Tutorials › Physiologically-Based Pharmacokinetics (PBPK) › Worked Construction of an IgG Tissue-Distribution Model

Biologics and Antibody Disposition in PBPK · Tutorial 470 of 799

Worked Construction of an IgG Tissue-Distribution Model

Build and check a numerical IgG tissue-distribution model, from compartment equations to time-step calculations and FcRn-dependent mass balance.

Advanced 12 min read

What You'll Learn

  • Define a coupled plasma, tissue-vascular, interstitial, and endosomal IgG model.
  • Calculate vascular exchange, convective filtration, lymph return, and FcRn fluxes with consistent units.
  • Advance compartment amounts over time using an explicit Euler step.
  • Verify that internal fluxes cancel in the whole-body mass balance.
  • Interpret FcRn recycling and degradation as competing endosomal exits.
  • Recognize numerical and physiological limitations of a compact tissue model.

From Mechanisms to a Checkable Model

A mechanistic antibody model is useful only if its fluxes connect consistently: antibody that leaves one compartment must enter another, and the whole-body amount should change only through explicitly represented elimination. This tutorial constructs a hypothetical one-organ IgG model that combines perfusion, transcapillary exchange, interstitial convection, lymph return, and FcRn-mediated recycling. The aim is not to represent every tissue in detail, but to make the accounting and numerical checks transparent.

The model follows the compartment and flux ideas introduced in the earlier material on antibody amount balance, transcapillary exchange, interstitial convection, and FcRn salvage. Here those pieces are assembled into one set of differential equations and evaluated with numerical values. All quantities and parameter values below are hypothetical and illustrative, not estimates for a particular antibody or organ.

Definition: Let \(A_p\), \(A_v\), \(A_i\), and \(A_e\) denote IgG amounts in central plasma, the organ vascular space, the organ interstitial space, and an endosomal compartment, respectively. Their volumes are \(V_p\), \(V_v\), and \(V_i\), so the corresponding concentrations are \(C_p=A_p/V_p\), \(C_v=A_v/V_v\), and \(C_i=A_i/V_i\). Amounts are measured in mg, volumes in L, concentrations in mg/L, flows and clearances in L/h, and first-order rate constants in h\(^{-1}\).

Model Structure and Assumptions

The central plasma and organ vascular spaces exchange blood at flow \(Q\). Perfusion moves IgG from plasma to organ vasculature at rate \(Q C_p\), and returns it at rate \(Q C_v\); the net transfer is therefore \(Q(C_p-C_v)\). Across the organ capillary, passive exchange is represented by \(PS(C_v-C_i)\), where \(PS\) is a permeability-surface area product. Positive values indicate net movement from organ vasculature to interstitium.

A filtration flow \(Q_f\) carries a fraction \(1-\sigma\) of the vascular concentration into the interstitium. Here \(\sigma\) is the vascular reflection coefficient: \(\sigma=1\) gives no convective IgG transfer in this representation, whereas a lower value permits more solvent-drag transfer. Interstitial fluid and its IgG leave at lymph flow \(Q_L\), with IgG flux \(Q_L C_i\), and lymph returns to central plasma. For the numerical example, filtration and lymph flows are equal, \(Q_f=Q_L\), so the assumed fluid balance is steady.

Interstitial IgG enters the endosomal compartment at first-order rate \(k_u A_i\). Endosomal IgG returns to central plasma at rate \(k_r A_e\), or is degraded at rate \(k_d A_e\). The model also includes systemic clearance from plasma, at rate \(CL C_p\). These are simplifying assumptions: concentrations are well mixed within each compartment, the listed parameters are constant, and no binding, spatial gradients within a compartment, or additional elimination pathways are represented.

Model equations: With \(J=PS(C_v-C_i)+(1-\sigma)Q_f C_v\) denoting net capillary transfer into the interstitium and \(L=Q_L C_i\) denoting lymph return, the balances are
$$ \begin{aligned} \frac{dA_p}{dt} &= -Q(C_p-C_v)+L+k_rA_e-CL\,C_p,\\ \frac{dA_v}{dt} &= Q(C_p-C_v)-J,\\ \frac{dA_i}{dt} &= J-L-k_uA_i,\\ \frac{dA_e}{dt} &= k_uA_i-(k_r+k_d)A_e. \end{aligned} $$

The terms are amount rates. For example, \(PS(C_v-C_i)\) has units \((\mathrm{L/h})(\mathrm{mg/L})=\mathrm{mg/h}\), as does \(k_u A_i\). This dimensional check helps catch errors such as multiplying a flow by an amount instead of a concentration.

Why the Whole-Body Balance Is a Strong Check

Add the four compartment equations. The perfusion terms cancel because the amount lost from plasma is gained by organ vasculature, and vice versa. Capillary transfer \(J\) cancels between vascular and interstitial compartments. Lymph loss from the interstitium cancels against lymph return to plasma. Endosomal uptake and recycling also cancel as internal transfers. What remains is

$$ \frac{d}{dt}(A_p+A_v+A_i+A_e)=-CL\,C_p-k_dA_e. $$

Thus total modeled IgG decreases only through the two explicit elimination routes: systemic clearance and endosomal degradation. This result is a bookkeeping identity, not a claim that those are the only biological elimination mechanisms. If a simulation shows an additional change in total amount, a flux has been omitted, duplicated, or assigned an inconsistent sign.

For a simple numerical solution, the explicit Euler update over a time step \(\Delta t\) is \(A_j(t+\Delta t)=A_j(t)+\Delta t\,(dA_j/dt)\), for each compartment \(j\). It approximates the amount change using the rates at the beginning of the step. It is most appropriate here for illustrating short steps; it is not an exact solution over a long interval.

Worked Example: Calculate Initial Fluxes and Advance One Step

Consider hypothetical values \(V_p=3.00\) L, \(V_v=0.150\) L, and \(V_i=0.450\) L. Let \(Q=0.600\) L/h, \(PS=0.0300\) L/h, \(Q_f=Q_L=0.0200\) L/h, and \(\sigma=0.900\). Use \(k_u=0.0800\) h\(^{-1}\), \(k_r=0.500\) h\(^{-1}\), \(k_d=0.0500\) h\(^{-1}\), and \(CL=0.0100\) L/h. Initially, \(A_p=150\) mg, \(A_v=7.50\) mg, and \(A_i=A_e=0\) mg. Calculate the initial rates and take one Euler step of \(0.100\) h.

The initial concentrations are \(C_p=150/3.00=50.0\) mg/L, \(C_v=7.50/0.150=50.0\) mg/L, and \(C_i=0/0.450=0\) mg/L. Perfusion has no net transfer initially because \(Q(C_p-C_v)=0.600(50.0-50.0)=0\) mg/h.

The passive capillary flux is \(PS(C_v-C_i)=0.0300(50.0-0)=1.50\) mg/h. The convective flux is \((1-\sigma)Q_fC_v=(1-0.900)(0.0200)(50.0)=0.100\) mg/h. Therefore \(J=1.60\) mg/h. Lymph return and endosomal uptake are initially zero, while systemic clearance is \(CL C_p=0.0100(50.0)=0.500\) mg/h.

Substituting these rates gives \(dA_p/dt=-0.500\) mg/h, \(dA_v/dt=-1.60\) mg/h, \(dA_i/dt=1.60\) mg/h, and \(dA_e/dt=0\) mg/h. For example, \(dA_v/dt=0-1.60=-1.60\) mg/h: the organ vascular space loses the capillary flux, with no net perfusion gain.

Applying the Euler formula gives \(A_p=150+0.100(-0.500)=149.95\) mg, \(A_v=7.50+0.100(-1.60)=7.34\) mg, \(A_i=0+0.100(1.60)=0.160\) mg, and \(A_e=0\) mg at \(t=0.100\) h. The total is \(157.45\) mg, down by \(0.0500\) mg from \(157.50\) mg. That decrease equals \(0.100\) h times the initial systemic clearance rate, \(0.500\) mg/h; the exchange fluxes have only redistributed IgG.

Checking the Coupled Fluxes at the Next Time Point

After the first step, the compartments no longer have equal concentrations. The next calculation checks that the model uses the changed concentrations, rather than reusing initial fluxes. This distinction matters in dynamic models: gradients and resulting fluxes evolve as amounts move between spaces.

Worked Example: Recalculate Rates and Check Mass Balance

Use the amounts at \(t=0.100\) h from the preceding calculation: \(A_p=149.95\) mg, \(A_v=7.34\) mg, \(A_i=0.160\) mg, and \(A_e=0\) mg. The concentrations are \(C_p=149.95/3.00=49.9833\) mg/L, \(C_v=7.34/0.150=48.9333\) mg/L, and \(C_i=0.160/0.450=0.35556\) mg/L.

Net perfusion into the organ vascular space is \(0.600(49.9833-48.9333)=0.6300\) mg/h. The passive capillary flux is \(0.0300(48.9333-0.35556)=1.4573\) mg/h, and the convective flux is \(0.100(0.0200)(48.9333)=0.09787\) mg/h. Thus \(J=1.5552\) mg/h. Lymph return is \(L=0.0200(0.35556)=0.007111\) mg/h, uptake is \(k_uA_i=0.0800(0.160)=0.01280\) mg/h, and systemic clearance is \(0.0100(49.9833)=0.49983\) mg/h.

The compartment rates are

$$ \begin{aligned} \frac{dA_p}{dt} &= -0.6300+0.007111-0.49983=-1.1227\ \mathrm{mg/h},\\ \frac{dA_v}{dt} &= 0.6300-1.5552=-0.9252\ \mathrm{mg/h},\\ \frac{dA_i}{dt} &= 1.5552-0.007111-0.01280=1.5353\ \mathrm{mg/h},\\ \frac{dA_e}{dt} &= 0.01280\ \mathrm{mg/h}. \end{aligned} $$

For a second \(0.100\)-h Euler step, the resulting amounts are \(149.8377\) mg in plasma, \(7.2475\) mg in organ vasculature, \(0.3135\) mg in interstitium, and \(0.001280\) mg in the endosome. Their sum is \(157.4000\) mg. The earlier total was \(157.4500\) mg, so the decrease is \(0.04998\) mg. The whole-body identity predicts the same decrease: \(0.100(0.49983+0)=0.04998\) mg, since \(A_e=0\) at the start of this step. Minor last-digit differences arise from rounding.

What FcRn Salvage Means in This Model

In this formulation, FcRn-mediated salvage is represented by an endosomal return flux competing with endosomal degradation. If an amount enters the endosome and remains there until one of these two first-order exits occurs, the probability of recycling is \(k_r/(k_r+k_d)\), and the probability of degradation is \(k_d/(k_r+k_d)\). These fractions follow from competing exponential exit rates; they apply to the simplified endosomal compartment, not necessarily to every biological route taken by IgG.

Worked Example: Quantify the Competing Endosomal Fates

Using \(k_r=0.500\) h\(^{-1}\) and \(k_d=0.0500\) h\(^{-1}\), the total endosomal exit rate is \(0.550\) h\(^{-1}\). The eventual recycling fraction is \(0.500/0.550=0.9091\), and the eventual degradation fraction is \(0.0500/0.550=0.09091\). If a hypothetical 1.00 mg enters this compartment and is followed until it exits, the expected amounts are 0.9091 mg recycled and 0.09091 mg degraded.

The mean time to either exit under these constant competing rates is \(1/(k_r+k_d)=1/0.550=1.818\) h. This does not mean every molecule resides for 1.818 h; it is the mean for the exponential residence-time model. Greater \(k_r\), holding \(k_d\) fixed, raises the recycling fraction; it does not make the endosomal amount itself a direct measure of whole-body clearance.

For a local check at \(t=0.100\) h in the preceding calculation, \(A_e=0\), so recycling and degradation fluxes are both zero at that instant. Uptake is \(0.01280\) mg/h, which exactly equals the endosomal accumulation rate. Once \(A_e\) becomes positive, both exit fluxes must be included in the amount balance.

Diagnostics, Alternatives, and Limitations

A useful numerical check is to calculate each transfer once, name its direction, and use that same value with opposite signs in the two affected compartments. Then sum all compartment rates and compare with the elimination terms. Also inspect concentrations and amounts at every step: negative amounts or unexpectedly large oscillations indicate that the numerical step may be too large, the signs may be wrong, or units may be inconsistent. Here the organ vascular space has a relatively rapid perfusion turnover, \(Q/V_v=4.00\) h\(^{-1}\), so the step size should be small enough to resolve its changes. A smaller Euler step or a suitable ODE solver can improve numerical integration, but neither can repair an incorrect model or flux definition.

This lumped model is one modeling choice, not a universal representation of antibody distribution. A model that omits explicit organ vascular space could combine vascular exchange differently; a model with multiple tissues would allocate perfusion and lymph return across organs. More detailed representations may separate endothelial and other cellular uptake, include binding or spatial gradients, or use concentration-dependent processes. Those alternatives need additional assumptions and data. The compact version here is valuable because its fluxes and mass balance are easy to inspect, but its predictions should not be treated as organ-specific evidence without appropriate parameterization and evaluation.

  • Sign error: Giving capillary transfer the same sign in vascular and interstitial balances creates or destroys antibody internally. The correction is to subtract \(J\) from \(A_v\) and add \(J\) to \(A_i\).
  • Wrong concentration for a flow: Lymph flow is multiplied by \(C_i\), not by \(A_i\). The correction follows from flow times concentration having units of amount per time.
  • Misreading salvage: Recycling returns an amount to plasma; it is not an additional elimination term. Only the degraded fraction contributes to \(k_dA_e\) in this model.
  • Overinterpreting a coarse step: Euler updates use beginning-of-step rates. A large step can misrepresent evolving gradients, even when the continuous equations are correctly written.
Key takeaway: A coupled tissue model is internally coherent when every transfer is paired across compartments, the units of each flux are amount per time, and the total modeled amount changes only by the explicitly specified elimination fluxes.

Check Your Understanding

Work through these questions, then compare your reasoning with the solutions.

  1. If \(C_v=C_i\), which part of the capillary flux is zero, and can the total capillary flux still be nonzero?
  2. For the hypothetical model above, what is the initial whole-body amount rate of change?
  3. If \(Q_L\) doubles while \(C_i\) remains fixed for an instant, what happens to the lymph return flux, and which compartment rates change directly?
  4. With \(k_r=0.300\) h\(^{-1}\) and \(k_d=0.100\) h\(^{-1}\), what fraction of endosomal IgG is eventually recycled in the competing-exit model?

Solutions. (1) The passive term \(PS(C_v-C_i)\) is zero, but the convective term can remain nonzero if \(Q_f>0\), \(\sigma<1\), and \(C_v>0\). (2) Initially \(A_e=0\) and \(C_p=50.0\) mg/L, so the whole-body rate is \(-CL C_p-k_dA_e=-0.0100(50.0)-0=-0.500\) mg/h. (3) \(L=Q_LC_i\) doubles at fixed \(C_i\); \(A_i\) loses that additional flux and \(A_p\) gains it, while the other compartment rates do not change directly through this term. (4) The recycled fraction is \(k_r/(k_r+k_d)=0.300/(0.300+0.100)=0.750\), or 75.0%.