the slope and intercept of plot between $1 / \mathrm{r}$ and $1 /[\mathrm{S}]$ are $10^{-2} \mathrm{s}$ and $10^{2} \mathrm{M}^{-1} \mathrm{s}$, respectively, If $\mathrm{E}_{0}=10^{-6} \mathrm{M}$ and $\frac{\mathrm{k}_{-1}}{\mathrm{k}_{2}}=1000$, the value of $\mathrm{k}_{1}$ will be close to (in units of $\mathrm{M}^{-1} \mathrm{s}^{-1}$ ) [r is the rate of reaction and $\mathrm{E}_{0}$ is initial concentration of enzyme
(a)$1 \times 10^{11}$
(b)$1 \times 10^{4}$
(c)$1 \times 10^{8}$
(d)$1 \times 10^{6}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
From the intercept, k₂E₀ = 10⁻², so k₂ = 10⁴ s⁻¹. From the slope, K_M = 10⁻⁴ M. k₁ = (k₋₁ + k₂)/K_M = 1001 × 10⁴/10⁻⁴ ≈ 1 × 10¹¹ M⁻¹ s⁻¹ (a).