Compartmental Accounting for IgG Disposition

Why Account for Antibody Pools Separately?

An observed decline in plasma IgG does not, by itself, reveal where the antibody went. It may have moved into tissues, entered cells, returned from an intracellular pathway, or been irreversibly degraded. A useful pharmacokinetic model must distinguish these possibilities: distribution changes location, whereas elimination removes antibody from the tracked system.

This tutorial builds a minimal accounting model for those processes. It is a bookkeeping framework, not yet a detailed model of FcRn binding or target-mediated disposition. Each compartment represents an amount of antibody, and each arrow represents a transfer with a specified direction and rate. The model is useful only if the units and signs of those transfers are consistent.

Definition: The central pool is antibody in the modeled vascular or plasma space; the peripheral pool is antibody in an exchangeable tissue space; the endosomal pool is antibody taken up into an intracellular trafficking space; and the catabolic pool is antibody committed to a modeled degradative pathway. These are model compartments, not necessarily directly measured anatomical volumes. Amounts in all pools are expressed in the same mass units, such as mg.

A Minimal Four-Pool Accounting Model

Let \(A_c\), \(A_p\), \(A_e\), and \(A_k\) denote the central, peripheral, endosomal, and catabolic amounts. Let \(V_c\) and \(V_p\) be the central and peripheral volumes, so \(C_c=A_c/V_c\) and \(C_p=A_p/V_p\). The model has reversible exchange between central and peripheral spaces, uptake from the central space into endosomes, return from endosomes to the central space, transfer from endosomes into the catabolic pool, and degradation out of that pool.

The distribution flow is represented by \(Q\), with units of volume per time. Endosomal uptake, endosomal return, transfer into catabolism, and degradation are represented by first-order rate constants \(k_u\), \(k_r\), \(k_t\), and \(k_d\), respectively; each has units of inverse time. \(R_{\mathrm{in}}(t)\) is the input rate into the central pool, in amount per time. Finally, \(A_{\mathrm{el}}\) is cumulative eliminated antibody, recorded in antibody-equivalent mass units.

\begin{aligned} \frac{dA_c}{dt} &= R_{\mathrm{in}}(t)-Q(C_c-C_p)-k_uA_c+k_rA_e,\\ \frac{dA_p}{dt} &= Q(C_c-C_p),\\ \frac{dA_e}{dt} &= k_uA_c-(k_r+k_t)A_e,\\ \frac{dA_k}{dt} &= k_tA_e-k_dA_k,\\ \frac{dA_{\mathrm{el}}}{dt} &= k_dA_k. \end{aligned}

The term \(Q(C_c-C_p)\) is the net distribution flux from central to peripheral space. It is positive when \(C_c>C_p\), negative when \(C_p>C_c\), and zero when the two concentrations are equal. The separate terms \(k_uA_c\), \(k_rA_e\), and \(k_tA_e\) specify transfer into endosomes, return to central space, and entry into the catabolic pool. The final term \(k_dA_k\) is the rate of irreversible elimination from the modeled pools.

Model assumptions: Amounts are well mixed within each compartment; the volumes are constant; the stated transfers are the only routes in this minimal model; and each first-order rate is proportional to the amount in its source compartment. The net distribution expression assumes equal forward and reverse exchange capacity \(Q\). Uptake is shown from the central pool only for simplicity. If tissue uptake is important, a separate peripheral uptake term and its matching loss from \(A_p\) must be added.

Why the Equations Conserve Mass

Internal transfers must appear once as a loss from their source and once as a gain to their destination. The distribution term subtracts from \(A_c\) and adds to \(A_p\). Uptake subtracts from \(A_c\) and adds to \(A_e\); return subtracts from \(A_e\) and adds to \(A_c\); and catabolic transfer subtracts from \(A_e\) and adds to \(A_k\). Only degradation leaves the four modeled pools, and it is recorded in \(A_{\mathrm{el}}\).

This can be checked directly by adding the five differential equations. The distribution, uptake, return, catabolic-transfer, and degradation terms cancel pairwise, leaving only input:

\frac{d}{dt}(A_c+A_p+A_e+A_k+A_{\mathrm{el}}) =R_{\mathrm{in}}(t).

Thus, with no input after a dose, the sum of antibody in the four modeled pools plus cumulative eliminated antibody is constant. This is a bookkeeping identity, not a claim that every compartment can be measured experimentally.

Worked Example: Checking Instantaneous Fluxes

Consider a hypothetical model state with \(V_c=3\) L, \(V_p=7\) L, \(A_c=120\) mg, \(A_p=70\) mg, \(A_e=10\) mg, and \(A_k=5\) mg. Suppose \(Q=0.6\) L/day, \(k_u=0.08\) day\(^{-1}\), \(k_r=0.50\) day\(^{-1}\), \(k_t=0.10\) day\(^{-1}\), and \(k_d=0.20\) day\(^{-1}\). There is no input at this instant.

First calculate the concentrations: \(C_c=120/3=40\) mg/L and \(C_p=70/7=10\) mg/L. The net distribution flux is \(0.6(40-10)=18\) mg/day from central to peripheral. Uptake is \(0.08(120)=9.6\) mg/day, return is \(0.50(10)=5.0\) mg/day, transfer into catabolism is \(0.10(10)=1.0\) mg/day, and degradation is \(0.20(5)=1.0\) mg/day.

Substituting these fluxes into the balances gives:

\begin{aligned} \frac{dA_c}{dt}&=-18-9.6+5=-22.6\ \text{mg/day},\\ \frac{dA_p}{dt}&=18\ \text{mg/day},\\ \frac{dA_e}{dt}&=9.6-5-1=3.6\ \text{mg/day},\\ \frac{dA_k}{dt}&=1-1=0\ \text{mg/day},\\ \frac{dA_{\mathrm{el}}}{dt}&=1\ \text{mg/day}. \end{aligned}

Adding the five rates gives \(-22.6+18+3.6+0+1=0\) mg/day, as required with no input. Central antibody is declining, but most of the instantaneous central loss is redistribution or uptake, not elimination. Only 1 mg/day is leaving the tracked pools through degradation in this state.

Worked Example: Following a Finite-Interval Mass Ledger

Suppose a hypothetical IV dose places 100 mg in the central pool at the start of an interval, with all other pools and cumulative elimination initially zero. Over the interval, 30 mg moves from central to peripheral space, 10 mg enters endosomes, 6 mg returns from endosomes to central space, 4 mg transfers from endosomes to the catabolic pool, and 2 mg is degraded from that pool. Treat these as integrated transfer amounts over the interval.

The ending central amount is \(100-30-10+6=66\) mg. The peripheral amount is \(0+30=30\) mg. The endosomal amount is \(0+10-6-4=0\) mg. The catabolic amount is \(0+4-2=2\) mg, and cumulative eliminated antibody is 2 mg.

The accounting check is \(66+30+0+2+2=100\) mg. The 30 mg transferred to peripheral space has not been eliminated: it is still in the modeled body pools. The 2 mg recorded in \(A_{\mathrm{el}}\), by contrast, has left those pools irreversibly.

Input, Initial Conditions, and Model Choice

For a hypothetical IV bolus of dose \(D\) into the central pool, one can specify \(A_c(0^+)=D\), \(A_p(0^+)=A_e(0^+)=A_k(0^+)=A_{\mathrm{el}}(0^+)=0\). For an infusion, \(R_{\mathrm{in}}(t)\) is nonzero over the infusion period. These choices describe how antibody enters the model; they do not alter the requirement that every internal transfer be balanced at both ends.

The four-pool representation is one level of structural detail, not the only valid choice. A two-compartment model can summarize central–peripheral distribution without explicitly representing intracellular handling. A one-pool model can summarize observed central concentrations while combining unresolved processes into fewer terms. The four-pool model is preferable when the distinction between tissue exchange, intracellular return, and catabolic commitment matters to the question. More compartments are not automatically better: each additional pool requires a defensible interpretation and enough data to constrain its behavior.

Worked Example: Separating Input From Loss

Use the state and parameters from the first example, but now suppose a hypothetical infusion supplies \(R_{\mathrm{in}}=20\) mg/day. The internal fluxes are unchanged at that instant. The central balance becomes \(20-18-9.6+5=-2.6\) mg/day; the peripheral, endosomal, catabolic, and eliminated rates remain \(18\), \(3.6\), \(0\), and \(1\) mg/day, respectively.

The sum is \(-2.6+18+3.6+0+1=20\) mg/day. The total accounted amount is increasing at exactly the infusion rate. In particular, the negative central rate does not mean the whole system is losing antibody at 2.6 mg/day: the other pools and input must be included to interpret the balance.

Diagnostics, Failure Modes, and Limitations

A useful implementation check is to evaluate the total-mass identity at every time point. With numerical integration, define the residual as the change in total accounted amount minus integrated input. A residual that is not close to zero relative to the amounts being tracked can indicate a sign error, a missing destination term, inconsistent units, or numerical integration error.

Other warning signs include negative compartment amounts, rates with units that do not yield amount per time, or a transfer that appears as a gain without a matching source loss. Negative amounts are not physically meaningful in this model; they may result from unsuitable numerical settings or an incorrectly specified equation. A fitted model that requires implausible rates or cannot distinguish several pool amounts from the observed data may also be too detailed for the available measurements.

Distribution is not elimination. A decrease in \(A_c\) caused by a positive central-to-peripheral flux is a location change, not loss from the total system.
Catabolic entry is not necessarily the moment of elimination. Here \(k_tA_e\) transfers antibody into \(A_k\); only \(k_dA_k\) leaves the tracked pools. A model that defines catabolic entry itself as the irreversible sink would use a different, simpler bookkeeping convention.
Do not mix concentration and amount terms. \(Q(C_c-C_p)\) has units of amount per time, whereas \(k_uA_c\) has the same units because \(k_u\) is inverse time.

The model assumes fixed volumes, well-mixed pools, and first-order transfers. Real tissue spaces can be heterogeneous, and additional uptake routes or transformations may matter. The catabolic pool is an accounting device, not a claim that degraded antibody persists as intact IgG. The equations also do not specify the molecular mechanism governing return from endosomes; they only represent a return flux. These abstractions should be expanded only when the scientific question and available information justify the added structure.

Key takeaway: Write one balance per compartment, put every internal transfer in both its source and destination equations, and reserve the term elimination for transfer out of the tracked antibody pools. Summing the balances is the fastest check that the accounting is coherent.

Check Your Understanding

Answer each question, then compare with the solutions below.

If \(C_c=C_p\), what is the net central–peripheral distribution flux in this model? Does this imply that antibody amounts in the two pools are equal?
At a particular instant, \(k_tA_e=3\) mg/day and \(k_dA_k=1\) mg/day. What are the contributions of these processes to the catabolic and eliminated amounts?
A modeler subtracts \(Q(C_c-C_p)\) from the central equation but forgets to add it to the peripheral equation. What mass-balance error results?
With no input, the modeled pool amounts change at rates \(-12\), \(+8\), \(+2\), and \(+1\) mg/day for \(A_c,A_p,A_e,A_k\), respectively. What must \(dA_{\mathrm{el}}/dt\) be for total mass to balance?

Solutions. 1. The net flux is \(Q(C_c-C_p)=0\). Equal concentrations do not require equal amounts, because \(A_c=V_cC_c\) and \(A_p=V_pC_p\), and the volumes may differ. 2. The catabolic pool gains 3 mg/day from endosomes and loses 1 mg/day through degradation, so \(dA_k/dt=2\) mg/day; cumulative eliminated antibody increases at 1 mg/day. 3. The central loss has no matching peripheral gain, so the sum of modeled amounts spuriously decreases by the omitted distribution flux when that flux is positive. 4. The four pool rates sum to \(-12+8+2+1=-1\) mg/day. Therefore \(dA_{\mathrm{el}}/dt\) must be \(1\) mg/day, making the total rate zero.