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 $K_D$.