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Receptor Binding and Target Engagement · Tutorial 212 of 799

Worked Calculations For Competitive Occupancy

Learn to calculate each ligand’s occupancy, total receptor engagement, and the conditions under which one competitor dominates.

Intermediate 11 min read

What You'll Learn

  • Derive the two-ligand equilibrium occupancy equations from mass action.
  • Calculate bound fractions and unoccupied receptor from free concentrations and dissociation constants.
  • Compare ligand engagement using concentration-to-affinity ratios.
  • Recalculate occupancy when concentrations change across scenarios.
  • Recognize when free-ligand or equilibrium assumptions limit the calculation.

Why Competitive Occupancy Requires a Shared Calculation

When two ligands can bind the same receptor site, each ligand competes for the receptor molecules that would otherwise be available to the other. The occupancy of either ligand therefore depends on both ligands’ concentrations and affinities. Calculating each ligand’s occupancy as if it were alone and then adding those values generally does not give a valid receptor allocation.

The calculation below describes equilibrium binding to one shared, mutually exclusive site. It predicts the fraction of receptors bound to each ligand and the fraction left unoccupied. These are measures of target engagement, not, by themselves, predictions of functional response: ligands with different efficacy can produce different responses at the same occupancy.

Definition: Let \(C_A\) and \(C_B\) denote the free concentrations of ligands A and B at the receptor, and let \(K_{D,A}\) and \(K_{D,B}\) denote their equilibrium dissociation constants for the same binding site. The fractional occupancies \(\theta_A\) and \(\theta_B\) are the fractions of the total receptor pool bound to A and B, respectively. The unoccupied fraction is \(\theta_0\), and total occupancy is \(\theta_{\mathrm{total}}=\theta_A+\theta_B\).

The model assumes that each receptor has one relevant site; A and B cannot occupy that site simultaneously; binding has reached equilibrium; and the concentrations used are the free concentrations at the receptor. It also assumes the receptor pool is represented by one binding population with fixed dissociation constants, with no cooperative or allosteric effects on this site. Under these assumptions, each ligand’s binding follows the ordinary mass-action relation.

Deriving the Two-Ligand Equations

Write \(R\) for unbound receptor, \(RA\) for receptor bound to A, and \(RB\) for receptor bound to B. For each ligand, the dissociation constant is the concentration of unbound receptor multiplied by free ligand concentration, divided by bound-receptor concentration. Thus \(K_{D,A}=[R]C_A/[RA]\), and \(K_{D,B}=[R]C_B/[RB]\). Rearranging expresses each bound state relative to unbound receptor.

$$ [RA]=[R]\frac{C_A}{K_{D,A}}, \qquad [RB]=[R]\frac{C_B}{K_{D,B}} $$

The total receptor concentration is the sum of the three mutually exclusive states: \([R]_{\mathrm{total}}=[R]+[RA]+[RB]\). Substituting the two mass-action expressions gives a common denominator for all three fractions. Dividing each state concentration by total receptor concentration yields the occupancy equations.

$$ D=1+\frac{C_A}{K_{D,A}}+\frac{C_B}{K_{D,B}}, \qquad \theta_0=\frac{1}{D}, \qquad \theta_A=\frac{C_A/K_{D,A}}{D}, \qquad \theta_B=\frac{C_B/K_{D,B}}{D} $$

The ratios \(C_A/K_{D,A}\) and \(C_B/K_{D,B}\) are dimensionless. Each is a concentration relative to the ligand’s own affinity scale. Their sum with 1 determines the shared denominator, so \(\theta_0+\theta_A+\theta_B=1\). Total occupancy is \(1-\theta_0\), or equivalently the sum of the two bound fractions.

Key result: A dominates receptor engagement over B when \(C_A/K_{D,A} > C_B/K_{D,B}\). Their occupancy ratio is \(\theta_A/\theta_B=(C_A/K_{D,A})/(C_B/K_{D,B})\), provided both concentrations are positive. Dominance is determined by concentration relative to affinity, not by concentration or affinity alone.

Worked Calculations Across Concentration Scenarios

Worked Example: Equal Affinity-Weighted Concentrations

Suppose hypothetical free concentrations are \(C_A=20\) nM and \(C_B=200\) nM, with \(K_{D,A}=10\) nM and \(K_{D,B}=100\) nM. Assume both ligands bind the same site at equilibrium and the stated values are free concentrations at that site.

The dimensionless weights are \(C_A/K_{D,A}=20/10=2.00\) and \(C_B/K_{D,B}=200/100=2.00\). The shared denominator is \(D=1+2.00+2.00=5.00\).

Therefore, \(\theta_A=2.00/5.00=0.400\), \(\theta_B=2.00/5.00=0.400\), and \(\theta_0=1/5.00=0.200\). Total occupancy is \(0.400+0.400=0.800\), or 80.0%.

Although B is present at ten times the free concentration of A, the two ligands occupy equal receptor fractions because B also has a tenfold larger \(K_D\), meaning lower affinity in this comparison. Neither ligand dominates engagement in this scenario.

Worked Example: The More Concentrated Ligand Does Not Dominate

Now let \(C_A=5\) nM and \(C_B=100\) nM, with \(K_{D,A}=1\) nM and \(K_{D,B}=50\) nM. These are hypothetical equilibrium conditions at a single shared receptor site.

The weights are \(5/1=5.00\) for A and \(100/50=2.00\) for B, so \(D=1+5.00+2.00=8.00\). The occupancies are \(\theta_A=5.00/8.00=0.625\), \(\theta_B=2.00/8.00=0.250\), and \(\theta_0=1/8.00=0.125\).

A occupies 62.5% of receptors and B occupies 25.0%; 12.5% are unoccupied, giving 87.5% total occupancy. B is present at 20 times the concentration of A, but A dominates because its concentration relative to its \(K_D\) is greater: \(5.00\) versus \(2.00\). In this comparison A’s \(K_D\) is 50 times smaller, while its concentration is 20 times smaller.

Worked Example: A Change in Concentration Reverses the Dominant Ligand

Consider hypothetical ligands with \(K_{D,A}=6\) nM and \(K_{D,B}=10\) nM, while \(C_B\) remains 30 nM. Compare two scenarios for A: \(C_A=12\) nM and \(C_A=24\) nM. The receptor site, affinity constants, and equilibrium assumptions are held fixed.

Scenario 1: \(C_A=12\) nM. The weights are \(12/6=2.00\) for A and \(30/10=3.00\) for B. Thus \(D=1+2.00+3.00=6.00\). A’s occupancy is \(2.00/6.00=0.333\), B’s is \(3.00/6.00=0.500\), and the unoccupied fraction is \(1/6.00=0.167\). B dominates engagement.

Scenario 2: \(C_A=24\) nM. A’s weight rises to \(24/6=4.00\), while B’s remains 3.00. Now \(D=1+4.00+3.00=8.00\), so \(\theta_A=4.00/8.00=0.500\), \(\theta_B=3.00/8.00=0.375\), and \(\theta_0=1/8.00=0.125\). A now dominates.

The change in A’s concentration doubles its weight, but it does not simply double its occupancy: the denominator also changes because the receptor allocation is shared. The boundary between the two ligands occurs when their weights are equal. Here that requires \(C_A/6=C_B/10=3.00\), or \(C_A=18\) nM when \(C_B=30\) nM. Below that boundary B has the larger occupancy; above it A does.

Interpreting and Checking a Calculation

For any scenario, calculate both weights before deciding which ligand dominates. The ligand with the larger weight has the larger bound fraction because both occupancy equations use the same positive denominator. To see this directly, \(\theta_A/\theta_B=(C_A/K_{D,A})/(C_B/K_{D,B})\); the denominator \(D\) cancels. This ratio compares the bound fractions, while the denominator is still needed to calculate either absolute fraction.

Several checks help identify calculation or input problems:

  • Each weight and the denominator must be dimensionless. Concentrations and dissociation constants must use compatible units before division.
  • Each occupancy must lie between 0 and 1, and \(\theta_0+\theta_A+\theta_B\) must equal 1, apart from rounding.
  • Total occupancy must not exceed 1. A sum above 1 usually signals that the ligands were calculated independently rather than as competitors.
  • If one ligand’s weight is much larger than the other’s, its bound fraction should also be larger, even if it is not necessarily close to 100% when the other weight or the unoccupied-state weight is substantial.
  • A result that changes substantially when total rather than free concentrations are substituted should prompt review of whether the free concentrations at the receptor are known or need to be estimated.

A frequent error is to calculate each ligand’s single-ligand occupancy, \(C/(K_D+C)\), and add the results. That expression applies when that ligand is considered alone with the receptor. For the first scenario above, the separate values would be \(20/(10+20)=0.667\) for A and \(200/(100+200)=0.667\) for B; their sum, 1.33, is impossible for fractions of one shared receptor pool. The correct competitive calculation gives 0.400 for each ligand and 0.800 total occupancy.

When This Calculation Is Appropriate

The two-ligand equilibrium expression is a compact way to compare engagement across concentration scenarios when free ligand concentrations and equilibrium dissociation constants are appropriate inputs. It is useful for separating two influences that can otherwise be conflated: how much ligand is available and how strongly it binds. For instance, equal weights produce equal occupancies even when concentrations differ, as the first example shows.

The principal alternative in a simple equilibrium setting is to calculate each ligand as though it binds alone. That approach can describe a deliberately separate single-ligand condition, but it does not allocate receptors correctly when the ligands compete simultaneously. If a more detailed model has several independent receptor populations, their occupancy calculations must be performed for their respective populations and then interpreted accordingly; a single common \(K_D\) pair should not be assumed to represent distinct affinities.

The formula also has limits. If ligand depletion is appreciable, the total ligand concentration may differ from the free concentration used in the equations; free concentrations then need to be found consistently with ligand and receptor mass balances. If binding has not equilibrated, an equilibrium calculation may not describe occupancy at the time of interest, and a kinetic binding model is more appropriate. Cooperative binding, allostery, receptor state changes, or more than two mutually exclusive ligands also require a model that represents those features rather than an unexamined reuse of this two-ligand formula.

Finally, occupancy identifies how receptors are allocated among binding states under the stated model. It does not establish that the ligand with the greater occupancy produces a greater response. Efficacy, receptor reserve, and downstream signaling can affect the relationship between engagement and function; those properties require information beyond this occupancy calculation.

Check Your Understanding

Use the competitive equilibrium model and show how the weights, denominator, and fractions support each answer.

  1. For \(C_A=8\) nM, \(K_{D,A}=4\) nM, \(C_B=15\) nM, and \(K_{D,B}=5\) nM, calculate both occupancies and the unoccupied fraction.
  2. For what value of \(C_A\) do the ligands have equal occupancies if \(K_{D,A}=2\) nM, \(C_B=12\) nM, and \(K_{D,B}=6\) nM?
  3. If both free concentrations are multiplied by the same positive factor while the \(K_D\) values remain fixed, does the ratio \(\theta_A/\theta_B\) change? Explain.
  4. A calculation gives \(\theta_A=0.70\) and \(\theta_B=0.55\). Identify a basic consistency problem and state one likely source.

Solutions. (1) The weights are \(8/4=2.00\) and \(15/5=3.00\), so \(D=1+2.00+3.00=6.00\). Thus \(\theta_A=2/6=0.333\), \(\theta_B=3/6=0.500\), and \(\theta_0=1/6=0.167\); B dominates. (2) Equal occupancy requires equal weights: \(C_A/2=12/6=2.00\), so \(C_A=4.00\) nM. (3) No. The ratio is \((C_A/K_{D,A})/(C_B/K_{D,B})\); multiplying both concentrations by the same factor cancels from the ratio, though the absolute occupancies and total occupancy can change. (4) The fractions sum to 1.25, exceeding the available receptor pool. The ligands may have been calculated independently rather than with a shared competitive denominator.

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