Competitive inhibition Information & Competitive inhibition Links at HealthHaven.com

Competitive inhibition is a form of enzyme inhibition where binding of the inhibitor to the enzyme prevents binding of the substrate and vice versa.

[edit] Mechanism

Diagram showing competitive inhibition

In competitive inhibition, the inhibitor binds only the free enzyme, and cannot bind when the substrate is bound (in other words, it cannot bind the enzyme-substrate complex). In virtually all cases, 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 cannot bind to the allosteric site when the substrate is bound. 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 (it means: 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.

[edit] Equation

$\text{apparent } K_m=K_m\times \left(1+\frac{[I]}{K_I}\right)$

$V max$ remains the same because the presence of the inhibitor can be overcome by higher substrate concentrations.

where $K I$ is the inhibitors dissociation constant and [I] is the substrate concentration.

[edit] Derivation

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

E + S <==> ES ---> E + P

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

EI + S <==> E + I + S <==> ES + I --> 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. To derive the equation describing the kinetics, first assign microscopic rate constants to each step:

k₁ = E + S --> ES

k₋₁ = ES --> E + S

k₂ = ES --> E + P

k₃ = E + I --> EI

k₋₃ = EI --> E + I

Just as with the derivation of the Michaelis-Menten equation, assume that the system is at steady-state, that is that the concentration of each of the enzyme species is not changing.

$\frac{dE}{dt} = \frac{d}{dt}ES = \frac{d}{dt}{EI} = 0.$

Furthermore, the known total enzyme concentration is E_T = E + ES + EI, 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:

eq 1:

E T = E + E S + E I

eq 2: $\frac{dE}{dt} = 0 = -k_1*E*S + k_{-1}*ES + k_2*ES -k_3*E*I + k_{-3}*EI$

eq 3: $\frac{dES}{dt} = 0 = k_1*E*S - k_{-1}*ES - k_2*ES$

eq 4: $\frac{dEI}{dt} = 0 = k_3*E*I - k_{-3}*EI$

where S, I and E_T are known. The initial velocity is defined as v = dP/dt = k₂ ES, so we need to define the unknown ES in terms of the knowns S, I and E_T.