Combining First- and Zero-Order Absorption

Why Combine the Two Input Processes?

An oral dose need not enter the body through a single absorption process. A fraction may be absorbed gradually at a rate proportional to the amount remaining in an absorption depot, while another fraction enters at a constant rate for a specified duration. When these inputs occur at the same time, their contributions to concentration add under linear disposition. The resulting profile can therefore differ from either a purely first-order or a purely zero-order profile.

This model is distinct from sequential zero-order and first-order absorption: here, both pathways begin at the same dosing time and operate concurrently. The central bookkeeping requirement is to allocate the administered dose consistently. If the two fractions represent the entire dose, they must sum to one.

Definition: Let \(D\) be the administered dose. Let \(f_1\) be the fraction assigned to first-order absorption and \(f_0\) the fraction assigned to zero-order absorption. The dose is fully partitioned when \(f_1\geq 0\), \(f_0\geq 0\), and \(f_1+f_0=1\). With a common bioavailability \(F\), the systemic amounts delivered by the two pathways are \(Ff_1D\) and \(Ff_0D\), respectively.

Assumptions and Concentration Profile

Assume a single dose is given at time zero, there is no lag time, and both pathways start immediately. The first-order pathway releases drug from an absorption depot with rate constant \(k_a\). The zero-order pathway supplies drug at a constant rate for duration \(T\). After reaching the systemic compartment, drug follows one-compartment disposition with volume \(V\) and first-order elimination rate constant \(k\). Bioavailability \(F\) is the same for both pathways. The model also assumes linear disposition: concentrations from separate inputs can be added.

For the first-order pathway, the systemic input rate at time \(t\) is \(Ff_1Dk_a e^{-k_at}\). The zero-order systemic input rate is \(Ff_0D/T\) from time zero through \(T\), and zero thereafter. These rates have units of amount per time. With \(V\) in volume units and rate constants in inverse-time units, the concentration contributions have units of amount per volume.

C(t)=\frac{FD}{V}\left[ f_1\frac{k_a}{k_a-k}\left(e^{-kt}-e^{-k_at}\right) + \begin{cases} \displaystyle \frac{f_0}{kT}\left(1-e^{-kt}\right), & 0\leq t\leq T,\\[6pt] \displaystyle \frac{f_0}{kT}\left(1-e^{-kT}\right)e^{-k(t-T)}, & t>T. \end{cases} \right].

The expression assumes \(k_a\ne k\). If \(k_a=k\), the first-order contribution is obtained by taking its limit, giving \(f_1 k_a t e^{-kt}\) inside the brackets. The zero-order term changes at \(T\) because input stops then, but the concentration does not jump: drug already in the body continues to be eliminated.

Deriving the Contributions

Under one-compartment disposition, an amount entering the systemic circulation at time \(s\) contributes a fraction \(e^{-k(t-s)}\) of that amount to the central amount at a later time \(t\). Dividing by \(V\) converts amount to concentration. Integrating each input rate over its active period gives its concentration contribution.

C_1(t)=\frac{Ff_1D}{V}\int_0^t k_a e^{-k_as}e^{-k(t-s)}\,ds =\frac{Ff_1D}{V}\frac{k_a}{k_a-k}\left(e^{-kt}-e^{-k_at}\right).

For zero-order input during its active period, the input rate is \(Ff_0D/T\). Integrating from zero to \(t\) gives the concentration while \(t\leq T\). After \(T\), the same input has acted only from zero to \(T\), and the resulting concentration decays from the end of input.

C_0(t)= \begin{cases} \displaystyle \frac{Ff_0D}{VT}\int_0^t e^{-k(t-s)}\,ds =\frac{Ff_0D}{VkT}\left(1-e^{-kt}\right), & t\leq T,\\[8pt] \displaystyle \frac{Ff_0D}{VT}\int_0^T e^{-k(t-s)}\,ds =\frac{Ff_0D}{VkT}\left(1-e^{-kT}\right)e^{-k(t-T)}, & t>T. \end{cases}

Adding \(C_1(t)\) and \(C_0(t)\) gives the stated total profile. The two zero-order branches agree at \(t=T\): substituting \(T\) into the first gives a factor \(1-e^{-kT}\), and substituting \(T\) into the second gives \((1-e^{-kT})e^0\). Thus the formula is continuous at the stopping time.

Worked Example: Concentration During and After Zero-Order Input

Suppose a hypothetical 100 mg oral dose has \(F=0.80\), \(V=20\) L, \(k=0.10\ \mathrm{h}^{-1}\), \(k_a=0.80\ \mathrm{h}^{-1}\), and \(T=4\) h. Let \(f_1=0.40\) and \(f_0=0.60\), so the fractions sum to one. Calculate the predicted concentration at 2 h and 6 h.

The common multiplier is \(FD/V=(0.80)(100)/20=4.00\ \mathrm{mg/L}\). At 2 h, zero-order input is still active:

C(2)=4\left[ 0.40\frac{0.80}{0.80-0.10}(e^{-0.20}-e^{-1.60}) +\frac{0.60}{(0.10)(4)}(1-e^{-0.20}) \right].

Using \(e^{-0.20}=0.8187\) and \(e^{-1.60}=0.2019\), the first-order term in brackets is \(0.2820\), and the zero-order term is \(0.2719\). Therefore, \(C(2)=4(0.5539)=2.216\ \mathrm{mg/L}\).

At 6 h, zero-order input has stopped, so its residual contribution includes decay from \(T=4\) h:

C(6)=4\left[ 0.40\frac{0.80}{0.70}(e^{-0.60}-e^{-4.80}) +\frac{0.60}{0.40}(1-e^{-0.40})e^{-0.20} \right].

The first-order term in brackets is \(0.2471\), and the zero-order term is \(0.4049\). Thus \(C(6)=4(0.6520)=2.608\ \mathrm{mg/L}\). The later concentration is higher in this example because the zero-order pathway continued supplying drug until 4 h; concentration need not peak before that input ends.

Dose Accounting, AUC, and a Second Check

With a common \(F\), the total systemic amount delivered is \(Ff_1D+Ff_0D=FD\). Under linear one-compartment elimination, the total area under the concentration-time curve after a single dose is therefore

\mathrm{AUC}_{0-\infty} =\frac{Ff_1D}{kV}+\frac{Ff_0D}{kV} =\frac{FD}{kV} =\frac{FD}{CL}, \qquad CL=kV.

The result does not depend on how the dose is divided, provided the full dose is allocated, the common bioavailability applies, and the model's linear disposition assumptions hold. The pathway split changes the timing and shape of the profile, not the total AUC in this model.

Worked Example: Verifying Systemic Amounts and AUC

For a hypothetical \(D=120\) mg dose, let \(F=0.75\), \(f_1=0.25\), \(f_0=0.75\), \(V=30\) L, \(k=0.15\ \mathrm{h}^{-1}\), and \(T=3\) h. Confirm the partition, calculate each systemic amount and the zero-order input rate, then calculate total AUC.

The fractions satisfy \(0.25+0.75=1\). The first-order systemic amount is \((0.75)(0.25)(120)=22.5\) mg. The zero-order systemic amount is \((0.75)(0.75)(120)=67.5\) mg. Their sum, 90.0 mg, equals \(FD=(0.75)(120)\).

The zero-order input rate is \(67.5/3=22.5\ \mathrm{mg/h}\). Clearance is \(CL=kV=(0.15)(30)=4.50\ \mathrm{L/h}\). Thus total AUC is \(90.0/4.50=20.0\ \mathrm{mg\cdot h/L}\). The pathway AUC contributions are \(22.5/4.50=5.00\ \mathrm{mg\cdot h/L}\) for first-order input and \(67.5/4.50=15.0\ \mathrm{mg\cdot h/L}\) for zero-order input; together they give \(20.0\ \mathrm{mg\cdot h/L}\).

Worked Example: Recovering a Dose Fraction From a Known Input Rate

In a hypothetical formulation, \(D=200\) mg, \(F=0.60\), and the zero-order input lasts \(T=5\) h. Suppose the specified constant rate is the systemic input rate and equals \(12\ \mathrm{mg/h}\). Find \(f_0\), \(f_1\), and the systemic amount assigned to each pathway.

The zero-order systemic amount is rate times duration: \((12\ \mathrm{mg/h})(5\ \mathrm{h})=60\) mg. Since this amount is \(Ff_0D\),

f_0=\frac{60}{(0.60)(200)}=0.50,\qquad f_1=1-f_0=0.50.

The first-order systemic amount is also \((0.60)(0.50)(200)=60\) mg. The two systemic amounts total 120 mg, equal to \(FD\), as required. This calculation would need adjustment if the stated rate were a release rate before bioavailability rather than a systemic input rate.

Diagnostics, Alternatives, and Limitations

Before interpreting a fitted profile, check the dose fractions, units, and input timing. Fractions outside zero to one or a sum other than one contradict a full-dose partition under this model. A zero-order rate should have units of amount per time, and \(T\) should have units of time. Verify that the zero-order branch is selected at the correct time and that its two expressions agree at \(T\). A discontinuity there usually signals an implementation or coding error.

Systematic residual patterns can indicate that the assumed input shape is inadequate. For example, consistent early underprediction and later overprediction may suggest that the fixed input mixture does not capture the observed timing. Such patterns do not uniquely identify the cause: misspecified disposition, inaccurate dose records, or sampling limitations can also contribute. A fitted fraction near zero or one may indicate that the data provide little support for one pathway, but it is not by itself proof that the pathway is absent. Estimates of \(f_0\), \(f_1\), \(T\), and \(k_a\) can trade off in their effects on the profile, especially when observations do not cover both the input period and the later decline.

The table contrasts this simultaneous mixture with simpler input choices. Each is appropriate only when its assumptions are plausible for the formulation and data.

Input model	Input timing and shape	What it can represent
First-order only	All allocated dose enters through an exponential absorption process	Gradual input that slows as the depot is depleted
Zero-order only	All allocated dose enters at a constant rate for a specified duration	Approximately constant-rate release over a finite interval
Sequential zero-order then first-order	The input processes act in sequence	A formulation with a staged rather than simultaneous input process
Combined first- and zero-order	Two dose fractions enter concurrently by the two specified processes	A mixture of gradual exponential and finite-duration constant-rate input

Important limitations follow directly from the assumptions. The model has one zero-order duration and one first-order absorption rate; real input may vary over time or across individuals. A common \(F\) makes the pathway fractions a partition of the administered dose with a total systemic amount \(FD\). If pathway-specific bioavailabilities are needed, the systemic amounts are not represented by this common-\(F\) model, and the simple total-AUC calculation based on \(FD\) no longer follows. The model also assumes one-compartment linear disposition, so it should not be treated as a universal profile formula.

Common mistake: If \(f_1=0.35\), it is correct that \(f_0=0.65\). It is not correct to report the zero-order systemic amount as \(0.52D\) unless a value \(F=0.80\) has been specified. With unspecified common bioavailability, the amount is \(F(0.65)D=0.65FD\). Keep the administered dose fractions separate from the systemic amounts.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

If \(f_1=0.35\) and the two pathways receive the whole dose, find \(f_0\) and express the zero-order systemic amount when \(F\) is unspecified.
A hypothetical 80 mg dose has \(F=0.75\) and \(f_0=0.40\). What systemic amount is assigned to zero-order input?
If the same zero-order systemic amount is supplied over twice the duration, how does its constant input rate change?
For \(D=150\) mg, \(F=0.80\), \(k=0.20\ \mathrm{h}^{-1}\), and \(V=25\) L, calculate total AUC under the model.
Why should the predicted concentration be continuous at \(t=T\), even though zero-order input stops at that time?

Solutions. 1. \(f_0=1-0.35=0.65\). The zero-order systemic amount is \(Ff_0D=0.65FD\); without a specified \(F\), it cannot be reduced to a numerical multiple of \(D\). 2. The amount is \(Ff_0D=(0.75)(0.40)(80)=24.0\) mg. 3. Rate equals amount divided by duration. Doubling the duration while keeping the amount fixed halves the rate. 4. \(CL=kV=(0.20)(25)=5.00\ \mathrm{L/h}\), and \(FD=(0.80)(150)=120\) mg, so \(\mathrm{AUC}_{0-\infty}=120/5.00=24.0\ \mathrm{mg\cdot h/L}\). 5. Stopping input prevents additional drug from entering through that route, but it does not instantaneously remove drug already in the body. The post-input expression starts with the same accumulated zero-order contribution at \(T\) and then decays, so the total concentration is continuous.