(cell) -0.7 MPa +0.9 MPa -1.6 MPa

Transcription

1 Water potential 1. If the cell sap of a plant cell has an osmotic potential of -2.0 MPa,what will the cell water potential and cell pressure potential be at (a) incipient plasmolysis, and (b) full turgor, assuming no change in the osmotic potential due to dilution? Water potential Pressure potential (a) -2.0 MPa 0 MPa (b) +0.0 MPa +2.0 MPa 2. An algal cell is moved from water to a hypertonic solution. What happens? Water moves out. How does one describe the cell's condition? Plasmolyzed. How do water potential, osmotic potential, and pressure potential change? All three drop, pressure potential to 0; the others become negative 3. A cell with a pressure potential of 0.8 MPa and an osmotic potential of -1.6 MPa is placed in a beaker of pure water. What is the water potential in the beaker and in the cell initially? Beaker, 0; cell, -0.8 MPa. In which direction will water flow? Into the cell. After equilibrium, what will be the component potentials and water potentials in the beaker and cell? What assumptions have you made in your calculations? Water potential Pressure potential Osmotic potential (beaker) 0.0 MPa 0 MPa 0 MPa (cell) 0.0 MPa +1.6 MPa -1.6 MPa assuming no significant change in cell volume. 4. If the cell in the previous question is placed in a beaker that contains a solution with an osmotic potential of -0.7 MPa, which direction will the water flow? Compute your answer both assuming the cell is used before and after equilibrium in water. Before, into the cell; after, out of the cell. What will be the values of water potential, pressure potential, and osmotic potential in the cell at equilibrium? Water potential Pressure potential Osmotic potential (cell) -0.7 MPa +0.9 MPa -1.6 MPa 5. Cell A has an osmotic potential of -2.0 MPa and a pressure potential of +0.6 MPa. Cell A is placed adjacent to Cell B, which has an osmotic potential of -1.6 MPa and a pressure potential of +1.2 MPa. In which direction will water flow? Cell A: water potential = -1.4 MPa; cell B: water potential = -0.4 MPa. Water flows from B to A. What further information would you need to calculate the final equilibrium potentials? How would osmotic and pressure potentials end up in both cells? 6. If cell A has a pressure potential of +0.5 MPa and contains 0.5 M glucose, and cell B has a pressure potential of +0.5 MPa and contains 0.5 M sucrose, in which direction will water flow? What assumptions did you make in answering this question? Assuming that the concentrations of other components are equal and that their environments are the

2 same, there should be no net flow of water from either cell to the other. (There might be flow from the cells to the environment.) 7. The primary walls of leaf mesophyl cells in some plants are much more elastic (that is, stretchy) than in other plants. For example, a willow, Salix lasiandra, which grows near streams, has more elastic cell walls than Ginkgo biloba trees, which grow in drier areas. (a) If I gave you a corn leaf and a sunflower leaf, how would you tell how elastic the cells of each were? Attach a force (weight) to an end of the leaf and measure the degree of length change as a fraction of the original length. (b) What adaptive advantage could greater elasticity (stretchiness) provide? Hint: consider that growth rate is related to Ψ p. Greater elasticity might lead to faster growth. (c) What disadvantage might be associated with greater elasticity? Greater elasticity might result is less support, less surface toward the light. 8. The eggs of the marine alga Fucus are shed directly into sea water without any cell walls. This makes it easier for the antherozoids (sperm) to fertilize them. A cell wall forms 4 to 8 hours after fertilization. About hours after fertilization, the cell starts to enlarge, forming a long, thin rhizoid. Discuss the components of water potential of the Fucus egg before fertilization, immediately after wall formation, and during the formation of the rhizoid. Assume that the sea water has a Ψ s of -0.7 MPa. Before fertilization, without a cell wall, the Ψ p of the egg must be close to 0; Ψ s of the egg will be close to -0.7 MPa. Immediately after wall formation, there need be no change in Ψ p (there could be); during the formation of the rhizoid, the Ψ p of the egg will have become positive, allowing the cell to stretch out the wall to form the rhizoid. 9. Discuss the reason why the addition of 1 molal sucrose to water lowers the water potential by 2.24 MPa. This refers to the end of Monday's lecture; sucrose lowers the activity of water (by reducing its concentration), or sucrose lowers the pressure in the solution (by impinging on the free surface). 10. A fresh lettuce leaf with a 90% relative water content is placed in a box at 20 o C that contains air of 90% relative humidity. Will the leaf lose water, absorb water, or stay the same? If the box is put in a refrigerator at 4 o C, will the leaf lose water, absorb water, or stay the same? "Fresh" suggests that the lettuce leaf is moist, i.e. Ψ w near 0. At 90% RH, 20 o C, water potential is quite negative, MPa; at 90% RH, 4 o C, water potential is still negative, MPa. At either temperature, the leaf loses water (but less at 4 o C). 11. Calculate Ψ w, Ψ s, and Ψ p for the cells in a thin slice of plant tissue, using data that show the gain and loss of weight, and the % plasmolysis, when the tissue is placed in different concentrations of sucrose. What errors are inherent in this method? Sucrose concentration Weight change (g) % of cells plasmolyzed (Molal) (= Ψ w )

3 If the tissue has no weight change, we assume that water potential, Ψ w, of the tissue is equal to the water potential of the solution. Interpolating, this occurs near M sucrose. Ψ w = -2.44* = MPa. At incipient plasmolysis, Ψ p = 0, and Ψ w = Ψ s, which occurs at about 0.24 M sucrose. Ψ s = -2.44*0.24 = MPa. Assuming that Ψ s is about the same in the two conditions (at no weight change; at incipient plasmolysis), Ψ s of the original tissue is equal to MPa. Ψ p in the original condition = Ψ w -Ψ s = ( ) = MPa. 12. When a small cut shoot is enclosed in a pressure bomb with the cut end protruding and sufficient pressure is applied to cause the sap to return to the level of the cut, what does this pressure equal? Explain in terms of water potential what happens when the shoot is cut and what happens in the pressure bomb. Sap in the xylem is generally under negative pressure, Ψ p < 0. When the leaf is cut, Ψ p = 0 at the cut end, higher than that in the xylem, and the sap is pushed/pulled into the xylem. In the pressure bomb, when the sap returns to the cut end of the shoot, Ψ p, xylem = 0 = Ψ p, original xylem + Ψ p, pressure bomb. Ψ p, original xylem = -Ψ p, pressure bomb. Transpiration 13. A small branch is cut from a cottonwood tree on an early spring morning, when the temperature is 10 C and there is a light fog. The branch is placed in a Scholander pressure bomb. It takes 1.0 MPa of pressure to push the xylem sap to the end of the branch so that it is just visible. (a) Is the system in the bomb at equilibrium with regard to water potential? (b) Assuming that Ψs(xylem) = 0, calculate Ψ(xylem) and Ψ(leaf cell) when the branch was on the tree and when the branch is in the bomb. (c) What differences would you expect if the experiment were repeated later in the day, when the temperature was 30 C, the fog had disappeared, and there was a brisk wind? (a) The system inside the bomb is at equilibrium. (b) In the bomb, Ψw(xylem) = 0 (it is at equilibrium with air outside the bomb), and Ψw(leaf cell) = 0; on the tree, Ψw(xylem) = - 1 MPa, and Ψw(leaf cell) = -1 MPa. (c) The plant will be transpiring, and both Ψw(xylem) and Ψw(leaf cell) would be < -1 MPa. On the plant (not in the bomb), Ψw(leaf cell) < Ψw(xylem), the amount of difference depending on the transpiration rate. On the plant, the system is not at equilibrium. 14. Given that the water column in a gymnosperm ruptures (i.e., a bubble forms) within one conducting element of the xylem, and that the smallest water-filled pores within the primary cell walls of the pits are 7 nanometers in diameter, describe the conditions inside

4 the element, in an adjacent element, and at the pits, both prior to rupture and after a new steady state has been re-established. Prior to rupture: Inside the conducting element: Water flow follows the xylem pressure gradient. Negative pressure (tension) is present through the element. Adjacent element: Water flow follows the xylem pressure gradient. Pits: Water flows through the pits of adjacent elements, along the pressure gradient. After new steady state: Inside the conducting element: The element is completely cavitated, and filled with a gas bubble. Some water may evaporate from adjacent elements into the cavitated element. Adjacent element: Water detours around the embolized element through adjacent elements. Pits: An AWI (Air-Water Interface) is formed at each of the pits between the affected element and the adjacent element. 15. Water has certain physiochemical properties; for example, strong cohesive forces exist between individual water molecules, and one result of these forces is that water has a high surface tension. Explain why these two properties are so important in terms of water movement in the plant. Cohesion prevents water columns in xylem from breaking. Surface tension prevents air from entering xylem through pits, keeps cells walls wet, and increases the water capacity of fine-grained soils. 16. What would be the effect of reducing the cohesion of water to 1/10 or 1/100 of its actual value to (a) a redwood seedling, (b) a tall redwood tree? The reduction in cohesion doesn't change the gradient in Ψp needed to raise water, it just limits the gradient in Ψp that can form before a water column breaks. There would be no effect on a seedling. In the 100-m tall trees, there is a tension of at least 1 MPa; some books say 2 MPa to account for overcoming resistance. The textbook suggests a maximum possible tension of -20 MPa before a water column breaks. This would suggest that reducing cohesion to 1/10 would limit trees to 100 m; reducing cohesion to 1/100 would limit them to 10 m. (Of course water columns break at lower tension, because dissolved air comes out of solution.) 17. Critically analyze the following statement: "Upon cutting a stem, the tension in the xylem is relieved and the water potential in the xylem, and therefore the leaf cells, becomes zero." When the stem is cut, a new air water interface (AWI) is formed at the cut site. The tension in the water column pulls this AWI up the cut tracheids and vessels. The AWI moves upward until it enters the primary cell wall in the pit. Once in the primary wall, the forces of adhesion and surface tension lead to a curved AWI. This new curved AWI generates a tensional force acting in the opposite direction of the tensional force produced

5 by the AWI in the mesophyll. The curvature of the new AWI will increase until the force it generates is equal and opposite (in direction) of the force generated by the mesophyll AWI. Therefore, upon the cutting of the stem, the local tension in the xylem is only relieved for an instant, and is re-established once the AWI enters the wall of the pit. Because this occurs rapidly, the tension in the rest of the xylem and in the leaf does not become zero at any point. Because the Ψp at the cut site is equal to the Ψp in the leaf, there is no Ψw gradient to drive water flow. Thus Δ Ψw becomes zero between the leaf and the stem, but the values of Ψw in both the xylem and the leaf remain negative. 18. Assume a situation in which roots are in a soil that has a matric plus osmotic potential of -0.1 MPa. The cortical root cells have an osmotic potential of -1.0 MPa and an initial pressure potential of +0.7 MPa. If the roots come to equilibrium with the soil, what will be the equilibrium potentials of the root? After equilibration, if the soil is irrigated with a salt solution that brings the total water potential of the soil to -0.5 MPa, what will happen to the root potentials immediately and after a new equilibrium is established? When the roots come to equilibrium potential with the soil, Ψ s = -1.0 MPa and Ψ p = 0.9 MPa. Immediately after the soil is irrigated, the water potential of the root compared to the soil will increase, as the root Ψ w is now more positive relative to the soil s lower Ψ w. After a new equilibrium, the root Ψ s = -1.0MPa and Ψ p = 0.5 MPa. 19. Would a tub of water maintained at 30 C lose water by evaporation more rapidly on a cold, clear winter day (temperature 0 C) or on a warm summer day (temperature 35 C)? Assume 40% (0.4) relative humidity at both temperatures. The rate of water loss will depend directly on dc/dx (Δc/Δx, considering a linear gradient across an unstirred boundary layer). Assume Δx is the same for both tubs. Δc will be the difference between c just above the tub (saturated at 30 C, mol/m3) and c in bulk air, 0.4 x = at 0 C and 0.4 x = at 35 C. The difference is greater at 0 C. 20. How high could atmospheric pressure alone be expected to push water up a tree? Assuming a maximum root pressure of 0.3 MPa and that root pressure is the only force involved in water transport, what would be the upper limit for plant growth? The pressure gradient needed to move water up a tree is 0.01 MPa/m.: this is due to gravity is (10 3 kg/m 2 ) x (9.8 m/s 2 ) x tree height. If you account for resistance, the total pressure difference needed is 0.02 MPa/m x tree height Atmospheric pressure is about 0.1 MPa, so alone it could raise water 5 or 10 meters, depending on resistance. Finally, 0.3/0.02 = 15 meters due to root pressure alone. 21. If the mean diameter of a xylem vessel is 100 µm, what would be the maximum height that capillarity could pull water up a tree? From question 20, the pressure difference needed is 0.02 MPa/m x tree height.

6 Ψp = -2σ/r = (-2 x 7.3 x 10-8 MPa m) / (50 x 10-6 m) = x 10-4 MPa So, x 10-4 MPa / (0.01 MPa / m) = 2920 x 10-4 m = 29.2 cm, 22. Provide a detailed description of why plants can use the energy from the sun to drive the flow of water through the plant. Under most circumstances, water is pulled through the plant by the "tug", exerted when water evaporates due to negative water potential of the air. The energy of the sun warms the leaf and the air in the leaf, which increases its saturation vapor concentration, which in turn increases the gradient between the vapor concentration in leaf and the outside air. More indirectly, water can be pushed into the xylem by "root pressure," which comes from accumulation of solutes in the xylem fluid. The energy of the sun is involved through photosynthesis, phloem transport, and respiration to power the accumulation. 23. What causes water uptake by roots: (a) when the plant is transpiring rapidly? (b) when transpiration rate is very low? Does uptake in either (a) or (b) depend on any particular structural feature of the root? a) Under rapid transpiration, water uptake is driven by the gradient in pressure potential between the xylem, the root tissues, and the soil. b) Under low transpiration rates, water uptake is driven by the root uptake of ions and transport into the xylem, which decreases the xylem osmotic potential. This phenomenon is known as root pressure. Uptake in both cases benefits the presence of from root hairs, as well as conducting cells. (b) also depends on ion pumping into the steele; the Casparian strip limits leakage out of the steele. 24. Why should (a) wind increase transpiration rate? (b) leaf hairs decrease leaf heating in sunlight? a) Wind can increase transpiration rate by reducing the leaf boundary layer, which will reduce the air resistance, and can also increase the rate of convective heating. b) Leaf hairs decrease leaf heating by both reflecting light, and shading the leaf surface, to decrease the net incoming radiation (Q). Leaf hairs can also stabilize the boundary layer, slowing convective heating in a warm-air environment. 25. Explain why water can be moved to the top of a tall tree while a mechanical vacuum pump is unable to draw water higher than about 10 meters. The energy expended is the same per unit water lifted. The major point of this question has to do with the pump being a vacuum pump pulling from the top. This means that the distance is limited by difference in air pressure vs vacuum (1 atmosphere, or 0.1 MPa). Energy balance and water stress

7 26. From Science magazine, "Laboratory measurement of foliar uptake of sulfur dioxide and ozone by red kidney beans demonstrated a strong effect of relative humidity on internal pollutant dose. Foliar uptake was enhanced two- to three-fold for sulfur dioxide and three- to four-fold for ozone by an increase in relative humidity from 35 to 75 percent. For the same exposure concentration, vegetation growing in humid areas (such as the eastern United States) may experience a significantly greater internal flux of pollutants than that in more arid regions." Discuss anatomical and physiological reasons why this might be true. Stomata tend to close under water stress, reducing their uptake of pollutants (as well as CO 2 ). In addition, under very dry conditions, the cuticle of a leaf will dry out and be less permeable to gases. 27. An article by Boyer (Science 218:773, 1982) demonstrated how plant breeding could increase crop yield. His point was that new varieties were better at extracting water from the soil and transferring it to the leaves. Discuss the genetic traits that might have led to increased water uptake ability. What other traits (besides those leading to increased water uptake ability) might affect average afternoon water potential of the leaves? Increased water uptake could be due to: more root branching, more root per shoot, more aquaporins in root cell membranes (reducing resistance), larger tracheid and vessel diameters (reducing resistance), more vessels compared to tracheids, more xylem in stems, petioles, and leaves. Other possible traits include: control of stomatal aperture in response to water stress or time-of-day. 28. Discuss the significance of the primary cell wall to water flow through the xylem elements of a plant. Discuss the associated advantages and disadvantages of tracheids and vessels to water flow in plants. Are there environmental conditions under which the plant might be better adapted with one conducting element over the other? The primary cell wall is reasonably permeable to water, so it represents a low resistance to water flow between tracheids (although more than the open connections between vessel elements). However, bubbles cannot pass through a primary cell wall, so bubbles forming in a tracheid are confined to that trached (whereas bubbles forming in a vessel element will expand into adjacent vessel elements. Under condition of high transpiration, where water supply to the leaves is important, vessels are better. Under conditions of very low temperature (particularly freezing), when bubbles may form, tracheids are better. 29. Water in the plant can be described as a system that is constantly experiencing a "tugof-war." Using the special properties of water and the anatomical features of the plant, explain this "tug-of-war" phenomenon. "Tug" at the top (water-air interface in the leaf) is due to the force exerted by surface tension in small capillaries (at the pits of xylem, in the cell walls of parenchymal cells). Another source of "tug"--inside the cell--is provided by the solute effect. Tug at the

8 bottom of the xylem is provided by gravity, by solute effect of stelar and cortical cells, and by matric potential of soil (also surface tension). The balance requires cohesion of water through intermolecular hydrogen bonding. 30. What prevents the columns of water in the xylem from breaking? Hydrogen bonds. Why might they be liable to break in any case? Extreme tension; conditions for bubble formation: high gas concentration and low solubility (i.e., low temperature). 31. Explain how the water potential, osmotic potential, and pressure potential of a leaf cell will vary during 24 hours on a summer day. All three will drop (become more negative or less positive as the sun rises and stomata open. The reverse will occur at sunset. 32. The following water content percentages were obtained for three soils: Clay Silt Sand Field capacity Permanent wilting percentage Similar plants were placed in a pot of each soil and watered until water drained from the bottom of the pot. The plants were then placed in the open air during a dry period and not watered further. Which plant would probably wilt first? Why? The plant in sand has less total water available and will probably wilt first. Explain why the permanent wilting percentage is relatively independent of the kind of plant used in its determination. The structures of different plants (cell walls, concentration of solutes in cytoplasm and vacuoles) are more similar than are the structures of different soils. Why do different soils have different permanent wilting percentages? The amount of water in a soil (fraction of the soil that is water) depends on the size of soil particles: smaller particles have more surface area that binds layers of water. 33. Two similar tobacco plants are growing in glass chambers, one at 20oC, the other at 30oC. The atmospheric moisture in each chamber is adjusted to give a water vapor concentration deficit (difference from saturation) of 0.5 mol/m3. When sunlight strikes the chambers, the leaf temperature rises to 5oC above air temperature in each chamber. Will the rates of transpiration also increase by the same amount in each of the two chambers? Assume that the controlling factor is c sat - c air. Before the lights were turned on, c sat - c air was 0.5 mol/m3. Then the lights were turned on. At the chamber temperature of 20oC, the leaf temperature was 25oC, and the vapor concentration inside the leaf was 1.28 mol/m3. Outside the leaf, at 20oC, the vapor concentration was = mol/m3. The difference is = 0.82 mol/m3. At the chamber temperature of 30oC, the leaf temperature was 35oC, and the vapor concentration inside the leaf was mol/m3. Outside the leaf, the vapor concentration was = mol/m3. The difference was = mol/m3. The effect of the lights was greater in the warmer chamber.

9 34. A small bean plant is growing under a bell jar in a saturated atmosphere at 25oC. As the sunlight strikes the leaf, its temperature rises 10oC above the air temperature. A similar bean plant is growing under a cloth shade nearby. Air temperature is 25oC, relative humidity is 70%, and leaf temperature is 25oC. Which plant can be expected to have the more rapid transpiration rate? Compute the rates, assuming Rs = 100 s m-1 and Ra = 50 s m-1. The plant in the bell jar transpires more rapidly. Bell jar: Δc = = mol/m3; rate = 20000*0.921/150 = mg/m2-s. Shade: Δc = (1.28) = mol/m3; rate = 20000*0.384/150 = 51.2 mg/m2-s. 35. What would be the influence on transpiration of raising the air temperature without increasing the temperature of the leaf? Why does an increase in the temperature of the air usually increase the rate of water loss from a leaf? Raising only the outside air temperature will raise the water vapor concentration. If water vapor concentration inside the leaf does not change, the difference will be smaller and the rate of transpiration should decrease. Generally, however, raising the air temperature will also raise the leaf temperature (and the vapor saturation pressure inside the leaf). 36. Some crop plants tend to grow mostly at night, while others grow more or less uniformly day and night. Discuss differences in the plants that could account for the differences in growth patterns, considering especially net water potential and stomatal control. Growth depends on Ψp and wall structure (yield threshold and response to pressure above the yield point). Plants that grow mostly at night are responding to higher Ψw and Ψp. Plants that grow uniformly must (a) be well watered (uniform Ψw), (b) close stomata to maintain a high Ψw, (c) adjust Ψs to keep Ψp constant, and/or (d) adjust their wall structure to vary the yield threshold or response. 37. A small herbaceous plant growing on a sparsely covered sand dune has narrow, leathery leaves, which are green on top but almost pure white underneath. Closer inspection shows a thick mat of white hairs on the bottom surface. What effect will these hairs have on the plant s mid-day energy balance, if (a) all the stomata are on the top surface; or if (b) all the stomata are on the bottom surface? Leaf hairs have two effects: reflecting light/ir radiation and stabilizing the boundary layer. In (a), they do not affect the boundary layer and transpiration, so their effect is to reduce +Q, with a generally cooling effect. In (b), they still reduce +Q, but they also may raise (make less negative) -V. The net effect depends on the relative values of +Q (from below) and -V. 38. The average bare-footed visitor to the Rec Pool in the summer knows that it is easier to walk on the lawn than the dry concrete apron around the pool. Explain why this is true,

10 using the concepts of radiation, transpiration, and convection. Assume that the heat flows of the lawn and the apron reach equilibrium, and their temperatures are constant. Both the grass and the concrete receive radiation at the same rate, +Qs are equal. The big difference is -V, transpiration cooling of the grass. Because the grass is cooler, -Q (radiative cooling) and -C (convective cooling) will be less for the lawn than for the concrete, but transpiration is the controlling factor. 39. Many desert plants--like cacti--are "CAM" plants, which close their stomates during the day and open them at night. Explain how this affects their heat balance, compared to desert plants--like mesquite--that open their stomates during daylight hours. CAM plants do not have the advantage of transpirational cooling during the day. Their temperatures will generally be higher than those of transpiring plants. What adaptations help them avoid overheating? Large heat sinks (as in barrel cacti); reflective hairs or cuticles (large -Q); living in cool regions. 40. Infrared cameras mounted on aircraft are now used to survey forests and croplands to find areas where the plants are undergoing water stress. How does this work? Infrared cameras measure outgoing radiation, essentially -Q. Under water stress, stomata close and -V (transpirational cooling) is reduced. Temperature goes up, and -Q rises to bring the system to equilibrium. 41. Oranges developing in orchards in winter are very sensitive to freezing. The problem can be countered when it occurs on clear, calm nights. There are several techniques. Smudgepots (burning oil, which produces thick smoke) were used in the past, but no longer because of the pollution they cause. Sprinklers are popular, as are large fans. How does each help the problem? The problem occurs on clear nights, because there is very little incoming Q from space (or air). The trees and ground radiate Q outward, and their temperature, and the temperature of the surface air, drops below freezing. The cold air, being denser, stays at the surface. Smudge pots produce smoke, which radiates Q to the ground (and absorbs and re-irradiates the Q from the trees and ground). Sprinklers, which increase the amount of water vapor in the air, do the same thing. In addition, water that freezes releases heat. Large fans remove the cold layer of air and warm the trees by convection with the warmer air from above. 42. Tobacco plants, like most plants, open their stomata during the day and close them at night. Kalenchoe, a so-called "CAM plant," opens stomata during the night and closes them during the day. What differences would you expect between the guard cells of the two different types of plants? In CAM plants, Ψp becomes more positive, because Ψs becomes more negative, at night, rather than in the day. Either the blue-light photoreceptor or circadian control, or both, at night stimulate the conversion of starch to malate, which serves as counter-ion for

11 accumulated K+. The control is, at least in part, manifested in the phosphorylation and activation of PEP carboxylase. In C3 plants, this activation occurs in the light. (In CAM plants, malate is also formed as a CO2 storage form in mesophyll cells.) 43. Discuss the main resistances or control points to water flow in the soil-plant-air continuum. Under what conditions is each likely to be the major controlling factor? Soil (water layer around soil particles)--major when soil is dry Xylem--major in tall trees, especially gymnosperms (all tracheids, no vessels) Leaves--generally major, movement requires evaporation Stomata--major when stomata are closed Boundary layer--major when thick, e.g. stomatal crypts, low wind speed, low radiative heating 44. It is thought the opening of stomata is due to (a) an uptake of potassium, (b) which develops sufficient osmotic force (c) to create a pressure inside the guard cells to open these cells. Describe some experimental methods that could test these three statements. (a) stain for potassium, or use K 40 (b) counter by adding impermeable osmotic agent (c) pressure probe 45. Explain the following observation: A row of beans and a row of pumpkin plants are adjacent in a Davis garden. During the course of each day, the pumpkin leaves wilt, but the beans leaves do not. We can assume that this is temporary wilt and thus that Ψw of soil is less negative than Ψs of leaf cells (Ψp of leaf cells is positive at night). The wilt indicates that the Ψw of the leaf apoplast falls below Ψs of the leaf cells during the day. This must occur in pumpkin but not bean, either because Ψs of pumpkin leaf cells is less negative than that of bean, or because of resistance to water flow up the pumpkin xylem or into the root stele is greater than in bean. If the Ψs of the bean cells is more negative than that of pumpkin cells, it may be so all the time, or it may adjust daily as Ψw of the soil falls. 46. List several ways in which water deficit can affect shoot growth. Which is permanent in effect even when sufficient water subsequently becomes available. Reduce Ψp of cells below yield threshold; change wall synthesis (increasing yield threshold--permanent); inhibit metabolism (protein synthesis, cell division--permanent); induce ABA accumulation; close stomata (inhibiting photosynthesis). 47. How does the green plant balance its need to conserve water against its other needs? Other needs: CO2 uptake: open stomata only when photosynthesis reduces [CO2] inside the leaf. Mineral uptake: grow roots when stomata are closed (Ψ and Ψp are higher)

12 48. Drought and soil salinity have somewhat similar effects on water uptake by plants. Explain. In both cases Ψ of soil is more negative. 49. A plant with roots in pure water may wilt temporarily when salts are added to the water, but will probably regain turgidity after a few hours. Explain. The addition of salt reduces Ψ (Ψs becomes negative) and may pull water from root cells if they have a less negative Ψ. If the root cells can take up the salts and transport them to the xylem, a gradient of Ψ can be reestablished. 50. Certain bacteria cause wilting of infected plants under conditions in which normal plants remain turgid. Suggest ways in which such wilting could be brought about. Plugging up xylem. Breaching the epidermis of leaves to allow more water loss. 51. The transpiration of a desert succulent plant is minimal at midday even though that is when evaporation from open water in the same area is maximal. Explain how and why this can be true. CAM plants close their stomata during the day (the how) in order to reduce water loss during the period when high temperature would maximize the water vapor concentration difference between leaf and air (the why). 52. Using the Zea mays stomatal guard cell system as your model, explain the events that may give rise to stomatal closure during a period of water stress. Start with Ψw of soil drops. Ψw of soil drops --> Ψw and Ψp of root cells drop --> ABA synthesis --> ABA exported to xylem sap and moves to leaf in transpiration stream [and] xylem sap ph rises from 6.3 to 7.2 (why? how?) --> ABA moves to guard cells (receptors in plasma membrane or in cytoplasm--see p. 549 ff) --> increased cytoplasmic Ca 2+, inhibition of H + efflux, opening of anion channels --> efflux of Cl - and malate - --> depolarization of membrane potential --> K + efflux --> less negative Ψs --> less positive Ψp --> elastic contraction of cell walls --> stoma closes Mineral Nutrition 53. List the major elements required by a living plant cell. For each element, give the ionic or molecular form taken up into the cell and used by the cell; state whether uptake is active or passive; give a biochemical reason for the requirement for each element (name a molecule that incorporates the element). See slide 2 of the Mineral Nutrition lecture. 54. Use the molecular definition of an essential element to discuss the roles played by Mg2+ in the physiological process of a generic plant. Consider how the plant might respond to changes in Mg2+ levels in the soil in terms of supplying Mg2+ to the body of the plant. Molecular definition: the element is required for the structure or biochemical activity of

13 an component that is essential for life. Mg2+ is essential for many enzymes involving DNA or RNA synthesis, ATP use, phosphate transfer; constituent of chlorophyll. (In contrast, the physiological definition is an element required for optimum growth, development, and reproduction.) In a situation where Mg2+ concentration is low, a plant might (a) induce carriers with greater affinity (if genes for such carriers are available); (b) mobilize Mg2+ from the lower, mature leaves to support further growth of roots and flowers. 55. Describe and explain the differences between the nutritional requirements of a plant and an animal. Plant requirements are all inorganic, including some oxidized ions that animals cannot reduce, like NO3- and Fe3+. Energy is obtained from light (although in culture, plants can obtain energy by respiring supplied carbohydrates, e.g. sucrose). Animals have some essential organic nutrients (humans: amino acids, some fatty acids, vitamins). 56. Provide a reason for the inability of plant physiologists to identify micronutrients in early nutritional studies. Boron has been recognized as an essential element only recently. Explain why it was so difficult to make this determination. Some elements are needed only in very low concentrations; some, like B, were contaminants in "purified" sources of other essential elements. 57. Why is it difficult to supply iron in nutrient solutions? What is the best way to overcome this difficulty? What characteristic of soil would make iron less available? What method(s) might a plant use to extract iron from soils. Iron precipitates as oxides and phosphates (text p. 128). In nutrient solutions, it is added with a chelator (e.g., EDTA). It is more soluble in acid soils than basic soils. Plants may acidify soils, possibly secreting chelators (e.g., citric acid). 58. Discuss the principle of an essential mineral nutrient with respect to the cations Ca2+ and Fe3+, and the anion NO3-. Ca2+ is essential for cell wall and membrane structures and for the activity of some enzymes, and it serves as a signal compound; a lack results in necrosis of growing regions. Fe3+ must be reduced to Fe2+, which is essential for chlorophyll synthesis and for some cofactors (cytochromes) in mitochondrial respiratory metabolism; a lack results in chlorotic young leaves. NO3- is a source of N, essential for the synthesis of any amino acid and nucleotide, but it must be reduced to NH4+ before it can be incorporated into glutamine and from there to other compounds; a lack results in chlorosis of older leaves, stunting, and brittleness from excess carbohydrates. 59. Boron has been recognized as an essential element only recently. Explain why it was so difficult to make this determination.

14 The concentration that is toxic is only 2-3-fold higher than the concentration below which one gets deficiency symptoms. This means that it is difficult to find an optimum level. That observation, coupled with the fact that B is a contaminant in some chemicals that are needed to make a nutrient solution for hydroponic plant growth, made it difficult to show that adding B had a promotive effect. 60. The fungus Helminthosporium maydis, cause of the corn blight, which destroyed most of the corn crop in the United States in 1970, starved" the host plants by severely restricting their rate of photosynthesis; this restriction was caused partly by inhibition of potassium uptake. Explain the relationship between potassium uptake and photosynthesis. K+ activates a number of enzymes of photosynthesis.