DIFFUSION AND OSMOSIS

Lab 5 DIFFUSION AND OSMOSIS OBJECTIVES Describe the process of diffusion at the molecular level; State the physical factors that determine the direction and rate of diffusion; Discuss why diffusion rates, cell volume and surface area are thought to place limits on cell size; Understand the nature of a semi-permeable membrane and why its presence is necessary for osmosis to occur; Recognize isotonic, hypotonic, and hypertonic environments and the effect they have on living organisms; Define the key terms listed in this exercise. INTRODUCTION The plasma membrane serves as a "gate keeper", a remarkable structure that allows water and dissolved gases such as oxygen and carbon dioxide to pass freely while differentiating between ions and small molecules. Some solutes enter and exit the cell while others are barred. This selective permeability of the plasma membrane is fundamental to maintaining homeostatic conditions within the cell necessary for life. Molecules that can permeate plasma membranes do so by either passive transport where molecules move down their concentration gradients, or by active transport where molecules are moved against their concentration gradient. Active transport requires cellular energy, which is provided by ATP. In today's lab, you will be introduced to two forms of passive transport diffusion and osmosis, which do not require energy. BEFORE COMING TO LAB: 1. Read Chapter 5 and your notes from lecture pertaining to diffusion and osmosis. 2. Read this lab carefully and be familiar with the words listed under Key Terms. KEY TERMS solvent hemolysis aqueous tonicity permeate solute diffusate homeosta sis ATP passive transport selectively permeable diffusion semipermeable concentration gradient isotonic hypertonic hypotonic osmosis turgor pressure plasmolysi s 1

I. THE DIFFUSION OF MOLECULES Regardless of the state they are in (gas, liquid, or solid) molecules are continually moving. If a concentration gradient exists, the collective movement of molecules can be predicted. Specifically, there will be a net movement of molecules from an area of high concentration to an area of low concentration. This net movement if defined as diffusion. The rate at which diffusion occurs is influenced by the temperature of the environment, the mass (atomic or molecular weight) of the particle and the magnitude to which a concentration gradient exists. A. Rate of Diffusion of Solutes Solutes move within a cell s cytoplasm largely because of diffusion. However, the rate of diffusion (the distance diffused in a given amount of time) is affected by such factors as temperature and the size of the solute molecules. In the following exercise, we will examine the effects of these two factors in gelatin, a substance much like the cell s cytoplasm and used to simulate it in this experiment. Prior to lab, six glass test tubes containing 5% gelatin were prepared; and 1 ml of a dye were added to each test tube. The identity of the solutes used and their molecular weights are provided in Table 5-1. Set 1 (tubes 1-3) were placed in a refrigerator at 5 C while set 2 (tubes 4-6) were maintained at room temperature (23 o C). Your lab instructor will inform you of the time at which the solutions were added to the gelatin tubes (T 0 ). TIME: Procedure 1. Remove set 1 from the refrigerator and compare the distance the dye has diffused in corresponding tubes of each set. 2. Be certain the cap to each tube is tight. Invert and hold each tube vertically in front of a white sheet of paper. 3. Use a metric ruler to measure to the nearest 1.0 mm how far each dye has diffused from the gelatin s surface. Record this distance in Table 5-1. 4. Return set 1 to the refrigerator immediately. 5. Determine the rate of diffusion for each dye by using the following formula: Rate of diffusion = distance elapsed time (hours) Table 5-1 The Effect of Temperature on Diffusion Rate of Various Solutes Elapsed Time: hrs Solute (dye) Molecular Weight Test Tube Number (5 C) Distance Diffused (mm) Diffusion Rate (mm/hr) Test Tube Number (Rm. Temp) Distance Diffused (mm) Diffusion Rate (mm/hr) Aniline Blue 737.74 1 4 Potassium Dichromate 294.21 2 5 Janus Green 511.09 3 6 2

Diffusion rate (mm/hr) Diffusion rate (mm/hr) 6. Use Figure 5-1 to construct linear graphs illustrating diffusion rates as a function of molecular weight at 5 C and room temperature. 5 C 23 C (RT) Molecular Weight Molecular Weight 3

Figure 5-1 Diffusion rates as a function of molecular weight at 5 C and room temperature. Questions 1. Which of the solutes diffused the slowest (regardless of temperature)? 2. Which diffused the fastest? 3. What effect did temperature have on the rate of diffusion? 4. What conclusions about the diffusion of a solute in a gel (relating the rate of diffusion to the molecular weight of the solute and to temperature) can you make concerning the rate at which diffusion occurs? B. Cell Size Limitations Based on Diffusion Rates and Cell Surface Area/Volume Ratios As cells undergo metabolism they take up O 2 and food molecules from their environment and release CO 2 and other waste products. All such exchanges occur across the surface of the plasma membrane and it is thought that the mathematical relationship between this surface area and the cytoplasm volume is a crucial factor in limiting cell size. To explore the relationship between surface area and cell volume it is useful to assume a hypothetical cube shaped cell as seen in Figure 5-2. L L L Figure 5-2 Hypothetical cube shaped cell. Example: Assuming each side of our cell is 4 microns (µ) in length, then: a. Surface Area = 6 L 2 = 6 (4 µ) 2 = 96 µ 2 b. Volume = L 3 = (4 µ) 3 = 64 µ 3 c. Thus the ratio between the surface area and volume in this particular example is 92/64 or 1.5 in its simplest form 4

Surface/Volume Procedure 1. To clearly see this relationship between the surface area and the volume of an object calculate the surface area/volume ratios for cube shaped cells having the dimensions found in Table 5-2. Table 5-2 Cell Surface Area/Volume Ratios Length (µ) Surface Area Volume Surface/Volume 1 2 3 4 96 64 96/64 = 1.5 5 6 2. Construct a line graph in Figure 5-3 using your data from Table 5-2. 6.0 1.0 1.0 Length ( ) 6.0 5

II. OSMOSIS Figure 5-3 Surface/volume ratio as a function of cell size. The principles that apply to diffusion also apply to osmosis except that the presence of a semi- permeable membrane is necessary. The nature of the membrane must be that it allows water to move freely but will not allow the movement of larger molecules. Thus, if we have a difference in tonicity (the total concentration of solute particles) between the two sides of the membrane, we will see a net movement of water form the side with a higher water concentration to the side with the lower water concentration. This net movement of water is termed osmosis and any fluid pressure that may build up as a result is called osmotic pressure. A. Osmosis Procedure 1. Obtain four 15 cm sections of dialysis tubing. Presoak them in dh 2 O. Recall that the dialysis tubing is permeable to water molecules but not to sucrose. 2. Fold one end of each bag over and tie it tightly with string. 3. Attach a string tag to the tied end of each bag and number them 1-4. 4. Place the open end of a numbered sac over the stem of a funnel (see Figure 5-4). Using a graduated cylinder to the measure the volume, fill with 10 ml of water or the appropriate sucrose solution as indicated in Table 5-3. Fill in order of increasing strength. 5. As each bag is filled squeeze out any excess air by squeezing the bottom end of the tube. 6. Fold the open end and tie it tightly with another piece of string, rinse with tap water and gently blot off the excess water with paper towel. Figure 5-4 Method for filling dialysis tubing. 10 ml graduated cylinder Dialysis bag 6

7. Weigh each bag to the nearest 0.1 g and record the weights in Table 5-3 in the column marked 0 Min. 8. Using a grease pencil, number four 600 ml beakers on the labeling tape. DO NOT MARK DIRECTLY ON THE BEAKERS. 9. Add 200 ml of d H 2 O to beakers 1-3 and 200 ml of 30% Sucrose to beaker 4. 10. Place bags 1-3 in their correspondingly numbered beakers. Place bag 4 in the beaker containing 30% sucrose. 11. After 15 minutes, remove each bag from its beaker, blot off the excess fluid, and weigh each bag to the nearest 0.01 g. 12. Record the weights of each bag in Table 5-3. Table 5-3 Weight Changes in Artificial Cells as a Consequence of Osmosis Sac Bag Contents/ 0 Min. 15 Min. 30 Min. 45 Min. 60 Min. No. Beaker Contents weight weight weight weight weight (g) (g) (g) (g) (g) 1 H 2 O/H 2 O 2 15% Sucrose/ H 2 O 3 30% Sucrose/ H 2 O 4 H 2 O /30 %Sucrose 13. After all of the bags are weighed, return them to their respective beakers. 14. Repeat steps 11-13 at 30, 45, and 60 minutes from time zero. 15. Use the information you have recorded in Table 5-3 to calculate the percentage weight change (final-initial/ initial x 100) as a function of time and enter this data in Table 5-4. Table 5-4 Percent Weight Changes in Artificial Cells as a Consequence of Osmosis Sac No. Bag Contents/ Beaker Contents 0 Min. % change 15 Min. % change 30 Min. % change 45 Min. % change 60 Min. % change 1 H 2 O/H 2 O 0% 2 15% Sucrose/ H 2 O 0% 3 30% Sucrose/ H 2 O 0% 7

Percent Weight Change 4 H 2 O/30% Sucrose 0% 16. Use Figure 5-5 to create line graphs of the results for each bag you have calculated % change for in Table 5-4. (+) 0 10 20 30 40 50 Time (min) (-) Figure 5-5 Percent weight changes of artificial cells over a 30 minute incubation period in various osmotic environments. III. SELECTIVE PERMEABILITY OF MEMBRANES Procedure 1. Obtain a 25-cm section of dialysis tubing that has been soaked in d H 2 O. 2. Fold over one end of the tubing and tie it securely with string to form a leak-proof bag (Figure 5-4). 3. Slip the open end of the bag over the stem of a funnel and fill the bag approximately half full with 25 ml of a solution of 0.1% starch in 1% sodium sulfate (Na 2 SO 4 ). 4. Remove the bag from the funnel; fold and tie the open end of the bag with another piece of string. 5. Rinse the tied bag with d H 2 O. 6. Pour 200 ml of a solution of 1% albumin (a protein) in 1% sodium chloride (NaCl) into a 400 ml beaker. 7. Place the bag into the fluid in the beaker. 8. Record the time: 9. With a grease pencil, label eight test tubes, numbering them 1-8. DO NOT WRITE DIRECTLY ON THE TEST TUBES. 10. Sixty minutes after the start of the experiment, pour 5 ml of the beaker contents into each of the first four test tubes. 8

11. Perform the following tests. Record your test results (+ or for each test) in Table 5-5. Your instructor will have a series of test tubes showing positive results for starch, sulfate and chloride ions, and proteins. Compare your results with the known positives. a. Test for starch Add several drops of iodine solution (I 2 KI) to test tube 1. If starch is present, the solution will turn blue-black. b. Test for sulfate ion Add several drops of barium chloride (Ba 2 Cl) to test tube 2. If sulfate ions (SO 4 -) are present, a white precipitate of barium sulfate (BaSO 4 ) will form. c. Test for chloride ion Add several drops of 2% silver nitrate (AgNO 3 ) to test tube 3. A milky white precipitate of silver chloride (AgCl) indicates the presence of chloride ions (Cl - ). d. Test for protein Add several drops of biuret reagent to test tube 4. If protein is present, the solution will change from blue to pinkish-violet. The more intense the violet hue, the greater the quantity of protein. Table 5-5 Results of Tests for Substances in Beaker Starch At Start of Experiment After 60 min. Sulfate Ion Chloride Ion + Albumin + 12. Thoroughly rinse the bag in d H 2 O. 13. Using scissors, cut the bag open and empty its contents into a clean 25-mL graduated cylinder. 14. Pour 5 ml samples into each of the four remaining test tubes. 15. Perform the tests for starch, sulfate ions, chloride ions, and protein on tubes 5-8, respectively. 16. Record the results of this series of tests in Table 5-6 Table 5-6 Results of Tests for Substances in Dialysis Tubing At Start of Experiment Starch + After 60 min. Sulfate Ion + Chloride Ion Albumin 9

Questions 1. To which substances was the dialysis tubing permeable? 2. What physical property of the dialysis tubing might explain why some substances could cross while others could not (differential permeability)? IV. OSMOTIC CHANGES IN ANIMAL AND PLANT CELLS A. Plasmolysis in Plant Cells Plant cells are surrounded by a rigid cell wall, composed primarily of the glucose polymer, cellulose. The presence of a cell wall and large fluid-filled central vacuole will affect the cell's response to environments of differing tonicities. Normally, the solute concentration within the cell's central vacuole is greater (hypertonic) than that of the external environment. Thus, in a hypotonic solution, water moves into the cell and ultimately into the cell's central vacuole, creating turgor pressure, which presses the cytoplasm against the cell wall. Such cells are said to be turgid. The following experiment demonstrates the effects of external solute concentrations (Figure. 5-6). HYPOTONIC MEDIUM Cell wall Water flow (a) Low turgor pressure (b) Increased turgor pressure (c) Greatest turgor pressure Figure 5-6 The effect of the cell wall and turgor pressure on the direction of net flow of water in a plant cell. 10

Procure 1. With a forceps, remove two young leaves from the tip of an Elodea plant. 2. Mount one leaf in a drop of dh 2 O on a microscope slide and the other in 20 % NaCl solution on a second microscope slide. 3. Place coverslips over both leaves. 4. Observe both leaves with the compound microscope using the high power objective. After several minutes, the cell in the 20% NaCl will have lost water. This process results in a loss of turgor pressure and called plasmolysis. 5. Make sketches of your observations in Figure 5-7. Label the sketches, indicating whether the cells are normal or plasmolyzed. Slide 1 Slide 2 Figure 5-7 Elodea cells. 6. After completing this part of the experiment: a. Remove the slide when finished. Rinse slides with tap water and dry with a paper towel. b. Remove the coverslip from the drain and dispose of it in the trash. c. Gently wipe with a piece of lens paper moistened with lens cleaner, the 10x and 40x objective lenses of your microscope. d. Return the microscope to its assigned cabinet. 11

Tonicity describes one s solute concentration compared to that of another solution. The solution containing the lower concentration of solute molecules than another is hypotonic relative to the second solution. Solutions containing equal concentrations of solute are isotonic to each other, while one containing a greater concentration of solute relative to a second one is hypertonic. Given these considerations, answer the following questions. Questions 1. Were the contents of the vacuole in the Elodea leaf in dh 2 O hypotonic, isotonic, or hypertonic compared to the dh 2 O? 2. Was the 20% NaCl solution hypertonic, isotonic, or hypotonic relative to the cytoplasm? 3. If a hypotonic and a hypertonic solution are separated by a selectively permeable membrane, in which direction will the water have moved? 4. Name two selectively permeable membranes that are present within Elodea cells and that were involved in the plasmolysis process. a. b. 12

B. Osmotic Changes in Red Blood Cells Animal cells lack the rigid cell wall of a plant. The external boundary of an animal cell is the selectively permeable plasma membrane. Thus, an animal cell tends to increase in volume as water enters the cell. However, since the plasma membrane is relatively fragile, it ruptures when too much water enters the cell. This is because of excessive pressure pushing out against the membrane. In red blood cells, the membrane is especially fragile and will hemolyze (rupture) if the cells are placed in a hypotonic environment. Conversely, if red blood cells are placed in a hypertonic environment they will lose water and take on a spiny appearance. This process is called crenation (see Figure 5-8 below). (a) RBCs in an isotonic solution ( normal ) (b) RBCs in a hypertonic solution ( crenate ) (c) RBCs in a hypotonic solution (hemolyzed) Figure 5-8 Scanning electron micrographs of red blood cells. Proced ure 1. Number three clean microscope slides using a grease pencil (slide # s 1-3). 2. Put one drop of blood on each of the three slides #1-3. 3. Prepare three blood slides according to the protocol below. a. To slide 1 add one drop of distilled H 2 O b. To slide 2 add one drop of 0.9% NaCl c. To slide 3 add one drop of 10 % NaCl 4. Add a cover slip to each of the slides and observe them under the high-dry objective. 13

5. In Figure 5-9, make a simple sketch of a few representatives from each of the three slides. Label the sketches indicating whether the cells are normal, plasmolyzed (crenate), or hemolyzed. Slide 1 Slide 2 Slide 3 Figure 5-9 Microscopic appearance of red blood cells in different osmotic environments. 6. After completing this part of the experiment: a. Dispose of your microscope slides and coverslips into the biohazard box labeled Used Blood Slides. b. Use lens paper moistened with lens cleaner, to wipe the 10x and 40x objective lenses of your microscope. c. Return the microscope to its assigned cabinet. Questions 1. Why do red blood cells burst when put in a hypotonic solution whereas Elodea leaf cells do not? 14

POST-LAB QUESTIONS 1. You want to dissolve a solute in water. Without shaking or swirling the solution, what might you do to increase the rate at which the solute would go into solution? Relate your answer to your method s effect on the motion of the molecules. 2. If a 10% sugar solution is separated from a 20% sugar solution by a selectively permeable membrane, in which direction will there be a net movement of water? 3. Based on your observations in this exercise, would you expect dialysis membrane to be permeable to sucrose? Why? 4. You are having a party and you plan to serve celery, but your celery has gone limp, and the stores are closed. What might you do to make the celery crisp (turgid) again? 5. Why don t plant cells undergo osmotic lysis? 6. This drawing represents a plant cell that has been placed in a solution. a. What process is taking place in the direction of the arrows? What is happening at the cellular level when a wilted plant is watered and begins to recover from the wilt? b. Is the solution in which the cells have been placed hypotonic, isotonic, or hypertonic relative to the cytoplasm? 7. A human lost at sea without fresh drinking water is effectively lost in an osmotic desert. Why would drinking salt water be harmful? 8. How does diffusion differ from osmosis? 9. Plant fertilizer consists of numerous different solutes. A small dose of fertilizer can enhance plant growth, but overfertilization can kill the plant. Why might overfertilization have this effect? 10.What does the word lysis mean? (Now does the name of the disinfectant Lysol make sense?) 15