Why Competition Changes an Occupancy Estimate
When two ligands can bind the same target site, each ligand competes with the other for the available target. Calculating each occupancy as if the other ligand were absent can therefore overstate both occupancies. A competitive estimate accounts for both ligands in a shared target balance, so the fractions bound to ligand A, bound to ligand B, and unoccupied must sum to one.
This tutorial uses free ligand concentrations and equilibrium dissociation constants to calculate those fractions. The calculation is useful only when the quantities and assumptions match the binding system: in particular, the ligands must compete for the same mutually exclusive site, and the system must be adequately described by equilibrium binding. The result is an occupancy estimate, not a prediction of functional response.
Model and Assumptions
The dissociation constant for A is \(K_{D,A}=[T][A]/[TA]\), where \([T]\) is free target concentration and \([TA]\) is target–A complex concentration. The corresponding definition applies to B. Thus \(x\) and \(y\) are dimensionless concentration-to-affinity ratios. A smaller dissociation constant indicates tighter binding, all else equal.
The formulas rely on these assumptions:
- Both ligands bind the same target population at the same site, and binding of one ligand excludes binding of the other to that target molecule.
- Binding is reversible and has reached equilibrium. The dissociation constants describe the relevant conditions, such as temperature, pH, and assay matrix.
- The binding sites are homogeneous and noncooperative, with one binding site per target molecule; there is no additional allostery or interaction between target molecules.
- The inputs are the free concentrations at the target site, not total concentrations, and the free concentrations are effectively specified rather than substantially depleted by target binding.
- The measured dissociation constants and concentrations apply to the same binding system. No unmodeled ligand or competing species occupies the site.
If these conditions are not plausible, the calculated fractions may be numerically precise but scientifically misleading. In particular, a concentration measured in plasma is not automatically the free concentration at a tissue target.
Deriving the Competitive Occupancy Fractions
At equilibrium, mass action gives \([TA]=[T]C_A/K_{D,A}=[T]x\) and \([TB]=[T]C_B/K_{D,B}=[T]y\). The total concentration of target sites is the sum of free target and the two mutually exclusive complexes:
Dividing each complex concentration by total target gives its fraction. For example, \([TA]/[T]_{\mathrm{total}}=[T]x/\{[T](1+x+y)\}\), so the free target concentration cancels and \(\theta_A=x/(1+x+y)\). The same calculation gives \(\theta_B=y/(1+x+y)\), while \(\theta_0=1/(1+x+y)\). These fractions sum to one because their numerators sum to the common denominator.
This derivation also explains why the free target concentration is not needed for the fractional estimate. It does not mean that target concentration is irrelevant to every aspect of the system: if binding appreciably depletes either ligand, then the free concentrations themselves must be determined from mass balance rather than treated as known inputs.
Worked Example: Calculating Both Ligands’ Occupancy
Suppose a hypothetical equilibrium system has free ligand A concentration \(C_A=20.0\) nM and \(K_{D,A}=10.0\) nM. Ligand B has free concentration \(C_B=30.0\) nM and \(K_{D,B}=15.0\) nM. Assume both ligands meet the model assumptions.
First calculate the dimensionless ratios:
The shared denominator is \(1+x+y=1+2.00+2.00=5.00\). Therefore:
Under these hypothetical conditions, an estimated 40.0% of target sites are occupied by A, 40.0% by B, and 20.0% are unoccupied. These are fractions of the same target population, not separate estimates that can be added without regard to competition.
How Competition Affects a Desired Occupancy
The same model can be rearranged to find the concentration of one ligand required to reach a specified occupancy while a competitor remains present. For a target occupancy \(q\) by A, solve \(q=x/(1+x+y)\). Rearranging yields \(x=q(1+y)/(1-q)\), for \(0\leq q<1\). The required free concentration is then \(C_A=K_{D,A}q(1+y)/(1-q)\). This is an algebraic target under the equilibrium model, not a dosing recommendation.
Worked Example: Concentration Needed for a Target Occupancy
In a hypothetical system, ligand B has \(C_B=20.0\) nM and \(K_{D,B}=10.0\) nM, so \(y=2.00\). Ligand A has \(K_{D,A}=8.00\) nM. What free concentration of A gives 70.0% occupancy by A at equilibrium?
Set \(q=0.700\). The required ratio is:
Thus the required free concentration is \(C_A=xK_{D,A}=7.00(8.00\ \mathrm{nM})=56.0\) nM. Checking in the original expression gives:
Without ligand B, the ratio required for 70.0% occupancy would instead be \(x=q/(1-q)=2.33\), corresponding to \(18.7\) nM A. The larger concentration in the competitive case follows from B’s contribution to the denominator. This comparison holds only with B’s free concentration maintained as specified.
Propagating Uncertainty in Concentration and Affinity
An occupancy estimate inherits uncertainty from its inputs. A useful local approximation is the delta method: for small input uncertainties, approximate the output variance by combining the model’s sensitivities with the input covariance matrix. Because concentrations and dissociation constants are positive and often uncertain multiplicatively, it is convenient to express sensitivities with respect to their natural logarithms.
For \(\theta_A=x/(1+x+y)\), differentiating with respect to the log inputs gives:
and
These signs have direct interpretations: increasing A’s free concentration raises its occupancy, while increasing A’s dissociation constant lowers it. Increasing B’s concentration reduces A’s share, while increasing B’s dissociation constant weakens B and raises A’s share. If the four log inputs are independent, the approximate variance is the sum of each squared sensitivity times its log-scale variance. If inputs are correlated, covariance terms must also be included; omitting meaningful correlations can misstate uncertainty.
Worked Example: Approximate Uncertainty in A’s Occupancy
Use the first hypothetical example, where \(\theta_A=\theta_B=0.400\). Suppose each of the four inputs \(C_A,K_{D,A},C_B,K_{D,B}\) has an independent log-scale standard deviation of 0.100. This represents approximately 10% multiplicative uncertainty on the log scale. The model and equilibrium assumptions are taken as fixed; this calculation does not account for model uncertainty.
At these occupancy values, the four derivatives with respect to the log inputs are \(+0.400(0.600)=+0.240\) for \(C_A\), \(-0.240\) for \(K_{D,A}\), \(-0.400(0.400)=-0.160\) for \(C_B\), and \(+0.160\) for \(K_{D,B}\). Independence gives:
The approximate standard error is \(\sqrt{0.001664}=0.0408\), in occupancy-fraction units. A normal, first-order approximation gives \(0.400\pm1.96(0.0408)\), or approximately 0.320 to 0.480. Expressed as percentages, this is about 32.0% to 48.0% occupancy by A. It is an uncertainty interval conditional on the stated input-uncertainty model and the competitive model; it is not automatically a confidence interval from experimental sampling.
The approximation is local. It may be inadequate when input uncertainty is large, when parameter distributions are strongly non-normal, or when the output distribution is notably asymmetric or close to its bounds of zero and one. In those cases, propagate an appropriate joint input distribution through the occupancy equations, for example by simulation, and report the resulting distribution or quantiles. Correlations should be retained in that propagation.
Diagnostics, Alternatives, and Limitations
Several checks can reveal calculation errors or a mismatch between the model and the system. First, verify the units: each \(C/K_D\) ratio must be dimensionless, so concentration units must match. Second, verify that all three fractions are between zero and one and sum to one. Third, check limiting behavior: if both ligand concentrations approach zero, the target is unoccupied; if \(C_A/K_{D,A}\) becomes very large while \(y\) stays finite, \(\theta_A\) approaches one. If the calculation violates these checks, revisit the inputs and formula.
A common mistake is to calculate each ligand’s single-ligand occupancy separately as \(x/(1+x)\) and \(y/(1+y)\), then interpret both as simultaneous fractions of the same target. In the first example, that calculation gives \(2/3\) for each ligand, totaling \(4/3\), or 133% occupancy. The contradiction occurs because each isolated calculation assumes the other ligand is absent. In the competitive model, both ratios enter the same denominator, giving 40.0% occupancy by each and 20.0% unoccupied target.
A different biological model may be appropriate if the ligands bind distinct sites and can occupy the target simultaneously, if target sites are heterogeneous, or if binding is cooperative. A simple independent-site calculation or a model with multiple site classes may then be considered, but it answers a different mechanistic question and should not be substituted without evidence. The earlier material on cooperative binding and heterogeneous target populations addresses those structures. If the system is not at equilibrium, an equilibrium fraction based only on concentrations and dissociation constants may also be unsuitable; time-dependent occupancy requires kinetic information and is treated separately.
Finally, the calculation is conditional on the free concentrations. If only total ligand concentrations are known and ligand binding substantially depletes free ligand, the free concentrations cannot simply be replaced by the totals. A mass-balance model is then needed to determine free and bound amounts together. Likewise, uncertainty propagation for the formula does not repair biased assay measurements, inappropriate affinity constants, unmodeled competitors, or an incorrect binding mechanism.
Check Your Understanding
Work through each question, then compare with the solutions.
- For \(x=1.00\) and \(y=3.00\), calculate \(\theta_A\), \(\theta_B\), and \(\theta_0\).
- If B’s free concentration increases while all other inputs stay fixed, what happens to A’s occupancy? Explain using the model.
- For a desired A occupancy \(q=0.600\) and competitor ratio \(y=1.00\), find the required value of \(x\).
- Give one reason why substituting total ligand concentration for free concentration may invalidate the estimate.
Solutions. (1) The denominator is \(1+1.00+3.00=5.00\), so \(\theta_A=1.00/5.00=0.200\), \(\theta_B=3.00/5.00=0.600\), and \(\theta_0=1/5.00=0.200\); they sum to one. (2) Increasing B’s free concentration increases \(y\), while \(\theta_A=x/(1+x+y)\) has an unchanged numerator and a larger denominator, so A’s occupancy decreases. (3) \(x=q(1+y)/(1-q)=0.600(2.00)/0.400=3.00\). (4) If target binding materially depletes ligand, total concentration is not the free concentration available for binding; a mass balance is needed to estimate the free concentration before applying the competitive occupancy formula.