Occupancy Under Pulsed Exposure

Why Pulse Timing Matters

A concentration profile with sharp peaks and drug-free intervals can produce an occupancy profile that differs substantially from the equilibrium occupancy evaluated at an average concentration. Binding takes time: occupancy rises during exposure and can persist after free concentration falls, with the persistence governed by dissociation. Thus, peak concentration alone does not describe target engagement, and two schedules with the same concentration-time exposure need not have the same occupancy peak, trough, or average.

The model below treats free concentration as an externally specified input and tracks the fraction of a single target population that is bound. It extends the standard reversible-binding rate equation to repeated, piecewise-constant concentration pulses. It does not assume that occupancy reaches equilibrium during each pulse or washout.

Definition: Let $O(t)$ denote fractional target occupancy at time $t$, with $0\leq O(t)\leq1$, and let $C(t)$ be free drug concentration. For reversible binding to a single class of independent sites, with association rate constant $k_{\mathrm{on}}$ and dissociation rate constant $k_{\mathrm{off}}$, occupancy follows $$ \frac{dO}{dt}=k_{\mathrm{on}}C(t)\{1-O(t)\}-k_{\mathrm{off}}O(t). $$ Here $k_{\mathrm{on}}$ has units of concentration$^{-1}$ time$^{-1}$, $k_{\mathrm{off}}$ has units of time$^{-1}$, and occupancy is dimensionless.

The association term is proportional to the unoccupied fraction $1-O(t)$; the dissociation term is proportional to the occupied fraction. The model assumes one-to-one reversible binding, constant rate constants, a homogeneous target population, and a known free concentration profile. It also assumes that binding does not appreciably deplete free drug. If free concentration is altered by binding, the concentration must instead be determined consistently with a binding mass balance.

Occupancy During One Constant-Concentration Segment

When concentration is constant at $C$, define the total rate $k_{\mathrm{on}}C+k_{\mathrm{off}}$ and the segment’s equilibrium occupancy $O_{\mathrm{eq}}(C)$. Rearranging the differential equation gives a first-order linear equation whose solution, starting at occupancy $O_0$, is:

O(t)=O_{\mathrm{eq}}(C)+\{O_0-O_{\mathrm{eq}}(C)\} e^{-(k_{\mathrm{on}}C+k_{\mathrm{off}})t}, \qquad O_{\mathrm{eq}}(C)=\frac{k_{\mathrm{on}}C}{k_{\mathrm{on}}C+k_{\mathrm{off}}}.

For a pulse of concentration $C_p$ lasting $t_{\mathrm{on}}$, the endpoint occupancy is $O_{\mathrm{peak}}=O_{\mathrm{eq}}(C_p)+\{O_{\mathrm{start}}-O_{\mathrm{eq}}(C_p)\}e^{-(k_{\mathrm{on}}C_p+k_{\mathrm{off}})t_{\mathrm{on}}}$. If concentration then becomes zero for $t_{\mathrm{off}}$, occupancy decays exponentially: $O_{\mathrm{next}}=O_{\mathrm{peak}}e^{-k_{\mathrm{off}}t_{\mathrm{off}}}$. In a single pulse, these endpoints are the pulse peak and the subsequent trough, provided the occupancy rises during exposure and declines during washout. They need not equal the peak and trough of a repeated schedule at steady cycling, because the occupancy at the start of each pulse depends on the preceding cycle.

Worked Example: One Pulse Followed by Washout

Suppose a hypothetical target has $k_{\mathrm{on}}=0.0200\ \mathrm{nM}^{-1}\mathrm{h}^{-1}$ and $k_{\mathrm{off}}=0.100\ \mathrm{h}^{-1}$. Occupancy is $0.200$ when free concentration rises to $10.0\ \mathrm{nM}$ for $4.00\ \mathrm{h}$, followed by $4.00\ \mathrm{h}$ at zero concentration. Calculate the pulse-end occupancy, the post-washout occupancy, and the average occupancy over the full 8.00 hours.

During the pulse, $k_{\mathrm{on}}C_p=0.0200(10.0)=0.200\ \mathrm{h}^{-1}$, so the total rate is $0.200+0.100=0.300\ \mathrm{h}^{-1}$, and $O_{\mathrm{eq}}=0.200/0.300=0.6667$. Therefore,

O_{\mathrm{peak}}=0.6667+(0.200-0.6667)e^{-0.300(4.00)} =0.6667-0.4667(0.3012)=0.5261.

During washout the concentration is zero, so the decay rate is $k_{\mathrm{off}}$: $O_{\mathrm{trough}}=0.5261e^{-0.100(4.00)}=0.5261(0.6703)=0.3527$. Occupancy rises during the pulse, from $0.200$ to $0.5261$, then declines during washout to $0.3527$.

For a constant-concentration segment of duration $T$, integrating the occupancy solution gives its occupancy-time area:

\int_0^T O(t)\,dt =O_{\mathrm{eq}}T+(O_0-O_{\mathrm{eq}}) \frac{1-e^{-\lambda T}}{\lambda}, \qquad \lambda=k_{\mathrm{on}}C+k_{\mathrm{off}}.

The pulse area is $0.6667(4.00)+(0.200-0.6667)(1-e^{-1.20})/0.300=1.580\ \mathrm{h}$. During washout, $O_{\mathrm{eq}}=0$ and $\lambda=0.100\ \mathrm{h}^{-1}$, giving area $0.5261093678(1-e^{-0.400})/0.100=1.734\ \mathrm{h}$. Thus the interval-average occupancy is $(1.580+1.734)/8.00=0.4143$. It is dimensionless and lies between the minimum occupancy, $0.200$, and the maximum, $0.5261$, over this interval.

Repeated Pulses and the Steady-Cycle Peak and Trough

Consider repeated pulses with concentration $C_p$ for $t_{\mathrm{on}}$, followed by zero concentration for $t_{\mathrm{off}}$. Let $A=e^{-(k_{\mathrm{on}}C_p+k_{\mathrm{off}})t_{\mathrm{on}}}$ be the fraction of the distance from pulse-start occupancy to pulse equilibrium that remains at pulse end, and let $B=e^{-k_{\mathrm{off}}t_{\mathrm{off}}}$ be the fraction of pulse-end occupancy remaining after washout. Write $Q=O_{\mathrm{eq}}(C_p)$. If $x$ is the trough at the start of a pulse and $y$ its peak at the end, then $y=Q+(x-Q)A$ and the next trough is $x_{\mathrm{next}}=By$.

At steady cycling, $x_{\mathrm{next}}=x$. Substitution and rearrangement give the periodic trough and peak:

x=B\{Q+(x-Q)A\} \quad\Longrightarrow\quad O_{\mathrm{trough}}=\frac{BQ(1-A)}{1-AB}, \qquad O_{\mathrm{peak}}=Q+(O_{\mathrm{trough}}-Q)A.

For positive pulse and washout durations with positive rates, $0<A<1$ and $0<B<1$, so $1-AB>0$ and the recurrence has a unique steady-cycle trough. The recurrence also explains how to handle the first cycles: apply the pulse and washout equations successively from the specified initial occupancy rather than assuming the steady-cycle values immediately.

Worked Example: Peak, Trough, and Average at Steady Cycling

Use the same hypothetical binding rates and pulse concentration as above, but now administer $10.0\ \mathrm{nM}$ for $2.00\ \mathrm{h}$, followed by $6.00\ \mathrm{h}$ at zero concentration, repeatedly. Find steady-cycle peak, trough, and average occupancy over the 8.00-hour cycle.

The pulse equilibrium remains $Q=0.6667$. Both exponential factors are $e^{-0.600}=0.5488$: $A=e^{-0.300(2.00)}$ and $B=e^{-0.100(6.00)}$. The steady trough is $0.5488(0.6667)(1-0.5488)/(1-0.5488^2)=0.2362$. The corresponding peak is $0.6667+(0.2362-0.6667)(0.5488)=0.4304$.

To find the cycle average, integrate separately over the pulse and washout. The pulse area is $0.6667(2.00)+(0.2362-0.6667)(1-0.5488)/0.300=0.6859\ \mathrm{h}$. The washout area is $0.4304(1-0.5488)/0.100=1.942\ \mathrm{h}$. Hence average occupancy over the cycle is $(0.6859+1.942)/8.00=0.3285$. The trough is the cycle-start occupancy, the occupancy rises during the pulse to its peak, and washout returns it to the trough.

The interval-average occupancy is the area under $O(t)$ divided by the interval duration. It is not generally the equilibrium occupancy at the time-averaged concentration. During a periodic cycle the occupancy and concentration change together, and the binding response retains information about the preceding exposure.

Comparing Schedules With Equal Exposure

The following comparison uses the same hypothetical rate constants and an 8-hour cycle. Schedule A has a $10.0\ \mathrm{nM}$, 2-hour pulse and a 6-hour washout; Schedule B has a $5.00\ \mathrm{nM}$, 4-hour pulse and a 4-hour washout. Both have concentration-time exposure of $20.0\ \mathrm{nM}\cdot\mathrm{h}$ per cycle. For each schedule, the steady-cycle formulas apply; no assumption of instantaneous equilibrium is made.

Worked Example: Equal Exposure, Different Pulse Schedules

Schedule A was calculated above: its steady-cycle trough is $0.2362$, peak is $0.4304$, and average is $0.3285$. For Schedule B, the pulse equilibrium is $Q=0.0200(5.00)/\{0.0200(5.00)+0.100\}=0.5000$. Its pulse and washout factors are $A=e^{-0.200(4.00)}=0.4493$ and $B=e^{-0.100(4.00)}=0.6703$.

The trough is $0.6703(0.5000)(1-0.4493)/(1-0.4493(0.6703))=0.2641$. The peak is $0.5000+(0.2641-0.5000)(0.4493)=0.3940$. The pulse area is $0.5000(4.00)+(0.2641-0.5000)(1-0.4493)/0.200=1.350\ \mathrm{h}$ using the displayed rounded inputs; with unrounded steady-cycle values, it is approximately $1.35052\ \mathrm{h}$. The washout area is $0.3940(1-0.6703)/0.100=1.299\ \mathrm{h}$. Thus Schedule B’s cycle-average occupancy is $(1.350+1.299)/8.00\approx0.3311$ using the displayed rounded areas; computed from unrounded areas, it is $0.3312$.

In this calculation, Schedule B has a lower peak and a higher trough than Schedule A, so its peak-to-trough fluctuation is smaller. Its average is slightly higher, $0.3312$ versus $0.3285$, despite equal concentration-time exposure. This numerical result is specific to these rates and schedules; equal exposure does not imply that one schedule will always have a higher average occupancy.

Steady-cycle quantity	Schedule A: 10 nM for 2 h	Schedule B: 5 nM for 4 h
Exposure per 8-hour cycle	20.0 nM·h	20.0 nM·h
Trough occupancy	0.2362	0.2641
Peak occupancy	0.4304	0.3940
Cycle-average occupancy	0.3285	0.3312

Assumptions, Diagnostics, and Alternatives

The calculation is only as appropriate as its assumptions. The concentration used must represent free concentration at the target site, not merely a dose or a total concentration that does not track the binding input. The rectangular-pulse approximation must reasonably represent the time course within each on- and off-period. Rate constants must be applicable across the modeled range, and the single-site reversible-binding equation must be an adequate description of the target population. The solution also presumes that target abundance and binding capacity do not change during the interval.

Warning signs include systematic disagreement between predicted and observed occupancy at pulse endpoints, or errors that grow over successive cycles. Such patterns can arise if concentration is not actually constant during the nominal pulse, if washout is incomplete, if free concentration is mischaracterized, or if the assumed binding structure is inadequate. Measurements only at pulse peaks may miss a poorly predicted trough and can make a schedule appear better characterized than it is. A useful diagnostic is to compare observations across both exposure and washout phases and across more than one cycle.

The equilibrium occupancy relation $O_{\mathrm{eq}}(C)$ is a simpler alternative when concentration changes slowly enough, relative to binding and dissociation, that occupancy remains close to equilibrium. It cannot generally represent lag, persistence after washout, or cycle-to-cycle accumulation. The kinetic pulse model retains this time dependence but requires rate constants and a concentration-time profile. If the concentration varies continuously rather than as a rectangular pulse, the same differential equation can be solved numerically using the time-varying $C(t)$; replacing a varying profile with pulse segments is an approximation, not a consequence of the binding model.

Common mistake: Treating the equilibrium occupancy calculated from the cycle-average concentration as the cycle-average occupancy, or treating each pulse as if occupancy instantly reaches its pulse equilibrium. The correction is to propagate occupancy through each segment with its exponential solution and integrate that trajectory over the cycle. The equilibrium value is an asymptote for a constant concentration, not the occupancy reached automatically during a finite pulse.

Check Your Understanding

Use the segment solution and the stated assumptions to answer each question. Solutions follow the questions.

At zero concentration, what is the occupancy after a washout lasting $t$, starting from occupancy $O_0$?
For $k_{\mathrm{on}}C=0.30\ \mathrm{h}^{-1}$, $k_{\mathrm{off}}=0.10\ \mathrm{h}^{-1}$, and a 2-hour pulse starting at occupancy $0.20$, calculate the pulse-end occupancy.
Why can two schedules with equal concentration-time exposure have different steady-cycle peaks and troughs?
What should be done if binding appreciably depletes free concentration?

Solutions. (1) With $C=0$, the equation is $dO/dt=-k_{\mathrm{off}}O$, so $O(t)=O_0e^{-k_{\mathrm{off}}t}$. (2) The pulse equilibrium is $0.30/(0.30+0.10)=0.750$, and the total rate is $0.400\ \mathrm{h}^{-1}$. Thus $O(2)=0.750+(0.200-0.750)e^{-0.800}=0.750-0.550(0.4493)=0.503$. (3) Occupancy responds dynamically to the timing and duration of concentration, not only to its integral; each schedule has different pulse and washout exponential factors. (4) Determine free concentration jointly with binding using an appropriate mass balance rather than imposing it as an unaffected input.