1. Diffusion:  C s  X - D s J s = difference in concentration distance diffusion coefficient flux of a solute in solution = (mass/surface area/time)

2. 1. Diffusion Fick (1855) discovered that the rate of solute transport is directly proportional to the concentration gradient and inversely proportional to distance traveled. Fick’s Law describes passive movement of molecules down a concentration gradient. Substances move from high [ ] to low [ ].

3. Diffusion:  C s  X - D s J s = difference in concentration distance diffusion coefficient flux of a solute in solution = (mass/surface area/time)

4. Diffusion: From Fick’s Law, can predict the time it takes for a substance to diffuse a particular distance  X ) 2 D t ½ =

5. diffusion in a cell: small molecule..............  X = 50  m  X ) 2 D t ½ = (0.00005 m) 2 10 -9 m 2 s -1 = = 2.5 sec Conclude: diffusion is rapid across small distances (i.e., within a cell).

6. Diffusion:  X ) 2 D t ½ = But … replace 50  m with 1 m and t ½ becomes 24 years! So … diffusion important only over very short distances. It can not possibly explain long distance movement within the plant.

7. Diffusion: describes water evaporation from stomata. C wv high C wv low  C wv  X - D wv J wv = the driving force the resistance

8. 2. Bulk flow (mass flow):  movement along pressure gradients (e.g., flow in a garden hose, flow in xylem or phloem)  long distance transport in plants volume rate of flow driving force ~ (cm 3 cm -2 s -1 )

9. Bulk flow (mass flow):  relationship determined experimentally by Jean-Louis Poiseuille in 1840  used glass capillary tubes (d = 0.01 – 0.3 mm)

10. Bulk flow (mass flow): Poiseuille Equation PP XX r2r2 88. J v = JvJv volume rate of flow per unit area rradius  viscosity PP change in hydrostatic pressure XX path length

11. Bulk flow (mass flow): Poiseuille Equation PP XX r2r2 88 J v = What is the pressure gradient necessary to cause flow in xylem vessels? rearranging:

12. Bulk flow (mass flow): Poiseuille Equation PP XX r2r2 88 J v = picking some “reasonable” values: r = 20  m  = 0.01 g cm -1 s -1 J v = ?

13. Bulk flow (mass flow): estimating sap velocity thermocouple voltage applied hardwoods (large diameter vessels) like oak, ash 20-25 conifers (narrow tracheids)2-4 m/hr heat pulse use 3.6 m/hr = 0.1cm/sec

14. Bulk flow (mass flow): Poiseuille Equation PP XX r2r2 88 J v = picking some “reasonable” values: r = 20  m  = 0.01 g cm -1 s -1 J v = 0.1 cm s -1

15. Bulk flow (mass flow): Poiseuille Equation PP XX r2r2 88 J v = What is the pressure gradient necessary to cause flow in xylem vessels? (0.1 cm s -1 )(0.08 g cm -1 s -1 ) (2 x 10 -5 ) 2 = = 0.2 bar m -1

16. This is the hydrostatic pressure gradient necessary to obtain flow in a horizontal tube of 20  m radius. P = 1 bar P = 3 bar 10 m P = -2 barP = -4 bar 0.2 bar m -1 overcomes the frictional resistances in the tube.

17. Water transport in plants: 1.diffusion: within a cell or tightly localized 2.bulk flow (mass flow): long distance; no membranes crossed 3.osmosis: cell to cell, crossing membranes

18. 0.2 bar m -1 This is the hydrostatic pressure gradient necessary to obtain flow in a horizontal tube of 20  m radius. P = 1 bar P = 3 bar 10 m P = -2 barP = -4 bar 0.2 bar m -1 overcomes the frictional resistances in the tube.

19. 10 m Gravity effect - 5 bars - 8 bars H 2 O at 10 m will move downward unless the force of gravity is opposed. So … not only need0.2 bar m -1 but also need:0.1 bar m -1 0.3 bar m -1 to move H 2 O against the force of gravity and through the frictional resistance of the xylem

21. 100 m !! in giant Sequoia: 30 bars more negative up here than here (in order to move water at rates similar to that observed in transpiring plants) i.e., 3 bars for every 10 m

22. Partial summary: 1. Diffusion H 2 O (or any substance) flows along concentration gradients from high [ ]  low [ ] Relatively rapid across short distances but can’t explain long-distance transport

23. Partial summary: 2. Mass flow (bulk flow) long-distance transport in plants flow is along pressure gradients (Poiseuille) supports rapid movement e.g., a transpiring sunflower leaf loses the equivalent of its entire leaf H 2 O content every 20 min

24. Water transport in plants: 3. Osmosis (van’t Hoff, 1887) movement of a solvent (e.g., H 2 O) across a semi-permeable membrane H 2 O will flow across membrane into solution where the chemical potential (free energy) of the H 2 O is lower

25. Water transport in plants: 3. Osmosis flow is spontaneous in response to a driving force high  w low  w

26. AB solute less concentrated solute more concentrated an osmotic pressure will develop in “B” =  = RTC s water van’t Hoff 1887

27. osmotic pressure =  = RTC s R = universal gas constant T = ° K C s = osmolality = moles of solute kg -1 H 2 O plant cell sap contains ~ 0.5  2.5 mol kg -1  = RTC s = R(293)(0.75 mol kg -1 ) = 18.3 bars !! (this is ~ 260 psi) (sea water is ~ 28 bars)

28. VAC CYT i.e, the pressure on the internal wall can easily be 18 to 20 bars This illustrates that plant cells have very substantial capacity to draw in and retain water.

29. Osmosis is critical in cell enlargement - expansion  role of elastic cell walls  they have sufficient structural rigidity to allow P to build up

30. Summary: H 2 O movement can be affected by: pressure concentration gravity

31. f (concentration) pure water Does this raise or lower the free energy content of the H 2 O molecules? add solute (e.g., sucrose)

32. .............. Solutes decrease the free energy of the H 2 O molecules. Therefore vapor pressure is decreased. The contribution of solutes to  w is always negative (always lowers  w ).

33. f (pressure) VAC CYT + P(i.e., turgor) on inner walls of living cells  pp  w = 0 then:  w =   +  p  w = - 12 + 4 = - 8 e.g.,    = -12  p = +4

34. f (pressure) -P(i.e., tension or negative hydrostatic pressure) in xylem cavity and cell walls  pp  w = 0 xylem

35. small radii surface tension of H 2 O adhesion and cohesion give rise to negative pressures and because the column of H 2 O is continuous, the -P is transmitted all the way to the roots

36. 95% RH (-69 bars) 60% RH (-700 bars) 50% RH (-950 bars) soil surface.... water-filled pore in the soil. A - 2 bars - 4 bars - 6 bars - 8 bars - 12 bars - 18 bars B. H 2 O is continuous from A  B

37. The lowered  in the transpiration pathway provides the driving force for the movement of H 2 O out of adjacent tissues such as: leaf mesophyll phloem root cortical cells The consequent loss of H 2 O from these tissues is what constitutes a water deficit. Key concept: