Fraction Absorbed Across Sequential Gut Segments

From Segmental Competition to Cumulative Absorption

A molecule in the gut lumen can be absorbed from its current segment or transit onward. The probability of absorption in one segment therefore depends not only on the absorption rate, but also on how quickly drug leaves that segment. Across several sequential segments, a molecule that escapes one segment may still be absorbed later. The overall fraction absorbed is consequently not the sum of the segmental absorption fractions conditional on entry: each later segment receives only the fraction that escaped all earlier segments.

The earlier material on transit-absorption competition establishes the single-segment result for constant, competing first-order absorption and transit rates. Here that result is extended across an ordered sequence of segments. The focus is the cumulative fraction absorbed by the end of the sequence, along with the amount absorbed in each segment and the amount that exits the modeled gut without being absorbed.

Definition: In segment $i$, let $k_{a,i}$ be the first-order absorption rate constant and $k_{tr,i}$ the first-order transit rate constant, both with units of inverse time. Conditional on entering that segment, the segmental absorption fraction is $a_i=k_{a,i}/(k_{a,i}+k_{tr,i})$, and the segmental escape fraction is $e_i=k_{tr,i}/(k_{a,i}+k_{tr,i})$. Thus $a_i+e_i=1$.

The word conditional matters. The segmental fraction $a_i$ is the fraction absorbed among the drug that enters segment $i$, not the fraction of the original oral dose absorbed there. If little drug reaches a later segment, even a high local absorption fraction there may correspond to a modest amount absorbed from the original dose.

Assumptions for the Product Rule

Consider $m$ segments in a fixed sequence, indexed $i=1,\ldots,m$. For the product calculation to apply as written, the drug considered is dissolved and available for absorption when it enters the first segment; all of it enters that segment; each segment has constant, nonnegative first-order absorption and transit rates; and transit from one segment leads to the next. Absorption removes drug from the modeled lumen, while transit preserves it in the luminal amount. Drug leaving the final segment without absorption is counted as unabsorbed in this model.

The calculation also assumes no degradation, secretion back into the lumen, return to an earlier segment, or other competing loss. The rates may differ across segments, but they are treated as fixed for the calculation. These assumptions distinguish the result from a model with time-varying conditions or a continuing input of drug into the gut. In those cases, one must account for the timing and amount of input rather than apply a single set of fixed segmental fractions to the whole dose.

Result: For a dose $D$ entering segment 1, the fraction of that dose escaping all $m$ segments is $\prod_{i=1}^{m}e_i$. The cumulative fraction absorbed by the end of segment $m$ is therefore $$ F_{\mathrm{abs}}=1-\prod_{i=1}^{m} e_i =1-\prod_{i=1}^{m}\frac{k_{tr,i}}{k_{a,i}+k_{tr,i}}. $$

Why the Escape Fractions Multiply

In one segment, the total rate of disappearance from the luminal amount is $k_{a,i}+k_{tr,i}$. Under constant competing first-order rates, the probability that a molecule remains in the segment for time $t$ is $\exp[-(k_{a,i}+k_{tr,i})t]$. The probability density that absorption is the first event at time $t$ is the probability of still being present multiplied by the absorption rate.

P(\text{absorption in segment }i\mid\text{entry}) =\int_0^\infty k_{a,i}\exp[-(k_{a,i}+k_{tr,i})t]\,dt =\frac{k_{a,i}}{k_{a,i}+k_{tr,i}}.

Replacing $k_{a,i}$ by $k_{tr,i}$ in the event density gives the probability of transit before absorption, $e_i=k_{tr,i}/(k_{a,i}+k_{tr,i})$. Because these are the only two possible first events under the stated assumptions, the probabilities sum to one.

To reach the end of the sequence unabsorbed, a molecule must escape segment 1, then escape segment 2 conditional on reaching it, and continue in this way through segment $m$. Multiplication of these successive conditional probabilities gives the overall escape fraction. The complementary event is absorption in at least one segment, so its probability is one minus the overall escape fraction.

P(\text{escape all }m\text{ segments}) =P(E_1)P(E_2\mid E_1)\cdots P(E_m\mid E_1,\ldots,E_{m-1}) =\prod_{i=1}^{m}e_i, \qquad F_{\mathrm{abs}}=1-\prod_{i=1}^{m}e_i.

The amount entering segment $i$ is $D\prod_{j=1}^{i-1}e_j$, where the product over no earlier segments for $i=1$ is defined to be 1. The amount absorbed in that segment is this entering amount multiplied by $a_i$. Summing those amounts yields the same cumulative result: since $a_i=1-e_i$, each term is the difference between the fraction escaping the previous segments and the fraction escaping through segment $i$. The sum therefore telescopes.

\frac{A_i}{D} =\left(\prod_{j=1}^{i-1}e_j\right)(1-e_i), \qquad \sum_{i=1}^{m}\frac{A_i}{D} =\sum_{i=1}^{m}\left(\prod_{j=1}^{i-1}e_j-\prod_{j=1}^{i}e_j\right) =1-\prod_{i=1}^{m}e_i.

Worked Calculations

Worked Example: Three Segments With Different Rates

Hypothetical data. Suppose 100 mg of dissolved drug enters the first of three segments. The absorption and transit rates, in min$^{-1}$, are respectively $0.30$ and $0.20$ in segment 1, $0.20$ and $0.30$ in segment 2, and $0.10$ and $0.40$ in segment 3. Assume the fixed-rate, sequential model described above.

For segment 1, $a_1=0.30/(0.30+0.20)=0.60$ and $e_1=0.20/0.50=0.40$. For segment 2, $a_2=0.20/0.50=0.40$ and $e_2=0.30/0.50=0.60$. For segment 3, $a_3=0.10/0.50=0.20$ and $e_3=0.40/0.50=0.80$.

The fraction escaping all three segments is $$ e_1e_2e_3=(0.40)(0.60)(0.80)=0.192. $$ Thus $F_{\mathrm{abs}}=1-0.192=0.808$, or 80.8% of the entering dose. For 100 mg, that is 80.8 mg absorbed and 19.2 mg leaving the modeled gut unabsorbed.

The segment amounts provide a check. Segment 1 absorbs $100(0.60)=60$ mg, leaving $100(0.40)=40$ mg to enter segment 2. Segment 2 absorbs $40(0.40)=16$ mg, leaving 24 mg to enter segment 3. Segment 3 absorbs $24(0.20)=4.8$ mg and leaves $24(0.80)=19.2$ mg. The total absorbed amount is $60+16+4.8=80.8$ mg, consistent with the product calculation.

Worked Example: A Two-Segment Dose Calculation

Hypothetical data. An 80 mg dissolved dose enters a two-segment sequence. In segment 1, $k_{a,1}=0.25$ min$^{-1}$ and $k_{tr,1}=0.75$ min$^{-1}$. In segment 2, $k_{a,2}=0.50$ min$^{-1}$ and $k_{tr,2}=0.50$ min$^{-1}$. The rates are constant and absorption and transit are the only competing exits in each segment.

The conditional absorption and escape fractions are $a_1=0.25/(0.25+0.75)=0.25$, $e_1=0.75$, $a_2=0.50/(0.50+0.50)=0.50$, and $e_2=0.50$. Overall escape is $0.75(0.50)=0.375$, so the cumulative absorbed fraction is $1-0.375=0.625$, or 62.5%.

In amounts, segment 1 absorbs $80(0.25)=20$ mg and passes $80(0.75)=60$ mg onward. Segment 2 absorbs $60(0.50)=30$ mg and passes $60(0.50)=30$ mg out of the modeled sequence. The total absorbed is $20+30=50$ mg, and $50/80=0.625$. The remaining 30 mg is the unabsorbed amount at the end of the sequence.

Worked Example: Using Absorption-to-Transit Rate Ratios

Hypothetical data. Suppose the absorption-to-transit rate ratios $k_{a,i}/k_{tr,i}$ are 0.5, 1, and 2 across three consecutive segments, and 90 mg enters the first segment. Each ratio is dimensionless because numerator and denominator have the same units. Assume constant competing first-order rates in each segment.

If $r_i=k_{a,i}/k_{tr,i}$, then $a_i=r_i/(1+r_i)$ and $e_i=1/(1+r_i)$. The three segmental absorption fractions are $1/3$, $1/2$, and $2/3$, while the escape fractions are $2/3$, $1/2$, and $1/3$. Their product is $(2/3)(1/2)(1/3)=1/9$, so the cumulative absorbed fraction is $1-1/9=8/9$, approximately 88.9%.

Segment 1 absorbs $90(1/3)=30$ mg and passes 60 mg. Segment 2 absorbs $60(1/2)=30$ mg and passes 30 mg. Segment 3 absorbs $30(2/3)=20$ mg and passes 10 mg. The total is 80 mg absorbed plus 10 mg unabsorbed, which equals the original 90 mg. This example also shows that rate ratios, rather than the absolute time units, determine these segmental fractions under the fixed-rate competition model.

Checks, Alternatives, and Limitations

A useful first diagnostic is to check each segment: $k_{a,i}$ and $k_{tr,i}$ must be nonnegative, their sum must be positive for the stated fraction formulas, and $a_i+e_i$ must equal 1. Then check the sequence: the product of escape fractions must lie between 0 and 1, and the cumulative fraction absorbed must also lie between 0 and 1. In amount terms, the sum absorbed across segments plus the amount escaping the final segment must equal the amount entering the first segment.

A result outside these bounds usually indicates an arithmetic or coding error, a mixing of rates and fractions, or a mismatch in the amount used as the denominator. In particular, adding the conditional segmental absorption fractions $a_i$ is generally wrong. A later segment's fraction applies only to the amount that reaches it. The correct segmental contribution relative to the entering dose is $(\prod_{j=1}^{i-1} e_j)a_i$.

For fixed segmental rates, the product formula is an exact shortcut to solving the corresponding sequence of mass balances. A direct numerical solution of those balances is a useful alternative when the rates or inputs vary with time, or when the model includes additional states or processes. A single empirical net absorbed fraction is another possible summary, but it does not show how the sequential segment assumptions produce that fraction. Neither alternative changes the definition of absorption; the appropriate approach depends on whether the fixed segmental competition assumptions are adequate for the question.

The product rule does not itself establish that the rate constants are known, constant, or identifiable from observed data. Nor does the luminal fraction absorbed equal the fraction reaching systemic circulation: presystemic loss after absorption is outside this calculation. A formulation that supplies drug gradually, time-varying physiology, or additional luminal processes may require a model that tracks amounts and timing explicitly. The sequential product is most informative when its assumptions are stated and its predictions pass the mass-balance checks.

Key takeaway: Multiply the conditional escape fractions to obtain the fraction of dose left unabsorbed after a sequential series. Subtract that product from 1 to obtain cumulative luminal absorption; do not add conditional segmental absorption fractions.

Check Your Understanding

Use the fixed-rate sequential model and show how the segmental fractions combine. Solutions follow the questions.

Two segments have escape fractions 0.8 and 0.25. What fraction is absorbed by the end of the second segment?
A dose of 60 mg enters a segment with $k_a=0.2$ min$^{-1}$ and $k_{tr}=0.3$ min$^{-1}$. How much enters the next segment?
For three segments, the conditional absorption fractions are 0.2, 0.3, and 0.4. Why is their sum not the cumulative fraction absorbed?
If all segmental escape fractions are between 0 and 1, what bounds must hold for the cumulative absorbed fraction?

Solutions. 1. Overall escape is $(0.8)(0.25)=0.20$, so the absorbed fraction is $1-0.20=0.80$, or 80%. 2. The escape fraction is $0.3/(0.2+0.3)=0.60$; therefore $60(0.60)=36$ mg enters the next segment. 3. Each fraction is conditional on entering its own segment. The correct cumulative fraction is $1-(0.8)(0.7)(0.6)=0.664$, or 66.4%, not the sum 0.9. 4. Since the product of escape fractions is between 0 and 1, $1-\prod_i e_i$ is also between 0 and 1, inclusive.