All question types · Kinetics
Enzyme kinetics from elementary rate constants
UKChO Round 2 · IChO · unreviewed
Enzyme mechanism with all three elementary rate constants, [E]₀ and [S] known; steady state on ES. Find the initial rate.
Worked example
For E + S ⇌ ES (k₁ = 2.4 × 10⁷ dm³ mol⁻¹ s⁻¹, k₋₁ = 3000 s⁻¹) and ES → E + P (k₂ = 2100 s⁻¹), find v₀ at [E]₀ = 4.1 × 10⁻⁸ mol dm⁻³ and [S] = 9.5 × 10⁻⁵ mol dm⁻³.
1. Add k₋₁ + k₂
= 5100 s⁻¹
2. Divide (k₋₁ + k₂) ÷ k₁ → K_M
= 0.2125 mmol dm⁻³
3. Multiply k₂ × [E]₀ → V_max
= 8.610 × 10⁻⁵ mol dm⁻³ s⁻¹
4. Multiply V_max × [S] → V_max[S]
= 0.008179 mol² m⁻⁶ s⁻¹
5. Add K_M + [S]
= 0.3075 mmol dm⁻³
6. Divide V_max[S] ÷ (K_M + [S]) → v₀
= 2.660 × 10⁻⁵ mol dm⁻³ s⁻¹
Answer: v₀ = 2.66 × 10⁻⁵ mol dm⁻³ s⁻¹
Method
Steady state on ES gives K_M = (k₋₁ + k₂)/k₁ and V_max = k₂[E]₀; then v₀ = V_max[S]/(K_M + [S]).
Common mistakes
- That uses K_S = k₋₁/k₁ (pre-equilibrium). With k₂ comparable to k₋₁, the steady-state K_M = (k₋₁ + k₂)/k₁.
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