For the enzyme-catalyzed conversion of a certain substrate, K_M = 0.038 mol · L^–1 at 25°C. When the substrate concentration is 0.156 mol · L^–1, the rate of the reaction is 1.21 mmol · L^–1s^–1. The maximum rate of the conversion reaction is reached at high substrate concentrations (see Exercise 7.14). Calculate the maximum rate of this enzyme-catalyzed reaction.

Exercise 7.14

From the following mechanism, derive Eq. 1a in Topic 7E, which Michaelis and Menten proposed to represent the rate of formation of products in an enzyme-catalyzed reaction.

Show that the rate is independent of substrate concentration at high concentrations of substrate.

where E is the free enzyme, S is the substrate, ES is the enzyme– substrate complex, and P is the product. Note that the steady-state concentration of free enzyme will be equal to the initial concentration of the enzyme less the amount of enzyme that is present in the enzyme–substrate complex: [E] = [E]₀ = [ES].

Transcribed Image Text:

E+S ES ESE + P k, k' K