Question

A fatty acid hydroxylase enzyme has a Km for palmitic acid of 7.31 micromolar and a turnover rate of 1,267.9 micromole substrate per minute per micromole enzyme. If 54.7 nanomolar enzyme is mixed with an unknown concentration of substrate and the reaction rate was 48.83 micromolar substrate per minute, how much substrate was in the reaction to start with? Report your answer in micromolar to the nearest 0.1 uM.

Answer 1

k_cat = Vmax/ [E]

Vmax = k_cat * [E]

           = 1267.9 micromole * 54.7 nanomole

           = 1267.9 micromole * (54.7 * 0.001)micromole

           = 69.354 micromole/ min

According to Michaelis-Menten Kinetics

V = Vmax [S]/Km + [S]

V (Km + [S]) = Vmax [S]

48.83 (7.31 + [S]) = 69.354 *[S]

[S]= 17.39 micrometer ~ 17.4 micrometer