An enzyme has $V_{ma x} = 80 μ M mi n^{- 1}$ and $K_{m} = 4.0 mM$ . A competitive inhibitor at $[I] = 2.0 mM$ with $K_{i} = 1.0 mM$ is added. Using $K_{m}^{a pp} = K_{m} (1 + [I] / K_{i})$ and $v = V_{ma x} [S] / (K_{m}^{a pp} + [S])$ , find $v$ in $μ M mi n^{- 1}$ at $[S] = 6.0 mM$ , to two decimals.

Numerical answer type, answer in uM/min, tolerance 0.02.

Answer: 26.67 uM/min

K_{m}^{a pp} = 4.0 (1 + 2.0/1.0) = 12.0 mM

. Then

v = 80 \times 6.0/ (12.0 + 6.0) = 480/18 = 26.667

, i.e. 26.67

μ M mi n^{- 1}