Depot Models For Intramuscular And Subcutaneous Dosing

Why Represent an Injection as a Depot?

An intramuscular (IM) or subcutaneous (SC) dose does not generally enter the systemic circulation instantaneously. Drug is deposited at an injection site and becomes available for absorption over time. A structural model can represent the amount still at that site as a depot compartment, then describe transfer from the depot into the systemic circulation with a first-order rate constant.

This distinction matters when interpreting early concentrations and the time course after dosing. An immediate-input model puts the entire dose into the central compartment at once; a depot model allows the input to build and decline over time. The equations below use a one-compartment disposition model so that absorption and elimination are easy to distinguish. The same depot input can instead drive a multi-compartment disposition model.

Definition: A depot compartment is a model compartment holding drug at an administration site before systemic absorption. With first-order absorption, the rate of transfer out of depot \(r\) is \(k_{a,r}A_{D,r}(t)\), where \(A_{D,r}(t)\) is the amount in that depot and \(k_{a,r}\) is its absorption rate constant, with units of inverse time.

Single-Route Depot Model

Consider one dose \(D\) placed at time zero into an IM or SC depot. Assume the entire modeled dose is initially in that depot, absorption is first-order, and there is no lag time. Let \(A_D(t)\) be the depot amount, \(A_C(t)\) the central amount, \(V\) the central volume, and \(CL\) the clearance. For the one-compartment disposition model, define \(k=CL/V\). The model is

\frac{dA_D}{dt}=-k_aA_D,\qquad A_D(0)=D

\frac{dA_C}{dt}=k_aA_D-kA_C,\qquad A_C(0)=0,\qquad C(t)=\frac{A_C(t)}{V}.

The model assumes that depot drug leaves at a rate proportional to the amount remaining, that absorbed drug enters the central compartment, and that central drug is eliminated with a first-order rate. It also assumes fixed parameters and no other drug input during this dose’s time course. These are structural assumptions, not guarantees that every injection behaves this way.

Key result: For \(k_a\ne k\) and \(t\ge 0\), the concentration after one dose is

C(t)=\frac{D}{V}\frac{k_a}{k_a-k}\left(e^{-kt}-e^{-k_at}\right).

The result describes the concentration contribution of this dose under the stated model. Its units are dose divided by volume, such as mg/L. Both rate constants must use the same time unit; for example, if time is in hours, \(k_a\) and \(k\) are in h\(^{-1}\).

Why the Concentration Formula Has This Form

The depot equation has solution \(A_D(t)=De^{-k_at}\). Substituting this amount into the central equation gives a linear differential equation with input \(k_aDe^{-k_at}\). Multiplying by the integrating factor \(e^{kt}\) yields

\frac{d}{dt}\left(e^{kt}A_C(t)\right)=k_aD e^{(k-k_a)t}.

Integrating from zero to \(t\), and using \(A_C(0)=0\), gives \(e^{kt}A_C(t)=k_aD\{e^{(k-k_a)t}-1\}/(k-k_a)\). Rearranging and dividing by \(V\) gives the displayed concentration formula. The sign is consistent for either ordering of \(k_a\) and \(k\): the combined expression gives a nonnegative concentration for \(t\ge0\).

If \(k_a=k\), the formula with denominator \(k_a-k\) is undefined as written, but the model itself remains well-defined. Taking the limit as \(k_a\) approaches \(k\) gives \(C(t)=(D/V)kte^{-kt}\). This special case avoids treating a removable algebraic singularity as a biological problem.

Worked Example: Concentration After One SC Dose

Suppose a hypothetical SC dose of 100 mg is placed in a depot. Let \(V=20\) L, \(k_a=0.80\) h\(^{-1}\), and \(k=0.20\) h\(^{-1}\). Assume no lag, one-compartment disposition, and first-order absorption and elimination. Find the concentration at 2 hours.

Since \(k_a\ne k\), use the single-dose formula:

C(2)=\frac{100\ \mathrm{mg}}{20\ \mathrm{L}} \frac{0.80\ \mathrm{h}^{-1}}{0.80\ \mathrm{h}^{-1}-0.20\ \mathrm{h}^{-1}} \left(e^{-(0.20)(2)}-e^{-(0.80)(2)}\right).

The dose-to-volume ratio is \(5\) mg/L and the rate-constant ratio is \(0.80/0.60=1.333\). Using \(e^{-0.40}=0.6703\) and \(e^{-1.60}=0.2019\),

C(2)=(5)(1.333)(0.6703-0.2019)=3.123\ \mathrm{mg/L}.

Thus the model predicts a concentration of approximately 3.12 mg/L at 2 hours. This is a model prediction under the assumptions above, not a statement about a real product or patient.

For \(k_a\ne k\), the time of the single-dose peak can be found by setting the derivative of \(C(t)\) to zero. The positive peak time is

t_{\max}=\frac{\ln(k_a/k)}{k_a-k}.

The ratio inside the logarithm is dimensionless, and the result has units of time. In the preceding example, \(t_{\max}=\ln(0.80/0.20)/(0.80-0.20)=\ln(4)/0.60=2.310\) h. This peak time follows from the model’s absorption and elimination rates; it is not simply the time at which half the dose has left the depot.

Multiple Routes and Multiple Doses

If a modeled dose is divided among routes, assign each route its own depot and absorption rate constant. For route \(r\), let \(D_r\) be the amount placed in its depot and \(k_{a,r}\) its first-order absorption rate constant. With a shared central compartment, the equations are

\frac{dA_{D,r}}{dt}=-k_{a,r}A_{D,r},\qquad \frac{dA_C}{dt}=\sum_r k_{a,r}A_{D,r}-kA_C.

Under linear disposition, the central concentration is the sum of the contributions from the route-specific depots. Each contribution has the single-dose form with that route’s dose and absorption rate. This setup allows different routes to have different absorption rates without requiring separate systemic compartments. The equations here treat each \(D_r\) as the amount assigned to that route’s modeled depot; they do not estimate a route-specific fraction reaching systemic circulation.

Worked Example: Simultaneous IM and SC Inputs

For a hypothetical dose, 60 mg is assigned to an IM depot and 40 mg to an SC depot at time zero. Let \(V=20\) L, \(k=0.20\) h\(^{-1}\), \(k_{a,\mathrm{IM}}=1.00\) h\(^{-1}\), and \(k_{a,\mathrm{SC}}=0.25\) h\(^{-1}\). Assume each depot follows first-order absorption and both inputs enter the same central compartment. Calculate total concentration at 2 hours.

The IM contribution is

C_{\mathrm{IM}}(2)=\frac{60}{20}\frac{1.00}{1.00-0.20} \left(e^{-0.40}-e^{-2.00}\right) =3(1.25)(0.6703-0.1353)=2.006\ \mathrm{mg/L}.

The SC contribution is

C_{\mathrm{SC}}(2)=\frac{40}{20}\frac{0.25}{0.25-0.20} \left(e^{-0.40}-e^{-0.50}\right) =2(5)(0.6703-0.6065)=0.6379\ \mathrm{mg/L}.

Adding the contributions gives \(C(2)=2.006+0.6379=2.644\) mg/L. The faster IM absorption contributes more at this particular time in this example. That comparison is time-specific: a route’s contribution changes over time as absorption and elimination proceed.

Repeated dosing is handled by adding the contribution of every administered dose, with each contribution evaluated at its own elapsed time since administration. This superposition is valid when the model is linear and its parameters do not change with time or dose. A dose given at the same route and with the same absorption rate has the same concentration-time contribution as any other such dose, shifted to its administration time.

Worked Example: Two Repeated SC Doses

A hypothetical 50 mg SC dose is given at time zero and again 12 hours later. Let \(V=25\) L, \(k_a=0.50\) h\(^{-1}\), and \(k=0.10\) h\(^{-1}\). Assume identical first-order depots and linear one-compartment disposition. Find the concentration at 14 hours, two hours after the second dose.

At 14 hours, the first dose has been in the system for 14 hours and the second for 2 hours. For either dose, the coefficient is

\frac{D}{V}\frac{k_a}{k_a-k} =\frac{50}{25}\frac{0.50}{0.50-0.10} =2(1.25)=2.5\ \mathrm{mg/L}.

The second dose contributes \(2.5(e^{-0.2}-e^{-1.0})=2.5(0.8187-0.3679)=1.127\) mg/L. The first contributes \(2.5(e^{-1.4}-e^{-7.0})=2.5(0.2466-0.0009119)=0.6142\) mg/L. Therefore,

C(14)=1.127+0.6142=1.741\ \mathrm{mg/L}.

Both doses contribute even though the earlier dose is largely eliminated. The calculation relies on linear superposition; it would not be justified if the model included dose-dependent absorption or changing parameters.

Assumptions, Diagnostics, and Model Choice

A depot model is useful only to the extent that its assumptions are plausible for the data and dosing process. Before interpreting its parameters, check that dose times, route, and modeled dose amounts are correctly represented. Then examine whether the observed concentration-time pattern is compatible with gradual input from the proposed depot. Sparse sampling near the expected rise and peak may make it difficult to distinguish absorption behavior from other features of the concentration profile.

Useful signs of possible structural mismatch include a rise that is consistently later or more prolonged than predicted, systematic residual patterns around the peak, or markedly different concentration-time shapes for doses assigned the same route and parameters. Such patterns do not by themselves identify the correct alternative. They motivate checking the dosing record and considering whether the release process needs a lag time, transit compartments, or a different input shape. The earlier material on absorption lag times, transit-compartment absorption, zero-order drug input, and sequential and parallel absorption describes those structures.

The main alternatives represent different input assumptions. An immediate input is appropriate when the dose enters the systemic model at once; a lag-time structure shifts the onset of input; transit compartments represent a sequence of absorption stages; and zero-order input delivers drug at a constant rate for a defined period. These are not interchangeable descriptions: choose among them based on the route, formulation, sampling information, and whether their predicted concentration-time profiles are consistent with the data.

The first-order depot model is intentionally compact. A single \(k_a\) describes an exponential decline in depot amount, so the model cannot represent every possible release pattern. It also does not, by itself, explain why absorption rates differ between individuals or occasions. Real injection-site behavior may depend on factors that are not included in this structural model. A good fit is evidence of compatibility with the observations, not proof that the mathematical depot is a literal anatomical compartment.

Common Mistakes

Treating the dose as an immediate central input. That removes the depot and predicts a different early concentration profile. Keep the dose in the depot at administration when modeling gradual absorption.
Adding route doses but using only one absorption rate. If routes are assigned distinct rates, calculate each depot’s contribution with its own \(k_{a,r}\), then sum the concentrations.
Using the single-dose formula for repeated dosing without time shifts. Evaluate each dose at the elapsed time since that dose was given, then add the contributions under the linear model.
Interpreting \(k_a\) as the fraction absorbed. \(k_a\) is a rate constant with units of inverse time; it is not a proportion or an amount. The depot amount declines as \(e^{-k_at}\).
Assuming a good fit proves the injection-site mechanism. Several input structures may describe observed profiles similarly. The depot is a model representation, and its adequacy should be assessed against the data and dosing context.

Key takeaway: A first-order IM or SC depot releases drug at a rate proportional to the amount remaining. Each depot creates a route-specific input into the systemic model; with linear disposition, concentrations from multiple routes and repeated doses are added after accounting for their individual dose times.

Check Your Understanding

Answer each question using the model assumptions stated in the question. Brief solutions follow.

A depot contains 80 mg at time zero and has \(k_a=0.30\) h\(^{-1}\). What amount remains after 2 hours?
Why is \(k_a\) not a fraction of the dose absorbed?
Two depots have different absorption rate constants but feed the same linear central compartment. How is total concentration calculated?
Why must each repeated dose be evaluated at its own elapsed time?
Name one concentration-time pattern that could suggest a first-order depot is inadequate, and one alternative input structure to investigate.

Solutions. (1) \(A_D(2)=80e^{-(0.30)(2)}=80e^{-0.60}=43.90\) mg. (2) \(k_a\) has units of inverse time and determines the rate of depot loss, \(k_aA_D\); it is not dimensionless. (3) Calculate each depot’s concentration contribution using its own dose and \(k_a\), then sum the contributions under linear disposition. (4) Each dose starts its concentration-time course at a different administration time, so its elapsed time determines the remaining depot amount and central concentration contribution. (5) A systematically delayed rise relative to model predictions may indicate mismatch; a lag-time or transit-compartment input could be considered.