Occupancy Under Changing Free-Ligand Concentrations

Why Changing Concentrations Require a Dynamic Calculation

When free-ligand concentration changes, receptor occupancy generally does not jump immediately to the equilibrium value associated with the new concentration. Binding and dissociation take time, so occupancy reflects both the current concentration and the receptor's prior state. This lag matters when exposure rises gradually, falls abruptly or lasts only briefly.

For a single reversible binding site, let $\theta(t)$ be the fraction of receptors occupied at time $t$, and let $L(t)$ be the free-ligand concentration. The association rate constant is $k_{\mathrm{on}}$, and the dissociation rate constant is $k_{\mathrm{off}}$. The concentration is treated as specified over time; it is not calculated from the binding process in this model.

Definition: Under one-site reversible binding with a known free-ligand profile, occupancy follows $$ \frac{d\theta}{dt} = k_{\mathrm{on}}L(t)\bigl(1-\theta(t)\bigr)-k_{\mathrm{off}}\theta(t). $$ Here $k_{\mathrm{on}}$ has units of concentration$^{-1}$time$^{-1}$, $k_{\mathrm{off}}$ has units of time$^{-1}$, and $\theta(t)$ is dimensionless. The first term is association to unoccupied receptors; the second is dissociation from occupied receptors.

This equation assumes a single class of independent binding sites, reversible binding, constant rate constants, and a free-ligand concentration known at each time. It also assumes that receptor availability is not changing through synthesis, degradation or internalization. The state is a fraction, so a physically meaningful initial condition satisfies $0\leq\theta(0)\leq1$. The calculation describes binding occupancy, not a downstream functional response.

Exact Update for Constant Concentration

If $L(t)=L$ is constant on an interval, the equation can be rearranged as $\frac{d\theta}{dt}=k_{\mathrm{on}}L-(k_{\mathrm{on}}L+k_{\mathrm{off}})\theta$. Define $\lambda=k_{\mathrm{on}}L+k_{\mathrm{off}}$, the rate at which occupancy approaches its interval-specific steady value. That value is $\theta_{\mathrm{eq}}=k_{\mathrm{on}}L/\lambda=L/(L+K_D)$, where $K_D=k_{\mathrm{off}}/k_{\mathrm{on}}$.

Result: For constant $L$ from time $t_0$ to $t_0+\Delta t$, the exact update is $$ \theta(t_0+\Delta t) =\theta_{\mathrm{eq}}+ \bigl(\theta(t_0)-\theta_{\mathrm{eq}}\bigr)e^{-\lambda\Delta t}. $$ The interval's starting occupancy is essential: the new equilibrium value is approached exponentially, not imposed instantly.

The update follows by subtracting $\theta_{\mathrm{eq}}$ from both sides of the differential equation. Since $k_{\mathrm{on}}L=\lambda\theta_{\mathrm{eq}}$, the difference $u(t)=\theta(t)-\theta_{\mathrm{eq}}$ satisfies $du/dt=-\lambda u$. Separating variables and integrating gives $u(t)=u(t_0)e^{-\lambda(t-t_0)}$, which yields the stated update. For nonnegative concentrations and rate constants, the update is a weighted average of the starting occupancy and $\theta_{\mathrm{eq}}$, with weights between zero and one; thus it remains between them.

Approximating a Concentration Ramp

A ramp can be evaluated by dividing time into intervals and approximating $L(t)$ within each interval by a representative concentration. For a midpoint approximation, use the concentration at the midpoint of each interval, then apply the exact constant-concentration update over that interval. The update is exact for the chosen piecewise-constant approximation, but only approximate for the actual ramp.

Worked Example: A Four-Hour Rising Ramp

Suppose a hypothetical ligand concentration rises linearly from 0 to 20 nM over 4 h, so $L(t)=5t$ nM. Let $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.10$ h$^{-1}$, and initial occupancy $\theta(0)=0.100$. Approximate the ramp with four one-hour midpoint intervals.

The midpoint concentrations are 2.5, 7.5, 12.5 and 17.5 nM. For each interval, calculate $\lambda=0.020L+0.10$ h$^{-1}$, $\theta_{\mathrm{eq}}=0.020L/\lambda$, and use the previous interval's ending occupancy as the next starting occupancy.

For the first interval, $L=2.5$ nM gives $\lambda=0.150$ h$^{-1}$ and $\theta_{\mathrm{eq}}=1/3$. Therefore, $$ \theta(1)=\frac{1}{3}+ \left(0.100-\frac{1}{3}\right)e^{-0.150(1)} \approx 0.132501. $$

For the second interval, $L=7.5$ nM gives $\lambda=0.250$ h$^{-1}$ and $\theta_{\mathrm{eq}}=0.600$. Thus, $$ \theta(2)=0.600+(0.132501-0.600)e^{-0.250} \approx 0.235912. $$

For the third interval, $L=12.5$ nM gives $\lambda=0.350$ h$^{-1}$ and $\theta_{\mathrm{eq}}=0.714286$. Hence, $$ \theta(3)=0.714286+(0.235912-0.714286)e^{-0.350} \approx 0.377181. $$

For the fourth interval, $L=17.5$ nM gives $\lambda=0.450$ h$^{-1}$ and $\theta_{\mathrm{eq}}=0.777778$. Therefore, $$ \theta(4)=0.777778+(0.377181-0.777778)e^{-0.450} \approx 0.522346. $$ The midpoint-discretization estimate is about 0.522, or 52.2% occupancy, at 4 h. It is below the last interval's equilibrium value because the concentration has not been held at its final level long enough for occupancy to reach that value.

The midpoint estimate should be checked for sensitivity to the interval width. Repeating the calculation with shorter intervals provides a practical convergence check: if the result changes materially, the original time grid was too coarse for the desired precision. A numerical ordinary differential equation solver can instead evaluate $L(t)$ continuously or at its recorded sampling times, subject to appropriate tolerances and a suitable representation of the concentration profile.

A Brief Pulse and Washout

A rectangular pulse has a constant concentration during exposure and a different concentration after exposure. Apply the constant-concentration update separately to each interval, carrying occupancy forward across the concentration change. During washout with $L=0$, the equilibrium occupancy is zero and occupancy decays at rate $k_{\mathrm{off}}$.

Worked Example: A Half-Hour Pulse Followed by Washout

Consider a hypothetical brief pulse with $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.14$ h$^{-1}$, and initial occupancy zero. Let $L=20$ nM for 0.5 h, followed immediately by complete washout ($L=0$) for 3 h.

During the pulse, $\lambda=(0.020)(20)+0.14=0.54$ h$^{-1}$, and $\theta_{\mathrm{eq}}=0.40/0.54=0.740741$. The occupancy at the end of the pulse is $$ \theta(0.5)=0.740741+(0-0.740741)e^{-0.54(0.5)} \approx 0.175274. $$

After washout, $\lambda=k_{\mathrm{off}}=0.14$ h$^{-1}$, and $\theta_{\mathrm{eq}}=0$. Three hours later, $$ \theta(3.5)=0+(0.175274-0)e^{-0.14(3)} \approx 0.115164. $$ Occupancy is about 17.5% at the end of exposure and 11.5% three hours after washout. It does not fall to zero at the instant the ligand is removed; dissociation takes time.

Interpreting Changes and Choosing a Method

For a general time-varying concentration, the equation is linear in $\theta$, even when $L(t)$ varies. Writing $\lambda(t)=k_{\mathrm{on}}L(t)+k_{\mathrm{off}}$, an integrating factor gives the exact expression $$ \theta(t)=\theta(0)e^{-\int_0^t\lambda(s)\,ds} +\int_0^t k_{\mathrm{on}}L(u) e^{-\int_u^t\lambda(s)\,ds}\,du. $$ The first term is the remaining contribution of the initial occupancy. The integral accumulates association over time, with earlier contributions discounted by subsequent binding and dissociation. For an arbitrary profile, this integral may not have a convenient closed form, so numerical integration is commonly used.

Worked Example: A Step-Down to a Lower Concentration

Suppose $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.10$ h$^{-1}$, and occupancy is $2/3$ just before concentration falls from a prior 10 nM exposure to 2 nM. Calculate occupancy after 3 h at 2 nM. Assume the free concentration remains fixed at 2 nM throughout those 3 h.

At 2 nM, $\lambda=(0.020)(2)+0.10=0.14$ h$^{-1}$, and $\theta_{\mathrm{eq}}=0.04/0.14=2/7$. The exact update is $$ \theta(3)=\frac{2}{7}+ \left(\frac{2}{3}-\frac{2}{7}\right)e^{-0.14(3)} =\frac{2}{7}+\frac{8}{21}e^{-0.42} \approx 0.536018. $$ Occupancy is approximately 0.536, or 53.6%, after 3 h. It has declined from $2/3$, but remains above the 2 nM equilibrium occupancy of $2/7$, because the receptors have not yet equilibrated at the lower concentration.

The equilibrium approximation uses the instantaneous concentration to calculate $L(t)/(L(t)+K_D)$ at each time. It is suitable when concentration changes slowly relative to the occupancy relaxation time $1/\lambda(t)$, so occupancy can track the changing equilibrium closely. The dynamic equation is needed when changes occur on a similar or shorter timescale, as in the pulse example. Equilibrium occupancy is an approximation to the dynamic state, not a separate update rule for a sudden concentration change.

Approach	What it calculates	When it is useful
Instantaneous equilibrium	Occupancy implied by the current concentration, without memory of the prior state	Slow concentration changes relative to binding and dissociation
Piecewise-constant exact updates	Exact dynamics for a concentration held constant within each interval	Steps, pulses, or numerical approximations to a ramp
Numerical ODE integration	Dynamic occupancy for a supplied time-varying concentration profile	Continuous profiles or fine-scale input data when the interval update is inconvenient

Diagnostics, Limitations, and a Common Pitfall

Check that the concentration and rate-constant units are compatible, that each interval uses its own concentration, and that the ending occupancy from one interval becomes the starting occupancy of the next. With nonnegative $L(t)$, $k_{\mathrm{on}}$, and $k_{\mathrm{off}}$, and an initial occupancy in $[0,1]$, the model should preserve that range. Numerical values outside it can signal an implementation error, a time step that is too large for the numerical method, or an unsuitable solver configuration.

A midpoint or other interval approximation can miss rapid concentration changes within a step. Compare results across finer grids or with a solver using the original profile. For measured concentration data, interpolation between observations is a modeling choice: a piecewise-constant profile represents steps, whereas linear interpolation represents ramps between measurements. The assumed profile can affect predicted occupancy, especially during short exposures.

A common mistake is to set occupancy equal to $L/(L+K_D)$ at every observed time and treat those values as the kinetic trajectory. That reasoning discards the prior occupancy and assumes instantaneous equilibration. The correction is to propagate $\theta$ using the time-dependent equation, or use the equilibrium approximation only when its timescale condition is justified. This single-site calculation also has limits: it does not account for ligand depletion, competing ligands, receptor turnover or multiple receptor states.

Key takeaway: Occupancy under changing exposure has memory. Use the time-dependent binding equation, propagate the state across concentration intervals, and check whether the time resolution and equilibrium assumptions match the concentration profile and binding timescale.

Check Your Understanding

Answer each question, then compare with the solutions below.

For constant $L$, what are $\lambda$ and $\theta_{\mathrm{eq}}$, and why does occupancy not generally jump to $\theta_{\mathrm{eq}}$?
A ligand is washed out completely. What is the occupancy after time $t$, given occupancy $\theta_0$ at washout?
Why should a midpoint ramp calculation be repeated with smaller time intervals?
Under what timescale condition is instantaneous equilibrium most plausible?

Solutions. (1) $\lambda=k_{\mathrm{on}}L+k_{\mathrm{off}}$ and $\theta_{\mathrm{eq}}=k_{\mathrm{on}}L/\lambda$; the exact update approaches this value exponentially from the prior occupancy. (2) With $L=0$, $\theta(t)=\theta_0e^{-k_{\mathrm{off}}t}$. (3) The midpoint method approximates the changing concentration as constant within each interval, so refinement checks whether that discretization materially affects the estimate. (4) The concentration must change slowly compared with the occupancy relaxation time $1/\lambda(t)$, allowing occupancy to track its changing equilibrium.