Why Equilibrium Binding Does Not Reveal Binding Speed
Reversible ligand–receptor binding is often summarized by an equilibrium dose–occupancy curve. That curve can reveal how much free ligand is needed to occupy receptors, but it does not by itself reveal how quickly ligand binds or dissociates. The distinction matters when a model is intended to describe both equilibrium response and the time course of drug action: a parameter combination can be identifiable from equilibrium measurements even when the individual parameters making up that combination are not.
The central result is that, under a single-site mass-action model, equilibrium occupancy depends on the association and dissociation rate constants only through their ratio. Equilibrium data can identify that ratio under suitable conditions, but any positive pair of rates with the same ratio gives the same equilibrium curve.
Mass-Action Model and Equilibrium Occupancy
Let \(R\) be free receptor, \(L\) free ligand, and \(RL\) the bound complex. The reaction is reversible: free receptor and ligand associate to form complex, while complex dissociates into its components. Write \(k_{\mathrm{on}}\) for the association rate constant and \(k_{\mathrm{off}}\) for the dissociation rate constant. Assume that free ligand concentration is held constant during the binding measurement and denote it by \(L\). Let total receptor concentration be \(R_T=[R]+[RL]\), which is constant.
Under mass-action kinetics, complex concentration changes at the association rate minus the dissociation rate. Replacing free receptor by \(R_T-[RL]\) gives
At equilibrium, the complex concentration is constant, so its time derivative is zero. Solving that balance for fractional occupancy gives
This is the one-site equilibrium occupancy curve. It rises from zero toward one as free ligand concentration increases. At \(L=K_D\), occupancy is one-half. This midpoint interpretation applies to fractional occupancy under the stated model; it does not mean that either rate constant equals the midpoint concentration.
What the Equilibrium Curve Identifies
Suppose free ligand concentration is known and a measured equilibrium occupancy is strictly between zero and one. Rearranging the occupancy equation yields
Thus, in the idealized model, one exact occupancy measurement at a known, positive \(L\) identifies \(K_D\). In practice, several concentrations are usually needed to assess whether the model describes the data and to estimate the curve with measurement error. The measurements should cover informative parts of the curve: data confined to very low occupancy or a nearly saturated plateau can provide weak information about the midpoint.
The result depends on assumptions. The receptor population is represented by one homogeneous binding site per receptor; binding is reversible and follows the stated mass-action model; the measurement has reached equilibrium; free ligand concentration is known; and occupancy is measured or can be normalized using known total receptor. Ligand depletion must be negligible or otherwise accounted for, since nominal total ligand concentration is not automatically the free concentration \(L\). The model also excludes additional binding states, cooperativity, and other processes that alter the occupancy relationship.
Worked Example: Estimating the Dissociation Constant From One Occupancy
In a hypothetical equilibrium assay, the free ligand concentration is \(12\) nM and the measured fractional occupancy is \(0.60\). Assume the single-site model and that equilibrium has been reached. Estimate \(K_D\).
Using the rearranged occupancy equation,
The estimated dissociation constant is \(8.00\) nM. This calculation identifies the ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\), not either rate constant separately. Its interpretation depends on the assumptions: if the reported \(12\) nM were total ligand in a depleted system rather than free ligand, the substitution would not be justified without a model for depletion.
Why the Individual Rates Are Not Identifiable at Equilibrium
The equilibrium equation contains \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\) only through \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\). A direct symmetry argument makes the resulting nonidentifiability explicit. For any positive scaling factor \(c\), define new rates \(k'_{\mathrm{on}}=c k_{\mathrm{on}}\) and \(k'_{\mathrm{off}}=c k_{\mathrm{off}}\). Then
Therefore, every equilibrium measurement at every ligand concentration is unchanged by this common rescaling. There is a continuum of rate pairs consistent with the same equilibrium curve. The rates are not separately structurally identifiable from these observations, even if the equilibrium curve were measured without noise at arbitrarily many concentrations. More measurements of the same equilibrium output cannot break a symmetry that leaves that output unchanged.
This is an instance of identifying a parameter combination rather than its components. The identifiable quantity is \(K_D\), while the separate positive values of \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\) remain ambiguous. A rate pair can be reported only if additional information or a different kind of observation constrains the rates.
Worked Example: Two Rate Pairs With the Same Equilibrium Curve
Consider two hypothetical parameter pairs, with ligand concentration measured in nM and time in minutes:
- Pair A: \(k_{\mathrm{on}}=0.020\ \mathrm{nM}^{-1}\mathrm{min}^{-1}\) and \(k_{\mathrm{off}}=0.10\ \mathrm{min}^{-1}\).
- Pair B: \(k_{\mathrm{on}}=0.060\ \mathrm{nM}^{-1}\mathrm{min}^{-1}\) and \(k_{\mathrm{off}}=0.30\ \mathrm{min}^{-1}\).
For Pair A, \(K_D=0.10/0.020=5.00\) nM. For Pair B, \(K_D=0.30/0.060=5.00\) nM as well. At \(L=5.00\) nM, both therefore predict
The same equality holds at every other free ligand concentration because both pairs have the same \(K_D\). Their rates differ threefold, but equilibrium occupancy data cannot distinguish them. The rates would imply different approach-to-equilibrium speeds; that information is absent from measurements taken only at equilibrium.
Justification From the Time-Dependent Equation
The model also shows why time-course information can, in principle, supply information that equilibrium observations lack. With constant free ligand, the mass-action equation can be written as
This linear equation has equilibrium value \(R_T k_{\mathrm{on}}L/(k_{\mathrm{on}}L+k_{\mathrm{off}})\) and a relaxation rate \(k_{\mathrm{on}}L+k_{\mathrm{off}}\). To see the latter, subtract the equilibrium value from both sides and let \(z(t)=[RL](t)-[RL]_{\mathrm{eq}}\). Then
Equilibrium occupancy supplies the ratio \(K_D\); a resolved time course can additionally inform the relaxation rate. Under the restrictive conditions of known constant \(L\), a suitable observed transient, and the same single-site model, these quantities can determine the two rates: \(k_{\mathrm{on}}=(k_{\mathrm{on}}L+k_{\mathrm{off}})/(L+K_D)\) and \(k_{\mathrm{off}}=K_D k_{\mathrm{on}}\). This comparison is not a guarantee that any time-course experiment will identify both rates; unknown initial states, sparse sampling, or other model uncertainties can create further ambiguity.
Diagnostics, Alternatives, and Limitations
Before interpreting an estimated \(K_D\), inspect whether the equilibrium model is plausible for the experiment. Useful checks include whether replicate measurements at the same free concentration agree, whether estimated occupancies remain between zero and one, and whether the curve is consistent with a single smooth rise toward saturation. Systematic departures may indicate that equilibrium was not reached, free ligand was mischaracterized, occupancy was not correctly normalized, or the one-site model is inadequate. A poor curve fit is not evidence that the rates have become separately identifiable; it may instead show that the assumed model does not describe the observations.
There are several possible approaches, but they answer different questions:
| Observation or approach | What it can address | Key limitation |
|---|---|---|
| Equilibrium occupancy across ligand concentrations | Estimates \(K_D\) under the one-site equilibrium assumptions | Cannot separate \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\) |
| Time-resolved binding or dissociation measurements | Can add information about kinetic rates when the time course is sufficiently informative | Requires appropriate timing, inputs, observations, and model assumptions |
| Independent information on one rate | Can combine with an identified \(K_D\) to calculate the other rate | Depends on the accuracy and relevance of the external rate information |
For example, if \(k_{\mathrm{on}}\) is independently known and \(K_D\) has been estimated, then \(k_{\mathrm{off}}=K_D k_{\mathrm{on}}\). This calculation does not mean that the equilibrium data estimated \(k_{\mathrm{off}}\) alone: the result relies on the added information about \(k_{\mathrm{on}}\).
A common pitfall is to fit an equilibrium curve and report precise estimates of both rate constants because the fitting software returned two numbers. The correction is to check what combinations the observations actually constrain. If the predicted equilibrium output depends only on \(k_{\mathrm{off}}/k_{\mathrm{on}}\), different rate pairs with the same ratio fit equally well in the ideal model. A numerical optimizer may select one pair because of its starting values, constraints, or regularization; that selection does not establish that equilibrium data distinguish the rates.
Check Your Understanding
Answer each question, then compare with the concise solutions below.
- At \(L=15\) nM, equilibrium occupancy is \(0.75\). What is \(K_D\), under the single-site model?
- Why do equilibrium observations at many ligand concentrations still fail to distinguish rate pairs that have the same \(k_{\mathrm{off}}/k_{\mathrm{on}}\)?
- If both rate constants are multiplied by four, what happens to the equilibrium occupancy curve?
- Name two assumptions needed to use the stated occupancy equation with measured ligand concentrations.
Solutions. (1) \(K_D=15(1-0.75)/0.75=5.00\) nM. (2) Every predicted equilibrium occupancy depends on the rates only through their ratio, so equal ratios give equal predictions at every \(L\). (3) The ratio and thus the equilibrium curve are unchanged. (4) For example, measurements must be at equilibrium, free ligand concentration must be known, and the one-site mass-action model must apply; any two of these are sufficient answers.