Subcutaneous Depot Absorption and Lag Time

From a Subcutaneous Dose to Systemic Input

A subcutaneous dose is deposited in tissue before it enters the systemic circulation. A PBPK model can represent that sequence with a depot compartment: drug is placed in the depot, then leaves it over time as an absorption process. This separates the amount still at the injection site from the rate at which drug is supplied to the systemic model.

The simplest representation assumes that the depot is well mixed and that its fractional rate of loss is constant. A fixed absorption lag can be added when the model assumes that absorption does not begin immediately. These equations specify the input to the rest of the model; they do not by themselves predict a plasma concentration.

Definition: Let \(A_{\mathrm{dep}}(t)\) be the amount of drug in a subcutaneous depot at time \(t\), in mg, and let \(k_a\) be the first-order absorption rate constant, in \(\mathrm{h}^{-1}\). In the basic model, the depot loses drug at rate \(k_a A_{\mathrm{dep}}(t)\), and that loss is the systemic input rate \(R_{\mathrm{sys}}(t)\), in mg/h. Let \(D_{\mathrm{dep}}\) denote the amount assigned to the depot at dosing.

The baseline model makes specific assumptions: the dose is deposited instantaneously; the depot behaves as one homogeneous compartment; \(k_a\) is constant over time; and no drug is lost at the site by degradation or another process. It also treats all drug leaving the depot as systemic input. These assumptions define a useful model, not a claim that every injection behaves this way.

Deriving the No-Lag Model

With a dose at time zero and no absorption lag, the depot amount decreases in proportion to the amount remaining. The differential equation is \(dA_{\mathrm{dep}}/dt=-k_a A_{\mathrm{dep}}\), with initial condition \(A_{\mathrm{dep}}(0)=D_{\mathrm{dep}}\). Separating variables and integrating gives the exponential amount remaining. The systemic input is the rate of depot loss.

\frac{dA_{\mathrm{dep}}(t)}{dt}=-k_a A_{\mathrm{dep}}(t), \qquad A_{\mathrm{dep}}(0)=D_{\mathrm{dep}}

A_{\mathrm{dep}}(t)=D_{\mathrm{dep}}e^{-k_a t}, \qquad R_{\mathrm{sys}}(t)=k_a A_{\mathrm{dep}}(t)=k_aD_{\mathrm{dep}}e^{-k_a t}, \qquad t\geq 0

The dimensions provide a quick check: \(k_a\) has units \(\mathrm{h}^{-1}\), so \(k_a A_{\mathrm{dep}}\) has units mg/h. Integrating the input rate from time zero to \(t\) gives the cumulative amount that has entered the systemic model:

A_{\mathrm{cum}}(t)=\int_0^t R_{\mathrm{sys}}(u)\,du =D_{\mathrm{dep}}\left(1-e^{-k_a t}\right)

This result checks the depot mass balance: the amount remaining plus the cumulative amount supplied equals the original depot amount. At any finite time some drug remains; as time tends to infinity, the residual tends to zero and the cumulative input tends to \(D_{\mathrm{dep}}\), under the model assumptions.

Worked Example: Residual Amount and Input Without a Lag

Suppose a hypothetical subcutaneous dose places \(60\) mg in the depot and \(k_a=0.40\ \mathrm{h}^{-1}\). Find the residual amount, systemic input rate, and cumulative input at \(t=4\) h.

The exponent is \(k_at=(0.40\ \mathrm{h}^{-1})(4\ \mathrm{h})=1.60\), so \(e^{-1.60}\approx0.201897\). The residual amount is

A_{\mathrm{dep}}(4)=60e^{-1.6}\approx12.11\ \mathrm{mg}.

Using the unrounded exponential value for the next calculation, the input rate is

R_{\mathrm{sys}}(4)=0.40\times60e^{-1.6}\approx4.846\ \mathrm{mg/h}.

The cumulative input is calculated from the original dose and the unrounded residual expression:

A_{\mathrm{cum}}(4)=60\left(1-e^{-1.6}\right)\approx47.89\ \mathrm{mg}.

The mass balance holds: \(12.11+47.89\approx60.00\) mg, with the displayed rounding. The input rate is an amount per unit time at \(t=4\) h, not the amount absorbed over the entire four hours.

Adding a Fixed Absorption Lag

A fixed lag \(t_{\mathrm{lag}}\) means that the model delays the start of absorption by that amount of time. For a dose assigned to the depot at time zero, the amount remains \(D_{\mathrm{dep}}\) during the lag; after the lag, it declines exponentially as a function of elapsed absorption time \(t-t_{\mathrm{lag}}\). The lag shifts the input profile in time without changing its shape after absorption begins.

A_{\mathrm{dep}}(t)= \begin{cases} D_{\mathrm{dep}}, & 0\leq t<t_{\mathrm{lag}},\\ D_{\mathrm{dep}}e^{-k_a(t-t_{\mathrm{lag}})}, & t\geq t_{\mathrm{lag}}, \end{cases}

R_{\mathrm{sys}}(t)= \begin{cases} 0, & 0\leq t<t_{\mathrm{lag}},\\ k_aD_{\mathrm{dep}}e^{-k_a(t-t_{\mathrm{lag}})}, & t\geq t_{\mathrm{lag}}, \end{cases} \qquad A_{\mathrm{cum}}(t)= \begin{cases} 0, & 0\leq t<t_{\mathrm{lag}},\\ D_{\mathrm{dep}}\left(1-e^{-k_a(t-t_{\mathrm{lag}})}\right), & t\geq t_{\mathrm{lag}}. \end{cases}

At \(t=t_{\mathrm{lag}}\), the depot amount is continuous and equals \(D_{\mathrm{dep}}\); the input rate changes from zero to \(k_aD_{\mathrm{dep}}\). This abrupt start is a consequence of the deterministic fixed-lag assumption. It is not a gradual transition. Setting \(t_{\mathrm{lag}}=0\) recovers the no-lag model.

Worked Example: Evaluating a Dose During and After the Lag

For a hypothetical \(25\) mg depot dose, let \(k_a=0.70\ \mathrm{h}^{-1}\) and \(t_{\mathrm{lag}}=1.5\) h. Evaluate the model at \(t=1.0\) h and \(t=3.5\) h.

At \(1.0\) h, the lag has not ended. Therefore, \(A_{\mathrm{dep}}(1)=25\) mg, \(R_{\mathrm{sys}}(1)=0\) mg/h, and \(A_{\mathrm{cum}}(1)=0\) mg.

At \(3.5\) h, absorption has proceeded for \(3.5-1.5=2.0\) h. Thus the exponential term is \(e^{-(0.70)(2.0)}=e^{-1.4}\approx0.246597\). The residual amount is

A_{\mathrm{dep}}(3.5)=25e^{-1.4}\approx6.165\ \mathrm{mg}.

The systemic input rate and cumulative input are

R_{\mathrm{sys}}(3.5)=0.70\times25e^{-1.4}\approx4.315\ \mathrm{mg/h}, \qquad A_{\mathrm{cum}}(3.5)=25(1-e^{-1.4})\approx18.84\ \mathrm{mg}.

The residual and cumulative amounts sum to \(25.00\) mg after rounding. The lag contributes no absorption before \(1.5\) h; it does not change the dose assigned to the depot.

Multiple Doses and Mass Balance

For several subcutaneous doses, represent each dose as its own depot contribution. If dose \(j\) is assigned amount \(D_j\) at time \(\tau_j\), with lag \(t_{\mathrm{lag},j}\), it begins absorption at \(\tau_j+t_{\mathrm{lag},j}\). After that time, its residual amount is \(D_j e^{-k_a(t-\tau_j-t_{\mathrm{lag},j})}\). The total input is the sum of the active dose-specific input rates. This addition follows from the model's linear first-order structure; contributions from doses still in their lag period have zero input.

R_{\mathrm{sys}}(t)= \sum_{j:\,t\geq\tau_j+t_{\mathrm{lag},j}} k_aD_j e^{-k_a(t-\tau_j-t_{\mathrm{lag},j})}

Worked Example: Summing Input From Two Lagged Doses

Suppose hypothetical depot doses of \(40\) mg at \(t=0\) h and \(30\) mg at \(t=8\) h each have \(k_a=0.50\ \mathrm{h}^{-1}\) and a \(1\) h lag. Find the total residual depot amount and systemic input at \(t=10\) h.

The first dose has been absorbing for \(10-0-1=9\) h, and the second for \(10-8-1=1\) h. Their residual amounts are \(40e^{-0.5(9)}\) and \(30e^{-0.5(1)}\), respectively. Therefore,

A_{\mathrm{dep,total}}(10)=40e^{-4.5}+30e^{-0.5} \approx0.4444+18.20=18.64\ \mathrm{mg}.

The total input rate is the sum of \(k_a\) times each residual amount:

R_{\mathrm{sys}}(10)=0.50\left(40e^{-4.5}+30e^{-0.5}\right) \approx9.320\ \mathrm{mg/h}.

The total amount assigned to the depots was \(70\) mg. The cumulative input by \(10\) h is \(70-18.6403\approx51.36\) mg, consistent with the summed residual amount and the dose mass balance.

Checks, Alternatives, and Limitations

A reliable implementation should satisfy several checks. The exponent must be dimensionless, so \(k_a\) must have inverse-time units matching the time variable. The input rate must have amount-per-time units, and the residual amount must not become negative. For a single dose, the residual should stay constant during a fixed lag and then decline; the cumulative input should be zero during the lag and rise afterward. At every time, the depot amount plus cumulative input should equal the amount initially assigned to that depot.

Compare this representation with an instantaneous systemic input: that alternative puts the entire amount into the systemic model at one time, so it has no continuing depot absorption phase. A zero-order input instead supplies a constant rate for a specified duration, rather than a rate proportional to the amount remaining. A distributed or transit-style delay can represent a gradual spread in onset times, whereas a fixed lag imposes one shared onset time. These alternatives answer different structural questions; the available data and physiology should guide the choice.

The fixed-lag, first-order model can be inadequate if the observed or mechanistic process has a gradual onset, changing absorption rate, incomplete depot release, or another loss at the injection site. Warning signs include systematic mismatches in the timing or shape of predicted input-related concentration profiles, or mass-balance calculations inconsistent with the intended model. Plasma data reflect both input and the body's handling of drug, so a discrepancy alone does not identify which assumption failed. The model also does not establish a value for the absorption parameters; those require suitable data or external information.

Common mistake: It is tempting to calculate \(R_{\mathrm{sys}}(t)=k_aD_{\mathrm{dep}}\) at every time, or to use \(t\) rather than \(t-t_{\mathrm{lag}}\) in the lagged exponential. The first ignores depletion of the depot, and the second starts depletion before the modeled lag has ended. Use \(k_aA_{\mathrm{dep}}(t)\) for the rate and, after the lag, use elapsed absorption time \(t-t_{\mathrm{lag}}\). Keep the residual, rate, and cumulative amount distinct, with units of mg, mg/h, and mg.

Key takeaway: In the basic subcutaneous depot model, first-order absorption makes systemic input proportional to the amount remaining at the site. A fixed lag shifts the start of that process; it does not alter the first-order decay after absorption begins.

Check Your Understanding

Try each problem, then compare your reasoning with the solutions below.

A hypothetical \(80\) mg dose has \(k_a=0.25\ \mathrm{h}^{-1}\) and no lag. Find the residual amount, input rate, and cumulative input at \(4\) h.
A hypothetical \(50\) mg dose has \(k_a=0.30\ \mathrm{h}^{-1}\) and a \(2\) h lag. State the residual amount and input rate at \(1\) h; then calculate the residual and cumulative amounts at \(5\) h.
Explain why the systemic input rate at a given time is not the same quantity as the cumulative amount absorbed by that time.
State two assumptions of the fixed-lag, first-order depot model and one diagnostic check for its mass balance.

Solutions. (1) \(e^{-0.25(4)}=e^{-1}\approx0.367879\). Thus \(A_{\mathrm{dep}}(4)=80e^{-1}\approx29.43\) mg, \(R_{\mathrm{sys}}(4)=0.25(80e^{-1})\approx7.358\) mg/h, and \(A_{\mathrm{cum}}(4)=80(1-e^{-1})\approx50.57\) mg. The two amounts sum to \(80.00\) mg within rounding. (2) At \(1\) h the lag has not ended, so the residual is \(50\) mg and input is \(0\) mg/h. At \(5\) h, absorption time is \(5-2=3\) h and \(e^{-0.30(3)}=e^{-0.9}\approx0.406570\). The residual is \(50e^{-0.9}\approx20.33\) mg and cumulative input is \(50(1-e^{-0.9})\approx29.67\) mg. (3) The rate has units mg/h and describes input at an instant; the cumulative amount has units mg and is the integral of that rate over time. (4) Examples of assumptions include instantaneous placement in one homogeneous depot, constant \(k_a\), no loss at the site, and a deterministic fixed lag. A mass-balance check is that residual depot amount plus cumulative input equals the initial depot amount.