For enzyme-catalyzed reactions that follow the mechanism $$\mathbf{E}+\mathbf{S} \rightleftharpoons \mathbf{E} \cdot \mathbf{S}$$ $$\mathrm{E} \cdot \mathrm{S} \rightleftharpoons \mathrm{E}+\mathrm{P}$$ a graph of the rate as a function of \([\mathrm{S}]\), the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

First, we need to understand the reaction mechanism. The enzyme (E) interacts with the substrate (S) to form an enzyme-substrate complex (E·S). This step is reversible, meaning the complex can either dissociate back into enzyme and substrate or proceed to the second step. In the second step, the enzyme-substrate complex (E·S) breaks down into enzyme (E) and product (P). This step is also reversible. The rate of the enzyme-catalyzed reaction depends on the concentration of the substrate [S].

At low substrate concentrations, the reaction rate increases as the substrate concentration increases. This is because there are more substrate molecules available for the enzymes to interact with, increasing the probability of the enzyme-substrate complex forming. However, at higher substrate concentrations, the rate no longer changes with [S]. This is because in high concentration of substrate, most enzyme molecules have already formed complexes with substrate molecules. This means that the enzyme is saturated, and the enzymes cannot take up any additional substrate molecules to react.

As we increase [S], there is a point where the enzymes are working at their maximum capacity and cannot work any faster. This plateauing of reaction rate is due to the saturation of enzyme with the substrate. Since all the enzyme molecules are occupied with substrate and forming enzyme-substrate complexes, adding more substrate does not result in any increase in the reaction rate because there are no free enzyme molecules available for additional substrates to interact with. To conclude, at higher substrate concentrations, the rate of the enzyme-catalyzed reaction no longer changes with [S] because the enzyme molecules are saturated with the substrate, and further increases in substrate concentration does not result in any increase in the reaction rate.