In the Michaelis-Menten mechanism for enzyme kinetics, the expression obtained is: v/([E]₀[S])=1.4 × 10¹²-(10⁴ v)/([E]₀) The values of k₃ (Kems, mol L⁻¹ s⁻¹) and K (Michaelis constant, mol L⁻¹), respectively are

Q62 · CSIR-NET Chemistry, June 2012

Paper: CSIR-NET June 2012 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Hard

In the Michaelis-Menten mechanism for enzyme kinetics, the expression obtained is:

\[ \frac{\mathrm{v}}{[\mathrm{E}]_{0}[\mathrm{S}]}=1.4 \times 10^{12}-\frac{10^{4} \mathrm{v}}{[\mathrm{E}]_{0}} \]

The values of $\mathrm{k}_{3}\left(\mathrm{K}_{\text {ems }}, \mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)$ and K (Michaelis constant, $\mathrm{mol} \mathrm{L}^{-1}$ ), respectively are

(a)$1.4 \times 10^{12}, 10^{4}$

(b)$1.4 \times 10^{8}, 10^{4}$

(c)$1.4 \times 10^{8}, 10^{-4}$

(d)$1.4 \times 10^{12}, 10^{-4}$

Answer

Answer: C ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation

Eadie form: v/([E]₀[S]) = k₃/K − v/(K[E]₀). Comparing terms gives 1/K = 10⁴, so K = 10⁻⁴ mol L⁻¹, and k₃ = 1.4 × 10¹² × 10⁻⁴ = 1.4 × 10⁸ (c).

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