Competitive inhibition

Competitive inhibition is a form of enzyme inhibition where binding of the inhibitor to the active site on the enzyme prevents binding of the substrate and vice versa.^[1][2]

Most competitive inhibitors function by binding reversibly to the active site of the enzyme.^[1] As a result, many sources state that this is the defining feature of competitive inhibitors.^[3][4] This, however, is a misleading oversimplification, as there are many possible mechanisms by which an enzyme may bind either the inhibitor or the substrate but never both at the same time.^[1] For example, allosteric inhibitors may display competitive, non-competitive, or uncompetitive inhibition.^[1]

Mechanism

Diagram showing competitive inhibition

In competitive inhibition, at any given moment, the enzyme may be bound to the inhibitor, the substrate, or neither, but it cannot bind both at the same time.

In virtually every case, competitive inhibitors bind in the same binding site as the substrate, but same-site binding is not a requirement. A competitive inhibitor could bind to an allosteric site of the free enzyme and prevent substrate binding, as long as it does not bind to the allosteric site when the substrate is bound. For example, strychnine acts as an allosteric inhibitor of the glycine receptor in the mammalian spinal cord and brain stem. Glycine is a major post-synaptic inhibitory neurotransmitter with a specific receptor site. Strychnine binds to an alternate site that reduces the affinity of the glycine receptor for glycine, resulting in convulsions due to lessened inhibition by the glycine.^[5]

In competitive inhibition, the maximum velocity ($V_\max$) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased (the $K_d$ dissociation constant is apparently increased). The change in $K_m$ (Michaelis-Menten constant) is parallel to the alteration in $K_d$. Any given competitive inhibitor concentration can be overcome by increasing the substrate concentration in which case the substrate will outcompete the inhibitor in binding to the enzyme.

Competitive inhibition can also be allosteric, as long as the inhibitor and the substrate cannot bind the enzyme at the same time.

Another possible mechanism for allosteric competitive inhibition.

Equation

Competitive inhibition increases the apparent value of the Michaelis-Menten constant, $K^\text{app}_m$, such that initial rate of reaction, $V_0$, is given by

$V_0 = \frac{V_\max\,[S]}{K^\text{app}_m + [S]}$

where $K^\text{app}_m=K_m(1+[I]/K_i)$, $K_i$ is the inhibitor's dissociation constant and $[I]$ is the inhibitor concentration.

$V_\max$ remains the same because the presence of the inhibitor can be overcome by higher substrate concentrations. $K^\text{app}_m$, the substrate concentration that is needed to reach $V_\max / 2$, increases with the presence of a competitive inhibitor. This is because the concentration of substrate needed to reach $V_\max$ with an inhibitor is greater than the concentration of substrate needed to reach $V_\max$ without an inhibitor.

Derivation

In the simplest case of a single-substrate enzyme obeying Michaelis-Menten kinetics, the typical scheme

$E + S \, \overset{k_1}\underset{k_{-1}} \rightleftharpoons \, ES \, \overset{k_2} {\longrightarrow} \, E + P$

is modified to include binding of the inhibitor to the free enzyme:

$EI + S \, \overset{k_{-3}}\underset{k_3} \rightleftharpoons \, E + S + I \, \overset{k_1}\underset{k_{-1}} \rightleftharpoons \, ES + I \, \overset{k_2} {\longrightarrow} \, E + P + I$

Note that the inhibitor does not bind to the ES complex and the substrate does not bind to the EI complex. It is generally assumed that this behavior is indicative of both compounds binding at the same site, but that is not strictly necessary. As with the derivation of the Michaelis-Menten equation, assume that the system is at steady-state, i.e. the concentration of each of the enzyme species is not changing.

$\frac{d[E]}{dt} = \frac{d[ES]}{dt} = \frac{d[EI]}{dt} = 0.$

Furthermore, the known total enzyme concentration is $[E]_0 = [E] + [ES] + [EI]$, and the velocity is measured under conditions in which the substrate and inhibitor concentrations do not change substantially and an insignificant amount of product has accumulated.

We can therefore set up a system of equations:

$[E]_0 = [E] + [ES] + [EI]\,\!$

(1)

$\frac{d[E]}{dt} = 0 = -k_1[E][S] + k_{-1}[ES] + k_2[ES] -k_3[E][I] + k_{-3}[EI]$

(2)

$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$

(3)

$\frac{d[EI]}{dt} = 0 = k_3[E][I] - k_{-3}[EI]$

(4)

where $[S]$, $[I]$ and $[E]_0$ are known. The initial velocity is defined as $V_0 = d[P]/dt = k_2 [ES]$, so we need to define the unknown $[ES]$ in terms of the knowns $[S]$, $[I]$ and $[E]_0$.

From equation (3), we can define E in terms of ES by rearranging to

$k_1[E][S]=(k_{-1}+k_2)[ES] \,\!$

Dividing by $k_1[S]$ gives

$[E] = \frac{(k_{-1}+k_2)[ES]}{k_1[S]}$

As in the derivation of the Michaelis-Menten equation, the term $(k_{-1}+k_2)/k_1$ can be replaced by the macroscopic rate constant $K_m$:

$[E] = \frac{K_m[ES]}{[S]}$

(5)

Substituting equation (5) into equation (4), we have

$0 = \frac{k_3[I]K_m[ES]}{[S]} - k_{-3}[EI]$

Rearranging, we find that

$[EI] = \frac{K_m k_3[I][ES]}{k_{-3}[S]}$

At this point, we can define the dissociation constant for the inhibitor as $K_i = k_{-3}/k_3$, giving

$[EI] = \frac{K_m[I][ES]}{K_i[S]}$

(6)

At this point, substitute equation (5) and equation (6) into equation (1):

$[E]_0 = \frac{K_m[ES]}{[S]} + [ES] + \frac{K_m[I][ES]}{K_i[S]}$

Rearranging to solve for ES, we find

$[E]_0 = [ES] \left ( \frac{K_m}{[S]} + 1 + \frac{K_m[I]}{K_i[S]} \right )= [ES] \frac{K_m K_i + K_i[S] + K_m[I]}{K_i[S]}$

$[ES] = \frac{K_i [S][E]_0}{K_m K_i + K_i[S] + K_m[I]}$

(7)

Returning to our expression for $V_0$, we now have:

$V_0 = k_2[ES] = \frac{k_2 K_i [S][E]_0}{K_m K_i + K_i[S] + K_m[I]}$

$V_0 = \frac{k_2 [E]_0 [S]}{K_m + [S] + K_m\frac{[I]}{K_i}}$

Since the velocity is maximal when all the enzyme is bound as the enzyme-substrate complex, $V_\max = k_2 [E]_0$. Replacing and combining terms finally yields the conventional form:

$V_0 = \frac{V_{\max}[S]}{K_m(1 + \frac{[I]}{K_i}) + [S]}$

(8)

To compute the concentration of competitive inhibitor $[I]$ that yields a fraction $f_{V{_0}}$ of velocity $V_0$ where $0 < f_{V{_0}} < 1$:

$[I]= (\frac{1}{f_{V{_0}}} -1)K_i(1+\frac{[S]}{K_m})$

(9)

Notes and references

1. "Types of Inhibition". NIH Center for Translational Therapeutics. Retrieved 2 April 2012.
2. "Competitive Inhibition". Retrieved 2 April 2012.
3. Ophardt, Charles. "Virtual Chembook". Elmhurst College. Retrieved 1 September 2015.
4. "Enzyme Inhibition". Retrieved 2 April 2012.
5. Dick RM (2011). "Chapter 2. Pharmacodynamics: The Study of Drug Action". In Ouellette R, Joyce JA. Pharmacology for Nurse Anesthesiology. Jones & Bartlett Learning. ISBN 978-0-7637-8607-6.

See also

• Schild regression for ligand receptor inhibition
• Non-competitive inhibition