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Passive Transport (Team 5)

Page history last edited by Jean-Pierre Jacob 10 years, 9 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

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 is made up of a phospholipid bilayer with proteins embedded in it. The basic function of the cell membrane is to protect the cell from its surroundings. The plasma membrane separates the internal environment from the external one. Cell membranes are involved in a variety of cellular processes such as cell adhesion, ion conductivity, cell signaling, and serve as the attachment surface for several extracellular structures. The plasma membrane is selectively permeable so only some things are able to pass through a process called simple diffusion. The proteins that are embedded in the membrane help transport molecules and ions in and out of the cell through a process called facilitated diffusion. Phospholipids have a polar and hydrophilic head and a nonpolar and hydrophobic tail.  The membrane is arranged with the hydrophobic heads on the outside and the hydrophilic tails on the inside. Nonpolar and small molecules are able to pass through but polar, charged, and large molecules cannot pass.  Cholesterol is important to the plasma membrane because it is required to build and maintain cell membranes; it regulates membrane permeability and fluidity over a wide range of temperatures. The plasma membrane also contain  carbohydrates , they are called glycoproteins and glycolipids. They are found on the extracellular surface of the plasma membrane. Proteins can be found on the membrane or embedded in it. The functions of membrane proteins include cell signaling, enzymatic activity, and transporting substances across the plasma membrane.

 

 

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

Introduce the concept of osmosis.  In your discussion, include the roles of solute gradients and their effect on "free"water, the spontaneous nature of osmosis, and equilibrium.  Terms like hypertonic, hypotonic, and isotonic should be used. Include a useful image for the process.  

 

     Osmosis is the spontaneous net movement of water molecules through a semi-permeable membrane to an area with a higher solute concentration in order to equalize the solute concentrations on both sides.  In osmosis, the movement of water requires no input of energy. In osmosis there has to be a selectively permeable membrane because it allows for only the passage of water molecules and not solute molecules through the membrane. Equilibrium is when the concentration of solute and solution on one side has the same concentration on the other side of the membrane. If there is an imbalance in the concentrations, then the water molecules will cross the membrane until equilibrium is reached. After equilibrium has been reached, the water molecules will continue to move from one side of the membrane to and another in equal amounts, a process called dynamic equilibrium. You cant control or predict when solute cross the membrane and since there is no energy input, the crossing is spontaneous. An isotonic solution is a solution that has the same concentration of solutes on either side of the membrane. There will be no new movement of water across the cell membrane, however, water will continue to move from one side to the other in equal amounts in order to maintain dynamic equilibrium. In a hypotonic solution, there is a lower concentration of solutes in the solution relative to the inside of the cell. In this case, water will move into the cell via osmosis. This will cause the cell to swell. A hypertonic solution is a solution that has a higher concentration of solutes in the solution relative to the inside of the cell. In this situation, water will move out of the cell, via osmosis, causing the cell to shrink and crenate. The first picture shows the process of osmosis. In the second picture, the diagram to the left shows a hypotonic solution, the middle diagram shows a hypertonic solution, and the diagram to the right shows an isotonic solution.

 

 

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. No
  2. What question are your trying to answer with your experimental design? How solute concentration affects the rate of osmosis, and how the depth of the concentration gradient affects rate. 
  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? 10 mL
  4. In what units will you measure the rate of water movement? mL/min (1 g=1mL) 
  5. What is the density of water? 1 g/cm^3
  6. How long will you allow for the experiment to take place?  60 minutes. 

Experimental Design: 

 

Experimental Procedure

 

1. Make four 10mL solutions

 

a. fill a graduated cylinder with 0% sucrose solution (basically 10mL of dh2O)

 

b. fill another with 5% sucrose solution (9.5 mL of dh20 and .5mL of sucrose solution)

 

c. fill another with 10% sucrose solution (9.0 mL of dh20 and 1.0 mL of sucrose solution)

 

d. fill another with 15% sucrose solution (8.5 mL of dh20 and 1.5mL of sucrose solution)

 

2. Make the dialysis tubing (do it as per the video)

 

a. pour each of the 4 sucrose solutions into four separate dialysis tubes

 

b. clip the ends with dialysis clip

 

c. Rinse off the outside of the tubing with dH20 and gently pat the tubing dry.

 

3. Weigh each of the 4 dialysis tubings on a scale (grams) and record the measurements. There is no need to weigh out the tubing separate from the tubing+substance because the amount of substance that leaves/enters the tube after undergoing osmosis will remain the same.

 

4. Fill 4 400mL breakers with 100mL of dh20 (no significance in the number, but it’ll give some continuity between the 4 different solutions).

 

5. Place the dialysis tubing into the beaker and submerge it completely.

 

6. Let the tubings sit in the beaker for about 60 minutes total, however to precisely monitor the rate of diffusion, we will weigh the test tubes every 20 minutes.

 

7. After 60 minutes, take the tubing out of the beaker, and gently pat the tubing dry.

 

8. Measure the final weight of each of the 4 dialysis tubings on a scale (grams) and record the measurements.

 

9. Indicate whether water moved into the cell or out of the cell.

 

a. If the water moved into the cell, then the weight of the tubing after the experiment is higher than the initial weight of the tubing.

 

b. If the water moved out of the cell, then the weight of the tubing after the experiment is lower than the initial weight of the tubing.

 

10. Convert the grams to mL. (1g=1mL) and divide by 60 to find the rate of osmosis in mL/min

 

11. Compare the rates of osmosis between each of the different sucrose solutions in order to determine how the depth of the concentration gradient affects the rate of osmosis (i.e. determine the sucrose solution with the highest rate of diffusion, and observe how high of a sucrose concentration it had in order to determine how steepness of the concentration gradient affects rate of osmosis).

a: Observe how the cell reacted when placed in an isotonic solution vs. when the cell is placed in a hypotonic solution.

 

 


Results:

 

 

 

 

 

 

 

 

 


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 explains the net amount of molecules from a region of higher concentration to one of lower concentration. Unlike osmosis,diffusion also occurs when there is no concentration gradient. Diffusive equilibrium is only reached when the concentrations  of the substance  in the two compartments that diffusion is happening across becomes equal. The second law of thermodynamics states that  that the  entropy of an  isolates system never decreases, because they spontaneously evolve toward thermodynamic equilibrium which is the state of maximum entropy. In diffusion, a system that has two solutions of different concentrations that is separated by a semipermeable membrane has less entropy than a similar system having two solutions of equal concentration. The system with the differing concentrations is said to be more ordered, and thus has less entropy. The second law of thermodynamics requires the presence of an osmotic flow that will take the system from an ordered state of low entropy to a disordered state of higher entropy. The thermodynamic equilibrium is only achieved when the entropy gradient between the two solutions becomes zero.  

 

 

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.


Results:

 

 

 

 

 

 

 

 

 

 

 

 


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    
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?

 

     When a red blood cell (RBC) is placed in a solution containing dH2O, it is placed in a hypotonic solution. dH2O is a hypotonic solution because it contains a lower concentration of solutes relative to the inside of the RBC. In this case, water will move into the cell via osmosis in order to equalize the solute concentrations on either side of the membrane. When the RBC is placed in a solution of 0.9% NaCl, the cell is being placed in a isotonic solution. 0.9% NaCl is a isotonic solution because the concentrations of the solutes on both sides of the plasma membrane of the cell are equal (RBC's have a solute concentration that is about 0.9% NaCl). Even though equilibrium has been reached, water will continue to move in both directions via osmosis, a process called dynamic equilibrium. When a RBC is placed in 10% NaCl, this cell is being placed in a hypertonic solution because this solution has a higher solute concentration relative to the inside of the cell. Water will move out of the cell via osmosis as an attempt to equalize the solute concentrations on both sides of the membrane. 

      

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

 

     When the cell is placed in the hypotonic solution containing dH2O, the cell will swell because water is diffusing into the cell in order to equalize the solute concentrations. In some cases, a cell may take up so much water that it ends up rupturing, a process called osmotic lysis. If the cell is placed in the isotonic solution containing 0.9% NaCl, the shape of the cell would remain the same. This would occur because dynamic equilibrium is occurring where water is moving equally in both directions (both in and out of the cell) in order to maintain the equilibrium that has already been established. Lastly, when a cell is placed in the hypertonic solution containing 10% NaCl, the cell will shrink because water is diffusing out of the cell in order to equilize the concentrations on both sides of the cell membrane. The process by which cells shrink, as a result of water diffusing out of them, is called 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.

     
IMAGE  #1
IMAGE  #2
IMAGE  #3

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).

 

 

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