Estimating Pathway Fractions and Absorption Rates

Why Estimating Two Pathways Can Be Difficult

A concentration profile after an oral dose may reflect drug entering the systemic circulation through more than one first-order pathway. The pathways can carry different fractions of the dose and have different absorption rates. Estimating these quantities is not just a matter of reading two slopes from a graph: the observed concentration is the sum of overlapping pathway profiles, and the parameters must usually be fitted together.

This tutorial considers a one-compartment model with first-order elimination and two parallel first-order absorption pathways. The dose fractions are $f$ and $1-f$, where $f$ is the fraction assigned to pathway 1. The total concentration scale $S=FD/V$, the elimination rate constant $k_e$, and the sampling times are treated as known in the examples. This isolates the estimation of the pathway fractions and absorption rates; when the scale or $k_e$ is also unknown, their joint estimation can add further uncertainty.

Definition: For a dose divided between two first-order absorption pathways, the concentration at time $t$ is $$ C(t)=S\left[fq(t;k_{a1})+(1-f)q(t;k_{a2})\right], \qquad q(t;k_a)=\frac{k_a}{k_a-k_e}\left(e^{-k_e t}-e^{-k_a t}\right). $$ Here $k_{a1}$ and $k_{a2}$ are the absorption rate constants, $k_e$ is the elimination rate constant, $S=FD/V$ is the concentration scale, and $0\leq f\leq 1$. Rate constants have units of inverse time; $q$ and $f$ are dimensionless. When $k_a=k_e$, the continuous limiting form is $q(t;k_e)=k_e t e^{-k_e t}$.

The model assumes linear disposition, first-order absorption and elimination, a common concentration scale for the two pathways, and fractions that sum to one. It also assumes that the recorded concentrations correspond to the specified dose and times. For least-squares estimation, a usual working error model is that measurement errors are independent, have mean zero, and have a common variance on the concentration scale. If those assumptions are unsuitable, the fitting criterion must reflect an appropriate error model instead.

From the Pathway Model to an Estimation Strategy

The pathway function $q(t;k_a)$ is the concentration shape produced by a unit fraction of the dose at absorption rate $k_a$. It follows by integrating the first-order absorption input against the one-compartment elimination response:

q(t;k_a)=\int_0^t k_a e^{-k_a s}e^{-k_e(t-s)}\,ds =e^{-k_e t}k_a\int_0^t e^{-(k_a-k_e)s}\,ds =\frac{k_a}{k_a-k_e}\left(e^{-k_e t}-e^{-k_a t}\right).

Multiplying each shape by its dose fraction and the common scale, then adding the contributions, gives $C(t)$. This derivation also shows why one observed concentration generally cannot determine all three unknowns $f,k_{a1},k_{a2}$: each observed value is a mixture of both pathway contributions.

With several observed concentrations $C_i$ at times $t_i$, a standard approach is nonlinear least squares. For known $S$ and $k_e$, choose $f,k_{a1},k_{a2}$ to minimize

\operatorname{SSE}(f,k_{a1},k_{a2}) =\sum_{i=1}^{n}\left\{C_i-S\left[fq(t_i;k_{a1})+(1-f)q(t_i;k_{a2})\right]\right\}^{2}.

The parameters are constrained by $0\leq f\leq1$ and positive absorption rates. A numerical optimizer evaluates the predicted profile repeatedly and searches for parameter values that make the predictions close to the observations. If the concentration errors have different variances, weighted least squares can instead minimize a sum of squared residuals divided by their variances.

There is an exact label symmetry: exchanging $k_{a1}$ and $k_{a2}$ while replacing $f$ by $1-f$ leaves every prediction unchanged. To report a unique labeling, one can impose $k_{a1}>k_{a2}$, call pathway 1 the faster pathway, and interpret $f$ as the faster-pathway fraction. This convention does not change the model or add information to the data.

Worked Example: Recovering a Fraction When the Rates Are Known

Suppose a hypothetical profile has $S=10.0\ \text{mg/L}$ and $k_e=0.20\ \text{h}^{-1}$. Consider $t=1.00\ \text{h}$, with known rates $k_{a1}=1.00\ \text{h}^{-1}$ and $k_{a2}=0.25\ \text{h}^{-1}$. The observed concentration is $4.180\ \text{mg/L}$. Estimate the fraction $f$ assigned to the faster pathway.

The pathway shapes are

q(1;1.00)=\frac{1.00}{1.00-0.20}(e^{-0.20}-e^{-1.00})=0.5636,

and

q(1;0.25)=\frac{0.25}{0.25-0.20}(e^{-0.20}-e^{-0.25}) =5(e^{-0.20}-e^{-0.25})=0.1996.

Rearrange the mixture equation to solve for $f$:

f=\frac{C(1)/S-q(1;k_{a2})}{q(1;k_{a1})-q(1;k_{a2})} =\frac{4.180/10.0-0.1996}{0.5636-0.1996} \approx 0.600.

Thus, under the stated model and known values, about 60.0% of the dose is assigned to the faster pathway. Using the more precise shapes, the model prediction at $f=0.600$ is approximately $10[0.600(0.563564)+0.400(0.199650)]=4.180\ \text{mg/L}$. With the rounded shapes displayed above, the arithmetic is $10[0.600(0.5636)+0.400(0.1996)]=4.180\ \text{mg/L}$ to the shown precision.

Estimating Both Rates From a Profile

When both absorption rates are unknown, the fractions cannot generally be estimated one time point at a time. The rates change the shape of each pathway over the entire profile, so the estimates must be obtained by fitting all observation times together. Useful information comes from the rise, peak region, and decline, provided the sampling times cover those features well enough.

Worked Example: Fitting a Two-Pathway Profile

Consider hypothetical concentrations generated using $S=10.0\ \text{mg/L}$, $k_e=0.20\ \text{h}^{-1}$, a faster-pathway fraction $f=0.600$, and rates $k_{a1}=1.00\ \text{h}^{-1}$ and $k_{a2}=0.25\ \text{h}^{-1}$. The values below are rounded to three decimal places.

Time (h)	Observed concentration (mg/L)
1	4.180
2	5.288
4	4.862
8	2.843

For example, at $t=2\ \text{h}$, the two shapes are $q(2;1.00)=1.25(e^{-0.40}-e^{-2.00})=0.6687$ and $q(2;0.25)=5(e^{-0.40}-e^{-0.50})=0.3189$. The predicted concentration is therefore

10.0\left[0.600(0.6687)+0.400(0.3189)\right] =5.288\ \text{mg/L}

to three decimal places. Applying the same formula at 1, 4, and 8 hours gives predictions of approximately $4.180$, $4.862$, and $2.843\ \text{mg/L}$, respectively. A constrained nonlinear least-squares fit to the four rounded observations returns estimates close to $f=0.600$, $k_{a1}=1.00\ \text{h}^{-1}$, and $k_{a2}=0.25\ \text{h}^{-1}$. The parameter values used to generate the profile give an SSE of about $2.0\times10^{-7}\ (\text{mg/L})^2$; this small nonzero value arises because the observations were rounded.

There are four observations for three fitted parameters, so the profile has some information beyond simply matching one point per parameter. That count alone does not guarantee that the parameters are well determined: the observation times and the distinctness of the two rate effects matter. In real analysis, estimated parameters should be accompanied by uncertainty summaries and assessed alongside the residual pattern, rather than judged only by the closeness of the fitted curve.

Why Similar Rates Cause Confounding

If the absorption rates are equal, the two pathway shapes are identical. The mixture then reduces exactly to one pathway:

$$ f q(t;k_a)+(1-f)q(t;k_a)=q(t;k_a). $$

Consequently, no concentration profile from this model can determine $f$ when the rates are exactly equal. When they are merely close, changing the fraction has only a small effect on the predicted profile. The data may then estimate a combined or average absorption shape reasonably well while providing little information about how the dose is divided.

A short expansion explains this weak identification. Let the rates be close to a common value $k$, write $k_{a1}=k+\delta_1$ and $k_{a2}=k+\delta_2$, and expand the smooth function $q(t;k_a)$ around $k$. The weighted first-order change is proportional to $f\delta_1+(1-f)\delta_2$. If this weighted change is held fixed, the profile changes only at higher order as the rates approach one another. At equal rates, the fraction disappears from the model altogether. Thus an optimizer may return precise-looking point estimates while a wide range of fractions and nearby rates fit almost equally well.

Worked Example: Two Nearly Indistinguishable Rates

At $t=1.00\ \text{h}$, let $k_e=0.20\ \text{h}^{-1}$, $S=10.0\ \text{mg/L}$, and $f=0.600$. Compare rates of $0.50$ and $0.45\ \text{h}^{-1}$ with a single pathway at $0.48\ \text{h}^{-1}$. The shapes for the two pathways are

q(1;0.50)=\frac{0.50}{0.30}(e^{-0.20}-e^{-0.50})=0.3537, \qquad q(1;0.45)=\frac{0.45}{0.25}(e^{-0.20}-e^{-0.45})=0.3260.

Their mixture shape is $0.600(0.3537)+0.400(0.3260)=0.3426$, giving a concentration of about $3.426\ \text{mg/L}$. For the single rate, $q(1;0.48)=\frac{0.48}{0.28}(e^{-0.20}-e^{-0.48})=0.3428$, giving about $3.428\ \text{mg/L}$. The difference at this time is only about $0.0017\ \text{mg/L}$. A full, well-sampled profile can provide more discrimination than this single comparison, but the example illustrates why close rates make fraction estimates sensitive to small changes in data or model assumptions.

Diagnostics, Alternatives, and Limitations

After fitting, inspect residuals $C_i-\widehat{C}(t_i)$ against time and against predicted concentration. A systematic run of positive or negative residuals, or a consistent mismatch around the peak, suggests that the fitted model is not capturing the observed profile. Very large uncertainty, strong dependence between estimated $f$ and the rates, estimates near a boundary, or substantially different solutions from different starting values are warnings that the parameters may be weakly identified or that numerical optimization has found a local rather than a useful minimum.

A practical check is to fit from several plausible starting values and compare the resulting SSE and parameter estimates. If substantially different parameter combinations have nearly identical SSE, the profile does not strongly distinguish those combinations. A profile-likelihood or other uncertainty analysis can make that ambiguity more visible than a single point estimate. Good residuals do not establish that the two pathways are biologically real; they show only that the fitted curve is compatible with these data and assumptions.

The main alternatives answer different questions. A single first-order absorption pathway uses one rate and can be more stable when the data do not support separating two rates, but it may miss a reproducible profile shape. A two-pathway model is more flexible, at the cost of additional parameters and possible confounding. Grid evaluation over candidate rates can help visualize a flat fitting region; numerical optimization is more efficient for locating a minimum but does not, by itself, establish that the minimum is unique or well identified.

These estimates also depend on the assumed concentration scale and elimination rate. If either is misspecified or is estimated imprecisely, the absorption parameter estimates can shift. Sparse early samples may fail to characterize the rising portion, while limited late observations can make it difficult to separate absorption from elimination effects. Finally, this model assumes both pathways are first order and share the same disposition and scale; systematic lack of fit may indicate that those assumptions are inadequate rather than that the optimizer needs more tuning.

Common mistake: interpreting a fitted fraction as well determined because the curve fits. A good fit does not imply that the two rates and fraction are separately identifiable. Check whether different parameter combinations produce almost the same predictions.
Common mistake: assigning pathway labels without a convention. Swapping the rates and replacing $f$ with $1-f$ gives the same model. State which pathway is called pathway 1, such as the faster pathway.
Common mistake: estimating the fraction from one concentration when the rates are unknown. At one time, changes in the rates can mimic changes in the fraction. Fit the full profile and state which quantities were treated as known.

Key takeaway: A two-pathway profile can estimate dose fractions and absorption rates only when the data contain enough information to distinguish the pathway shapes. If the rates are equal, the fraction is unidentifiable; if they are close, expect weak separation and report that limitation.

Check Your Understanding

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

With $S$ and $k_e$ known, what profile features help estimate two absorption rates rather than just a dose fraction?
What happens to the concentration model when $k_{a1}=k_{a2}$?
Why is it useful to impose $k_{a1}>k_{a2}$ when reporting fitted parameters?
A two-pathway fit has small residuals, but many different fractions and nearby rates give nearly the same SSE. What should be concluded?

Solutions. (1) Observations across the rising, peak, and declining portions can reveal the distinct time-course shapes; one concentration generally cannot identify all three parameters. (2) The mixture becomes $S q(t;k_a)$, independent of $f$, so the fraction cannot be estimated from the profile. (3) The model is unchanged by swapping the two rates and replacing $f$ with $1-f$; ordering the rates makes the labels consistent. (4) The fit may describe the data well, but the separate fraction and rates are weakly identified, so report their uncertainty and avoid treating the point estimates as uniquely determined.