INVESTIGATION : Determining Osmolarity of Plant Tissue AP Biology This lab investigation has two main components. In the first component, you will learn about the osmolarity of plant tissues and the property of plant tissues known as water potential, a critical property that explains why water is able to move from soil into roots (and subsequently, leaves). In the second component, you will observe what occurs when a plant cell is exposed to the three types of osmotic solutions: hypotonic, hypertonic, and isotonic. BACKGROUND INFORMATION Part One: Osmolarity of Plant Tissues and Water Potential In this part of the lab, you will use a plant tissue of your choice placed in sucrose solutions of varying molar concentration to determine the water potential of plant cells. But what does water potential even mean? When you hear the word potential, think potential energy : the energy in a system that is available to do work. The work that water does in biological systems is that of dissolving solutes such as salts, proteins, and sugars--all of which are polar molecules. When a polar solute is present in an aqueous solution, the water molecules in the solution will form hydrogen bonds with the polar (or charged) portions of the solute. If an ionic solute (such as salt, NaCl) is present in the solution, the solute particles will dissociate into their constituent ions, and the water molecules will surround the individual ions, as shown in the diagram at right. You can see that the positive portions of the water molecules are attracted to the negative chloride ions, and the negative portions of the water molecule are attracted to the positive sodium ions. In biological molecules, the same thing happens--water is attracted to the polar components of those molecules and will thus dissolve them. The diagrams below show that water is attracted to the polar components of glucose (a carbohydrate) and a schematic drawing of a protein.
Notice that if water molecules are attracted to and are binding to polar portions of biological molecules, they are unable to perform the work of dissolution since they are already occupied since they have formed hydrogen bonds with the solute. Therefore, their potential to do work as they across the membrane is reduced. By adding solutes to an aqueous solution, the water potential of that solution is reduced. Water potential is affected by two physical factors. One factor is the addition of solute to water. The reason that adding solute to water reduces water potential is thi s: as water molecules form hydration shells around ions or solute molecules, their movement is restricted due to the formation of hydrogen bonds with these ions or solute molecules. The energy that would otherwise be available to do work is tied up in the hydrogen bonds formed between solute and water. This restricted movement means that the potential for them to do the work of dissolution is reduced and thus the water potential is lowered. The relationship between water potential and solute concentration is an inverse one. By convention, the water potential of pure water at atmospheric pressure (1 atm) is defined as being zero. The reason for this is that pure water has no solutes in it, so water molecules will not form hydration shells and thus are free to do the work of dissolution--their free energy is high. The potential energy of water molecules is at its highest in pure water.
The other factor is pressure potential (mechanical pressure). An increase in pressure raises the water potential of the solution. The pressure potential can also be thought of as the turgor pressure exerted by the cell membrane on the cell walls of plant cells. If a great deal of water is present in the plant cell s central vacuole, it will exert pressure on the cell wall, causing turgor pressure to increase. Conversely, if the central vacuole of a plant cell experiences a decrease in its volume, turgor pressure drops and the plant cell becomes flaccid. The relationship between pressure potential and water potential is a direct one. Movement of water into and out of a cell is influenced by the solute potential (relative concentration of solute) on either side of the cell membrane. If water moves out of the cell, the cell will shrink. If water moves into an animal cell, it will swell and may even burst. In plant cells, the presence of a cell wall prevents cells from bursting as water enters the cells, but pressure eventually builds up inside the cell and affects the net movement of water. As water enters a dialysis bag or a cell with a cell wall, pressure will develop inside the bag or cell as water pushes against the bag or cell wall. The pressure would cause, for example, the water to rise in an osmometer tube or increase the pressure on a cell wall. Water will always move from an area of higher water potential to an area of lower water potential. Water potential, then, measures the tendency of water to leave one place in favor of another place. You can picture the water diffusing down a water potential gradient, as shown at left. Movement of water into and out of a cell is also influenced by the pressure potential (physical pressure) on either side of the cell membrane. Water movement is directly proportional to the pressure on a system. For example, pressing on the plunger of a water-filled syringe causes the water to exit via any opening. In plant cells this physical pressure can be exerted by the cell pressing against the partially elastic cell wall. Pressure potential is usually positive in living cells: in dead xylem elements it is often negative.
How is Water Potential Actually Used? Botanists use the term water potential when predicting the movement of water into or out of plant cells. Water potential is abbreviated by the Greek letter psi ( Ψ ) and it has two components; a physical pressure component, pressure potential Ψ p, and the effects of solutes, solute potential Ψ s. Ψ = Ψ p + Ψ s Water potential = Pressure potential + Solute potential Mathematically, this is represented as: Ψ = -icrt where i = ionization constant (for organic molecules like sugar that do not ionize in solution, this is always 1, for other ionic compounds, it is equal to the number of ions the compound dissociates into) C = concentration of solution (in mol) R = 0.0821 L-bars/mol-K T = temperature in degrees K (so temperature in C + 273K) It is important for you to be clear about the numerical relationships between water potential and its components, pressure potential and solute potential. The water potential value can be positive, zero, or negative. Remember that water will move across a membrane in the direction of lower water potential. An increase in pressure potential results in a more positive value and a decrease in pressure potential (tension or pulling) results in a more negative value. In contrast to pressure potential, solute potential is always negative ; since pure water has a water potential of zero, any solutes will make the solution have a lower (more negative) water potential. Generally, an increase in solute potential makes the water potential value more negative and an increase in pressure potential makes the water potential more positive. To illustrate the concepts discussed above, we will look at a sample system using the figures at right. When a solution, such as that inside a potato cell, is separated from pure water by a selectively permeable cell membrane, water will move (by osmosis) from the surrounding water where water potential is higher, into the cell where water potential is lower (more negative) due to the solute potential (ψs). In the picture at the right (picture a) the pure water potential is 0 (ψ=0) and the solute potential is 3 (ψs = -3.) We will assume, for purposes of explanation, that the solute is not diffusing out of the cell. By the end of the observation, the movement of water into the cell causes the cell to swell and the cell contents push against the cell wall to produce an increase in pressure potential (turgor) (ψ=3). Eventually, enough turgor pressure builds up to balance the negative solute potential of the cell. When the water potential of the cell equals the water potential of the pure water outside the cell (ψ of cell = ψ of pure water = 0), a dynamic equilibrium is reached and there will be no NET movement of water (picture b). If you were to add solute to the water outside the potato cells, the water potential of the solution surrounding the cells would decrease. It is possible to add just enough solute to the water so that the water potential outside the cell is the same as the water potential inside the cell. In this case, there will be no net movement of water. This does not mean, however, that the solute concentrations inside and outside the cell are equal, because water potential inside the cell results from the combination of both pressure potential and solute potential. If enough solute is added to the water outside the cells, water will leave the cells, moving from an area of higher
water potential to an area of lower water potential. The loss of water from the cells will cause the cells to lose turgor. A continued loss of water will eventually cause the cell membrane to shrink away from the cell wall, known as plasmolysis. Now, you will investigate the concept of water potential by placing plant tissues of uniform size in sucrose solutions of varying concentration. You will have the opportunity to select one of several different types of plant tissues provided to you. Regardless of which plant tissue your group selects, the procedure for carrying out the investigation will be the same. SAFETY CONSIDERATIONS: Take caution when cutting with either the scalpel or the cork borer as both are sharp instruments. Do not eat any part of the lab. MATERIALS various plant tissues (e.g. potato, sweet potato, carrot, beet, radish, turnip, apple, squash) Unknown sucrose solutions, ranging from 0M to 1M cork borer electronic balance scalpel 2 ounce souffle cups with lids marker DIRECTIONS 1. Get six souffle cups and their lids. Label the outside of each cup with the color of unknown sucrose solution that will go inside it. You also need to label the cups with your lab group s initials and class period. 2. As a group, decide which plant tissue you would like to work with. Obtain it from the supply table. 3. Depending on which tissue you choose, you may need to cut a 3 cm thick slice of the tissue before cutting out the cylinders of tissue for your experiment. 4. You will need to cut 4 cylinders that measure each 2 cm in length for each of the souffle cups. You will need a total of 24 cylinders of plant tissue. Do not include any of the skin of the plant in your cylinder. 5. Using an electronic balance, mass each of the cups containing plant tissue cylinders, writing the initial mass on the outside of the souffle cup. Make sure you record the initial mass of the plant tissue cylinder in the data table provided. Be sure you account for the mass of the cup (zero out the mass of the cup on the balance). Make qualitative observations about your plant tissues, including descriptions and photos of your specimens. 6. Cover each cup with a lid until you pour the appropriate sucrose solution over each one. This will help prevent the tissues from drying out. 7. Fill each cup with the appropriate unknown sucrose solution. Fill the cups about halfway (about 30 ml) so that the plant cylinders are covered with liquid. 8. Once you have covered each cup of plant tissues with sucrose solution, replace the lid and stack the cups on top of one another where they will not be disturbed. Your plant tissues must sit overnight. 9. The next day, you will collect the data. When collecting data, do the following: a. Pour off the sucrose solution in the cup. b. Carefully blot the cylinders of plant tissue as dry as you can.
c. Using the same electronic balance as you did when you set up the lab, weigh the cylinders IN THE CUP and record the mass in the data table provided. Be sure you account for the mass of the cup. d. Again, make qualitative observations about your plant tissues, including descriptions and photos of your specimens. e. When you are done collecting data, throw away the plant tissues and cup/lid. PROCESSING YOUR DATA 1. Calculate the percent change in mass. To do this: 2. Plot your data on a graph. Think: a. What was your independent variable? b. What was your dependent variable? c. Determine the molar concentration of the unknown color sucrose solutions. 3. Now, determine the molar concentration of the sucrose in the plant tissue your group selected. This would be the sucrose molarity in which the mass of the tissue does not change. To do this: a. To find this, draw a line of best fit. In Excel, this is your trendline, which you can add after plotting your data. The point at which this line crosses the x-axis represents the molar concentration of sucrose with a water potential that is equal to the plant tissue water potential. At this concentration there is no net gain or loss of water from the tissue. This is also known as the isotonic point. Determine what the isotonic point is for your group s plant tissue and record it under the data table provided. WHAT YOU WILL TURN IN: THE DELIVERABLE Your final product for this lab will be turned in as individuals--this is not a group assignment. You will turn in a CER for this lab that includes the following: Your research question. The hypothesis you tested. Your claim about the plant tissue you investigated. Evidence, to include: Processed data (properly constructed graphs with explanation of data, statistical analysis) Your statistical analysis should be done by comparing your data to that of another group who used the same plant tissue as your group did, even if they are in another class period. Reasoning that supports your claim which includes showing the relationship between the outcome of your investigation and the biological concepts explored. You will submit your work as a Google Doc in Google Classroom. Be sure you have shared it appropriately so your teacher can view your work.
GROUP DATA Paste this into your BILL along with a printed copy of your graph. Plant Tissue Selected: Unknown Color Solution Sucrose Solution (M) Initial Mass of Plant Tissue (g) Final Mass of Plant Tissue (g) Mass Difference % change in mass Clear Yellow Green Blue Red Orange Isotonic Point: CLASS DATA Sucrose Solution (M) % change in mass 1 2 3 4 5 6 7 8 0M.2M.4M.6M.8M 1M