91Ó°ÊÓ

We are given a two-step enzyme-catalyzed reaction mechanism: \[ \begin{aligned} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{E} \cdot\mathrm{S} \\ \mathrm{E} \cdot\mathrm{S} \rightleftharpoons\mathrm{E}+\mathrm{P} \end{aligned} \] Here, E represents the enzyme, S represents the substrate, and P represents the product. The reaction consists of two reversible steps, where the enzyme (E) binds with the substrate (S) to form the enzyme-substrate (E · S) complex. This complex then transforms into the product (P) and releases the enzyme back into the system.

The graph of the reaction rate as a function of [S], the concentration of the substrate, depicts an initial increase in the reaction rate as the substrate concentration increases. However, at high substrate concentrations, the reaction rate plateaus and does not change further with an increase in [S].

As the substrate concentration increases, the reaction rate initially increases due to more frequent collisions between substrate molecules and enzyme molecules, resulting in the formation of enzyme-substrate complexes. This increases the rate of product formation. However, as the substrate concentration continues to increase, enzymes become saturated or fully occupied by the substrate molecules because there are already sufficient substrate molecules available for interaction. When all available enzyme molecules are bound to the substrate, no further increase in reaction rate is observed, as the enzyme is working at its maximum capacity. In this case, adding more substrate molecules does not affect the reaction rate because the enzyme concentration becomes the limiting factor. In conclusion, the reason for the plateau in the reaction rate at higher substrate concentrations is that the enzyme becomes fully saturated with substrate molecules and is working at its maximum capacity. Further increases in substrate concentration do not affect the reaction rate, as the enzyme concentration becomes the limiting factor.