From Membrane Flux to Compartment Equations
A membrane-flux law describes the rate at which drug crosses a boundary. A tissue model, however, needs equations for the amounts—or concentrations—inside its compartments. The translation is simple in principle: a net amount leaving the extracellular space enters the intracellular space, and each compartment’s amount balance also includes its own other inputs and losses. The signs and the conversion from amount to concentration must be handled consistently.
The equations below treat extracellular and intracellular spaces as well-mixed compartments. Each has a single concentration, and the membrane is represented by a finite-rate net flux. The flux law is taken as given; its permeability and area scaling are not derived here. The concentration used in that law must be the concentration form to which the membrane law applies, such as unbound drug concentration if that is what the law specifies.
A common linear finite-exchange law uses a conductance \(G_m\) and an equilibrium partition coefficient \(K_{i/e}\):
Here \(G_m\) has units of volume per time, so \(F_{e\to i}\) has units of amount per time. The dimensionless \(K_{i/e}\) is defined so that equilibrium under this law satisfies \(C_i=K_{i/e}C_e\). If the relevant concentrations have equal values at equilibrium, \(K_{i/e}=1\). The sign of the expression in parentheses determines the direction of net transfer: the flux is positive when \(C_e>C_i/K_{i/e}\), zero when they are equal, and negative when \(C_e<C_i/K_{i/e}\).
Writing the Amount Balances
The membrane flux is an outflow from the extracellular compartment and an equal inflow to the intracellular compartment. With the sign convention above, the amount balances are therefore:
These equations keep the two kinds of processes distinct. For example, blood delivery or extracellular elimination belongs in \(R_e\); intracellular metabolism or another intracellular loss belongs in \(R_i\). The exchange term appears once in each equation, with opposite signs. It should not be added as a loss from both spaces or as a gain to both.
The balance is first written for amount because the membrane flux has units of amount per time. Division by the compartment volume converts that rate into a concentration rate. Consequently, identical exchange rates do not generally produce identical concentration changes: a given amount rate changes concentration more in the smaller compartment.
Worked Example: Converting an Inward Flux to Concentration Rates
Suppose \(V_e=3.0\) L and \(V_i=1.0\) L, with \(C_e=4.0\) mg/L and \(C_i=1.0\) mg/L. Let \(K_{i/e}=2.0\), \(G_m=0.60\) L/h, and assume there are no other inputs or losses. Calculate the flux and the instantaneous concentration rates.
First evaluate the driving difference and flux:
The positive sign means drug is moving net from extracellular to intracellular space. With \(R_e=R_i=0\), the amount rates are \(dA_e/dt=-2.1\) mg/h and \(dA_i/dt=+2.1\) mg/h. Dividing by each volume gives:
The intracellular concentration rises three times as fast in magnitude because its volume is one-third as large. These are instantaneous rates at the stated concentrations, not predictions that the concentrations continue changing at these rates indefinitely.
Checking Direction and Including Other Processes
A signed flux avoids writing separate equations for inward and outward cases. If \(F_{e\to i}\) is negative, the same balance equations still work: \(-F_{e\to i}\) is then positive in the extracellular equation, and \(+F_{e\to i}\) is negative in the intracellular equation. There is no need to switch the signs by hand when the direction reverses.
Worked Example: A Negative Flux and Opposite Concentration Changes
Let \(V_e=2.0\) L, \(V_i=0.50\) L, \(C_e=1.0\) mg/L, and \(C_i=4.0\) mg/L. Take \(K_{i/e}=2.0\) and \(G_m=0.25\) L/min, with no other processes. Determine the signed flux and concentration rates.
The driving difference is \(1.0-4.0/2.0=-1.0\) mg/L, so:
The negative flux indicates net movement from intracellular to extracellular space. Applying the amount balances gives \(dA_e/dt=-(-0.25)=+0.25\) mg/min and \(dA_i/dt=-0.25\) mg/min. Therefore:
The flux sign and the concentration-rate signs are consistent with outward movement: extracellular amount increases and intracellular amount decreases. The two concentration rates differ because the volumes differ.
The terms \(R_e\) and \(R_i\) allow the exchange equations to be embedded in a larger tissue model without changing the membrane sign convention. For example, an extracellular input of \(q\) mg/min contributes \(+q\) to \(R_e\), while an extracellular elimination rate of \(l\) mg/min contributes \(-l\). If both occur, \(R_e=q-l\). This makes clear whether a term represents membrane exchange or an independent delivery, metabolism, or removal process.
Worked Example: Combining Extracellular Input, Loss, and Exchange
At a particular instant, suppose \(V_e=2.0\) L, \(V_i=1.0\) L, and the membrane flux is \(F_{e\to i}=0.10\) mg/min. An extracellular delivery rate is \(0.50\) mg/min and an extracellular loss rate is \(0.20\) mg/min. Intracellularly, drug is removed at \(0.05\) mg/min. If \(C_e=3.0\) mg/L and \(C_i=1.0\) mg/L, calculate the amount and concentration rates.
The non-membrane net rates are \(R_e=0.50-0.20=0.30\) mg/min and \(R_i=-0.05\) mg/min. The amount balances give:
Dividing by the respective volumes:
The extracellular concentration rises even though the membrane flux is outward from that space, because delivery exceeds the combined extracellular loss and exchange. The intracellular amount rises because the incoming membrane flux exceeds intracellular removal. Looking only at the exchange direction would not be enough to determine the sign of either total concentration rate.
Derivation, Volume Changes, and Diagnostics
The concentration equations follow directly from the definition of amount. For a fixed extracellular volume, \(A_e=V_eC_e\), so differentiation gives \(dA_e/dt=V_e\,dC_e/dt\). Substituting the amount balance and dividing by \(V_e\) yields \(dC_e/dt=(R_e-F_{e\to i})/V_e\). The intracellular equation follows in the same way from \(A_i=V_iC_i\) and \(dA_i/dt=R_i+F_{e\to i}\).
If a compartment volume changes with time, the product rule is required. For either compartment, \(A=VC\) implies \(dA/dt=V\,dC/dt+C\,dV/dt\), and hence:
Thus, amount balance alone does not give the concentration rate when volume is changing. Expansion can dilute concentration even if amount is increasing. This correction is only appropriate when the model actually represents a changing compartment volume; it should not be inserted into a fixed-volume model.
Worked Example: A Changing Extracellular Volume
Suppose an extracellular compartment has \(V_e=2.0\) L and \(C_e=3.0\) mg/L at an instant. Its net amount rate, after accounting for delivery, loss, and membrane exchange, is \(dA_e/dt=0.30\) mg/min. Its volume is expanding at \(dV_e/dt=0.020\) L/min. Find the instantaneous concentration rate.
Apply the product-rule result:
The volume-change term is \(0.060\) mg/min, so:
The amount is increasing at \(0.30\) mg/min, but the concentration increases more slowly because expansion contributes a dilution effect. Omitting the volume term would incorrectly give \(0.15\ \mathrm{mg\,L^{-1}min^{-1}}\).
Several checks expose common equation errors. First, the dimensions of \(F_{e\to i}\), \(R_e\), and \(R_i\) must all be amount per time before they are added. Second, dividing an amount rate by volume must produce concentration per time. Third, a positive flux must lower extracellular amount and raise intracellular amount when other processes are absent. Finally, the exchange law should give zero flux at its stated equilibrium condition \(C_i=K_{i/e}C_e\). These checks test the bookkeeping without requiring a numerical simulation.
An alternative is to write separate positive inward and outward fluxes instead of one signed net flux. That can be useful when the underlying mechanisms are distinct, but for a single reversible membrane law it adds bookkeeping and creates more opportunities to use inconsistent signs. A second alternative is to write concentration equations directly. This is compact only if the volume divisors are included from the start; writing amount balances first makes the transfer accounting easier to audit. Neither formulation changes the physical model when its definitions and signs are applied consistently.
These equations assume each compartment is well mixed, the flux law applies to the concentrations used in it, and transfer across the membrane is represented as equal and opposite amount rates. They do not represent drug temporarily accumulating within the membrane, spatial gradients inside either compartment, or a changing volume unless those features are added explicitly. A failed direction or unit check is a reason to inspect definitions and signs before interpreting a fitted concentration profile.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- With \(F_{e\to i}=0.40\) mg/h, \(V_e=2.0\) L, \(V_i=0.50\) L, and no other processes, what are the two concentration rates?
- If \(C_i=K_{i/e}C_e\), what is the membrane flux under the stated linear law?
- For \(F_{e\to i}=-0.30\) mg/min and no other processes, which compartment gains amount?
- A fixed-volume compartment has amount rate \(0.24\) mg/min and volume \(1.2\) L. What is its concentration rate?
- Why is \(dC/dt=(dA/dt)/V\) not generally valid when \(V\) changes with time?
Solutions. 1. The extracellular rate is \(-0.40/2.0=-0.20\ \mathrm{mg\,L^{-1}h^{-1}}\), and the intracellular rate is \(+0.40/0.50=+0.80\ \mathrm{mg\,L^{-1}h^{-1}}\). 2. The driving difference is zero, so \(F_{e\to i}=0\). 3. A negative signed flux is net movement from intracellular to extracellular space, so the extracellular compartment gains amount. 4. The rate is \(0.24/1.2=0.20\ \mathrm{mg\,L^{-1}min^{-1}}\). 5. Since \(A=VC\), differentiation gives \(dA/dt=V\,dC/dt+C\,dV/dt\); the volume-change term must be accounted for unless \(dV/dt=0\).