Worked Example: A Drug Crossing Into Tissue Cells

From Membrane Exchange to a Concentration Profile

A tissue-cell concentration need not follow the extracellular concentration immediately. If drug crosses the cell membrane at a finite rate, intracellular concentration can lag behind a changing extracellular profile. The two-compartment model makes that lag explicit: the membrane flux is driven by the concentration difference, and the intracellular amount changes as a result.

Here the extracellular concentration is treated as a specified input, \(C_e(t)\). This is useful when the extracellular profile is known from a surrounding reservoir or supplied by another part of a larger model. The intracellular concentration is the response to that input.

Definition: Let \(V_e\) and \(V_i\) be the extracellular and intracellular volumes, \(C_e(t)\) and \(C_i(t)\) their drug concentrations, and \(A_e(t)=V_eC_e(t)\) and \(A_i(t)=V_iC_i(t)\) their amounts. Let \(PS\) be the membrane permeability-surface-area product. With equal free-drug concentrations at passive equilibrium, the exchange flux from extracellular to intracellular space is \(J(t)=PS[C_e(t)-C_i(t)]\). The intracellular balance is \(dA_i/dt=J(t)\), or, for constant \(V_i\), \(dC_i/dt=(PS/V_i)[C_e(t)-C_i(t)]\).

The model assumes well-mixed extracellular and intracellular compartments, fixed volumes, a constant \(PS\), and passive exchange proportional to the concentration difference. It also assumes no intracellular metabolism, binding, active transport, or other gain or loss terms. In particular, the equilibrium concentration ratio is one. If a tissue partition relationship requires a different ratio, the flux law must represent that relationship rather than silently using this form.

Define the exchange rate constant \(k=PS/V_i\), with units of inverse time. For a specified extracellular input and initial intracellular concentration \(C_i(0)\), the governing equation is

\frac{dC_i}{dt}+kC_i=kC_e(t).

Solution for a Changing Extracellular Input

Multiplying the differential equation by the integrating factor \(e^{kt}\) gives

\frac{d}{dt}\left(e^{kt}C_i(t)\right)=k e^{kt}C_e(t).

Integrating from time zero to \(t\), then dividing by \(e^{kt}\), yields the general response:

C_i(t)=C_i(0)e^{-kt} +k\int_0^t e^{-k(t-s)}C_e(s)\,ds.

The first term is the remaining contribution from the initial intracellular concentration. The integral accumulates the effects of extracellular concentrations at earlier times, with older inputs attenuated more strongly. This expression also shows why the intracellular profile is generally not obtained by simply multiplying the current extracellular concentration by a constant.

For constant \(C_e\), the solution reduces to \(C_i(t)=C_e+[C_i(0)-C_e]e^{-kt}\). Thus \(C_i\) approaches \(C_e\) exponentially, with characteristic exchange time \(1/k=V_i/PS\). A larger \(PS\), or a smaller intracellular volume, produces a faster response.

Worked Example: Uptake During a Declining Extracellular Profile

Suppose \(V_e=1.00\) L, \(V_i=0.500\) L, and \(PS=0.300\) L/h. A hypothetical extracellular reservoir imposes \(C_e(t)=8.00e^{-0.400t}\) mg/L, with time in hours. Initially \(C_i(0)=0\). Find \(C_i\) and both compartment amounts at 2.00 h.

First, \(k=PS/V_i=(0.300\ \text{L/h})/(0.500\ \text{L})=0.600\ \text{h}^{-1}\). The intracellular balance is therefore \(dC_i/dt+0.600C_i=4.80e^{-0.400t}\), where the right-hand side has units mg/L/h.

Using the general solution, or integrating directly,

C_i(t)=\frac{0.600(8.00)}{0.600-0.400} \left(e^{-0.400t}-e^{-0.600t}\right) =24.0\left(e^{-0.400t}-e^{-0.600t}\right)\ \text{mg/L}.

At \(t=2.00\) h, \(e^{-0.800}=0.449328964\ldots\) and \(e^{-1.200}=0.301194212\ldots\). Thus \(C_i(2.00)=24.0(0.449328964\ldots-0.301194212\ldots)\approx3.55523\) mg/L. The intracellular amount is \(A_i=V_iC_i=(0.500\ \text{L})(3.55523\ \text{mg/L})=1.77762\) mg.

The imposed extracellular concentration is \(C_e(2.00)=8.00(0.449329)=3.59463\) mg/L, so \(A_e=(1.00\ \text{L})(3.59463\ \text{mg/L})=3.59463\) mg. The instantaneous flux is \(J=0.300(3.59463-3.55523)=0.01182\) mg/h, still directed into the cells. The result is consistent with a small positive concentration difference at this time.

The total amount in the two compartments is \(3.59463+1.77762=5.37225\) mg, less than the initial extracellular amount of \(8.00\) mg. This is not a violation of mass conservation: \(C_e(t)\) was imposed by an external reservoir, which removes drug as its concentration declines. The two compartments alone are not a closed system.

Washout and the Direction of Membrane Flux

A changing extracellular concentration can reverse the direction of exchange. When \(C_e>C_i\), the flux is inward; when \(C_e<C_i\), stored intracellular drug flows outward. A declining extracellular profile therefore does not imply that intracellular concentration declines at every instant: intracellular concentration can continue rising while it catches up with the current extracellular concentration.

Worked Example: Intracellular Washout After Extracellular Drug Is Removed

Use \(V_i=0.500\) L and \(PS=0.300\) L/h, so \(k=0.600\ \text{h}^{-1}\). Suppose a hypothetical washout holds \(C_e=0\) and the initial intracellular concentration is \(5.59044631\) mg/L. Find the intracellular concentration and amount after 2.00 h.

With zero extracellular concentration, the equation becomes \(dC_i/dt=-0.600C_i\), whose solution is \(C_i(t)=C_i(0)e^{-0.600t}\). Therefore,

C_i(2.00)=5.59044631e^{-1.200} =5.59044631(0.3011942119) \approx 1.68381007\ \text{mg/L}.

Rounded to six decimal places, \(C_i(2.00)=1.683810\) mg/L. The intracellular amount is \(A_i(2.00)=(0.500\ \text{L})(1.68381007\ \text{mg/L})=0.841905\) mg, or \(0.8419\) mg to four significant figures. The amount has decreased from \((0.500)(5.59044631)=2.79522\) mg. Since \(C_e=0<C_i\), \(J=PS(C_e-C_i)<0\), correctly indicating net outward flux. In this setup the surrounding sink removes the drug leaving the cells.

Constant Exposure and Limiting Behavior

A constant extracellular concentration provides a useful check on the model. The cell concentration approaches that constant value, but reaches it only asymptotically when \(PS\) is finite. In contrast, if \(PS/V_i\) is very large relative to the rate at which \(C_e(t)\) changes, \(C_i\) tracks \(C_e\) closely. That is the fast-exchange limit behind an instantaneous-equilibration approximation; it should not be assumed when the exchange time is comparable to the exposure timescale.

Worked Example: Response to a Constant Extracellular Concentration

Suppose \(V_i=0.500\) L, \(PS=0.300\) L/h, \(C_e=6.00\) mg/L is held constant, and \(C_i(0)=1.00\) mg/L. Calculate the intracellular concentration and amount at 1.00 h.

Here \(k=0.600\ \text{h}^{-1}\). Substitution into the constant-input solution gives \(C_i(t)=6.00+(1.00-6.00)e^{-0.600t}\) mg/L. At 1.00 h,

C_i(1.00)=6.00-5.00e^{-0.600} =6.00-5.00(0.548812) =3.25594\ \text{mg/L}.

The intracellular amount is \((0.500\ \text{L})(3.25594\ \text{mg/L})=1.62797\) mg, an increase of \(1.12797\) mg from the initial \(0.500\) mg. The exchange flux at 1.00 h is \(0.300(6.00-3.25594)=0.82322\) mg/h, positive as expected. Under the maintained constant extracellular concentration, the limiting value is \(C_i\to6.00\) mg/L and \(J\to0\) as \(t\to\infty\).

Amount Accounting, Diagnostics, and Model Choice

The exchange term transfers equal and opposite amounts between compartments. If both compartments are closed and exchange is the only process, their equations are \(dA_e/dt=-J\) and \(dA_i/dt=J\). Adding them gives \(d(A_e+A_i)/dt=0\), so total amount is conserved. When \(C_e(t)\) is prescribed instead, \(A_e=V_eC_e\) is constrained by that input and need not obey the closed-system equation.

For a prescribed extracellular profile, let \(R_e(t)\) denote the external source rate into the extracellular compartment, with positive values adding drug. Its balance is \(dA_e/dt=R_e-J\). Since \(A_e=V_eC_e\) for fixed volume, the external source required to enforce the prescribed profile is

R_e(t)=V_e\frac{dC_e}{dt}+J(t).

Then \(d(A_e+A_i)/dt=R_e\): the change in the two-compartment total is exactly the external input or removal. This distinction is important when checking a numerical implementation. A changing prescribed \(C_e\) is a boundary condition, not a second closed-compartment prediction.

The finite-permeability model differs from a perfusion-limited tissue model, which assumes tissue concentration adjusts rapidly enough to remain in equilibrium with blood under its specified partition relationship. It also differs from instantaneous membrane equilibrium, which eliminates the exchange lag. Finite-rate exchange is appropriate when membrane transfer itself can limit equilibration; the alternatives may be adequate when exchange is demonstrably much faster than the surrounding concentration changes.

Useful diagnostics follow directly from the equations. The sign of \(J\) must agree with the sign of \(C_e-C_i\). For constant \(C_e\), \(C_i\) must move monotonically toward \(C_e\) from its initial value. For a closed pair, the sum \(A_e+A_i\) must remain constant. A negative concentration from nonnegative initial concentrations and nonnegative inputs, a persistent amount residual in a closed pair, or an inward flux when \(C_i>C_e\) indicates a sign, unit, or implementation error. Also check that \(PS/V_i\) has units of inverse time and \(PS(C_e-C_i)\) has units of amount per time.

Common pitfall: A learner may calculate \(A_e+A_i\) under a prescribed, declining extracellular profile and conclude that mass conservation has failed because the total falls. The correction is to identify the external reservoir explicitly: it enforces \(C_e(t)\), so the total in the two modeled compartments changes by the net external input or removal. Conservation applies to the closed system that includes that reservoir, not automatically to the two compartments alone.

The model is a deliberate simplification. It does not represent spatial gradients within either compartment, changing membrane properties, active transport, intracellular binding, or metabolism. If those processes matter, the concentration difference alone is not a sufficient description of exchange or intracellular disposition. The model is most useful when its assumptions are plausible and its predicted exchange timescale can be compared with the timescale of the extracellular input.

Check Your Understanding

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

If \(PS\) doubles while \(V_i\) stays fixed, what happens to the exchange time \(V_i/PS\)?
For \(C_e=4.0\) mg/L and \(C_i=6.0\) mg/L, what is the direction of membrane flux?
With constant \(C_e=5.0\) mg/L, \(C_i(0)=1.0\) mg/L, and \(k=0.5\ \text{h}^{-1}\), calculate \(C_i(2.0\ \text{h})\).
Why can \(A_e+A_i\) change when \(C_e(t)\) is specified externally?

Solutions. (1) It halves, because \(V_i/PS\) is inversely proportional to \(PS\). (2) \(J=PS(4.0-6.0)<0\), so net flux is outward, from cells to extracellular space. (3) \(C_i(t)=5.0+(1.0-5.0)e^{-0.5t}\), so \(C_i(2.0)=5.0-4.0e^{-1}=5.0-4.0(0.367879)=3.52848\) mg/L. (4) The external reservoir may add or remove drug to impose the specified profile; the two-compartment total changes by that net external exchange.