diffusion of water (osmosis)
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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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
![Page 2: Diffusion of Water (Osmosis)](https://reader033.vdocuments.us/reader033/viewer/2022061518/5681454f550346895db22135/html5/thumbnails/2.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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°)
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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
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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
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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.
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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
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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