Tutorials › Physiologically-Based Pharmacokinetics (PBPK) › Adding Endosomal Exchange To A Tissue Model

Permeability-Limited and Multi-Compartment Tissues · Tutorial 76 of 799

Adding Endosomal Exchange To A Tissue Model

Learn to represent reversible cell–endosome exchange with finite membrane fluxes and assemble the resulting extracellular, intracellular and endosomal amount balances.

Intermediate 9 min read

What You'll Learn

  • Define amount and concentration state variables for three coupled tissue compartments.
  • Write passive, finite-rate fluxes across the extracellular–cell and cell–endosome membranes.
  • Derive coupled amount balances by tracking the direction of each flux.
  • Convert amount rates to concentration rates when compartment volumes are fixed.
  • Check flux signs, units and limiting cases to detect model errors.

Why Add an Endosomal Compartment?

A tissue model with extracellular and intracellular spaces can represent entry into cells, but it cannot distinguish drug in the cell from drug that has moved into an endosomal space. That distinction matters whenever transfer across the cell–endosome membrane takes time: the endosomal amount can change gradually rather than appearing immediately in a fixed proportion to the intracellular amount. The model therefore needs a separate endosomal state and a finite flux connecting it to the intracellular state.

The earlier material on choosing tissue compartments and deriving membrane fluxes supplies the basic ideas: compartments have amounts and concentrations, and a permeability–area product converts a concentration difference into an amount rate. Here those ideas are assembled into one three-compartment model. The purpose is to write the balances and make each flux direction explicit, not to assume that the compartments are instantaneously equilibrated.

Definition: Let \(A_e\), \(A_i\), and \(A_{en}\) be drug amounts in the extracellular, intracellular, and endosomal compartments. Let their fixed volumes be \(V_e\), \(V_i\), and \(V_{en}\), respectively, and define concentrations \(C_e=A_e/V_e\), \(C_i=A_i/V_i\), and \(C_{en}=A_{en}/V_{en}\). A membrane conductance \(G\) is the product of permeability and membrane area; with permeability in length/time and area in length squared, \(G\) has units of volume/time.

For clarity, the model below uses passive exchange on a common concentration basis: each membrane flux is conductance multiplied by the concentration difference across that membrane. This corresponds to equal partitioning on the chosen concentration basis. If a model requires a membrane partition factor or a more complex transport law, the driving-force expression must be modified consistently; it should not be added as an unexplained multiplier. Conductances are nonnegative and finite. Volumes are fixed, and the balances initially omit metabolism, binding reactions, and elimination within these three spaces.

Define Fluxes Before Writing Balances

Choose a positive direction for each flux once and use it consistently. Let \(J_{ec}\) be positive from extracellular space into the cell, and let \(J_{ie}\) be positive from the cell into the endosome. With conductances \(G_{ec}\) and \(G_{ie}\), the passive fluxes are:

$$ J_{ec}=G_{ec}(C_e-C_i), \qquad J_{ie}=G_{ie}(C_i-C_{en}). $$

Each flux has units of amount/time. For example, a positive \(J_{ie}\) means that the intracellular concentration exceeds the endosomal concentration, so the net flux is into the endosome. If the concentration difference reverses, the flux is negative and the net direction reverses. A negative value is not a negative amount in a compartment; it is a signed rate under the direction convention just defined.

Let \(R_e(t)\) denote any net non-membrane input to the extracellular compartment, in amount/time. It may represent a prescribed delivery rate or the net result of a more detailed vascular exchange model. It is not itself a membrane flux in the three-compartment system. Setting \(R_e(t)=0\) gives a closed model with no external delivery or removal.

Derive the Coupled Amount Balances

For each compartment, the rate of change of amount is the sum of its inflows minus the sum of its outflows. The extracellular compartment receives \(R_e(t)\) and loses \(J_{ec}\). The intracellular compartment gains \(J_{ec}\) and loses \(J_{ie}\). The endosomal compartment gains \(J_{ie}\). Applying this bookkeeping gives:

$$ \frac{dA_e}{dt}=R_e(t)-J_{ec}, \qquad \frac{dA_i}{dt}=J_{ec}-J_{ie}, \qquad \frac{dA_{en}}{dt}=J_{ie}. $$

Substituting the flux expressions produces the coupled model directly:

$$ \frac{dA_e}{dt}=R_e(t)-G_{ec}(C_e-C_i), $$ $$ \frac{dA_i}{dt}=G_{ec}(C_e-C_i)-G_{ie}(C_i-C_{en}), $$ $$ \frac{dA_{en}}{dt}=G_{ie}(C_i-C_{en}). $$

These are amount balances, so their left-hand sides and right-hand sides have units of amount/time. Because \(C_j=A_j/V_j\), fixed compartment volumes also allow the model to be written as concentration-rate equations:

$$ \frac{dC_e}{dt}=\frac{R_e(t)-J_{ec}}{V_e}, \qquad \frac{dC_i}{dt}=\frac{J_{ec}-J_{ie}}{V_i}, \qquad \frac{dC_{en}}{dt}=\frac{J_{ie}}{V_{en}}. $$

The volume division is essential: an amount rate is not a concentration rate. The same flux can produce a larger concentration change in a smaller compartment. If volumes change with time, the simple conversion \(dC_j/dt=(dA_j/dt)/V_j\) no longer suffices; differentiating \(A_j=V_jC_j\) also introduces the changing-volume term.

Model statement: Given initial amounts, fixed positive compartment volumes, nonnegative finite conductances, and a specified external rate \(R_e(t)\), the three equations above describe the stated passive exchange model. They do not imply instantaneous equilibration, and they do not include additional pathways unless those pathways are written as explicit terms in the appropriate balances.

Worked Example: Compute the Three Compartment Rates

Consider hypothetical values \(V_e=0.20\) L, \(V_i=0.50\) L, and \(V_{en}=0.010\) L, with \(C_e=8.0\) \(\mu\)mol/L, \(C_i=2.0\) \(\mu\)mol/L, and \(C_{en}=1.0\) \(\mu\)mol/L. Let \(G_{ec}=0.12\) L/h, \(G_{ie}=0.030\) L/h, and \(R_e=0.20\) \(\mu\)mol/h.

First calculate the membrane fluxes:

$$ J_{ec}=(0.12\ {\rm L/h})(8.0-2.0\ {\rm \mu mol/L}) =0.72\ {\rm \mu mol/h}, $$ $$ J_{ie}=(0.030\ {\rm L/h})(2.0-1.0\ {\rm \mu mol/L}) =0.030\ {\rm \mu mol/h}. $$

Both are positive under the defined directions. The amount rates are therefore:

$$ \frac{dA_e}{dt}=0.20-0.72=-0.52\ {\rm \mu mol/h}, \qquad \frac{dA_i}{dt}=0.72-0.030=0.69\ {\rm \mu mol/h}, \qquad \frac{dA_{en}}{dt}=0.030\ {\rm \mu mol/h}. $$

For fixed volumes, divide each rate by its volume. The resulting concentration rates are \(-0.52/0.20=-2.6\) \(\mu\)mol/L/h in extracellular space, \(0.69/0.50=1.38\) \(\mu\)mol/L/h intracellularly, and \(0.030/0.010=3.0\) \(\mu\)mol/L/h in the endosome. Thus, despite a smaller endosomal amount rate, its concentration rises fastest because its volume is smallest.

Direction Reversal and Membrane Scaling

A finite flux is signed, so neither membrane should be treated as a one-way transfer unless the biology specifically requires a one-way process. In a passive model, the concentration difference determines the net direction at each instant. This makes the same equations usable during uptake and return without switching formulas.

Worked Example: Endosomal Return When Concentrations Reverse

Suppose there is no external input, \(R_e=0\), and the hypothetical concentrations are \(C_e=1.0\), \(C_i=4.0\), and \(C_{en}=2.0\) \(\mu\)mol/L. Let \(G_{ec}=0.10\) L/h and \(G_{ie}=0.050\) L/h.

The extracellular-to-cell flux is \(J_{ec}=0.10(1.0-4.0)=-0.30\) \(\mu\)mol/h. Its negative sign means net movement from the cell to extracellular space. The cell-to-endosome flux is \(J_{ie}=0.050(4.0-2.0)=0.10\) \(\mu\)mol/h, still directed into the endosome.

The balances give \(dA_e/dt=-J_{ec}=+0.30\) \(\mu\)mol/h, \(dA_i/dt=J_{ec}-J_{ie}=-0.30-0.10=-0.40\) \(\mu\)mol/h, and \(dA_{en}/dt=J_{ie}=+0.10\) \(\mu\)mol/h. The intracellular amount falls through two distinct outgoing routes, while extracellular and endosomal amounts rise through their respective incoming routes.

If permeability \(P_{ie}\) and membrane area \(S_{ie}\) are supplied separately, set \(G_{ie}=P_{ie}S_{ie}\) after checking their units. This gives a useful scale check before integrating the model.

Worked Example: Calculate a Cell–Endosome Conductance

Take hypothetical \(P_{ie}=0.040\) cm/h and \(S_{ie}=25\) cm\(^2\). Their product is \(1.0\) cm\(^3\)/h, which is \(0.0010\) L/h because \(1000\) cm\(^3=1\) L. If \(C_i-C_{en}=10\) \(\mu\)mol/L, then:

$$ J_{ie}=(0.0010\ {\rm L/h})(10\ {\rm \mu mol/L}) =0.010\ {\rm \mu mol/h}. $$

For the same permeability and concentration difference, an area of \(100\) cm\(^2\) gives \(G_{ie}=4.0\) cm\(^3\)/h \(=0.0040\) L/h and \(J_{ie}=0.040\) \(\mu\)mol/h. The fourfold area increase produces a fourfold flux in this linear model. This calculation also illustrates why converting cm\(^3\)/h to L/h incorrectly can create a thousandfold error.

Assumptions, Checks, and Alternatives

The equations rely on a well-mixed concentration in each compartment, fixed volumes, passive linear fluxes, and the stated common concentration basis. A compartment concentration is therefore treated as representative of its entire volume. Conductance is assumed not to change over the calculation interval. If transport is saturable, active, or directionally asymmetric, the flux law must reflect that biology; replacing it with a passive concentration-difference term would be a model assumption, not a neutral simplification.

Several checks help identify a model that is being implemented incorrectly. At equal concentrations across a membrane, its passive flux must be zero. If the source-side concentration exceeds the destination-side concentration, the flux should be positive under the chosen direction. Every membrane term must appear as an outflow in one balance and an inflow in the adjacent balance with the same magnitude. Conductance times concentration must have units of amount/time, and concentration rates require division by volume. Finally, amounts should not become negative in a physically valid simulation; negative values often indicate an inconsistent sign, unsuitable numerical step size, or an invalid input rather than a meaningful prediction.

An explicit endosomal compartment is more informative than a single intracellular compartment when the amount in the endosome is a desired state or when finite transfer is important to the question. A one-cell-compartment model is simpler, but it cannot report a distinct endosomal amount. Conversely, an instantaneous partitioning approximation replaces dynamic exchange with an algebraic constraint and removes the endosomal transfer rate as a dynamic process. That approximation is only appropriate when its assumptions are justified for the application; it is not the same model as finite-rate exchange.

The three-compartment structure is still a simplification. It does not specify vascular delivery, binding, pH-dependent ionization, endosomal processing, or intracellular metabolism unless those processes are added explicitly. Nor does the presence of an endosomal state establish that available data can estimate its conductance. The model provides a clear bookkeeping structure; deciding whether its additional detail is needed and supported remains a separate modeling judgment.

Common pitfall: It is incorrect to write \(J_{ie}=G_{ie}(C_{en}-C_i)\) while retaining the balance \(dA_{en}/dt=J_{ie}\) and calling positive \(J_{ie}\) cell-to-endosome uptake. That expression makes the flux positive when the endosomal concentration is higher, reversing the intended direction. Either define the flux as cell-to-endosome and use \(G_{ie}(C_i-C_{en})\), or reverse both the definition and every balance sign consistently.

Check Your Understanding

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

  1. If \(C_i=C_{en}\), what is \(J_{ie}\) in the passive model, and what does that imply for the endosomal amount rate if no other process acts on it?
  2. Let \(J_{ec}=0.40\) \(\mu\)mol/h, \(J_{ie}=0.15\) \(\mu\)mol/h, and \(R_e=0\). Find all three amount rates.
  3. A flux is \(-0.08\) \(\mu\)mol/h when positive is defined as extracellular-to-cell. State its net direction and its contribution to the extracellular amount rate.
  4. Why can a smaller endosomal amount rate produce a larger concentration rate than the intracellular amount rate?

Solutions. (1) Equal concentrations give \(J_{ie}=G_{ie}(C_i-C_{en})=0\), so \(dA_{en}/dt=0\) in the stated model. (2) The rates are \(dA_e/dt=-0.40\) \(\mu\)mol/h, \(dA_i/dt=0.40-0.15=0.25\) \(\mu\)mol/h, and \(dA_{en}/dt=0.15\) \(\mu\)mol/h. (3) The negative sign indicates net cell-to-extracellular transfer; because \(dA_e/dt\) includes \(-J_{ec}\), this flux contributes \(+0.08\) \(\mu\)mol/h to extracellular amount. (4) For fixed volume, \(dC/dt=(dA/dt)/V\). A smaller compartment volume can therefore yield a larger concentration rate even when its amount rate is smaller.