AP Biology
Diffusion and Osmosis in Model Systems
In parts 1 and 2 of this lab, you will have the opportunity to investigate the processes of diffusion and
osmosis in model membrane systems. You will also investigate the effect of solute concentration on
water potential as it relates to living plant tissues.
Objectives: At the completion of this AP laboratory, you should be able to:
1. Describe the mechanisms of diffusion and osmosis.
2. Describe how solute size and molar concentration affect the process of diffusion through a
selectively-permeable membrane.
3. Describe the relationship between solutions that are hypotonic, hypertonic, or isotonic.
4. Design an experiment to demonstrate water potential.
5. Relate osmotic potential to solute concentration and water potential.
6. Describe how pressure potential could affect the water potential of a solution.
7. Describe the effects of water gain or loss in animal and plant cells.
8. Calculate the water potential of living plant cells from experimental data.
Background Information
Part I. Osmosis.
In this part of the lab, you will use dialysis tubing filled with different molarities of sucrose to
investigate the relationship between solute concentration and the movement of water through a
selectively permeable membrane (the process of osmosis).
When a dialysis bag containing a sucrose solution is placed in fresh water, the bag accumulates water
as a result of osmosis. Because there is a higher concentration of water outside the bag than inside the
bag, water diffuses into the bag. The water outside the bag is said to be hypotonic to the solution in
the bag; the solution inside the bag is hypertonic relative to the water outside the bag. Water always
moves through a selectively permeable membrane (sucrose cannot move) from hypotonic to
hypertonic solutions.
In theory, if you were to add solute to the water outside the bag, you could decrease the water
concentration out there. It should be possible to add just the right amount of solute to the water so that
the concentration of dissolved substances outside the bag is the same as the concentration of dissolved
substances inside the bag.. Such solutions are said to be isotonic, and no net movement of water will
occur. If you were to continue to add solute to the water outside the bag, you would decrease the
concentration of water outside the bag until the solution outside became hypertonic to the solution
inside. Then, the net movement of water would be in the opposite direction: from inside to outside,
and the bag would shrink.
Part 2. Determining the Water Potential of Potato Cells.
In this part of the lab, you will use cores of potato tissue placed in different molar concentrations of
sucrose in order to determine the water potential of potato cells. First, however, let’s explore what we
mean by the idea of water potential!
In animal cells, movement of water into and out of the cell is influenced by the relative concentration
of solute on either side of the cell membrane. If water moves out of the cell, the cell will shrink,
(crenulate), and if the water moves into the cell it will swell, and may even burst (cytolize). In plant
cells, the presence of a rigid wall prevents cells from bursting as water enters the cells, but pressure
eventually builds up inside the cell and affects the process of osmosis.
In predicting which direction water will move through living plant tissues, a quantity known as water
potential is used. Water potential, abbreviated by the Greek letter SI refers to the potential energy of
water. It has many components, but the two on which we will focus are osmotic potential which
expends on solute concentration, and pressure potential ,which results from the exertion of pressure
(positive or negative) on a solution. We express this as:
ψ
water potential
=
ψp
= pressure potential
ψs
+
+
osmotic potential
Water will always move from an area of higher water potential (higher potential energy) to an area of
lower water potential (lower potential energy). Stated another way, water potential measures the
tendency of water to leave one place in favor of another place.
The water potential of pure water in a beaker open to the atmosphere is “0” ( ψ = 0 ), because both
the osmotic and pressure potentials are taken to be zero ( ψs = 0; ψp = 0). An increase in positive
pressure raises the pressure potential, and therefore raises the water potential. The addition of solute
to water lowers the osmotic potential (makes negative), and therefore lowers the water potential.
Therefore, any solution at atmospheric pressure (ψp = 0) will always have a negative water
potential. For example, a 0.1 M sucrose solution at atmospheric pressure (ψp = 0) has an osmotic
potential (ψs) of –2.3 bars due to the solute, and thus a total water potential of –2.3 bars ( 0 + -2.3 =
-2.3). A bar is a metric unit of pressure, measured with a barometer, that is about the same as one
atmosphere.
When a solution, such as that inside a potato cell, is separated from pure water by the selectively
permeable cell membrane, water will move by osmosis from the surrounding area where water
potential is higher (ψ = 0 ) into the cell where water potential is lower due to its dissolved solutes (ψ
is negative). In a case where the solute cannot leave the cell, the movement of water into the cell
causes the cell to swell, and the cell membrane pushes against the cell wall. This creates positive
turgor pressure in the cell.
Eventually, enough positive turgor pressure builds up to oppose the more negative osmotic pressure of
the cell. This will continue until the water potential of the cell equals the water potential of the pure
water outside the cell (ψ cell = ψ outside the cell = 0 ). At this point, a dynamic equilibrium is
reached and NET movement of water will cease.
If you add solute to the water outside the potato cells, you decrease the water potential of the solution
surrounding the cells. It should be possible to add just the right amount of solute to the water so that
the water potential outside the cell is the same as the water potential inside the cell. Therefore, net
water movement will cease. At this point, the water potential of the solution is equal to the water
potential of the turgid potato cells. (It is this information that you will use to calculate the water
potential of cells in the potato cores, which have been soaking in different molarities of sucrose.)
Eventually, if enough solute is added to the solution in the beaker, the water potential of the cell will
be greater (less negative) than the water potential of the solution in the beaker because of the turgor
that still exists within the cells. Water will diffuse out of the potato cells in response to a pressure
gradient, from the area of higher water potential (inside) to the area of lower water potential (outside).
If enough water was lost, the cell membrane would shrink and fall away from the cell wall, and the
cell would become plasmolyzed.
By weighing some potato cores and immersing them in different sucrose solutions, then re-weighing,
you should be able to pinpoint the sucrose solution which corresponds to the water potential of the
potato tissue!
Lab: Diffusion and Osmosis in Model Systems
Part 1 – SUGAR WATER
Objective: To determine the sucrose concentration of six unknown sugar water solutions
using osmosis.
Materials:
Dialysis tubing
Sucrose solutions (0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
Cups
Distilled water
Procedure:
You will be responsible to designing you own procedure. Make sure you use enough detail
that would allow someone to repeat your experiment. Here is some help to get you started:
Dialysis tubing (which allow water to penetrate, but not sucrose):
1. The tubing must be cut into strips about 15 cm in length and soaked in fresh distilled
water before you can open it.
2. Open one of the strips by rubbing it between your fingers. Then, twist one end of the
tubing, fold over a small portion of the twisted section and firmly tie it closed with a piece
of string to make a water-tight seal.
3. Make sure you record the amount of solution you put in each bag.
4. Remove most of the air from the bag by drawing the dialysis bag between two fingers. To
close the bags, tie a knot at the other end, thereby sealing the solution within the bag. BE
SURE that you have left 1.5 to 2 times as much empty space as that taken up by the
volume of the solution in the bag. (This leaves enough unfilled space within the bag to
accommodate the possible accumulation of water.)
5. Carefully rinse off and blot dry the outside of each dialysis bag. Measure and record the
initial mass of each of your bags.
6. Place each bag in one of the beakers of distilled water and label the beakers to indicate the
molarity of the solution in the dialysis bags.
7. Let stand overnight.
Part 2 – WATER POTENTIAL OF POTATO CELLS
Objective: To determine the sucrose solution from PART A that is iso-osmolar to the potato
(the same sucrose concentration) an calculate the water potential of the potato cells.
Materials per group:
Fresh potato cubes
Balance
Cups
Sucrose solutions (0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
Procedure:
1. Design an experiment to determine which of the six sucrose solutions is iso-osmolar to the
potato.
Graphs To Include in Your Report
1. Graph 1: Table 1 (Part 1) data.
2. Graph 2: Table 2 (Part 2) data.
Data Analysis, Water Potential of the Potato Cells
1. Determine the iso-osmolar concentration of the potato tissue and the sucrose solution by
examining Graph 2. Note the point on the curve corresponding to zero change in potato
mass (y-axis); drop a perpendicular line to find the corresponding point on the x-axis.
This point will represent the concentration of sucrose iso=osmolar to the potato issue. At
this concentration there is no net gain or loss of water from the tissue.
(From graph) Iso-osmolar concentration = ____________M.
2. The osmotic potential (ψ) of iso-osmolar sucrose solution can be calculated from the
molarity of the solution by applying this formula:
ψ = iCRT
where:
i =
ionization constant (for sucrose, we can use a value of 1 since sucrose does not ionize
in water).
C=
the iso=osmolar concentration (usually expressed in molality; read off the graph)
R=
pressure constant (handbook value R = 0.0821 bar/mole deg K for water)
T=
temperature, degree Kelvin (273 + degrees C) of solution.
What is the osmotic potential of the sucrose solution found (by your graph) to be iso-osmolar
to the potato tissue?
3. Since the pressure potential (ψp) of our solutions was zero, we can calculate the water
potential of the solution from the equation for water potential given in the pre-lab. The water
potential of the solution at equilibrium will be equal to the water potential of your potato cells.
What, then is the water potential of your potato cells?
Questions (Discussion Section)
1. You have calculated the water potential of your potato cells. Earlier you learned that pure
water at standard pressure and temperature has a water potential of 0. If you threw your
potato into a jar of pure H20, would the potato shrink, or swell? Explain.
2. We said the water potential (ψ) of the potato cells is equal to the water potential of the
solution, at equilibrium. Is the osmotic potential (ψs) of your potato cells the same as that
of the solution? Why or why not?
3. If a potato cylinder is allowed to dehydrate by sitting in the open air, would the potato
cells’ water potential become higher or lower? Explain.
Data
Table 1, Part 1 Data
Contents of
Dialysis Bags
Initial
Mass
Final
Mass
Percent Change
in Mass
Class Ave.
% Change
Initial
Mass
Final
Mass
Percent Change
in Weight
Class Ave.
% Change
0.0 M Sucrose
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
Table 2; Part 2 Data
Sucrose Solutions
Soaking Potato Cores
0.0 M Sucrose
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
HINT: % Change = Final Mass – Initial Mass
Initial Mass