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

Water potential is a measurement that combines the effects of solute concentration and pressure Ψ = ΨP + ΨS Water potential determines the direction of movement of water Water flows from regions of higher water potential to regions of lower water potential

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

How Solutes and Pressure Affect Water Potential Both pressure and solute concentration affect water potential The solute potential (ΨS) of a solution is proportional to the number of dissolved molecules Solute potential is also called osmotic potential

Pressure potential (ΨP) 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

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

The addition of solutes reduces water potential Fig. 36-8a ψ = −0.23 MPa (a) 0.1 M solution Pure water H2O ψP = 0 ψS = 0 ψP = 0 ψS = −0.23 ψ = 0 MPa The addition of solutes reduces water potential Figure 36.8a Water potential and water movement: an artificial model

Physical pressure increases water potential Fig. 36-8b (b) Positive pressure H2O ψP = 0.23 ψS = −0.23 ψP = 0 ψS = 0 ψ = 0 MPa ψ = 0 MPa Physical pressure increases water potential Figure 36.8b Water potential and water movement: an artificial model

Further Physical pressure increases water potential more Fig. 36-8c ψP =   ψS = −0.23 (c) Increased positive pressure H2O ψ = 0.07 MPa ψP = 0 ψS = 0 ψ = 0 MPa 0.30 Further Physical pressure increases water potential more Figure 36.8c Water potential and water movement: an artificial model

Negative pressure decreases water potential Fig. 36-8d (d) Negative pressure (tension) H2O ψP = −0.30 ψS = ψP = ψS = −0.23 ψ = −0.30 MPa ψ = −0.23 MPa Negative pressure decreases water potential Figure 36.8d Water potential and water movement: an artificial model

Fig. 36-9a If a flaccid cell is placed in an environment with a higher solute concentration, the cell will lose water and undergo plasmolysis (a) Initial conditions: cellular ψ > environmental ψ ψP = 0 ψS = −0.9 ψP = 0 ψS = −0.7 ψ = −0.9 MPa ψ = −0.7 MPa 0.4 M sucrose solution: Plasmolyzed cell Initial flaccid cell: Figure 36.9 Water relations in plant cells

Fig. 36-9b If the same flaccid cell is placed in a solution with a lower solute concentration, the cell will gain water and become turgid ψP = 0 ψS = −0.7 Initial flaccid cell: Pure water: ψP = 0 ψS = 0 ψ = 0 MPa ψ = −0.7 MPa ψP = 0.7 ψS = −0.7 ψ = 0 MPa Turgid cell (b) Initial conditions: cellular ψ < environmental ψ Figure 36.9 Water relations in plant cells

solute potential (ΨS) ΨS = - 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°)

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

ΨS = - (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K) You try it Calculate Ψ of a 0.15M solution of of NaCl at atmospheric pressure (ΨP = 0) at 25 °C. Note: NaCl breaks into 2 pieces so i = 2. ΨS = - (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K) ΨS = - 7.43 bars Ψ = 0 + (-7.43 bars) Ψ = -7.43 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? ΨS = - (2)(0.10 mole/liter)(0.0831 liters bars/mole-K)(298 K) = - 4.95 bars (solution) ΨS = - (2)(0.15mole/liter)(0.0831 liters bars/mole-K)(298 K) = - 7.43 bars (cell) Water will move from solution to cell.

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