Why Fit the Whole Time Course?
A single occupancy measurement describes a state at one time; a time course can also reveal how rapidly that state is approached and how quickly it is lost after ligand removal. Those rates matter when comparing target engagement over time, but they are only interpretable if the kinetic model is appropriate and the observations support its assumptions. A curve that looks plausible by eye can still have structured residuals that reveal a poor fit.
Here, the model is a reversible, one-site interaction observed as fractional receptor occupancy. During association, the free ligand concentration is treated as constant. During washout, free ligand is assumed to have been removed sufficiently that rebinding is negligible. Assay signal is assumed to have been calibrated to occupancy, rather than to an uncalibrated biomarker.
The model assumes a homogeneous population of equivalent, reversibly binding sites; a well-mixed system; known and effectively constant free ligand during association; negligible ligand rebinding after washout; and no material change in receptor abundance during the experiment. It also assumes that the measured occupancy is not distorted by time-varying assay calibration. Processes such as internalization, recycling, irreversible binding, and multiple receptor subtypes require additional states or mechanisms and should not be silently absorbed into a one-site rate estimate.
Association and Washout Solutions
For association at constant \(C\), define \(k_{\mathrm{obs}}=k_{\mathrm{on}}C+k_{\mathrm{off}}\). The steady occupancy \(B_{\mathrm{ss}}\) is the value at which the net rate is zero: \(k_{\mathrm{on}}C(1-B_{\mathrm{ss}})-k_{\mathrm{off}}B_{\mathrm{ss}}=0\). Rearranging gives \(B_{\mathrm{ss}}=k_{\mathrm{on}}C/(k_{\mathrm{on}}C+k_{\mathrm{off}})\). If the initial occupancy is \(B_0\), the solution is \(B(t)=B_{\mathrm{ss}}+(B_0-B_{\mathrm{ss}})e^{-k_{\mathrm{obs}}t}\). When association begins from an unoccupied state, this reduces to \(B(t)=B_{\mathrm{ss}}(1-e^{-k_{\mathrm{obs}}t})\).
After washout, set \(C=0\). The model becomes \(dB/dt=-k_{\mathrm{off}}B\), with solution \(B(t)=B(0)e^{-k_{\mathrm{off}}t}\). Thus the log of the occupancy ratio declines linearly with time. These solutions give practical fitting methods, but the transformed association fit has a limitation: taking logarithms changes the error scale. If the measurement errors are additive and roughly constant on the occupancy scale, nonlinear fitting of the original occupancy observations is generally preferable.
Worked Fitting Examples
Worked Example: Estimate Dissociation from Washout
Suppose a hypothetical calibrated assay records occupancy during washout, and the log occupancy ratios relative to the washout start are as follows. The washout starts at time zero, and the ligand is assumed to be sufficiently removed that rebinding is negligible.
| Time (h) | \(\ln[B(t)/B(0)]\) |
|---|---|
| 1 | -0.600 |
| 2 | -1.199 |
| 4 | -2.392 |
| 6 | -3.558 |
Under the washout model, \(w_i=-k_{\mathrm{off}}t_i\). Fitting a line through the origin gives
The estimated dissociation rate is therefore \(0.5950\ \mathrm{h}^{-1}\). For example, at 2 h the model predicts an occupancy ratio of \(e^{-0.59498(2)}\approx0.3042\). A rate constant is not itself a half-life; the corresponding half-life under this exponential washout model is \(\ln(2)/0.59498\approx1.165\) h. This calculation presumes the log ratios are based on positive, quantified occupancies and that the washout measurements reflect dissociation rather than continued rebinding or receptor loss.
Worked Example: Estimate Association and Combine the Rates
In a second hypothetical run, association begins at zero occupancy, the free ligand concentration is \(10.0\) nM, and an independently measured plateau is \(B_{\mathrm{ss}}=0.667\). The observed occupancies are rounded to three decimals.
| Time (h) | Observed occupancy | \(z=-\ln(1-B/B_{\mathrm{ss}})\) |
|---|---|---|
| 1 | 0.301 | 0.600 |
| 2 | 0.466 | 1.199 |
| 4 | 0.606 | 2.392 |
| 6 | 0.649 | 3.612 |
For example, at 2 h, \(z=-\ln(1-0.466/0.667)=-\ln(0.30135)\approx1.199\). Applying the through-origin estimator gives
Combining this estimate with the washout estimate from the preceding example gives \(k_{\mathrm{on}}C=k_{\mathrm{obs}}-k_{\mathrm{off}}=0.6007-0.5950=0.0057\ \mathrm{h}^{-1}\). At \(C=10.0\) nM, the implied association rate constant is \(k_{\mathrm{on}}=0.0057/10.0=0.00057\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\). The plateau and rate estimates are rounded, so the final digits are approximate. The necessary consistency condition is \(k_{\mathrm{obs}}>k_{\mathrm{off}}\) for positive \(k_{\mathrm{on}}C\); a fitted difference at or below zero would indicate sampling uncertainty, incompatible assay conditions, or model inadequacy, not a physically meaningful negative association rate.
The transformed estimates are useful for transparent calculations and initial values for fitting. They do not remove the need to inspect predictions on the occupancy scale. Residuals are defined as observed minus predicted occupancy, \(r_i=B_i-\widehat{B}(t_i)\). Positive residuals mean the observation exceeds the prediction; negative residuals mean it falls below it. Their sequence over time is more informative than a single summary of their size.
Worked Example: Read Residuals for Systematic Mismatch
Consider a separate hypothetical association data set with observations \(0.220\), \(0.466\), \(0.550\), and \(0.640\) at 1, 2, 4, and 6 h. Check a candidate model with \(B_{\mathrm{ss}}=0.667\) and \(k_{\mathrm{obs}}=0.600\ \mathrm{h}^{-1}\), using \(\widehat{B}(t)=0.667(1-e^{-0.600t})\). This is a residual check for that candidate, not a claim that these data generated the association fit above.
| Time (h) | Observed | Predicted | Residual |
|---|---|---|---|
| 1 | 0.220 | 0.301 | -0.081 |
| 2 | 0.466 | 0.466 | 0.000 |
| 4 | 0.550 | 0.606 | -0.056 |
| 6 | 0.640 | 0.649 | -0.009 |
At 6 h, for example, the prediction is \(0.667(1-e^{-0.600(6)})=0.648775\), which rounds to \(0.649\), and the residual is \(0.640-0.648775=-0.008775\), which rounds to \(-0.009\). The 1 h and 4 h residuals are notably negative, while the 2 h observation is close to its prediction. This pattern warns that a candidate curve may be too high at some times; it is not justified to call the fit adequate merely because one point matches.
As an arithmetic check at a different rate, if \(k_{\mathrm{obs}}=0.595\ \mathrm{h}^{-1}\), the 2 h prediction is \(0.667(1-e^{-0.595(2)})=0.464084\), or \(0.4641\) to four significant figures. For an observed occupancy of \(0.466\), the residual is \(0.466-0.464084=0.001916\), not \(0.003\).
Residual Diagnostics and Alternatives
A good fit should be judged against the data scale and experimental precision. Plot residuals against time and, where relevant, against predicted occupancy or experimental run. For an adequate model with an appropriate error model, residuals should fluctuate around zero without a persistent curve, run of one sign, or change in spread. A smooth trend can indicate a missing process, a wrong plateau, an incorrect starting occupancy, or a time-dependent concentration. A widening spread may indicate that constant-variance errors are unsuitable. A single large residual may reflect a recording or assay problem, but it should be investigated rather than discarded automatically.
A practical fitting sequence is:
Confirm units, occupancy calibration, concentration history, initial occupancy, and the timing of washout. Verify that the measured time window can reveal both the rise and the plateau, if estimating both.
Use the exponential solutions for transparent initial estimates. For final estimates, fit the original occupancy observations by nonlinear least squares when additive, roughly constant errors are plausible, or use a likelihood matched to the assay error model.
Plot observations and fitted curves, then examine residuals versus time. Check parameter bounds and the kinetic requirement \(k_{\mathrm{obs}}>k_{\mathrm{off}}\) when deriving a positive \(k_{\mathrm{on}}C\).
Structured residuals motivate checking assumptions or adding a justified mechanism; they do not by themselves identify which mechanism is missing.
The principal alternative to the transformed straight-line calculations is direct nonlinear fitting of \(B_{\mathrm{ss}}(1-e^{-k_{\mathrm{obs}}t})\) and \(B(0)e^{-k_{\mathrm{off}}t}\) to the occupancy observations. It preserves the original scale and can accommodate a shared parameter structure, but requires numerical optimization and an explicit error model. A log-linear washout fit is simple and interpretable when multiplicative errors are reasonable; it can be distorted by additive assay background, especially near the lower limit of quantification. Direct fitting is not automatically better if its error assumptions are wrong, and neither method can recover information absent from the sampling design.
Rate estimates can be weakly identified when the association samples cover only the early rise or never approach a plateau: \(B_{\mathrm{ss}}\) and \(k_{\mathrm{obs}}\) may trade off. A background offset can mimic residual curvature, while ligand depletion makes the assumed constant \(C\) false. Washout can appear slower than dissociation if free ligand remains and rebinds. Receptor turnover or internalization can also change occupancy independently of the binding rates. The earlier material on receptor turnover and internalization distinguishes those biological processes; here they are limitations of the simple fitting model, not terms to add without supporting data.
Common Mistakes
- Using the wrong washout slope sign. The slope of \(\ln[B(t)/B(0)]\) against time is negative; \(k_{\mathrm{off}}\) is its negative.
- Calling \(k_{\mathrm{obs}}\) the association rate constant. During association, \(k_{\mathrm{obs}}=k_{\mathrm{on}}C+k_{\mathrm{off}}\). It is not \(k_{\mathrm{on}}\), and it depends on ligand concentration.
- Ignoring residual patterns. A small average residual can conceal a run of positive or negative errors. Inspect residuals against time, not only a single fit statistic.
- Using the log transform without considering its error consequences. The transformed regression emphasizes deviations differently from fitting occupancy on its original scale.
- Forcing an impossible parameter estimate. If estimated \(k_{\mathrm{obs}}\leq k_{\mathrm{off}}\), do not report a negative \(k_{\mathrm{on}}\); check uncertainty, assay conditions, and model fit.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- In a washout experiment, \(\ln[B(t)/B(0)]\) has a fitted slope of \(-0.30\ \mathrm{h}^{-1}\). What is the estimated \(k_{\mathrm{off}}\)?
- Why is \(k_{\mathrm{obs}}\) not generally equal to \(k_{\mathrm{on}}\)?
- At a fixed concentration, an association fit gives \(k_{\mathrm{obs}}=0.40\ \mathrm{h}^{-1}\) and a washout fit gives \(k_{\mathrm{off}}=0.25\ \mathrm{h}^{-1}\). What is \(k_{\mathrm{on}}C\)?
- Residuals are positive early in association and negative late in association. What does this pattern suggest, and does it identify a unique missing mechanism?
- Why can a log-linear washout analysis be unreliable when occupancy is near an additive assay background?
Solutions. (1) \(k_{\mathrm{off}}=-(-0.30)=0.30\ \mathrm{h}^{-1}\). (2) Because \(k_{\mathrm{obs}}=k_{\mathrm{on}}C+k_{\mathrm{off}}\), so it includes dissociation and depends on \(C\). (3) \(k_{\mathrm{on}}C=0.40-0.25=0.15\ \mathrm{h}^{-1}\). (4) It suggests systematic mismatch, such as an incorrect curve shape or condition; residuals alone do not identify a unique cause. (5) Taking the logarithm of a signal containing an additive background does not produce the assumed linear log-occupancy relation, and low occupancies can be especially distorted.