Why Heterogeneous Binding Changes the Occupancy Curve
A single-site equilibrium model assigns one dissociation constant to every binding site. But a measured target population can contain distinct classes of sites, for example, because sites occupy different molecular environments. At the same free drug concentration, each class can then have a different fractional occupancy. The observed total occupancy combines those class-specific fractions, weighted by how much target belongs to each class.
This distinction matters because a mixture can have a broad or shoulder-shaped transition across concentrations, even when binding within every class follows a simple one-site equilibrium relationship. Fitting one hyperbola to such a curve can obscure the underlying affinity classes and can make a single apparent affinity depend on which concentration range was measured.
The weights refer to fractions of target sites, not fractions of drug or fractions of the measured signal unless those quantities are proportional to target amount. A class with many sites can contribute substantially to total occupancy even if its individual sites have relatively low occupancy at a given concentration.
Deriving the Weighted Occupancy Curve
Assume each class has reversible, one-to-one binding at equilibrium, and that sites within a class share the same \(K_{D,j}\). Assume the classes bind independently and experience the same free concentration \(C\). The concentration must be free drug concentration; if depletion is material, total concentration cannot simply be substituted for \(C\). The target-class amounts and their weights are taken as fixed over the conditions being modeled.
For class \(j\), the fraction bound is \(\theta_j(C)=C/(K_{D,j}+C)\). The amount bound in that class is \(B_j=T_j\theta_j(C)\). Adding bound amounts across classes and dividing by total target gives:
This is the weighted occupancy curve. Its assumptions concern a mixture of fixed classes, not a continuously varying population of affinities. The model can also be used when class amounts are reported in other consistent units, provided their ratios correctly represent the fractions of binding sites.
Worked Example: Calculating Occupancy Across Two Classes
Consider a hypothetical target population with 70% of its sites in a high-affinity class with \(K_{D,1}=1\) nM and 30% in a lower-affinity class with \(K_{D,2}=100\) nM. Assume equilibrium, independent binding, and that the stated concentrations are free concentrations. Calculate total occupancy at \(C=1\), 10, and 100 nM.
At 1 nM, class-specific occupancies are \(1/(1+1)=0.5000\) and \(1/(100+1)=0.009901\). Thus:
Total occupancy is 0.3530, or 35.30%. At 10 nM:
Total occupancy is 66.36%. At 100 nM:
Total occupancy is 84.31%. The high-affinity class is nearly saturated by 100 nM, while the lower-affinity class is only half occupied. The weighted total therefore remains below full occupancy at that concentration.
How the Mixture Departs from a Single-Site Curve
A single-site model has the form \(C/(K_D+C)\). For a heterogeneous population, the total curve approaches zero as \(C\) approaches zero and approaches one as \(C\) becomes very large. Those shared limits do not make the curves interchangeable: intermediate concentrations depend on all the class affinities and weights.
A short low-concentration expansion shows why a mixture is generally not itself a single-site hyperbola. When \(C\) is small relative to the class dissociation constants, expansion of each class curve gives:
Write \(A=\sum_j w_j/K_{D,j}\) and \(B=\sum_j w_j/K_{D,j}^2\). A single-site curve that matches the mixture's initial slope \(A\) must have \(K_D=1/A\). Its second-order coefficient is then \(A^2\), whereas the mixture's coefficient is \(B\). Since the weights sum to one, the weighted-square inequality gives:
Equality holds only when all classes with positive weight have the same dissociation constant. Therefore, if affinities differ, matching the initial slope does not also match the curvature. More generally, a mixture with distinct positive-weight affinities cannot be exactly represented by one hyperbola over the full concentration range.
The concentration producing half total occupancy is also a mixture property. It solves \(\sum_j w_j C/(K_{D,j}+C)=0.5\); it is not generally the arithmetic mean of the \(K_{D,j}\) values. It can be reported as an apparent half-occupancy concentration, but it does not by itself identify the component affinities.
Worked Example: Finding the Half-Occupancy Concentration
Use the 70% and 30% classes from the first example. Find the free concentration at which total occupancy is 50%. The assumptions remain equilibrium, independent binding, and fixed class weights.
Set the weighted curve equal to 0.5 and multiply by the positive denominator \((1+C)(100+C)\):
Expanding and collecting terms gives \(C^2+70.3C=50+50.5C+0.5C^2\), or \(C^2+39.6C-100=0\). The positive root is:
The other root is negative and is not a physically meaningful concentration. Thus the apparent half-occupancy concentration is about 2.382 nM, not the weighted arithmetic mean \(0.70(1)+0.30(100)=30.7\) nM. The high-affinity class dominates the mixture's occupancy near this point.
Worked Example: Translating Fractional Occupancy into Bound Target
A hypothetical sample contains 150 pmol of target: 120 pmol belongs to a class with \(K_D=4\) nM, and 30 pmol belongs to a class with \(K_D=40\) nM. At free concentration \(C=8\) nM, calculate total fractional occupancy and the amount of bound target. Assume the class amounts are binding-site amounts and all sites meet the equilibrium assumptions.
The weights are \(120/150=0.80\) and \(30/150=0.20\). The class occupancies are \(8/(4+8)=0.6667\) and \(8/(40+8)=0.1667\). Therefore:
The total bound amount is \(150\ \mathrm{pmol}\times0.5667=85.0\ \mathrm{pmol}\). As a check, the high-affinity class contributes \(120(0.6667)=80.0\) pmol bound, and the lower-affinity class contributes \(30(0.1667)=5.00\) pmol; their sum is 85.0 pmol.
Reading Curve Shape and Choosing a Model
When the affinities are well separated, one part of the population can become substantially occupied before another part does. The aggregate curve can then show a shoulder or a broad transition. When affinities are similar, the weighted curve may be difficult to distinguish from a single-site curve over a limited concentration range. A visibly non-hyperbolic pattern is a reason to examine the model, not proof of heterogeneous binding: other mechanisms and measurement problems can produce departures too.
| Approach | What it represents | Useful interpretation | Main caution |
|---|---|---|---|
| Single-site hyperbola | One affinity for all sites | A single \(K_D\) under the model | Can misrepresent a mixture's transition |
| Weighted affinity classes | Several fixed classes, each with its own \(K_{D,j}\) | Class-specific affinities and target fractions, if supported | More classes add parameters and may not be distinguishable from limited data |
| Hill-type curve | A flexible non-hyperbolic concentration-response shape | A descriptive slope and half-response concentration | Its shape parameter does not, by itself, establish affinity classes or a binding mechanism |
The Hill-type alternative is useful for describing a non-hyperbolic curve, but it is not a substitute for a mechanistic class mixture when the scientific question concerns distinct target-site affinities. Conversely, choosing a weighted mixture does not establish that the proposed classes exist; the model is a mechanistic hypothesis that must be assessed against the measurements and experimental design.
Diagnostics, Limitations, and a Common Mistake
Inspect residuals across concentration rather than relying only on an overall fit statistic. A systematic run of residuals on one side of zero at low concentrations and the opposite side at higher concentrations can indicate that one hyperbola is missing structure. A shoulder that remains unexplained by the fitted mixture can indicate that the chosen number of classes or their affinities are inadequate. Before attributing such patterns to affinity heterogeneity, check that concentrations represent free drug, that equilibrium is plausible, that the measurement is proportional to occupancy, and that the target amount has not changed across conditions.
A finite-class model is a simplification. Real targets may not divide naturally into a small number of classes, and distinct class parameters can be difficult to estimate if the observed concentration range does not span their transitions. A good fit alone does not guarantee that weights and affinities have unique or precise interpretations. Independent evidence about target classes and measurements across an informative concentration range strengthen interpretation.
Common mistake: “The mixture has a half-occupancy concentration of 2.382 nM, so every site has \(K_D=2.382\) nM.” This confuses an aggregate curve summary with a class-specific dissociation constant. The correction is to distinguish the apparent concentration giving 50% total occupancy from the component \(K_{D,j}\) values; the aggregate half-point generally does not equal any one component affinity.
Check Your Understanding
Answer each question, then compare with the solutions.
- A target has two classes with weights 0.60 and 0.40 and dissociation constants 2 nM and 20 nM. What is total occupancy at \(C=2\) nM?
- Why must the weights in the total occupancy equation sum to one?
- For equal weights and dissociation constants 1 nM and 100 nM, calculate total occupancy at 10 nM.
- Under what condition does the weighted curve reduce exactly to a single-site hyperbola?
- Give one reason a shoulder in measured occupancy should not automatically be taken as proof of heterogeneous affinities.
Solutions. (1) The class occupancies are \(2/(2+2)=0.5000\) and \(2/(20+2)=0.09091\), so total occupancy is \(0.60(0.5000)+0.40(0.09091)=0.3364\), or 33.64%. (2) Each weight is that class's fraction of total target, so the class fractions must account for the whole target population: \(\sum_j w_j=1\). (3) Total occupancy is \(0.50(10/11)+0.50(10/110)=0.4545+0.04545=0.5000\). (4) If every class with positive weight has the same \(K_D\), their weighted occupancy equals that common single-site curve. (5) A shoulder can also reflect mechanisms or measurement issues, such as non-equilibrium, concentration depletion, or an assay signal that is not proportional to occupancy; those possibilities should be checked before assigning an affinity-mixture explanation.