osmosis LAb(2)
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Feb 20, 2024
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AP Lab 4: Diffusion, Osmosis, and Water Potential
Before you begin this lab, review the relevant information in the content and in your textbook. Also prepare by
reading over this lab in the AP Biology Investigative Lab Manual on pages S51-S54.
Prelab (IMPORTANT): Watch the following video throughout these labs
https://www.youtube.com/watch?v=LeS2-6zHn6M
Diffusion demo: (min 2:42 of video). Watch the above video carefully, record data below, and answer
the analysis questions.
In this ac±vity, we will witness the diffusion of different par±cles across a selec±vely permeable membrane. We use the
term “selec±vely permeable” to discuss how a membrane lets some par±cles go in or out, but not all par±cles. This
membrane is an ar±ficial plas±c tube, not a real cell membrane. It selects par±cles based on their size – if a par±cle is
small enough to go through the small holes of the tubing, then it can cross. If a par±cle is too large, then it does not
cross. We will have several par±cles inside or outside the cell in this experiment, and we will be able to see what they do
over the course of 20-30 minutes.
Particles:
IKI – has a orange-brown color unless it is in the presence of starch – then it turns blue-black
starch
glucose – we will use test strips that turn different colors if glucose is present
Data:
Put “Xs” in the table below if the par±cle is present in that loca±on
Start of the experiment
Particle Is it outside the cell? Is it inside the cell?
IKI
starch
glucose
End of the experiment
Particle Is it outside the cell? Is it inside the cell?
IKI
\
glucose
Conclusion:
1.
Which par±cles were able to cross the cell membrane? Which par±cles were NOT able to cross the cell membrane?
How do you know (cite evidence from our observa±ons)?
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2.
How do par±cles naturally move? You cannot say that par±cles always move from outside to inside or inside to
outside – what is the rule demonstrated here?
3.
For the par±cles that moved, did they ALL move to the other side? What is the rule here?
Lab 1: Effects of Osmotic Potential Differences Across a Membrane
Materials:
6 ~ 20 cm strips of dialysis tubing
0.0 M sucrose solution – distilled water
0.2 M sucrose solution
0.4 M sucrose solution
0.6 M sucrose solution
0.8 M sucrose solution
1.0 M
sucrose solution
Distilled water
Timer
Digital scale
6 clear plastic cups
To make sucrose solutions: Add the number of grams of sucrose according to the table below
and fill to the 100mL line with distilled water. You may need to heat the water to get the higher
concentrations of sucrose to dissolve. You’ll use this for BOTH labs.
Molarity Grams sucrose per 100 mL
solution
0.2 6.8
0.4 13.7
0.6 20.5
0.8 27.4
1 34.2
Lab 1 Hypothesis
: Think about what your variables are and construct a hypothesis predicting the relationship
between them.
________________________________________________________________________________________
________________________________________________________________________________________
Procedure: You will conduct the experiment using the virtual lab here (don’t worry about recording
this data, you’ll be provided data):
https://video.esc4.net/video/assets/Science/Biology/Gateway%20Resources/cell%20homeostasi
s %20virtual%20lab%20-%20activity/index.html
Answer the questions after the virtual lab here.
1. Which dialysis tubes had li²le or no change in mass a³er a 24-hour period?
2. Why was pure water used as a control group?
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3. Why didn’t the mass of dialysis tubes “A” and “C” change over the 24-hour ±me frame?
Data
Concentration of sucrose solution inside dialysis bags
(outside always dH2O)
0.0 M (water) 0.2 M 0.4 M 0.6 M 0.8 M 1.0 M
Class mean
1.80% 7.8% 13.8% 18.9% 23.4% 26.2%
Class standard deviation
8.0% 8.9% 9.9% 12.2%
14.6% 7.6%
Class sample size (n)
27 27 27 27 27 24
Graph the above data (remember what your variables are and where they go!) and paste
your graph below:
Analysis Ques±ons
1. What is the general trend that you see between variables in the graph? Why does this occur?
2. Iden±fy a case where solu±ons where isotonic to each other in the above data. Also iden±fy one case where a
solu±on is hypertonic (be clear which solu±on you are referring to).
3. Iden±fy the general rule for predic±ng water movement if the solute concentra±ons on both sides are known (and
the solute is impermeable).
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4. These bags all spent the same amount of ±me in the dis±lled water (20 mins). So you could divide the % changes in
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mass by 20 minutes to calculate a rate of osmosis. As the concentra±on gradient increases between outside and
inside, what happens to the rate of osmosis?
Lab 2: Determining the Water Potential of Potato Cells
In
animal
cells,
the
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. If water
moves into the cell, the cell may swell or even burst. In plant cells, the presence of a
cell wall prevents
the cells from bursting, but pressure does eventually build up inside the cell and
affects the process of
osmosis. When the pressure inside the cell becomes large enough, no additional water will accumulate
in the cell even the though cell still has a higher solute concentration than does pure water. The
movement of water through the plant tissue cannot be predicted simply through knowing the relative
solute concentrations on either side of the plant cell wall. Instead, the concept of water potential is
used to predict the direction in which water will diffuse through living plant tissues.
In a general sense,
water potential is the tendency of water to diffuse from one area to another
under a given set of parameters
. Water potential is expression in bars, a metric unit of pressure
equal to about 1 atmosphere and measured with a barometer. Water potential is abbreviated by the
Greek letter psi (Ψ
)
and has two major components:
Materials:
0.0 M sucrose solution – distilled water
0.2 M sucrose solution
0.4 M sucrose solution
0.6 M sucrose solution
0.8 M sucrose solution
1.0 M sucrose solution
Distilled water
Procedure: Watch the video starting at 5:35
Timer
Digital scale (food scale works great) or
triple-beam balance
Potato
Potato peeler
6 clear plastic cups
Plastic wrap
Hypothesis: __________________________________________________________
Data
Concentration of sucrose solution outside the potatoes
(inside unknown solute concentration)
0.0 M (water) 0.2 M 0.4 M 0.6 M 0.8 M 1.0 M
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Class mean
21.88% 7.25% -7.91% -11.82% -25.88% -30.95%
Class standard deviation
15.6% 25.4%
8.8% 21.8% 19.5% 6.5%
Class sample size (n)
15 13 13 14 9 10
Analysis:
Prepare a corresponding graph of your findings. Be sure your graph indicates positive and negative
changes in mass as indicated above and below the x axis. Remember to follow good graphing
procedures by labeling each axis including units and giving the graph a title. Use a best fit line to
determine where the line crosses the X axis.
The black lines represent the x and y axes.
This point
of intersection represents the point at which the potato cells do not change in mass, meaning that the
solute concentration of the potato cells is isotonic to the external solution.
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Determine the molar concentration of the potato cube. This would be the sucrose molarity in which the
mass of the potato cube does not change.
To find this, look at graph above and find the point at
which your data line crosses the x-axis. This represents the molar concentration of sucrose
with a water potential that is equal to the potato tissue water potential.
At this concentration there
is no net gain or loss of water from the tissue. Indicate this concentration of sucrose in the space
provided below.
Molar concentration of sucrose =
___________
M
Calculation of Water Potential from Experimental Data:
The solute potential of the sucrose solution can be calculated using the following
formula:
Ψ
S
= -iCRT
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i = Ionization constant
(for sucrose this is 1.0 because sucrose does not ionize in water)
C = Molar concentration
(determined above)
R = Pressure constant
(R = 0.0831 liters bars/mole K)
T = Temperature K
(273 +
o
C of solution)
Please take a moment to review help information on page 52-53 of your student lab manual as well as
the recorded synchronous session on water potential to help you make the following calculations.
First calculate the solute potential (Ψ
S
) of the sucrose solution.
Show answers and work here.
Knowing the solute potential (Ψ
S
) of the sucrose solution and that the pressure potential of the solution
is zero (Ψ
P
= 0) allows you to calculate the water potential of the solution. The water potential will be
equal to the solute potential of the solution.
Ψ = 0 + Ψ
S
or Ψ = Ψ
S
The water potential of the solution at equilibrium will be equal to the water potential of the potato cells.
What is water potential of the potato cells?
Ψ
potato cells
=
___________
bars
Reasoning paragraph – Please construct a reasoning paragraph explaining the
following:
1. Identify your independent and dependent variables
2. What is the relationship between variables (discuss the trend you see)
3. Why does this trend occur?
4.
Discuss the concept of water potential and the two physical factors that affect it.
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