Coupling Gastric Emptying to Intestinal Drug Input

From Gastric Emptying to Intestinal Input

A gastric emptying model describes how drug leaves the stomach; it does not, by itself, describe how much drug has crossed the intestinal wall. In a PBPK model, the amount leaving the stomach becomes an input to the intestinal lumen. Absorption is a separate process that removes drug from that lumen. Keeping these rates separate is essential for interpreting the model and for checking that no drug has been created, lost, or counted twice.

The earlier material on fed-state emptying establishes ways to represent gastric release, including first-order, lagged, and distributed emptying. Here, the emptying model is treated as a specified transfer process and coupled to an intestinal compartment. The focus is the connection between its output and intestinal absorption, not the prediction of downstream plasma concentration-time summaries.

Definition: Let \(A_g(t)\) be the drug amount in the stomach, \(A_l(t)\) the drug amount in the intestinal lumen, and \(A_{abs}(t)\) the cumulative amount absorbed across the intestinal wall by time \(t\). The gastric emptying flux \(J_{ge}(t)\) is the amount transferred from stomach to intestinal lumen per unit time. The intestinal absorption flux \(J_{abs}(t)\) is the amount crossing from intestinal lumen per unit time. Both fluxes have units of amount per time, such as mg/h.

A minimal coupled model has no degradation or transit out of the intestinal lumen:

\frac{dA_g}{dt}=-J_{ge}(t), \qquad \frac{dA_l}{dt}=J_{ge}(t)-J_{abs}(t), \qquad \frac{dA_{abs}}{dt}=J_{abs}(t).

The same \(J_{ge}(t)\) that leaves the stomach enters the intestinal lumen. It is not automatically the absorption flux. For example, if drug accumulates in the intestinal lumen, the amount entering that compartment can exceed the amount absorbed over the same interval.

Assumptions and Mass Balance

The basic equations assume that the tracked drug amount is expressed on a consistent basis in all compartments, that gastric transfer and intestinal absorption are the only transfers in the model, and that the fluxes are nonnegative and physically appropriate for the formulation. Initial amounts must also be specified. For an initially gastric dose \(D\), a common setup is \(A_g(0)=D\), \(A_l(0)=0\), and \(A_{abs}(0)=0\). If there is an initial lag, the emptying flux may be zero during that lag; this changes the time course, not the accounting identity.

Adding the three balance equations cancels the internal transfers:

\frac{d}{dt}\left(A_g+A_l+A_{abs}\right) =-J_{ge}+J_{ge}-J_{abs}+J_{abs}=0.

Thus, under these assumptions, the total tracked amount is constant and equals the initial dose. If intestinal transit, degradation, or another loss is represented, it must appear as an explicit outflow and be included in the balance. For example, let \(J_{tr}(t)\) be a transit loss from the modeled intestinal segment, and let \(A_{tr}(t)\) be its cumulative amount. Then \(dA_l/dt=J_{ge}-J_{abs}-J_{tr}\) and \(dA_{tr}/dt=J_{tr}\), so \(A_g+A_l+A_{abs}+A_{tr}=D\).

Key distinction: Gastric emptying is a transfer into the intestine; intestinal absorption is a transfer out of the intestinal lumen. In a model with no other losses, the amount absorbed by time \(t\) is \(D-A_g(t)-A_l(t)\), not \(D-A_g(t)\).

First-Order Coupling and Its Solution

For a simple illustration, suppose the dose is immediately available for gastric emptying and both processes are first-order, with constant rate constants \(k_g\) and \(k_a\), each in h\(^{-1}\). This assumes a well-mixed gastric amount with an emptying rate proportional to \(A_g\), and a well-mixed intestinal amount with absorption proportional to \(A_l\). There is no lag, transit loss, dissolution limitation, or other removal.

J_{ge}=k_g A_g,\qquad J_{abs}=k_a A_l,

\frac{dA_g}{dt}=-k_gA_g,\qquad \frac{dA_l}{dt}=k_gA_g-k_aA_l,\qquad \frac{dA_{abs}}{dt}=k_aA_l.

The gastric equation gives \(A_g(t)=D e^{-k_g t}\). To obtain the intestinal amount, multiply its differential equation by the integrating factor \(e^{k_a t}\). With \(A_l(0)=0\), integration gives, when \(k_a\ne k_g\):

A_l(t)=D\frac{k_g}{k_a-k_g}\left(e^{-k_g t}-e^{-k_a t}\right).

Cumulative absorption then follows either by integrating \(k_aA_l\) or by applying the mass balance. The balance expression is:

A_{abs}(t)=D-A_g(t)-A_l(t).

This form is useful as a check, but it does not mean that absorption occurs at the instant of gastric emptying. It accounts for the interval during which transferred drug remains in the lumen. When \(k_a=k_g=k\), the limiting solution is \(A_l(t)=Dkt e^{-kt}\); this follows by taking the limit of the unequal-rate expression, or by solving the intestinal equation with \(A_g=D e^{-kt}\).

Worked Example: Separating Emptying and Absorption

Suppose a hypothetical 40 mg dose is initially in the stomach. Let \(k_g=0.25\) h\(^{-1}\) and \(k_a=0.50\) h\(^{-1}\), with no lag, transit, degradation, or other loss. Calculate the amounts in the stomach and lumen, the cumulative absorbed amount, and the amount emptied from the stomach at \(t=1\) h.

The model assumptions support the first-order equations above, and the initial amounts are \(A_g(0)=40\) mg and \(A_l(0)=A_{abs}(0)=0\). At one hour, use the unrounded exponential values \(e^{-0.25}=0.7788007831\) and \(e^{-0.50}=0.6065306597\).

The gastric amount is:

A_g(1)=40e^{-0.25}=31.1520313\ \text{mg}.

For the intestinal amount, \(k_g/(k_a-k_g)=0.25/0.25=1\), so:

A_l(1)=40\left(e^{-0.25}-e^{-0.50}\right) =40(0.1722701234)=6.89080494\ \text{mg}.

Using the unrounded exponential values gives \(A_l(1)=6.89080494\) mg. The cumulative absorbed amount is the remaining mass:

A_{abs}(1)=40-31.1520313-6.89080494=1.95716376\ \text{mg}.

The amount emptied from the stomach is \(40-A_g(1)=8.84796867\) mg. It equals the amount still in the intestinal lumen plus the amount absorbed: \(6.89080494+1.95716376=8.84796870\) mg, with the small difference in the last digits due to displayed rounding. To three significant figures, the gastric amount is 31.2 mg, the intestinal amount is 6.89 mg, the cumulative absorbed amount is 1.96 mg, and the emptied amount is 8.85 mg. This verifies that most of the drug emptied by one hour has not yet been absorbed.

Allowing Intestinal Transit

A modeled intestinal segment may also have an outflow representing transit to a downstream region. With first-order absorption and transit, let \(k_{tr}\) be the transit rate constant in h\(^{-1}\). The intestinal amount then leaves through two distinct routes:

\frac{dA_l}{dt}=k_gA_g-(k_a+k_{tr})A_l,\qquad \frac{dA_{abs}}{dt}=k_aA_l,\qquad \frac{dA_{tr}}{dt}=k_{tr}A_l.

Here \(A_{tr}\) is cumulative amount that has left the modeled segment by transit, not an absorbed amount. Adding all four states again gives a constant total dose. If the intestinal segment is intended to represent the entire lumen, transit can instead be represented by transfer to another intestinal compartment; that downstream compartment must then be included in the mass balance.

Worked Example: Absorption Competing With Transit

Consider a hypothetical \(D=100\) mg dose with \(k_g=0.30\) h\(^{-1}\), \(k_a=0.40\) h\(^{-1}\), and \(k_{tr}=0.10\) h\(^{-1}\). The initial amount is entirely gastric, and there are no other processes. Find the compartment and cumulative amounts at \(t=2\) h. The total intestinal exit rate constant is \(q=k_a+k_{tr}=0.50\) h\(^{-1}\), so the equations satisfy the stated first-order assumptions.

The gastric amount is \(A_g(2)=100e^{-0.60}=54.8812\) mg. The intestinal amount uses \(q\) in place of \(k_a\) as the total removal rate:

A_l(2)=100\frac{0.30}{0.50-0.30} \left(e^{-0.60}-e^{-1.00}\right) =27.1398\ \text{mg}.

Integrating each outflow \(kA_l(t)\) gives the cumulative amount for rate constant \(k\):

A_k(t)=D\frac{k_g k}{q-k_g} \left[ \frac{1-e^{-k_g t}}{k_g} -\frac{1-e^{-qt}}{q} \right].

The bracket at two hours is \((1-e^{-0.60})/0.30-(1-e^{-1.00})/0.50=0.239720095\) h. Therefore:

A_{abs}(2)=100\frac{0.30(0.40)}{0.20}(0.239720095) =14.3832\ \text{mg},

A_{tr}(2)=100\frac{0.30(0.10)}{0.20}(0.239720095) =3.59580\ \text{mg}.

The balance is \(54.8812+27.1398+14.3832+3.59580=100.0000\) mg, up to rounding. The two cumulative outputs must not be merged: one is absorption across the wall, while the other is transit out of this modeled segment.

Implementation and Diagnostic Checks

In a numerical PBPK implementation, compute the gastric outflow once and use that same flux with opposite signs in the gastric and intestinal balances. Then compute absorption from the intestinal state, not from the gastric state. A simple explicit time-step update illustrates the bookkeeping. For a step of duration \(\Delta t\), evaluate rates at the beginning of the step and transfer the corresponding amounts:

\Delta M_{ge}=J_{ge}\Delta t,\qquad \Delta M_{abs}=J_{abs}\Delta t,

A_g^{new}=A_g-\Delta M_{ge},\qquad A_l^{new}=A_l+\Delta M_{ge}-\Delta M_{abs},\qquad A_{abs}^{new}=A_{abs}+\Delta M_{abs}.

Worked Example: Checking a Time-Step Update

A hypothetical 50 mg dose starts in the stomach. Let \(k_g=0.40\) h\(^{-1}\) and \(k_a=0.60\) h\(^{-1}\), with no transit or other loss. Use explicit rate evaluation at the start of each \(\Delta t=0.50\) h step. This is a numerical approximation, so the values need not equal the continuous-time solution exactly; the purpose is to check the transfer accounting.

Initially, \(A_g=50\) mg and \(A_l=0\) mg. In the first step, gastric transfer is \((0.40)(50)(0.50)=10\) mg, while absorption is \((0.60)(0)(0.50)=0\) mg. The updated amounts are \(A_g=40\) mg, \(A_l=10\) mg, and \(A_{abs}=0\) mg. Their sum is 50 mg.

At the start of the second step, transfer is \((0.40)(40)(0.50)=8\) mg and absorption is \((0.60)(10)(0.50)=3\) mg. Thus \(A_g=32\) mg, \(A_l=10+8-3=15\) mg, and \(A_{abs}=3\) mg. Again, \(32+15+3=50\) mg. If the 8 mg gastric transfer were incorrectly added directly to cumulative absorption, absorption would be overstated and the intestinal intermediate amount would be omitted.

Useful diagnostics follow directly from the balances. Amounts should remain nonnegative, fluxes should have the stated units, and the dose total should be conserved to numerical tolerance when every destination is represented. A negative amount or a balance discrepancy that grows over time can indicate a sign error, inconsistent units, mismatched transfer terms, or a time step too large for the integration method. For explicit updates, a step that removes more drug than is present is especially concerning; reducing the step size or using a suitable positivity-preserving solver may be necessary.

Model Choices, Limitations, and a Common Pitfall

The coupled-compartment model is a transparent choice when an intestinal amount is needed and transfer rates are adequately represented by the chosen equations. A direct input function to an absorption process can be a useful reduced alternative when intestinal residence is not modeled explicitly, but it must still preserve the distinction between material entering the intestine and material absorbed. Transit-time distributions or multiple intestinal compartments offer richer descriptions of distributed residence and regional processes, at the cost of additional states and parameters. A single first-order intestinal compartment instead assumes a well-mixed amount and constant rate constants; it cannot represent every spatial or formulation-dependent process.

A further limitation is that this accounting tracks drug amount, not necessarily dissolved, permeable, or systemically available drug. If dissolution, precipitation, gut metabolism, or transport processes are relevant, they require their own terms and destinations. In particular, cumulative intestinal absorption is not automatically identical to systemic input: subsequent presystemic processes can alter how much reaches systemic circulation.

Common mistake: “The amount that has left the stomach is the amount absorbed.” This treats gastric-to-intestinal transfer as though it crossed the intestinal wall immediately.
Correction: The emptied amount is \(D-A_g\). In the no-transit model it equals \(A_l+A_{abs}\); only the cumulative absorption term represents amount that has crossed the intestinal wall.
Diagnostic failure: Adding gastric outflow to intestinal absorption while also putting that outflow into the intestinal compartment counts the same drug twice. Use one gastric flux as an internal transfer, and include absorption only as the intestinal outflow.

Key takeaway: Connect gastric emptying to the intestinal lumen through a transfer flux, define absorption separately from the intestinal amount, and include every modeled destination in the mass balance.

Check Your Understanding

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

In a model without transit loss, what does \(D-A_g(t)\) represent, and why is it generally not equal to \(A_{abs}(t)\)?
A gastric flux is 4 mg/h and intestinal absorption flux is 1.5 mg/h at a given instant. What is the instantaneous rate of change in intestinal amount if there are no other intestinal processes?
With intestinal transit, which cumulative amount must be added to the mass balance besides gastric, intestinal, and absorbed amounts?
For \(k_a=k_g=k\), why should the unequal-rate formula not be evaluated by directly substituting \(k_a=k_g\)? State the limiting intestinal-amount solution.

Solutions. 1. \(D-A_g\) is the amount emptied from the stomach. It includes both drug still in the lumen and drug absorbed, so \(D-A_g=A_l+A_{abs}\) under the stated no-transit assumptions. 2. \(dA_l/dt=J_{ge}-J_{abs}=4-1.5=2.5\) mg/h. 3. Add the cumulative amount that has left the modeled intestinal segment by transit, \(A_{tr}\). 4. The unequal-rate expression contains \(k_a-k_g\) in its denominator and is undefined under direct substitution, although its limit exists. The limiting solution is \(A_l(t)=Dkt e^{-kt}\).