Why Pulse History Matters
A ligand exposure may rise and fall repeatedly rather than remain at one concentration. If receptor binding and dissociation take time, occupancy at the start of a new pulse depends on what happened during the previous pulse and the intervening washout. Two exposures with the same peak concentration can therefore produce different occupancy profiles if their durations or spacing differ.
The earlier material on occupancy under changing free-ligand concentrations gives the time-dependent binding equation. Here, the task is to use that equation efficiently when the free-ligand concentration is specified in successive constant-concentration intervals. No pharmacokinetic model is derived: the concentration-time input is supplied.
Model for One Pulse or Washout Interval
Let \(\theta(t)\) be the fraction of receptors occupied at time \(t\), with \(0\leq\theta(t)\leq1\). For a single reversible binding site, let \(k_{\mathrm{on}}\) be the association rate constant and \(k_{\mathrm{off}}\) the dissociation rate constant. If free-ligand concentration \(C\) is constant during an interval, the occupancy equation is
The first term describes association onto unoccupied receptors; the second describes dissociation from occupied receptors. The derivation and use below assume one reversible binding site, constant rate constants, a prescribed free-ligand concentration that does not change in response to binding, and no receptor turnover or other processes that change the available receptor pool. The initial occupancy must be known or specified. Concentration and time units must be consistent with the units of \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\).
This interval formula applies during a pulse with \(C>0\) and during washout with \(C=0\). In washout, \(\theta_{\mathrm{eq}}(0)=0\) and \(\lambda(0)=k_{\mathrm{off}}\), so occupancy decays exponentially rather than disappearing immediately.
Why the Interval Formula Works
Rearrange the binding equation for constant \(C\):
The second equality follows because \(\lambda(C)\theta_{\mathrm{eq}}(C)=k_{\mathrm{on}}C\). Subtract the equilibrium value, \(u(t)=\theta(t)-\theta_{\mathrm{eq}}(C)\). Since \(C\) is constant over the interval, \(\theta_{\mathrm{eq}}(C)\) is constant, and
Substituting \(u(0)=\theta_{\mathrm{in}}-\theta_{\mathrm{eq}}(C)\) and evaluating at \(t=\Delta t\) gives the stated formula. Thus occupancy moves exponentially toward the equilibrium value for that interval; it does not jump to that value when concentration changes.
Worked Example: One Pulse Followed by Washout
Suppose a hypothetical ligand has \(k_{\mathrm{on}}=0.0200\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\) and \(k_{\mathrm{off}}=0.100\ \mathrm{h}^{-1}\). Starting from \(\theta=0\), free concentration is held at \(20.0\ \mathrm{nM}\) for \(2.00\ \mathrm{h}\), then set to zero for \(3.00\ \mathrm{h}\). Calculate occupancy at the end of the pulse and at the end of washout.
First, \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}=0.100/0.0200=5.00\ \mathrm{nM}\). During the pulse, \(\theta_{\mathrm{eq}}=20.0/(20.0+5.00)=0.800\), and \(\lambda=(0.0200)(20.0)+0.100=0.500\ \mathrm{h}^{-1}\). Therefore,
During washout, \(C=0\), so occupancy decays with \(k_{\mathrm{off}}=0.100\ \mathrm{h}^{-1}\):
About 37.5% of receptors remain occupied at the start of the next pulse in this hypothetical example. The zero concentration does not imply zero occupancy because dissociation takes time.
Propagating Occupancy Across Repeated Pulses
For a sequence of constant-concentration intervals, calculate the end occupancy of one interval and use it as the starting occupancy of the next. Occupancy remains continuous at a concentration change: the concentration input changes, but the fraction already bound cannot instantaneously jump under this model.
For each interval, record the free concentration \(C_j\) and duration \(\Delta t_j\). These are inputs, not quantities inferred by the occupancy calculation.
For each \(C_j\), calculate \(\theta_{\mathrm{eq}}(C_j)\) and \(\lambda(C_j)\) using the same \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\).
Apply the interval formula, carrying \(\theta_{\mathrm{out}}\) forward as the next interval's \(\theta_{\mathrm{in}}\). Use \(C_j=0\) for a zero-concentration washout interval.
Worked Example: A Pulse, Washout, and Lower-Concentration Pulse
For a hypothetical ligand, let \(k_{\mathrm{on}}=0.100\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\), \(k_{\mathrm{off}}=0.200\ \mathrm{h}^{-1}\), and initial occupancy \(\theta=0.200\). The specified input is \(10.0\ \mathrm{nM}\) for \(1.00\ \mathrm{h}\), zero concentration for \(2.00\ \mathrm{h}\), then \(5.00\ \mathrm{nM}\) for \(1.00\ \mathrm{h}\). Find the occupancy at the end of each interval.
For the \(10.0\ \mathrm{nM}\) pulse, \(\theta_{\mathrm{eq}}=10.0/(10.0+2.00)=0.8333\), since \(K_D=0.200/0.100=2.00\ \mathrm{nM}\). The rate is \(\lambda=(0.100)(10.0)+0.200=1.20\ \mathrm{h}^{-1}\). Thus,
For the \(2.00\ \mathrm{h}\) washout, \(\theta_{\mathrm{eq}}=0\) and \(\lambda=0.200\ \mathrm{h}^{-1}\):
For the final \(5.00\ \mathrm{nM}\) interval, \(\theta_{\mathrm{eq}}=5.00/(5.00+2.00)=0.7143\) and \(\lambda=(0.100)(5.00)+0.200=0.700\ \mathrm{h}^{-1}\). Starting from the washout occupancy,
The final occupancy is approximately 57.4%. It is below the final interval's equilibrium occupancy of 71.4% because the \(5.00\ \mathrm{nM}\) interval lasts only \(1.00\ \mathrm{h}\), and it starts below that equilibrium value.
Repeated Identical Cycles and Periodic Occupancy
When the same pulse and washout pattern repeats, occupancy at a particular point in each cycle follows a recurrence. For a pulse at concentration \(C_p\) lasting \(\Delta t_p\), followed by washout of duration \(\Delta t_w\), define
If \(x_n\) is occupancy just before pulse \(n\), occupancy at the end of that pulse is \(E+(x_n-E)a\). Washout then gives the next pre-pulse occupancy:
A periodic steady state is a value \(x_*\) unchanged from one cycle to the next. Setting \(x_{n+1}=x_n=x_*\) and solving gives
For positive pulse and washout durations and positive dissociation rate, \(0\leq a<1\), \(0\leq b<1\), so \(0\leq ab<1\). The denominator is therefore positive. Further, subtracting \(x_*\) from the recurrence yields \(x_{n+1}-x_*=ab(x_n-x_*)\), and hence \(x_n-x_*=(ab)^n(x_0-x_*)\). Because \((ab)^n\) tends to zero, occupancy converges to the periodic value from any initial occupancy in the model's range.
Worked Example: Occupancy at Periodic Steady State
Use the hypothetical rates \(k_{\mathrm{on}}=0.100\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\) and \(k_{\mathrm{off}}=0.200\ \mathrm{h}^{-1}\). Each cycle consists of \(10.0\ \mathrm{nM}\) for \(1.00\ \mathrm{h}\), followed by zero concentration for \(2.00\ \mathrm{h}\). Calculate the pre-pulse occupancy at periodic steady state and the occupancy at the end of the pulse.
The pulse equilibrium is \(E=10.0/(10.0+2.00)=0.8333\). The pulse rate is \(1.20\ \mathrm{h}^{-1}\), giving \(a=e^{-1.20}=0.3010\). For the washout, \(b=e^{-(0.200)(2.00)}=0.6703\). Substitute into the periodic formula:
The end-of-pulse occupancy at periodic steady state is
Thus, after many identical cycles, occupancy is about 48.9% just before a pulse and 73.0% just after it. The pre-pulse value is not zero because dissociation during each washout is incomplete.
Assumptions, Diagnostics, and Alternatives
The calculations are useful when binding is well described by a single reversible site and the supplied free-ligand input is a reasonable representation of concentration at the receptors. In a quantitative systems pharmacology model, this can isolate the impact of binding kinetics from the process generating the concentration profile. It does not establish that the specified profile is biologically correct.
Check the calculation and its assumptions by asking whether concentrations and rate-constant units match, whether the occupancy remains between zero and one, and whether occupancy moves toward the appropriate interval equilibrium. During a zero-concentration interval, it must decrease monotonically as \(e^{-k_{\mathrm{off}}\Delta t}\). During a constant positive-concentration interval, the solution approaches \(\theta_{\mathrm{eq}}(C)\) without crossing it. If computed occupancy jumps at a pulse boundary, exceeds one, or moves away from the interval equilibrium, check the recurrence, signs, units, and ordering of the intervals. If measurements disagree systematically with the predicted carryover, possible causes include an inaccurate concentration input, changing rate constants, or processes excluded from this model.
The interval solution is preferable to an equilibrium approximation when pulse durations are not long relative to \(1/\lambda(C)\), or when washout is incomplete before the next pulse. An equilibrium approximation replaces each interval's occupancy with its equilibrium value; it can be useful when the interval is long enough for occupancy to approach equilibrium, but it loses transient and carryover information. Numerical integration is an alternative for concentration profiles that vary continuously or for more complex equations. For piecewise-constant inputs governed by this linear one-site equation, the exact interval update is direct and avoids numerical time-step error. Neither approach repairs a misspecified exposure input or omitted biology.
A common mistake is to restart each pulse at zero occupancy because the free concentration was set to zero between pulses. That confuses ligand concentration with receptor occupancy: zero free concentration stops association in this model, but bound ligand still dissociates at the finite rate \(k_{\mathrm{off}}\). Carry the final occupancy of each interval into the next one.
Check Your Understanding
Try each question, then compare with the solutions below.
- A ligand has \(k_{\mathrm{off}}=0.0500\ \mathrm{h}^{-1}\). If occupancy is 0.600 before a \(4.00\ \mathrm{h}\) zero-concentration interval, what is occupancy afterward?
- For a constant concentration interval, explain why occupancy does not jump to \(\theta_{\mathrm{eq}}(C)\) at the moment concentration changes.
- Suppose the pulse multiplier is \(a=0.400\), the washout multiplier is \(b=0.500\), and pulse equilibrium occupancy is \(E=0.750\). Calculate periodic pre-pulse occupancy.
- State one condition under which replacing each interval's occupancy with its equilibrium value is a poor approximation.
Solutions. 1. With zero concentration, \(\theta_{\mathrm{out}}=\theta_{\mathrm{in}}e^{-k_{\mathrm{off}}\Delta t}=0.600e^{-(0.0500)(4.00)}=0.600e^{-0.200}=0.4912\). 2. Occupancy is governed by a differential equation with finite association and dissociation rates, so it evolves continuously; the new concentration changes the rate of change, not the current fraction bound. 3. \(x_*=b(1-a)E/(1-ab)=(0.500)(0.600)(0.750)/[1-(0.500)(0.400)]=0.225/0.800=0.2813\). 4. The approximation can be poor when a pulse or washout is short relative to the binding relaxation time, so occupancy has not approached that interval's equilibrium by its end.