$v_{\max }$ and $\mathrm{K}_{\mathrm{m}}$ for an enzyme catalyzed reaction are $2.0 \times 10^{-3} \mathrm{Ms}^{-1}$ and $1.0 \times 10^{-6} \mathrm{M}$, respectively. The rate of the reaction when the substrate concentration is $1.0 \times 10^{-6} \mathrm{M}$ is
(a)$3.0 \times 10^{-3} \mathrm{s}^{-1}$
(b)$1.0 \times 10^{-3} \mathrm{s}^{-1}$
(c)$2.0 \times 10^{-3} \mathrm{s}^{-1}$
(d)$0.5 \mathrm{s}^{-1}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
v = V_max[S]/(K_M + [S]) = 2×10⁻³ × (10⁻⁶/2×10⁻⁶) = 1×10⁻³ M s⁻¹.