An enzyme obeys v = dfracV_max[S]K_m+[S] with V_max = 120 muM min^-1 and K_m = 3.2 mM. Find v (in muM min^-1) at [S] = 5.6 mM, to two decimals.

JEE Main levelNumerical answerDifficulty 2 of 3sc084

An enzyme obeys $v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]}$ with $V_{ma x} = 120 μ M mi n^{- 1}$ and $K_{m} = 3.2 mM$ . Find $v$ (in $μ M mi n^{- 1}$ ) at $[S] = 5.6 mM$ , to two decimals.

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

v = 120 \times 5.6/ (3.2 + 5.6) = 672/8.8 = 76.364

, i.e. 76.36

μ M mi n^{- 1}