Occupancy After a Step Change in Free Concentration

Why Occupancy Does Not Jump with Concentration

A sudden change in free drug concentration immediately changes the rate at which unbound target becomes bound. It does not, by itself, instantaneously change how much target is already bound. After the change, association and dissociation continue together, and occupancy moves toward a new equilibrium at a rate determined by both processes. This distinction matters whenever response is linked to binding over a timescale comparable to the concentration change.

The model below treats free ligand concentration as an externally specified input. It is therefore appropriate when free concentration can be regarded as known and the target does not appreciably deplete ligand. The step is an idealization: concentration switches instantly from one fixed value to another, then remains at the new value.

Definition: Let $\theta(t)$ be the fraction of target bound by ligand at time $t$, with $0\leq\theta(t)\leq1$. For reversible one-to-one binding, let $k_{\mathrm{on}}$ be the association rate constant and $k_{\mathrm{off}}$ the dissociation rate constant. Their units are concentration$^{-1}$ time$^{-1}$ and time$^{-1}$, respectively. If free ligand concentration is $L$, the mass-action equation is $$ \frac{d\theta}{dt}=k_{\mathrm{on}}L(1-\theta)-k_{\mathrm{off}}\theta. $$

The first term is the rate of binding: it is proportional to free ligand concentration and the unoccupied target fraction $1-\theta$. The second term is the rate of dissociation, proportional to the bound fraction. The model assumes a homogeneous target population, fixed rate constants, reversible binding, and a constant free concentration during the period being modeled. It also assumes that the concentration input is prescribed rather than calculated from a binding mass balance.

Solution After a Step Change

Suppose concentration changes at time zero to a constant value $L_1$, and let the occupancy at that instant be $\theta_0$. For $t\geq0$, the concentration is $L_1$, so the equation becomes a linear differential equation with constant coefficients.

\frac{d\theta}{dt} =k_{\mathrm{on}}L_1-\left(k_{\mathrm{on}}L_1+k_{\mathrm{off}}\right)\theta.

At equilibrium, the derivative is zero. Define $\theta_{\mathrm{ss}}$ as the equilibrium occupancy under $L_1$, and define $\lambda_1$ as the positive rate constant governing the approach to it:

\theta_{\mathrm{ss}} =\frac{k_{\mathrm{on}}L_1}{k_{\mathrm{on}}L_1+k_{\mathrm{off}}} =\frac{L_1}{L_1+K_D}, \qquad \lambda_1=k_{\mathrm{on}}L_1+k_{\mathrm{off}}, \qquad K_D=\frac{k_{\mathrm{off}}}{k_{\mathrm{on}}}.

The expression for $\theta_{\mathrm{ss}}$ is the fractional occupancy at equilibrium; $K_D$ is the dissociation constant. The rate $\lambda_1$ has units of inverse time. Subtracting the equilibrium occupancy from both sides of the differential equation gives $d(\theta-\theta_{\mathrm{ss}})/dt=-\lambda_1(\theta-\theta_{\mathrm{ss}})$. Thus the deviation from the new equilibrium decays exponentially.

Result: Following a step to constant free concentration $L_1$, occupancy is $$ \theta(t)=\theta_{\mathrm{ss}}+ \left(\theta_0-\theta_{\mathrm{ss}}\right)e^{-\lambda_1t}, \qquad t\geq0. $$ The occupancy is continuous at the step, since $\theta(0)=\theta_0$. Its initial slope after the step is $k_{\mathrm{on}}L_1(1-\theta_0)-k_{\mathrm{off}}\theta_0$.

Justification. Set $u(t)=\theta(t)-\theta_{\mathrm{ss}}$. Because $\theta_{\mathrm{ss}}$ is constant after the step, $du/dt=-\lambda_1u$. Separating variables and integrating from time zero to time $t$ gives $\ln|u(t)|-\ln|u(0)|=-\lambda_1t$, or $u(t)=u(0)e^{-\lambda_1t}$. Substituting $u(0)=\theta_0-\theta_{\mathrm{ss}}$ yields the stated solution. If $u(0)=0$, the solution is constant at equilibrium, as the same formula implies.

When the pre-step occupancy was at equilibrium with an old free concentration $L_0$, it can be calculated as $\theta_0=L_0/(L_0+K_D)$. That is an initial condition, not a requirement of the time-course solution: the formula also applies when occupancy at the step is not at equilibrium.

Worked Examples

Worked Example: Increasing Free Concentration

Suppose free concentration increases abruptly from $10$ nM to $40$ nM. Let $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$ and $k_{\mathrm{off}}=0.300$ h$^{-1}$. Assume occupancy was at equilibrium before the increase. Find occupancy one hour after the step.

First calculate $K_D=k_{\mathrm{off}}/k_{\mathrm{on}}=0.300/0.020=15.0$ nM. The initial occupancy is therefore $$ \theta_0=\frac{10}{10+15.0}=0.400. $$

At the new concentration, the equilibrium occupancy and rate constant are $$ \theta_{\mathrm{ss}}=\frac{40}{40+15.0}=0.7273, \qquad \lambda_1=(0.020)(40)+0.300=1.10\ \mathrm{h}^{-1}. $$

Substitute these values into the step-response solution: $$ \theta(1)=0.7273+(0.400-0.7273)e^{-(1.10)(1)} =0.7273-0.3273(0.3329) \approx0.6183. $$

The predicted occupancy after one hour is about $0.618$, or $61.8\%$. It is between the initial occupancy of $40.0\%$ and the new equilibrium occupancy of $72.7\%$, as expected for an increase in concentration under this model.

Worked Example: Reducing Concentration to Zero

A ligand has $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$ and $k_{\mathrm{off}}=0.200$ h$^{-1}$. Before a step to zero free concentration, the concentration is $30$ nM and occupancy is at equilibrium. Calculate occupancy two hours after the step.

Here $K_D=0.200/0.020=10.0$ nM, so the occupancy just before and at the step is $$ \theta_0=\frac{30}{30+10.0}=0.750. $$

With $L_1=0$, the new equilibrium occupancy is zero and $$ \lambda_1=(0.020)(0)+0.200=0.200\ \mathrm{h}^{-1}. $$

Consequently, $$ \theta(2)=0+(0.750-0)e^{-(0.200)(2)} =0.750e^{-0.400} =0.750(0.6703) \approx0.5027. $$

The model predicts about $50.3\%$ occupancy after two hours. With no free ligand, the association term is zero, but bound ligand still dissociates; occupancy therefore declines rather than dropping instantly to zero.

Worked Example: Starting Away from Equilibrium

Suppose concentration steps to $20$ nM, with $k_{\mathrm{on}}=0.010$ nM$^{-1}$min$^{-1}$, $k_{\mathrm{off}}=0.200$ min$^{-1}$, and occupancy at the step equal to $0.300$. How long does it take for occupancy to reach $0.450$?

The new equilibrium occupancy and approach rate are $$ \theta_{\mathrm{ss}}=\frac{(0.010)(20)}{(0.010)(20)+0.200}=0.500, \qquad \lambda_1=(0.010)(20)+0.200=0.400\ \mathrm{min}^{-1}. $$

Use the general solution with the given, non-equilibrium initial condition: $$ 0.450=0.500+(0.300-0.500)e^{-0.400t} =0.500-0.200e^{-0.400t}. $$

Rearranging gives $e^{-0.400t}=0.050/0.200=0.250$. Taking natural logarithms, $$ t=\frac{-\ln(0.250)}{0.400} =\frac{\ln(4)}{0.400} \approx3.466\ \mathrm{min}. $$

The predicted time to reach $45.0\%$ occupancy is about $3.47$ minutes. The target value is reachable because it lies between the starting occupancy, $0.300$, and the new equilibrium, $0.500$.

Checking the Model and Choosing an Approach

The exact solution is often preferable to simply substituting the new concentration into the equilibrium formula. That substitution gives $\theta_{\mathrm{ss}}$, not occupancy at an arbitrary time after the step. It is appropriate only when the system has had enough time to approach equilibrium for the question being asked. For a finite time, the exponential solution accounts for the initial occupancy and the rates of both binding and dissociation.

Numerical integration is another option, especially when concentration varies continuously or the binding model includes additional processes. For a single ideal step with fixed rates, the exact expression is simpler and avoids numerical integration error. The alternatives answer different questions: equilibrium analysis describes the long-run value, the exact step solution describes this particular transient, and numerical methods can handle more complex specified equations.

Before interpreting a calculated time course, check whether its assumptions are plausible. Free concentration must be approximately constant after the step, rate constants must remain applicable over the modeled conditions, and the one-site reversible model must be a reasonable description. If ligand depletion is substantial, free concentration cannot simply be treated as an externally fixed input. If the concentration changes gradually rather than abruptly, the step solution does not represent the actual input history.

A useful basic diagnostic is the solution’s direction and bounds. Since $\lambda_1>0$, the exponential term tends to zero as time increases, and occupancy approaches $\theta_{\mathrm{ss}}$. For a change from an initial occupancy below equilibrium, the predicted occupancy rises monotonically; from above equilibrium, it falls monotonically. Under the stated model it remains between $\theta_0$ and $\theta_{\mathrm{ss}}$. A calculation that produces an overshoot or a value outside $0$ to $1$ signals an arithmetic, coding, or model-implementation problem.

In data, systematic departures from the predicted exponential pattern can indicate that one or more assumptions are inadequate. For example, persistent curvature inconsistent with a single exponential may suggest that the input was not a true step, the target population is not well represented by one kinetic process, or the assumed rate constants do not describe the conditions. Such patterns do not identify a unique cause; they are prompts to examine the concentration history, data quality, and model structure rather than to force an equilibrium interpretation.

Common Pitfall

Wrong reasoning: “The free concentration jumped to $40$ nM, so occupancy immediately became $40/(40+K_D)$.” This confuses the new equilibrium with the occupancy at the instant of the step. The equilibrium formula sets the derivative to zero; it does not describe an instantaneous change in bound target.

Correction: Preserve the pre-step occupancy as $\theta_0$, calculate the new $\theta_{\mathrm{ss}}$ and $\lambda_1$, and evaluate $\theta(t)=\theta_{\mathrm{ss}}+(\theta_0-\theta_{\mathrm{ss}})e^{-\lambda_1t}$. At $t=0$, this gives $\theta(0)=\theta_0$; only later does occupancy move toward the new equilibrium.

Key takeaway: After a step to a constant free concentration, occupancy approaches the new equilibrium exponentially. The new equilibrium is set by concentration relative to $K_D$; the speed of approach is set by $k_{\mathrm{on}}L_1+k_{\mathrm{off}}$. The starting occupancy determines the trajectory’s initial displacement from equilibrium.

Check Your Understanding

Use the step-response formula and state the reasoning for each answer. Solutions are provided below the questions.

For a step to $L_1=10$ nM, $k_{\mathrm{on}}=0.010$ nM$^{-1}$min$^{-1}$, and $k_{\mathrm{off}}=0.100$ min$^{-1}$, calculate $\theta_{\mathrm{ss}}$ and $\lambda_1$.
If $\theta_0=0.200$ and $\theta_{\mathrm{ss}}=0.600$, what is the occupancy at the instant of the concentration step? Explain why.
For $L_1=0$, what does the step-response formula reduce to?
Can occupancy reach $0.800$ if $\theta_0=0.300$ and $\theta_{\mathrm{ss}}=0.700$ under this constant-concentration model? Explain.

Solutions. (1) $\theta_{\mathrm{ss}}=(0.010)(10)/[(0.010)(10)+0.100]=0.500$, and $\lambda_1=(0.010)(10)+0.100=0.200$ min$^{-1}$. (2) It is $0.200$, because the step changes the binding rate, not the amount already bound at that instant. (3) $\theta_{\mathrm{ss}}=0$ and $\lambda_1=k_{\mathrm{off}}$, so $\theta(t)=\theta_0e^{-k_{\mathrm{off}}t}$. (4) No. The solution moves monotonically from $0.300$ toward $0.700$, so it does not reach $0.800$.