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Assume we have a cell which contains nonpermeating molecules (P; not ionized) at a concentration of 300 mM/L cell water. Assume the Initial cell volume is l.0 nanoliter. What will be the final (i.e. equilibrium) volume when the cell is immersed in a large volume of the following solutions?

A 300 mOsm sucrose (sucrose cannot penetrate the membrane)
B 150 mOsm sucrose
C 600 mOsm sucrose
D. 300 rnOsm glycerol (glycerol can penetrate the membrane)
E pure water
F 600 mOsm glycerol
G 300 mOsm sucrose & 300 mOsm glycerol

You’ll note that in 3 of these exercises we have introduced something new: The idea of a permeating solute (glycerol). The cell membrane is no longer perfectly semi-permeable (i.e. permeable only to solvent), and, in fact, all biological membranes are permeable to some solutes. The point to note in the problems is that the glycerol contributes nothing to equilibrium volume control; the final volume of the cell is the same as if tne glycerol were not present; all that matters is the non-permeating species ~(sucrose). And this leads to definition of a new term: tonicity. In the examples, both 300 mM sucrose and 300 mM glycerol are isosmotic with the cell contents - they each contain 300 mM of particles per liter of solution. But while the colligative properties of these solutions are identical, the biological properties clearly are not. (The cell was quite happy in sucrose, but swelled and burst in glycerol.) Thus we define a solution as isotonic if it does not affect cell volume. On this basis, the 300 mM sucrose is isotonic but the 300 mM glycerol is hypotonic since it causes the cell to swell (as if the cells were exposed to a hypo-osmotic solution of non-permeating solute . To be certain you understand, go back and label each test solution in terms of its osmolarity and tonicity.

Time Course of Volume Changes. In the problems above we were concerned only with the initial and final (equilibrium) states of the cell. Now let’s briefly return to example F (600 mOsm glycerol) and consider what happens to cell volume immediately after the cell is exposed to the glycerol solution. The solution is hyperosmotic to the cell, and so initially water will leave the cell. Glycerol of course will enter the cell. The cell volume, of course, depends only upon the movement of water. Thus, initially, the cell will shrink. Then, as glycerol con-tinues to enter the cell, the internal osmolarity increases, and so the movement of water will reverse its direction and the cell will swell and burst. So the change in volume with time might be graphed like this:

What would the curve for solution G look like?
The point is that the rate of change of volume can tell us something about the way solutes cross the membrane. The exact time course of volume change will depend on how easily solute can cross the membrane. If it crosses slowly (relative to water), the cell will stay shrunken for a longer period, although eventually it must swell and burst. By using solutes which have different molecular sizes, it is possible to estimate the size of the pores in the membrane, since small molecules will pass through the pores more easily (and therefore faster) than large molecules. Such experiments suggest that pores are about 6 Å in diameter.
Finally, in the experimental exercises above, we assumed that the composition of the external bathing solution was constant (i.e., that the cells were bathed in a large volume of ECF). However, in the body, this condition is not met, and movement of substances into or out of cells can change the composition of the ECF. This added complexity arises because the total ECF volume is not large compared to the ICF volume. Specifically, in a healthy young adult male:

Total body water = 42 liters
ICF = 23 liters
Exchangeable*ECF = 14.5 liters
*The rest of the ECF, 4.5 1, is inexchangeable, being sequestered from blood and interstitial fluid in bone, GI tract, bladder, etc.

Now assume that the normal Na+ and K+ concentrations of ICF and ECF are as given below, and that the man suffers a severe heart attack, which (as you will study later) can cause his cells to lose a little K+ and gain a little Na+:

Specifically, assume that the 4 mM/L K+ loss from the cells is exactly balanced by a gain of Na+. Thus, no other significant fluxes occur. What will be the ECF concentrations of Na+ and K+?