Q20 · CSIR-NET Chemistry, December 2017

Paper: CSIR-NET December 2017 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Medium

The slope and intercept obtained from (1/Rate) against (1/substrate concentration) of an enzyme catalyzed reaction are 300 and $2 \times 10^{5}$, respectively. The Michaelis-Menten constants of the enzyme in this reaction is
(a)$5 \times 10^{6}\,\mathrm{M}$
(b)$5 \times 10^{-6} \mathrm{M}$
(c)$1.5 \times 10^{3}\,\mathrm{M}$
(d)$1.5 \times 10^{-3} \mathrm{M}$
Answer
Answer: D ✓ checked by 4AB · confidence high

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

Explanation
Lineweaver–Burk: slope = K_M/V_max = 300, intercept = 1/V_max = 2×10⁵ ⇒ K_M = 300/2×10⁵ = 1.5×10⁻³ M.

Study loop for Chemical Kinetics

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