Parallel Zero- and First-Order Pathways

Why Combine These Pathways?

A concentration-time profile can reflect more than one absorption process. Some of a dose may enter the systemic circulation gradually at a nearly constant rate for a finite period, while the rest enters according to first-order absorption, with its input rate declining over time. A combined model can represent this mixture without treating either pathway as if it were the whole dose.

The earlier material on parallel absorption pathways establishes the dose-allocation idea: fractions assigned to parallel pathways must account for the intended dose. Here, the focus is the specific combination of a finite zero-order input and a first-order input, and how their separate concentration contributions add up.

Definition: Let \(D\) be the administered dose, \(F\) the common bioavailability, \(f_0\) the fraction assigned to finite zero-order absorption, and \(f_1\) the fraction assigned to first-order absorption, with \(f_0+f_1=1\). The zero-order pathway delivers its systemic dose \(Ff_0D\) uniformly over duration \(T\); the first-order pathway absorbs its systemic dose \(Ff_1D\) with absorption rate constant \(k_a\).

Assume a one-compartment disposition model with volume \(V\) and first-order elimination rate constant \(k_e\). Absorption begins immediately at time zero, the zero-order pathway stops at \(T\), and both pathways contribute to the same central compartment. The model is linear, so the concentration from the combined input is the sum of the concentrations from the two inputs. This requires dose-proportional disposition over the range considered and no saturation or interaction between pathways.

Input Functions and Concentration Contributions

The constant zero-order input rate is systemic dose divided by delivery duration. The first-order pathway’s input rate is largest at time zero and decays exponentially. Their total input rate is therefore piecewise: the zero-order rate is present only through \(T\), while the first-order rate continues after that time.

R_0(t)= \begin{cases} \dfrac{Ff_0D}{T}, & 0\leq t<T,\\ 0, & t\geq T, \end{cases} \qquad R_1(t)=Ff_1Dk_a e^{-k_at},\quad t\geq 0.

For any systemic input rate \(R(t)\), the amount in the central compartment at time \(t\) is the input accumulated up to that time, with each earlier increment reduced by elimination. Thus \(C(t)\) is the convolution of the input with the elimination response \(e^{-k_e(t-s)}\), divided by \(V\). Applying this result to each pathway gives the following contributions.

C_0(t)= \begin{cases} \dfrac{Ff_0D}{TVk_e}\left(1-e^{-k_et}\right), & 0\leq t\leq T,\\[6pt] \dfrac{Ff_0D}{TVk_e}\left(1-e^{-k_eT}\right)e^{-k_e(t-T)}, & t>T, \end{cases}

C_1(t)= \begin{cases} \dfrac{Ff_1Dk_a}{V(k_a-k_e)} \left(e^{-k_et}-e^{-k_at}\right), & k_a\ne k_e,\\[6pt] \dfrac{Ff_1Dk_a}{V}\,t e^{-k_at}, & k_a=k_e. \end{cases} \qquad C(t)=C_0(t)+C_1(t).

The zero-order concentration rises while input continues: input accumulates in the compartment even as elimination removes some of it. Once input stops, that pathway’s concentration declines by elimination. The first-order contribution rises initially when absorption exceeds elimination, then eventually declines. The total profile reflects both contributions; it need not look like either component alone.

Key distinction: “Zero-order” describes the constant absorption input rate during \(0\leq t<T\), not the elimination process. Elimination remains first-order in this model, including after the zero-order input ends.

Worked Calculations

Worked Example: Contributions at Two Times

Suppose a hypothetical 300 mg dose has \(F=0.80\), with \(f_0=0.40\) assigned to zero-order absorption over \(T=4\) h and \(f_1=0.60\) assigned to first-order absorption with \(k_a=0.75\) h\(^{-1}\). Let \(V=30\) L and \(k_e=0.15\) h\(^{-1}\). Calculate the pathway inputs and concentrations at 2 h and 4 h.

The systemic dose is \(FD=0.80(300)=240\) mg. The pathway doses are \(Ff_0D=96\) mg and \(Ff_1D=144\) mg. Thus the zero-order input rate is \(96/4=24\) mg/h, and the first-order input rate is \(144(0.75)e^{-0.75t}=108e^{-0.75t}\) mg/h.

At \(t=2\) h, the first-order input is \(108e^{-1.5}=24.10\) mg/h. The zero-order concentration is \(C_0(2)=\{24/[30(0.15)]\}(1-e^{-0.15(2)})=5.3333(1-e^{-0.3})=1.382\) mg/L. The first-order contribution is \(C_1(2)=\{144(0.75)/[30(0.75-0.15)]\}(e^{-0.3}-e^{-1.5})=6(0.740818-0.223130)=3.106\) mg/L. Therefore, \(C(2)=1.382+3.106=4.488\) mg/L.

At \(t=4\) h, the zero-order concentration has reached \(C_0(4)=5.3333(1-e^{-0.6})=2.406\) mg/L. The first-order contribution is \(C_1(4)=6(e^{-0.6}-e^{-3})=6(0.548812-0.049787)=2.994\) mg/L. The total is \(C(4)=2.406+2.994=5.400\) mg/L. The first-order input rate at 4 h is \(108e^{-3}=5.377\) mg/h; immediately before the zero-order input ends, the total input is \(24+5.377=29.38\) mg/h.

Worked Example: A Peak at the End of Zero-Order Input

Use the parameters from the preceding example to determine whether the concentration is rising or falling at the end of the zero-order input. For a central amount \(A=VC\), the amount balance is \(dA/dt=R_0+R_1-k_eA\), so \(dC/dt=(R_0+R_1-k_eVC)/V\).

Just before \(T=4\) h, \(C=5.400\) mg/L and the total input rate is 29.38 mg/h. The elimination rate is \(k_eVC=0.15(30)(5.400)=24.30\) mg/h. Hence \(dC/dt=(29.38-24.30)/30=0.169\) mg/L/h: concentration is still rising just before the zero-order input stops.

Immediately after \(T\), the zero-order rate is zero, while the first-order rate is still 5.377 mg/h. The concentration is continuous at \(T\), but its derivative becomes \(dC/dt=(5.377-24.30)/30=-0.631\) mg/L/h. In this example the concentration therefore reaches a peak at \(T\). To check that it was rising throughout the preceding interval, its derivative there can be simplified to \(4.5e^{-0.75t}-0.1e^{-0.15t}\) mg/L/h. It is positive for \(0\leq t\leq4\), since \(45e^{-0.60t}\geq45e^{-2.4}>1\). After \(T\), the zero-order component declines; the first-order component is also declining because its peak time is \(\ln(k_a/k_e)/(k_a-k_e)=\ln(5)/0.60=2.682\) h, earlier than \(T\). Thus both components decline after \(T\), confirming the peak.

Worked Example: Equal Absorption and Elimination Rates

Consider a hypothetical 100 mg dose with \(F=0.90\), \(f_0=f_1=0.50\), zero-order duration \(T=2\) h, \(V=20\) L, and \(k_a=k_e=0.20\) h\(^{-1}\). Find the total concentration at \(t=3\) h.

Each pathway delivers \(0.90(0.50)(100)=45\) mg systemically. The zero-order rate is \(45/2=22.5\) mg/h. Since \(t>T\), \(C_0(3)=\{22.5/[20(0.20)]\}(1-e^{-0.20(2)})e^{-0.20(3-2)}\) \(=5.625(1-e^{-0.4})e^{-0.2}=1.518\) mg/L.

The equal-rate expression gives \(C_1(3)=\{45(0.20)/20\}(3)e^{-0.20(3)}=0.45(3)(0.548812)=0.741\) mg/L. Therefore \(C(3)=1.518+0.741=2.259\) mg/L. At this time the zero-order input has ended, while the first-order input remains \(45(0.20)e^{-0.6}=4.939\) mg/h.

Why the Equal-Rate Formula Is Well-Behaved

The expression for \(C_1(t)\) when \(k_a\ne k_e\) appears to have a zero denominator when \(k_a=k_e\). This is a removable limit, not a failure of the model. To see why, let \(k_a\) approach \(k_e=k\). The difference of exponentials in the numerator approaches zero at the same rate as \(k_a-k\).

e^{-kt}-e^{-(k+\delta)t} =e^{-kt}(1-e^{-\delta t}) \approx e^{-kt}\delta t \quad\text{as }\delta\to0,

where \(\delta=k_a-k\) and \(1-e^{-\delta t}\approx\delta t\) for small \(\delta\). Substituting this approximation into the unequal-rate expression, \(k_a/(k_a-k)\) times the exponential difference approaches \(k t e^{-kt}\). Multiplying by \(Ff_1D/V\) yields the equal-rate formula. This limit also shows why direct numerical evaluation near equal rates can lose precision: it subtracts two nearly equal exponentials before dividing by a small difference.

Assumptions, Checks, and Modeling Pitfalls

The formulas depend on specific assumptions: dose fractions are nonnegative and sum to one; \(F\), \(V\), \(k_e\), and \(k_a\) are positive; zero-order input is uniform for the stated duration; first-order input starts at time zero; and disposition is linear and shared by both pathways. If pathway-specific bioavailability is needed, each pathway’s systemic dose must be written separately rather than silently using a common \(F\). A delayed start, nonlinear elimination, or changing input rate requires a different input specification.

A practical check is dose accounting. Over all time, the zero-order input integrates to \(Ff_0D\), and the first-order input integrates to \(Ff_1D\):

\int_0^T \frac{Ff_0D}{T}\,dt=Ff_0D, \qquad \int_0^\infty Ff_1Dk_ae^{-k_at}\,dt=Ff_1D.

These amounts sum to \(FD\), not \(D\), when \(F<1\). A second check is continuity: both component concentrations, and therefore the total concentration, remain continuous at \(T\). The derivative can change abruptly there because the zero-order input rate drops from \(Ff_0D/T\) to zero.

Putting the whole dose on both pathways. This counts the administered dose twice. Allocate fractions with \(f_0+f_1=1\), then multiply each allocated fraction by \(FD\).
Using a zero-order input beyond its duration. The zero-order rate must become zero after \(T\); only the concentration already in the compartment remains and is eliminated.
Calling the end of input the peak by default. Concentration may still be rising or may already be falling at \(T\). Compare total input with elimination, \(R_0+R_1\) with \(k_eVC\), on both sides of the switch.
Ignoring near-equal rates in computation. When \(k_a\) is very close to \(k_e\), use the limiting expression or a numerically stable equivalent rather than subtracting nearly equal exponentials without care.

Compared with a single zero-order model, the combined model can represent an early or ongoing first-order input in addition to the finite constant input. Compared with two parallel first-order pathways, it assigns one pathway a flat input rate with a defined stopping time. A sequential model instead places one process after another; that is not the same as two pathways operating in parallel. The combined model is still a simplification: a good fit does not establish that these are the unique biological mechanisms producing the profile.

Key takeaway: Calculate each pathway’s systemic input and concentration separately, then add the concentrations under linear disposition. Interpret the shape using the total input rate and elimination rate, with special attention to the instant the finite zero-order input stops.

Check Your Understanding

Answer each question, then compare with the solutions below.

A 200 mg dose has \(F=0.75\), \(f_0=0.30\), and zero-order duration 3 h. What is the zero-order systemic input rate?
For \(f_0=0.35\), what first-order fraction \(f_1\) completes the dose allocation?
At the end of zero-order input, what determines whether total concentration is rising or falling?
Why does the first-order concentration formula require a special expression when \(k_a=k_e\)?
Does a continuous concentration at the end of zero-order input imply a continuous concentration slope? Explain.

Solutions. (1) The systemic zero-order amount is \(Ff_0D=0.75(0.30)(200)=45\) mg, so the rate is \(45/3=15.0\) mg/h. (2) \(f_1=1-f_0=1-0.35=0.65\). (3) Compare the total input \(R_0+R_1\) with elimination \(k_eVC\): concentration rises if input exceeds elimination and falls if input is smaller. (4) The unequal-rate expression has \(k_a-k_e\) in its denominator, but its exponential difference also tends to zero; taking the limit gives \(Ff_1Dk_at e^{-k_at}/V\). (5) No. The concentration is continuous because the amount already in the compartment does not jump, but the input rate changes abruptly at the endpoint, so its derivative can change abruptly too.