Question

An erythrocyte is placed in a hypotonic solution of 100 mOsm. What will happen to the cell?

A) The cell swells to three times its original volume, Vf = 3Vo.

B) The cell shrinks to one half its original volume, Vf = 1/2 Vo.

C) The cell swells to twice its original volume, Vf = 2Vo.

D) The cell volume increases by one third, Vf = 4/3 Vo.

E) The cell volume remains unchanged.

Answer 1

In a hypotonic solution of 100 mOsm, an erythrocyte will swell as water enters the cell, potentially leading to cell rupture, or hemolysis. The closest listed outcome is that the cell swells to twice its original volume.

Step-by-step explanation:

If an erythrocyte, or red blood cell, is placed in a hypotonic solution of 100 mOsm, the cell will swell due to the water entering the cell in an attempt to equalize solute concentration on both sides of the cell membrane. Given that the solution is hypotonic compared to the cell's internal environment, water will flow into the cell, potentially causing it to swell and possibly rupture, a process known as hemolysis. Therefore, none of the provided options are strictly correct, but the closest option would be (C), if we consider only the initial swelling without reaching the point of rupture, which is the cell swells to twice its original volume, Vf = 2Vo.