35\. The ability of enzymes to catalyze reactions can be hindered by inhibitor
molecules. One of the mechanisms by which an inhibitor molecule works is by
competing with the substrate molecule for binding to the active site of the
enzyme. We can include this inhibition reaction in a modified Michaelis-Menton
mechanism for enzyme catalysis.
$$
\begin{aligned}
&\mathrm{E}+\mathrm{S} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}
\mathrm{ES} \\
&\mathrm{E}+\mathrm{I} \stackrel{\stackrel{k_{2}}{\rightleftharpoons{k}_{-2}}}
\mathrm{EI}
\end{aligned}
$$
$$
\mathrm{ES} \stackrel{k_{3}}{\longrightarrow} \mathrm{E}+\mathrm{P}
$$
In Equation \(2, \mathrm{I}\) is the inhibitor molecule and EI is the enzyme-
inhibitor complex. We will consider the case where reaction (2) is always in
equilibrium. Determine the rate laws for [S], [ES], [EI], and [P]. Show that
if the steady-state assumption is applied to ES, then
$$
[\mathrm{ES}]=\frac{[\mathrm{E}][\mathrm{S}]}{K_{m}}
$$
where \(K_{m}\) is the Michaelis constant, \(K_{m}=\left(k_{-1}+k_{3}\right) /
k_{1} .\) Now show that material balance for the enzyme gives
$$
[\mathrm{E}]_{0}=[\mathrm{E}]+\frac{[\mathrm{E}][\mathrm{S}]}{K_{m}}+[\mathrm{E}][\mathrm{I}]
K_{\mathrm{I}}
$$
where \(K_{\mathrm{I}}=[\mathrm{EI}] /[\mathrm{E}][\mathrm{I}]\) is the
equilibrium constant for step (2) of the above reaction mechanism. Use this
result to show that the initial reaction rate is given by
$$
v=\frac{d[\mathrm{P}]}{d
t}=\frac{k_{3}[\mathrm{E}]_{0}[\mathrm{~S}]}{K_{m}+[\mathrm{S}]+K_{m}
K_{\mathrm{I}}[\mathrm{I}]} \approx
\frac{k_{3}[\mathrm{E}]_{0}[\mathrm{~S}]_{0}}{K_{m^{\prime}}+[\mathrm{S}]_{0}}
$$
where \(K_{m^{\prime}}=K_{m}\left(1+K_{1}[\mathrm{I}]\right)\). Note that the
second expression in Equation 4 has the same functional form as the Michaelis-
Menton equation. Does Equation 4 reduce to the expected result when
\([\mathrm{I}] \rightarrow 0 ?\)
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