AP Biology
James
Pre-Lab: 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:
water potential
=
+
= pressure potential
+
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” (
= 0 ),
because both the osmotic and pressure potentials are taken to be zero (
= 0;
= 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
(
= 0) will always have a negative water potential. For example, a 0.1 M sucrose
solution at atmospheric pressure (
= 0) has an osmotic potential (
) 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 (
= 0 ) into the cell where water potential is lower due to its
dissolved solutes (
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 ( cell = 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 reweighing, you should be able to pinpoint the sucrose solution which corresponds to the water
potential of the potato tissue!
AP Biology
N
D
P
Lab: Diffusion and Osmosis in Model Systems
Refer to the Pre-Lab for lab OBJECTIVES and BACKGROUND INFORMATION. These
model systems for testing principles of diffusion and osmosis are probably new to you, so you
are not being asked to make an hypothesis for this lab.
Part I. Osmosis.
(a) Materials per lab group:
three 15 cm strips of pre-soaked dialysis tubing
three sucrose solutions, 15 ml each (available: 0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
three 250 ml cups
distilled water
(b) Procedure:
1. Obtain three 15-cm strips of pre-soaked dialysis tubing, and keep in fresh distilled water
in use.
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. Repeat with the other strip of dialysis tubing. Now you have two dialysis bags, each open
at one end.
4. Pour 10 ml of one of your test solutions into one bag, and 10 ml of your other test solution
in the other bag.
5. 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, as before (step 2), 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.)
6. Fix three beakers with distilled water.
7. Carefully rinse off and blot dry the outside of each dialysis bag. Measure and record the
initial mass of each of your two bags in Table 1. (Are you keeping track of which bag
contains which solution?)
8. 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.
9. Let stand overnight.
10. At the end of 24 hours, remove the bags from the water and carefully blot and re-weigh
them. Record the final masses in Table 1 and calcualte % change in mass. Pool class
data.
11. Graph the % change in mass as a function of dialysis bag sucrose solution molarity.
Part 2. Determining the Water Potential of Potato Cells.
(a) Materials per group:
Cork borer tool, inner diamter (id) = 5 mm.
Fresh potato
Razor blade or scalpel
Metric ruler
Balance
six cups
three sucrose solutions, 100 ml each (available: 0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
(b) Procedure:
1. Obtain 100 ml of each of two sucrose solutions assigned to your group. Pour each
solution into a separate, labeled 250 ml beaker.
2. Make 4 potato cylinders! Use a = 5 mm inner diameter cork borer to cut across the
potato. Trim the cylinders to 3 cm in length. Keep the potato cylinders in a covered
beaker or plastic bag until it is your turn to use the balance.
3. Mass 4 of the cylinders together; record this in Table 2 as the initial mass of the cylinders.
Place all 4 cylinders together into one of your two sucrose solutions (label!).
4. Weigh the other 4 potato cylinders, record in Table 2 their initial mass, and place them in
the other sucrose solution assigned to your group (label!).
5. Cover the beakers of sucrose to prevent evaporation, and let stand overnight.
6. Record the temperatures of the sucrose solutions after they have been out overnight.
7. Remove the 4 potato cylinders from one sucrose solution; blot them gently on a paper
towel, and reweigh them (the four as a unit). Use the same balance as you did the day
before, if at all possible. Repeat with the other set of 4 cylinders.
8. Record the final weights of both sets of potato cylinders in Table 2, and calculate %
change in mass as in Part 1. Pool class data.
9. Graph the % change in mass of the potato cylinders as a function of sucrose solution
molarity.
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 (
) of iso-osmolar sucrose solution can be calculated from the
molarity of the solution by applying this formula:
= 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.
Sample problem: What is
of a 1.0 molal sucrose solution at 30 degrees C under
standard pressure conditions?
What is the osmotic potential of the sucrose solution found (by your graph) to be iso-osmolar
to the potato tissue? SHOW WORK.
Answer: ___________ bars.
3. Since the pressure potential (
) 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?
Answer: __________________bars.
Questions (in lie of a 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 (
) of the potato cells is equal to the water potential of the
solution, at equilibrium. Is the osmotic potential (
) 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.
4. (See figures below.) A dialysis bag is attached to the end of a glass tube to make a simple
osmometer. If the pressure potential ( ) is 0. The bag is 1.0 M sucrose solution, with
of –35 bars.
(a) What is the osmotic potential (
) of the sucrose solution in the bag?
(b) If the osmometer is placed in a beaker of pure water and osmosis occurs, what will
be the values of , , and, at equilibrium for both the solution in the bag and the
water in the beaker?
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