1. Plants, Isomolar Point, and Water Potential Chapters: 36

2. 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.

3. 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

4. Finding the Isomolar Point of Tissues Incubate tissues in liquids with increasing molarity (salts, sugars, etc.) Mass before and after

5. 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

6. 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…

7. Water Potential Ψ Water concentration; direction of water flow Problem: a 0.5 M sucrose solution at 20°C under normal pressure (open system).

8. Water Potential Ψ Ψ = Ψ pressure + Ψ solute Ψ solute = -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 Ψ solute (s) = -(1)(.5)(.0831)(273+20) Ψ s = -12.17 bars

9. Water Potential Ψ Open system = no outside pressure or vacuum applied  Ψ pressure (p) = 0 Ψ = -12.17 bars + 0 = -12.17 bars Water potential = -12 bars