PLB 111 Fall, 2012 FIRST MIDTERM EXAM

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Aubrey Carpenter
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1 FIRST MIDTERM EXAM 1. (20) Complete the following table, giving reasonable estimates for the components of water potential (in MPa) at the points of a soil-tree-air-system that I have listed. Assume that the measurements of Ψ w given for the air and soil are correct and that the tree is 10 meters tall. Place Ψ w Ψ s Ψ p bulk air -20 air inside leaf leaf palisade cell leaf guard cell leaf xylem root xylem cortical root cell soil near root -0.2 (5) Does Ψ w = Ψ s + Ψ p for each entry? (5) Are the signs correct? Ψ w negative; Ψ s negative; Ψ p negative in xylem, positive in cells. (5) Are the values of the numbers reasonable? The values of Ψ w should be between -0.2 and -20. The values of Ψ s should be close to 0 in xylem, more negative in cells, more negative in guard cell than (leaf) palisade cell. The values of Ψ p should be more negative in leaf xylem than root xylem; +0.5 to +1.5 MPa in cells. (5) Is the gradient for Ψ w correct? The values for leaf regions should be at least 1 MPa more negative than values for root regions.

2 2. (25) The tallest redwood tree is 113 meters tall. Using the concepts of water potential, Ψ w, and its components,ψ s, Ψ p, and Ψ g, explain why trees are limited to this height. (5) For water to move to the top of the tree, there must be a gradient of Ψ w, and this means that Ψ w at the top of the tree must be less (more negative) than Ψ w at the bottom. (5) In order for cells at the top of the tree to be turgid, important for rigidity and, in the meristems to expand the cells, the cells must have sufficiently positive Ψ p (say 0.25 MPa). (5) At the top of the tree Ψ g is more positive (2), approximately 1.1 MPa (3), than at the ground level (and even more relative to subterranean roots). (5) But this means that Ψ s must be more negative than Ψ p + Ψ g. It must be even more negative (perhaps 2x) in order to move water to the top of the tree, overcoming the resistance of pits between xylem cells. (5) Ψ s = 2.44 Σc i. Thus the concentration of solutes in the cells must be greater than 1.1/2.44 = 0.45 M, probably much greater. It is not unreasonable that at this concentration, interactions between cell components (i.e. metabolism) will be inhibited.

3 3. (15) Name five mineral elements required by plants (not C, H, or O) and briefly state the role each one plays in plant cell physiology. (1.5 points each correct element; 1.5 for reasonable role) N K S P Ca Fe Mg Mn Cl B Zn Cu Co Mo Na Ni B Si Proteins, nucleic acids Activating ribosomes, 40 other enzymes; osmoticum, charge balancer Structure of proteins (cysteine), coenzymes: thiamin, biotin, CoA, lipoic acid Nucleic acids, NTPs, P-lipids Cell walls; signalling re gene expression/activation Heme, ferridoxin Chlorophyll, ATP complex; activating ATPases, kinases, Rubisco, ribosomes Kreb cycle, PS oxygen evolution PS oxygen evolution Carbohydrate translocation Alcohol, glutamate dehydrogenases, carbonic anhydrase, etc. Phenoloxidases, ascorbate oxidase, plastocyanin, etc. N fixation N fixation, N reduction, xanthine dehydrogenase C4, CAM photosynthesis Urease Cell walls Some cell walls

4 4. (19) A plant physiologist placed samples of green algal cells collected from pond water in chambers containing solutions with different concentrations of sucrose. After a sufficient time for equilibration, he measured the gain in weight of each sample and counted the proportion of cells that were plasmolyzed. The graph below shows his data. What were the original solute potential and the pressure potential of the cells when they were in pond water? Explain your logic. (5) Ψ s = MPa (5) Ψ p = MPa Explanation: (2) The abscissa point where weight change is 0 is the original Ψ cell of the tissue. (2) The point of incipient plasmolysis (estimated point at which no cells are plasmolyzed) is where Ψ p = 0 and the abscissa indicates Ψ s. (2) Assuming that Ψ s is about the same under the original conditions, you can calculate the original Ψ p as Ψ cell - Ψ s. (3) Ψ cell = -2.44(0.4) = MPa; Ψ s = -2.44(0.6) = MPa, Ψ p = MPa. The calculations assume that the tissue is homogeneous, that the tissue is at equilibrium when the observations are made, and that the tissue has not changed during the assay.

5 5. (21) In PLB 105, a potted plant was placed on a sensitive balance and under a flood light; changes in the weight of the plant over time were used to measure the rate of transpiration. The class first measured the rate without wind; then a fan was directed at the plant to simulate wind, and the rate was measured again. (a) Why was the flood light thought to be necessary to observe transpiration? (5) Blue light is a signal for opening stomata; blue and red light stimulate photosynthesis, which reduces [CO 2 ], another control factor. If stomata are closed, the rate of transpiration will be very low. (4) The light increases net Q, thus T leaf, and thus C wv,leaf. (b) The TA told the class that the fan would increase the rate of transpiration. What was his explanation for why this would occur? (6) the boundary layer of unstirred air represents a significant resistance to transpiration. The fan would reduce the unstirred layer and resistance and thus increase the transpiration rate. (c) The class complained that the experiment did not work: instead, the fan decreased the rate of transpiration. Why did this occur? (6) The light had heated the leaf and increased the saturation vapor pressure inside, thus increasing the driving force for transpiration. But the fan increased convective cooling, thus lowering the driving force, C wv,leaf - C wv,air, apparently more than it lowered the resistance. (2-3) Alternative: light heating closed the stomata

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