Diffusion/Osmosis/Water Potential
Lab
Directions
• Read through the background information covered in the next few slides.
This is essential knowledge to complete this assignment.
• Write the your answers to the questions in your CLASS NOTEBOOK.
• Be sure to separate questions by section. For example:
– Part 1 Diffusion
1.
2.
3.
–
Part 2 Osmosis
1.
•
•
•
•
•
Answer
A.) blah…. B.) ehhh. C.) huummm..
A.)… B.)…C)…
……
Must be neatly organized and written.
You do not need to rewrite the questions.
YOU DO need to recreate the entire chart for Part 3 and generate a
graph.
Show your work for all math problems (rewrite equations, ect.)
Questions based on the concepts and skills in this assignment/lab will
be part of your Chapter 4/5 Test)
• Diffusion
Molecules are in constant motion and tend to move from regions
where they are in higher concentration to regions where they
are less concentrated. Diffusion is the net movement of
molecules down their concentration gradient. Diffusion can
occur in gases, in liquids, or through solids. An example of
diffusion in gases occurs when a bottle of perfume is opened at
the front of a room. Within minutes people further and further
from the source can smell the perfume.
• Osmosis is a specialized case of diffusion that involves
the passive transport of water. In osmosis water
moves through a selectively permeable membrane
from a region of its higher concentration to a region of
its lower concentration. The membrane selectively
allows passage of certain types of molecules while
restricting the movement of others.
• The solute concentration in the beaker is
higher than that in the bag, and thus the
water concentration is lower in the beaker
than in the bag. This causes water to move
from the bag (left) into the beaker (right).
• There are often several different types of molecules in a solution. The
motion of each type of molecule is random and independent of other
molecules in the solution. Each molecule moves down its own
concentration gradient, from a region of its high concentration to a region
of its low concentration.
• Though the net movement of molecules is down their concentration
gradient, at any time molecules can move in both directions as long as
the membrane is permeable to the molecule. Keep this in mind while you
take a closer look at the beaker on the next slide
• Notice that the starch molecules are too large to pass
through the pores in the membrane. The iodine molecules
move across the membrane in both directions, but their net
movement is from the bag, where their concentration is
higher, into the beaker, where their concentration is lower.
The iodine combines with starch to form a purplish-colored
compound.
• The net movement of water is into the beaker
Movement of Molecules Inside a Cell
• Like dialysis bags, cell membranes are selectively
permeable. As you view the next animation, watch for
the selective property of the cell membrane and the
two-way diffusion of molecules. Finally, notice the net
movement of the molecules.
• The movement of water is influenced by the solute
concentrations of the solutions. Let's review the
different types of solutions.
Types of Solutions Based on Solute Concentration:
The terms hypotonic, hypertonic, and isotonic are used to
compare solutions relative to their solute concentrations
• In the illustration, the solution in the bag
contains less solute than the solution in
the beaker. The solution in the bag is
hypotonic (lower solute concentration)
to the solution in the beaker.
• The solution in the beaker is hypertonic
(higher solute concentration) to the one
in the bag. Water will move from the
hypotonic solution into the hypertonic
solution
• In this illustration the two
solutions are equal in their
solute concentrations. We say
that they are isotonic to each
other.
Water Potential
• YOU MUST UNDERSTAND the concept of
water potential.
– Biologists use this term to describe the tendency
of water to leave one place in favor of another.
Water always moves from an area of
higher water potential to an area
of lower water potential.
Water Potential
• Water potential is affected by two factors: pressure
and the amount of solute.
– For example, imagine a red blood cell dropped into
distilled water. Water will move into the red blood cell
and cause the cell to expand, stretching the flexible
membrane. At some point, the pressure of the
incoming water will cause the cell to pop, just like an
over-filled balloon.
Water Potential
• If a plant cell is placed in distilled water, water
will enter the cell and the cell contents will
expand. However, the elastic cell wall exerts a
back pressure, which will limit the net gain of
water.
Water potential is calculated using the following formula:
Water potential () =
pressure potential (p) + solute potential (s)
Solute potential (s):
The effect of solute concentration. Pure water
at atmospheric pressure has a solute potential
of zero. As solute is added, the value for solute
potential becomes more negative. This causes
water potential to decrease also. In sum, as
solute is added, the water potential of a
solution drops, and water will tend to move
into the solution
Pressure potential (p)
In a plant cell, pressure exerted by the rigid cell
wall that limits further water uptake.
For this ‘lab’ we will be using bars as the unit of measure for water potential; 1 bar =
approximately 1 atmosphere.
Water Potential
• Other units of measurement: megapascals
(Mpa)
1 MegaPascal = 10 atm = 145.1 psi
• Water potential for pure water: 0
• Anything that lowers the “free energy”of
water lowers it potential.
-dissolved solutes
Water potential = pressure potential + solute potential
Clarifying Water Potential Values
• (2) Factors to consider:
p = pressure potential (outside & inside)
s = solute potential
system = p + s
SO……
 p results in
“+” value
 p results in “-” value
Water Potential Values
• High water potential (+Value):
- less solute
- more water
- (hypotonic)
• Zero (0) Value:
- Pure water
• Low water potential (-Value):
- More solute
- less water
- (hypertonic)
****Water will move across a membrane in the direction of the lower water potential****
Calculating Solute Potential
Ψs = – iCRT
• Variables involved: i, C, R, T
i = ionization constant: NaCl = 2.0 (Na+ & Cl-)
**for sucrose it will be 1.0 (it doesn’t ionize)
C = Molar concentration
T = Temperature: ° K
R = pressure constant (R = 0.0831 liter bars/mole K)
R will always be the same.
T will be given to you, but must be in ° K.
T = temperature in Kelvin (273 + °C)
Calculating the Solute Potential (s)
• s = - iCRT
• Sample Calc.
A 1.0 M sugar solution @ 22° C under
standard atmospheric conditions:
s = -(1)(1.0 mol)(0.0831 L · bar )(295K)
L
s = -24.22 bars
mol · K
Water Balance-Review
• Osmoregulation~ control
of water balance
– Hypertonic~ higher
concentration of solutes
– Hypotonic~ lower
concentration of solutes
– Isotonic~ equal
concentrations of solutes
• Cells with Walls:
– Turgid (very firm)
– Flaccid (limp)
– Plasmolysis~ plasma
membrane pulls away from
cell wall
Dialysis Tubing Experiment
An Artificial Cell
Permeable to: monosaccharides & water
Impermeable to: Disaccharides
The Lab
QUESTIONS START HERE
For a more detailed, or another explanation, you can visit:
http://www.phschool.com/science/biology_place/labbench/lab1/intro.html
Part 1 Diffusion
• Materials: 500 ml beaker; Dialysis bags; Glucose indicating strips;
Glucose/starch solution; dilute iodine solution (actual lab uses IKI, potassium
iodide solution)
• In this activity, you fill a dialysis bag with a sugar/starch solution and
immerse the bag in a dilute iodine solution. Water, sugar, starch, and
iodine molecules will all be in motion, and each molecule will move to a
region of its lower concentration, unless the molecule is too large to pass
through the membrane. Your task is to determine relative size of the various
molecules and gather evidence of molecular movement.
– Hint: One piece of information that will help you is to recall that when iodine
comes in contact with starch, it changes from an orange-brown color to blueblack.
1.
2.
3.
Create an illustration of this lab set-up.
Based on what you have learned, generate a hypothesis for the movement
of (a) glucose, (b) starch, and (c) iodine across the membrane.
What physical evidence (the data you need to collect) will you use to
determine the movement of (a) glucose, (b) starch, and (c) iodine across
the membrane.
Part 2 Osmosis
• Materials: 6x500 ml beaker; 6 Dialysis bags; Scale; Distilled water;
Sucrose (molar mass = 342 g/mol);
• In this activity you investigate the relationship between solute
concentration and water movement by filling six different dialysis
bags with increasing concentrations of sucrose (pure water; 0.2 M; 0.4
M; 0.6 M; 0.8 M; & 1.0 M) and placing the bags into distilled water.
• After the time for the experiment has elapsed, will need to determine
and quantify the movement of water for each of the beaker/dialysis
tube set-ups in order to explain the relationship between solute
concentration and water movement.
1. Illustrate/draw the set-up for this experiment.
2. Generate a hypothesis for this experiment.
3. How will you quantify/measure net water movement for each
beaker/dialysis tube set-ups?
4. Knowing that the molar mass of sucrose is 242 g/mol, how would one
go about making a liter of 1 M sucrose solution?
Part 3
• This activity is very similar to Part 2, except that you use cores from
potatoes instead of dialysis bags. You submerge the cores in solutions of
varying sucrose concentrations.
1. On the next slide is a chart filled in with data for you. You need to
recreate this chart. Using the data given, calculate the change in mass,
& % change (Massf-Massi/Massi x 100) for each potato/solution set-up.
2. Graph the % Change in Mass (X axis) against the sucrose molarity (Y
axis).
– B. Identify the independent and the dependent variable.
3.
Using your graph, what is the molar concentration of the potato core?
Explain how you know.
– Since you know that the pressure potential of the surrounding solution in
an open beaker is zero, you can now calculate the water potential. (use
this in part 4)
Solution Temperature
Potato Cylinders
Solution
Initial
Temp
Final
Temp
Initial
Mass
Final Mass Change in
Mass (g)
Water
21
21
1.5
2.1
0.2 M
21
21
1.5
1.6
0.4 M
21
21
1.5
1.3
0.6 M
21
21
1.5
1.2
0.8 M
21
21
1.6
1.1
1.0 M
21
21
1.5
0.9
% Change
in Mass
Part 4: Calculation of Water Potential
from Experimental Data
•
(Since you know that the pressure potential of the surrounding solution in
an open beaker is zero, you can now calculate the water potential. )
1. In this exercise, use the value for the molar
concentration of the potato cores that you
obtain in Exercise 3 to determine the water
potential for the potato cells.
– Ψs = – iCRT
– Ψ = Ψp + Ψs
Post-Questions
1. What is the symbol for water potential?
2. What is the symbol for pressure potential?
3. What is the symbol for solute potential (aka
osmotic potential)?
4. What is the equation for calculating water
potential?
5. What is the equation to find solute potential?
Post-Questions
6. Suppose you have an artificial cell that was
permeable to monosaccharides and
impermeable to disaccharides. What would
happen to the cell if it had 0.80 M maltose and
0.85 M fructose in it and was placed in a
solution containing 0.45 M glucose, 0.65 M
fructose, and 0.40 M sucrose.
a.) Which direction would the water flow?
b.) Which area has a higher water potential?
c.) What would happen to the concentration of the
maltose inside the cell (increase, decrease, remain the
same)?
7. What is the ionization constant (i) for sucrose?
Post-Questions
8. Why don't red blood cells pop in the bloodstream?
9. The molar concentration of a sugar solution in an
open beaker has been determined to be 0.3M.
Calculate the solute potential at 27 degrees. Round
your answer to the nearest hundredth.
10.The pressure potential of a solution open to the air
is zero. Using the solute potential of the solution
from the question above, calculate the water
potential.
Post-Questions
11. If inside of a cell has a water potential of -24 bars and outside the
cell has a water potential of -5 bars, how will water move?
12. If a cell’s ΨP = 3 bars and its ΨS = -4.5 bars, what is the resulting Ψ?
13. The cell from question above is placed in a beaker of sugar water
with ΨS = -4.0 bars. In which direction will the net flow of water
be?
14. The original cell from question # 12 is placed in a beaker of sugar
water with ΨS = -0.15 MPa (megapascals). We know that 1 MPa =
10 bars. In which direction will the net flow of water be?
Post-Questions
15.The value for Ψ in root tissue was found to be -3.3
bars. If you take the root tissue and place it in a 0.1
M solution of sucrose at 20°C in an open beaker,
what is the Ψ of the solution, and in which direction
would the net flow of water be?
16.NaCl dissociates into 2 particles in water: Na+ and Cl-.
If the solution in question 4 contained 0.1M NaCl
instead of 0.1M sucrose, what is the Ψ of the
solution, and in which direction would the net flow of
water be?