From Gastric Amount to Emptying Model
A gastric emptying model describes how drug leaves the stomach over time. Its output is the amount transferred out of the stomach, not the amount absorbed into the systemic circulation. That distinction matters in a PBPK model: gastric emptying supplies drug to a downstream region, where intestinal dissolution, permeability, and metabolism may affect what happens next.
The earlier material on solid and liquid gastric emptying time courses motivates representing emptying as a changing amount rather than a single completion time. Here, two compact representations are formulated: continuous first-order emptying from the start, and first-order emptying after a fixed lag. Both are useful only as simplified descriptions of a specified dose or gastric drug amount under stated conditions.
For the basic model, assume \(A_0\geq 0\), \(k_g\geq 0\), and time is measured from the time the modeled amount is placed in the stomach. Assume no additional drug enters this gastric amount after \(t=0\), and that gastric disappearance of the modeled drug occurs only by emptying. In particular, this model does not include gastric degradation, precipitation, or a separate gastric loss pathway. A real formulation may require additional processes; they are not implicitly represented by \(k_g\).
First-Order Emptying
First-order emptying means that the instantaneous amount leaving the stomach per unit time is proportional to the amount currently present. The proportionality constant is \(k_g\). Thus, a larger gastric amount produces a larger emptying rate, while the fraction emptied per unit time remains constant. This is not a constant-amount-per-hour model.
Separating variables and integrating gives the amount remaining. For \(A_g(t)>0\), the separation is direct; the resulting expression also applies when \(A_0=0\).
The emptying rate is the positive amount flux out of the stomach, \(R_{\mathrm{empty}}(t)=k_gA_g(t)\). Under the stated mass-balance assumption, cumulative emptying is the initial amount minus the amount still in the stomach.
For \(k_g>0\), the half-time—the time at which half of \(A_0\) remains—is \(\ln(2)/k_g\). If \(k_g=0\), the differential equation gives \(A_g(t)=A_0\) for all \(t\), and there is no finite half-time. This boundary case is important: a formula involving division by \(k_g\) cannot be used at zero.
Worked Example: First-Order Amount Remaining and Transfer
Suppose a hypothetical dose has \(A_0=120\) mg in the stomach and \(k_g=0.35\) h\(^{-1}\). Calculate the gastric amount remaining, cumulative amount emptied, and instantaneous emptying rate at \(t=2\) h.
The first-order assumptions are that no additional amount enters this gastric pool, the emptying rate is proportional to its current amount, and emptying is its only loss. The units are consistent because \(k_gt=(0.35\ \text{h}^{-1})(2\ \text{h})=0.70\), a dimensionless exponent.
Amount remaining:
Cumulative amount emptied:
Instantaneous emptying rate:
The amount balance checks: \(59.59\ \text{mg}+60.41\ \text{mg}=120.00\ \text{mg}\), up to rounding. The rate is an amount per time at \(t=2\) h; it is not the cumulative amount that has already emptied.
Adding a Fixed Lag
A lagged first-order model supposes that the modeled amount does not begin emptying until a specified delay has elapsed. Once the delay ends, the remaining gastric amount follows the same first-order rule. This is a deterministic delay: all of the modeled amount shares the same \(T_{\mathrm{lag}}\). It is not a distribution of different emptying times.
The time argument after the lag is \(t-T_{\mathrm{lag}}\), not \(t\): first-order emptying has been active for only that elapsed time. At \(t=T_{\mathrm{lag}}\), the expressions give \(A_g=A_0\) and \(E=0\). The amount is continuous at that point, although the model’s predicted rate changes abruptly from zero to \(k_gA_0\) when \(k_gA_0>0\).
A short justification follows from solving the same differential equation on the interval after the delay. At \(t=T_{\mathrm{lag}}\), the initial condition for that interval is \(A_g(T_{\mathrm{lag}})=A_0\). Replacing elapsed time from the start of emptying by \(u=t-T_{\mathrm{lag}}\) gives \(dA_g/du=-k_gA_g\), so the first-order solution is \(A_0e^{-k_gu}\). Before the lag, the assumption of no emptying makes the amount constant.
For either model, the mass balance can be checked by differentiation. During active first-order emptying, \(dA_g/dt=-k_gA_g\) and \(dE/dt=k_gA_g\), so \(d(A_g+E)/dt=0\). Before a fixed lag, both derivatives are zero. Since \(A_g(0)+E(0)=A_0\), the total stays \(A_0\) throughout. If \(k_g=0\), the amount does not empty even after the lag; it has no finite half-time. If \(k_g>0\), the lagged model reaches half of the initial amount at \(T_{\mathrm{lag}}+\ln(2)/k_g\).
Worked Example: Calculating Emptying After a Lag
Consider a hypothetical gastric amount \(A_0=200\) mg, \(T_{\mathrm{lag}}=1.5\) h, and \(k_g=0.40\) h\(^{-1}\). Calculate the amount remaining and cumulative amount emptied at \(t=1.0\) h and \(t=3.0\) h.
At \(t=1.0\) h, time is before the lag because \(1.0<1.5\). Thus no amount has yet emptied:
At \(t=3.0\) h, first-order emptying has been active for \(3.0-1.5=1.5\) h. The exponent is dimensionless: \((0.40\ \text{h}^{-1})(1.5\ \text{h})=0.60\).
Therefore,
The result satisfies the mass balance to rounding: \(109.8+90.24\approx200.0\) mg. Applying \(e^{-k_gt}\) with \(t=3.0\) h would incorrectly count the 1.5 h lag as active emptying time.
Comparing the Models and Checking Their Behavior
The models differ in when emptying begins, not in the post-onset shape. With the same \(A_0\) and \(k_g\), for \(0\leq t\leq T_{\mathrm{lag}}\) the lagged model has at least as much drug remaining as the immediate first-order model. It has strictly more when \(A_0>0\), \(k_g>0\), \(T_{\mathrm{lag}}>0\), and \(0<t<T_{\mathrm{lag}}\). After the lag, its curve follows the same exponential form shifted to the right by \(T_{\mathrm{lag}}\).
| Feature | First-order from time zero | Fixed-lag first-order |
|---|---|---|
| Emptying onset | At \(t=0\) | At \(t=T_{\mathrm{lag}}\) |
| Amount before onset | Decreases exponentially | Remains at \(A_0\) |
| Post-onset amount | \(A_0e^{-k_gt}\) | \(A_0e^{-k_g(t-T_{\mathrm{lag}})}\) |
| Additional parameter | None beyond \(A_0\) and \(k_g\) | \(T_{\mathrm{lag}}\) |
| Key limitation | Requires emptying to start immediately | Assumes a sharp, common delay for all the modeled amount |
Worked Example: Isolating the Effect of the Lag
For a hypothetical \(A_0=100\) mg and \(k_g=0.50\) h\(^{-1}\), compare immediate and fixed-lag first-order emptying at \(t=2\) h when the lagged model has \(T_{\mathrm{lag}}=1\) h. Assume both models have the same initial amount and post-onset rate constant.
For immediate first-order emptying:
For the lagged model, active emptying time is \(2-1=1\) h:
At 2 h, the fixed lag leaves approximately \(100(e^{-0.5}-e^{-1})=23.87\) mg more in the stomach and has transferred approximately \(23.87\) mg less. These are alternative predictions for the same initial amount, not amounts to add together.
Several checks help detect implementation or interpretation errors:
- Bounds: with \(A_0\geq0\) and \(k_g\geq0\), both models must give \(0\leq A_g(t)\leq A_0\) and \(0\leq E(t)\leq A_0\).
- Mass balance: \(A_g(t)+E(t)=A_0\). A violation usually indicates a sign, time-shift, or bookkeeping error under these model assumptions.
- Units: \(k_gt\) and \(k_g(t-T_{\mathrm{lag}})\) must be dimensionless; mixing hours with a rate in min\(^{-1}\) without conversion gives invalid predictions.
- Boundary behavior: setting \(T_{\mathrm{lag}}=0\) makes the lagged model identical to immediate first-order emptying. Setting \(k_g=0\) gives no emptying in either model.
- Rate versus amount: \(R_{\mathrm{empty}}\) has units of amount per time, while \(E(t)\) is the accumulated amount transferred. Substituting one for the other can produce incorrect downstream inputs.
A common mistake is to calculate lagged emptying as \(A_0e^{-k_gt}\) after the lag has elapsed. That formula counts the delay as time during which emptying occurred. The correction is to use the time since onset, \(t-T_{\mathrm{lag}}\), and to use the no-emptying branch when \(t<T_{\mathrm{lag}}\).
Interpretation and Limitations
A first-order rate constant summarizes the speed of proportional emptying: a larger \(k_g\) means faster decline once emptying is active. A fixed lag independently specifies the time before that decline starts. The two parameters therefore have distinct roles in the equations. Neither parameter, by itself, determines how much drug ultimately reaches systemic circulation; emptying only transfers drug from the modeled stomach pool to a downstream compartment.
The fixed-lag model is useful when a delayed onset is an adequate approximation, but its abrupt start is a strong assumption. A gradual onset or a spread of emptying times across material would not be represented faithfully by one shared \(T_{\mathrm{lag}}\). Likewise, a single first-order compartment may be inadequate if emptying changes over time or if the modeled pool has multiple components with different behavior. More flexible descriptions are alternatives when these features matter; they should not be mimicked by choosing a lag without physiological or data-based justification.
A practical diagnostic is to inspect predicted gastric amount and transfer over the full time course, not just at one selected time. Look for a lag inconsistent with observed timing, an implausibly abrupt switch in rate, or persistent disagreement in the declining portion of the curve. Such patterns indicate that the chosen time-course assumptions may not describe the process. They do not, on their own, identify which alternative mechanism is responsible.
Check Your Understanding
Work each question using the model assumptions stated above. Solutions are included below the questions.
- A hypothetical \(80\) mg gastric amount empties with \(k_g=0.20\) h\(^{-1}\) and no lag. Find the amount remaining and amount emptied at \(t=5\) h.
- A hypothetical \(150\) mg amount has \(T_{\mathrm{lag}}=2\) h and \(k_g=0.30\) h\(^{-1}\). How much remains at \(t=1\) h and at \(t=4\) h?
- For \(k_g=0\), what are the amount remaining and finite half-time in the immediate and lagged models?
- Explain why a lagged-model calculation at \(t=5\) h with \(T_{\mathrm{lag}}=2\) h uses three hours of first-order emptying, rather than five hours.
Solutions. (1) \(A_g(5)=80e^{-(0.20)(5)}=80e^{-1}=29.43\) mg, so \(E(5)=80-29.43=50.57\) mg. (2) At \(1<2\) h, \(A_g=150\) mg and \(E=0\). At 4 h, active time is \(4-2=2\) h, so \(A_g(4)=150e^{-(0.30)(2)}=150e^{-0.6}=82.32\) mg. (3) In both models, \(A_g(t)=A_0\) and \(E(t)=0\) for all \(t\); there is no finite half-time. (4) The first two hours are the specified no-emptying delay, leaving \(5-2=3\) hours during which the first-order equation applies.