Deriving Profiles From Parallel First-Order Pathways

From Dose Allocation to Concentration

When a dose is divided between two absorption pathways, each pathway supplies drug to the systemic circulation over time. The concentration at any time is therefore the sum of the concentrations produced by the two inputs, provided both inputs enter the same linear disposition system. The key is to derive each contribution using its own absorption rate, then add them without confusing dose fractions with rate constants.

Assume an oral dose \(D\) is split into fractions \(f\) and \(1-f\), where \(0\leq f\leq1\). Each fraction enters a separate, initially populated depot and is absorbed by first-order kinetics. Both pathways have the same bioavailability \(F\), apparent volume of distribution \(V\), and first-order elimination rate constant \(k\), but their absorption rate constants \(k_{a1}\) and \(k_{a2}\) may differ. The concentration is measured after the dose at time \(t\geq0\).

Definition: A parallel two-pathway first-order absorption model assigns fractions \(fD\) and \((1-f)D\) to two absorption depots. The amounts in those depots decline at rates \(k_{a1}\) and \(k_{a2}\), respectively, and their absorbed drug enters a shared, linearly eliminating central compartment. The parameters \(k\), \(k_{a1}\), and \(k_{a2}\) have units of inverse time; \(F\) and \(f\) are dimensionless; \(D\) is an amount; and \(V\) is a volume.

The model assumes complete first-order depletion of each depot, no lag time, no dose loss beyond the common bioavailability \(F\), and linear one-compartment disposition. It also assumes that the same central compartment receives both inputs and that elimination is independent of the input pathway. These assumptions are what permit the two concentration contributions to add directly.

Deriving the Two Contributions

Let \(A_1(t)\) and \(A_2(t)\) denote the amounts remaining in the two depots, and let \(A_c(t)\) be the amount in the central compartment. Initially, \(A_1(0)=fD\), \(A_2(0)=(1-f)D\), and \(A_c(0)=0\). The absorbed amounts are multiplied by the common bioavailability \(F\) on entering systemic circulation.

\frac{dA_1}{dt}=-k_{a1}A_1,\qquad \frac{dA_2}{dt}=-k_{a2}A_2,\qquad \frac{dA_c}{dt}=F k_{a1}A_1+F k_{a2}A_2-kA_c,\qquad C(t)=\frac{A_c(t)}{V}.

Solving the depot equations gives \(A_1(t)=fD e^{-k_{a1}t}\) and \(A_2(t)=(1-f)D e^{-k_{a2}t}\). To see how one depot contributes to the central amount, consider a general depot with initial amount \(D_i\) and absorption rate \(k_{ai}\). Its input rate is \(F k_{ai}D_i e^{-k_{ai}t}\). Multiplying the central-compartment equation by the integrating factor \(e^{kt}\) and integrating from 0 to \(t\) gives the following result when \(k_{ai}\ne k\).

A_{c,i}(t) =e^{-kt}\int_0^t F k_{ai}D_i e^{(k-k_{ai})s}\,ds =F D_i\frac{k_{ai}}{k_{ai}-k} \left(e^{-kt}-e^{-k_{ai}t}\right).

Dividing by \(V\), then substituting \(D_1=fD\) and \(D_2=(1-f)D\), produces the total concentration profile.

C(t)=\frac{FD}{V}\left[ f\frac{k_{a1}}{k_{a1}-k}\left(e^{-kt}-e^{-k_{a1}t}\right) +(1-f)\frac{k_{a2}}{k_{a2}-k}\left(e^{-kt}-e^{-k_{a2}t}\right) \right].

The bracketed terms are the two pathway-specific contributions. Each contribution is zero at \(t=0\), rises as drug is absorbed, and eventually declines as elimination dominates. The dose fraction scales its pathway's contribution; it does not change that pathway's absorption rate. The overall profile is the sum, not the average, of the two contributions.

Key result: Under common linear disposition and a shared bioavailability, the concentration from parallel inputs is additive. The formula above requires \(k_{a1}\ne k\) and \(k_{a2}\ne k\) as written. If an absorption rate equals \(k\), use the limiting expression \(k t e^{-kt}\) for that pathway's rate-dependent term.

Worked Calculations

Worked Example: Calculating Both Pathway Contributions

Suppose a hypothetical dose is \(D=100\) mg, with \(F=0.80\), \(V=20\) L, \(f=0.60\), \(k=0.10\ \text{h}^{-1}\), \(k_{a1}=1.20\ \text{h}^{-1}\), and \(k_{a2}=0.20\ \text{h}^{-1}\). Find the concentration at \(t=2\) h. The fractions are between 0 and 1, and both absorption rates differ from \(k\), so the stated model and formula apply.

First, \(FD/V=(0.80)(100\ \text{mg})/(20\ \text{L})=4.00\ \text{mg/L}\). For pathway 1, the rate factor is \(1.20/(1.20-0.10)=1.09091\), and \(e^{-0.10(2)}-e^{-1.20(2)}=e^{-0.20}-e^{-2.40}=0.728013\). Its concentration contribution is

C_1(2)=4.00(0.60)(1.09091)(0.728013)=1.906\ \text{mg/L}.

For pathway 2, the rate factor is \(0.20/(0.20-0.10)=2.00\), and \(e^{-0.20}-e^{-0.40}=0.148411\). Thus

C_2(2)=4.00(0.40)(2.00)(0.148411)=0.4749\ \text{mg/L}.

Adding the contributions gives \(C(2)=1.906+0.4749=2.381\ \text{mg/L}\). The faster pathway contributes more at this time for these parameter values, but that is a result of the particular fractions and rates, not a universal rule.

Worked Example: Checking Total Exposure With AUC

Consider a hypothetical dose \(D=120\) mg with \(F=0.75\), \(V=30\) L, \(k=0.15\ \text{h}^{-1}\), and pathway fractions \(f=0.25\) and \(1-f=0.75\). Let \(k_{a1}=0.90\ \text{h}^{-1}\) and \(k_{a2}=0.30\ \text{h}^{-1}\). Both absorption rates are positive and distinct from \(k\), and both pathways are completely absorbed according to the model.

Integrating the contribution from one pathway over time gives

\int_0^\infty C_i(t)\,dt =\frac{F D_i}{V}\frac{k_{ai}}{k_{ai}-k} \left(\frac{1}{k}-\frac{1}{k_{ai}}\right) =\frac{F D_i}{Vk}.

The equality follows because \(\frac{k_{ai}}{k_{ai}-k}(\frac{1}{k}-\frac{1}{k_{ai}})=1/k\). Thus the two pathway AUCs are proportional to the allocated doses: \(fFD/(Vk)=0.25(0.75)(120)/(30(0.15))=5.00\ \text{mg}\cdot\text{h/L}\), and \(0.75FD/(Vk)=15.0\ \text{mg}\cdot\text{h/L}\). The total is \(20.0\ \text{mg}\cdot\text{h/L}\). Equivalently, \(FD/(Vk)=90\ \text{mg}/(4.50\ \text{L/h})=20.0\ \text{mg}\cdot\text{h/L}\).

This check shows that, with complete absorption and common linear disposition, changing the absorption rates changes the time course but not the total AUC. The fractions determine how that AUC is divided between the pathway contributions.

Worked Example: When an Absorption Rate Equals Elimination

Suppose pathway 1 has \(k_{a1}=k=0.20\ \text{h}^{-1}\), while pathway 2 has \(k_{a2}=0.50\ \text{h}^{-1}\). The two absorption rates remain distinct. Let \(FD/V=5.00\ \text{mg/L}\), \(f=0.30\), and evaluate at \(t=3.00\) h. The formula with denominator \(k_{a1}-k\) cannot be used directly for pathway 1 because that denominator is zero.

Taking the limit as \(k_{a1}\) approaches \(k\), or integrating its input directly, gives the pathway term \(k t e^{-kt}\). Therefore pathway 1 contributes

C_1(3)=5.00(0.30)(0.20)(3.00)e^{-0.60} =0.4939\ \text{mg/L}.

For pathway 2, the rate factor is \(0.50/(0.50-0.20)=1.66667\), and \(e^{-0.60}-e^{-1.50}=0.325681\). Its contribution is

C_2(3)=5.00(0.70)(1.66667)(0.325681) =1.8998\ \text{mg/L}.

The total is \(0.4939+1.8998=2.394\ \text{mg/L}\). The limiting expression avoids a division by zero and gives a finite, positive contribution.

Interpretation, Checks, and Limitations

Several checks help prevent errors when applying the profile:

Check the starting value. At \(t=0\), each difference of exponentials is zero, so \(C(0)=0\) for an immediate first-order input with no initial central amount.
Check units. Each rate constant has units \(\text{h}^{-1}\), each exponential argument is dimensionless, and \(FD/V\) has units of concentration. The whole expression therefore has concentration units.
Check signs. For positive rates, the expression for a contribution remains nonnegative. If \(k_{ai}<k\), both \(k_{ai}-k\) and \(e^{-kt}-e^{-k_{ai}t}\) are negative at \(t>0\), so their ratio is positive.
Check limiting dose allocations. If \(f=0\), pathway 1 contributes nothing; if \(f=1\), pathway 2 contributes nothing. The model reduces to a single first-order pathway in either case.
Check total AUC. The sum is \(FD/(Vk)\), independent of the pathway rates, under the stated assumptions. A different result usually signals an algebraic error or a change in assumptions.

The concentration peak is not generally the peak of either pathway considered alone. Because the total is the sum of two time-varying contributions, its peak time must be found from the derivative of the total profile (or by numerical evaluation); it is not generally obtained by averaging the two absorption rates. Similarly, the two-pathway curve does not usually reduce to a single first-order absorption curve with one “average” absorption rate. A single-pathway model has only one absorption timescale, whereas the parallel model can combine a relatively rapid and a relatively slow input.

The model's additivity depends on its assumptions. If pathways have different bioavailabilities, each contribution needs its own bioavailability factor. If disposition is nonlinear, the concentration response to the sum of inputs need not equal the sum of separate concentration responses. The profile alone also does not guarantee that the pathway parameters can be estimated distinctly; that is an inference question, not a reason to change the derivation.

Common Mistakes

Using the same absorption rate for both pathways. This discards the defining feature of the model when the pathways have distinct rates. Keep \(k_{a1}\) and \(k_{a2}\) in their respective terms.
Forgetting the dose fraction. The first term must be scaled by \(f\), and the second by \(1-f\). Omitting either factor implicitly assigns the full dose to that pathway.
Subtracting exponentials in the wrong order. The rate denominator and exponential difference must occur together as \((e^{-kt}-e^{-k_{ai}t})/(k_{ai}-k)\). Changing only one sign can produce an impossible negative concentration.
Dividing by zero when \(k_{ai}=k\). The quotient has a finite limit, \(k t e^{-kt}\); use that limit rather than the unsimplified expression.
Assuming the faster pathway always dominates concentration. Its contribution also depends on the fraction allocated to it and on time. Compare the full terms at the time of interest.

Key takeaway: For two parallel first-order pathways entering the same linear central compartment, derive each contribution with its own fraction and absorption rate, then add them. The result preserves total AUC under complete absorption and common linear disposition, while allowing the concentration-time shape to reflect two distinct absorption timescales.

Check Your Understanding

Answer each question using the assumptions and equations developed above. Solutions follow the questions.

If \(f=0\), which pathway contributes to concentration, and why?
Why does changing \(k_{a1}\) alter the concentration-time profile but not total AUC under the stated assumptions?
At \(t=0\), what concentration does the model predict?
For a pathway with \(k_{ai}=k\), what expression replaces the rate-dependent term in the profile?
Why is replacing \(k_{a1}\) and \(k_{a2}\) by their arithmetic mean generally not equivalent to the two-pathway model?

Solutions. (1) Pathway 1 contributes zero because its dose fraction is \(f=0\); pathway 2 receives the entire dose. (2) The integrated contribution from pathway \(i\) is \(FD_i/(Vk)\), which contains no \(k_{ai}\), so changing that rate redistributes concentration over time without changing its integrated exposure. (3) \(C(0)=0\), since both exponential differences are zero at \(t=0\). (4) The term \(k_{ai}(e^{-kt}-e^{-k_{ai}t})/(k_{ai}-k)\) is replaced by \(k t e^{-kt}\). (5) The two contributions contain different exponentials and are weighted by separate dose fractions; their sum generally cannot be represented by one exponential absorption timescale.