2. Water Potential Notes Osmotic issues in plants cells.

3. What is Water Potential (Ψ w )? The tendency of water to move across a membrane due to the combined forces of osmotic potential and pressure potential; Ψ w is the same as osmosis (in animal cells). In plant cells, we have to consider the cell wall and the pressure it puts on the internal solution.

4. Ψ w is made of 2 components Ψ w = Ψ p + Ψ S 1.Pressure Potential (Ψ p ); the tendency of water to move through a membrane due to hydrostatic pressure; as pressure increases, Ψ p increases; Ψ p =usually zero (unless it’s in a plant cell) 2.Osmotic Potential (Ψ S ): The tendency of water to move through a membrane due to the solute in a solution; Ψ s = - iCRT (this is a calculation for osmosis) a)i = ionization constant (often 1.0); b)C = Molar concentration (from your experimental data) c) R = Pressure constant = 0.0831 liter bar/mole K d)T = temperature (K) = 273 + °C of solution

5. How is Ψ w a useful measurement? Used for water- deficit stress in plants To determine drought tolerance in plants Irrigation needs in different crops

6. 5 important rules to follow: 1.Water always moves across a selectively permeable membrane toward a region of lower (more negative) water potential 2.The addition of solutes lowers water potential

7. 3. Water potential and solute concen- tration are inversely related

8. 4. If plant is in solution with -Ψ, it loses Water; plasmolysis

9. 5. If plant in solution with +Ψ, it gains water; turgor

10. What are aquaporins? Channels in the plasma membrane where water mainly diffuses though the membrane

11. Practice Problems A flaccid plant cell has a water potential of -0.6 Mpa. Fill in the water potential equation for the cell. Ψ p = +Ψ S = Ψ w = Answer: Ψ p = 0Mpa +Ψ S =-0.6Mpa Ψ w = -0.6Mpa

12. Practice Problems The cell is then placed in a beaker of distilled water. Fill out equation for the cell after it reaches equilibrium. Explain. Ψ p = +Ψ S = Ψ w = Answer: Ψ p = 0.6Mpa +Ψ S =-0.6Mpa Ψ w = 0Mpa Explanation: Pressure of cell wall (pushing in) offsets pressure of water (pushing out). Water potential of both cell and solution = 0.

13. Practice Problems What would happen to this cell, if placed in a solution with a water potential of - 0.8 Mpa. Explain. Ψ p = +Ψ S = Ψ w = Answer: Ψ p = 0Mpa +Ψ S = -0.8Mpa Ψ w = -0.8Mpa Explanation: Cell would plasmolyze, losing water until solute potential of cell (Ψ S ) would equal the surrounding solution (- 0.8). No turgor pressure.

14. Practice Problems (more detail on Ψ S ) Equation: Ψ S =-iCRT i=ionization constant C=concentration R=pressure constant (0.0831liter bars/mol K) T=temperature in K (Celsius + 273) i=how many ions formed in water (NaCl=2, sucrose =1) C=given in question R=given constant T=given in Celsius, you figure out Kelvin

15. Practice Problems (more detail on Ψ S ) Equation: -iCRT i=ionization constant C=concentration R=pressure constant (0.0831liter bars/mol K) T=temperature in K (Celsius + 273) The molar concentration of sucrose in an open beaker has been determined to be 0.2M. Calculate solute potential (Ψ S ), at 22 degrees Celsius.

16. Practice Problems (more detail on Ψ S ) Equation: Ψ S= -iCRT Ψ S =- (1) (0.2 moles/liter)(0.0831liter/bars/moleK) (295K) Ψ S = -5 bars The molar concentration of sucrose in an open beaker has been determined to be 0.2M. Calculate solute potential (Ψ S ), at 22 degrees Celsius.

17. Practice Problems (more detail on Ψ S ) It’s in a beaker, so Ψ p =0. Ψ p = 0 +Ψ S =-5 Ψ w = -5 What is the overall water potential? Ψ p = +Ψ S = Ψ w =