BIOL 105 Osmosis and Diffusion Lab 1

BIOL 105 Osmosis and Diffusion Lab 1 Lab 2: Osmosis and Diffusion Objectives: To become acquainted with some of the physical processes important to cellular functioning, and to understand the principles of these processes and how they relate to the functioning of cells. The Passage of Water across a Differentially Permeable Membrane A water molecule, by virtue of its properties, passes readily through the cell membrane. This passage of water through a differentially permeable membrane is termed osmosis. Water moves across the membrane in both directions, but the direction of greatest flow or net movement is determined by the mole fraction of water, and thus the free energy of water, on either side of the membrane. In a situation where the internal environment (cytosol) and the external environment of the cell have equal mole fractions of water, the water will pass readily through the differentially permeable membrane in both directions at the same rate. This condition of the environment, which causes no change in cell volume, is considered isotonic to that of the cell. The vacuole membrane if also differentially permeable and water exchange occurs between the vacuole and cytosol in the same manner. Under conditions where there is a lower mole fraction of water in the external environment, water will rapidly move out of the cell, causing a decrease in cell volume and a shrinking of the protoplast. This transfer of water will continue until a point is reached where the mole fractions of water within the cell and within the external environment are equal or until the cell shrinks too severely to function with any degree of efficiency. Under these conditions the environment is termed hypertonic to the cell. The shrinkage of cells as a result of exposure to hypertonic conditions is a property of both plant and animal cells. The terminology for the same phenomenon, however, differs in reference to the two types of cells. Shrinking of plant cells is termed plasmolysis, while shrinking of animal cells is called crenation. Plant cells are surrounded by a rigid cell wall, which remains unaltered when the protoplast within shrinks. The presence of the rigid cell wall provides a point of reference, indicating the original size of the cell, for comparison of its size after plasmolysis. The reverse situation may also exist. An external solution that has a mole fraction of water greater than that of the cell will cause water to move rapidly across the membrane into the cell. As a result the protoplast will swell and become turgid. If an equilibrium is reached, the isotonic condition is restored and the passage of water into and out of the cell will again occur at equal rates. If, however, the difference in mole fractions of water is too great the membrane may reach the limit of its elasticity before equilibrium is reached and the cell will burst. Conditions which cause either a swelling or a bursting of cells are termed hypotonic to the cell. When a plant cell is exposed to a hypotonic environment, the protoplast takes in water and begins to swell. The plant cell, however, is limited in its ability to increase in size because of the cell wall, and when the cell membrane begins pressing against the cell wall a pressure potential increases until water uptake ceases even though the protoplast is not isotonic with the environment. For this reason, rupture

BIOL 105 Osmosis and Diffusion Lab 2 or bursting of the cell membrane in a hypotonic solution is essentially a characteristic of animal cells. (Because red blood cells are standardly used for this type of experiment, the term hemolysis is generally accepted for reference to bursting of animal cells). Determination of the Isotonic Point A variety of plant tissues may be used for this experiment; such as a leaf of Eloeda or moss, or thin sections of tissue from the pigmented lower epidermis of Zebrina or Rhoeo, or pigmented cells of red onion. You will be supplied with one of these tissue sources. Distilled water and molal (m) concentrations of NaCl and mannitol are available in dropper bottles to use to determine the isotonic point of the tissue. The molal (m) concentrations of NaCl and mannitol are the same (O.l5m; 0.30m; 0.45m; 0.60m; 0.75m). Other concentrations may be added or substituted if deemed desirable by your instructor. NaCl is an electrolyte with a formula weight of 58.4 and mannitol is a non-electrolyte with a formula weight of 182.2. The isotonic point of a tissue represents a condition of equality of the mole fraction or free energy of water between the tissue and the external bathing solution. The rate of water movement in and out of the cell is the same and the cells will remain almost the same in appearance. An external bathing solution is considered to be isotonic with the tissue when 50% of the cells in the microscopic field show very slight plasmolysis and the other 50% show no plasmolysis at all. This condition is known as "incipient plasmolysis" and may be difficult to recognize because the slight plasmolysis is very subtle. Look for a slight shrinking of the protoplast away from the cell wall. Often this will only occur in the corner of a cell or at irregularly space intervals rather than along the entire cell wall. Use the high power field of the microscope for these examinations. Plasmolysis is easier to observe in tissues with pigmented cells (e.g., Rhoeo, Zebrina, red onion). 1. Make a wet mount of the tissue using one of the NaCl or mannitol solutions. Allow about 5 minutes for water exchange. Observe the tissue during this time and at the end of 5 minutes and determine the approximate percentage of plasmolyzed cells. plasmolyzed cells X total cells in field 100 = percent plasmolysis 2. Enter your data in Table I. Wash the slide thoroughly or use another clean slide. With a fresh section of tissue repeat the procedure using a different concentration of mannitol or NaCl. It is important that each section of tissue you use be as nearly the same size and thickness as possible. It is also important that you be consistent as to the portion of each tissue you examine to make your counts of plasmolyzed cells. If you examine cells close to the edge of your tissue section then do so with all tissue examined. If the field you chose to examine is in the center of the tissue, then examine all tissue in the center. 3. Repeat the procedure until you have determined the percentage of plasmolysis for all the solution concentrations. 4. Prepare a graph of % Cells Plasmolyzed vs. Molal Concentration. Plot two curves, one representing the results with mannitol and the other with NaCl. Determine the concentration of each that represents the isotonic point.

BIOL 105 Osmosis and Diffusion Lab 3 5. Mannitol remains in molecular form in solution, while NaCl dissociates into Na + and Cl - ions. How does this dissociation affect the number of particles in solution of NaCl compared with mannitol? How is the difference between the number of particles in solution in the NaCl and mannitol reflected in the isotonic points of each? At equal concentrations (e.g., 0.30 m, etc.) which solution (mannitol or sodium chloride) would have the greatest mole fraction of water? Table 1: Isotonic Point Mannitol (m) D. W. 0.15 0.30 0.45 0.60 0.75 % Plasmolysis NaCl (m) D.W. 0.15 0.30 0.45 0.60 0.75 % Plasmolysis Notes: Measuring the Rate of Osmosis Each group prepares 5 dialysis bags from presoaked dialysis tubing. The tubing material is permeable to water, but is not permeable to sucrose molecules. Each piece of tubing should be closed by folding the end over and tying with thread or dental floss or by using a clamp. Dispense l5 mls of liquid into each bag according to this schedule: Bag Number Liquid (15 ml) 1 Distilled Water 2 0.25m Sucrose 3 0.50m Sucrose 4 0.75m Sucrose 5 Distilled Water

BIOL 105 Osmosis and Diffusion Lab 4 After each bag has been filled, remove the air by gently squeezing below to bring the liquid near the top of the bag. Press the side of the bag together to prevent reentry of air. Fold the end of the bag over about 2 cm and tie or clamp to tightly close the end. Weigh each bag separately to the nearest 0.0lg or 0.lg (depending on the balance available) and record the weights in Table 2 at 0 time. Place bags 1, 2, 3 and 4 in separate beakers of distilled water, and place bag 5 in a beaker of 0.75 m sucrose solution. Have enough water or 0.75m sucrose solution in the beakers to cover the bags with about l cm of liquid. At 15 minute intervals (i.e., 15, 30, 45 and 60 minutes) remove each bag from its beaker, quickly wipe off surface water and quickly weigh. Immediately return each bag to respective immersion liquid. Always place a weighing paper or tray on the balance before weighing the bags to protect the balance pan surface. Record all weights in Table 3a. Determine the change in weight of the bags and record that in Table 3b. Construct a graph and plot the changes in weight of each bag vs. time. Calculate the initial mole fraction of water in the distilled water and sucrose solutions used in both the dialysis bags and the beakers and record in Table 2. All the sucrose solutions are molal (m), the molecular weight of sucrose is 342, and the molal concentration of one liter of pure water is 55.6m. Calculate the initial osmotic potential (π) of each solution using the van't Hoff formula. Sucrose is a non-electrol yte, so consider the ionization coefficient (i) to be 1.0. Take the temperature of the solutions, or if that is not possible consider the temperature to be 22 C. Compare the mole fraction of water in the dialysis bag with the mole fraction of the bathing solution in the beaker in each case. According to your mole fraction gradient calculations, which direction should water flow in each case. Does the weight change data for the bags agree with the mole fraction gradients? Is there a difference in the rate at which water enters or leaves a bag? What conclusions, regarding osmosis and the rate of osmosis, can be drawn from your data? Table 2 : Osmosis Values at start of dialysis (calculated) Bag number: 1 2 3 4 5 Mole fract. of H 2 O in bag Mole fract. of H 2 O in beaker π of D. W. or sol 'n in bag π of D. W. or sol 'n in beaker NOTE: See Section B (Basic Information Review; Osmosis Information) for how to calculate: Mole fractions see page 8 π see page 6

BIOL 105 Osmosis and Diffusion Lab 5 Table 3a : Osmosis Data (weight of dialysis bag) Time (minutes) Weight of Bag (grams) 1 2 3 4 5 0 15 30 45 60 Notes: Table 3b : Osmosis Data (CHANGE in weight of dialysis bag) Time (minutes) Change of Weight of Bag (grams) 1 2 3 4 5 0 1 5 3 0 4 5 6 0 NOTE: Plot the data in Table 3b the change in weight of the dialysis bag vs. time

BIOL 105 Osmosis and Diffusion Lab 6 Factors Affecting the Rate of Diffusion This demonstration is designed to illustrate some of the factors that affect the rate of diffusion including concentration of the diffusing substance, temperature, molecular weight and molecular size of the diffusing substance. The apparatus includes glass tubing approximately 3 inches long containing gelatin. The tubing is inserted into test tubes containing small amounts of various dyes. The test tubes are kept upright so that only one end of the gelatin tubes are immersed in the dyes. Diffusion of each dye may occur up through the gelatin. The tubes are kept under controlled temperature conditions during the time diffusion is taking place. Observe the initial setup of this demonstration and then observe and interpret the final results when the tubes are displayed. Record the results in Table 4. Conditions of demonstration: Copper sulfate at 2% and 20% concentrations and both held at 20 C. 0.00lM meth ylene blue at both 10 C and 20 C. The dyes in the following list all have different molecular weights as indicated in parenthesis. Some, but not all, of these will be selected to provide a range of molecular weights for the demonstration. All dyes are at a 0.00lM concentration, and all that are used will be kept at the same temperature (20 C). Dyes not listed, may at times be substituted if those on the list are not available. Evans blue (MW 961) erythrosin B ( MW 880) methyl blue (MW 800) eosine yellow ( MW 692) eosine B (MW 624) methyl green ( MW 517) methyl orange (MW 327) methylene blue (MW 320). Congo red is at a concentration of 0.00lM and kept at 20 C, but is colloidal in size.

BIOL 105 Osmosis and Diffusion Lab 7 Table 4: Rat e of Diffusion D ye MW Concentration Temp C Distance Diffused CuSO 4 (ions ) 1 6 0 2 % 2 0 CuSO 4 (ions ) 1 6 0 2 0 % 2 0 Methylene Blue 3 2 0 0. 0 0 1 M 1 0 Methylene Blue 3 2 0 0. 0 0 1 M 2 0 Evans Blue 9 6 1 0. 0 0 1 M 2 0 Erythrosin B 8 8 0 0. 0 0 1 M 2 0 Methyl Blue 8 0 0 0. 0 0 1 M 2 0 Eosin Yellow 6 9 2 0. 0 0 1 M 2 0 Eosin B 6 2 4 0. 0 0 1 M 2 0 Methyl Green 5 1 7 0. 0 0 1 M 2 0 Methyl Orange 327 0. 001 M 20 Congo Red ( colloidal ) 697 0. 001 M 20 Notes: The effect of temperature on physical or chemical processes is frequently analyzed by means of temperature coefficients. The temperature coefficient, or Q 10, is defined as the ratio of the rates of a process at a given temperature and at a temperature 10 C lower. Q 10 = VT V T-10 C V T = velocity of a reaction a t a give n temperature ( C). V T - 1 0 C = velocity of the same reaction a t a temperature 1 0 C lower.

BIOL 105 Osmosis and Diffusion Lab 8 In general, physical processes such as diffusion, usually have a Q l0 ranging between 1.0 and 1.5, while chemical reactions usually have a Q l0 of 2.0 or greater. Thus, the determination of the Q l0 for a biological process may make it possible to determine whether the process is primarily physical or chemical in nature. Raising the temperature 10 C will approximately double the rate of a biological process (within a biological temperature range), if it primarily involves chemical reactions, but will only slightly raise the rate if it is primarily physical. Calculate the Q l0 for the diffusion of 0.00lM methylene blue at 10 C and 20 C. Is the diffusion of methylene blue primarily chemical or physical? The process of diffusion is basic to both the physical and biological world and must be understood. Almost all physiological functions in organisms depend at one stage or another on this fundamental process. SAMPLE OSMOTIC PROBLEMS: 1. What would the osmotic potential be (in bars) for a sucrose solution made up by placing 1.75 g of sucrose in 150 ml of water? Molecular weight of sucrose = 342. Sucrose is a nonelectrolyte. Consider temperature to be 17 C. 2. The osmotic pressure of blood is 7.65 atmospheres at 37 C. What weight of glucose should be added to 1000 ml of water to prepare an intravenous injection that will have the same osmotic pressure as blood? Molecular weight of glucose = 180. Glucose is a non-electrolyte.