From Membrane Permeability to a Concentration-Change Rate
A membrane permeability is not, by itself, a rate at which a tissue concentration changes. The amount crossing the membrane also depends on how much membrane is available, while the resulting concentration change depends on the volume into which that amount enters or from which it leaves. Keeping these quantities separate makes it possible to see why identical permeability values can produce different tissue equilibration rates.
The starting point is the membrane flux relation from the earlier material on deriving membrane flux: for a homogeneous membrane with constant permeability, net exchange is proportional to membrane area and to the concentration difference across the membrane. Here, consider an extracellular compartment and an intracellular compartment. Assume the concentrations are expressed on the same basis, so equal concentrations represent equilibrium for the species being modeled.
The equations below assume that each compartment is well mixed, \(P\), \(A_m\), and compartment volumes remain constant, and membrane exchange is the only process changing the two concentrations. They also assume no binding, metabolism, or unequal equilibrium partitioning between the compartments. These assumptions isolate the effect of permeability, area, and volume; a model with additional processes must include them separately.
The signs follow the direction of the concentration difference. If \(C_e>C_i\), net transfer is into the cell: \(C_e\) decreases and \(C_i\) increases. If \(C_i>C_e\), the difference reverses and both signs of change reverse. The rate coefficients \(P A_m/V_e\) and \(P A_m/V_i\) need not be equal, because the same exchanged amount is distributed across different volumes.
Why Area and Volume Must Be Considered Together
The key quantity for a compartment receiving membrane flux is the ratio of membrane area to that compartment’s volume. For the intracellular concentration, the coefficient is \(P A_m/V_i=P(A_m/V_i)\). A larger membrane area increases exchange per unit time; a larger intracellular volume dilutes the same amount of exchange over more space. Permeability alone cannot capture either effect.
For a spherical cell of radius \(r\), surface area is \(4\pi r^2\) and volume is \(4\pi r^3/3\). Therefore, its surface-area-to-volume ratio is \(3/r\). If the extracellular concentration is held fixed, the intracellular concentration difference from that fixed value changes according to a first-order rate coefficient \(P A_m/V_i=3P/r\). This particular result assumes that the cell is spherical, its whole surface is accessible for exchange, and the surrounding concentration does not change appreciably.
Thus, at the same permeability, a smaller spherical cell has more membrane area per unit intracellular volume and a larger volume-normalized rate coefficient. This does not mean that a small cell has a greater membrane permeability: \(P\) is unchanged. The difference arises from geometry.
A Short Derivation of the Volume-Normalized Coefficient
The coefficient follows directly from a concentration mass balance. Over a short time interval, the net amount entering the cell is \(Q_m\,dt\). For fixed intracellular volume \(V_i\), that amount changes the intracellular concentration by \(dC_i=Q_m\,dt/V_i\). Dividing by \(dt\) and substituting the membrane flux relation gives the intracellular equation:
Here \(A_i\) denotes intracellular amount, not membrane area. The analogous extracellular balance has the opposite sign and uses \(V_e\). This derivation also shows why units provide a useful check: \(P A_m\) has units of volume per time, and dividing by \(V_i\) gives inverse time. Multiplying that coefficient by a concentration difference gives concentration per time, as required for \(dC_i/dt\).
Worked Example: Calculating Flux and Both Concentration Rates
Consider a hypothetical tissue region with \(P=2.00\times10^{-5}\) cm/s, \(A_m=150\) cm\(^2\), \(V_e=0.300\) mL, and \(V_i=0.200\) mL. At one instant, let \(C_e=10.0\) mg/L and \(C_i=2.00\) mg/L. Assume constant volumes and the exchange-only conditions stated above. Calculate the amount-transfer rate and both concentration-change rates.
First calculate the exchange capacity:
The concentration difference is \(10.0-2.00=8.00\) mg/L. Therefore, the amount-transfer rate into the cell is:
Convert the compartment volumes to liters: \(V_e=3.00\times10^{-4}\) L and \(V_i=2.00\times10^{-4}\) L. The concentration rates are then:
The same amount leaves the extracellular compartment and enters the intracellular compartment, but the smaller intracellular volume undergoes the larger concentration change. These are instantaneous rates at the stated concentrations; they are not constant over time because the concentration difference will change.
Cell Size and Surface Area per Unit Volume
For individual cells, the spherical result makes the size effect explicit. At fixed \(P\), doubling radius halves \(A_m/V_i\), and therefore halves the intracellular rate coefficient when the extracellular concentration is held constant. The following example calculates that change and a corresponding first-order half-time. The half-time here is for approach toward a fixed extracellular concentration under the stated simplified conditions.
Worked Example: Comparing Two Spherical Cell Sizes
Suppose two hypothetical cell populations have the same membrane permeability, \(P=1.00\times10^{-5}\) cm/s, and spherical radii of \(10.0\) and \(20.0\) μm. The surrounding concentration is held fixed. Convert the radii to centimeters, calculate the surface-area-to-volume ratios and intracellular rate coefficients, and compare the half-times.
Since \(1\) μm \(=10^{-4}\) cm, the radii are \(0.00100\) cm and \(0.00200\) cm. For the smaller cells:
For the larger cells:
For a fixed extracellular concentration, the concentration difference decays exponentially with rate coefficient \(k_i\), so its half-time is \(\ln(2)/k_i\). The smaller-cell half-time is \(0.693/0.0300=23.1\) s; the larger-cell half-time is \(0.693/0.0150=46.2\) s. Under these assumptions, doubling the cell radius doubles this half-time because it halves the area-to-volume ratio.
From Cell Geometry to Tissue Exchange Capacity
A tissue contains many cells, so the relevant membrane area is the total accessible area, not the area of one representative cell unless the model is explicitly scaled to one cell. Consider a hypothetical tissue with a fixed total intracellular volume made of spherical cells. If the cells are smaller, more cells fit into that volume; the total membrane area consequently increases. The next example calculates the aggregate effect.
Worked Example: Same Cell Volume, Different Total Membrane Area
Two hypothetical tissues each contain \(1.00\) cm\(^3\) of intracellular volume. In one tissue, all cells have radius \(10.0\) μm; in the other, all have radius \(20.0\) μm. Assume spherical cells and that every cell surface is accessible. Let \(P=1.00\times10^{-5}\) cm/s. Calculate total membrane area and the intracellular rate coefficient for each tissue.
For spherical cells, the total area divided by total cell volume is \(3/r\), regardless of the number of identical cells. Thus, for \(r=0.00100\) cm, total area is:
For \(r=0.00200\) cm, total area is:
The corresponding exchange capacities are \(P A_{m,\text{total}}=0.0300\) cm\(^3\)/s and \(0.0150\) cm\(^3\)/s. Dividing each by the same \(1.00\) cm\(^3\) intracellular volume gives \(k_i=0.0300\) s\(^{-1}\) and \(0.0150\) s\(^{-1}\), respectively. The smaller-cell tissue has twice the total area and twice the volume-normalized rate coefficient, despite having the same total intracellular volume and the same permeability.
Choosing a Scaling Approach and Checking the Model
There are several ways to represent membrane exchange, but they answer different questions. Keeping \(P\) and \(A_m\) separate is useful when permeability and geometry are known or may change independently. Using \(P A_m\) as a single exchange capacity is convenient when only their product is identifiable. Using \(P A_m/V_i\) is useful in a concentration-based equation because it directly multiplies a concentration difference to give a concentration rate. These forms are equivalent when units and definitions are consistent; the volume-normalized coefficient is not a replacement for the underlying membrane capacity.
| Quantity | Units | Role in the model |
|---|---|---|
| \(P\) | length/time | Permeability of the membrane material or effective membrane pathway |
| \(A_m\) | area | Effective membrane area available for exchange |
| \(P A_m\) | volume/time | Amount-transfer capacity per unit concentration difference |
| \(P A_m/V_i\) | 1/time | Coefficient multiplying the concentration difference in the intracellular equation |
A perfusion-based description is a different model choice: it scales exchange with blood flow and an assumed tissue-blood relationship, whereas the equations here scale membrane transfer with \(P A_m\). One should not substitute a flow rate for \(P A_m\) merely because both have units of volume per time. They represent different mechanisms. Which description is appropriate depends on whether membrane passage is the limiting process and on whether the relevant area and permeability can be represented adequately.
Useful checks include confirming that permeability and area units multiply to volume per time, converting all volumes to units compatible with that product, and verifying that the sign of each concentration rate agrees with the direction of the gradient. A large value of \(P A_m\) does not necessarily imply a large concentration rate if the receiving volume is also large. Likewise, a measured or calculated cell radius should use the same length units as \(P\).
The geometric calculation also has limits. Real cells are not perfect spheres; membrane folds, cell contacts, and inaccessible surfaces can make effective area differ from geometric area. A tissue-level \(A_m\) may be an aggregate over many cells, and scaling it as though every surface were exposed can overstate exchange. In addition, the constant-\(P\) assumption can fail if permeability varies with conditions or membrane state. When these features matter, use an effective area or permeability justified for the model rather than treating a geometric estimate as exact.
- Using permeability alone. The reasoning “\(P\) is high, so exchange must be fast” omits membrane area and compartment volume. Calculate \(P A_m\), then divide by the relevant volume for a concentration-rate coefficient.
- Confusing total area with area-to-volume ratio. A larger tissue can have more total area simply because it contains more cells. To compare concentration changes fairly, relate accessible area to the volume being changed.
- Mixing units. Multiplying cm/s by cm\(^2\) gives cm\(^3\)/s, not L/min. Convert before combining with a volume or concentration expressed in other units.
- Assuming equal concentration rates in both compartments. Equal and opposite amount-transfer rates do not imply equal concentration changes. The rates differ when \(V_e\) and \(V_i\) differ.
Check Your Understanding
Use the assumptions and equations in this tutorial. The solutions are included below so you can check each calculation.
- A membrane has \(P=4.00\times10^{-6}\) cm/s and \(A_m=250\) cm\(^2\). What is \(P A_m\) in cm\(^3\)/s?
- If \(P A_m=0.0600\) mL/min and \(V_i=0.300\) mL, what is the intracellular rate coefficient in min\(^{-1}\)?
- For spherical cells with the same permeability, what happens to \(A_m/V_i\) when radius is tripled?
- Two compartments have the same membrane exchange capacity, but one has twice the volume of the other. For the same concentration difference, how do their concentration-change magnitudes compare?
Solutions. (1) \(P A_m=(4.00\times10^{-6}\ \text{cm/s})(250\ \text{cm}^2)=1.00\times10^{-3}\) cm\(^3\)/s. (2) \(k_i=(0.0600\ \text{mL/min})/(0.300\ \text{mL})=0.200\) min\(^{-1}\). (3) Since \(A_m/V_i=3/r\), tripling \(r\) reduces the ratio to one-third its original value. (4) Concentration-change magnitude is exchange rate divided by volume. The compartment with twice the volume has half the concentration-change magnitude for the same amount-transfer rate and concentration difference; the direction is determined by whether the compartment is gaining or losing drug.