Deriving Transit-Absorption Competition

Why Absorption and Transit Compete

Dissolved drug in a gut segment can have two fates in a segmental model: it can be absorbed across the segment wall, or it can leave the segment by transit. The fraction absorbed therefore depends not only on the absorption rate but also on how quickly transit removes drug from the segment. This tutorial derives that competition for a single cohort of dissolved drug. It does not require the segment to receive drug continuously; an initial amount is followed as it is absorbed, transited, or remains in the lumen.

The relevant rates are the segmental first-order absorption rate constant \(k_{\mathrm{abs}}\) and the first-order transit rate constant \(k_{\mathrm{tr}}\), both with units of inverse time. The earlier material on mapping effective permeability to segmental absorption rates provides the context for \(k_{\mathrm{abs}}\). Here, treat the rate constants as given and ask what their competition implies for the amount absorbed.

Definition: In a single, well-mixed segment, let \(D(t)\) be the dissolved drug amount in the lumen at time \(t\), and let \(D_0\) be its amount at \(t=0\). Under constant first-order absorption and transit, the absorption flux is \(k_{\mathrm{abs}}D(t)\) and the transit flux out of the segment is \(k_{\mathrm{tr}}D(t)\). The eventual absorbed fraction is the amount absorbed over all time divided by \(D_0\), assuming the cohort is followed until no drug remains in the segment.

Assumptions and Mass-Balance Derivation

The derivation uses a constant-volume, well-mixed segment for this dissolved cohort: at any instant, absorption and transit rates are proportional to the same amount \(D(t)\). Both rate constants are constant over the period considered, absorption and transit are the only removal routes from this dissolved pool, and the model is linear in drug amount. There is no additional dissolution or incoming luminal input during the calculation. These are model assumptions, not claims that real gut contents are perfectly mixed or that physiological rates never vary.

Because the two routes remove drug from the same pool, their fluxes add in the mass balance. Define \(A(t)\) as the cumulative amount absorbed by time \(t\), and \(T(t)\) as the cumulative amount transited out by time \(t\). The governing equations are:

\frac{dD}{dt}=-(k_{\mathrm{abs}}+k_{\mathrm{tr}})D,\qquad \frac{dA}{dt}=k_{\mathrm{abs}}D,\qquad \frac{dT}{dt}=k_{\mathrm{tr}}D

The initial conditions are \(D(0)=D_0\), \(A(0)=0\), and \(T(0)=0\). Solving the first equation gives the remaining dissolved amount. Integrating the two fluxes then gives cumulative absorption and transit:

D(t)=D_0e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t}

A(t)=D_0\frac{k_{\mathrm{abs}}}{k_{\mathrm{abs}}+k_{\mathrm{tr}}} \left(1-e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t}\right),\qquad T(t)=D_0\frac{k_{\mathrm{tr}}}{k_{\mathrm{abs}}+k_{\mathrm{tr}}} \left(1-e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t}\right)

The common factor \(1-e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t}\) is the fraction that has left the dissolved pool by time \(t\). The two rate ratios divide that removed amount between the absorption and transit routes. At finite time, some drug may still be in the segment, so the cumulative absorbed fraction is not generally the eventual absorbed fraction.

Result: For \(D_0>0\), the cumulative absorbed fraction by time \(t\) is \(F_{\mathrm{abs}}(t)=A(t)/D_0\). If \(k_{\mathrm{abs}}+k_{\mathrm{tr}}>0\), the eventual absorbed fraction is \(k_{\mathrm{abs}}/(k_{\mathrm{abs}}+k_{\mathrm{tr}})\), and the eventual transited fraction is \(k_{\mathrm{tr}}/(k_{\mathrm{abs}}+k_{\mathrm{tr}})\).

A short mass-balance check justifies these expressions and is useful when implementing the model. Adding the two cumulative fluxes to the remaining amount gives:

A(t)+T(t)+D(t) =D_0\left(1-e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t}\right) +D_0e^{-(k_{\mathrm{abs}}+k_{\mathrm{tr}})t} =D_0

Thus the amount is conserved: drug either remains in the segment, has been absorbed, or has transited out. As \(t\) tends to infinity, the exponential remaining fraction tends to zero, and the two eventual fractions sum to one.

Write both removal fluxes.
Absorption removes \(k_{\mathrm{abs}}D\) per unit time and transit removes \(k_{\mathrm{tr}}D\) per unit time.

Combine them in the dissolved-pool balance.
The total first-order loss rate is \(k_{\mathrm{abs}}+k_{\mathrm{tr}}\), so \(D(t)\) decays exponentially at that rate.

Integrate the route-specific flux.
Integrating each flux gives its cumulative amount; division by \(D_0\) gives its fraction.

Check the accounting.
At each time, the remaining, absorbed, and transited fractions must sum to one.

Calculating Cumulative Fractions and Amounts

Worked Example: A Two-Hour Segment Exposure

Suppose a hypothetical segment initially contains \(D_0=50.0\) mg of dissolved drug. Let \(k_{\mathrm{abs}}=0.400\ \mathrm{h}^{-1}\), \(k_{\mathrm{tr}}=0.200\ \mathrm{h}^{-1}\), and evaluate the cohort after \(t=2.00\) h. The total removal rate is \(0.400+0.200=0.600\ \mathrm{h}^{-1}\), so the remaining fraction is:

\frac{D(2.00)}{D_0}=e^{-(0.600)(2.00)} =e^{-1.20}=0.3011942\ldots\approx 0.3012

The cumulative absorbed fraction is the absorption share of total removal multiplied by the fraction removed:

F_{\mathrm{abs}}(2.00) =\frac{0.400}{0.600}\left(1-e^{-1.20}\right) =\frac{2}{3}(0.6988058\ldots) =0.4658705\ldots\approx 0.4659

The cumulative transited fraction is:

F_{\mathrm{tr}}(2.00) =\frac{0.200}{0.600}\left(1-e^{-1.20}\right) =\frac{1}{3}(0.6988058\ldots) =0.2329353\ldots\approx 0.2329

The corresponding amounts are \(50.0(0.4658705\ldots)=23.2935\ldots\) mg absorbed, or 23.29 mg to two decimal places; \(50.0(0.2329353\ldots)=11.6468\ldots\) mg transited, or 11.65 mg; and \(50.0(0.3011942\ldots)=15.0597\ldots\) mg remaining, or 15.06 mg. The rounded amounts total \(50.00\) mg, consistent with the mass balance. The absorbed fraction is cumulative at 2.00 h, not the eventual fraction.

How Absorption Rate Changes the Competition

At a fixed transit rate, a larger \(k_{\mathrm{abs}}\) increases the share of the total loss assigned to absorption. This comparison assumes the same segment, transit rate, and initial dissolved amount; only the absorption rate changes. In an application, a change in permeability can alter \(k_{\mathrm{abs}}\), while the mapping from permeability to rate depends on the segmental model inputs. The table illustrates the rate competition without making further claims about that mapping.

Worked Example: Comparing Three Absorption Rates

Consider a hypothetical segment with \(k_{\mathrm{tr}}=0.200\ \mathrm{h}^{-1}\). Suppose three permeability levels, represented by the supplied absorption rates, give \(k_{\mathrm{abs}}=0.0500\), \(0.400\), and \(1.20\ \mathrm{h}^{-1}\). For each case, calculate the eventual absorbed fraction using \(k_{\mathrm{abs}}/(k_{\mathrm{abs}}+k_{\mathrm{tr}})\).

Illustrative level	\(k_{\mathrm{abs}}\) (\(\mathrm{h}^{-1}\))	Calculation	Eventual absorbed fraction
Lower	0.0500	\(0.0500/(0.0500+0.200)\)	0.2000
Intermediate	0.400	\(0.400/(0.400+0.200)\)	0.6667
Higher	1.20	\(1.20/(1.20+0.200)\)	0.8571

The arithmetic shows that increasing \(k_{\mathrm{abs}}\) from \(0.0500\) to \(0.400\ \mathrm{h}^{-1}\) increases the eventual fraction from 0.2000 to 0.6667; increasing it to \(1.20\ \mathrm{h}^{-1}\) increases the fraction to 0.8571. These are fractions of the dissolved cohort eventually absorbed under the stated two-route model. They do not say that the entire administered oral dose is absorbed, because the calculation concerns only the specified dissolved amount in this one segment.

The finite-time fractions also depend on the combined removal rate in the exponential. At \(t=2.00\) h, the lower-rate case gives \(0.2000(1-e^{-0.500})=0.07869\); the intermediate case gives \(0.4659\), as calculated above; and the higher-rate case gives \((1.20/1.40)(1-e^{-2.80})=0.8050\). In each case the finite-time absorbed fraction is below its eventual value because some drug remains in the segment or has transited out by that time.

Worked Example: A Different Exposure Time and Initial Amount

A hypothetical cohort has \(D_0=12.0\) mg, \(k_{\mathrm{abs}}=0.300\ \mathrm{h}^{-1}\), \(k_{\mathrm{tr}}=0.150\ \mathrm{h}^{-1}\), and is followed for \(1.50\) h. The combined rate is \(0.450\ \mathrm{h}^{-1}\), and its rate-time product is \((0.450)(1.50)=0.675\). Thus:

e^{-0.675}=0.5091564\ldots,\qquad F_{\mathrm{abs}}(1.50)=\frac{0.300}{0.450}(1-e^{-0.675}) =\frac{2}{3}(0.4908436\ldots)=0.3272291\ldots

The absorbed amount is \(12.0(0.3272291\ldots)=3.92675\ldots\) mg, or 3.927 mg. The transit fraction is \((0.150/0.450)(1-e^{-0.675})=0.1636145\ldots\), corresponding to \(1.96337\ldots\) mg, or 1.963 mg. The remaining fraction is \(0.5091564\ldots\), corresponding to \(6.10988\ldots\) mg, or 6.110 mg. Before rounding, the three amounts sum to \(12.0\) mg.

The total amount initially present changes the route amounts in milligrams but not the fractions, provided the rate constants and observation time remain the same. The finite-time absorbed fraction is determined by the two rates and elapsed time, not by \(D_0\).

Interpretation, Checks, and Alternatives

The competition result can be interpreted as a first-event calculation: at any moment, both routes act on the same dissolved pool, and the ratio of the absorption rate to the total rate determines absorption's share of material leaving that pool. It is equivalent to integrating the two fluxes in the mass-balance equations. The interpretation depends on constant, first-order rates and the specified two-route pool; additional removal or input processes require an altered mass balance.

If \(k_{\mathrm{tr}}=1/\tau\), where \(\tau\) is the mean residence time for a first-order transit process, the eventual absorbed fraction can also be written \(k_{\mathrm{abs}}\tau/(1+k_{\mathrm{abs}}\tau)\). This is a reformulation of the competing-rate result, not a different residence-time assumption. It differs from a model that gives every molecule a fixed, deterministic residence time \(\tau\). Under fixed residence time and first-order absorption alone, the fraction absorbed by the time of exit would be \(1-e^{-k_{\mathrm{abs}}\tau}\). A fixed exit time and an exponentially distributed first-order transit process are not interchangeable assumptions.

Several checks help identify calculation or implementation errors:

Units: \(k_{\mathrm{abs}}t\) and \(k_{\mathrm{tr}}t\) must be dimensionless. If rates are in \(\mathrm{h}^{-1}\), time must be in hours.
Fractions: cumulative absorbed, cumulative transited, and remaining fractions must each be between zero and one and sum to one, apart from rounding.
Direction: holding transit fixed, increasing absorption rate must increase the eventual absorbed fraction. Holding absorption fixed, increasing transit rate must decrease it.
Limits: with no absorption (\(k_{\mathrm{abs}}=0\)), the absorbed fraction is zero. With no transit and positive absorption, the eventual absorbed fraction is one. At \(t=0\), both cumulative fractions are zero.
Numerical accuracy: use the unrounded exponential in subsequent calculations and round only reported results. A small rounding error in the remaining fraction can propagate into both route amounts.

Common mistake: It is tempting to calculate \(1-e^{-k_{\mathrm{abs}}\tau}\) for the absorbed fraction, using \(\tau\) as the mean residence time of a first-order transit process. That expression assumes a fixed exposure duration \(\tau\) with no random first-order exit during that duration. With first-order absorption competing against first-order transit, the eventual fraction is instead \(k_{\mathrm{abs}}/(k_{\mathrm{abs}}+k_{\mathrm{tr}})\), or \(k_{\mathrm{abs}}\tau/(1+k_{\mathrm{abs}}\tau)\) when \(k_{\mathrm{tr}}=1/\tau\). Use the formula that matches the transit assumption.

The closed-form expressions are useful because they expose how the two rates jointly determine uptake, and they provide a benchmark for checking numerical solutions to the same one-segment equations. They are limited to the stated cohort and constant-rate assumptions. Time-varying rates, additional pathways, or continuing luminal input change the model equations and should not be handled by applying this formula unchanged.

Check Your Understanding

Work through each question using the two-route model. Answers follow the questions so that each result can be checked directly.

For \(k_{\mathrm{abs}}=0.300\ \mathrm{h}^{-1}\), \(k_{\mathrm{tr}}=0.200\ \mathrm{h}^{-1}\), and \(D_0=10.0\) mg, what is the eventual absorbed amount?
For \(k_{\mathrm{abs}}=0.200\ \mathrm{h}^{-1}\), \(k_{\mathrm{tr}}=0.300\ \mathrm{h}^{-1}\), what fraction remains after \(2.00\) h?
Why is \(1-e^{-k_{\mathrm{abs}}\tau}\) not generally the eventual absorbed fraction when \(\tau\) is the mean residence time of first-order transit?
If \(k_{\mathrm{abs}}=0\) and \(k_{\mathrm{tr}}>0\), what are the eventual absorbed and transited fractions?

Solutions. (1) The eventual absorbed fraction is \(0.300/(0.300+0.200)=0.6000\). Therefore, the amount absorbed is \(10.0(0.6000)=6.00\) mg. (2) The remaining fraction is \(e^{-(0.200+0.300)(2.00)}=e^{-1.00}=0.3679\). Thus 36.79% remains after 2.00 h. (3) First-order transit gives a random, exponentially distributed exit time with mean \(\tau=1/k_{\mathrm{tr}}\), rather than a fixed exit time for every molecule. Absorption and transit compete, so the eventual absorbed fraction is \(k_{\mathrm{abs}}/(k_{\mathrm{abs}}+k_{\mathrm{tr}})\). (4) With no absorption, the eventual absorbed fraction is zero and the eventual transited fraction is one; the dissolved pool is eventually removed by transit under the stated positive transit rate.