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One-, Two- and Three-Compartment Structural Models · Tutorial 38 of 797

Three-Compartment Microconstants and Exchange Pathways

Learn to read each microconstant as a directional, amount-based flow rate and use the five constants to calculate elimination, exchange, and compartment amount changes.

Beginner 10 min read

What You'll Learn

  • Identify central elimination and the four directional exchange rate constants.
  • Interpret each rate constant using its donor compartment and amount.
  • Calculate instantaneous amount flows from rate constants and compartment amounts.
  • Derive the three-compartment amount-balance equations from the pathways.
  • Check units, signs, and total amount balance in worked calculations.
  • Distinguish pathway-specific microconstants from system-level exponential rates.

From Compartments to Directional Amount Flows

A three-compartment model is not defined just by having three amounts. Its structure also specifies which compartments exchange amount and where elimination occurs. In the mammillary structure, compartment 1 is central and connects separately to peripheral compartments 2 and 3. Elimination occurs from the central compartment; there is no direct exchange pathway between compartments 2 and 3.

Each pathway is described by a first-order rate constant. The constant gives the fraction of the amount in its donor compartment that flows out per unit time. Thus, a rate constant alone is not an amount flow: to calculate the flow, multiply it by the amount currently in the donor compartment. This distinction is essential when interpreting the model or checking its amount-balance equations.

Definition: Let \(A_1\), \(A_2\), and \(A_3\) denote the amounts of drug in the central compartment and two peripheral compartments. The central elimination constant is \(k_{10}\). The four directional exchange constants are \(k_{12}\) (central to peripheral 2), \(k_{21}\) (peripheral 2 to central), \(k_{13}\) (central to peripheral 3), and \(k_{31}\) (peripheral 3 to central). A constant \(k_{ij}\) describes movement from compartment \(i\) to compartment \(j\), at flow rate \(k_{ij}A_i\). Each constant has units of inverse time; each flow has units of amount per time.

The subscripts encode direction: the first index identifies the compartment supplying the amount, and the second identifies its destination. For elimination, the 0 in \(k_{10}\) denotes the outside of the modeled system, not another compartment with an amount. The amount eliminated per unit time is \(k_{10}A_1\).

Writing the Amount-Balance Equations

An amount-balance equation adds all flows entering a compartment and subtracts all flows leaving it. For the central compartment, elimination and both outward exchange pathways remove amount; the two reverse pathways return amount. Peripheral compartment 2 receives amount from compartment 1 and returns it to compartment 1, with the same structure applying to compartment 3.

$$ \begin{aligned} \frac{dA_1}{dt} &= -(k_{10}+k_{12}+k_{13})A_1+k_{21}A_2+k_{31}A_3+R_{\mathrm{in}}(t),\\ \frac{dA_2}{dt} &= k_{12}A_1-k_{21}A_2,\\ \frac{dA_3}{dt} &= k_{13}A_1-k_{31}A_3. \end{aligned} $$

Here \(R_{\mathrm{in}}(t)\) is an input rate into the central compartment, such as an infusion, measured in amount per time. When there is no ongoing input, set \(R_{\mathrm{in}}(t)=0\). These equations assume that each pathway follows linear first-order kinetics over the modeled conditions, that the rate constants are fixed, and that the stated pathways are the only pathways for exchange and elimination. The constants are nonnegative; ordinarily, each modeled pathway has a positive constant.

The equations follow directly from the pathway definitions. For example, \(k_{12}A_1\) leaves compartment 1 and enters compartment 2, while \(k_{21}A_2\) leaves compartment 2 and returns to compartment 1. The same flow must appear with opposite signs in the two compartments it connects. It would violate amount balance to subtract \(k_{12}A_1\) from the central equation but omit it from the peripheral-2 equation.

Key takeaway: A microconstant is tied to a particular direction and donor amount. The instantaneous flow for \(i\) to \(j\) is \(k_{ij}A_i\), not \(k_{ij}\) by itself. The model has one elimination pathway, \(k_{10}A_1\), and four exchange pathways.

Worked Flow Calculations

Worked Example: Calculating All Five Pathway Flows

Consider a hypothetical model at one instant. Let \(A_1=100\) mg, \(A_2=30\) mg, and \(A_3=20\) mg. Suppose \(k_{10}=0.12\) h\(^{-1}\), \(k_{12}=0.40\) h\(^{-1}\), \(k_{21}=0.20\) h\(^{-1}\), \(k_{13}=0.15\) h\(^{-1}\), and \(k_{31}=0.05\) h\(^{-1}\). There is no infusion at this instant.

Multiply each constant by the amount in its donor compartment:

  • Elimination from compartment 1: \(k_{10}A_1=(0.12\ \text{h}^{-1})(100\ \text{mg})=12\) mg/h.
  • Exchange from 1 to 2: \(k_{12}A_1=(0.40)(100)=40\) mg/h.
  • Exchange from 2 to 1: \(k_{21}A_2=(0.20)(30)=6\) mg/h.
  • Exchange from 1 to 3: \(k_{13}A_1=(0.15)(100)=15\) mg/h.
  • Exchange from 3 to 1: \(k_{31}A_3=(0.05)(20)=1\) mg/h.

The amount derivatives are therefore \(\frac{dA_1}{dt}=-(12+40+15)+6+1=-60\) mg/h, \(\frac{dA_2}{dt}=40-6=34\) mg/h, and \(\frac{dA_3}{dt}=15-1=14\) mg/h. At this instant the central amount is decreasing, while both peripheral amounts are increasing. These are instantaneous rates, not amounts that will necessarily change by exactly these values over an entire hour; the amounts, and hence the flows, can change over time.

A short balance check verifies that the signs and pathways are consistent. Add the three derivatives. Each exchange flow appears once as an outflow and once as an inflow, so those terms cancel:

$$ \frac{d(A_1+A_2+A_3)}{dt} = -k_{10}A_1 + R_{\mathrm{in}}(t). $$

In the example, the total derivative is \(-60+34+14=-12\) mg/h, which equals the negative elimination flow. Exchange redistributes amount among compartments but does not change the total amount in the modeled system. With no input, only elimination reduces that total.

Worked Example: Opposite Directions Need Not Have Equal Flows

Suppose a hypothetical subject has \(A_1=80\) mg in the central compartment and \(A_2=50\) mg in peripheral compartment 2. Let \(k_{12}=0.30\) h\(^{-1}\) and \(k_{21}=0.10\) h\(^{-1}\). Calculate the flows in both directions and the net contribution of this pair to \(A_2\).

The flow from 1 to 2 is \((0.30\ \text{h}^{-1})(80\ \text{mg})=24\) mg/h. The reverse flow is \((0.10\ \text{h}^{-1})(50\ \text{mg})=5\) mg/h. Hence the net change in \(A_2\) due to this pair is \(24-5=19\) mg/h into compartment 2; its derivative contribution is \(+19\) mg/h. The corresponding net contribution to \(A_1\) is \(-19\) mg/h.

Although \(k_{12}\) is three times \(k_{21}\), the net flow is not found by subtracting the constants. The two constants multiply different donor amounts, so both amounts are needed to calculate the directional flows.

Worked Example: Equal Exchange Flows Do Not Mean No Exchange

For a hypothetical central-to-peripheral-3 pair, let \(A_1=50\) mg, \(A_3=100\) mg, \(k_{13}=0.20\) h\(^{-1}\), and \(k_{31}=0.10\) h\(^{-1}\). The forward flow is \((0.20)(50)=10\) mg/h, and the reverse flow is \((0.10)(100)=10\) mg/h. The net exchange contribution to each amount is zero at this instant.

There is still continuous movement in both directions; it is the two flows that balance, not the flows themselves disappearing. If \(k_{10}=0.08\) h\(^{-1}\), elimination is \((0.08)(50)=4\) mg/h. With no input, compartment 1 is therefore losing 4 mg/h overall, while compartment 3 has zero instantaneous net change. Equal exchange flows do not imply that the entire model is at steady state.

Reading the Constants Without Confusing Their Roles

The five constants describe pathway-specific behavior, not the amount currently moving. For fixed amounts, doubling \(k_{12}\) doubles the instantaneous flow from compartment 1 to compartment 2. But a large \(k_{12}\) does not by itself establish that this flow exceeds the reverse flow: the reverse flow depends on \(k_{21}A_2\). Nor does a particular \(k_{12}\) describe the whole model’s observed time course by itself. The three-compartment equations jointly determine how amounts evolve.

This is also why a directional exchange constant should not be mistaken for a net transfer rate. For the pair connecting compartments 1 and 2, the net contribution to \(A_2\) is \(k_{12}A_1-k_{21}A_2\). It can be positive, negative, or zero, depending on both amounts and both constants. The same reasoning applies to the pair connecting compartments 1 and 3.

The structural arrangement is important. In a mammillary model, compartments 2 and 3 do not exchange directly. Their amounts can still influence the central amount through the return flows \(k_{21}A_2\) and \(k_{31}A_3\), and each can receive amount from the central compartment. Adding a direct flow between compartments 2 and 3 would define a different model and require additional terms and parameters.

Derivation, Diagnostics, and Common Pitfalls

The amount equations are obtained by applying one rule to every pathway: add a flow to its destination and subtract it from its source. For example, \(k_{13}A_1\) contributes \(-k_{13}A_1\) to \(dA_1/dt\) and \(+k_{13}A_1\) to \(dA_3/dt\). Summing all three equations cancels each exchange term, leaving only input and elimination. This cancellation is a direct justification of the total-amount balance, not an extra assumption about exchange.

Several checks can reveal a mistaken interpretation or equation:

  • Check units. Multiplying an inverse-time constant by an amount must give amount per time. A term such as \(k_{12}A_2\) has the wrong donor amount for the 1-to-2 pathway.
  • Check the direction and sign. A flow leaving a compartment is negative in that compartment’s balance; the same flow is positive in its destination’s balance.
  • Check the total balance. With no input, the sum of amount derivatives must be \(-k_{10}A_1\), not a quantity that includes uncancelled exchange terms.
  • Check the structure. A direct \(k_{23}\) or \(k_{32}\) term is not part of the mammillary model specified here.
  • Check the assumptions. If rates change with concentration, or if an additional elimination or exchange pathway is present, the five-constant linear model does not represent those processes as written.

A common mistake is to say that \(k_{12}=0.30\) h\(^{-1}\) means that 0.30 mg/h moves from compartment 1 to compartment 2. That treats a rate constant as a flow and ignores the donor amount. The correction is to calculate \(k_{12}A_1\); for example, with \(A_1=80\) mg, the flow is \(24\) mg/h. A second mistake is to assume that the larger of \(k_{12}\) and \(k_{21}\) identifies the net direction. The correction is to compare \(k_{12}A_1\) with \(k_{21}A_2\), because the constants act on different amounts.

The microconstants should also be distinguished from the exponential rates used to describe a fitted concentration-time curve. The microconstants label the model’s individual pathways; the exponential rates arise from the combined system of amount-balance equations. In contrast to estimating pathway flows directly from a single amount snapshot, describing the observed curve with exponential terms summarizes the system’s joint behavior. Those summaries do not change the directional meaning of \(k_{10}\), \(k_{12}\), \(k_{21}\), \(k_{13}\), or \(k_{31}\).

The model is useful because its constants have interpretable pathway roles and its equations preserve amount across exchange. Its limitations follow from the same simplicity: first-order rates are assumed, elimination is restricted to the central compartment, and the two peripheral compartments connect only through the central one. A good fit to data does not by itself prove those structural assumptions; unusual residual patterns or systematic disagreement with observed profiles can indicate that the chosen structure or kinetic assumptions may be inadequate.

Check Your Understanding

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

  1. If \(A_1=40\) mg and \(k_{13}=0.25\) h\(^{-1}\), what is the instantaneous flow from compartment 1 to compartment 3?
  2. What does \(k_{31}\) represent, and which amount is multiplied by it to calculate its flow?
  3. If exchange flows are \(k_{12}A_1=8\) mg/h and \(k_{21}A_2=11\) mg/h, what is the net contribution to \(dA_2/dt\) from this pair?
  4. With no input, what should the sum \(dA_1/dt+dA_2/dt+dA_3/dt\) equal?
  5. Why is it not enough to know that \(k_{12}>k_{21}\) to conclude that amount is moving net from compartment 1 to compartment 2?

Solutions. (1) \(k_{13}A_1=(0.25\ \text{h}^{-1})(40\ \text{mg})=10\) mg/h. (2) \(k_{31}\) represents flow from peripheral compartment 3 into the central compartment; its flow is \(k_{31}A_3\), using the donor amount \(A_3\). (3) The net contribution is \(8-11=-3\) mg/h, so this pair causes a net loss of 3 mg/h from compartment 2. (4) It equals \(-k_{10}A_1\), because all exchange flows cancel in the sum. (5) The constants multiply different donor amounts. The direction of net exchange depends on comparing \(k_{12}A_1\) with \(k_{21}A_2\), not on comparing the constants alone.