The mechanism of enzyme catalysed reaction is given by Michaelis and Menten as:
step I: E + S ⇌(K₁, K₋₁) ES (fast)
step II: ES →(K₂) P + E (slow)
The rate of product formation may be given as

Question

The mechanism of enzyme catalysed reaction is given by Michaelis and Menten as:
step I: E + S ⇌(K₁, K₋₁) ES (fast)
step II: ES →(K₂) P + E (slow)
The rate of product formation may be given as: +dP/dt = (K₁ K₂ [E]₀ [S]) / (K₋₁ + K₂ + K₁ [S]), where [E]₀ is the total enzyme concentration. For an enzyme-substrate system obeying simple Michaelis and Menten mechanism, the rate of product formation when the substrate concentration is very large, has the limiting value 0.02 mol/dm³. At a substrate concentration of 250 mg/dm³, the rate is half this value. The value of K₁/K₋₁ (in dm³/kg), assuming that K₂<

Prashant Kotian · Accepted Answer

Correct answer: 4000. The rate equation can be rearranged by dividing the numerator and denominator by K₁: Rate = (K₂ [E]₀ [S]) / ((K₋₁ + K₂)/K₁ + [S]). The limiting rate (V_max) occurs at very high [S] and is V_max = K₂[E]₀ = 0.02 mol/dm³. The Michaelis constant is K_m = (K₋₁ + K₂)/K₁. Since K₂ << K₋₁, K_m ≈ K₋₁ / K₁. When the rate is half of V_max, the substrate concentration equals K_m. Thus, K_m = 250 mg/dm³ = 0.25 g/dm³ = 0.25 × 10⁻³ kg/dm³. So, K₋₁ / K₁ = 0.25 × 10⁻³ kg/dm³. We need K₁ / K₋₁ = 1 / (0.25 × 10⁻³) = 4000 dm³/kg. Therefore, correct answer is 4000.

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