Water Potential of Potato Cell Lab

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Water Potential of Potato Cell Lab Biology 11 AP November 28, 2018 AP BIOLOGY Lab: Determining the Water Potential of Potato Cells Background In animal cells, the movement of water into and out of the cell is in±uenced by the 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 the cell, the cell may swell or even bursts. In plant cells, the presence of a cell wall prevents the cells from bursting, but pressure does eventually build up inside the cell and a²ects the process of osmosis. When the pressure inside the cell becomes large enough, no additional water will accumulate in the cell even though the cell still has a higher solute concentration than does pure water. So the movement of water through the plant tissue cannot be predicted simply through knowing the relative solute concentrations on either side of the plant cell wall. Instead, the concept of water potential is used to predict the direction in which water will di²use through living plant tissues. In a general sense, the water potential is the tendency of water to di²use from one area to another under a given set of parameters. Water potential is expressed in bars, a metric unit of pressure equal to about 1 the atmosphere and measured with a barometer. Water potential is abbreviated by the Greek letter psi and has two major components: solute potential, which is dependent on solute concentration and pressure potential, which results from the exertion of pressure - either positive or negative - on a solution. We express this as: In Figure 1.1. A potato cell is placed in pure water. Initially, the water potential outside the cell is 0 and is higher than the water potential inside the cell (-3). Under these conditions, there will be a net movement of water into the cell. The pressure potential inside the cell will increase until the cell reaches a state of equilibrium.

Pre-Lab Questions 1. What would happen if you applied saltwater to the roots of a plant? Why? (2 marks) Due to the fact that the ionization constant of salt water is -2, this creates a solution with a lower water potential. Water would move out of the cell into the hypertonic solution (from higher to lower potentials), causing the roots of the plant to wilt as a result. 2. Will water moves into or out of a plant cell if the plant cell has higher water potential than the surrounding environment? (1 mark) The water will move out of the plant cell into the surrounding environment, trying to reach a state of equilibrium. Directions: 1. You will be provided with 5 sucrose solutions of unknown molarity (Each unknown is one of the following: 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0M). 2. Slice a potato into 5 equal cylinders. 3. Pour enough of each unknown solution into a test tube to cover the potato. Mark each test tube. 4. Determine the initial mass of the 5 potato cylinders and record. 5. Place the potato cylinders into the test tubes with the solution and cover with plastic wrap. Leave overnight. 6. Remove the cylinders from the test tubes and carefully blot o² any excess solution. Record the room temperature in Celsius. 7. Determine the ³nal mass of the potato cylinders and record. 8. Calculate the % change. Record. 9. Determine the molarity of the unknown solutions. This step will require some thought. 10. Graph the results. In order to do so, the 0 axis line should actually be in the middle of your graph. The y-axis above this line should be labelled % increase in

mass while the y-axis below this line should be labelled % decrease. The x-axis is the sucrose molarity within the test tube. Data Table 1: Mass of potato before and after exposure in solutions of di²erent molarities Potato in Solution Initial Mass (g) Final mass (g) Calculations Di±erence in Mass (g) Change (%) A 4.9 3.79 3.79 - 4.9 -1.11 -23 B 5.05 4.25 4.25 - 5.05 -0.8 -16 C 4.39 4.04 4.04 - 4.39 -0.35 -8 D 4.31 4.17 4.17 - 4.31 -0.14 -3 E 4.38 4.44 4.44 - 4.38 0.06 1 Room temperature = 22 degrees Celsius = 295 degrees Kelvin Analysis and Graph: (5 marks – 1 for the title, 1 for labelling x-axis, 1 for labelling y-axis, 2 for points)

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1. Determine the molar concentration of each unknown potato cores. This would be the sucrose molarity in which the mass of the potato core does not change. To ³nd this, draw the straight line on your graph that best ³ts 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 potato tissue water potential. At this concentration, there is no net gain or loss of water from the tissue. (1 mark for a line of best ³t) 2. What are the Molar concentration of Solution A, B, C, D, E? (1 mark) A - 1.0M B - 0.8M C - 0.6M D - 0.4M E - 0.2M 3. What is the Molar concentration of sucrose when it crosses the x-axis? (1 mark) The molar concentration of sucrose when it crosses the x-axis is 0.32 mol/l 4. Calculate the solute potential for the sucrose solution at this point, show your work: (2 marks) Ѱₛ =-iCRT Ѱₛ = -(1)(0.32 mol/L)(0.0831 k*bars/mol*L)(295k) Ѱₛ = -(1)(0.32)(0.0831)(295) Ѱₛ = -7.84 bars POST-LAB QUESTIONS

1. If a potato core is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells be higher or lower than the environment? Why? (2 marks) Since the cell is dehydrating, water is ±owing out of the cell into its environment making the water potential of the environment lower than that of the potato cell (hypertonic). 2. If a plant cell has a lower water potential than its surrounding environment is the cell hypertonic (in terms of solute concentration) or hypotonic to its environment? Will the cell gain or lose water? Explain. (3 marks) The cell is hypotonic. Since the water will ±ow from high to low and the plant cell has a lower water potential than its surrounding environment than the water will be ±owing from the environment to the cell. The cell will be gaining water and will become hypotonic. 3. If the water potential for a sucrose solution in a dialysis bag is -6.25 bars and it is immersed in a cup of sucrose solution having a water potential of -3.25 bars will the bag gain or lose mass? Explain your answer. (2 marks) The bag will gain weight as it ³lls with more of the solution because the cup of sucrose solution is hypotonic to the sucrose solution in the bag. In order to reach a state of equilibrium, the sucrose solution will have to enter the dialysis bag.

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