Name: Bio A.P. Lab Diffusion & Osmosis

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1 Name: Bio A.P. Lab Diffusion & Osmosis BACKGROUND: Many aspects of the life of a cell depend on the fact that atoms and molecules are constantly in motion (kinetic energy). This kinetic energy results in molecules bumping into and rebounding off each other and moving in new directions. One result of this molecular motion is the process of diffusion. Diffusion is the random movement of molecules from and area of higher concentration to an area of lower concentration. For example, if one were to open a bottle of perfume (or hydrogen sulfide which smells like rotten eggs), in one corner of a room, it would not be too long before someone in the opposite corner would perceive the smell. The bottle contains a higher concentration of the perfume or hydrogen sulfide molecules than the room does, therefore the vapor diffuses from an area of higher concentration to an area of lower concentration. Eventually DYNAMIC equilibrium will be reached where the concentration is equal throughout the room and no NET movement of molecules will occur from one side of the room to the other. Osmosis is a special case of diffusion. Osmosis is the diffusion of water through a selectively permeable membrane (a membrane that allows for diffusion of selected solutes and water) from a region of higher water potential to a region of lower water potential until equilibrium is reached. Water potential is the measure of free energy of water in a solution. Where diffusion and osmosis do not suffice to explain the movement of ions or molecules into and out of cells, we may have to invoke properties of the living system itself. Active transport is the name for the process of using protein carrier molecules AND energy derived from ATP to move substances through the cell membrane- against a concentration gradient (from regions of low concentration to regions of higher concentration). 1

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4 LEARNING OBJECTIVES: To describe the mechanism of diffusion and osmosis To describe how solute size and molar concentration affect the process of diffusion through a selectively permeable membrane To relate osmotic potential to solute concentration and water potential Describe how pressure affects the water potential of a solution Calculate the water potential of a living plant cell from experimental data GENERAL SAFETY PRECAUTIONS: You must wear safety glasses or goggles and lab aprons. Some of the reagents used in this are stains and precautions must be taken to not get them on your skin or clothing. Do not work in the laboratory without your teacher s supervision. Talk to your teacher if you have any questions or concerns about the experiments. THE INVESTIGATIONS: This investigation consists of two major parts. In Part I you will use dialysis tubing to investigate the relationship between solute concentration and the movement of water through a selectively permeable membrane by the process of osmosis. In Part II you will use cores of potato tissue placed in different molar concentrations of sucrose in order to determine the water potential of potato cells. 4

5 INTRODUCTION: Distinguish between the terms Hypotonic, hypertonic and isotonic Water potential, osmotic concentration and pressure potential Describe what will happen to an animal cell AND a plant cell when placed in a hypotonic environment, a hypertonic environment and an isotonic environment. Be sure to use the terms, turgid, flaccid. lyses, plasmolysis. Will water move into or out of a plant cell if the cell has a higher water potential than the surrounding environment? Part I In this experiment you will use dialysis tubing to investigate the relationship between solute concentration and the movement of water through a selectively permeable membrane by the use of osmosis. When a dialysis bag containing a sucrose solution of placed in distilled water the bag will accumulate water as a result of osmosis. Because there is higher water potential outside the bag than inside the bag, water will diffuse into the bag. In the absence of pressure, the solution outside the bag is hypotonic relative to the solution inside the bag. If the solute is added to the water outside the bag, you decrease the water potential of he solution outside the bag. It should be possible to add just the right amount of solute to the water so the water potential outside the bag is the same as the solution inside the bag. The two solutions are isotonic and no NET water movement will occur. If you continue to add more solute to the water solution outside the bag, the water potential outside the bag will decrease to the point where the solution outside the bag is hypotonic to the solution inside the bag. The NET movement of water will then be in opposite direction, from inside (where the water potential is higher) to outside the bag (where the water potential is lower). PROCEDURE: 1. Obtain six 20 to 30 cm strips of presoaked dialysis tubing. 2. Tie off on end of each piece of tubing to form six bags. Pour 10 ml of each of the following solutions into separate bags. a) Distilled water b) 0.2 M sucrose c) 0.4M sucrose d) 0.6 M sucrose e).8 M sucrose f) 1.0 M sucrose. 3. Tie off the other end of the bag leaving ENOUGH room for the bag to expand. 4. Carefully blot the outside of each bag and record the initial mass (grams) of each bag in your data table. 5. Fill six 250 ml beakers or cups two-thirds full with distilled water. 6. Immerse each bag in one of the beakers of distilled water and label the beaker to indicate the molarity (M) of the solution the dialysis bag. 7. Let each set up stand for exactly 30 minutes. 8. At the end of 30 minutes remove the bags from the water. Carefully blot and determine the mass of each bag. 9. Record you data. Calculate the percent change in mass for each solution. 10. Graph your results. % Change = Final Mass-Initial Mass x 100 Initial Mass 5

6 Contents of Dialysis Tube 0.0 M sucrose (distilled water) 0.2 M sucrose Initial Mass Final Mass Percent change in Mass 0.4 M sucrose 0.6 M sucrose 0.8 M sucrose 1.0 M sucrose 1. Predict what would happen in an experiment of all the bags were placed in a 0.4 M sucrose solution instead of distilled water. Defend your answer. Include a labeled diagram showing the contents of the beaker, dialysis bag and draw an arrow to show the NET movement of water. Part II: Determining the Water Potential of Potato Cells In this exercise you will use the cores of potato tissue placed in different molar concentrations sucrose in order to determine the water potential of potato cells. 6

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8 PROCEDURE: Work in Groups. You will be s\assigned one or more of the beaker contents listed in Part I of the investigation: 1. Pour 100 ml of the assigned liquid into a labeled 250 ml beaker 2. Use a cork borer to cut two to four potato cylinders. Cut each cylinder to approximately 3 cm in length. Do not include any skin on the cylinders. 3. Keep your potato cylinders in a covered container until it is your turn to use the balance. 4. Determine the combined mass of the cylinders and record. 5. Put the cylinders in the beaker. Cover the beaker with parafilm or plastic wrap to prevent evaporation. 6. Let stand overnight. 7. The next day record the temperature of the liquid in the beaker. It will be used later in the calculation of osmotic potential. 8. Remove the cores from the beakers, blot them gently on a paper towel, and determine their mass. Record the final mass. 9. Obtain the results from other member of your group 10. Calculate the percent change in mass % Change = Final Mass-Initial Mass x 100 Initial Mass Temperature: Contents of beaker Initial Mass Final Mass Percent change in Mass 0.0 M sucrose (distilled water) 0.2 M sucrose 0.4 M sucrose 0.6 M sucrose 0.8 M sucrose 1.0 M sucrose 11. In order to determine the osmotic potential of the potato core you need to determine the osmotic molar concentration of the potato core. This will be when the potato core is isotonic to the sucrose solution. Graph the percent change in mass. Note that on this graph you will be plotting both positive and negative percent changes in mass. The potato core and the sucrose solution are isotonic at the exact point your plotted line intersects 0% mass change. Mark this point on your graph. This value is the osmotic molar concentration of the potato. 8

9 ANALYSIS OF RESULTS: 1. Determine the osmotic molar concentration of the potato from your graph: 2. Calculate the osmotic potential of the solution (potato core) using the formula Osmotic Potential is equal to: Ψs = icrt Where: i = ionization constant C = molar concentration R = pressure constant (R = liter. bars/mole. K) T = temperature in Kelvin (273 + o C) 3. Knowing the osmotic potential of the solution (Ψs) and knowing that the pressure potential of the solution is zero (ψ P =0) calculate the water potential of the solution (potato core). Water Potential = Pressure Potential + Solute Potential (Turgor Pressure) (Osmotic Potential) ψ = ψ P + ψ S 4. If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cell become higher or lower? Defend your answer. 9