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Lab 6: Passive Transport (EF, PG)

Page history last edited by Emily Fu-Sum 12 years, 7 months ago

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


The plasma membrane separates the internal contents of cells from their external environments. It is involved with the selective uptake and export of ions and molecules. Phospholipids make up the main framework of the membrane and are arranged in a bilayer. They are amphipathic molecules, with their hydrophobic tails forming the interior of the membrane and their hydrophilic heads forming the exterior of the membrane. If the fatty acid tails of the phospholipids are unsaturated and kinked, the bilayer is more fluid. Cellular membranes also contain proteins that function in transporting ions and molecules, enzymatic activities, cell signaling, and cell-to-cell adhesion. Cholesterol is also present in membranes, and tends to stabilize them. At high temperatures, cholesterol helps to lower the fluidity of the membrane while at high temperatures, it helps to increase the fluidity. Carbohydrates can be attached to proteins or lipids in the membrane, and are said to serve as recognition signals for other proteins. It also plays a role in determining blood type. The arrangement of the amphipathic phospholipids plays a role in the selective functions of the membrane. Due to the hydrophobic interior of the membrane, hydrophilic molecules cannot diffuse at an effective pace across the membrane. This is where proteins that aid in transportation across the membrane come into play.

Plasma Membrane Structure

Plasma Membrane's Components




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 flow of a liquid or gaseous substance through a semi-permeable membrane. A semi-permeable membrane is a membrane through which only specific kinds of molecules can readily pass. Osmosis is tremendously important in processes where water moves in and out of cells, because all living cell membranes are semi-permeable. Osmosis keeps the amount of water in all the cells of an animal or a plant approximately equal and helps to prevent certain cells from drying out. An example is the movement of urea. If a urea molecule is added to one membrane side, it won’t be able to diffuse because it is polar and large in size. It interacts with other polar molecules such as water. The interaction reduces the number of “free” water on that side. With less free water molecules on the side with urea, there is a net movement of water molecules down their gradient to the urea molecule side. If the osmotic concentrations of two solutions are equal, the solutions are isotonic. When the solutions have unequal concentrations, the solution with the higher concentration is hypertonic and the solution with the lower concentration is hypotonic.



This picture shows how osmosis works. http://www.occc.edu/biologylabs/Images/Cells_Membranes/osmosis.gif


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.




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 cylinders

balance to measure weight



  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? 

Experimental Design: 


Hypothesis: The dialysis tube that has the highest solute concentration will have the highest rate of water movement, and thus the highest weight. 


1. Prepare dialysis Tube 1 with 10 mL of distilled water to act as the control then place this tube into a solution of distilled water. Measure the weight of the tube before and after placing it in the solution.

2. Fill Tubes 2 with 10 mL of 3% sucrose solution. Prepare this by adding 1 mL of 30% sucrose solution to 9 mL of deionized water.

3. Fill Tubes 3 with 10 mL of 15% sucrose solution. Prepare this by adding 5 mL of 30% sucrose solution to 5 mL of deionized water.

4. Fill Tube 4 with 10 mL of the 30% sucrose solution.

5. Place each tube in a beaker of deionized water.

6. Measure the weight of each tube before placing it in the solution then measure it again every 15 minutes for an hour.

  Tube 1
Tube 2
Tube 3
Tube 4
15.13 g
`15.25 g
15 min.
30 min.
45 min.
60 min.


E.  Movement of Solutes Across a Selectively Permeable Membrane:

Solute diffusion occurs across a selectively permeable membrane across a concentration gradient from high concentration to low concentration. A concentration gradient is a difference between concentrations in a space. The second law of thermodynamics says the transfer of energy from one form to another increases entropy or degree of disorder of a system. Entropy is the measure of the randomness of molecules in a system.  Diffusion is spontaneous because there is an increased amount of free energy due to the highly ordered molecules. This energy causes molecules to go from a higher concentration with free energy to a lower concentration.  Chemical equilibrium is when the rate of formation of products equals the rate of formation of reactants. When equilibrium is reached there is no longer a concentration gradient.  


diffusion2.gif (640×480)

The above animation shows the basic process of diffusion across a plasma 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.



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.






F.  Effect of Osmosis on Cells

When a plant/bacteria/fungi/algae cell is placed in a hypotonic solution, a small amount of water may enter the cell. However, the cell wall prevents any major expansion due to the cell wall. Alternatively, if the plant/bacteria/fungi/algae cell is placed in a hypertonic solution, water will leave the cell and the plasma membrane will pull away from the cell wall, a process called plasmolysis. In plant cells, turgor pressure (the hydrostatic pressure required to stop the new flow of water across a membrane) pushes the plasma membrane against the rigid cell wall.


When an animal cell is placed in a hypotonic solution, cells swell and may undergo osmotic lysis because the water is taken into the cell. Because there is no cell wall, nothing is protecting against major expansion. So when water enters an animal cell, it can rupture much more easily than a cell with a cell wall. When an animal cell is placed in a hypertonic solution, cells undergo shrinkage (crenation) because water exits the cell.


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


Condition Hypertonic Hypotonic Isotonic
0.9% NaCl    
10% NaCL    


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?

dH2O: water would move into the cell

0.9% NaCl: there would be no net movement of water

10% NaCl: water would move out of the cell


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

dH2O: the cell would swell/expand, possibly to the point that it ruptures (osmotic lysis)

0.9% NaCl: the cell would maintain its shape

10% NaCl: the cell will shrink/shrivel (crenation)



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.

0.9% Slide at 1000x (Dirt from Ocular Lens) 0.9% Slide at 1000x  10% Slide at 1000x 

0.09% Solution (Hypotonic)
Cells lysed due to immense amount of water

that entered the cells to balance the concentrations

inside and outside the cell.

0.9% Solution (Isotonic)
Cells remained the same, as

they were placed in a solution that

had the same solute concentrations as

them and thus osmosis did not need

to occur in order to balance the

concentrations inside and outside the cell.

10% solution (Hypertonic)
Cells shriveled up due to water leaving

the cell, since the solute concentration

outside the cell was higher.


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).




Comments (1)

Derek Weber said

at 5:20 am on Nov 11, 2011

Again, nice use of a graph and outstanding job on conclusions in both videos.

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