Worked Example of Kinetic Occupancy After Dosing

Why Binding Kinetics Matter After Dosing

An equilibrium calculation answers what occupancy would be if binding had enough time to settle at a fixed concentration. After a dose, however, concentration changes over time, and occupancy may lag behind it. A kinetic model describes that lag: it separates the rate of association from the rate of dissociation and lets us predict occupancy during both exposure and washout.

The worked analysis below uses a deliberately simple model. Its purpose is to show how a binding time course can provide rate estimates, and how those estimates inform occupancy predictions when concentration changes. The calculations are conditional on the model assumptions; they do not establish that those assumptions hold for every target or assay.

Definition: Let $\theta(t)$ be the fraction of the accessible target population occupied at time $t$, and let $C(t)$ be the free drug concentration at the target. In a one-site, reversible binding model, association occurs at rate $k_{\mathrm{on}}C(t)(1-\theta)$, and dissociation occurs at rate $k_{\mathrm{off}}\theta$. Thus, $$ \frac{d\theta}{dt}=k_{\mathrm{on}}C(t)(1-\theta)-k_{\mathrm{off}}\theta, $$ where $k_{\mathrm{on}}$ has units of concentration$^{-1}$ time$^{-1}$, and $k_{\mathrm{off}}$ has units of time$^{-1}$.

This model assumes one homogeneous class of independent binding sites, reversible binding, and a known free concentration at the target. It also assumes that the measured occupancy corresponds to $\theta(t)$, and that target availability is constant over the period considered. Drug depletion by binding, changing target abundance, internalization, and other binding mechanisms are not represented. These assumptions should be plausible for the assay and time window before the fitted rates are interpreted.

Constant-Concentration Solution

When $C(t)=C$ is constant, collect the terms involving $\theta$:

\frac{d\theta}{dt}=k_{\mathrm{on}}C- (k_{\mathrm{on}}C+k_{\mathrm{off}})\theta.

Define $k_{\mathrm{obs}}=k_{\mathrm{on}}C+k_{\mathrm{off}}$, the observed first-order rate constant at this concentration, and $\theta_{\mathrm{ss}}=k_{\mathrm{on}}C/k_{\mathrm{obs}}$, the model's steady-state occupancy. The differential equation becomes $d\theta/dt=-k_{\mathrm{obs}}(\theta-\theta_{\mathrm{ss}})$. Its solution, with initial occupancy $\theta_0$, is:

\theta(t)=\theta_{\mathrm{ss}}+ (\theta_0-\theta_{\mathrm{ss}})e^{-k_{\mathrm{obs}}t}.

For $\theta_0\ne\theta_{\mathrm{ss}}$, this follows by separating variables and integrating:

\int_{\theta_0}^{\theta(t)} \frac{d\theta}{\theta-\theta_{\mathrm{ss}}} = -k_{\mathrm{obs}}\int_0^t du, \qquad \ln\left|\frac{\theta(t)-\theta_{\mathrm{ss}}} {\theta_0-\theta_{\mathrm{ss}}}\right| =-k_{\mathrm{obs}}t.

Exponentiating and rearranging gives the stated solution. If instead $\theta_0=\theta_{\mathrm{ss}}$, the differential equation gives $d\theta/dt=0$ at that occupancy, so the trajectory remains at $\theta_{\mathrm{ss}}$; the same solution formula follows directly. This separate case matters because the separation and logarithm above are not defined when the initial deviation is zero.

At zero concentration, the association term vanishes and occupancy decays as $\theta(t)=\theta_0e^{-k_{\mathrm{off}}t}$. At a constant nonzero concentration, the time course has an exponential approach to $\theta_{\mathrm{ss}}$. The equilibrium dissociation constant implied by this model is $K_D=k_{\mathrm{off}}/k_{\mathrm{on}}$, so $\theta_{\mathrm{ss}}=C/(C+K_D)$. This equilibrium expression describes the long-run level, not how rapidly it is reached.

Worked Example: Fitting Rates to a Synthetic Binding Time Course

Suppose a hypothetical in vitro assay holds free drug concentration at $2.00\ \mathrm{nM}$, starts with unoccupied target, and measures occupancy at several times. The following synthetic, noise-free values are generated by a one-site reversible model; displayed values are rounded.

Time (h)	Occupancy, $\theta$
0	0.000000
1	0.314775
2	0.505696
4	0.691732
6	0.760170

Fit the constant-concentration trajectory $\theta(t)=\theta_{\mathrm{ss}}(1-e^{-k_{\mathrm{obs}}t})$, since $\theta_0=0$. For example, at $t=2\ \mathrm{h}$, the fitted values $\theta_{\mathrm{ss}}=0.800$ and $k_{\mathrm{obs}}=0.500\ \mathrm{h}^{-1}$ predict:

\theta(2)=0.800(1-e^{-(0.500\ \mathrm{h}^{-1})(2\ \mathrm{h})}) =0.800(1-e^{-1}) =0.5057.

The prediction agrees with the tabulated $0.505696$. Fitting the exponential to the full time course by nonlinear least squares, that is, minimizing the sum of squared differences between measured and predicted occupancies, gives approximately $\theta_{\mathrm{ss}}=0.800$ and $k_{\mathrm{obs}}=0.500\ \mathrm{h}^{-1}$. The displayed data are rounded, so the estimates are approximate.

At a known constant concentration, the fitted asymptote and rate identify the two binding rates under this model. Rearranging their definitions gives:

k_{\mathrm{on}}=\frac{\theta_{\mathrm{ss}}k_{\mathrm{obs}}}{C} =\frac{(0.800)(0.500\ \mathrm{h}^{-1})}{2.00\ \mathrm{nM}} =0.200\ \mathrm{nM}^{-1}\mathrm{h}^{-1}, $$ $$ k_{\mathrm{off}}=(1-\theta_{\mathrm{ss}})k_{\mathrm{obs}} =(1-0.800)(0.500\ \mathrm{h}^{-1}) =0.100\ \mathrm{h}^{-1}.

The resulting $K_D$ is $0.100/0.200=0.500\ \mathrm{nM}$. The parameters are interpretable only insofar as the assumed model and the concentration used in the fit are appropriate. In noisy data, uncertainty and correlation between the fitted asymptote and rate should also be assessed rather than treating rounded point estimates as exact.

Predicting Occupancy During Exposure and Washout

For a piecewise-constant concentration, apply the constant-concentration solution separately over each interval, using the occupancy at the end of one interval as the starting occupancy for the next. Occupancy is continuous at a concentration change in this model; its rate of change can change immediately.

Worked Example: A Four-Hour Exposure Followed by Washout

Use the fitted rates from the preceding example. A hypothetical exposure holds $C=2.00\ \mathrm{nM}$ for $4.00\ \mathrm{h}$, beginning at $\theta_0=0$, then concentration falls to zero for $6.00\ \mathrm{h}$. During exposure, $k_{\mathrm{obs}}=(0.200)(2.00)+0.100=0.500\ \mathrm{h}^{-1}$, and $\theta_{\mathrm{ss}}=0.800$. Therefore:

\theta(4)=0.800(1-e^{-(0.500)(4.00)}) =0.800(1-e^{-2}) =0.6917.

At washout, $C=0$, so only dissociation remains. Six hours later, predicted occupancy is:

\theta(10)=0.6917e^{-(0.100\ \mathrm{h}^{-1})(6.00\ \mathrm{h})} =0.6917e^{-0.600} =0.3796.

Thus the model predicts approximately $69.2\%$ occupancy at the end of exposure and $38.0\%$ six hours into washout. The concentration is zero during washout, but occupancy does not instantly become zero: bound drug dissociates at a finite rate. This prediction assumes that washout really makes free concentration at the target negligible and that $k_{\mathrm{off}}$ remains applicable during that interval.

Worked Example: Re-Exposure Before Equilibrium

Starting from the predicted washout occupancy $0.3796$, suppose concentration rises to a constant $1.00\ \mathrm{nM}$. How much target is occupied after $2.00\ \mathrm{h}$? Here $k_{\mathrm{obs}}=(0.200)(1.00)+0.100=0.300\ \mathrm{h}^{-1}$, and $\theta_{\mathrm{ss}}=(0.200)(1.00)/0.300=0.6667$. The starting occupancy is not zero, so retain it in the trajectory:

\theta(2)=0.6667+(0.3796-0.6667)e^{-(0.300)(2.00)} =0.6667-(0.2871)(0.5488) =0.5091.

The predicted occupancy is $50.9\%$, below the $66.7\%$ steady-state value at $1.00\ \mathrm{nM}$. Replacing the dynamic prediction with the equilibrium value would ignore the finite time available to approach that level. The previous occupancy also matters: starting from an empty target would give a different prediction at the same concentration and time.

Comparing Kinetic and Equilibrium Estimates

For the fitted rates, equilibrium occupancy at $2.00\ \mathrm{nM}$ is:

\theta_{\mathrm{eq}}=\frac{C}{C+K_D} =\frac{2.00}{2.00+0.500}=0.800.

That is the same asymptote estimated from the time course, as it should be under the model. It is not the predicted occupancy after a finite exposure: at four hours the kinetic prediction is $0.6917$, rather than $0.800$. At zero concentration, equilibrium occupancy is zero, but after six hours of washout the kinetic prediction remains $0.3796$. Equilibrium calculations are useful when binding is sufficiently rapid relative to concentration changes and the observation time; they can be misleading when those conditions do not hold.

A kinetic fit makes a stronger use of time-course data than an equilibrium estimate, but it also requires informative measurements over time. A single late measurement near a plateau can help estimate the asymptote but provides little information about the approach rate. Conversely, early measurements alone may constrain the initial slope without clearly identifying the long-run level. Measurements spanning both the rising portion and the plateau are more informative for estimating both quantities.

Diagnostics, Failure Modes, and Limitations

Inspect the observed time course against the fitted exponential, not just the estimated rates. A systematic pattern of residuals, such as early observations consistently above predictions and late observations below them, suggests that one exponential may not describe the data. Replicate variability, assay calibration, and whether occupancy measurements are comparable across time should also be considered. A poor fit may reflect measurement problems, an incorrect concentration history, or binding behavior not represented by the model; the curve alone does not identify which explanation is correct.

The estimated parameters can also be checked for plausibility. Under the stated model, $k_{\mathrm{on}}$ and $k_{\mathrm{off}}$ should be nonnegative, the occupancy should remain between zero and one, and $k_{\mathrm{obs}}=k_{\mathrm{on}}C+k_{\mathrm{off}}$ should increase linearly with concentration if rates are measured at multiple concentrations. A parameter estimate that violates these implications is a warning about the fit, data, or assumptions. Even a visually acceptable fit at one concentration does not prove the one-site model is correct.

The main alternative is an equilibrium occupancy calculation using $K_D$ or an estimated affinity. It is simpler and can be adequate when equilibrium is established at each relevant concentration. The kinetic model instead predicts transient occupancy and washout, but needs time-course information and a concentration history. If free concentration is not known, or changes substantially within a sampling interval, predictions based on a constant concentration for that interval may be inaccurate. More complex biological mechanisms may also require a different model; the one-site calculation should not be extended to them without justification.

Common pitfall: Treating $\theta_{\mathrm{ss}}$ as the occupancy immediately reached after a dose, or treating zero concentration as immediate zero occupancy. In the kinetic model, occupancy approaches its steady-state value over time and decays during washout according to $k_{\mathrm{off}}$. Use the equilibrium value only when the equilibration assumption is defensible; otherwise calculate the trajectory with the initial occupancy and elapsed time.

Key takeaway: At fixed concentration, the occupancy asymptote and observed rate jointly determine $k_{\mathrm{on}}$ and $k_{\mathrm{off}}$ under the one-site reversible model. Those rates predict occupancy after concentration changes, provided the model assumptions and concentration history are adequate.

Check Your Understanding

Answer each question, then compare with the solutions below.

At $C=1.00\ \mathrm{nM}$, use $k_{\mathrm{on}}=0.200\ \mathrm{nM}^{-1}\mathrm{h}^{-1}$ and $k_{\mathrm{off}}=0.100\ \mathrm{h}^{-1}$ to calculate $k_{\mathrm{obs}}$ and $\theta_{\mathrm{ss}}$.
If occupancy is $0.600$ when concentration falls to zero, what occupancy does the model predict after $5.00\ \mathrm{h}$?
Why is equilibrium occupancy not necessarily the correct prediction two hours after concentration rises?
Name one assumption of the one-site kinetic model whose failure could make the fitted rates misleading.

Solutions. (1) $k_{\mathrm{obs}}=(0.200)(1.00)+0.100=0.300\ \mathrm{h}^{-1}$, and $\theta_{\mathrm{ss}}=(0.200)(1.00)/0.300=0.6667$. (2) With $C=0$, $\theta(5)=0.600e^{-(0.100)(5.00)}=0.600e^{-0.500}=0.3639$. (3) The system may not have had enough time to approach the new steady state; the prediction depends on the starting occupancy and binding rates as well as the elapsed time. (4) Examples include a homogeneous class of independent sites, negligible drug depletion, a known free concentration at the target, reversible binding, or constant target availability.