Enzymes are often described as following the two-step mechanism: $$ \begin{array}{c}{\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \text { (fast) }} \\ {\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad(\text { slow })}\end{array}$$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate,\(\mathrm{ES}=\) enzyme-substrate complex, and \(\mathrm{P}=\) product.(a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

The given two-step mechanism of the enzyme-catalyzed reaction is: \[ \begin{array}{c} \text{E + S} \rightleftharpoons \text{ES} \quad (\text{fast}) \\ \text{ES} \longrightarrow \text{E + P} \quad (\text{slow}) \end{array} \]

Since the second step is the slow (rate-limiting) step, the overall rate of the reaction depends on the rate of the ES complex formation. The rate law for the second step can be written as follows: \[ \text{Rate} = k_2[\text{ES}] \] To determine the rate law for the overall reaction, we need to express [ES] in terms of [E] and [S]. To do this, we will look at the first step and assume that it is in a rapid equilibrium state, which means that the forward and reverse rates are equal: \[ k_1[\text{E}][\text{S}] = k_{-1}[\text{ES}] \] Solving for [ES], we get: \[ [\text{ES}] = \frac{k_1}{k_{-1}} [\text{E}][\text{S}] \] Now we can substitute this expression for [ES] in the rate law we derived before: \[ \text{Rate} = k_2\left(\frac{k_1}{k_{-1}} [\text{E}][\text{S}]\right) \] Finally, we define \(\alpha = \frac{k_1k_2}{k_{-1}}\), the rate law for the reaction is: \[ \text{Rate}=\alpha [\text{E}][\text{S}] \]

Now we need to add an additional step accounting for the interaction of enzyme E with the inhibitor I. Since the inhibitor binds to the active site of the enzyme, it competes with the substrate, and the resulting enzyme-inhibitor complex EI is not converted into the product. The additional elementary step could be written as: \[ \text{E + I} \rightleftharpoons \text{EI} \quad (\text{rapid equilibrium}) \] With this additional step, the enzyme's interaction with the inhibitor is now incorporated into the mechanism.