Effect of competitive inhibitor on substrate inhibition

Question

In an enzyme that undergoes substrate inhibition, how would the presence of a competitive inhibitor affect said substrate inhibition? Would the substrate concentration at which substrate inhibition begins be affected? Would the rate of enzyme activity decrease with increasing substrate be affected?

Answer 1

We need to deal with the mechanism of competitive inhibition+substrate inhibition: \begin{align} \ce{E + S <=> ES} \quad &K_\text{M} = \frac{\ce{[E]}\ce{[S]}}{\ce{[ES]}} \tag{R1/1} \\ \ce{ES -> E + P} \quad &r = k[\ce{ES}] \tag{R2/2} \\ \ce{E + I <=> EI} \quad &K_\text{I} = \frac{\ce{[E]}\ce{[I]}}{\ce{[EI]}} \tag{R3/3} \\ \ce{ES + S <=> ES2} \quad &K_\text{S} = \frac{\ce{[ES]}\ce{[S]}}{\ce{[ES2]}} \tag{R4/4} \end{align} where:

• $\ce{E}$ is the enzyme.
• $\ce{ES}$ is the enzyme-substrate complex.
• $\ce{I}$ is the inhibitor.
• $\ce{EI}$ is the enzyme-inhibitor complex.
• $\ce{ES2}$ is the compound generated by the reaction of the enzyme and the enzyme-substrate complex, that is taking away the substrate the enzyme needs.
• The $K$'s are the equilibrium constants, e.g., $K_\text{M}$ is the Michaelis's one. In biochemistry, it is standard practice to write them upside down.

Our task is to find an expression for the rate of formation of the product $\ce{P}$, in terms of the substrate concentration $\ce{S}$, the variable we can easily track.

Enzyme balance The initial concentration of enzyme is equal to \begin{align} C_\text{E0} &= \underbrace{[\ce{E}]}_1 + [\ce{ES}] + \underbrace{[\ce{EI}]}_3 + \underbrace{[\ce{ES2}]}_4 \tag{5} \\ \end{align} The underbraced parts (1), (3), and (4) are obtained via Eqs. (1), (3), and (4) \begin{align} K_\text{M} = \frac{\ce{[E]}\ce{[S]}}{\ce{[ES]}} &\rightarrow \ce{[E]} = \frac{K_\text{M}\ce{[ES]}}{\ce{[S]}} \tag{6} \\ K_\text{I} = \frac{\ce{[E]}\ce{[I]}}{\ce{[EI]}} &\rightarrow \ce{[EI]} = \left(\frac{\ce{[I]}}{K_\text{I}}\right)\ce{[E]} \tag{7} \\ K_\text{S} = \frac{\ce{[ES]}\ce{[S]}}{\ce{[ES2]}} &\rightarrow \ce{[ES2]} = \frac{\ce{[ES]}\ce{[S]}}{K_\text{S}} \tag{8} \end{align} and combining Eqs. (5-8) \begin{align} C_\text{E0} &= \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} + [\ce{ES}] + \left(\frac{\ce{[I]}}{K_\text{I}}\right)\color{blue}{\ce{[E]}} + \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{9} \\ \end{align} In Eq. (9) the enzyme concentration apears again in blue, so we combine Eq. (9) and Eq. (6) again \begin{align} C_\text{E0} &= \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} + [\ce{ES}] + \left(\frac{\ce{[I]}}{K_\text{I}}\right) \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} + \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\ C_\text{E0} &= \color{blue}{\frac{K_\text{M}[\ce{ES}]}{[\ce{S}]}} + [\ce{ES}] + \color{blue}{\frac{K_\text{M}\ce{[ES]}}{[\ce{S}]} \left(\frac{\ce{[I]}}{K_\text{I}}\right)} + \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\ C_\text{E0} &= \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} \underbrace{\left(1 + \frac{[\ce{I}]}{K_\text{I}}\right)}_\alpha + [\ce{ES}] + \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{10} \\ \end{align} where we factored out the repeating term in both blue parts of the equation.

The main issue with Eq. (10), is that $\ce{[I]}$ is still there. We could use an inhibitor balance, but unfortunately, we can prove that we can't have a simple final expression. If you want further math just say in the comments. I will name underbraced part with $\alpha$, and name it the inhibitor power. We will explore its consequences further. Eq. (10) now becomes \begin{align} C_\text{E0} &= \frac{[\ce{ES}]}{[\ce{S}]}K_\text{M} \alpha + [\ce{ES}] + \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\ C_\text{E0} &= [\ce{ES}] \left( \frac{K_\text{M} \alpha}{[\ce{S}]} + 1 + \frac{[\ce{S}]}{K_\text{S}}\right) \\ C_\text{E0} &= [\ce{ES}] \left( \frac{K_\text{M}\alpha + [\ce{S}]}{[\ce{S}]} + \frac{[\ce{S}]}{K_\text{S}}\right) \\ C_\text{E0} &= [\ce{ES}] \left( \frac{K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2} {K_\text{S}[\ce{S}]}\right)\\ [\ce{ES}] &= \left(\frac{K_\text{S}[\ce{S}]} {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2}\right)C_\text{E0} \tag{11} \\ \end{align} Combining Eqs. (2) and (11) \begin{equation} r = \frac{kK_\text{S}[\ce{S}]C_\text{E0}} {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2} \rightarrow \boxed{\frac{r}{kC_\text{E0}} = \frac{[\ce{S}]} {K_\text{M}\alpha + [\ce{S}] + \dfrac{[\ce{S}]^2}{K_\text{S}}}} \tag{12} \\ \end{equation} Eq. (12) is exactly what we want, a rate which is only a function of the substrate concentration. I also put it in a dimensionless form, which will be easier to analyze. The plot of Eq. (12) is the following, where I employed for simplicity $K_\text{S} = K_\text{M} = 1 \; \pu{mol/dm3}$:

Observations:

• If $\alpha = 1$, the inhibitor power is zero, and $\ce{R3}$ just vanishes. We end with only substrate inhibition.
• As $\alpha$ goes up, $\ce{R3}$ is more aggressive. This means that more enzime is consumed and be inactivated. Thus, the reaction goes down for every concentration of substrate.
• Any operation with too high concentrations of a substrate is discouraged, since $\ce{R4}$ is favored, and the substrate inhibition is too predominant.
• For all cases, we have an optimum value of substrate concentration, where the rate is maximum. This is really obvious in Eq. (12). In the numerator we have a linear function, that will win when $\ce{[S]}$ is low, and in the denominator a quadratic function that will when when $\ce{[S]}$ is high. This is a typical characteristic that substrate inhibition displays. We can prove that this happens in $$ \frac{\mathrm{d}(r/kC_\text{E0})}{\mathrm{d}[S]} = 0 \rightarrow \boxed{[\ce{S}]_\text{opt} = \sqrt{K_\text{S}K_\text{M}\alpha}} \tag{13} $$ This is the curve in magenta, thus, the best "ideal" operation will be the one that follows this curve for every instant in time in the biochemical reactor, disregarding the spatial effects.