Question

The rates of enzyme catalysis can be lowered by the presence of inhibitor molecules I, which bind to the active site of the enzyme. This adds the following additional step to the reaction mechanism considered in Section 18.7

$\text{E+IEI}\underset{{\mathrm{k}}_2}{\overset{{\mathrm{k}}_1}{\rightleftarrows \mathrm{EI}}}$

Determine the effect of the presence of inhibitor at total concentration $\left[I_0\right]=\left[I\right]+\left[EI\right]$ on the rate expression for formation of products derived at the end of this chapter.

Short answer

The rate expression is given as:

$\mathrm{Rate}=\frac{{\mathrm{k}}_2\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[{\mathrm{I}}_0\right]+{\mathrm{k}}_1\left[\mathrm{I}\right]\right)}{\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).}$

Step-by-step solution

Net rate of ES concentration

The overall enzyme-catalyzed reaction mechanism in the presence of inhibitor molecules I is given by

$$\begin{gathered} \mathrm{E}+\mathrm{S}\underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\rightleftharpoons }}\mathrm{ES} \\ \mathrm{ES}\stackrel{{\mathrm{k}}_2}{\to }\mathrm{E}+\mathrm{P} \\ \mathrm{E}+\mathrm{I}\underset{{\mathrm{k}}_3}{\overset{{\mathrm{k}}_3}{\leftrightharpoons }}\mathrm{EI} \end{gathered}$$

Where E is the enzyme,S is the substrate,I is the inhibitor,ES is the enzyme-substrate complex and EI is the enzyme inhibitor complex.

The rate expression for the above reaction is given as

$\mathrm{Rate}={\mathrm{k}}_2\left[\mathrm{ES}\right]........\left(1\right)$

In the above mechanism ES is formed at the rate of and is lost at the rate${\mathrm{k}}_1\left[\mathrm{ES}\right]$and is lost at the rate of ${\mathrm{k}}_{-1}\left[\mathrm{ES}\right]\mathrm{and}{\mathrm{k}}_2\left[\mathrm{ES}\right]$Therefore the net rate of change of $\left[\mathrm{ES}\right]$is given as:

$\frac{\mathrm{d}\left[\mathrm{ES}\right]}{\mathrm{dt}}={\mathrm{k}}_1\left[\mathrm{E}\right]\left[\mathrm{S}\right]-{\mathrm{k}}_{-1}\left[\mathrm{ES}\right]-{\mathrm{k}}_2\left[\mathrm{ES}\right]........\left(2\right)$

Steady-state approximation

In certain reaction mechanism, some are slow steps ,while others are fast steps. In order to predict the rate law , one uses the steady state approximation principle, which states that the intermediate formed in the reaction ie ES remains constant throughout the reaction .In this reaction $\left[\mathrm{ES}\right]$is considered as reaction intermediate and hence

$\frac{\mathrm{d}\left[\mathrm{ES}\right]}{\mathrm{dt}}=0$

Rate of the reaction

From equation 2 , enzyme concentration is given as $\left[\mathrm{E}\right]$, finally total enzyme concentration is given as $\left[{\mathrm{E}}_{\mathrm{T}}\right]$and it is sum of free enzyme concentration, $\left[E_T\right]$enzyme substrate and enzyme inhabitor concentration ie $\left[\mathrm{ES}\right]\mathrm{and}\left[\mathrm{EI}\right]$Concentration.

$$\begin{gathered} \left[{\mathrm{E}}_{\mathrm{T}}\right]=\left[\mathrm{E}\right]+\left[\mathrm{ES}\right]+\left[\mathrm{EI}\right]...........\left(3\right) \\ \mathrm{or}\left[\mathrm{E}\right]=\left[\mathrm{E}\right]+\left[\mathrm{ES}\right]+\left[\mathrm{EI}\right] \end{gathered}$$

Substituting the value of $\left[\mathrm{E}\right]$ in equation-2

$$\begin{gathered} \frac{\mathrm{d}\left[\mathrm{ES}\right]}{\mathrm{dt}}={\mathrm{k}}_1\left(\left[{\mathrm{E}}_{\mathrm{T}}\right]-\left[\mathrm{ES}\right]-\left[\mathrm{EI}\right]\right)\left[\mathrm{S}\right]-{\mathrm{k}}_{-1}\left[\mathrm{ES}\right]-{\mathrm{k}}_2\left[\mathrm{ES}\right] \\ ={\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]\left[\mathrm{S}\right]-{\mathrm{k}}_1\left[\mathrm{ES}\right]\left[\mathrm{S}\right]-{\mathrm{k}}_{-1}\left[\mathrm{ES}\right]-{\mathrm{k}}_2\left[\mathrm{ES}\right] \\ \frac{\mathrm{d}\left[\mathrm{ES}\right]}{\mathrm{dt}}=\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[\mathrm{EI}\right]\right)-\left[\mathrm{ES}\right]\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).....\left(4\right) \end{gathered}$$

Comparing eqaution 3 and 5

$$\begin{gathered} 0=\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[\mathrm{EI}\right]\right)-\left[\mathrm{ES}\right]\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right) \\ \left[\mathrm{ES}\right]=\frac{\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[\mathrm{EI}\right]\right)}{\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).}.........\left(5\right) \end{gathered}$$

But here

localid="1662096631238" $$\begin{gathered} \left[{\mathrm{I}}_0\right]=\left[\mathrm{I}\right]+\left[\mathrm{EI}\right]......\left(6\right) \\ \mathrm{or} \\ \left[{\mathrm{I}}_0\right]-\left[\mathrm{I}\right]=\left[\mathrm{EI}\right]...\left(6a\right) \end{gathered}$$

Substitute the value of $\left[\mathrm{EI}\right]$ in eqaution-5

localid="1662096663307" $$\begin{gathered} \left[\mathrm{ES}\right]=\frac{\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[\mathrm{EI}\right]\right)}{\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).}.........\left(5\right) \\ \left[{\mathrm{I}}_0\right]-\left[\mathrm{I}\right]=\left[\mathrm{EI}\right] \\ \left[\mathrm{ES}\right]=\frac{\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[{\mathrm{I}}_0\right]-\left[\mathrm{I}\right]\right)}{\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).}.......\left(7\right) \end{gathered}$$

Substituting the value of $\left[\mathrm{ES}\right]$in eqaution-2

localid="1662096687748" $\mathrm{Rate}=\frac{{\mathrm{k}}_2\left[\mathrm{S}\right]\left({\mathrm{k}}_1\left[{\mathrm{E}}_{\mathrm{T}}\right]-{\mathrm{k}}_1\left[{\mathrm{I}}_0\right]+{\mathrm{k}}_1\left[\mathrm{I}\right]\right)}{\left({\mathrm{k}}_1\left[\mathrm{S}\right]+{\mathrm{k}}_{-1}+{\mathrm{k}}_2\right).}......\left(8\right)$

From the expression it is concluded that rate of enyzme catalysed reaction decreases on increasing the concentration of inhabitor.