Q24 · CSIR-NET Chemistry, December 2014

Paper: CSIR-NET December 2014 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Medium

For an enzyme-substrate reaction, a plot between $\dfrac{1}{v}$ and $\dfrac{1}{[S_0]}$ yields a slope of $40\,\mathrm{s}$. If the enzyme concentration is $2.5\,\mu\mathrm{M}$, then the catalytic efficiency of the enzyme is
(a)$40\,\mathrm{L\,mol^{-1}\,s^{-1}}$
(b)$10^{-4}\,\mathrm{L\,mol^{-1}\,s^{-1}}$
(c)$10^{7}\,\mathrm{L\,mol^{-1}\,s^{-1}}$
(d)$10^{4}\,\mathrm{L\,mol^{-1}\,s^{-1}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Lineweaver–Burk slope = Km/Vmax = Km/(kcat[E]) = 40 s → kcat/Km = 1/(40 × 2.5×10⁻⁶) = 10⁴ L mol⁻¹ s⁻¹ (d).

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