Name KEY Block Date AP Water Potential Practice 1. If a cell's Yp = 3 bars and its Y's = - 4.5 bars, what is the resulting Y'? Y = 3 bars + (-4.5 bars) = - 1.5 bars 2. The cell from question #1 is placed in a beaker of sugar water with Y's = - 4.0 bars. In which direction will the net flow of water be? -1.5 bars is higher than -4.0 bars so water will move OUT from cell to beaker 3. The original cell from question # 1 is placed in a beaker of sugar water with Y's = - 0.15 MPa (megapascals). We know that 1 MPa = 10 bars. In which direction will the net flow of water be? -0.15 MPa x 10 = - 1.5 bars -1.5 bars is equal to -1.5 bars, so there is no net flow (equilibrium) 4. The value for Y' in root tissue was found to be -3.3 bars. If you take the root tissue and place it in a 0.1 M solution of sucrose at 20°? in an open beaker, what is the Y' of the solution, and in which direction would the net flow of water be? Find solute potential first using: Y's = - CRT Y's =- (1)(0.1)(0.0831)(273 +20) Y's = - 2.43483 = ~ - 2 bars in the solution Then find water potential Y' = Y's + 4'p Y = - 2 bars + 0 bars = - 2 bars Since water moves from an area of HIGHER water potential to LOWER water potential, - 2 bars in the beaker is HIGHER than -3.3 bars in the root tissue (LOWER) so water will move from the beaker into the root tissue 5. NaCl dissociates into 2 particles in water: Na+ and Cl . If the solution in question 4 contained 0.1M NaCl instead of 0.1M sucrose, what is the Y' of the solution, and in which direction would the net flow of water be? Y's = - (2)(0.1)(0.0831)(293) = - 5 bars Y = - 5 bars + 0 bars = - 5 bars -5 bars in the beaker is LOWER than -3.3 bars in the root tissue (HIGHER) so water will move out of the root tissue into the beaker
6. A plant cell with a Y's of -7.5 bars keeps a constant volume when immersed in an open-beaker solution that has a Y's of -4 bars. What is the cell's Y'p? The plant cell keeps a constant volume because of the buildup of turgor pressure inside the cell. Y's + Y'p of beaker = Y's + 4'p of cell -7.5 bars + 0 = - 4 bars + 4'p Y'p = 3.5 bars The Y'p at equilibrium would be the difference between the two solute potentials, which is 3.5 bars. 7. At 20°C, a cell containing 0.6M glucose is in equilibrium with its surrounding solution containing 0.5M glucose in an open container. What is the cell's Yp? Surrounding solution: Y