An enzyme obeys $E + S k X_{1} k X_{- 1} ES k X_{2} E + P$ with $k_{1} = 1.0 \times 1 0^{6} M^{- 1} s^{- 1}$ , $k_{- 1} = 5.0 \times 1 0^{3} s^{- 1}$ , $k_{2} = 5.0 \times 1 0^{3} s^{- 1}$ . Applying the steady-state approximation to ES, with $[E]_{0} = 2.0 \times 1 0^{- 6} M$ and $[S] ≫ [E]_{0}$ equal to $1.0 \times 1 0^{- 2} M$ , the rate of product formation in $M s^{- 1}$ is

Answer: D

Steady state for ES gives

[ES] = [E]_{0} [S] / (K_{M} + [S])

with

K_{M} = (k_{- 1} + k_{2}) / k_{1} = 1.0 \times 1 0^{- 2} M

. Rate

= k_{2} [ES] = (5.0 \times 1 0^{3}) (2.0 \times 1 0^{- 6}) (1.0 \times 1 0^{- 2}) / (1.0 \times 1 0^{- 2} + 1.0 \times 1 0^{- 2}) = 5.0 \times 1 0^{- 3} M s^{- 1}

. The value

1.0 \times 1 0^{- 2}

V_{m a x} = k_{2} [E]_{0}

, and using

K_{M} = k_{- 1} / k_{1}

would give

6.7 \times 1 0^{- 3}