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A plant cell has an osmotic (or solute) potential of -2.1 MPa. Neglecting the changes

A plant cell has an osmotic (or solute) potential of -2.1 MPa. Neglecting the changes in volume due to turgor pressure, what will its internal pressure be when allowed to come into equilibrium with each of the following solutions:

jocelyn-cleveland
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A plant cell has an osmotic (or solute) potential of -2.1 MPa. Neglecting the changes

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  1. A plant cell has an osmotic (or solute) potential of -2.1 MPa. Neglecting the changes • in volume due to turgor pressure, what will its internal pressure be when allowed • to come into equilibrium with each of the following solutions: • Water (Answer: cells internal pressure after incubation in water is Ψp = 2.1) • -2.1 MPa mannitol (Answer: Ψp = 0) • -2.5 MPa mannitol (Answer: Ψp = -0.4; plasmolysis will likely occur) • -1.0 MPa triethyl citrate to which the cell is highly permeable (Answer: Ψp = 2.1)

  2. A wind-borne pollen grain with an osmotic (solute) potential of -1.5 MPa has been • drying out while aloft. Thus, it has lost all turgor pressure. It lands on a stigma, • whose cells have an osmotic potential of -2.0 MPa and an internal pressure of • 0.8 MPa. In answering the following questions, assume that the stigma is massive • compared with the pollen grain: • In which direction will water flow? Why? • What will be the internal pressure of the pollen be after it has been on the stigma • for quite some time? • A fresh pollen grain with the same osmotic potential but an internal pressure of • -1.3 MPa is placed on the same stigma. In which direction will water flow? Why? • Answers: • Ψw of stigma = Ψs + Ψp (Ψw = -2.0 MPa + 0.8 MPa; Ψw = -1.2 MPa • Ψw of pollen in air = Ψs + Ψp (Ψw = -1.5 MPa + 0 MPa; Ψw = -1.5 MPa • Water will flow from stigma (Ψw = -1.2 MPa ) to pollen (Ψw = -1.5 MPa) because • the water in the stigma has more energy (more free water in stigma) than in • the pollen. • b. At equilibrium, Ψw of pollen will be equal to that of the stigma (Ψw = -1.2 MPa). • Ψw of pollen on stigma = Ψs + Ψp (-1.2 MPa = -1.5 MPa + ? MPa); Ψp = 0.3 MPa • c. Ψw of fresh pollen = Ψs + Ψp (Ψw = -1.5 MPa + 1.3 MPa; Ψw = -0.2 MPa • Water will flow from pollen (Ψw = -0.2 MPa) to stigma (Ψw = -1.2 MPa ) because • the water in the pollen has more energy (more free water in fresh pollen) than in • the stigma.

  3. A plant 47 meters tall is transpiring at a rate of 14 liters per hour. Suddenly, the rate drops to 7 liters per hour. What factors could have caused the rate to decrease? Give as many factors as you can. Answers could include: cloud cover increased (reduced temperature-lower driving force); wind speed decreased (increased boundary layer and increased resistance); reduced photosynthetic rate in guard cells- to affect solute production and water potential In guard cell, which affects size of stomatal opening and decreases conductivity; Increase in humidity (rain), which reduces driving force

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