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

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

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

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

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

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

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

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

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

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

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

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

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

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 = liter bars/ mole-K T = = 298 K Ψ S = - (1)(.15M)( 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

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)( liters bars/mole-K)(298 K) Ψ S = bars Ψ = 0 + (-7.43 bars) Ψ = bars

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)( liters bars/mole-K)(298 K) = bars (solution) Ψ S = - (2)(0.15mole/liter)( liters bars/mole-K)(298 K) = bars (cell) Water will move from solution to cell.

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 = (-7.43) Ψ = -4.95

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