Choosing a Structure for a Biphasic Profile
A concentration profile with an early rise and a later shoulder, second rise, or slow decline may be consistent with more than one absorption structure. A fast and a slow first-order pathway can produce a broad or two-component peak. An immediate pathway combined with a lagged pathway can produce early and delayed input. A first-order pathway combined with a zero-order pathway can produce prompt input alongside input spread over time. The task is not to name a mechanism from the curve alone; it is to compare defensible candidate structures against the observations while respecting how the dose is allocated.
The earlier material on parallel absorption pathways establishes the component-profile idea: the predicted concentration is the sum of the contributions from the pathways. This workshop uses that idea to make a practical comparison. The examples use small, hypothetical data sets and illustrative model predictions; they are not results from a real study.
For comparison, let \(S=FD/V\) denote the concentration scale associated with dose \(D\), bioavailability \(F\), and apparent distribution volume \(V\). A first-order absorption contribution with rate \(k_a\), elimination rate \(k_e\), and no lag is
Here \(k_a\) and \(k_e\) have units of inverse time, and \(A(t;k_a)\) has concentration units. If the rates are equal, the continuous limiting expression is \(A(t;k_a)=S k_a t e^{-k_a t}\). A lagged version is \(A((t-\tau)_+;k_a)\), where \((t-\tau)_+=\max(t-\tau,0)\) and \(\tau\geq0\); it contributes no absorbed-dose concentration before the lag has elapsed.
Three Candidate Structures
The structures differ in which input features they allow to vary. A fast–slow model gives each pathway a first-order absorption rate. An immediate–lagged model uses a first-order profile for an immediate pathway and a time-shifted profile for the other. A first-order–zero-order model combines a first-order profile with input occurring at a constant rate over a specified duration. In each case, a fraction weights the contribution of each component.
In these expressions \(0\leq f\leq1\), \(k_{a,f}>k_{a,s}>0\) for the fast–slow candidate, and \(Z(t)\) is the concentration contribution per unit fraction from the specified zero-order input. The shared fraction \(f\) is a convenient way to express a complete allocation between two pathways. If only a fraction \(F\) of the administered dose is available to these pathways, the corresponding absolute bioavailable fractions are \(Ff\) and \(F(1-f)\), not two unrelated fractions that may sum to more than \(F\).
A candidate is meaningful only with its assumptions stated. These examples assume a known dose history, the same disposition scale and elimination rate across the pathways being compared, nonnegative pathway inputs, and concentration observations made at the listed times. They also assume the candidate profile equations are suitable for the route and formulation under consideration. A visually biphasic curve does not establish any of these assumptions; they are modeling choices to be checked against study design and pharmacologic knowledge.
Worked Example: Comparing Three Candidate Predictions
Suppose hypothetical concentrations are observed at 0.5, 1, 2, and 4 hours. The following values are illustrative predictions from three candidate structures evaluated at their fitted parameter settings. Compare their residual sums of squares, using observed minus predicted concentration as the residual.
| Time (h) | Observed (mg/L) | Fast–slow (mg/L) | Immediate–lagged (mg/L) | First-order–zero-order (mg/L) |
|---|---|---|---|---|
| 0.5 | 4.800 | 4.613 | 3.878 | 5.604 |
| 1 | 6.300 | 6.173 | 4.936 | 8.034 |
| 2 | 7.000 | 6.473 | 7.366 | 7.790 |
| 4 | 5.100 | 5.074 | 5.368 | 5.718 |
For the fast–slow predictions, the residuals are \(0.187,\ 0.127,\ 0.527,\) and \(0.026\) mg/L. Thus,
For the immediate–lagged predictions, the residuals are \(0.922,\ 1.364,\ -0.366,\) and \(-0.268\) mg/L, giving \(0.922^2+1.364^2+(-0.366)^2+(-0.268)^2=2.91636\;(\mathrm{mg/L})^2\). For the first-order–zero-order predictions, they are \(-0.804,\ -1.734,\ -0.790,\) and \(-0.618\) mg/L. Using the displayed rounded predictions gives \(0.804^2+1.734^2+0.790^2+0.618^2=4.659196\;(\mathrm{mg/L})^2\); calculations from unrounded predictions give approximately \(4.659\;(\mathrm{mg/L})^2\).
On this residual criterion, the fast–slow candidate fits these four points best. That is a numerical ranking of these particular fits, not proof that the real input consisted of a fast and a slow pathway. It is also not a fair complexity-adjusted comparison unless the parameter counts, constraints, and fitting procedures are taken into account.
Estimating and Constraining a Pathway Fraction
When the component predictions are fixed, a pathway fraction can be estimated directly by least squares. Write the two-component prediction at observation \(i\) as \(C_i(f)=S_i+fD_i\), where \(S_i\) is the second pathway’s contribution and \(D_i=F_i-S_i\) is the difference between the first and second pathway predictions. For observed concentration \(y_i\), minimize \(\sum_i[y_i-S_i-fD_i]^2\). Differentiating with respect to \(f\), setting the derivative to zero, and solving gives the unconstrained minimizer below.
This expression follows because the derivative of the sum of squared residuals is \(-2\sum_iD_i(y_i-S_i-fD_i)\). Equating it to zero yields \(\sum_iD_i(y_i-S_i)=f\sum_iD_i^2\). If \(\sum_iD_i^2>0\), division gives the stated result. For a two-pathway complete allocation, the admissible fraction is \(0\leq f\leq1\). If the unconstrained solution falls outside this interval, the constrained least-squares solution is the nearest endpoint, 0 or 1. This is a boundary constraint, not evidence that a negative or greater-than-one dose fraction is biologically meaningful.
Worked Example: Checking a Fitted Fraction Against Its Bounds
Consider hypothetical observed concentrations \(y=(4.800,6.300,7.000,5.100)\) mg/L. For the fast and slow components, suppose the slow-pathway predictions are \(S=(2.101,3.537,5.041,5.233)\) mg/L and the fast-minus-slow differences are \(D=(3.865,4.057,2.203,-0.244)\) mg/L. These are component predictions at the same four observation times.
The numerator is \(3.865(2.699)+4.057(2.763)+2.203(1.959)+(-0.244)(-0.133)\), because \(y-S=(2.699,2.763,1.959,-0.133)\) mg/L. The displayed products sum to \(25.989255\). The denominator is \(3.865^2+4.057^2+2.203^2+(-0.244)^2=36.310219\). Therefore,
The estimate is within \([0,1]\), so it is admissible as a dose fraction: approximately 0.716 of the modeled dose goes to the fast component, and 0.284 goes to the slow component. The fractions sum to 1, and both are nonnegative. This calculation estimates the fraction conditional on the supplied component predictions; it does not show that the component shapes or rates are uniquely determined by the data.
Checking the First-Order–Zero-Order Calculation
A component-based calculation is useful for checking whether a fitted prediction actually respects its stated pathway fraction. In the following hypothetical parameterization, the first-order contribution per unit fraction is \(A(t)\), and the zero-order contribution per unit fraction is \(Z(t)\). The zero-order component has duration 2 hours and the specified scale gives, after that input ends, \(Z(t)=50(1-e^{-0.2})e^{-0.1(t-2)}\) mg/L. Thus its contributions at 2 and 4 hours are calculated from the endpoint value and the subsequent elimination factor. These component values are part of this illustrative parameterization; they should not be transferred to a different dose scale or duration.
Worked Example: Verifying the Two-Hour and Four-Hour Predictions
Set \(f=0.70\). The zero-order component has duration 2 hours, so its contribution at 2 hours is its endpoint value, \(Z(2)=50(1-e^{-0.2})=9.0635\) mg/L. At 4 hours, it has undergone 2 hours of elimination:
The first-order component values at those observation times are \(A(2)=7.2441\) mg/L and \(A(4)=4.9880\) mg/L. Combining the components with the stated fraction gives
These values show why component contributions and the pathway fraction should be checked before calculating residuals. At the four observed times, the first-order–zero-order predictions in the comparison table give residuals \(-0.804,\ -1.734,\ -0.790,\) and \(-0.618\) mg/L, with an SSR of approximately \(4.659\;(\mathrm{mg/L})^2\) using the unrounded predictions. A small arithmetic error in one component can alter both the residual pattern and the model ranking.
Diagnostics, Alternatives, and Limits
A sum of squared residuals is a transparent first comparison when candidate predictions are evaluated at the same observations and on the same concentration scale. It is not a complete selection rule. The candidates may have different numbers of parameters, different boundary constraints, or different abilities to fit a sparse set of points. When an appropriate likelihood and observation-error model are specified, likelihood-based criteria can account for fit and complexity; the details depend on the model and design. A lower SSR by itself does not justify adding pathways.
Inspect residuals against time as well as their sum. A run of positive residuals followed by negative residuals suggests systematic profile mismatch; a single large residual may dominate SSR; and a pattern concentrated near the peak may indicate that the predicted timing or shape is wrong. Check whether fitted pathway fractions remain within their allowed bounds and whether the lag, rates, and duration are scientifically plausible. Parameters that move substantially when one time point is omitted, or candidate predictions that are nearly indistinguishable over the sampled times, signal weak information in the data. Sparse early samples are particularly limiting for distinguishing rapid input from an immediate contribution.
The main alternatives are not only these three structures. A single first-order input may be adequate if the apparent second phase is noise or sampling variation; a transit or other delayed-input structure may be preferable when the design and mechanism support it. Comparing plausible alternatives is more informative than selecting a complex model solely because its curve can reproduce a shoulder. No profile-shape comparison alone establishes a physical absorption mechanism, and estimates can be sensitive to dose history, sampling schedule, assay error, and assumptions about disposition.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- If two pathways represent the full modeled dose and one has fraction \(f=0.63\), what is the other fraction, and are both admissible?
- Why can a smaller SSR fail to establish that one pathway structure is biologically correct?
- For observations \(y=(5.0,7.0)\), slow-component predictions \(S=(2.0,4.0)\), and fast-minus-slow differences \(D=(2.0,1.0)\), calculate the unconstrained least-squares fraction.
- Give one residual pattern that would suggest a candidate profile is systematically inadequate.
- At 4 hours, an observed concentration is 5.100 mg/L and a candidate predicts 4.984 mg/L. State the residual using observed minus predicted, with units.
Solutions. (1) The other fraction is \(1-0.63=0.37\); both are nonnegative and sum to 1. (2) SSR does not by itself account for parameter count, uncertainty, or mechanistic nonuniqueness; similar profiles may arise from different structures. (3) Here \(y-S=(3,3)\), so \(\hat f=[2(3)+1(3)]/[2^2+1^2]=9/5=1.8\). This is outside \([0,1]\), so the constrained estimate is \(f=1\). (4) For example, residuals that are consistently positive early and negative late indicate a time-dependent mismatch. (5) The residual is \(5.100-4.984=+0.116\) mg/L.