Why Competition Changes Occupancy
A ligand’s affinity and free concentration determine how strongly it occupies a target when it binds alone. When several ligands compete for the same binding site, however, each ligand’s occupancy also depends on the concentrations and affinities of the others. An inhibitor can therefore reduce the fraction of target bound by a drug even if the drug’s own free concentration and affinity remain unchanged.
The key calculation is a set of equilibrium fractions: the proportion of total target that is unbound and the proportion bound to each ligand. These fractions apply to one homogeneous target population with a single binding site shared exclusively by the ligands. They use free ligand concentrations. If appreciable target binding depletes the ligands, total concentrations cannot simply be substituted for free concentrations; the binding mass balance must also be considered, as in the earlier material on ligand depletion and the binding mass balance.
Deriving the Competition Fractions
At equilibrium, each ligand follows the mass-action definition of its dissociation constant:
Rearranging gives \([RL_i]=[R]C_i/K_{D,i}\). Since the target is either unbound or bound to exactly one of the mutually exclusive ligands, total target is the sum of those states:
Dividing each state concentration by \(R_T\) gives the occupancy fractions:
Each ratio \(C_i/K_{D,i}\) is dimensionless and represents ligand \(i\)’s relative contribution to binding in this equilibrium model. The common denominator means that raising one ligand’s ratio reduces the fractions available to all other states. The fractions are nonnegative and sum to one:
For two ligands, A and B, the bound fractions are \(C_A/K_{D,A}\) and \(C_B/K_{D,B}\), respectively, divided by \(1+C_A/K_{D,A}+C_B/K_{D,B}\). If B is absent, its ratio is zero and the expression for A reduces to the familiar single-ligand equilibrium occupancy \(C_A/(K_{D,A}+C_A)\). This limiting case is a useful check on the algebra.
Worked Example: Two Ligands Competing for One Target
Suppose two hypothetical ligands are at equilibrium with a shared target site. The free concentration of A is \(6\) nM and its \(K_D\) is \(2\) nM. The free concentration of inhibitor B is \(4\) nM and its \(K_D\) is \(1\) nM. Calculate the fractions of target in each state.
Check the model inputs. Both concentrations are specified as free concentrations, and all values are in nM. Assume one homogeneous site, mutually exclusive binding, equilibrium, and no cooperative interaction.
The dimensionless ratios are \(C_A/K_{D,A}=6/2=3\) and \(C_B/K_{D,B}=4/1=4\). The common denominator is \(1+3+4=8\). Therefore:
Thus, 12.5% of the target is unbound, 37.5% is bound to A, and 50.0% is bound to B under these hypothetical conditions. The fractions sum to \(0.125+0.375+0.500=1.000\). B occupies a larger fraction than A here because its concentration-to-\(K_D\) ratio is larger, not simply because its concentration is larger.
How Much Inhibitor Halves the Competing Ligand’s Occupancy?
For a tracer or drug A competing with inhibitor I, define \(x=C_A/K_{D,A}\) and \(y=C_I/K_{D,I}\). The fraction of target bound to A is \(x/(1+x+y)\). In the absence of inhibitor, that fraction is \(x/(1+x)\). The inhibitor concentration that halves A’s bound fraction can be derived by equating the occupied fraction in the presence of inhibitor to one-half its no-inhibitor value:
This result assumes that the free concentration of A is held fixed while inhibitor is added, and that both free concentrations are known. It describes a halving of A’s bound fraction relative to its own no-inhibitor value; it does not mean that half of all target is bound by inhibitor. The inhibitor’s own occupancy at that concentration is obtained from the full competition formula.
Worked Example: Inhibitor Concentration for Half-Displacement
A hypothetical tracer A has a free concentration of \(2\) nM and \(K_{D,A}=2\) nM. An inhibitor has \(K_{D,I}=0.5\) nM. Find the free inhibitor concentration that halves A’s bound fraction compared with the no-inhibitor condition.
Check the model inputs. The two binding partners are assumed to compete for the same single site at equilibrium. The free tracer concentration is held at \(2\) nM, and depletion is assumed negligible or otherwise accounted for in specifying the free concentration.
Here \(x=C_A/K_{D,A}=2/2=1\). The derived expression gives:
Without inhibitor, A’s bound fraction is \(1/(1+1)=0.500\). At \(C_I=1.00\) nM, the inhibitor ratio is \(1/0.5=2\), so A’s fraction is \(1/(1+1+2)=0.250\). This is half of \(0.500\), as required. The inhibitor-bound fraction is \(2/4=0.500\), and the remaining \(0.250\) is unbound.
More Than Two Competing Ligands
The same expression applies when any number of ligands compete for the site: add each ligand’s free-concentration-to-\(K_D\) ratio to the denominator, then use the ratio for the ligand of interest as its numerator. This makes it possible to account for background endogenous ligands or several co-administered compounds without treating their effects as separate, independent occupancies.
Worked Example: Three Ligands Sharing a Target
Consider three hypothetical ligands with the following free concentrations and dissociation constants: A, \(C_A=1.5\) nM and \(K_{D,A}=0.5\) nM; B, \(C_B=2.0\) nM and \(K_{D,B}=2.0\) nM; and C, \(C_C=0.5\) nM and \(K_{D,C}=1.0\) nM. Calculate the fraction of target bound to each ligand and the unbound fraction.
Check the model inputs. All three ligands are assumed to compete exclusively for one site at equilibrium, with the stated values representing free concentrations. The ratios are \(3.0\), \(1.0\), and \(0.5\), respectively, so the denominator is \(1+3.0+1.0+0.5=5.5\).
Approximately 54.55% of target is bound to A, 18.18% to B, 9.091% to C, and 18.18% is unbound. The fractions total \(1.000\), allowing for rounding. A contributes the largest occupancy because it has the largest \(C_i/K_{D,i}\) ratio. The occupancy of B is not calculated by considering only B and the unbound target; A and C also compete for the same target pool.
Assumptions, Diagnostics, and Alternatives
The formula is appropriate when the target is represented by a homogeneous population of single binding sites, binding is reversible and at equilibrium, and ligands are mutually exclusive at that site. The \(K_D\) values must describe the same target state and conditions as the concentrations used. The calculation also presumes that each ligand’s free concentration is known. When target binding materially changes free ligand concentrations, concentrations in the formula must be obtained using the binding mass balance rather than assumed equal to total concentrations.
Several checks help reveal a misuse or a poor fit. First, the calculated fractions must lie between zero and one and sum to one. Second, increasing one ligand’s free concentration while holding the others fixed must increase its own fraction and decrease the unbound fraction; it must also decrease the fractions of its competitors. Third, predicted displacement should be compatible with the measured conditions and the stated \(K_D\) values. If the model cannot account for observed occupancy across conditions, consider whether free concentrations were misrepresented, equilibrium was not attained, target sites are not homogeneous, or the ligands do not share a single exclusive site. A mismatch does not by itself identify which assumption failed.
This competitive model differs from an independent-site model. For distinct, noncompeting sites, binding at one site does not exclude binding at another, and simultaneous occupancy can occur; the mutually exclusive state accounting used above would then be inappropriate. Likewise, a system with cooperative binding or multiple target populations requires a different model rather than simply adding more terms to this denominator. The current calculation’s strength is its transparent equilibrium interpretation; its limitation is that it is not a general model for every binding mechanism.
A frequent mistake is to calculate each ligand’s single-ligand fraction \(C_i/(K_{D,i}+C_i)\) separately and treat those values as simultaneous occupancies. This ignores competition: those separate calculations each allocate the target as though the other ligands were absent, so their fractions can sum to more than one. The correction is to use one shared denominator containing every competing ligand’s \(C_i/K_{D,i}\) term.
Check Your Understanding
For each question, calculate using the stated assumptions. Solutions follow the questions.
- Ligand A has \(C_A=3\) nM and \(K_{D,A}=3\) nM. Inhibitor B has \(C_B=2\) nM and \(K_{D,B}=1\) nM. What are the unbound, A-bound, and B-bound fractions?
- A ligand has free concentration \(4\) nM and \(K_D=2\) nM. With no competitor, what fraction of target is bound?
- A tracer has \(C_A=1\) nM and \(K_{D,A}=1\) nM. An inhibitor has \(K_{D,I}=0.25\) nM. What free inhibitor concentration halves the tracer’s bound fraction?
- Why is it generally incorrect to substitute total ligand concentrations into the competitive occupancy formula when target binding depletes free ligand?
Solutions. (1) The ratios are \(1\) and \(2\), giving denominator \(1+1+2=4\). Thus \(\theta_0=1/4=0.250\), \(\theta_A=1/4=0.250\), and \(\theta_B=2/4=0.500\). (2) The single-ligand fraction is \(4/(2+4)=4/6=0.667\), or 66.7%. (3) Here \(C_A/K_{D,A}=1\), so \(C_I=0.25(1+1)=0.500\) nM. (4) The formula is derived using free concentrations in the mass-action relation. If binding appreciably removes ligand from solution, total concentration is not the concentration available to bind; the free concentration must be determined with the binding mass balance.