Question

For enzyme-catalyzed reactions that follow the mechanism

$$\begin{gathered} E+S\rightleftharpoons E.S \\ E+S\rightleftharpoons E.P \end{gathered}$$

a graph of the rate as a function of [S], the concentration

of the substrate, has the following general appearance:

Note that at high substrate concentrations, the rate oflonger changes with [S]. Suggest a reason for this.

Short answer

When enzyme saturation is completed with the substrate. The concentration of enzyme-substrate complex does not increase which leads to the rate of reaction remaining constant.

Step-by-step solution

Describing the enzyme catalysis and enzyme-substrate complex

This process takes place in two steps. In 1st step, the substrate reacts with the enzyme to form an enzyme-substrate complex. In the second step, this complex breaks down to produce the products and enzymes again.

$$\begin{gathered} Enzyme–substratecomplexformation \\ E+S\rightleftharpoons ES \\ Enzyme+substrate\rightleftharpoons Enzyme-Substrate \end{gathered}$$

$$\begin{gathered} BreakageofE.Scomplexbond \\ ES\rightleftharpoons E+S \\ Enzyme-SubstrateComplex\rightleftharpoons E+P \end{gathered}$$

Describing the rate of the enzyme-catalyzed reaction using the enzyme-substrate complex

The rate of enzyme-catalyzed reactions depends upon the concentration of the enzyme-substrate complex .

The rate of the reaction increases slowly as the concentration of the substrate increases.

The concentration of the enzyme-substrate complex remains constant at the higher substrate concentrations. The enzyme is completely saturated with the substrate complex and does not increase.