From an Oral Dose to Systemic Input
An oral dose does not usually appear in the systemic circulation all at once. A compartment model can represent the administered dose as an amount in an absorption depot, then describe how quickly material leaves that depot and how much of it reaches the systemic circulation. Keeping these quantities distinct is essential: the rate of depot disappearance is not always the rate of systemic input.
This tutorial defines the depot amount, absorption flux, and bioavailability fraction, and uses them to calculate systemic input. It does not derive a concentration-time profile. That requires combining the input with the disposition model and is a separate calculation.
For this simple model, \(0\leq F\leq 1\), and \(k_a\geq 0\). A dose \(D\) administered into the depot gives \(A_g(0)=D\), assuming no drug is already present. The depot follows the first-order amount balance
The minus sign indicates that the amount in the depot decreases. The model assumes that a fraction \(F\) of the material leaving the depot becomes systemic input; the remaining fraction, \(1-F\), represents loss before systemic entry. Thus the systemic input rate is \(F\) times the depot disappearance rate, not the full disappearance rate unless \(F=1\).
Rates, Amounts, and Mass Balance
A flux is a rate, so it must include a time unit. If \(A_g(t)\) is in mg and \(k_a\) is in \(\mathrm{h}^{-1}\), then \(k_aA_g(t)\) and \(R_{\mathrm{in}}(t)\) are in mg/h. By contrast, \(F\) is dimensionless. The amount entering the systemic circulation over an interval is the integral of the flux over that interval:
When \(F\) and \(k_a\) are constant and the depot follows the stated balance, the integrated input can also be calculated from the change in depot amount. Since \(dA_g/dt=-k_aA_g\), rearranging gives \(k_aA_g=-dA_g/dt\). Therefore:
This is a mass-balance result: the systemic amount is the bioavailable fraction of the amount that disappeared from the depot over the interval. If the depot is eventually depleted completely, the total systemic input is \(FD\). This conclusion does not require calculating a concentration profile. It does require the assumptions above, including a constant \(F\) and no other process removing drug from the depot.
Worked Example: Calculating an Instantaneous Absorption Flux
Suppose a hypothetical model has \(F=0.75\), \(k_a=0.30\ \mathrm{h}^{-1}\), and a depot amount of \(A_g(t)=160\ \mathrm{mg}\) at a particular time. Calculate the depot disappearance rate and systemic absorption flux at that time.
The depot disappearance rate is
Only the fraction \(F=0.75\) reaches the systemic circulation in this model. Thus,
The model therefore has a depot disappearance rate of 48.0 mg/h and a systemic input rate of 36.0 mg/h. The difference, 12.0 mg/h, is the modeled non-systemic fraction of the depot loss at that instant.
Worked Example: Systemic Input From a Change in Depot Amount
In a hypothetical interval, the amount in an oral depot decreases from 500 mg to 320 mg. Suppose \(F=0.80\) is constant and all depot loss follows the simple model. How much drug enters the systemic circulation during the interval?
The depot loss is \(500-320=180\ \mathrm{mg}\). By mass balance,
Thus, 144 mg enters the systemic circulation during the interval. The remaining 36 mg of the 180 mg depot loss does not enter systemically under this model: \((1-0.80)(180\ \mathrm{mg})=36\ \mathrm{mg}\). The 320 mg still in the depot is not part of either interval amount.
Two Equivalent Ways to Apply Bioavailability
Model implementations can place \(F\) in different parts of the bookkeeping. One convention puts the full dose \(D\) into the depot and multiplies the depot-loss flux by \(F\), as above. Another defines a bioavailable depot amount \(A_b(t)=F A_g(t)\), so that its input flux is \(k_aA_b(t)\). If \(F\) is constant, these conventions give the same systemic flux:
A third equivalent bookkeeping choice is to initialize a model depot with \(FD\) and then use \(k_aA_b(t)\) as the systemic flux. In that convention, the depot amount represents only the bioavailable portion. The conventions are alternatives, not multipliers to combine: applying \(F\) to a depot already initialized at \(FD\) would count bioavailability twice.
Worked Example: Avoiding a Double Application of Bioavailability
Consider a hypothetical oral dose of 300 mg with \(F=0.60\), \(k_a=0.20\ \mathrm{h}^{-1}\), and a specified time at which the full-dose depot convention has \(A_g(t)=100\ \mathrm{mg}\). Calculate the systemic flux under two equivalent conventions.
Full-dose depot convention. The flux is
Bioavailable depot convention. The corresponding bioavailable depot amount is \(A_b(t)=F A_g(t)=(0.60)(100\ \mathrm{mg})=60.0\ \mathrm{mg}\). Its flux is
Both conventions give 12.0 mg/h. If the 60.0 mg bioavailable amount were multiplied by \(F\) again, the calculation would give \((0.60)(0.20)(60.0)=7.20\ \mathrm{mg/h}\), which is too small because \(F\) has been applied twice.
Assumptions, Checks, and Alternatives
The calculations are valid for the stated simple input model. In particular, they assume a first-order disappearance rate proportional to the current depot amount, a constant bioavailability fraction, and immediate systemic entry of the bioavailable portion as it leaves the depot. They also assume that no separate process—such as degradation or transit to another unmodeled location—changes the depot amount. These are modeling assumptions, not conclusions that follow just from observing an oral dose.
Several checks can reveal a bookkeeping or model problem:
- Check the units. \(k_a A_g\) must have units of amount per time, while \(F\) has no units. A result in mg for an instantaneous flux indicates a missing time factor.
- Check the range of \(F\). In this simple fraction interpretation, \(F\) must be between 0 and 1. A value outside this range signals a coding, parameterization, or interpretation issue.
- Check cumulative mass balance. Over an interval, systemic input cannot exceed the depot loss when \(0\leq F\leq 1\). If the depot is fully depleted, total systemic input should be \(FD\), not \(D+FD\).
- Check what the depot amount means. An amount defined as the bioavailable portion should not have \(F\) applied a second time. The model definition must match the implementation.
- Check whether the assumptions fit the intended process. A lag before input, a nearly constant input rate over a period, or a distributed delay may call for a different input structure. The first-order depot equations should not be treated as a universal description of oral absorption.
The alternatives differ in what they assume about input. A first-order depot makes the instantaneous disappearance rate proportional to the amount currently in the depot. A constant-rate input instead specifies a rate over a defined period. Models with a delay or a sequence of transit compartments represent delayed or distributed input differently. Regardless of the chosen structure, the model must specify what amount is represented and how the systemic input rate is obtained; \(F\) should be applied consistently with that definition.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A hypothetical model has \(A_g(t)=90\ \mathrm{mg}\), \(k_a=0.40\ \mathrm{h}^{-1}\), and \(F=0.50\). What are the depot disappearance rate and systemic flux at that time?
- A depot amount falls from 240 mg to 150 mg over an interval, with constant \(F=0.70\). How much enters the systemic circulation during the interval?
- Under the full-dose depot convention, what is the total systemic input if a 600 mg dose is completely depleted and \(F=0.65\)?
- A model initializes a bioavailable depot at \(FD\) and uses \(k_a A_b(t)\) as its systemic flux. Should the flux also be multiplied by \(F\)? Explain.
Solutions. (1) The depot disappearance rate is \((0.40\ \mathrm{h}^{-1})(90\ \mathrm{mg})=36.0\ \mathrm{mg/h}\). The systemic flux is \((0.50)(36.0)=18.0\ \mathrm{mg/h}\). (2) The depot loss is \(240-150=90\ \mathrm{mg}\), so systemic input is \((0.70)(90\ \mathrm{mg})=63.0\ \mathrm{mg}\). (3) Complete depletion gives \(FD=(0.65)(600\ \mathrm{mg})=390\ \mathrm{mg}\). (4) No. The depot already represents the bioavailable portion, so multiplying its flux by \(F\) again would count bioavailability twice.