Transcript Plants, Isomolar Point, and Water Potential

Plants, Isomolar Point, and
Water Potential
Chapters: 36
What you need to know!
• The role of diffusion (osmosis),
active transport, and bulk flow in the
movement of water and nutrients in
plants.
• How water potential explains
transpiration.
Plant cells have three stages
(conditions)
1. Turgid: vacuole filled to max, cell is
very firm, high pressure, plant appears
upright
2. Flaccid: vacuole is not filled to max,
cell gets dehydrated, plant starts wilting
3. Plasmolysis: extreme loss of water,
vacuole is very small, plasma
membrane detaches from cell wall,
plant is wilted, dries up
Turgor Pressure: Water pressure in plant
cells, regulated through opening/closing
of stomata
Finding the Isomolar Point of
Tissues
• Incubate tissues in liquids with
increasing molarity (salts, sugars,
etc.)
• Mass before and after
Finding the Isomolar Point of
Tissues
• If tissue gains mass: outside was
hypotonic, inside was hypertonic so
water flowed into tissue
• If tissue mass stayed the same:
outside and inside are isotonic; the
molarity of the solution = the
molarity of tissue (Isomolar Point)
• If tissue lost mass: outside was
hypertonic, inside was hypotonic so
water flowed out of the tissue
Finding the Isomolar Point of
Tissues
Important:
Isomolar point has to be identified
using a graph. Use the point where
your line intersects with the zero
change mass line.
Example on board…
Water Potential Ψ
• Water concentration; direction of
water flow
• Problem: a 0.5 M sucrose solution
at 20°C under normal pressure
(open system).
Water Potential Ψ
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Ψ = Ψpressure + Ψsolute
Ψsolute = -i x C x R x T
i = ionization constant (sucrose = 1)
C = molar concentration of solute
(.5)
R = Pressure constant (0.0831)
T = temperature in Kelvin = 273 +
°C
Ψsolute (s) = -(1)(.5)(.0831)(273+20)
Ψs = -12.17 bars
Water Potential Ψ
• Open system = no outside pressure
or vacuum applied
Ψpressure (p) = 0
• Ψ = -12.17 bars + 0 = -12.17 bars
• Water potential = -12 bars