Download Water potential 1. If the cell sap of a plant cell has an osmotic

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?
(a)
(b)
Water potential
-2.0 MPa
+0.0 MPa
Pressure potential
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
(beaker)
0.0 MPa
0 MPa
(cell)
0.0 MPa
+1.6 MPa
assuming no significant change in cell volume.
Osmotic potential
0 MPa
-1.6 MPa
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?
(cell)
Water potential
-0.7 MPa
Pressure potential
+0.9 MPa
Osmotic potential
-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
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 Ψp.
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 15-25 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 Ψs of -0.7 MPa. Before
fertilization, without a cell wall, the Ψp of the egg must be close to 0; Ψs of the egg will be
close to -0.7 MPa. Immediately after wall formation, there need be no change in Ψp
(there could be); during the formation of the rhizoid, the Ψp 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).
o
10. A fresh lettuce leaf with a 90% relative water content is placed in a box at 20 C that
contains air of 90% relative humidity. Will the leaf lose water, absorb water, or stay the
o
same? If the box is put in a refrigerator at 4 C, will the leaf lose water, absorb water, or
stay the same? "Fresh" suggests that the lettuce leaf is moist, i.e. Ψw near 0. At 90%
RH, 20oC, water potential is quite negative, -14.1 MPa; at 90% RH, 4oC, water potential
is still negative, -13.3 MPa. At either temperature, the leaf loses water (but less at 4oC).
11. Calculate Ψw, Ψs, and Ψp 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
(Molal) (= Ψw)
0
Weight change (g)
% of cells plasmolyzed
+0.53
0
0.05
+0.50
0
0.1
+0.40
0
0.15
+0.15
0
0.2
-0.05
0
0.25
-0.20
10
0.3
-0.40
90
If the tissue has no weight change, we assume that water potential, Ψw, of the tissue is
equal to the water potential of the solution. Interpolating, this occurs near 0.1875 M
sucrose. Ψw = -2.44*0.1875 = -0.4575 MPa. At incipient plasmolysis, Ψp = 0, and Ψw =
Ψs, which occurs at about 0.24 M sucrose. Ψs = -2.44*0.24 = -0.5856 MPa. Assuming
that Ψs is about the same in the two conditions (at no weight change; at incipient
plasmolysis), Ψs of the original tissue is equal to -0.5856 MPa. Ψp in the original
condition = Ψw -Ψs = -0.4575-(-0.5856) = 0.1281 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, Ψp < 0. When the leaf is cut, Ψp =
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, Ψp, xylem =
0 = Ψp, original xylem + Ψp, pressure bomb. Ψp, original xylem = -Ψp, 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
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
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, Ψs = -1.0 MPa and Ψp = 0.9
MPa. Immediately after the soil is irrigated, the water potential of the root compared to
the soil will increase, as the root Ψw is now more positive relative to the soil’s lower Ψw.
After a new equilibrium, the root Ψs = -1.0MPa and Ψp = 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, 1.687 mol/m3) and c in bulk air, 0.4 x 0.269 = 0.108 at 0°C and
0.4 x 2.201 = 0.880 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 (103 kg/m2) x (9.8 m/s2) 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.
Ψp = -2σ/r = (-2 x 7.3 x 10-8 MPa m) / (50 x 10-6 m)
= -29.2 x 10-4 MPa
So, -29.2 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
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
CO2). 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
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
38
22
9
Permanent wilting percentage
18
11
3
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 csat - cair. Before the lights were turned on, csat cair 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 0.961- 0.5 = 0.461
mol/m3. The difference is 1.28-0.46 = 0.82 mol/m3. At the chamber temperature of
30oC, the leaf temperature was 35oC, and the vapor concentration inside the leaf was
2.201 mol/m3. Outside the leaf, the vapor concentration was 1.687-0.5 = 1.187 mol/m3.
The difference was 2.201-1.187 = 1.014 mol/m3. The effect of the lights was greater in
the warmer chamber.
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 = 2.201 - 1.28 = 0.921 mol/m3; rate = 20000*0.921/150 = 122.8 mg/m2-s.
Shade: Δc = 1.28 - 0.7(1.28) = 0.384 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,
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
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 K40
(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)
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 Ca2+, 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
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.
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.