Question

Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E, and ends with the release of product P from the enzyme.

E + S --> ES --> EP --> E + P

In many circumstances,

Km = [E][S]/[ES]

A. What proportion of enzyme molecules is bound to substrate when [S] = Km?
B. Recall that when [S] = Km, the reaction rate is ½Vmax. Does your answer to part A make sense in the light of this rate information?

Answer 1

When the substrate concentration equals the Michaelis-Menten constant (Km), 50% of the enzyme's active sites are occupied, corresponding to the reaction rate being at half of its maximum velocity (½Vmax). This proportion of enzyme-substrate complex formation logically reflects the enzyme's capacity at ½Vmax according to the Michaelis-Menten model.

Step-by-step explanation:

When the substrate concentration [S] equals the Michaelis-Menten constant (Km), it is indicative that half of the enzyme's active sites are bound with the substrate, meaning 50% of the enzyme molecules are in the enzyme-substrate complex (ES) form. This is because Km represents the substrate concentration at which the reaction rate is at half its maximum velocity (½Vmax). Therefore, when [S] = Km, 50% of the enzyme is bound to the substrate.

This finding aligns with the Michaelis-Menten model, which states that at ½Vmax, the enzyme is working at half of its capacity since half of the active sites are occupied. Since the maximum reaction rate (Vmax) occurs when all available enzyme active sites are saturated with the substrate, having 50% of the enzyme-substrate complexes at ½Vmax makes logical sense.