Parallel Lagged and Immediate Absorption Pathways

Why Split a Dose Between Immediate and Delayed Input?

A single oral dose can produce an early concentration rise and a later contribution to the profile if some of the dose enters the body promptly while the rest becomes available only after a delay. A useful structural model represents these as parallel pathways: one fraction is input immediately, and the complementary fraction begins first-order absorption after a fixed lag. The model makes the timing assumptions explicit and lets us calculate how changing the delayed fraction redistributes concentrations.

The fixed lag is a deterministic waiting time, not an absorption rate. After the lag, the delayed portion is absorbed gradually at rate constant \(k_a\). By contrast, the immediate pathway enters the central compartment at time zero. The distinction matters: a fixed-lag pathway does not imply that the entire delayed fraction appears in plasma at the lag time.

Definition: Let \(D\) be a single dose, \(f\) the fraction assigned to the delayed pathway, and \(1-f\) the fraction assigned to the immediate pathway, where \(0\leq f\leq1\). The immediate fraction enters the central compartment at time zero. The delayed fraction begins first-order absorption at fixed lag \(T_L\). Let \(V\) be the apparent central volume, \(k_e\) the first-order elimination rate constant, and \(k_a\) the delayed-pathway absorption rate constant. Time and \(T_L\) use the same units, and \(k_a\) and \(k_e\) have inverse-time units.

For the calculations here, assume complete systemic availability, so the fractions sum to the administered dose. Disposition is one-compartment and linear, elimination is first-order, and the immediate input is an instantaneous central-compartment dose. There is a single dose, no other input, and no concentration-dependent absorption or elimination. These assumptions isolate the consequence of splitting the dose by input timing; they are not claims that every formulation follows this model.

Concentration Profile and Its Derivation

The immediate fraction contributes a bolus concentration that declines by elimination from time zero. For \(t\leq T_L\), the delayed pathway has not yet begun, so the profile consists only of that immediate contribution. For \(t>T_L\), delayed drug absorbed at each time after the lag is eliminated between its absorption time and the observation time.

C(t)= \begin{cases} \dfrac{(1-f)D}{V}e^{-k_et}, & 0\leq t\leq T_L,\\[6pt] \dfrac{(1-f)D}{V}e^{-k_et} +\dfrac{fD}{V}\dfrac{k_a}{k_a-k_e} \left[e^{-k_e(t-T_L)}-e^{-k_a(t-T_L)}\right], & t>T_L,\; k_a\ne k_e. \end{cases}

The delayed contribution follows by integrating over absorption times \(s\). At \(s\geq T_L\), its input rate is \(fDk_a e^{-k_a(s-T_L)}\); drug entering at \(s\) contributes a factor \(e^{-k_e(t-s)}\) to concentration at time \(t\). Thus

C_{\mathrm{delayed}}(t) =\frac{1}{V}\int_{T_L}^{t} fDk_a e^{-k_a(s-T_L)}e^{-k_e(t-s)}\,ds =\frac{fD}{V}\frac{k_a}{k_a-k_e} \left[e^{-k_e(t-T_L)}-e^{-k_a(t-T_L)}\right].

This expression applies for \(t>T_L\). At \(t=T_L\), the delayed contribution is zero, so concentration is continuous across the lag. If \(k_a=k_e=k\), the displayed ratio has a finite limit; the delayed contribution becomes \((fD/V)k(t-T_L)e^{-k(t-T_L)}\). The two rate constants may therefore be equal, although the ordinary formula must then be evaluated using this limit rather than by dividing by zero.

Key interpretation: Before the lag, increasing \(f\) reduces concentration because it assigns less dose to immediate input. After the lag, the delayed pathway adds concentration, but the net effect of increasing \(f\) depends on time: it replaces immediate input with delayed input rather than simply adding another dose.

Worked Concentrations With a Fixed Lag

Worked Example: Concentrations Before and After the Lag

Consider a hypothetical single dose of \(D=100\) mg, with \(f=0.30\), \(V=20\) L, \(k_e=0.20\ \mathrm{h}^{-1}\), \(k_a=1.00\ \mathrm{h}^{-1}\), and \(T_L=2.00\) h. Assume the model conditions stated above. The immediate fraction is \(0.70\), corresponding to 70 mg, and the delayed fraction is 30 mg.

At \(t=1\) h, the lag has not elapsed. Therefore only the immediate fraction contributes:

C(1)=\frac{0.70(100\ \mathrm{mg})}{20\ \mathrm{L}}e^{-(0.20\ \mathrm{h}^{-1})(1\ \mathrm{h})} =3.5\ \mathrm{mg/L}\times0.8187308 =2.866\ \mathrm{mg/L}.

At \(t=2\) h, the delayed pathway has just begun and has contributed no absorbed amount yet. Thus \(C(2)=3.5e^{-0.4}=2.346\ \mathrm{mg/L}\). There is no concentration jump at the lag.

At \(t=3\) h, the time since the lag is \(1\) h. The immediate contribution is \(3.5e^{-0.6}=1.920841\ \mathrm{mg/L}\). The delayed contribution is

\frac{30\ \mathrm{mg}}{20\ \mathrm{L}}\frac{1.00}{1.00-0.20} \left(e^{-0.20(1)}-e^{-1.00(1)}\right) =1.875(0.4508513) =0.845346\ \mathrm{mg/L}.

Using the unrounded components, \(C(3)=1.9208407+0.8453462=2.7661869\ \mathrm{mg/L}\), or 2.766 mg/L. The concentration is higher at 3 h than it would be if the full dose had entered immediately: that comparison profile is \(5e^{-0.6}=2.744\ \mathrm{mg/L}\). The delayed input has shifted some concentration later, despite reducing the initial concentration.

How the Delayed Fraction Changes the Profile

At fixed \(D,V,k_e,k_a,\) and \(T_L\), the concentration is linear in the pathway fraction \(f\). Before the lag, \(C(t)=(1-f)(D/V)e^{-k_et}\), which decreases as \(f\) increases. After the lag, the profile is a weighted combination of the full-dose immediate response and the full-dose delayed response. Consequently, the delayed fraction does not have one fixed direction of effect at all times.

Worked Example: Comparing Two Delayed Fractions

Use the same hypothetical dose and kinetic parameters as above, but compare \(f=0.30\) with \(f=0.60\). At 1 h, both profiles are before the 2 h lag. With \(f=0.60\), the immediate fraction is \(0.40\), so

C(1)=\frac{0.40(100)}{20}e^{-0.20} =2(0.8187308) =1.637\ \mathrm{mg/L}.

This is lower than the \(2.866\ \mathrm{mg/L}\) at \(f=0.30\), because the larger delayed fraction leaves less dose in the immediate pathway.

At 3 h, the immediate contribution for \(f=0.60\) is \(2e^{-0.6}=1.097623\ \mathrm{mg/L}\). The delayed contribution is

\frac{0.60(100)}{20}\frac{1.00}{0.80}(0.4508513) =3.75(0.4508513) =1.690692\ \mathrm{mg/L}.

The total is \(2.7883157\), or 2.788 mg/L, slightly above the \(2.766\ \mathrm{mg/L}\) for \(f=0.30\). At 6 h, the time since the lag is 4 h. The immediate and delayed contributions are \(2e^{-1.2}=0.602388\) and \(3.75(e^{-0.8}-e^{-4})=1.616300\ \mathrm{mg/L}\), respectively. Thus \(C(6)=2.218688\), or 2.219 mg/L, compared with \(1.862\ \mathrm{mg/L}\) for \(f=0.30\). In this example the larger delayed fraction gives lower early concentration but higher concentrations at the later comparison times.

These comparisons are specific to the selected parameters and times. They do not imply that every increase in \(f\) raises concentration at every time after the lag.

Assumptions, Checks, and Alternatives

Before applying the equations, check that \(f\) is a fraction between zero and one, that \(T_L\geq0\), and that rate constants and time units are consistent. The delayed absorption rate is defined after \(T_L\); treating the lag as though absorption proceeds during the waiting period changes the model. Under the stated model, the delayed contribution starts at zero, rises as absorption supplies drug, and eventually declines as elimination dominates. The full profile is the sum of that contribution and the declining immediate component.

A fixed-lag parallel model is not automatically supported merely because a profile has a late concentration contribution. If observations before the lag are absent, the immediate fraction and the timing of input may be difficult to distinguish from other parameter choices. Sparse samples near the lag can also leave the onset time weakly informed. A substantial observed rise before the specified lag, a sharp jump in concentration at the lag, or a late profile inconsistent with the assumed first-order decline would conflict with this model's structure. Such patterns call for reconsidering the assumptions or input representation, not silently altering the formula.

Input model	Input timing	What it represents
Immediate-only	All dose enters at time zero	A single bolus input; no delayed fraction
Fixed-lag-only first-order	Absorption starts at \(T_L\)	A delayed pathway without an immediate component
Parallel immediate and fixed-lag	One fraction at zero; the remainder after \(T_L\)	A dose divided between prompt and delayed input
Transit absorption	Input is distributed over a sequence of transit stages	A distributed delay rather than one deterministic lag

The parallel model is useful when it is scientifically reasonable to represent two distinct input components and a fixed onset time is plausible. Its simplicity is also a limitation: real delay may vary across doses or individuals, and first-order absorption may not describe the post-lag input. A transit representation instead spreads the delay, while a single immediate or single delayed pathway removes one component. These alternatives encode different assumptions; they are not interchangeable labels for the same profile.

Common Mistake

A frequent error is to treat the delayed fraction as an additional dose on top of the immediate dose. That would use \(D\) for the immediate input and then add \(fD\) later, effectively giving a total dose of \((1+f)D\). The correct allocation is \((1-f)D\) immediately and \(fD\) through the lagged pathway, which sums to \(D\). Another error is to add the delayed contribution before \(T_L\); it is exactly zero before the lag. Always check both dose accounting and the time condition before calculating a concentration.

Key takeaway: The delayed fraction shifts dose from the immediate component to a first-order input that begins at \(T_L\). Its effect is therefore time-dependent: it reduces pre-lag concentrations, while the later contribution can make concentrations exceed those from a larger immediate fraction.

Check Your Understanding

Answer the questions using the model assumptions and parameters stated in each problem. Full solutions follow.

For \(D=80\) mg, \(f=0.25\), \(V=16\) L, \(k_e=0.10\ \mathrm{h}^{-1}\), and \(T_L=3\) h, what is \(C(2)\)?
At \(t=T_L\), is the delayed-pathway concentration contribution positive, zero, or discontinuous? Explain.
For a dose of 120 mg with \(f=0.40\), how many milligrams are assigned to each pathway?
Why can a larger delayed fraction lower an early concentration but raise a later one?

Solutions. (1) At 2 h the lag has not elapsed, so \(C(2)=[(1-0.25)80/16]e^{-0.10(2)}=3.75(0.8187308)=3.070\ \mathrm{mg/L}\). (2) The contribution is zero and concentration is continuous: the delayed expression has the difference \(e^0-e^0=0\) at the lag. (3) The immediate amount is \((1-0.40)120=72\) mg, and the delayed amount is \(0.40(120)=48\) mg; they sum to 120 mg. (4) Increasing the delayed fraction reduces the dose entering immediately, lowering the pre-lag contribution. After the lag, it also increases the delayed input; depending on the time and kinetic parameters, that later contribution can exceed the immediate input that was displaced.