A.  Learning Objectives:

In this lab, students will:

• study the effect of temperature and molecular weight on the rate of diffusion.

• investigate how concentration gradient influences the direction of net water flow during osmosis.

• observe the selective diffusion of various substances across a selectively permeable membrane.

• observe the effects of water movement due to osmosis in plant cells.

 

B. Textbook Correlation: 

Please review Sections 5.1 and 5.3 of Chapter 5: Membrane Structure, Synthesis, and Transport when preparing for the lab.

 

C.  Introduction

Describe the structure/function of the plasma membrane.  In your discussion, include the roles of phospholipids, cholesterol, proteins, and carbohydrates in the function of a membrane.  Discuss how the membrane acts as a selective barrier based on the arrangement of phospholipids.  Also include an image of a phospholipid. 

  

The plasma membrane consists of two layers of phospholipids, which are molecules with hydrophobic (non-polar) tails and a hydrophilic (polar) head, membrane proteins, lipids embedded in the plasma membrane, and carbohydrates attached to lipids and proteins. The plasma membrane is also known as the phospholipid bilayer because of the presence of the two layers (these layers are also called leaflets) of phospholipids. The phospholipids orient themselves in such a way that their hydrophilic heads are facing the outside and their hydrophobic tails are facing inside. This creates a bilayer with a hydrophilic exterior and a hydrophobic interior. The function of the plasma membrane is to separate a cell from its external environment, thereby compartmentalizing it. It also serves to selectively import and export vital substances. It is able to be selectively and semi permeable because of the phospholipids in the phospholipid bilayer. Because of the hydrophobic interior of the phospholipid bilayer, the plasma membrane acts as a barrier to the diffusion of hydrophilic solutes. This means that small, uncharged, non-polar molecules will cross the most permeability, while large, charged, polar molecules will have the least permeability. In this way the plasma membrane can "choose" which molecules it lets in and out. It also has other functions that are achieved via the molecules embedded in the cell membrane. For example, the carbohydrates that are attached to the different proteins and lipids in the cell membrane are used in the process of cell-signaling which allows one cell to be differentiated from another cell. Cholesterol is another example of a molecule that has a different function in the plasma membrane; it keeps the plasma membrane fluid. Proteins that are embedded in the cell membrane allow molecules to cross the plasma membrane that would not usually be allowed across. For example, transport proteins in the cell membrane allow for sodium and potassium ions to cross the cell membrane, which is an essential process for the cell to establish an electrochemical gradient. 

 

     Representation of a phospholipid bilayer 

 

Representation of a phospholipid

                                                                              

 

 

The experiments today investigate several aspects of diffusion and osmosis including: factors that affect the rate of diffusion, the role of concentration gradients as the driving force for osmosis, the selective movement of different substances across a selectively permeable membrane, and the effects of osmosis in two different living cells. 

 

D.  Osmosis

Osmosis is the spontaneous movement of solvent molecules through a semipermeable membrane, in an attempt to equalize the solute concentrations on either side of the membrane. 

A solute gradient is created as unequal levels of solute exist on either side of a membrane. As the solute concentration increases on one side of the membrane, the solute begins to take up "free" water. Free water exists when the solvent is water, and there are open H₂O molecules that do not have a solute bonded to them. As the solute levels go up, the solute will begin to bind to the free water, making those molecules no longer free. If you reach a point where there is no remaining free water, the solution is considered saturated. The less free water there is, the less stable the solution is. It is more organized, but less stable. Due to the second law of thermodynamics, we know that the universe tends to move to a less organized but more stable state.

 

Osmosis is the process by which water will move through a membrane to try to equalize the free water and solute levels on both sides to a membrane. This creates a state where both sides can equally share a level of disorganization, and hence become more stable. This is why osmosis is spontaneous. An isotonic solution is one in which the solute equilibrium is even. A hypertonic solution is one in which there is a greater concentration of solute on the outer side of the membrane. For example, a cell is in a hypertonic solution is the amount of solvent is higher outside of the cell compared to the inside. The water levels will adjust to this by having water rush out of the cell, to create equal proportions of free water. The other type of solution is hypotonic, the opposite of hypertonic. In a hypotonic solution, the solute concentration is lower on the outer side. Take the previous example, but imagine a higher concentration of solute inside the cell. To fight this, water would rush into the cell instead of out. 

 

Using the above picture as an example, we see that the initial state has unequal levels of solute concentration. There proportion of solute to free water is higher on the left side than the right. To combat this, the water will rush to that side of the semipermeable membrane, therefore regulating the proportion of the solute to free water. It is an awesome process because it almost seems to defy gravity, and will happen spontaneously!

 

 

 

Your goal is to design an experiment to demonstrate how concentration gradients effect the rate of water movement across a membrane and if this rate is impacted by the depth of the gradient.  We will recreate a selective membrane using dialysis tubing.  This 15mm dialysis tubing has small pores that allows only for the passage of water and not solutes.  Dialysis clips are utilized to close off each end of the tube and prevent the loss of solution.

 

Preparing dialysis tubing.  This video demonstrates how to fill the dialysis tubing with solution. After the tubing is filled, we have created an artifical cell that contains a solution of cytoplasm. Click on this link if your video won't load.

 

 

Materials

15 mm dialysis tubing (anywhere from 3-6 tubes are available) that hold 10mL of sucrose solution

30% stock solution of sucrose

 

400 mL beakers containing DI-water

dialysis clips

graduated cyliners

balance to measure weight

 

Hints:

1. View the video above about filling the dilaysis tubing.
2. What question are your trying to answer with your experimental design?
   
3. You will be responsible for any dilutions of your 30% stock.  Think about how many different concentrations you want to test.  What is the final volume of your dilutions?
   
4. In what units will you measure the rate of water movement? 
5. What is the density of water?
6. How long will you allow for the experiment to take place? 

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Experimental Design: 

 

1. Place a dialysis tube and two dialysis clips onto the balance. Record their mass.

2. Then take a dialysis tube and roll one end of the tube tightly for about an inch and then clip the dialysis clip onto the rolled up end. Repeat this process three more times.

3. Measure out 10 mL of DI Water in a graduated cylinder.

4. Take a dialysis tube that is closed by a dialysis clip at one end and insert the funnel into the open end, rolling the sides of the dialysis tube as you go. Then slowly pour the solution into the funnel.

5. After the dialysis tube has been filled, remove the funnel from it. Then push out any air from the bag and roll up the open end. Once end is closed, secure it with a dialysis clip.  This tube will be known as Tube A.

6. Measure out 10 mL of 30% stock solution of sucrose in a graduated cylinder.

7. Steps 4 and 5 will be repeated for this solution. Except this tube will be known as Tube B Instead of Tube A.

8. Measure out 5 mL of the 30% stock solution of sucrose in a graduated cylinder

9. Measure out 5 mL of DI Water in a graduated cylinder, and then gently mix the 5 mL of the 30% stock solution of sucrose into it

10. Steps 4 and 5 will be repeated for this solution. Except this tube will be known as Tube C instead of Tube A.

11. Measure out 2 mL of the 30% stock solution of sucrose in a graduate cylinder

12. Measure out 8 mL of DI water in a graduated cylinder, and then gently mix the 2 mL of the 30% stock solution of sucrose into it

13. Steps 4 and 5 will be repeated for this solution. Except this tube will be known as Tube D instead of Tube A.

14. Place each dialysis tube onto the balance. Record their respective masses.

15. Calculate the mass of each solution by subtracting the mass of dialysis tube and two dialysis clips from the mass of each dialysis tube filled with solution.

16. Then place each filled dialysis tube inside a separate beaker with 400 mL of DI Water for 15-20 minutes.

17. After 15-20 minutes have passed, take out each tube and weigh them, separately, again

18. Repeat step 15

19. Calculate the density of the solution in Tube A after it has been taken out of the beaker. (Density= mass/volume) 

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Results:

 

 

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E.  Movement of Solutes Across a Selectively Permeable Membrane:

Introduce the concept of diffusion of solutes.  In your discussion, include the roles of concentration gradients, the spontaneous nature of diffusion due to the second law of thermodynamics, and equilibrium.  Include a useful image for the process.  

 

Diffusion is the movement of a solute from a region of high concentration to a region of low concentration down the concentration gradient. There are two types of diffusion: simple diffusion and facilitated diffusion. Solutes in simple diffusion do not need the aid of a transport protein in order to pass through a membrane, while solutes in facilitated diffusion do need the aid of transport proteins as a passageway for the solute to diffuse across a membrane. Diffusion is spontaneous due to the second law of thermodynamics, which states that entropy, the degree of disorder in a system, is always increasing. Since diffusion is a form of passive transport and does not use any energy, and it distributes the solute particles more evenly, the 2^nd law of thermodynamics applies to this situation. Diffusion continues until equilibrium is reached, when the concentration of solute particles is spread out evenly. If the diffusion is taking place over a membrane, the concentration of solute must be the same on both sides of the membrane. 

 

 

 

In this experiment, you will recreate a cell and its extracellular environment.  Cell membranes are selectively permeable to solutes based on size, charge and polarity.  In our experiment, we will use dialysis tubing to recreate the cell membrane.  In our experimental system, the membrane is only permeable based on size.   We will first fill our artificial cell with a solution to recreate the cytoplasm of the cell.  We will then place the artificial cell into a beaker of solution that represents the extracellular fluid.  The goal of this experiment is to determine the direction of solute movement based on size and the presence of a concentration gradient.

 

Procedure: 

Part I – Setting up the artificial cells and the extracellular environment

1. Locate the 25-cm length of dialysis tubing.  Fold over and close off one end with a dialysis clip.

2. Place the open end of the dialysis bag over the stem of a clean funnel and fill with 25mL of the starch/Na₂SO₄ solution.

3. Fold the open end of each dialysis bag, squeeze from the tied end to remove as much air as possible, and close with a second dialysis clip.

4. Rinse each bag off in the pan of dH₂O, gently pat dry them a paper towel.

5. Submerge the dialysis bag into the beaker solution (extracellular fluid).

6. Record starting time.

7. Allow the experiment to run for 60 minutes.

 

Part II – Determining the direction of solute movement

8. Use the china marker to label the test tubes 1-8.

9. After 60 minutes pour 20 mL from the beaker into the 25-mL graduated cylinder.

10. Then pour 5 mL of this solution into each of test tubes 1-4.

11. Clean and dry the graduated cylinder.

12. Remove the dialysis bag from the beaker solution, rinse it off, and cut open one end.

13. Pour 20 mL of the bag solution into the graduated cylinder.

14. Then pour 5 mL of this solution into each of test tubes 5-8.

15. Perform the starch test on test tubes 1 and 5.

a. Add several drops of Lugol’s solution to each test tube.

b. If starch is present, the test tube solution will turn a dark blue-black color.

c. If the solution turns blue-black, record as a + test result. If there is no color change (other than the brown color of Lugol’s solution), record as a – test result.

d. Results from the beaker solution are record in Table 3. Results from the bag solution are recorded in Table 4.

16. Perform the sulfate ion test on test tubes 2 and 6.

a. Add several drops of 2% BaCl2 solution to each test tube.

b. If sulfate ions are present, a white precipitate (barium sulfate) will form.

c. If the precipitate forms, record as a + test result. If there is no precipitate, record as a – test result.

d. Results from the beaker solution are record in Table 3. Results from the bag solution are recorded in Table 4.

17. Perform the chloride ion test on test tubes 3 and 7.

a. Add several drops of silver nitrate to each test tube.

b. If chloride ions are present, a milky-white precipitate (silver chloride) will form.

c. If the precipitate forms, record as a + test result. If there is no precipitate, record as a – test result.

d. Results from the beaker solution are record in Table 3. Results from the bag solution are recorded in Table 4.

18. Perform the protein test on test tubes 4 and 8.

a. Add several drops of Biuret reagent to each test tube.

b. If protein is present, the solution will turn light lavender.

c. If the solution turns light lavender, record as a + test result. If there is no color change (other than the bright blue color of Biuret’s reagent), record as a – test result.

d. Results from the beaker solution are record in Table 3. Results from the bag solution are recorded in Table 4.

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Results:

 

 

 

 

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F.  Effect of Osmosis on Cells

Explain the impact of water balance on cells that contain cell walls (plants and bacteria) and cells that have no cell wall (animal cells).  Predict what would happen in the following scenarios.

 

Imagine we placed an animal cell like a red blood cell into the following solutions:

 

Condition	Hypertonic	Hypotonic	Isotonic
dH2O		x
0.9% NaCl			x
10% NaCL	x

 

1.  In the table above, place an x in the box that best describes each condition compared to the cell.  (Note:  Red blood cells have a solute concentration roughly equal to a 0.9%NaCl solution).

 

2.  Which direction would water move in each scenario?

 

In a hypotonic solution, the concentration of solute is higher inside the cell than outside the cell. The red blood cell with a 0.09% concentration of NaCl would have a higher concentration of solute than the deionized water solution. Therefore, water would flow into the cell to balance out the concentrations.

In an isotonic solution, the concentrations of solute inside the cell and outside the cell are equal. The red blood cell has the same concentration of solute as the 0.09% NaCl solution it is placed in. Therefore, the rate of water diffusion is the same going into the cell and coming out of the cell.

In a hypertonic solution, the concentration of solute is higher outside the cell than inside the cell. In this example, the 10% NaCl solution outside the cell is higher in concentration than the 0.9% concentration inside the cell. Therefore, the water would flow out of the cell in order to balance out the concentrations. 

 

  

3.  What would happen to the shape of the cell in each case?

 

In the hypotonic solution, water will flow into the cell, causing it to swell. In some cases, the cell may swell so much that it ruptures. This is called osmotic lysis.

In the isotonic solution, the cell shape will stay the same because an equal amount of water is flowing into the cell and out of the cell.

In the hypertonic solution, water will leave the cell and so the cell will shrink and crenate. 

 

 

I will provide sheeps blood that has been treated under the three conditions above on a prepared slide.  Please view each at 400x total magnification.  Take video of each specimen and narrate what is happening in each environment.

IMAGE  #1	IMAGE  #2	IMAGE  #3

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Presentation:

Record two presentations using the previous format (Introduction, Experimental Design/Execution, Results and Conclusions):

     1.  The diffusion of solutes across a permeable membrane (Section E).

     2.  Osmosis and its effects on cells (Sections D and F).