Diffusion of Water (Osmosis)

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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 Presentation

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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

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