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 = 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
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 (ΨS) 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 (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
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 Msolution
Purewater
H2O
ψP = 0
ψS = 0ψP = 0ψS = −0.23
ψ = 0 MPa
The addition of solutes reduces water potential
Fig. 36-8b
(b)Positivepressure
H2O
ψP = 0.23
ψS = −0.23
ψP = 0
ψS = 0ψ = 0 MPa ψ = 0 MPa
Physical pressure increases water potential
Fig. 36-8c
ψP = ψS = −0.23
(c)
Increasedpositivepressure
H2O
ψ = 0.07 MPa
ψP = 0
ψS = 0ψ = 0 MPa
0.30
Further Physical pressure increases water potential more
Fig. 36-8d
(d)
Negativepressure(tension)
H2O
ψP = −0.30ψS =
ψP =ψS = −0.23
ψ = −0.30 MPa ψ = −0.23 MPa
0 0
Negative pressure decreases water potential
Fig. 36-9a
(a) Initial conditions: cellular ψ > environmental ψ
ψP = 0 ψS = −0.9
ψP = 0 ψS = −0.9
ψP = 0ψS = −0.7
ψ = −0.9 MPa
ψ = −0.9 MPa
ψ = −0.7 MPa0.4 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
ψP = 0ψS = −0.7
Initial flaccid cell:
Pure water:ψP = 0ψS = 0ψ = 0 MPa
ψ = −0.7 MPa
ψP = 0.7ψS = −0.7ψ = 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 (ΨS)
ΨS = - iCRTi is the ionization constantC is the molar concentrationR 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 °Ccalculate Ψ1st use ΨS = - iCRT to calculate ΨS
i = 1 (sucrose does not ionize) C = 0.15 mole/literR = 0.0831 liter bars/ mole-KT = 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
You try itCalculate 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 againCalculate 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
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
