Equilibrium Occupancy in a Reversible Binding Step

From Reversible Binding to Occupancy

For a reversible 1:1 binding step, the equilibrium amount of complex depends on how readily the partners associate and how readily the complex dissociates. The bound fraction, or occupancy, turns that amount into an interpretable fraction of the available binding sites. It is a basic link between a molecular binding step and the extent to which a target is occupied.

Consider \(A+B \rightleftharpoons AB\), where \(A\) is a ligand, \(B\) is a binding site, and \(AB\) is the complex. Let \([A]\), \([B]\), and \([AB]\) denote their concentrations. The association rate constant is \(k_{\mathrm{on}}\), and the dissociation rate constant is \(k_{\mathrm{off}}\). For this tutorial, occupancy means the fraction of all \(B\) sites that are bound by \(A\).

Definition: For the reversible 1:1 binding step \(A+B \rightleftharpoons AB\), the dissociation constant is \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\). At equilibrium, the occupancy of \(B\) by \(A\) is \(\theta=[AB]/([B]+[AB])\), where the denominator is the total concentration of free and bound \(B\) sites.

This definition assumes one homogeneous class of independent binding sites, one-to-one binding, and a system at chemical equilibrium for this reaction. The association and dissociation rates follow mass-action rate laws: association is proportional to \([A][B]\), while dissociation is proportional to \([AB]\). The concentrations in the occupancy formula are concentrations of the species present, not automatically the total amounts added to a mixture.

Deriving the Occupancy Formula

At equilibrium for this reversible step, the association and dissociation fluxes are equal. Using the mass-action rate laws gives

k_{\mathrm{on}}[A][B]=k_{\mathrm{off}}[AB].

Rearrange this equality and use \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\):

\frac{[AB]}{[B]} =\frac{k_{\mathrm{on}}[A]}{k_{\mathrm{off}}} =\frac{[A]}{K_D}.

The total concentration of \(B\) sites is \([B]+[AB]\). Substituting the ratio above into the definition of occupancy yields

\theta =\frac{[AB]}{[B]+[AB]} =\frac{[A]/K_D}{1+[A]/K_D} =\frac{[A]}{[A]+K_D}.

Key result: For a single 1:1 binding step at equilibrium, the bound fraction of \(B\) is \(\theta=[A]/([A]+K_D)\), where \([A]\) is the free concentration of \(A\). This expression requires the free concentration; it does not in general hold if \([A]\) is replaced by the total amount of \(A\).

The ratio \([A]/K_D\) is dimensionless, as required for a fraction. Since both concentrations are nonnegative, \(0\leq\theta\leq1\). When \([A]=K_D\), the formula gives \(\theta=1/2\): the free ligand concentration equals the dissociation constant at half-occupancy under these assumptions. If \([A]\) is much smaller than \(K_D\), occupancy is approximately \([A]/K_D\); if \([A]\) is much larger than \(K_D\), occupancy approaches one. A smaller \(K_D\) therefore means greater occupancy at the same free ligand concentration.

The expression can also be rearranged to give the occupancy odds:

\frac{\theta}{1-\theta}=\frac{[A]}{K_D}, \qquad [A]=K_D\frac{\theta}{1-\theta}.

This form is useful when a desired occupancy is specified and the free ligand concentration needed to produce it is sought. It does not by itself specify how much ligand must be added to a system, because added and free concentrations can differ.

Worked Example: Occupancy From Free Ligand Concentration

Suppose a hypothetical binding system has \(K_D=10\ \mathrm{nM}\), and the free ligand concentration is \([A]=30\ \mathrm{nM}\). Calculate the equilibrium occupancy of \(B\).

The one-site equilibrium formula applies because the quantities provided are a dissociation constant and a free ligand concentration, and the calculation assumes the system has reached equilibrium:

\theta=\frac{[A]}{[A]+K_D} =\frac{30\ \mathrm{nM}}{30\ \mathrm{nM}+10\ \mathrm{nM}} =\frac{30}{40} =0.750.

The predicted occupancy is 0.750, or 75.0%. In this hypothetical system, three quarters of the \(B\) sites are bound at equilibrium. This is a statement about occupancy under the stated conditions, not a claim that 75.0% of the total ligand is bound.

Using Rate Constants and Checking Units

The previous result can be used once \(K_D\) is known. If it is not provided, the equilibrium constant can be obtained from the rate constants for the same reversible step. The units provide a useful check: \(k_{\mathrm{off}}\) has units of inverse time, and \(k_{\mathrm{on}}\) has units of concentration\(^{-1}\) time\(^{-1}\), so their ratio has units of concentration.

Worked Example: Occupancy From Association and Dissociation Rates

In a hypothetical system, let \(k_{\mathrm{on}}=2.0\times10^6\ \mathrm{M}^{-1}\mathrm{min}^{-1}\), \(k_{\mathrm{off}}=0.040\ \mathrm{min}^{-1}\), and free ligand concentration \([A]=5.0\ \mathrm{nM}\). Calculate \(K_D\) and occupancy.

First calculate the dissociation constant, retaining the rate units in the ratio:

K_D=\frac{k_{\mathrm{off}}}{k_{\mathrm{on}}} =\frac{0.040\ \mathrm{min}^{-1}} {2.0\times10^6\ \mathrm{M}^{-1}\mathrm{min}^{-1}} =2.0\times10^{-8}\ \mathrm{M} =20\ \mathrm{nM}.

The time units cancel, and the result has concentration units. Now apply the occupancy formula using nM for both concentrations:

\theta=\frac{5.0\ \mathrm{nM}}{5.0\ \mathrm{nM}+20\ \mathrm{nM}} =\frac{5.0}{25.0} =0.200.

The predicted equilibrium occupancy is 20.0%. The free ligand concentration is one quarter of \(K_D\), which is consistent with occupancy below one half.

When Total Concentrations Matter

The simple occupancy formula is expressed in terms of free ligand. If ligand is present in large excess over binding sites, binding may use only a small proportion of the ligand, making free and total ligand concentrations close. If the binding sites use a substantial proportion of the ligand, however, the free concentration can be much lower than the total concentration. In that case, substituting the total ligand concentration into the simple formula can overestimate occupancy.

For a closed system with total concentrations \(A_T=[A]+[AB]\) and \(B_T=[B]+[AB]\), let \(x=[AB]\). At equilibrium, the definition of \(K_D\) gives

K_D=\frac{[A][B]}{[AB]} =\frac{(A_T-x)(B_T-x)}{x}.

This relation can be solved for \(x\), with the physically meaningful solution satisfying \(0\leq x\leq\min(A_T,B_T)\). Then the occupancy of \(B\) is \(x/B_T\). This finite-total calculation still uses the same equilibrium binding relationship; it accounts explicitly for ligand and binding sites being depleted from their free pools.

Worked Example: Correcting for Ligand Depletion

Consider a hypothetical closed system with \(A_T=B_T=100\ \mathrm{nM}\) and \(K_D=10\ \mathrm{nM}\). Find the occupancy of \(B\), allowing for ligand depletion.

Set \(x=[AB]\). Substituting the totals into the equilibrium relationship gives

10=\frac{(100-x)(100-x)}{x},

where concentrations in this calculation are in nM. Multiplying by \(x\) and expanding gives

x^2-210x+10000=0, \qquad x=\frac{210\pm\sqrt{210^2-4(10000)}}{2} =\frac{210\pm\sqrt{4100}}{2}.

The two roots are approximately \(137.016\ \mathrm{nM}\) and \(72.984\ \mathrm{nM}\). The larger root is impossible because it exceeds both total concentrations; it would imply a negative free concentration. Thus \(x=72.984\ \mathrm{nM}\), and

\theta=\frac{x}{B_T} =\frac{72.984\ \mathrm{nM}}{100\ \mathrm{nM}} =0.7298.

The equilibrium occupancy is approximately 73.0%. The free concentrations are each \(100-72.984=27.016\ \mathrm{nM}\). Checking the equilibrium relationship gives \((27.016)(27.016)/72.984\approx10.00\ \mathrm{nM}\), as required. By contrast, substituting the total \(A_T\) into the simple formula would give \(100/(100+10)=0.9091\), or 90.9%. That estimate is too high because much of \(A\) is bound and is therefore not free ligand.

Assumptions, Diagnostics, and Alternatives

The occupancy formula is useful because it reduces a reversible 1:1 binding step to a simple relationship among free ligand, \(K_D\), and bound fraction. Its interpretation depends on the model assumptions: one binding site per \(B\), no competing ligands or additional binding states, and the equilibrium condition for this reaction. If these conditions are not appropriate, the formula may not describe the measured system.

Several checks can reveal an inappropriate calculation or a mistaken interpretation:

Check the concentration used. The simple formula requires free \([A]\). If only total concentrations are available and depletion may matter, use the total-concentration relationship rather than silently substituting \(A_T\).
Check dimensions and bounds. \(K_D\) and \([A]\) must use the same concentration units, and the resulting occupancy must lie between zero and one.
Check the limiting behavior. Zero free ligand gives zero occupancy, \([A]=K_D\) gives one-half occupancy, and increasing free ligand moves occupancy toward one.
Check the binding model. Multiple site classes, cooperativity, competition, or additional reactions can change the relationship between concentration and occupancy. A one-site formula should not be presumed valid for those systems.

For a system in which free ligand is known or effectively unchanged by binding, \(\theta=[A]/([A]+K_D)\) is the direct calculation. For a closed system with appreciable depletion, solving for complex concentration from the total amounts is the relevant alternative. Both calculations use the equilibrium dissociation relationship, but they answer different data situations. The dissociation constant is the free ligand concentration associated with half-occupancy in this model; it is not necessarily the total concentration that must be added to obtain half-occupancy.

Common pitfall: “The total ligand concentration is \(K_D\), so occupancy must be 50%.” This reasoning substitutes total ligand for free ligand without checking depletion. Half-occupancy follows when the free ligand concentration equals \(K_D\); with substantial binding, the free concentration can be far below the total.

Check Your Understanding

Use the one-site equilibrium model unless a question states that total concentrations must be used. Solutions are included below the questions.

If \(K_D=8\ \mathrm{nM}\) and free \([A]=8\ \mathrm{nM}\), what is the occupancy of \(B\)?
If \(K_D=12\ \mathrm{nM}\) and occupancy is 0.600, what free ligand concentration is required?
For \(k_{\mathrm{on}}=5.0\times10^5\ \mathrm{M}^{-1}\mathrm{s}^{-1}\) and \(k_{\mathrm{off}}=0.010\ \mathrm{s}^{-1}\), calculate \(K_D\) in nM.
Why can the simple occupancy formula overestimate occupancy when total ligand is substituted for free ligand?

Solutions. (1) \(\theta=8/(8+8)=0.500\), so occupancy is 50.0%. (2) Rearranging gives \([A]=K_D\theta/(1-\theta)\), so \([A]=(12\ \mathrm{nM})(0.600)/(0.400)=18.0\ \mathrm{nM}\). (3) \(K_D=0.010/(5.0\times10^5)=2.0\times10^{-8}\ \mathrm{M}=20\ \mathrm{nM}\); the seconds cancel. (4) Binding removes ligand from the free pool. If depletion is appreciable, free ligand is lower than total ligand, and occupancy depends on the lower free concentration.