Deriving the One-Site Equilibrium Occupancy Equation

From Reversible Binding to Equilibrium Occupancy

A ligand can associate with a receptor and later dissociate. The balance between those opposing processes determines the fraction of receptors occupied once binding has reached equilibrium. For a single binding site, that balance leads to a compact equation relating occupancy to free-ligand concentration. Its parameter, the dissociation constant, has a direct interpretation: it is the free-ligand concentration that gives half occupancy under the model assumptions.

The equation is useful because it connects a measurable concentration to the fraction of available receptors bound. It describes an equilibrium relationship, however, not how quickly the system approaches that relationship. Applying it therefore requires attention to what concentration is used and whether the binding system can reasonably be treated as a single reversible site at equilibrium.

Model and Assumptions

Let \(L\) denote free ligand, \(R\) free receptor, and \(RL\) the ligand–receptor complex. Write their concentrations as \([L]\), \([R]\), and \([RL]\), respectively. The one-site reversible binding scheme is

L+R \underset{k_{\mathrm{off}}}{\overset{k_{\mathrm{on}}}{\rightleftharpoons}} RL.

Here \(k_{\mathrm{on}}\) is the association rate constant, with units of concentration\(^{-1}\) time\(^{-1}\), and \(k_{\mathrm{off}}\) is the dissociation rate constant, with units of time\(^{-1}\). In this model, association occurs at a rate proportional to both free-ligand and free-receptor concentrations; dissociation occurs at a rate proportional to complex concentration.

Model assumptions: There is one class of identical, independent receptor sites, each of which can bind at most one ligand molecule. Binding is reversible and follows the mass-action rates shown above. Total receptor concentration is conserved during the binding process. The system is at equilibrium when occupancy is calculated, and \([L]\) is the free-ligand concentration at that equilibrium. The model does not include competing ligands, cooperativity, receptor synthesis or loss, or an appreciable difference between free and nominal ligand concentration.

The equilibrium equation can be derived without assuming that all receptors are occupied or that ligand is abundant. It does require that the free-ligand concentration used in the equation is the concentration available to bind receptors. When ligand depletion is substantial, total ligand concentration cannot simply be substituted for free ligand.

Deriving the Equilibrium Equation

Let \(R_T\) be the total concentration of receptor sites. Conservation gives \(R_T=[R]+[RL]\). Define fractional occupancy \(\theta\) as the fraction of those sites in the bound state:

\theta=\frac{[RL]}{R_T}=\frac{[RL]}{[R]+[RL]}.

The binding rate is \(k_{\mathrm{on}}[L][R]\), and the unbinding rate is \(k_{\mathrm{off}}[RL]\). At equilibrium the net rate of change of complex concentration is zero, so the two rates are equal. Rearranging that equality defines the dissociation constant \(K_D\):

\begin{aligned} 0 &= k_{\mathrm{on}}[L][R]-k_{\mathrm{off}}[RL],\\ k_{\mathrm{on}}[L][R] &= k_{\mathrm{off}}[RL],\\ K_D &\equiv \frac{k_{\mathrm{off}}}{k_{\mathrm{on}}} =\frac{[L][R]}{[RL]}. \end{aligned}

Because \(k_{\mathrm{off}}\) has units of time\(^{-1}\) and \(k_{\mathrm{on}}\) has units of concentration\(^{-1}\) time\(^{-1}\), \(K_D\) has units of concentration. To obtain occupancy, solve the equilibrium relation for the bound-to-free receptor ratio, then use receptor conservation:

\frac{[RL]}{[R]}=\frac{[L]}{K_D}, \qquad \theta =\frac{[RL]}{[R]+[RL]} =\frac{[L]/K_D}{1+[L]/K_D} =\frac{[L]}{K_D+[L]}.

One-site equilibrium occupancy: For a single class of reversible binding sites at equilibrium, the fraction of sites occupied is \(\theta=[L]/(K_D+[L])\), where \([L]\) is free-ligand concentration and \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\) is the dissociation constant. The unoccupied fraction is \(1-\theta=K_D/(K_D+[L])\).

The equation gives values between zero and one for nonnegative \([L]\) and positive \(K_D\). At zero free ligand, \(\theta=0\); as free-ligand concentration becomes very large relative to \(K_D\), occupancy approaches one. These limits are consistent with the interpretation of \(\theta\) as a fraction of sites.

Worked Example: Occupancy at a Specified Free-Ligand Concentration

Suppose a hypothetical receptor has \(K_D=12\ \mathrm{nM}\). At equilibrium, the free-ligand concentration is \(6\ \mathrm{nM}\). Calculate fractional occupancy and the percentage of receptors unoccupied.

The model uses free ligand, so the stated \(6\ \mathrm{nM}\) is the concentration substituted into the equation. The occupancy is

\theta=\frac{[L]}{K_D+[L]} =\frac{6\ \mathrm{nM}}{12\ \mathrm{nM}+6\ \mathrm{nM}} =\frac{6}{18} =0.3333.

Thus, the predicted equilibrium occupancy is \(0.3333\), or \(33.33\%\). The unoccupied fraction is \(1-\theta=0.6667\), so \(66.67\%\) of the receptor sites are predicted to be unoccupied. The units cancel in the ratio, as required for a fraction.

Why \(K_D\) Gives Half Occupancy

Set occupancy to one-half in the equilibrium equation. Solving for free-ligand concentration shows precisely why \(K_D\) is often described as the concentration for half occupancy:

\begin{aligned} \frac{1}{2} &= \frac{[L]}{K_D+[L]},\\ K_D+[L] &= 2[L],\\ [L] &= K_D. \end{aligned}

This result also works in reverse: if the free-ligand concentration and equilibrium fractional occupancy are known under the model assumptions, then

K_D=[L]\frac{1-\theta}{\theta}, \qquad 0<\theta<1.

A lower \(K_D\) means that a lower free-ligand concentration is required for half occupancy in this model. It does not, by itself, say whether association is fast or dissociation is slow: \(K_D\) is the ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\), and the same ratio can result from different pairs of rate constants.

Worked Example: Finding the Half-Occupancy Concentration

For a hypothetical ligand–receptor pair, let \(k_{\mathrm{on}}=0.20\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\) and \(k_{\mathrm{off}}=0.60\ \mathrm{h}^{-1}\). Calculate \(K_D\), then state the free-ligand concentration giving half occupancy.

First calculate the dissociation constant from the ratio of rate constants:

K_D=\frac{k_{\mathrm{off}}}{k_{\mathrm{on}}} =\frac{0.60\ \mathrm{h}^{-1}} {0.20\ \mathrm{nM}^{-1}\mathrm{h}^{-1}} =3.0\ \mathrm{nM}.

The time units cancel and the remaining unit is concentration. Since \(\theta=0.5\) when \([L]=K_D\), the half-occupancy concentration is \(3.0\ \mathrm{nM}\) of free ligand. This conclusion is about equilibrium occupancy, not how long the system takes to reach it.

Worked Example: Inferring \(K_D\) From an Occupancy Measurement

In a hypothetical equilibrium experiment, free ligand is measured as \(8.0\ \mathrm{nM}\), and the occupied fraction is \(0.80\). Assuming the one-site model applies, infer \(K_D\).

Rearrange \(\theta=[L]/(K_D+[L])\) to isolate \(K_D\):

K_D=[L]\frac{1-\theta}{\theta} =8.0\ \mathrm{nM}\times\frac{1-0.80}{0.80} =8.0\ \mathrm{nM}\times 0.25 =2.0\ \mathrm{nM}.

The inferred dissociation constant is \(2.0\ \mathrm{nM}\). A consistency check gives \(8.0/(2.0+8.0)=0.80\), recovering the stated occupancy. This calculation depends on the measured concentration being free ligand and on the equilibrium, one-site assumptions; the arithmetic alone cannot establish those conditions.

Checking the Model and Interpreting Its Limits

The equation is a direct consequence of the specified rate balance and receptor conservation. Its simplicity makes it useful, but also means that a good numerical fit does not prove that the assumptions are biologically correct. Before interpreting a calculated \(\theta\) or \(K_D\), check the experimental quantity and the model structure.

Check concentration type. The equation requires free-ligand concentration. If the reported value is total added ligand and appreciable ligand binds to receptors or other components, using it as \([L]\) can overstate predicted occupancy.
Check equilibrium. The derivation sets association and dissociation rates equal. If the system has not reached equilibrium, the equation need not describe the current occupancy.
Check the binding-site assumption. A single \(K_D\) presumes one homogeneous class of independent sites. Systematic departures from the predicted concentration–occupancy relationship may indicate that this idealization is inadequate, though measurement error and concentration error are also possible explanations.
Check the measurement scale. \(\theta\) is a fraction from zero to one. If data are reported as a percentage, convert to a fraction before using the inverse equation; for example, \(80\%\) becomes \(0.80\).

The equilibrium occupancy equation is not a substitute for a time-dependent description when the question concerns the approach to equilibrium, or for a model that accounts for ligand mass balance when free and total ligand differ materially. Those approaches answer related but distinct questions. Likewise, a fitted \(K_D\) summarizes the equilibrium relationship in this one-site model; it is not a complete description of all receptor biology or of downstream response.

Common pitfall: “The total concentration added was \(10\ \mathrm{nM}\), so I can substitute \([L]=10\ \mathrm{nM}\) and calculate occupancy.” That substitution is justified only when \(10\ \mathrm{nM}\) represents free ligand at equilibrium, or when the difference between total and free concentration is negligible. The correction is to identify the free-ligand concentration required by the model and assess whether it is known or reasonably approximated.

Check Your Understanding

Answer each question using the one-site equilibrium model. Numerical answers are followed by concise solutions so the calculations can be checked.

A hypothetical system has \(K_D=4.0\ \mathrm{nM}\) and free ligand \(12.0\ \mathrm{nM}\). What is the predicted equilibrium occupancy?
At what free-ligand concentration is occupancy one-half if \(K_D=0.75\ \mathrm{\mu M}\)?
For free ligand \(5.0\ \mathrm{nM}\) and occupancy \(0.50\), calculate \(K_D\).
Why is it generally incorrect to use a nominal total ligand concentration in place of free-ligand concentration without checking?
Can two systems with the same \(K_D\) necessarily be assumed to have the same association and dissociation rate constants? Explain.

Solutions. (1) \(\theta=12.0/(4.0+12.0)=0.750\), or \(75.0\%\). (2) Half occupancy occurs at \([L]=K_D=0.75\ \mathrm{\mu M}\). (3) \(K_D=[L](1-\theta)/\theta=5.0(0.50)/(0.50)=5.0\ \mathrm{nM}\). (4) The equation uses free ligand; total ligand can differ when binding or other processes remove ligand from the free pool. (5) No. \(K_D\) fixes only the ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\); multiple rate-constant pairs can have the same ratio.