ABC26PH0887 · Chemical Kinetics

Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Exam: NET JUNE 2012 · 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
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Explanation
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