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Choosing the Structural Model · Tutorial 107 of 599

Transit-Compartment Absorption

Learn to construct and interpret a transit-compartment absorption model, including how the number of compartments and mean transit time shape systemic drug input.

Intermediate 10 min read

What You'll Learn

  • Define the transit rate constant from the number of transit compartments and mean transit time.
  • Derive the systemic input-rate function for a chain of transit compartments.
  • Calculate input rates and cumulative absorbed amounts from hypothetical doses.
  • Explain how transit number changes the timing and dispersion of absorption.
  • Compare transit models with first-order absorption, lag-time, and zero-order input models.
  • Recognize assumptions, diagnostics, and common calculation errors.

Why Represent Absorption With Transit Compartments?

Oral absorption may begin after a delay and continue over a range of times rather than entering the systemic circulation all at once. A single first-order absorption process gives a simple declining input rate from the moment the dose is available for absorption. A lag time can shift that process later, while zero-order input can represent a constant rate over a specified interval. A chain of transit compartments offers a different description: drug moves through successive stages, each with first-order transfer, before reaching the systemic circulation. The resulting input is delayed and dispersed, with a shape that can be adjusted through the number of stages.

The model is a structural description of the absorption process, not a claim that the gastrointestinal tract literally consists of identical, well-mixed compartments. Its practical value is that it can represent a gradual rise and fall in systemic input with a small number of interpretable parameters.

Definition: In a transit-compartment absorption model, a dose is placed in the first of \(N\) transit compartments and moves through the chain at a common first-order rate constant \(k_{\mathrm{tr}}\). Drug leaves the final transit compartment and enters the systemic circulation at the same rate. The mean transit time (MTT) is the expected time from placement of the dose into the chain until systemic entry. With this convention, \(N=0\) is the usual first-order absorption model.

Model and Input-Rate Function

Assume a single oral dose \(D\) is available at time zero, a fixed fraction \(F\) reaches the systemic circulation, and the transit stages all share the same positive rate constant \(k_{\mathrm{tr}}\). There are no competing losses from the transit chain. If \(A_j(t)\) denotes the amount in transit compartment \(j\), the chain for \(N\geq 1\) is

$$ \frac{dA_1}{dt}=-k_{\mathrm{tr}}A_1,\qquad \frac{dA_j}{dt}=k_{\mathrm{tr}}A_{j-1}-k_{\mathrm{tr}}A_j \quad (j=2,\ldots,N), \qquad R_{\mathrm{in}}(t)=F k_{\mathrm{tr}}A_N(t). $$

The initial conditions are \(A_1(0)=D\) and \(A_j(0)=0\) for \(j>1\). When \(N=0\), the dose is instead placed in a first-order absorption depot, giving \(R_{\mathrm{in}}(t)=FDk_{\mathrm{tr}}e^{-k_{\mathrm{tr}}t}\).

The model has \(N+1\) successive exponential transfer times: one for each transit compartment and one for transfer from the last compartment into the systemic circulation. Each has mean \(1/k_{\mathrm{tr}}\). Therefore the mean time to systemic entry is

$$ \mathrm{MTT}=\frac{N+1}{k_{\mathrm{tr}}}, \qquad k_{\mathrm{tr}}=\frac{N+1}{\mathrm{MTT}}. $$

Solving the chain gives a compact expression for the systemic input rate:

$$ R_{\mathrm{in}}(t) = FD\,k_{\mathrm{tr}}e^{-k_{\mathrm{tr}}t} \frac{(k_{\mathrm{tr}}t)^N}{N!}, \qquad t\geq 0. $$

The factor \(k_{\mathrm{tr}}\) is essential: \(FD\) has units of amount, and \(R_{\mathrm{in}}\) must have units of amount per time. For \(N=0\), the formula reduces to first-order input. For \(N>0\), input starts at zero, rises to a peak, and then declines. Differentiating the rate with respect to time shows that its peak occurs at \(t=N/k_{\mathrm{tr}}\), or, after substituting the MTT parameterization, at \(t=N\,\mathrm{MTT}/(N+1)\).

Worked Example: Calculating the Input Rate

Suppose a hypothetical oral dose is \(D=300\) mg, \(F=0.750\), there are \(N=2\) transit compartments, and \(\mathrm{MTT}=6.00\) h. Find the input rate at \(t=4.00\) h.

First calculate the common transfer rate:

$$ k_{\mathrm{tr}}=\frac{N+1}{\mathrm{MTT}} =\frac{3}{6.00\ \mathrm{h}}=0.500\ \mathrm{h}^{-1}. $$

Here \(k_{\mathrm{tr}}t=0.500(4.00)=2.00\), and \(FD=300(0.750)=225\) mg. Substitution into the input-rate equation gives

$$ \begin{aligned} R_{\mathrm{in}}(4.00) &=225\ \mathrm{mg}\,(0.500\ \mathrm{h}^{-1})e^{-2.00} \frac{(2.00)^2}{2!}\\ &=225\ \mathrm{mg}\,(0.500\ \mathrm{h}^{-1})e^{-2.00}(2.00)\\ &=30.45\ \mathrm{mg/h}. \end{aligned} $$

For \(N=2\), the peak is at \(t=N/k_{\mathrm{tr}}=2/0.500=4.00\) h, so this calculation is also the peak systemic input rate. It is an input rate, not a plasma concentration; predicting concentration additionally requires a disposition model.

What Transit Number and MTT Control

The MTT sets the average time to systemic entry, while \(N\) controls how dispersed the entry times are around that average. The total transit time is the sum of \(N+1\) independent exponential transfer times, each with rate \(k_{\mathrm{tr}}\). Its variance is \((N+1)/k_{\mathrm{tr}}^2\), so its coefficient of variation is \(1/\sqrt{N+1}\). At fixed MTT, increasing \(N\) therefore makes the distribution of transit times relatively narrower. It also requires a larger \(k_{\mathrm{tr}}\), because \(k_{\mathrm{tr}}=(N+1)/\mathrm{MTT}\).

This change in shape is distinct from simply increasing the MTT. Increasing MTT at fixed \(N\) slows every transfer and shifts absorption later. Increasing \(N\) at fixed MTT adds stages while adjusting their rate so that the average total transit time remains unchanged. The input then becomes more concentrated around that average.

Worked Example: Comparing Transit Numbers at the Same MTT

Use a hypothetical dose with \(FD=225\) mg and \(\mathrm{MTT}=6.00\) h. Compare the input rates at \(t=4.00\) h for \(N=0\), \(N=1\), and \(N=2\). The dose and MTT are held fixed; \(k_{\mathrm{tr}}\) changes with \(N\).

For \(N=0\), \(k_{\mathrm{tr}}=1/6=0.1667\ \mathrm{h}^{-1}\). Thus

$$ R_{\mathrm{in}}(4) =225(0.1667)e^{-0.1667(4)} =37.50e^{-0.6667} =19.25\ \mathrm{mg/h}. $$

For \(N=1\), \(k_{\mathrm{tr}}=2/6=0.3333\ \mathrm{h}^{-1}\), and \(k_{\mathrm{tr}}t=1.333\). Therefore

$$ \begin{aligned} R_{\mathrm{in}}(4) &=225(0.3333)e^{-1.333}\frac{1.333}{1!}\\ &=100.0e^{-1.333}\\ &=26.36\ \mathrm{mg/h}. \end{aligned} $$

For \(N=2\), the rate is the value calculated in the previous example, \(30.45\ \mathrm{mg/h}\). The rates at this selected time increase across these three models, but that does not mean that increasing \(N\) must increase the rate at every time. The whole profile changes shape. The peak times are \(0\) h for \(N=0\), \(3.00\) h for \(N=1\), and \(4.00\) h for \(N=2\). In particular, the \(N=0\) model begins at its largest input rate and then declines; it does not have the delayed rise produced by positive \(N\).

Mass Balance and a Cumulative-Input Calculation

A useful check on the model is that the total systemic amount entered over all time equals \(FD\), provided the chain has no competing loss. Integrating the input-rate equation and using \(u=k_{\mathrm{tr}}t\) gives

$$ \int_0^\infty R_{\mathrm{in}}(t)\,dt = FD\int_0^\infty e^{-u}\frac{u^N}{N!}\,du =FD. $$

The final equality follows from \(\int_0^\infty e^{-u}u^N\,du=N!\) for nonnegative integer \(N\). Thus \(F\) scales the total amount entering, whereas \(N\) and MTT determine the timing and shape of entry.

Worked Example: Amount Entered by One Mean Transit Time

Consider a hypothetical dose with \(D=400\) mg, \(F=0.800\), \(N=1\), and \(\mathrm{MTT}=5.00\) h. How much drug has entered the systemic circulation by \(t=5.00\) h?

Here \(FD=320\) mg and \(k_{\mathrm{tr}}=2/5.00=0.400\ \mathrm{h}^{-1}\). For \(N=1\), integrating the input rate from zero to \(t\) gives \(FD[1-e^{-k_{\mathrm{tr}}t}(1+k_{\mathrm{tr}}t)]\). At \(t=5.00\) h, \(k_{\mathrm{tr}}t=2.00\), so

$$ \begin{aligned} A_{\mathrm{entered}}(5.00) &=320\left[1-e^{-2.00}(1+2.00)\right]\ \mathrm{mg}\\ &=320\left[1-3(0.1353)\right]\ \mathrm{mg}\\ &=190.1\ \mathrm{mg}. \end{aligned} $$

This is about \(59.40\%\) of the \(320\) mg that will eventually enter, not \(59.40\%\) of the administered dose before accounting for \(F\). The remaining systemic input after 5.00 h is approximately \(320-190.1=129.9\) mg.

Assumptions, Diagnostics, and Alternatives

The chain assumes a common first-order transfer rate at every stage, a specified initial placement of the dose, and no loss within the chain other than transfer onward. The input equation describes systemic entry; it does not by itself specify distribution or elimination after entry. When concentrations are modeled, the transit input must be connected to an appropriate disposition model. The interpretation of a fitted MTT also depends on the assumed chain: it is the mean time to entry under this model, not necessarily a directly measured physiological time.

Model adequacy should be assessed against the observed concentration-time data and the complete structural model. Useful warning signs include systematic residual patterns around the rising or peak portion of the profile, an estimated MTT or transit number that cannot reproduce the observed timing, and strong sensitivity of the estimated absorption shape to the assumed disposition model or sampling schedule. The integer \(N\) is often selected by comparing plausible candidate models; it is not ordinarily a continuously varying parameter. A more complex chain is not automatically better, especially if the data do not identify its shape.

The main alternatives describe different assumptions. First-order absorption is the \(N=0\) case and is parsimonious, but its input starts immediately at its maximum. A lag-time model shifts the start of an absorption process without, by itself, creating a gradual transit delay. Zero-order input imposes a constant rate for a specified duration, which differs from the rising-and-falling transit profile. Transit compartments can represent a smooth delayed input, but their extra shape flexibility can be weakly identified from sparse concentration measurements. These models should be compared as alternative structural descriptions, not treated as interchangeable explanations.

Common mistake: Omitting \(k_{\mathrm{tr}}\) from \(R_{\mathrm{in}}(t)\) produces \(FD e^{-k_{\mathrm{tr}}t}(k_{\mathrm{tr}}t)^N/N!\), which has units of amount rather than amount per time. The correct input-rate equation includes the leading factor \(k_{\mathrm{tr}}\). Check both the units and the total-input identity \(\int_0^\infty R_{\mathrm{in}}(t)\,dt=FD\).
Key takeaway: Given a transit number \(N\) and mean transit time MTT, calculate \(k_{\mathrm{tr}}=(N+1)/\mathrm{MTT}\), then use the transit input-rate function. MTT sets the mean timing; \(N\) changes the dispersion and shape of the absorption profile.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. For \(N=3\) and \(\mathrm{MTT}=8.00\) h, calculate \(k_{\mathrm{tr}}\) and the time of peak input.
  2. For \(N=0\), explain why the input rate does not rise from zero after the dose.
  3. For a hypothetical dose with \(FD=100\) mg, \(N=1\), and \(k_{\mathrm{tr}}=0.500\ \mathrm{h}^{-1}\), calculate \(R_{\mathrm{in}}(2.00\ \mathrm{h})\).
  4. What does increasing \(N\) at fixed MTT do to the coefficient of variation of transit time?
  5. State one diagnostic sign that a transit model may not describe observed absorption adequately.

Solutions. (1) \(k_{\mathrm{tr}}=(3+1)/(8.00\ \mathrm{h})=0.500\ \mathrm{h}^{-1}\). The peak is at \(N/k_{\mathrm{tr}}=3/0.500=6.00\) h. (2) With \(N=0\), \(R_{\mathrm{in}}(t)=FDk_{\mathrm{tr}}e^{-k_{\mathrm{tr}}t}\), which is largest at \(t=0\) and declines thereafter. (3) Here \(k_{\mathrm{tr}}t=1\), so \(R_{\mathrm{in}}=100(0.500)e^{-1}(1)=18.39\ \mathrm{mg/h}\). (4) The coefficient of variation is \(1/\sqrt{N+1}\), so it decreases as \(N\) increases. (5) One sign is a systematic residual pattern around the observed concentration peak or rising phase, suggesting that the modeled timing or profile shape is inadequate.

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