Kayla Romeo
September 12, 2011
Lab 1
Objectives:




Understand diffusion between two solutions
Understand movement through selectively permeable membranes (osmosis)
Understand water potential
Describe plasmolysis
Section 1A
Background:
In this experiment, we explored diffusion through dialysis tubing. We placed a
bag of glucose and starch into a beaker filled with distilled water and Lugol’s solution (IKI) and
observed the diffusion of molecules out and into the bag.
Procedure:
1. Obtain a 20-cm piece of 2.5-cm dialysis tubing that has been soaking in water.
Tie off one end of the tubing to form a bag. To open the other end of the bag,
rub the end between your fingers until the edges separate.
2. Test the 15% glucose/1% starch solution for the presence of glucose. Your
teacher may have you do a Benedict’s test or use glucose Testape® or
Clinistix®. Record the results in Table 1.1.
3. Place 15 mL of the 15% glucose/1% starch solution in the bag. Tie off the
other end of the bag, leaving sufficient space for the expansion of the contents
in the bag. Record the color of the solution in Table 1.1.
4. Fill a 250-mL beaker or cup two-thirds full with distilled water. Add
approximately 4 mL of Lugol’s solution to the distilled water and record the
color of the solution in Table 1.1. Test this solution for glucose and record the
results in Table 1.1.
5. Immerse the bag in the beaker of solution.
6. Allow your setup to stand for approximately 30 minutes or until you see a
distinct color change in the bag or in the beaker. Record the final color of the
solution in the bag, and of the solution in the beaker, in Table 1.1.
7. Test the liquid in the beaker and in the bag for the presence of glucose. Record
the results in Table 1.1.
Kayla Romeo
September 12, 2011
Lab 1
Table 1.1
Bag
Beaker
Initial
Contents
15% glucose
and 1%
starch
Water and
IKI
Solution Color
Initial
Final
Cloudy
Bluish
white
purple
Presence of Glucose
Initial
Final
Yes (1000
Yes (5000+
mg/dL)
mg/dL)
Golden (like
apple juice)
None
Golden (like
apple juice)
Yes (250
mg/dL)
Results
1. Iodine seemed to enter the bag since the bag changed colors, due to the fact that
iodine is a starch indicator. The contents of the bag went from a cloudy white to a
bluish purple after being immersed in the water and IKI solution. Glucose is
leaving the bag since there was a presence of glucose in the water and IKI
solution after the bag was immersed in it. We tested for glucose with a strip and it
showed the presence of 250 mg/dL of glucose.
2. Since there was more solute in the bag, the water concentration was less in the
bag than it was in the outside solution. Due to the selectively permeable
membrane, glucose, iodine, and water could diffuse across the membrane. It
seems as though starch can’t pass through because if it could have, the solution in
the beaker would have changed colors.
3. This experiment could be modified by measuring the dialysis bags before and
after we put them into the water/IKI solution. This would have allowed for us to
see if water entered the bag or not.
4. Based on our observations, water molecules are the smallest. The next biggest is
glucose molecules, followed by IKI molecules, membrane pores, and starch
molecules.
5. If the experiment started with glucose and IKI solution inside the bag and only
starch and water outside, I would expect the contents of the bag to remain white
while the contents outside of the bag changed to blue. The iodine would diffuse
across the membrane and react with the starch molecules to create a blue-purple
color.
Conclusion
Based on our data that we collected, it seems as though iodine moved into the
dialysis bag and glucose moved out of the bag. I think that our final test for the presence
of glucose in the bag was a little off though. It seems as though the glucose shouldn’t
have gone up, but it did quite significantly. We can’t tell if water moved into or out of the
bag, but we can assume that water diffused into the bag.
Kayla Romeo
September 12, 2011
Lab 1
Section 1B
Background
In this experiment, we will examine the movement of water, otherwise known as
osmosis, through a selectively permeable membrane. Water tends to move from an area of low
concentration to an area of high concentration. In this experiment, water will most likely move
from a hypotonic solution to a hypertonic solution.
Figure 1.1
blue- hypertonic
yellow- hypotonic
Procedure
1. Obtain six 30-cm strips of presoaked dialysis tubing.
2. Tie a knot in one end of each piece of dialysis tubing to form 6 bags. Pour
approximately 15-25 mL of each of the following solutions into separate bags:
a. Distilled water
b. 0.2 M sucrose
c. 0.4 M sucrose
d. 0.6 M sucrose
e. 0.8 M sucrose
f. 1.0 M sucrose
Remove most of the air from each bag by drawing the dialysis bag between two
fingers. Tie off the other end of the bag. Leave sufficient space for the expansion
of the contents in the bag. (The solution should fill only about one-third to onehalf of the piece of tubing.)
3. Rinse each bag gently with distilled water to remove any sucrose spilled during
the filling.
4. Carefully blot the outside of each bag and record in Table 1.2 the initial mass of
each bag, expressed in grams.
Kayla Romeo
September 12, 2011
Lab 1
5. Place each bag in an empty 250-mL beaker or cup and label the beaker to indicate
the molarity of the solution in the dialysis bag.
6. Now fill each beaker two-thirds full with distilled water. Be sure to completely
submerge each bag.
7. Let them stand for 30 minutes.
8. At the end of 30 minutes remove the bags from the water. Carefully blot and
determine the mass of each bag.
9. Record your group’s data in Table 1.2.
Table 1.2
Contents in
dialysis bag
Distilled water
0.2 M sucrose
0.4 M sucrose
0.6 M sucrose
0.8 M sucrose
Initial Mass
Final Mass
18.8 g
14.4 g
17.9 g
13.1 g
21.3 g
1.0 M sucrose
17.0 g
19.2 g
15.2 g
20.7 g
15.0 g
Not tied tightly
enough
21.2 g
Mass Difference Percent Change
in Mass
.4 g
2.1%
.8 g
5.6%
2.8 g
15.6%
1.9 g
14.5%
N/A
N/A
4.1 g
24.1%
Graph 1.1
Percent Change in Mass of Dialysis Bags
30.00%
25.00%
20.00%
15.00%
Percent Change in Mass of
Dialysis Bags
10.00%
5.00%
0.00%
Distilled
Water
0.2 M
sucrose
0.4 M
sucrose
0.6 M
sucrose
0.8 M
sucrose
1.0 M
sucrose
Kayla Romeo
September 12, 2011
Lab 1
Results
1. As the molarity of the sucrose went up, the change in mass of the bags went up
also. This happened because the net movement of water was from the distilled
water (hypotonic solution) into the dialysis bags filled with different amounts of
sucrose (hypertonic solution).
2. If all of the bags were placed in a 0.4 M sucrose solution rather than distilled
water, the mass of the bags would increase if they were containing solutions less
than 0.4 M sucrose, and they would decrease if the bags were containing solutions
more than 0.4 M sucrose. This happens because the solution with more solute is
hypertonic and the solution with less solute is hypotonic, therefore the net
movement of water would be from the hypotonic into the hypertonic solution.
3. We calculated the percent change in mass rather than the change in mass because
all of the bags contained different amounts of the solution. So there was no
uniform increase or decrease in just the mass. With the percent change in mass,
we can actually compare the bag to another bag.
4. 18 g- 20 g= -2 g
-2 g/ 20 g= -0.1 g
-0.1 g x 100= -10%
Percent change in mass= -10%
5. Hypertonic
Conclusion
Based on our results, water will always move from a hypotonic solution to a
hypertonic solution. As the molarity of the sucrose went up, the percent change in mass
also went up. This demonstrates that water was diffusing into the bags.
Section 1C
Background
In this experiment we will explore water potential. Water potential is affected by
pressure and the amount of solute. Water potential is when water leaves one area in favor of
another area. In plant cells, the cell will expand as water enters the cell, but the cell wall will
push back as the cell becomes too saturated.
Procedure
1. Pour 100 mL of the assigned solution into a labeled 250-mL beaker. Slice a potato
into discs that are approximately 3 cm thick.
Kayla Romeo
September 12, 2011
Lab 1
2. Use a cork borer (approximately 5 mm in inner diameter) to cut four potato
cylinders. Do not include any skin on the cylinders. You need four potato
cylinders for each beaker.
3. Keep your potato cylinders in a covered beaker until it is your turn to use the
balance.
4. Determine the mass of the four cylinders together and record the mass in Table
1.4. Put the four cylinders into the beaker of sucrose solution.
5. Cover the beaker with plastic wrap to prevent evaporation.
6. Let it stand overnight.
7. Remove the cores from the beakers, blot them gently on a paper towel, and
determine their total mass.
8. Record the final mass in Table 1.4. Calculate the percentage change as you did in
Exercise 1B.
9. Graph your individual data for the percent change in mass in Table 1.4.
Table 1.4
Contents in
Beaker
Initial Mass
Final Mass
Mass
Difference
Distilled
Water
0.2 M
sucrose
0.4 M
sucrose
0.6 M
sucrose
0.8 M
sucrose
1.0 M
sucrose
1.3 g
1.7 g
0.4 g
Percent
Change in
Mass
30.8%
1.3 g
1.4 g
0.1 g
7.7%
2.0 g
2.0 g
0g
0%
1.0 g
0.7 g
-0.3 g
-30%
0.7 g
0.8 g
0.1 g
14.3%
1.2 g
0.9 g
-0.3 g
-25%
Graph 1.2
Kayla Romeo
September 12, 2011
Lab 1
Percent Change in Mass of Potato Cores at
Different Molarities of Sucrose
40.00%
30.00%
20.00%
10.00%
0.00%
-10.00%
-20.00%
-30.00%
-40.00%
Percent Change in Mass of
Potato Cores at Different
Molarities of Sucrose
Distilled
Water
0.2 M
sucrose
0.4 M
sucrose
0.6 M
sucrose
0.8 M
sucrose
1.0 M
sucrose
30.80%
7.70%
0%
-30%
14.30%
-25%
10. Molar concentration of sucrose= 0.4 M sucrose
Conclusion
Based on our analysis of results, 0.4 M sucrose was the molar concentration of the
potato cores. For anything below 0.4 M sucrose, the mass increased whereas for anything above
0.4 M sucrose the mass decreased. This means that the water potential of anything below 0.4 M
sucrose was higher than the potato cores and the water potential of anything above 0.4 M sucrose
was lower than the potato cores.
Section 1D
Questions
2. –(1.0)(0.4)(0.0831)(295)
= - 9.8058 bars
1. The water potential of the dehydrated potato core will decrease because all of the
water is gone. There is nothing left.
2. The plant cell is hypertonic to its surrounding environment; therefore the cell will
gain water because water moves from an area of high concentration to areas of
lower concentration.
3. The pressure potential of the system is 0.
4. The greatest water potential is in the dialysis bag.
Kayla Romeo
September 12, 2011
Lab 1
5. Water will diffuse out of the bag because water moves from a hypotonic solution
to a hypertonic solution and the solution in the bag would be considered the
hypotonic solution.
Graph 1.3
Percent Change in Mass of Zucchini Cores
% increase/decrease in mass
of cores
30%
20%
10%
0%
-10%
-20%
-30%
-40%
Percent Change in Mass of
Zucchini Cores
Distilled
Water
0.2 M
sucrose
0.4 M
sucrose
0.6 M
sucrose
0.8 M
sucrose
1.0 M
sucrose
20%
10%
-3%
-17%
-25%
-30%
7b. approximately 0.35 M sucrose
8a. (-1.0)x(0.35)x(0.0831)x(295)
= - 8.58 bars
8b. Water potential= 0 + -8.58 bars
Water potential= -8.58 bars
9. Adding solute raises the solute potential of that solution.
10a. Distilled Water
10b. Distilled Water
10c. Water would diffuse into the red blood cell and the red blood cell would most likely
burst from the pressure of water pushing out through the inside of the cell.
Section 1E
Background
Kayla Romeo
September 12, 2011
Lab 1
In this experiment, we investigated plasmolysis. In a plant cell, the cytoplasm
shrinking due to water moving out of the cell into a hypertonic solution surrounding the cell is
called plasmolysis. We used the epidermis of a red onion for our experiment.
Procedure
1. Prepare a wet mount of a small piece of the epidermis of an onion. Observe under
100X magnification. Sketch and describe the appearance of the onion cells.
The cell is a purple color and they are arranged semi-uniformly against the
other cells.
2. Add 2 or 3 drops of NaCl to one edge of the cover slip. Draw this salt solution
across the slide by touching a piece of paper towel to the fluid under the opposite
edge of the cover slip. Sketch and describe the onion cells. Explain what has
happened.
The cell has shriveled up and pulled away from the cell wall because of
the water diffusing out of the cell into the NaCl solution.
3. Remove the cover slip and flood the onion epidermis with fresh water. Observe
under 100X. Describe and explain what happened.
The cells are gradually regaining their original shape because water is
being diffused back into the cell. However, the cells won’t fully return to
their original size because of the NaCl they absorbed.
Results
1. Plasmolysis is when a cell shrivels up and the plasma membrane pulls away from
the cell wall because water is being diffused out of the cell into a hypertonic
solution outside of the cell.
2. The onion cells plasmolyzed because they were exposed to sodium chloride
(NaCl) and the water from inside the cell was diffusing out to try and balance the
solution.
3. In winter, grass often dies near roads that have been salted to remove ice because
the water from inside the grasses cells is being diffused outside of the cell and
into the salt to balance the solution.
Conclusion
Kayla Romeo
September 12, 2011
Lab 1
Based on our results, the onion cells plasmolyzed when we exposed them to
sodium chloride. The cells pulled away from the cell wall leaving the purple part
shriveled up. When we re-added water, the cells gradually returned back to their starting
shape, but they never quite made it back to their starting position because they had
already absorbed some of the sodium chloride.