An enzyme has $K_{m} = 2.0 mM$ . In the presence of a competitive inhibitor ( $[I] = 3.0 mM$ , $K_{i} = 1.5 mM$ ) the apparent Michaelis constant is $K_{m}^{a pp} = K_{m} (1 + [I] / K_{i})$ and $V_{ma x}$ is unchanged. At $[S] = 6.0 mM$ , the ratio of the rate with inhibitor to the rate without inhibitor is

Answer: D

K_{m}^{a pp} = 2.0 (1 + 3.0/1.5) = 6.0 mM

. With inhibitor

v / V_{ma x} = 6/ (6 + 6) = 0.5

. Without inhibitor

v / V_{ma x} = 6/ (2 + 6) = 0.75

. The ratio is

0.5/0.75 = 2/3