Derivation of Hill Langmuir Equation

A simple reaction scheme for the interaction of an agonist [A] with its receptor [R] can be described as:

Rate of forward reaction: k_{+1}

Rate of reverse reaction: k_{-1}

If the system is at equilibrium and [A] >> [R] (in other words, there isn’t any ligand depletion), then the rate of reaction is proportional to the product of the concentration of the reactants, i.e.

[A][R]k_{+1} = [AR]k_{-1}, so let’s find [AR]

If  \frac{k_{-1}}{k_{+1}} = K_D, then

K_D = \frac{[A][R]}{[AR]} or [AR] = \frac{[A][R]}{K_D}, but we can’t measure [R], so let’s simplify further

We know that [R_T] = [R] + [AR]  or [R] = [R_T] - [AR]

Thus [AR] = \frac{[A]([R_T]-[AR])}{K_D}

Simplify for [AR],

[AR]K_D = [A][R_T] - [A][AR]

[AR](K_D + [A]) = [A][R_T]

[AR] = \frac{[A][R_T]}{(K_D + [A])}

which can also be written as –

\frac{[AR]}{[R_T]} = \frac{[A]}{(K_D+[A])} or \delta_{[AR]} = \frac{[A]}{(K_D+[A])}

where \delta_{[AR]} represents the proportion of occupied receptors

We can use this equation to understand the relationship between ligand concentration [A] and receptor occupancy –

At 50% receptor occupancy \delta_{[AR]} = \frac{[A]}{K_D + [A]} = 0.5

0.5K_D+0.5[A] = [A]

or 0.5[A] = 0.5K_D,

in other words, at 50% receptor occupancy, the concentration of the ligand [A] is equal to it’s K_D .