Bio 11: Second Exam Handout

1. The pod trichomes of most soybean varieties are white, but some varieties have tan-colored trichomes. Suppose that the effects of trichome density on beetle feeding were observed only in tan-haired trichomes. What might this finding suggest about how these trichomes deter beetles?

- The finding might suggest that the tan-colored trichomes deter the beetles by some means other than physically obstructing the beetles. Perhaps they contain a chemical that is harmful or distasteful to the beetles.

2. How does the vascular tissue system enable leaves and roots to function together in supporting growth and development of the whole plant?

- The vascular tissue system connects leaves and roots, allowing sugars to move from leaves to roots in the phloem and allowing water and minerals to move to the leaves in the xylem.

3. What plant structure is each of the following? (a) brussel sprouts; (b) celery; (c) onions; (d) carrots?- (a) large axillary buds; (b) petioles; (c) a bulb, an underground shoot with a small stem and large storage

leaves; (d) storage roots4. If humans were photoautotrophs, making food by capturing light energy for photosynthesis, how might

our anatomy be different?- To get sufficient energy from photosynthesis, we would need lots of surface area exposed to the sun. This

large surface-to-volume ratio, however, would create a new problem— evaporative water loss. We would have to be permanently connected to a water source—the soil, also our source of minerals. In short, we would probably look and behave very much like plants.

5. Explain how central vacuole and cellulose cell walls contribute to plant growth?- As plant cells enlarge, they typically form a huge central vacuole that contains a dilute watery sap. Central

vacuoles enable plant cells to become large with only a minimal investment of new cytoplasm. The orientation of the cellulose microfibrils in plant cell walls affects the growth pattern of cells.

6. Distinguish between primary and secondary growth.- Primary growth arises from apical meristems and involves production and elongation of organs.

Secondary growth arises from lateral meristems and adds to the thickness of roots and stems.7. Cells in lower layers of your skin divide and replace dead cells sloughed from the surface. Are such regions

of cell division comparable to a plant meristem?- Your dividing cells are normally limited in the types of cells they can form. In contrast, the products of cell

division in a plant meristem can differentiate into all the types of plant cells.8. Roots and stems grow indeterminately, but leaves do not. How might this benefit the plant?- The largest, oldest leaves would be lowest on the shoot. Since they would probably be heavily shaded,

they would not photosynthesize much regardless of their size.9. Suppose a gardener uproots some carrots after one season and sees they are too small. Carrots are

biennials, and so the gardener leaves the remaining plants in the ground, thinking their roots will grow larger during their second year. Is this a good idea?

- No, the carrot roots will probably be smaller at the end of the second year because the food stored in the root will be used to produce flowers, fruits, and seeds.

10. Why are not the terms pith and cortex used to describe the ground tissue of monocot stem?- Pith and cortex are defined, respectively, as ground tissue that is internal and ground tissue that is

external to vascular tissue. Since vascular bundles of monocot stems are scattered throughout the ground tissue, there is no clear distinction between internal and external relative to the vascular tissue.

11. Contrast primary growth in roots and shoots.

- In roots, primary growth occurs in three successive stages, moving away from the tip of the root: the zones of cell division, elongation, and differentiation. In shoots, it occurs at the tip of apical buds, with leaf primordia arising along the sides of an apical meristem. Most growth in length occurs in older internodes below the shoot tip.

12. If a plant species has vertically oriented leaves, would you expect its mesophyll to be divided into spongy and palisade layers?

- No. Because vertically oriented leaves, such as maize, can capture light equally on both sides of the leaf, you would expect them to have mesophyll cells that are not differentiated into palisade and spongy layers. This is typically the case. Also, vertical leaves usually have stomata on both leaf surfaces.

13. How are root hairs and microvilli analogous structures?- Root hairs are cellular extensions that increase the surface area of the root epidermis, thereby enhancing

the absorption of minerals and water. Microvilli are extensions that increase the absorption of nutrients by increasing the surface area of the gut.

14. How does the vascular cambium cause some tissues to rupture?- The vascular cambium produces growth that increases the diameter of a stem or root. The tissues that are

exterior to the vascular cambium cannot keep pace with the growth because their cells no longer divide. As a result, these tissues rupture.

15. A sign is hammered into a tree 2 m from the tree’s base. If the tree is 10 m tall and elongates 1 m each year, how high will the sign be after 10 years?

- The sign will still be 2 m above the ground because this part of the tree is no longer growing in length (primary growth); it is now growing only in thickness (secondary growth).

16. Stomata and lenticels are both involved in exchange of CO2and O2. Why do stomata need to be able to close, but lenticels do not?

- Stomata must be able to close because evaporation is much more intensive from leaves than from the trunks of woody trees as a result of the higher surface-to-volume ratio in leaves.

17. Would you expect a tropical tree to have distinct growth rings? Why or why not? - Since there is little temperature variation in the tropics, the growth rings of a tree from the tropics would

be difficult to discern unless the tree came from an area that had pronounced wet and dry seasons.18. If a complete ring of bark is removed around a tree trunk (a process called girdling), the tree usually dies.

Explain why.- Girdling removes an entire ring of secondary phloem (part of the bark), completely preventing transport

of sugars and starches from the shoots to the roots.19. Most of the growth of a plant body is the result of

a.cell differentiation b.morphogenesis. c.cell division d.cell elongation e.reproduction20. The innermost layer of the root cortex is the

a.core b.pericycle. c.endodermis d.pith e.vascular cambium. 21. Heartwood and sapwood consist of

a.bark. b.periderm. c.secondary xylem. d.secondary phloem. e.cork. 22. Which of the following arise, directly or indirectly, from meristematic activity? a.secondary xylem b.leaves

c.dermal tissue d.tubers e.all of the above23. Which of the following would not be seen in a cross-section through the woody part of a root?

a.sclerenchyma cells b.parenchyma cells c.sieve-tube elements d.root hairs e. vessel elements24. The phase change of apical meristem from juvenile to mature vegetative phase is often revealed by

a.a change in the morphology of the leaves produced. b. the initiation of secondary growth. c.the formation of lateral roots.

d.a change in the orientation of preprophase bands and cytoplasmic microtubules in lateral meristems. e.the activation of floral meristem identity genes.

25. Describe at least three specializations in plant organs and plant cells that are adaptations to life on land.- Here are a few examples: The cuticle of leaves and stems protects these structures from desiccation.

Collenchyma and sclerenchyma cells have thick walls that provide support for plants. Strong, branching root systems help anchor the plant in the soil.

26. How does branching differ in roots versus stems?- Lateral roots emerge from the pericycle and destroy plant cells as they emerge. In stems, branches arise

from axillary buds and do not destroy any cells.27. What advantages did plants gain from the evolution of secondary growth?- With the evolution of secondary growth, plants were able to grow taller and shade competitors.28. By what mechanism do plant cells tend to elongate along one axis instead of expanding like a balloon in all

directions?- The orientation of cellulose microfibrils in the innermost layers of the cell wall causes this growth along

one axis. Microtubules play a key role in regulating the plane of cell expansion. It is the orientation of microtubules in the cell’s outermost cytoplasm that determines the orientation of cellulose microfibrils.

29. With your finger, trace the progression of leaf emergence, starting with leaf number 29. What is the pattern?

- The leaves are being produced in a counterclockwise spiral.30. Would a higher leaf area index always increase the amount of photosynthesis? Explain. - A higher leaf area index will not necessarily increase photosynthesis because of upper leaves shading

lower leaves.31. Assume that a plant cell has all four of the plasma membrane transport proteins shown above. Assume

also that you have specific inhibitors for each of the four transport proteins. Then predict what effect the individual application of each inhibitor would have on the cell’s membrane potential.

- A proton pump inhibitor would depolarize the membrane potential because fewer H+ ions would be pumped out across the plasma membrane. The immediate effect of an inhibitor of the H/sucrose transporter would be to hyperpolarize the membrane potential because fewer H+ ions would be leaking back into the cell through these cotransporters. An inhibitor of the H/NO3cotransporter would have no effect on the membrane potential because the simultaneous cotransport of a positively charged ion and a negatively charged ion has no net effect on charge difference across the membrane. An inhibitor of the Kion channels would decrease the membrane potential because additional positively charged ions would not be accumulating outside the cell.

32. How does the Casparian strip force water and minerals to pass through the plasma membranes of endodermal cells?

- Casparian strip blocks water and minerals from moving between endodermal cells or moving around an endodermal cell via the cell’s wall. Therefore, water and minerals must pass through an endodermal cell’s plasma membrane.

33. Does phloem sap contain more sugar near sources than sinks?- Because the xylem is under negative pressure (tension), excising a stylet that had been inserted into a

tracheid or vessel element would probably introduce air into the cell. No xylem sap would exude unless positive root pressure was predominant.

34. Why is long-distance transport important for vascular plants? - Vascular plants must transport minerals and water absorbed by the roots to all the other parts of the

plant. They must also transport sugars from sites of production to sites of use. 35. What architectural features influence self-shading?

Many features of plant architecture affect self-shading, including leaf arrangement, leaf and stem orientation, and leaf area index.

36. Some plants can detect increased levels of light reflected from leaves of encroaching neighbors. This detection elicits stem elongation, production of erect leaves, and reduced lateral branching. How do these responses help the plant compete? Increased stem elongation would raise the plant’s upper leaves. Erect leaves and reduced lateral branching would make the plant less subject to shading by the encroaching neighbors.

37. If you prune a plant’s shoot tips, what will be the short-term effect on the plant’s branching and leaf area index? Pruning shoot tips removes apical dominance, resulting in axillary buds growing into lateral shoots (branches). This branching produces a bushier plant with a higher leaf area index

38. Explain how fungal hyphae provide more surface area for nutrient absorption.Fungal hyphae are long, thin filaments that form a large interwoven network in the soil. Their high surface-to-volume ratio is an adaptation that enhances the absorption of materials from the soil.

39. If a plant cell immersed in distilled water has a ψs of -0.7 MPa and a ψof 0 MPa, what is the cell’s ψP? If you put it in an open beaker of solution that has a ψ of -0.4 MPa, what would be its ψp at equilibrium?The cell’s ψPis 0.7 MPa. In a solution with a ψof –0.4 MPa, the cell’s ψp at equilibrium would be 0.3 MPa.

40. How would a reduction in the number of aquaporin channels affect a plant cell’s ability to adjust to new osmotic conditions?The cells would still adjust to changes in their osmotic environment, but their responses would be slower. Although aquaporins do not affect the water potential gradient across membranes, they allow for more rapid osmotic adjustments.

41. How would the long-distance transport of water be affected if tracheids and vessel elements were alive at maturity? Explain.If tracheids and vessel elements were alive at maturity, their cytoplasm would impede water movement, preventing rapid long-distance transport.

42. What would happen if you put plant protoplasts in pure water? Explain.The protoplasts would burst. Because the cytoplasm has many dissolved solutes, water would enter the protoplast continuously without reaching equilibrium. (When present, the cell wall prevents rupturing by excessive expansion of the protoplast.)

43. How do xylem cells facilitate long-distance transport?Because water-conducting xylem cells are dead at maturity and form essentially hollow tubes, they offer little resistance to water flow, and their thick walls prevent the cells from collapsing from the negative pressure inside.

44. A horticulturalist notices that when Zinnia flowers are cut at dawn, a small drop of water collects at the surface of the stump. However, when the flowers are cut at noon, no drop is observed. Suggest an explanation. At dawn, a drop is exuded because the xylem is under positive pressure due to root pressure. At noon, the xylem is under negative pressure tension when it is cut and the xylem sap is pulled away from the cut surface up into the stem. Root pressure cannot keep pace with the increased rate of transpiration at noon.

45. A scientist adds a water-soluble inhibitor of photosynthesis to roots of a transpiring plant, but photosynthesis is not reduced. Why? The endodermis regulates the passage of water-soluble solutes by requiring all such molecules to cross a selectively permeable membrane. Presumably, the inhibitor never reaches the plant’s photosynthetic cells.

46. Suppose an Arabidopsis mutant lacking functional aquaporin proteins has a root mass three times greater than that of wild-type plants. Suggest an explanation. Perhaps greater root mass helps compensate for the lower water permeability of the plasma membranes.

47. How are the Casparian strip and tight junctions similar? The Casparian strip and tight junctions both prevent movement of fluid between cells.

48. The symplast transports all of the following except a.sugars. d.proteins. b.mRNA. c.DNA. e.viruses. 49. Which of the following is an adaptation that enhances the uptake of water and minerals by roots?

a.mycorrhizae b.cavitation c.active uptake by vessel elements d.rhythmic contractions by cortical cells e.pumping through plasmodesmata

50. Which structure or compartment is part of the symplast? a.the interior of a vessel element b.the interior of a sieve tube c.the cell wall of a mesophyll cell d.an extracellular air space e.the cell wall of a root hair

51. Movement of phloem sap from a source to a sink a.occurs through the apoplast of sieve-tube elements. b.depends ultimately on the activity of proton pumps. c.depends on tension, or negative pressure potential. d.depends on pumping water into sieve tubes at the source. e.results mainly from diffusion.

52. Photosynthesis ceases when leaves wilt, mainly because a.the chlorophyll in wilting leaves is degraded. b.flaccid mesophyll cells are incapable of photosynthesis. c.stomata close, preventing CO2 from entering the leaf. d.photolysis, the water-splitting step of photosynthesis, cannot occur when there is a water deficiency. e.accumulation of CO2 in the leaf inhibits enzymes.

53. What would enhance water uptake by a plant cell? a.decreasing the ψ of the surrounding solution b.increasing the pressure exerted by the cell wall c.the loss of solutes from the cell d.increasing the ψ of the cytoplasm e.positive pressure on the surrounding solution

54. A plant cell with a ψs of –0.65 MPa maintains a constant volume when bathed in a solution that has a ψs of –0.30 MPa and is in an open container. The cell has a a.ψp of 0.65 MPa. d.ψp of 0.30 MPa. b.ψ of –0.65 MPa. c.ψp of 0.35 MPa. e.ψ of 0 MPa.

55. Compared with a cell with few aquaporin proteins in its membrane, a cell containing many aquaporin proteins will a.have a faster rate of osmosis. b.have a lower water potential. c.have a higher water potential. d.have a faster rate of active transport. e.accumulate water by active transport.

56. Which of the following would tend to increase transpiration? a.a rainstorm d.higher stomatal density b.sunken stomata c.a thicker cuticle e.spiny leaves

57. What are the stimuli that control the opening and closing of stomata? Stomatal opening at dawn is controlled mainly by light, CO2concentrations, and a circadian rhythm. Environmental stresses such as drought, high temperature, and wind can stimulate stomata to close during the day. Water deficiency can trigger release of the plant hormone abscisic acid, which signals guard cells to close stomata.

58. The pathogenic fungus Fusicoccum amygdali secretes a toxin called fusicoccin that activates the plasma membrane proton pumps of plant cells and leads to uncontrolled water loss. Suggest a mechanism by which the activation of proton pumps could lead to severe wilting. The activation of the proton pump of stomatal cells would cause the guard cells to take up K. The increased turgor of the guard cells would lock the stomata open and lead to extreme evaporation from the leaf.

59. If you buy cut flowers, why might the florist recommend cutting the stems underwater and then transferring the flowers to a vase while the cut ends are still wet?After the flowers are cut, transpiration from any leaves and from the petals (which are modified leaves) will continue to draw water up the xylem. If cut flowers are transferred directly to a vase, air pockets in xylem vessels prevent delivery of water from the vase to the flowers. Cutting stems again underwater, a

few centimeters from the original cut, will sever the xylem above the air pocket. The water droplets prevent another air pocket from forming while placing the flowers in a vase.

60. Explain why the evaporation of water from leaves lowers their temperature.Water molecules are in constant motion, traveling at different rates. The average speed of these particles depends on the water’s temperature. If water molecules gain enough energy, the most energetic molecules near the liquid’s surface will impart sufficient speed, and therefore sufficient kinetic energy, to cause water molecules to propel away from the liquid in the form of gaseous molecules or, more simply, as water vapor. As the particles with the highest kinetic energy levels evaporate, the average kinetic energy of the remaining liquid decreases. Because a liquid’s temperature is directly related to the average kinetic energy of its molecules, the liquid cools as it evaporates.

61. Compare and contrast the forces that move phloem sap and xylem sap over long distance. In both cases, the long-distance transport is a bulk flow driven by a pressure difference at opposite ends of tubes. Pressure is generated at the source end of a sieve tube by the loading of sugar and resulting osmotic flow of water into the phloem, and this pressure pushessap from the source end to the sink end of the tube. In contrast, transpiration generates a negative pressure potential (tension) as a force thatpullsthe ascent of xylem sap.

62. Identify plant organs that are sugar sources, organs that are sugar sinks, and organs that might be either. Explain.The main sources are fully grown leaves (by photosynthesis) and fully developed storage organs (by breakdown of starch). Roots, buds, stems, expanding leaves, and fruits are powerful sinks because they are actively growing. A storage organ may be a sink in the summer when accumulating carbohydrates, but a source in the spring when breaking down starch into sugar for growing shoot tips.

63. Why can xylem transport water and minerals using dead cells, whereas phloem requires living cells?Positive pressure, whether it be in the xylem when root pressure predominates or in the sieve-tube elements of the phloem, requires active transport. Most long-distance transport in the xylem depends on bulk flow driven by negative pressure potential generated ultimately by the evaporation of water from the leaf and does not require living cells.

64. Apple growers in Japan sometimes make a nonlethal spiral slash around the bark of trees that are destined for removal after the growing season. This practice makes the apples sweeter. Why?

65. The spiral slash prevents optimal bulk flow of the phloem sap to the root sinks. Therefore, more phloem sap can move from the source leaves to the fruit sinks, making them sweeter.

66. How do plasmodesmata differ from gap junctions? Plasmodesmata, unlike gap junctions, have the ability to pass RNA, proteins, and viruses from cell to cell.

67. Nerve-like signals in animals are thousands of times faster than their plant counterparts. Suggest a behavioral reason for the difference. Long-distance signaling is critical for the integrated functioning of all large organisms, but the speed of such integration is much less critical to plants because their responses to the environment, unlike those of animals, do not typically involve rapid movements.

68. Suppose plants were genetically modified to be unresponsive to viral movement proteins. Would this be a good way to prevent the spread of infection? Explain.Although this strategy would eliminate the systemic spread of viral infections, it would also severely impact the development of the plants.