Diffusion, Osmosis, and Water PotentialSamantha A. Price

Noble Group A.P. Biology

10-8-09 Lab #2

Introduction:

If a substance is placed in a hypotonic solution it will lose in its concentration through diffusion.If a substance is placed in a hypertonic solution it will gain in its concentration though diffusion.If a substance is placed in an isotonic solution the concentrations will remain the same.If the membrane of a substance is not permeable by the molecules in a sucrose solution it will gain or

loss water depending on its concentration levels though osmosis.

Many aspects of the life of a cell depend on the fact that atoms and molecules have kinetic energy and are constantly in motion (Lab One Diffusion and Osmosis worksheet). Atoms and molecules that have intrinsic kinetic energy are caused to bump into one another and move in new directions (thermal motion). An example of this is Diffusion (Reece Campbell Biology 6th Edition).

Diffusion is the tendency for molecules of any substance to spread out into the available space. In order for diffusion to occur the membrane pores have to be large enough to allow the molecule though to the other side in an effort to equalize the concentration on each side of the membrane.

Though if the membrane is not large enough to allow all molecules though then water will move from one side of the membrane to the other to lower concentration.

In order to understand osmosis and diffusion one must understand the concentration gradient and the water potential rules. Hypertonic solutions, those that have more molecules, have a low water potential. This means that they will gain water from the surrounding environment if it has a higher water potential, fewer molecules, and the molecules can’t move, meaning that it is hypotonic. Water always moves from high to low potential areas if particles cannot move, and concentrations always move from hypertonic to hypotonic if they can permeate the membrane. Though if something is in an isotonic solution no changes will occur due to the levels of concentration and potential being equal to one another. This changes based on the type of cell used, as seen in below figure.

This lab tests the process if diffusion and osmosis in a model membrane system and the effect of solute concentration on water potential as it relates to living plant tissues.

Materials & Procedure:Note: Bullets are the materials, and the numbers are the procedure, for each exercise.

Diffusion Exercise:

20cm piece of dialysis tubing 15% glucose, 1% starch solution Beaker Distilled water

1. Obtain a 20-cm piece of dialysis tubing that has been soaking in water. Tie off one end of the tubing of from a gag. 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. Record the Results.3. Place 15mL 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.4. Fill a 250mL beaker or cup 2/3 full with distilled water. Add approximately 4mL of Lugol’s

solution to the distilled water and record the color of the solution. Test this solution for glucose and record the results.

5. Immerse the bag in the beaker of solution.6. Allow your setup to stand for 30 min 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.7. Test the liquid in the beaker and in the bag for the presence of glucose. Record the results.

Osmosis Exercise:

3 20cm piece of dialysis tubing 0.6M sucrose solution 0.8M sucrose solution 1.0M sucrose solution 3 beaker Distilled water Paper towel Scale that measures in grams

1. Obtain 3 20cm piece of dialysis tubing.2. Tie a knot in one end of each piece of dialysis tubing to form bags. Pour approximately 15 mL of each

of the following solutions into separate bags:a. Distilled water

b. 0.2M sucrosec. 0.4M sucrosed. 0.6M sucrosee. 0.8M sucrosef. 1.0M 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 for the contents in the bad. (Note: the solution should only fill 1/3-1/2 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 bad and record the initial mass of each bag expressed in grams.5. Place each bag in an empty 250mL beaker or cup and label the beaker to indicate the morality of the

solution in the dialysis bag. 6. Fill each beaker 2/3 full with distilled water. Be sure to completely submerge each bag.7. Let the stand for 30 min.8. At the end of 30 min remove the bags form the water. Carefully blot and determine the mass of each

bag.9. Record group’s data. Obtain data from other lab groups in class to compare.10. Graph.

Water Potential Exercise:

8 potato core samples 8 sweet potato core samples 0.8M sucrose solution 1.0M sucrose solution 2 beaker Paper towel Scale that measures in grams Plastic wrap

1. Pour 100mL of assigned solution (0.8M and 1.0M sucrose solutions) into a labeled beaker. 2. Take the 4 potato and 4 sweet potato cores Ketcham cut and mass together by type of potato and record

data. Put the cylinders into the beaker of sucrose solution.3. Cover with plastic wrap to prevent evaporation.4. Let stand overnight.5. Remove the cores, blot gently, mass together like in step 2.6. Record final mass and calculate the % change.7. Graph.

Results, Data Collection & Analysis:

Initial contentsSolution Color Presence of Glucose

Initial Final Initial FinalBag 15% glucose & 1% starch Clear Indigo > 2000 mg/dL 500 mg/dL

Beaker H2O & IKl Amber Golden yellow 100 mg/dL 175 mg/dL

Figure 1 (Diffusion Exercise)

Contents in Dialysis Bag Initial Mass Final MassMass

Difference

Percent Change in

Massa) 0.0M Distilled waterb) 0.2M Sucrose (Red)

c) 0.4M Sucrose (Green)d) 0.6M Sucrose (Blue) 16.49 17 0.51 3e) 0.8M Sucrose (Clear) 17.11 20.2 3.09 18.05f) 1.0M Sucrose (Purple) 16.91 20 4.09 24.189

Figure 2 (Group’s Data on Osmosis Exercise)

Percent Change in Mass of Dialysis BagsTotal

Class Average Book ResultsNoble Tag Dempsey Auger Sanderson

a) 0.0M Distilled water - - 0 0.76 - 0.79 0.38 1.2b) 0.2M Sucrose

(Red) - - 4.4 2.8 4.15 11.38 3.78 3.1c) 0.4M Sucrose

(Green) - - 7.2 11.24 12.71 31.14 10.38 7.7d) 0.6M Sucrose

(Blue) 3 5.55 - - - 8.55 4.27 11e) 0.8M Sucrose

(Clear) 18.05 4.93 - - 17.94 40.92 13.64 14.8f) 1.0M Sucrose

(Purple) 24.189 33.82 - - - 58.02 29.01 18.2

Figure 3 (Class Data on Osmosis Exercise)

Figure 4 (graph of class data on Osmosis Exercise)

Contents in Dialysis Bag Initial Mass Final MassMass Difference

Percent Change in Mass

a) 0.0M Distilled water        b) 0.2M Sucrose (Red)        c) 0.4M Sucrose (Green)        d) 0.6M Sucrose (Blue)        e) 0.8M Sucrose (Clear) 20.17 16.35 -3.82 -18 f) 1.0M Sucrose (Purple) 20.03 14.69 -5.34 -28.75

Figure 5 (Potato in Water Potential Exercise)

Contents in Dialysis Bag Initial Mass Final MassMass Difference

Percent Change in Mass

a) 0.0M Distilled water        b) 0.2M Sucrose (Red)        c) 0.4M Sucrose (Green)        d) 0.6M Sucrose (Blue)        e) 0.8M Sucrose (Clear) 18.57 18.61 0.04 0.21 f) 1.0M Sucrose (Purple) 16.69 15.52 -1.17 -6.41

Figure 6 (Sweet Potato in Water Potential Exercise)

 Percent Change in Mass of Potato Cores

TotalClass

AverageNoble Tag Dempsey Auger Sandersona) 0.0M Distilled water   17.52      17.52 17.52b) 0.2M Sucrose (Red)     4.12    4.12 4.12c) 0.4M Sucrose (Green)         -7.11 -7.11 -7.11d) 0.6M Sucrose (Blue)       -18.12  -18.12 -18.12e) 0.8M Sucrose (Clear) -18        -18 -18 f) 1.0M Sucrose (Purple) -28.75        -28.75 -28.75

Figure 7 (Class Data for Potato in Water Potential Exercise)

 Percent Change in Mass of Sweet Potato Cores

TotalClass

AverageNoble Tag Dempsey Auger Sandersona) 0.0M Distilled water   19.5      19.5 19.5b) 0.2M Sucrose (Red)     12.02    12.02 12.02c) 0.4M Sucrose (Green)         7.46 7.46 7.46d) 0.6M Sucrose (Blue)       3.6  3.6 3.6e) 0.8M Sucrose (Clear) 2.1        2.1 2.1 f) 1.0M Sucrose (Purple) -6.41        -6.41 -6.41

Figure 8 (Class Data for Sweet Potato in Water Potential Exercise)

Figure 9 (graph of Class Data for Water Potential Exercise)Discussion/Conclusion:

Diffusion Exercise:The hypothesis that states that when a substance is placed in a hypertonic solution it will gain in

its concentration though diffusion is proven correct in this exercise. Glucose entered the beaker and left the bag in an effort to even out the amount of glucose in the two solutions. This is evident when Figure 1 is taken into account. The solution within the bag started out with over 2000mg/dL of glucose and ended with 500mg/dL, where as the beaker started with 100mg/dL glucose and ended with 175mg/dL. This reaction would have kept going until each solution was isotonic to one another if more time was given for the reaction. The fact that glucose traveled from the bag to the beaker shows that the membrane had pores large enough to allow glucose to flow with the concentration gradient, into the beaker.

Osmosis Exercise:

Questions:

Diffusion Exercise:1) Which substance(s) are entering the bag and which are leaving the bag? What experimental evidence

supports your answer?a) Glucose is entering the beaker and leaving the bag trying to even out the amount of glucose in the two

solutions. The evidence that supports this is that the amount of glucose in the bag decreased while the amount in the beaker increased.

2) Explain the results you obtained. Include the concentration differences and membrane pore size in your discussion.a) The membrane was large enough for glucose and water to pass through the pores.

3) Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag?a) If one was to mass the bag before and after or to measure the amount of ml in the bag before and after.

4) Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water molecules, IKI molecules, membrane pores, starch molecules.a) IKI, glucose, water, membrane, starch.

5) What results would you expect if the experiment started with a glucose and IKI solution inside the bag and only starch and water outside? Why?a) One would expect that glucose and IKI would defuse out of the bag, into the beaker. Water will defuse

into the bag. Starch is too large to pass through the membrane pores.

Osmosis Exercise:1) Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags.

a) As molarity increases the percent change will have a positive correlation to the molarity.2) Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4 M

sucrose solution instead of distilled water. Explain your response.a) The mass would be greater due to small sucrose molecules being able to penetrate the membrane pore

easier than water molecules.

3) Why did you calculate the percent change in mass rather than simply using the change in mass?a) In order to get a better comparison one wants a reading that tells what part of the total was either

increased or decreased. If another group were to use different volumes than we could still compare results.

4) A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag’s initial mass is 20 g and its final mass is 18 g. Calculate the percent change of mass, showing your calculations here.a) 20-18=2 <-change change/initial = 2/10 = 0.1 = 10%

Water Potential Exercise:1) If the solute potential of this sucrose solution can be calculated using the following formula:2) Water potential values are useful because they allow us to predict the direction of the flow of water. Recall

from the discussion that water flows from an area of higher water potential to an area of lower water potential.For the sake of discussion, suppose that a student calculates that the water potential of a solution inside a bag is -6.25 bar (Ψs = -6.25, Ψp = 0) and the water potential of a solution surrounding the bag is –3.25 bar (Ψs = -3.25, Ψp = 0). In which direction will the water flow?a) Water will flow into the bag. This occurs because there are more solute molecules inside the bag

(therefore a value further away from zero) than outside in the solution.3) If a potato core is allowed to dehydrate by sitting in the open air, would the water potential of the potato

cells decrease or increase? Why?a) The water potential would decrease because of evaporated water leaving the potato.

4) If a plant cell has a lower water potential than its surrounding environment and if pressure is equal to zero, is the cell hypertonic (in terms of solute concentration) or hypotonic to its environment? Will the cell gain water or lose water? Explain your response.a) The plant cell is hypertonic and will gain water because water potential is low but concentration is high

and water moves from hyper to hydro.

Figure 10

5) In Figure 10 the beaker is open to the atmosphere. What is the pressure potential (Ψp) of the system?a) Zero due to it being an open system

6) In figure 10 where is the greatest water potential? (Correct answer is bold)a) Beakerb) Dialysis Bag

7) Water will diffuse _______ the bag. Why? (Correct answer is bold)a) Intob) Out Ofc) Because water moves from areas of less concentration to areas of greater concentration to decrease the

concentration of the greater concentration.

Other1) Zucchini cores placed in sucrose solutions at 27˚C resulted in the following percent changes after 24 hours:

%Change in Mass

Sucrose Morality

20% 010% 0.2-3% 0.4-17% 0.6-25% 0.8-30% 1

What is the molar concentration of solutes within the zucchini cells? __≈ 0.35___

2) Refer to the procedure for calculating water potential from experimental data above.a) Calculate solute potential (Ψs) of the sucrose solution in which the mass of the zucchini cores does not

change. Show your work here:i) Ψs = (-1)(0.36)(0.0831)(295)

Ψs = -8.825b) Calculate the water potential (Ψ) of the solutes within the zucchini cores. Show your work here:

i) Ψ = 0+ Ψs

Ψ = 0+ -8.825

Ψ = -8.8253) What effect does adding solute have on the solute potential component (Ψs) of that solution? Why?

a) It decreases it because it increases concentration in a negative format causing a smaller Ψs. It decreases its solute potential because the more solute the less potential to add more.

4) Consider what would happen to a red blood cell (RBC) placed in distilled water:a) Which would have the higher concentration of water molecules? (correct answer is bold)

i) Distilled H2Oii) RBC

b) Which would have the higher water potential? (correct answer is bold)i) Distilled H2Oii) RBC

c) What would happen to the red blood cell? Why?i) The RBC would lyse (burst) due to an extreme on take of H2O.