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JEE Main levelNumerical answerDifficulty 2 of 3sc085

For an enzyme with $K_{m} = 3.2 mM$ obeying $v = V_{ma x} [S] / (K_{m} + [S])$ , at what substrate concentration (in mM) is $v = 0.40 V_{ma x}$ ? Give two decimals.

Numerical answer type, answer in mM, tolerance 0.02.

Answer: 2.13 mM

0.40 = [S] / (K_{m} + [S])

gives

[S] = 0.40 K_{m} /0.60 = 3.2 \times 2/3 = 2.1333

, i.e. 2.13 mM.

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