzyme concentration in the assay). (5) Calculate kcat = Vmax / [E]_total. Example: Vmax = 2.4 × 10⁻⁷ M/s; [E] = 5 × 10⁻⁹ M: kcat = 2.4 × 10⁻⁷ / 5 × 10⁻⁹ = 48 s⁻¹. Note: [E]_total must be active enzyme concentration — determined by active site titration (not just protein concentration, which may include inactive forms).
In the simple Michaelis-Menten mechanism: E + S ⇌ ES → E + P. Rate constants: k₁ (association), k₋₁ (dissociation), k₂ (catalytic step). kcat = k₂ (the rate constant for the chemical step). Km = (k₋₁ + k₂)/k₁. In a more complex mechanism with multiple steps after ES: kcat = 1 / (Σ 1/kᵢ) for each sequential step → kcat is limited by the slowest step. Product release is often rate-limiting: for DNA polymerase and others, kcat is limited by the rate at which the product dissociates from the enzyme. Significance: a mutation that accelerates the chemical step may still not increase kcat if product release is still rate-limiting. Directed evolution and rational enzyme design often aim to improve the rate-limiting step.