Lab 6: Passive Transport (Team 1)

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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.

Every cell is surrounded by a layer called the plasma membrane of the phospholipid bilayer. As the name suggests, the membrane is composed of two layers of phospholipids, which are amphipathic molecules. From the attached picture you can see that they consist of a polar region (enclosed by the glycerol head and phosphate group) and a nonpolar region (enclosed by the two fatty acyl tails). These phospholipids arrange themselves so the hydrophobic regions are being protected by the hydrophilic regions. The plasma membrane is usually referred to the fluid-mosaic model due to the variety of proteins, cholesterol and carbohydrates found scattered within the membrane. The macromolecules that have the most variety in function are the proteins. Within the plasma membrane there are three general types of proteins. First is a transmembrane protein, the most common type. These proteins have one or more parts that are inserted into the hydrophobic interior of the bilayer. Second is the lipid-anchored protein, which is a protein that is covalently attached to a lipid of the membrane, since the lipid is inserted into the hydrophobic region, it is a stable protein. Finally the third type of protein is a peripheral protein. These proteins, unlike the other two, don't directly interact with the hydrophobic region of the cell, rather they are noncovalently (usually by hydrogen or ionic bonds) bound to either the polar heads of the phospholipids or the part of the integral proteins that aren't within the nonpolar region. Most of these proteins act in order to bring materials into and out of the cell, some of these are channels that function in facilitated diffusion while other are pumps which are used along with ATP to bring about active transport. Carbohydrates usually attach to the polar region of an integral protein or directly to a phospholipid. They usually act to identify materials in the extracellular space and also they can anchor proteins in certain cells (such as anchoring a molecule of a hormone to its receptor). In addition they can also help build up or form the membrane through oligosaccharides. Finally there's the cholesterol. Cholesterol controls the fluidity of the plasma membrane. It will stabilize the membrane, depending on the temperature. For example if the temperature is low, then cholesterol will make the membrane more fluid and prevent it from freezing, while in opposite conditions it will make the membrane less fluid. All together these components create the permeability of the plasma membrane. Since the interior of the cell membrane is hydrophobic, that prevents the passage of any polar molecule, in addition size is also a factor. Large particles would not be able to fit in between the phospholipids.

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 movement of water across a semipermeable membrane(the diffusion of water). Water travels to the place with a higher solute concentration. When there is a different in solute concentration on the sides of a membrane, there is a solute gradient, and solutes will tend to move towards the side with a lower concentration. That is diffusion and i wont speak of that. Osmosis is just the movement of water, and it is affected when solutes can not pass through the membrane, so that there is always a solute gradient with a higher concentration on one side. Osmosis is spontaneous, meaning that it does not require additional energy to work, and water is always passing through the membrane, trying to reach a state of equilibrium where there are the same number of free water molecule on both side of the membrane. When there are solutes in the water, some water molecules attach to the solute (due to their polarity) and lose their ability (effectively) to cross the membrane. when there is an imbalance of solutes (a gradient) then this also causes an imbalance of water, because as the water molecules attach to the solutes and become trapped, there are less free water molecules on that side, and water from the other side has to come over to reach equilibrium again. For example, lets say that there are 100 water molecules on both sides of a membrane. Then i put 100 solutes on one side of the membrane, and each of them attract 1 water molecule. Now I have 100 free water molecules on one side, and 100 molecules on the other side that are attached to a solute. The free water molecules cross the membrane to reach equilibrium, and now there are 50 free water molecules on both sides, but the side with the solute also has 100 attached water molecules. The total water is 150 molecules on one side and 50 on the other, due to osmosis. If the solute concentration is equal on both sides, the it is an isotonic solution and there will be no net water movement. If there is a higher solute concentration on one side, then that side is a hypertonic solution, and there will be net water movement to that side. In a hypotonic solution, there will be a lower solute concentration, and there will be net water movement away from the hypotonic side. If one side is hypotonic, then the other side is hypertonic compared to that side.

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:

View the video above about filling the dilaysis tubing.
What question are your trying to answer with your experimental design?
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?
In what units will you measure the rate of water movement?
What is the density of water?
How long will you allow for the experiment to take place?

Experimental Design:

In this experiment, the main issue we are trying to answer is whether concentration gradients will initiate osmosis--diffusion of water. We will test 4 different concentrations, so we have a range of different concentrations, to get many results. A 30% solution will be needed, along with a 20%, 10%, and a 0% solution. The total amount of stock solution we will be using is 20 mL, ans we will be using 20mL of DI water. We will require 10 mL of stock solution for the 30% solution, 6.66 mL for the 20% solution, and 3.33 mL for the 10% percent stock solution; we need 3.33 mL of DI water for the 20% stock solution, 6.66 mL for the 10% solution, and 10 mL for the 0% solution. The rate of water movement should be measured in mL/minute, as it will take 60 minutes for the experiment to fully cycle, and we are only using 10 mL for each solution, so measuring in mL is reasonable. The density of water is one, so one gram of water is equivalent to one mL. When we weigh the bags, if the bag weigh any amount of grams less, that same number of grams is the amount of mL of water that has flowed. The control is the 0% solution, because we know that osmosis will not occur. The osmosis should occur in 60 minutes, and the following tests should take 25 minutes.

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.

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.

Cells that have a cell wall and cells that lack one typically react the same in an isotonic solution, by remaining the same. When a cell with a cell wall is placed in a hypertonic solution, it pulls away from the cell wall, becoming flacid. When a cell without a cell wall is placed in a hypertonic solution, it shrinks and shrivels and is called crenation. The difference comes in the hypotonic solution, when water rushes into the cell, in cells that lack a cell water, the plasma membrane isn't able to hold all the incoming water, causing the cell to burst. However, cells with a cell wall are strong enough to hold the water in the cell, a position called turgid. For many plant cells this is the preferred state in order to maximize the amount of water within the cell.

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 hypertonic solution, since there is a higher concentration outside the cell than inside, the water would move from inside the cell to out side faster than it comes in, causing a shrinkage or a crenation of the cell.

In an isotonic solution, since the concenntration of solutes is equal on both sides of the membrane, then the water balanced has reached an equilibrium and water would continue to move in both direction, but at an equal rate.

Finally in a hypotonic solution, since the concentration of solutes is greater inside the cell than outside, the extracellular water would rush into the cell faster than it leaves, causing a lysis.

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

In a hypertonic solution, since the water is moving out of the cell, the cell will shrivel up.

In an isotonic solution, since there is not distinct change in water movement on either side of the membrane, the cell remains in its normal state.

In a hypotonic solution, since water is moving into the cell, the cell will either be blown up to the maximum extent or lyse.

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.