Cells and Their Environment
Experiment A
Examine the diagram to the right, and read the description of the experiment.
The solutions in the arms of the U-tube (at right) are separated by a selectively permeable
membrane that is permeable to water and solute A but not to solute B. Forty grams (g) of
solute A and 20g of solute B have been added to the water on side 1 of the U-tube.
Twenty grams (g) of solute A and 40g of solute B have been added to the water on side 2
of the U-tube.
Answer the following questions related to Experiment A.
1. Draw the U-tube and label it showing the solute amounts on both sides of the selectively (semi) permeable
membrane.
40g solute A
20g solute B
20g solute A
40g solute B
2. Initially (before anything moves), the solution on side 1 is ISOTONIC to the solution on side 2 of the membrane. (The
question is asking about the solution, so you look at the total solute per volume of water.)
Assume that after a period of time, equilibrium has been reached.
3a. How many grams of solute A will be in solution on side 1 of the U-tube? 30g
3b. How many grams of solute A will be in solution on side 2 of the U-tube? 30g
3c. Explain your answer in depth. There is more solute on side 1 and it will diffuse down its concentration gradient into
side 2 since the membrane is permeable to solute A. There will be a net movement of solute A into side 2 until it reaches
equilibrium.
4a. How many grams of solute B will be in solution on side 1 of the U-tube? 20g
4b. How many grams of solute B will be in solution on side 2 of the U-tube? 40g
4c. Explain your answer in depth. The membrane is impermeable to solute B so even though there is a concentration
gradient, it cannot diffuse across the membrane.
5a. What has happened to the water level in the U-tube? The water has decreased on side 1 and increased on side 2.
5b. Explain your answer in depth. Since 10g solute A moved into side 2, side 2 is now hypertonic to side 1 (or side 1 is
hypotonic to side 2). Water will move from the hypotonic solution (side 1 ) into the hypertonic solution (side 2) since
there is more “free and available” water in a hypotonic solution.
*Remember: “hypo” and “hyper” refer to the amount of solute, not the amount of available water! Hypotonic
has less solute, therefore more available water (available to leave).
Experiment B
Read the description of the experiment below and answer the following questions about what will happen to the red
blood cells in each flask.
Flasks X, Y, and Z contain solutions with different concentrations of the solute NaCl. Flask X has 0.9% NaCl, Flask Y has
0.5%, and Flask Z has 1.5% NaCl. Red blood cells (0.9% NaCl) will be placed in each flask.
6. Draw the red blood cells in each of the flasks and label each of the flasks and all concentrations.
0.9%
NaCl
0.9%
Flask “X”
0.5%
NaCl
0.9%
Flask “Y”
1.5%
NaCl
0.9%
Flask “Z”
7. Predict what will happen to the red blood cells in Flask X. In your prediction, be sure to include the osmotic state
(isotonic, hypertonic, hypotonic) of the solution and the red blood cell, as well as direction (if any) of water movement,
and effect on the cell’s size.
The cell is isotonic to the solution so there won’t be any net water movement and the cell will stay the same size.
8. Predict what will happen to the red blood cells in Flask Y. In your prediction, be sure to include the osmotic state
(isotonic, hypertonic, hypotonic) of the solution and the red blood cell, as well as direction (if any) of water movement,
and effect on the cell’s size.
The cell is hypertonic to the solution (or the solution is hypotonic to the cell) so there will be net water movement into
the cell and the cell will swell in size.
9. Predict what will happen to the red blood cells in Flask Z. In your prediction, be sure to include the osmotic state
(isotonic, hypertonic, hypotonic) of the solution and the red blood cell, as well as direction (if any) of water movement,
and effect on the cell’s size.
The cell is hypotonic to the solution (or the solution is hypertonic to the cell) so there will be net water movement out of
the cell and the cell will shrink in size.