= Receptor binding occupancy
{title2=$\theta$}
<Receptor binding occupancy> is the fraction of binding sites occupied by a <ligand>. Consider reversible, noncooperative binding at <chemical equilibrium>, with homogeneous independent sites: every site has the same binding properties, and binding at one site does not change binding at another. Let $L$ denote free <ligand>, $R$ an unoccupied <receptor> site and $RL$ the bound complex. In this dilute <concentration> model,
$$
L+R\rightleftharpoons RL,\qquad K_d=\frac{[L][R]}{[RL]},\qquad R_T=[R]+[RL],\qquad \theta=\frac{[RL]}{R_T}=\frac{[L]}{K_d+[L]}.
$$
Square brackets denote <concentrations>; $R_T$ is the total site <concentration>. Here the <dissociation constant> $K_d$ has <concentration> units. Use free <ligand> in excess so binding negligibly depletes it; only under that approximation may the supplied total <ligand> <concentration> substitute for $[L]$. See https://pmc.ncbi.nlm.nih.gov/articles/PMC5521016/[Analysis of Receptor–Ligand Interactions: equilibrium binding and saturation].
For an illustrative <dissociation constant> $K_d=10\,\mathrm{nM}$, where $1\,\mathrm{nM}=10^{-9}\,\mathrm{mol\,L^{-1}}$, free <ligand> <concentrations> $1$, $10$ and $100\,\mathrm{nM}$ give $\theta=1/11\approx0.091$, $\theta=1/2=0.5$ and $\theta=10/11\approx0.909$, respectively. At $[L]=K_d$, exactly half the sites are occupied in this model. Raising $[L]$ tenfold from $10$ to $100\,\mathrm{nM}$ raises occupancy from $50\%$ to about $91\%$, rather than tenfold: the finite pool of sites causes saturation.
\Image[/receptor-binding-occupancy.png]
{title=<Receptor binding occupancy> for a toy <dissociation constant> of $10\,\mathrm{nM}$}
{description=The same independent-site <chemical equilibrium> model on linear and logarithmic <ligand> <concentration> axes. Marked examples are $1/11$, $1/2$ and $10/11$. The logarithmic axis changes the apparent curve shape without introducing cooperative binding. The vertical axis is fraction occupied, not cellular response.}
A short derivation uses the <law of mass action>: at <chemical equilibrium>, association balances dissociation, giving $[RL]=[L][R]/K_d$. Substitute this into $\theta=[RL]/([R]+[RL])$ and cancel $[R]$ to obtain the displayed formula. The limiting values follow directly: zero free <ligand> gives zero occupancy, and increasing $[L]$ far above $K_d$ drives occupancy toward one.
<Receptor binding occupancy> is not automatically the fraction of maximal response. <Signal amplification>, differing effects of bound <ligands>, and downstream saturation can change the relationship. Consequently, $[L]=K_d$ specifies half occupancy here; it does not universally specify either half-maximal or maximal cellular response. These numbers describe a toy binding model, not measured responses of every <receptor>. See https://pmc.ncbi.nlm.nih.gov/articles/PMC11267233/[Higham and Colquhoun, 2024: the affinity–efficacy problem].
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