1. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Diffusion of Water (Osmosis) To survive, plants must balance water uptake and loss Osmosis determines the net uptake or water loss by a cell and is affected by solute concentration and pressure

2. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Water potential is a measurement that combines the effects of solute concentration and pressure – Ψ = Ψ P + Ψ S Water potential determines the direction of movement of water Water flows from regions of higher water potential to regions of lower water potential

3. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Water potential is abbreviated as Ψ and measured in units of pressure called megapascals (MPa) Ψ = 0 MPa for pure water at sea level and room temperature

4. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings How Solutes and Pressure Affect Water Potential Both pressure and solute concentration affect water potential The solute potential (Ψ S ) of a solution is proportional to the number of dissolved molecules Solute potential is also called osmotic potential

5. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Pressure potential (Ψ P ) is the physical pressure on a solution Turgor pressure is the pressure exerted by the plasma membrane against the cell wall, and the cell wall against the protoplast

6. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Measuring Water Potential Consider a U-shaped tube where the two arms are separated by a membrane permeable only to water Water moves in the direction from higher water potential to lower water potential

7. Fig. 36-8a ψ = − 0.23 MPa (a) 0.1 M solution Pure water H2OH2O ψ P = 0 ψ S = 0 ψ P = 0 ψ S = − 0.23 ψ = 0 MPa The addition of solutes reduces water potential

8. Fig. 36-8b (b) Positive pressure H2OH2O ψ P = 0.23 ψ S = − 0.23 ψ P = 0 ψ S = 0 ψ = 0 MPa Physical pressure increases water potential

9. Fig. 36-8c ψ P = ψ S = − 0.23 (c) Increased positive pressure H2OH2O ψ = 0.07 MPa ψ P = 0 ψ S = 0 ψ = 0 MPa 0.30 Further Physical pressure increases water potential more

10. Fig. 36-8d (d) Negative pressure (tension) H2OH2O ψ P = − 0.30 ψ S = ψ P = ψ S = − 0.23 ψ = − 0.30 MPaψ = − 0.23 MPa 0 0 Negative pressure decreases water potential

11. Fig. 36-9a (a) Initial conditions: cellular ψ > environmental ψ ψ P = 0 ψ S = − 0.9 ψ P = 0 ψ S = − 0.7 ψ = − 0.9 MPa ψ = − 0.7 MPa 0.4 M sucrose solution: Plasmolyzed cell Initial flaccid cell: If a flaccid cell is placed in an environment with a higher solute concentration, the cell will lose water and undergo plasmolysis

12. Fig. 36-9b ψ P = 0 ψ S = − 0.7 Initial flaccid cell: Pure water: ψ P = 0 ψ S = 0 ψ = 0 MPa ψ = − 0.7 MPa ψ P = 0.7 ψ S = − 0.7 ψ = 0 MPa Turgid cell (b) Initial conditions: cellular ψ < environmental ψ If the same flaccid cell is placed in a solution with a lower solute concentration, the cell will gain water and become turgid

13. solute potential (Ψ S ) Ψ S = - iCRT i is the ionization constant C is the molar concentration R is the pressure constant (0.0831 liter bars/mole-K) T is the temperature in K (273 + C°)

14. Calculating Water potential Say you have a 0.15 M solution of sucrose at atmospheric pressure (Ψ P = 0) at 25 °C calculate Ψ 1 st use Ψ S = - iCRT to calculate Ψ S i = 1 (sucrose does not ionize) C = 0.15 mole/liter R = 0.0831 liter bars/ mole-K T = 25 + 273 = 298 K Ψ S = - (1)(.15M)(0.0831 liter bars/mole-K)(298 K) = -3.7 bars 2 nd use Ψ = Ψ P + Ψ S to calculate Ψ Ψ = Ψ P + Ψ S = 0 + (-3.7bars) = -3.7 bars

15. You try it Calculate Ψ of a 0.15M solution of of NaCl at atmospheric pressure (Ψ P = 0) at 25 °C. Note: NaCl breaks into 2 pieces so i = 2. Ψ S = - (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K) Ψ S = - 7.43 bars Ψ = 0 + (-7.43 bars) Ψ = -7.43 bars

16. You try it again Calculate the solute potential of a 0.1 M NaCl solution at 25 °C. If the NaCL concentration inside a plant cell is 0.15 M, which way will the water diffuse if the cell is placed into the 0.1 M NaCl solution? Ψ S = - (2)(0.10 mole/liter)(0.0831 liters bars/mole-K)(298 K) = - 4.95 bars (solution) Ψ S = - (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K) = - 7.43 bars (cell) Water will move from solution to cell.

17. What must Turgor Pressure (Ψ P )equal if there is no net diffusion between the solution and the cell? Ψ P in cell must equal 2.49 Goal to make Ψ of cell = Ψ of solution (-4.95) Ψ = Ψ P + Ψ S (of cell) Ψ = 2.49 + (-7.43) Ψ = -4.95

18. Diffusion & Osmosis Lab Read the background material for Lab 4 – Diffusion and Osmosis Procedure 1 – Plasmolysis Procedure 2 – Osmosis & Diffusion on Plant Tissue Procedure 3 – Inquiry