1. 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. 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

3. 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. 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

5. 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

6. 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. 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

8. Physical pressure increases water potential
Fig. 36-8b
(b)
Positive pressure
H2O
ψP = 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

9. Further Physical pressure increases water potential more
Fig. 36-8c
ψP =   ψS = −0.23
(c)
Increased positive pressure
H2O
ψ = 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

10. 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

11. 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 = 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

12. 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 = ψS = −0.7
ψ = 0 MPa
Turgid cell
(b) Initial conditions: cellular ψ < environmental ψ
Figure 36.9 Water relations in plant cells

13. solute potential (ΨS) ΨS = - 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°)

14. 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 = liter bars/ mole-K T = = 298 K
ΨS = - (1)(.15M)( 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

15. 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)( liters bars/mole-K)(298 K)
ΨS = bars
Ψ = 0 + (-7.43 bars)
Ψ = 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?
ΨS = - (2)(0.10 mole/liter)( liters bars/mole-K)(298 K)
= bars (solution)
ΨS = - (2)(0.15mole/liter)( liters bars/mole-K)(298 K)
= bars (cell)
Water will move from solution to cell.

17. 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 = (-7.43)
Ψ = -4.95