Answers to the Problem of Osmotic Balance

What solutions exist to this apparently fatal problem? There are three basic strategies that have developed in different types of cells. First, the problem could be eliminated by making the cell membrane impermeable to water. This turns out to be quite difficult to do and is not a commonly found solution to the problem of osmotic balance. However, certain kinds of epithelial cells have achieved very low permeability to water. A second strategy is commonly found and was likely the first solution to the problem. Here, the basic idea is to use brute force: build an inelastic wall around the cell membrane to physically prevent the cell from swelling. This is the solution used by bacteria and plants. The third strategy is that found in animal cells: achieve osmotic balance by making the cell membrane impermeable to selected extracellular solutes. This solution to the problem of osmotic balance works by balancing the concentration ofnon-permeating molecules inside the cell with the same concentration of nonper-meating solutes outside the cell.

To see how the third strategy works, it will be useful to work through some examples using a simplified model animal cell whose membrane is permeable to water. Suppose the model cell contains only one solute: non-permeating protein molecules, P, dissolved in water at a concentration of 0.25 M. We will then perform a series of experiments on this model cell by placing it in various extracellular fluids and deducing what would happen to its volume in each case. Assume that the initial volume of the cell is one-billionth of a liter (1 nanoliter, or 1 nl) and that the volume of the ECF in each case is infinite. This latter assumption means that the concentration of extracellular solutes does not change during the experiments, because the infinite extracellular volume provides an infinite reservoir of both water and external solutes.

The first experiment will be to place the cell in a 0.25 M solution of sucrose, which does not cross cell membranes. This is shown in Figure 3-4a. In this situation, only water can cross the cell membrane. For water to be at equilibrium, the internal osmolarity must equal the external osmolarity, or:

Because the internal and external osmolarities are both 0.25 Osm, this condition is met. Thus, there will be no net diffusion of water, and cell volume will not change.

Figure 3-4 Effects of various extracellular fluids on the volume of a simple model. (a) The ECF contains an impermeant solute (sucrose), and the osmolarity is the same as that inside the cell. (b) The ECF contains an impermeant solute, and the osmolarity is lower than that inside the cell. (c) The ECF contains a permeant solute (urea) and external and internal osmolarities are equal. (d) The ECF contains a mixture of permeant and impermeant solutes.

Initial volume = 1 nl

Initial volume = 1 nl

sucrose

sucrose

Final volume = 1 nl

Final volume = 1 nl

sucrose

sucrose

Initial volume = 1 nl

Initial volume = 1 nl

0.125 M

sucrose

0.125 M

sucrose

Final volume = 2 nl

Final volume = 2 nl

0.125 M

sucrose

0.125 M

sucrose

Initial volume = 1 nl

Initial volume = 1 nl

0.25 M urea

Initial volume = 1 nl

Initial volume = 1 nl

0.25 M urea

0.25 M urea

Final volume = 1 nl

Final volume = 1 nl

sucrose +

0.25 M urea

sucrose +

0.25 M urea

In the second example, shown in Figure 3-4b, the cell is placed in 0.125 M sucrose rather than 0.25 M sucrose. Again, only water can cross the membrane, and Equation (3-3) must be satisfied for equilibrium to be reached. In 0.125 M sucrose, however, the internal osmolarity (0.25 Osm) is greater than the external (0.125 Osm), and water will enter the cell until internal osmolarity falls to 0.125 M. This will happen when the cell volume is twice normal, that is, 2 nl. What would the equilibrium cell volume be if we placed the cell in 0.5 M sucrose rather than 0.125 M?

The point of the previous two examples is that water will be at equilibrium if the concentration of impermeant extracellular solute is the same as the concentration of impermeant internal solute. To see that the external solute must not be able to cross the cell membrane, consider the example shown in Figure 3-4c. In this case, the model cell is placed in 0.25 M urea, rather than sucrose. Unlike sucrose, urea can cross the cell membrane, and thus we must take into account both urea and water in determining diffusion equilibrium. In equation form, equilibrium will be reached when these two relations hold:

Here, Equation (3-4) specifies diffusion equilibrium for urea, and Equation (3-5) applies to diffusion equilibrium for water. Because the external volume is infinite, [urea]o will be 0.25 M at equilibrium, and according to Equation (3-4) [urea] must also be 0.25 M at equilibrium. Together, Equations (3-4) and (3-5) require that [P]i must be zero at equilibrium. Thus, the equilibrium volume is infinite, and the cell will swell until it bursts. Qualitatively, when the cell is first placed in 0.25 M urea, there will be no net movement of water across the membrane because internal and external osmolarities are both 0.25 Osm. But as urea enters the cell down its concentration gradient, internal osmolarity rises as urea accumulates. Water will then begin to enter the cell down its concentration gradient. The cell begins to swell and continues to do so until it bursts. Thus, an extracellular solute that can cross the cell membrane cannot help a cell achieve osmotic balance.

An interesting example is shown in Figure 3-4d. In this experiment, the model cell is placed in mixture of 0.25 M urea and 0.25 M sucrose. The equilibrium for urea will once again be given by Equation (3-4), and water will be at equilibrium when

Both Equation (3-4) and Equation (3-6) will be satisfied when [P] = 0.25 M, which is the initial condition. Therefore, in this example, the cell volume at diffusion equilibrium will be the normal volume, 1 nl. The point is that even if some extracellular solutes can cross the cell membrane, the presence of a nonpermeating external solute at the same concentration as the nonpermeating internal solute allows the cell to achieve diffusion equilibrium for water and thus to maintain its volume. This is the strategy taken by animal cells to avoid bursting. As shown in Table 2-1, the impermeant extracellular solute in the case of real cells is sodium.

In all the examples of osmotic equilibrium we just worked through, the answer was arrived at using just one rule: For each permeating substance (including water), the inside concentration must equal the outside concentration at equilibrium.

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