97
A plant has a measured pressure potential Ψp = 0.21MPa and a solute potential Ψs =-3.50MPa. The soil is saturated with water because it rained. How will the water move? After three months of dry weather, the soil has dried out. How will the water potential of the soil compare to the water potential measured immediately before the rain? How will the stomata respond to the change in weather?
  1. The water will move from the plant to the soil. Dry soil has a lower water potential than wet soil. Under drought conditions, the stomata close to conserve water and leaves may also be shed if the drought continues.
  2. The water will move from the soil to the plant. Dry soil has a higher water potential than wet soil. Under drought conditions, the stomata close to conserve water and leaves may also be shed if the drought continues.
  3. The water will move from the soil to the plant. Dry soil has a lower water potential than wet soil. Under drought conditions, the stomata open its pores wider in order to perform a better rate of transpiration.
  4. The water will move from the soil to the plant. Dry soil has a lower water potential than wet soil. Under drought conditions, the stomata close to conserve water and leaves may also be shed if the drought continues.
98

Table with six columns and three rows. Topmost header row reads: Transpiration rate versus temperature. Second row, left header reads: Temperature (°C). Second row, second cell reads: 20. Second row, third cell reads: 23. Second row, fourth cell reads: 27. Second row, fifth cell reads: 28. Second row, sixth cell reads: 30. Third row, left header reads: Transpiration rate (mmol/m2sec). Third row, second cell reads: 1.5. Third row, third cell reads: 3. Third row, fourth cell reads: 5. Third row, fifth cell reads: 4.5. Third row, sixth cell reads: 4.

Plants lose water from their aboveground surfaces in the process of transpiration. Most of this water is lost from stomata. Excess loss of water has severe consequences and may be fatal for the plant. The table shows data collected on a sunny day. What is the best explanation for the transpiration rates leveling off and declining at temperature higher than 27°C?

  1. The plant ran out of water.
  2. The plant needs less water as temperature increases, so transpiration slows down to limit water uptake by the roots.
  3. Stomata close to conserve water, slowing down transpiration.
  4. The amount of water in the leaves decreases at high temperature and less is available for evaporation.
99
Humidity is an environmental factor that affects transpiration rate. Which statement accurately explains the shape of the curve obtained when increasing humidity is plotted against constant temperature to find the rate of transcription?
  1. Increasing humidity leads to reduced evaporation rates due to increased difference in water vapor pressure between leaf and atmosphere.
  2. Increasing humidity leads to reduced evaporation rates due to decreased difference in water vapor pressure between leaf and soil.
  3. Increasing humidity leads to reduced evaporation rates due to decreased difference in water vapor pressure between leaf and atmosphere.
  4. Increasing humidity leads to increased evaporation rates due to decreased difference in water vapor pressure between leaf and atmosphere.
100

The graph plots hormone levels on the y-axis versus time of drought on the x-axis. There are four plotted lines for auxin, abscisic acid, cytokinin, and gibberellins. The auxin line starts at a hormone level just above 3 and gradually lowers to a hormone level just below 3 over time. The abscisic acid line starts at a hormone level of 1, jumps to a hormone level of 3, and then gradually rises to above 4 over time. The cytokinin line starts at a hormone level of 2 and gradually rises to a hormone level of about 2.5 over time. The gibberellins line starts at a hormone level of 3, rises slightly above 3, and then lowers to slightly below 3 over time.

Plants sense drought through the decrease in water potential in the ground. This graph shows concentrations of several hormones that were measured during a drought period and plotted versus time. According to the data in the graph, which hormone shows the strongest response to drought?

  1. auxin
  2. abscisic acid
  3. cytokinin
  4. gibberellins
101

The graph plots hormone levels on the y-axis versus time of drought on the x-axis. There are four plotted lines for auxin, abscisic acid, cytokinin, and gibberellins. The auxin line starts at a hormone level just above 3 and gradually lowers to a hormone level just below 3 over time. The abscisic acid line starts at a hormone level of 1, jumps to a hormone level of 3, and then gradually rises to above 4 over time. The cytokinin line starts at a hormone level of 2 and gradually rises to a hormone level of about 2.5 over time. The gibberellins line starts at a hormone level of 3, rises slightly above 3, and then lowers to slightly below 3 over time.

When drought conditions are forecast, fields are sprayed with a hormone that will promote a stress response. According to the graph, which hormone should be sprayed and why?

  1. Gibberellins, to promote plant growth before the plants are damaged
  2. Abscisic acid, to promote plant growth before the plants are damaged
  3. Abscisic acid, to promote protective response to drought before the plants are damaged.
  4. Gibberellins, to promote protective response to drought before the plants are damaged.
102

Table with four columns and two rows. Top, left header reads: Plate A. Top, middle header reads: Plate B. Top, right header reads: Plate C. Second row, left header reads: Percentage germination. Second row under Plate A reads: 95%. Second row under Plate B reads: 50%. Second row under Plate C reads: 52%.

Seeds were germinated in the dark on three plates. Plate A was irradiated with a short pulse of red light; plate B was irradiated with a short pulse of red light followed by a pulse of far-red light; and plate C was the control and was maintained in the dark. After three days, the plates were scored for percentage of germination, as shown in this table. What conclusion can be drawn from the experiment?

  1. Darkness inhibits germination.
  2. Red light promotes germination.
  3. Far-red light promotes germination.
  4. Germination is independent from light irradiation.
103

Table with five columns and two rows. The top headers read: Plate A, Plate B, Plate C, and Plate D. Second row, left header reads: Percentage germination. Second row under Plate A reads: 95%. Second row under Plate B reads: 50%. Second row under Plate C reads: 96%. Second row under Plate D reads

Seeds were germinated in the dark on three plates. Plate A was irradiated with a short pulse of red light; plate B was irradiated with a short pulse of red light followed immediately by a pulse of far-red light; plate D was irradiated by a short pulse of red light followed one hour later by a pulse of far-red light; and plate C was the control and was maintained in the dark. After three days, the plates were scored for percentage of germination, as shown in this table. What hypothesis do the results suggest about the mechanism of action of red light?

  1. Red light converts the phytochrome to its active form Pr which can be converted to the inactive form Pfr by far red light. After one hour, cascade of events initiated by Pfr has already begun promoting germination and hence, it cannot be reversed even by the pulse of far light.
  2. Red light converts the phytochrome to its active form Pfr, which can be converted to the inactive form Pr by far-red light. After one hour, cascade of events initiated by Pr has already begun promoting germination and, hence, it cannot be reversed even by the pulse of far light.
  3. Far red light converts the phytochrome to its active form Pfr, which can be converted to the inactive form Pr by red light. After one hour, the cascade of events initiated by Pr has already begun promoting germination and, hence, it cannot be reversed even by the pulse of far light.
  4. Red light converts the phytochrome to its active form Pfr which can be converted to the inactive form Pr by far red light. After one hour, the cascade of events initiated by Pfr has already begun promoting germination and, hence, it cannot be reversed even by the pulse of far light.
104
After branches of woody saplings were trimmed, half of the cuts were covered with a sealant and the other half were left untouched. The plants with sealed cuts fared much better after several weeks. What is the likely reason?
  1. The sealant stopped evaporation.
  2. The plants with sealed cuts grew new branches.
  3. The plants with unsealed cuts were infected by pathogens that entered through the cuts.
  4. The plants with unsealed cuts lost photosynthates through bleeding of sap.
105

Table with three columns and three rows. Top, left header reads: Mutant plants (Ja–). Top, right header reads: Normal plants (Ja+). Second row, left header reads: Average size of wound from fungal infection. Second row under Mutant plants reads: 10 mm. Second row under Normal plants reads: 4 mm. Third row, left header reads: Weight of moth larvae. Third row under Mutant plants reads: 80 mg. Third row under Normal plants reads: 55 mg.

Jasmonate is produced in plants as a response to injury. Researchers compared the response to infection of mutant plants that were unable to produce jasmonate (Ja-) with the response of normal plants (Ja+) from the same species. Leaves were inoculated with spores from pathogenic molds. The size of the wounds was examined 48 hours after application. The plants were also infected with moths and the weight of the larvae was determined aafter 48 hours. This table shows the results. According to the results of the experiment, what conclusion can the researchers draw about the specificity of jasmonate protection?

  1. Jasmonate protects against infection from a variety of pathogens.
  2. Jasmonate protects against infection from one pathogen.
  3. Jasmonate cannot provide protection against infection.
  4. Jasmonate provides specific defense in winters and the defense is non-specific in summers.
106
In the Northern Hemisphere, a florist grows shrubs of the same species of woody plant under two different light schedules for three weeks. The first set is maintained under 15 hours of light and 9 hours of dark daily. The second set is maintained under 9 hours of light followed by 14 hours of dark daily. The first set of plants does not form flowers, but the second set of plants blooms. What can you conclude about these plants?
  1. This species of shrub does not flower if the day is short.
  2. They bloom early in the year (around February).
  3. They bloom mid-summer (around June).
  4. The critical dark period is 9 hours.
107

The graph plots the circular motion of sunflower seedlings on the y-axis versus time from 0 to 96 hours on the x-axis. The plot line forms a wave with peaks at approximately 12, 36, 60, and 84 hours and valleys at approximately 24, 48, 72, and 96 hours. The peaks gradually decline over time, while the valleys remain relatively the same.

Heliotropism is the description of a response to the light of the sun. Seedlings of sunflowers were exposed to sunlight for 15 days. Following the 15 days of exposure to sunlight, the seedlings were transferred to complete darkness and their movement was monitored. This graph plots the movement of the seedlings in the dark versus time. What conclusion can be drawn about the light dependence of the movement of sunflowers from the graph?

  1. The movement does require light once it is set but it will eventually slow down, suggesting that a “clock” molecule is degraded over time.
  2. The movement does not require light once it is set and it will keep showing this upward and downward trend in the same manner.
  3. The movement does not require light once it is set and it will eventually slow down, suggesting that a “clock” molecule never degrades.
  4. The movement does not require light once it is set and it will eventually slow down, suggesting that a “clock” molecule is degraded over time.
108

Table with three columns and five rows. Top, left header reads: Dish A. Top, right header reads: Dish B. Second row header reads: Germinated seeds. Second row under Dish A reads: 12. Second row under Dish B reads: 20. Third row header reads: Green-leaved seedlings. Third row under Dish A reads: 0. Third row under Dish B reads: 15. Fourth row header reads: Yellow-leaved seedlings. Fourth row under Dish A reads: 12. Fourth row under Dish B reads: 5. Fifth row header reads: Mean length of stem below 1st leaves. Fifth row under Dish A reads: 8 mm. Fifth row under Dish B reads: 3 mm.

A student randomly chose 40 tobacco seeds of the same species from a packet. He placed 20 seeds on moist paper towels in each of two petri dishes. He wrapped dish A completely in an opaque cover to exclude all light. He did not wrap dish B. He placed the dishes equidistant from a light source set to a cycle of 14 hours of light and 10 hours of dark. All other conditions were the same for the two dishes. He examined the dishes after 7 days, and permanently removed the opaque cover from dish A. This table shows the student’s data. The most probable cause for the difference in mean stem length between plants in dish A and plants in dish B is ____.

  1. shortening of cells in the stem in response to the lack of light
  2. elongation of the stem in response to the lack of light
  3. enhancement of stem elongation by light
  4. genetic differences between the seeds
109

The graph plots internode length (mm) on the y-axis versus gibberellin (µg/ml) on the x-axis. The plot line begins at 0, 0 and then arcs upward before it reaches a plateau.

Groups of 20 seedlings from the same plant species were treated with gibberellins. Each group received a different concentration of hormone. The seedlings were grown under the same environmental conditions. After 15 days of growth, the internode distances between the first and second sets of leaves were measured in each group of seedlings. On this graph, the mean internode distance for each group is plotted against the concentration of gibberellins that the group received. According to the results, why is this effect of gibberellins on internode length used in agriculture to spray grapes with oversized fruit?

  1. to lengthen the internode distance and accommodate larger fruit
  2. to shorten the internode distance and accommodate larger fruit
  3. to lengthen the internode distance and accommodate more flowers
  4. to shorten the internode distance and accommodate smaller fruit