For an enzyme catalyzed reaction, a Lineweaver-Burk plot gave the following data: slope =40 s, intercept =4 (mmol dm⁻³ s⁻¹)⁻¹. If the initial concentration of enzyme is 2.5 × 10⁻⁹ mol dm⁻³, what is the catalytic efficiency (in dm⁻³.mol⁻¹ s⁻¹) of the reaction?

Q103 · CSIR-NET Chemistry, June 2011

Paper: CSIR-NET June 2011 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Medium

For an enzyme catalyzed reaction, a Lineweaver-Burk plot gave the following data: slope $=40 \mathrm{s}$, intercept $=4\left(\mathrm{mmol} \mathrm{dm}^{-3} \mathrm{s}^{-1}\right)^{-1}$. If the initial concentration of enzyme is $2.5 \times 10^{-9} \mathrm{mol} \mathrm{dm}^{-3}$, what is the catalytic efficiency (in $\mathrm{dm}^{-3} \left.\mathrm{mol}^{-1} \mathrm{s}^{-1}\right)$ of the reaction?

(a)$10^{5}$

(b)$10^{6}$

(c)$10^{7}$

(d)$10^{4}$

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

V_max = 1/intercept = 0.25 mmol dm⁻³ s⁻¹ = 2.5×10⁻⁴ M s⁻¹; k_cat = V_max/[E]₀ = 1×10⁵ s⁻¹; K_M = slope × V_max = 40 × 2.5×10⁻⁴ = 0.01 M. Catalytic efficiency k_cat/K_M = 10⁵/0.01 = 10⁷ dm³ mol⁻¹ s⁻¹.

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