ExamView - AP Biology Botany MC - Weeblyrhsscience.weebly.com/uploads/1/2/4/9/12490351/ap...Version A 5 9. An experiment is performed where plants are exposed to a mixture of gases

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(1)Test Questions are Copyright © 1984-2002 by College Entrance Examination Board, Prenceton, NJ. All rights reserved. For face-to-fact teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited. (2) AP ® is registered traemark of the College Entrance Examination Borard. The College Entrance Examination Board was not involved in the production of and does not endorse this product. Permissionis granted of individual classroom teahcers to reproduce the activity sheets and illustation for their own classroom use. Any other type of reproduction of these material is strictly prohibited.

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AP* Biology: Botany

Directions: Each of the questions or incomplete statements below is followed by four suggested answers or completions. Select the one that is best in each case and enter the appropriate letter in the corresponding space on the answer sheet.

Questions 1 - 3

Seeds given various treatments are planted in small pots. The height of each seedling is measured at 4-day intervals beginning with day 4 and ending with day 16. The graph below is an illustration of the data obtained. A key to the seed type and treatment is shown below the graph.

The Effect of Plant Hormones on Plant Growth

D = Dwarf pea plants seeds -- no treatment DG = Dwarf pea plant seeds soaked in gibberellic acid (gibberellin) DGA = Dwarf pea plant seeds soaked in gibberellic acid and indoleacetic acid (auxin)

T = Tall, nondwarf pea plant seeds--no treatment

1. According to the data, at day 8 of the experiment, all of the following statements are correct EXCEPT:

A) The T, DG, and DGA seedlings are very similar in height.B) The D seedlings are less than half as tall as the other seedlings. C) The T, DG, and DGA seedlings appear to be in a more rapid growth phase than the D

seedlings.D) The eventual greater height of the T seedlings over the DG seedlings is already

predictable.

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2. A hypothesis these data tend to support would be that

A) D pea seedlings lack amounts of gibberellin sufficient for achieving tall statureB) gibberellin and auxin both stimulate cell elongationC) tallness in this particular variety of pea plants is dominant over dwarfnessD) auxin promotes cell division

3. After examination of the data, there is an indication that the

A) D seedlings will not mature and produce new seedsB) auxin partially counteracts the action of gibberellin on the growth of dwarf seedlingsC) stems of the T plants are not as strong as those of the DG plantsD) DGA pea plants have longer internodes than the other plants

4. Most plants and some fungi display alternation of generations which includes both a multicellular diploid stage and a multicellular haploid stage. Which of the following represents the first cell of the sporophyte generation in the life cycle of a pine tree?

A) microspore mother cellB) pollen grainC) zygoteD) polar nuclei

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Questions 5 - 7

In 1928, Fritz Went investigated how a growth factor causes a coleoptile to grow toward the light. Coleoptiles were placed in the dark, their tips were then removed and placed on slabs of agar. It was predicted that the growth factor would be absorbed by the slab of agar. Next, the slabs were either placed centered on top of the coleoptile to distribute the growth factor evenly or placed offset to increase the concentration to one side. Additionally, a slab of agar without any growth hormone was centered on one coleoptile. The results are shown below.

5. Which of the following coleoptiles serve as a control for this experiment?

A) AB) A, BC) A, ED) A, E, and F

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6. The work of Fritz Went was based on experiments conducted by Charles Darwin and Francis Darwin in 1880. These experiments were extended by Peter Boysen Jenson in 1913. The results of the experiments are shown below.

Based on the work of Darwin, Boysen-Jensen and Went, the next logical step in the investigation of this particular growth factor of plants would include which of the following?

A) Determing where the growth factor is made in the plantB) Determing which side of the tip is making the growth factorC) Determing what is inhibiting mitosis on the dark side of the plant D) Determing what is serving as the stimulus for the bending of the plant

7. Which of the following best represents the results of the experiment?

A) The coleoptile bent away from the side with the increased concentration of growth factor.B) The coleoptile bent toward the side with the increased concentration of growth factor.C) The centered agar block without any growth factor will cause the coleoptile to increase in

height.D) The centered agar block with growth factor will cause the coleoptile to bend in either

direction.

8. A certain short-day plant flowers only when days are less than 12 hours long. Which of the following would cause it to flower?

A) A 9-hour night and 15-hour day with 1 minute of darkness after 7 hoursB) An 8-hour day and 16-hour night with a flash of white light after 8 hoursC) A 13-hour night and 11-hour day with 1 minute of darkness after 6 hoursD) A 12-hour day and 12-hour night flash of red light after 6 hours

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9. An experiment is performed where plants are exposed to a mixture of gases for several hours and then analyzed. One of the gases included in the mixture is radioactive CO2 that contains the C-14 isotope of carbon rather than the C-12 isotope. Which of the following structures would contain the highest concentration of organic molecules containing the C-14 isotope?

A) xylemB) phloemC) epidermisD) root hairs

10.

A team of researchers monitors the nitrogen content present in the soil of crop fields over a number of years. The soil has been tilled and planted with crops such as corn annually. The graph above depicts the findings of that study. Which of the following farming practices presents the most practical solution for restoring nitrogen to the soil?

A) Allow the fields to lay fallow every other year. B) Rotate the species of corn planted every other year.C) Burn the remains of the corn stalks once the corn has been harvested.D) Alternate the crops every other year with a legume such as soy beans.

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11. As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the ground. Fifteen years later you return and discover the tree has grown to a height of 30 meters. How far above the ground is the nail in this mature tree?

A) 1.5 metersB) 3.0 metersC) 15.0 metersD) 28.5 meters

12. The genus Atriplex (Family Chenopodiaceae), saltbush, is found along the shoreline at the ocean’s edge worldwide. Which of the following is an adaptation that would aid Atriplex in surviving such saline conditions?

A) Thin leaves with large surface area to excrete excess salt.

B) Leaves with waxy cuticle so that transpiration will not occur.

C) Glands that actively secrete salt into vacuoles which is then released into the environment.

D) Xylem which has reversed the movement of water. Moving water down into the root to dilute the saline solution.

13. What adaptations should one expect of the seed coats of angiosperm species whose seeds are dispersed by frugivorous (fruit-eating) animals, as opposed to angiosperm species whose seeds are dispersed by other means?

A) The seed coat, upon its complete digestion, should provide vitamins or nutrients to animals.

B) The seed coat should contain secondary compounds that irritate the lining of the animal's mouth.

C) The seed coat should be resistant to the animals' digestive enzymes.

D) The exterior of the seed coat should have barbs or hooks.

14. You are studying a plant from the arid southwestern United States. Which of the following adaptations is most likely to have evolved in response to water shortages?

A) The closing of stomata at night to prevent evaporative water loss.

B) The reduction in surface area of leaves to reduce water loss.

C) The formation of a fibrous root system spreading over a small area to prevent dehydration.

D) The association of mycorrhizae with the root system to prevent water loss.

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15.

A nitrogen fixing pathway for certain plants is shown above. Which of the following best describes how nitrogen is utilized within a plant?

A) Nitrogen is required to make glucose which is used for energy.B) Nitrogen is required to make cellulose which is used for cell walls.C) Nitrogen is required to make DNA which is used for heredity.D) Nitrogen is required to make triglycerides which are used for energy and serve as the

energy resource in seeds.

16. The value of the water potential, , in a sample of root tissue is determined to be –0.15 MPa. Which of the following best describes the net flow of water that occurs when a sample of the root tissue is placed in a 0.1 M solution of sucrose ( = –0.23 MPa)?

A) Water flows from the root tissue into the sucrose solution.

B) Water flows from the sucrose solution into the root tissue.

C) Water flow will be in both directions thus the concentrations would remain unchanged.

D) Water flow occurs only as ATP is hydrolyzed in the tissue.

17. When the water vapor in the atmosphere is lower than that in the apoplast of a leaf, water vapor will diffuse from the intercellular spaces of the leaf through the stomata to the outside air in a process known as transpiration. Which of the following factors is most important in the movement of water up a tall tree?

A) air pressureB) capillarity in the phloemC) cohesion of water moleculesD) active transport in the xylem

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18. The seeds of orchids are among the smallest known, with virtually no endosperm and with miniscule seed leaves. Consequently, what should one expect to be true of such seeds?

A) They require extensive periods of dormancy during which the embryo develops.

B) The developing embryo within is dependent upon the gametophyte for nutrition.

C) They are surrounded by brightly colored, sweet fruit.

D) They germinate very soon after being released from the ovary.

19. A plant is growing in soil where the concentration of a certain nutrient is 50 ppm (parts per million). A sample is taken from the plant’s root tissue and the concentration of this nutrient is determined to be 200 ppm. Which of the following mechanisms is most likely responsible for the rapid uptake of this nutrient into the plant’s root tissue?

A) osmosisB) facilitated transportC) active transportD) passive diffusion

20.

Botanists use aphids to study the transport of materials in plants. Aphids are insects that feed by inserting their stylet (needle-like mouth part) under the epidermis of herbaceous plants (Diagrams A and B). A laser is used to separate the stylet inserted under the eipdermis of a plant from the body of an anesthetized aphid (Diagram C). The stylet remains in the plant tissue allowing sap to ooze from it. This observation best supports which of the following?

A) The material moving through the xylem is under positive pressure.B) The material moving through the xylem is under negative pressure.C) The material moving through the phloem is under positive pressure.D) The material moving through the phloem is under negative pressure.

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21.

The germinating seed shows its radical with root hairs. The function of root hairs is

A) to provide increased surface area for the absorption of water and minerals.B) to provide increased strength in anchoring the plant.C) to provide increased storage for carbohydrates produced in the leaves. D) to provide increased surface area for the interaction with bacteria to obtain nutrients.

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Questions 22 and 23

22. The apparatus pictured above is used to measure changes in pressure. A porous clay pot is attached to the top of a thin tube whose lower end is suspended in a beaker of mercury. Water is trapped in the tube above the mercury. Atmospheric pressure can support a column of mercury (Hg) 760 mm tall.

Water is evaporated from the porous clay pot. As water evaporates, the column of water pulls the mercury up the tube. The experiment is repeated with a plant replacing the clay pot. Which of the following best represents the results of this experiment?

A) Water moves up xylem as a result of transpirational pull. B) Water moves up xylem as a resuult of root pressure.C) Water moves up xylem as a result as explained by the bulk flow hypothesis.D) Water moves up xylem as a result of capillary action.

23. What are the pores in the clay pot analogous to in the plant used in the experiment?

A) root hairsB) xylem found in the rootC) guard cellsD) stomata

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24. Phosphate Fertilizer Added

(g)Mycorrhizae Present Seedling

Dry Weight (g)

0 No 1 0 Yes 1012 No 2812 Yes 16624 No 2024 Yes 210

An expermeint was conducted to determine the effect of phosphate fertilizer, mycorrhizal fungi or both on the growth of lemon seedlings. Each plant was subjected to the same amount of water and sunlight for 14 days. The data is shown above. Which of the following conclusions is best represented by the data?

A) Phosphates alone do not increase the mass of seedlings significantly.B) The presence mycorrhizae significantly increases the plant’s ability to obtain phosphate.C) There is a linear relationship between the amount of phosphate contained in the soil and

the dry mass of the seedlings.D) It is the mycorrhizae and not the phosphate that is responsible for the increase in dry mass

of the seedlings.

25. The cycads, a mostly tropical phylum of gymnosperms, evolved about 300 million years ago and were dominant forms during the Age of the Dinosaurs. Though their sperm are flagellated, their ovules are pollinated by beetles. These beetles obtain nutrients from the pollen and shelter from the microsporophylls. The beetles transfer pollen to the exposed ovules when the beetles visit the megasporophylls. In cycads, pollen cones and seed cones are borne on different plants. Cycads synthesize neurotoxins, especially in the seeds, that are effective against most animals, including humans. Which feature of cycads makes them similar to many angiosperms?

A) They have exposed ovules.B) They have flagellated sperm.C) They are pollinated by animals.D) They produce cones.

Questions 26 and 27

In onion (Allium), cells of the sporophyte have 16 chromosomes within each nucleus. Match the number of chromosomes present in each of the onion tissues listed.

26. How many chromosomes should be in the tube cell nucleus?

A) 4B) 8C) 16D) 24

27. How many chromosomes should be in an endosperm nucleus?

A) 4B) 8C) 16D) 24

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28. Most of the water a plant absorbs by the roots is lost by evaporation, typically from stomata on the plant’s leaves and stems. This evaporation of water from above ground plant parts, particularly through the stomata, is called transpiration. Which of the following is a factor in the mechanism of transpiration?

A) Transpiration causes a positive pressure on continuous columns of water in the plant’s xylem.

B) Water molecules are held together by cohesion of hydrogen bonds so the movement of one molecule causes a pull on them all.

C) The tension extending from leaves that may be hundreds of feet in the air is enhanced by gas bubbles forming from dissolved gases coming out of solutions.

D) The collective strength of the hydrogen bonds between water molecules causes them to form droplets on the leaf margins.

29. While studying transpiration, a scientist used a dendrometer to record the small daily changes in the diameter of a tree trunk at two different heights (2 meters and 3 meters) above the ground at the same time. The diameter decreased in the daytime and first occurred at the 3 m location.

Which of the following would most likely cause this decrease in diameter?

A) An overcast day without much wind.B) A cool eveningC) Increased transpiration D) A humid afternoon

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30. Accumulation of K+ by guard cells results in water uptake and the turgid condition of the cells keeps the stomata open. Which of the following stimulate stomatal opening?

A) The presence of low levels of O2 in the leaf. B) Exposure of the leaf to sunlight.C) A decrease in humidity in the leaf’s environment.D) Exposure of the leaf to green light.

31. Seven groups of 200 lettuce seeds each were incubated in water for 16 hours in the dark. One group was then exposed to white light for 1 minute. A control group remained in the dark. Five other groups were exposed to red (R) light or a combination of red (R) and far-red (FR) light for 1 minute each. All the seeds were then returned to darkness for 2 more days. The number of germinating seeds for each group was recorded and the following data were obtained.

Group Condition Seeds Germinated

1 White light 1992 Dark 173 R 1964 R then FR 1085 R then FR then R 2006 R then FR then R then FR 867 R then FR then R then FR then R 198

According to the data, which of the following can be concluded about the photoreceptors involved and their role in seed germination?

A) The length of time these seeds were exposed to red and far-red light adversely affected the rate of germination.

B) A large percentage of lettuce seeds will germinate when exposed to long periods of night. C) Red and far-red light reverse each other’s effects.D) Lettuce seeds should be spread on top of the soil to allow exposure to light in order to

germinate.

32. Assume a particular chemical interferes with the establishment and maintenance of proton gradients across the membranes of plant cells. Which of the following processes would be most affected by the disruption of these proton gradients?

A) phloem transportB) water absorption by the rootsC) stomatal openingD) xylem transport

(1) Test Questions are Copyright © 1984-2012 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited. (2) AP® is a registered trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of and does not endorse this product. Permission is granted for individual classroom teachers to reproduce the activity sheets and illustrations for their own classroom use. Any other type of reproduction of these materials is strictly prohibited.

AP Biology Botany Unit Exam

Part B Directions: These three questions require numeric answers. Calculate the correct answer for each question, and enter your answer on the grid following each question. Examples of correct entry for the grid-in questions are shown below. The actual questions for this exam begin on the next page. Integer Answer Integer Answer Decimal Answer Fraction Answer 502 502 −4.13 −2/10

5 0 2 5 0 2 − 4 . 1 3 − 2 / 1 0

1.

A student was investigating the cumulative loss of water from two plants of the same species using a potentiometer as shown above. One plant served as a control and was kept at room temperature in still air with no wind.

The second plant was placed in front of a fan used to simulate wind as shown above. The data collected over the 30 minutes is shown below.

Next, the total leaf surface area of both plants was determined:

Surface area of control plant = 0.0012 m2 Surface area of plant with wind = 0.0014 m2

Calculate the amount of water lost by the control plant after 30 minutes per unit of surface area. Give your answer to the nearest hundredth.

2. Using the data presented in question number 1,calculate the average rate of water loss for the20 to 30 minute time interval for the plant receivingthe wind treatment. Give our answer to the nearest

hundredth 2

mLm

min.

Time (minutes)

Treatment 0 10 20 30

No wind 0.0 mL 0.0026 mL 0.0055 mL 0.0079 mL

Wind 0 mL 0.0063 mL 0.0106 mL 0.0154 mL

3. A student was asked to determine the number of stomata per cm2 on the lower leaf surface of Chlorophytum comosum more commonly known as the airplane plant. The student measured the diameter of the field of view

of the microscope at 100× as shown below on the left and determined it to be 1.5 mm. The student then carefully removed the lower epidermis of the leaf and viewed the tissue under the microscope at a total magnification of

450×. A photograph of the plant’s epidermis is shown below on the right.

Use the data associated with these two micrographs to calculate the estimated number of stomata per cm2.

Give your answer to the nearest whole number. (The area of a circle is = π r2 where r is the radius and π = 3.142)

Name:____________________________Period___________________

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AP* Biology: Botany


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Dry Weight (g)





of the seedlings.







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A) to provide increased surface area for the absorption of water and minerals.

B) to provide increased strength in anchoring the plant.

C) to provide increased storage for carbohydrates produced in the leaves.

D) to provide increased surface area for the interaction with bacteria to obtain nutrients.

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Questions 22 and 23






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Questions 25 and 26



A) 4B) 8C) 16D) 24


A) 4B) 8C) 16D) 24

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A) The presence of low levels of O2 in the leaf. B) Exposure of the leaf to sunlight.C) A decrease in humidity in the leaf’s environment.D) Exposure of the leaf to green light.

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AP* Biology: BotanyAnswer Section

MULTIPLE CHOICE

1. ANS: CThe correct answer is C. The material in the xylem will have negative pressure in the absence of root pressure and provided the stomata are open. The material moving through the phloem is under positive pressure. Students need to know that the phloem is closer to the epidermis than the xylem and that the mechanism for moving material through the phloem is positive pressure.

Graphics: http://www.grin.com/en/doc/253502/transcriptional-profiling-of-aphid-resistant-and-susceptible-melon-cucumishttp://web.uconn.edu/mcbstaff/graf/BuAp/Baphidsym.htm

OBJ: LO 4.8 NAT: SP 3.3 2. ANS: C

A short day plant requires a long night. The 13-hour night is the only uninterrupted night longer than 12 hours. The other regimes do not have 13 hours of uninterrupted darkness.

OBJ: LO 2.36 NAT: SP 6.1 3. ANS: D

All of the statements that appear in the answer choices can be obtained from the graph. At day eight The T, DG, and DGA seedlings are very similar in height. It is not possible to predict that the eventual greater height of the T seedlings over the DG seedlings is already predictable. In fact, at day eight the DG seedlings are taller than the T seedlings. The D seedlings are less than half as tall as the other seedling. At day eight the D seedling is approximately 4 cm tall and the shortest other seedling (DGA) is approximately 8.5 cm tall. It is apparent that the T, DG, and DGA seedlings are in a more rapid growth phase than the D seedlings.

DIF: 80% correct OBJ: LO 2.35 NAT: SP 4.2 NOT: 1986 #102 4. ANS: A

To answer this question students need to know that auxins cause cell elongation, and that gibberellins cause an increase in cell division. The only hypothesis that the data supports is that dwarf (D) pea seedlings lack amounts of gibberellin sufficient for achieving tall stature.

B) Gibberellin causes an increase is cellular reproduction and auxins caused cell elongation.C) There is no indication that tallness is dominant over dwarfness.D) Auxins promotes cell elongation.

DIF: 51% correct OBJ: LO 2.35 NAT: SP 4.2 NOT: 1986 #103

ID: B

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5. ANS: BThe best answer is that auxin partially counteracts the action of gibberellin in the growth of dwarf seedlings. According to the graph the seeds that were soaked in a solution of auxins and gibberellins produced seedlings that did not grow as tall as if it were just soaked in gibberellins.

A) D seedlings should mature and produce seeds, otherwise, where did the D seedlings come from in the first place?.

C) There is no indication that the stems of the T plants are not as strong as those of the DG plants.D)There was no measurement between the internodes.

DIF: 84% correct OBJ: LO 2.35 NAT: SP 4.2 NOT: 1986 #104 6. ANS: A

The coleoptile bent away from the side with increased chemical growth factor concentration. (auxin)

B) The coleoptile did not bend toward the side with the increased concentration of growth factor. It bent in the opposite direction.C) The centered agar block without any growth factor did not cause an increase in height.D) The centered agar block with growth factor did not bend in either direction


The correct answer is D. Coleoptiles A, E, and F are controls. A is a control as it ensures that the coleoptile does not bend on its own. E is a control as it ensures that the agar slab by itself does not cause the coleoptile to bend. F is a control as it ensures that the agar slab offset by itself does not cause the coleoptile to bend.

OBJ: LO 2.36 | LO 2.2.1 NAT: SP 4.1 | SP 6.1 8. ANS: B

The next logical step in this investigation is to determine which side of the tip of the plant is making the growth factor.

A) Darwin’s experiments already demonstrated that the growth factor is made in the tip of the plant. C) Mitosis does not occur in this response but rather it is the enlarging of the cells on the dark side of the plant. I f mitosis were inhibted on the dark side and not the light side then the plant would bend in the opposite direction. D) The stimulus for the curvature of the is light as shown by Darwin’s experiment

Graphic is from Campbell.

OBJ: LO 2.36 | LO 2.21 NAT: SP 6.1 | SP 4.1

ID: B

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9. ANS: DAlternating crops with legumes such a soybeans is a common practice used by farmers to restore nitrogen to the soil. Legumes such as soybeans have nodules on their roots. Root nodules are a swelling on the root of a leguminous plant, such as the pea or clover, that contains bacteria of the genus Rhizobium, capable of nitrogen fixation. A symbiotic relationship between the roots and nitrogen fixing bacteria is thus established. Excess fixed-nitrogen is returned to the soil as a by-product of this relationship,

A) Allowing the fields to lay fallow every other year would not change the nitrogen content.B) Rotating the species of corn planted every other year would continue to decrease the amount of nitrogen in the soil. C) Burning the remains of the corn stalks would not increase the nitrogen content of the soil.

Graphic http://passel.unl.edu/pages/informationmodule.php?idinformationmodule=1130447042&topicorder=2&maxto=8&minto=1

OBJ: LO 2.40 NAT: SP 7.2 10. ANS: B

The surface area of mycorrhizae significantly increases the plants ability to obtain phosphate. With both the phosphate and mycorrhizae the plant’s weight increased the most.

A) Compare the plants without mycorrhizae and phosphates added and then plants with neither plants or mycorrhizae. It clearly shows the mass of the plants increase with phosphate.C) A linear relationship does not exist between the amount of phosphate added to the soil and the dry mass of the seedlings. When the amount of phosphate added to soil is doubled, the dry weight of the plants do not double. D) The mycorrhizae obtains the phosphate and without the phosphate, there will be no, or very little, increase in the dry weight of the seedling

Data Hillis


The data table indicates that the red and far-red light cancel out each other’s effects. If the alternation sequence ends with red light, roughly the same number of seeds germinate as in the light. If the sequence ends with far-red light, the number of germinating seeds is greatly reduced.

A) The time of exposure to each type of light is the same for each group.B) The length of time in darkness does not seem to be a variable in determining seed germination.D) Indeed, lettuce seeds should be spread on top of the soil to allow exposure to light in order to germinate, but this is not necessarily in response to the red and far-red lights.

OBJ: LO 2.21 NAT: SP 4.1

ID: B

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12. ANS: C

This is a diagram of the alternation of generation of plants. The first cell of the sporophyte generation is the zygote.

A) the microspore mother cells is haploidB) the pollen grain or male gametophyte is haploidD) the polar nuclei are haploid


The concentrations given in the quesiton require that the nutrient is moving against a concentration gradient. This would require active transport.

A) Osmosis refers to movement of waterB) Facilitated transport moves material in the direction of a concentration gradient.D) Passive transport moves material move material in direction of a concentration gradient.


The stomatal opening involves a proton gradient and a potassium ion (K+) pump.

A) The bulk flow mechanism is responsible for movement of material through phloem. B) Osmosis is responsible for water absorption by the roots. D) Transpiration pull is the mechanism for movement of material through xylem.


ID: B

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15. ANS: AIt is important to realize you are being asked for the NET flow of water. Water always moves into the direction of the lowest (more negative) water potential. Since the water potential of the sucrose solution is lower (more negative) than that of the root tissue, water will move out of the tissue into the solution.

B) Water, with regard to water potential, moves from an area of higher water potential to an area of lower water potential. The water potential of the sucrose is lower (more negative) than that of the root.C) Water, with regard to water potential, moves from an area of higher water potential to an area of lower water potential. The water potential of the sucrose is lower than that of the root.D) ATP is not involved in this movement.

OBJ: LO 2.25 | LO 2.12 NAT: SP 6.2 | SP 1.4 16. ANS: C

The driving force of the movement of water through xylem is transpiration pull. Transpiration is the loss of water through the stomata of the leaves due to water potential in the atmosphere being less than in the tissue of the leaf. In addition, the reason that water moves up the stem involves the hydrogen bonding between water molecules (cohesion). This makes the water molecules attracted to one another, or cohesive. When one molecule moves, it causes other water molecules to move because of the hydrogen bonding. This creates tension in the column of water. Lastly, water molecules are attracted to the sides of the xylem vessels because of the cellulose walls, which are polar. The acronym to help you remember how water moves up xylem tubes is the T.A.C.T theory. (T.ranspiration pull A.dehsion C.ohesion T.ension).

A) Air pressure is the atmospheric pressure as indicated by a barometer. Not a plausible answer.B) Capillary action is the movement of water up a small tube due to the polarity of the water molecules attracted to the polarity of the tube. It is found in xylem, not phloem. Again, this is not enough pressure to move water up a 15-meter tree. D) Xylem is dead tissue so active transport cannot be a factor as it requires ATP.


The carbon-14 from the radioactive carbon dioxide is incorporated into glucose molecules and other organic molecules in cells such as mesophyll cells . Once the carbon is fixed it can be transported to other parts of the plant. Most organic molecules are transported through the phloem.

A) The xylem transports water and inorganic material, thus it would have very little carbon.C) Epidermal cells do not have chloroplasts, thus cannot conduct photosynthesis and fix the carbon. D) Root hairs do not have chloroplasts, thus cannot conduct photosynthesis and fix the carbon and are also involved in absorbing water and inorganic ions from the soil.


The correct answer is C. It is not unusual for halophytes to actively secrete salt out of the plant via salt secreting glands.A. The leaves of halophytes tend to be small and thick to better store water. This water can be used to dilute the saline solution. B. The leaves do have waxy cuticles but there are still stomata. Transpiration can still occur.D. The xylem always move water from the roots to the leaves.


ID: B

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19. ANS: DEndosperm supplies the nutrients that seed embryos need to survive periods of dormancy. Without endosperm, the seed will have to germinate very quickly in order to survive.

A) If seeds lacked endosperm, then there would not be nutrients for the embryo to use for long term storage during dormancy.B) The seed embryo is dependent upon the endosperm for germination.C) Brightly colored fruit is not a characteristic of seeds of orchids.

OBJ: LO 1.2 NAT: SP 2.2 | SP 5.3 20. ANS: C

The seed coat should be resistant to the animals’ digestive enzymes else the seed will not be intact in excrement.

A) The fruit may provide vitamins and nutrients to animals.B) If the plant had secondary compounds that irritated the lining of the animals mouth, it is very likely that the animal would not eat the fruit.D) If the exterior of the seed coat had barbs or hooks, the seed would be very hard to eat.

OBJ: LO 1.2 NAT: SP 2.2 | 5.3 21. ANS: A

The function of root hairs is to increase surface area for the absorption of water and minerals.

B) Root hairs are extensions of epidermal cells and do not provide support for the root.C) Carbohydrates are stored in the stele of the roots.D) Roots of legumes form nodules.

Graphic http://bio1152.nicerweb.com/Locked/media/ch35/root.html

OBJ: LO 2.7 NAT: SP 6.2 22. ANS: A

The clay pot represents the leaves of a plant. Evaporation through the leave’s stomata is called transpiration. Transpiration pull is the mechanism by which water is pulled up a stem.

b. The plant in the apparatus does not have roots.c and d. The tube has only water and is not involved in movement up the phloem.


The pores in the clay part are analogous to the stomata found in the leaves of the plant. Both allow water vapor to exit the apparatus.

There are no roots in this setup. Guard cells regulate the size of the stoma but they are not the pores themselves.


ID: B

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24. ANS: C


Tube cells are haploid and should have n number of chromosomes; and n = 8.


The endosperm is a result of double fertilization. Endosperm is triploid or should have 3n and n = 8 or 24 chromosomes.


More than you want to know...

When conditions are conducive to stomatal opening (e.g., high light intensity and high humidity), a proton pump drives protons (H+) from the guard cells. This means that the cells' electrical potential becomes increasingly negative. The negative potential opens potassium voltage-gated channels and so an uptake of potassium ions (K+) occurs.

A) Low levels of oxygen do not necessarily stimulate stomatal opening but high levels of CO2 can stimulate the opening of the stomata.C) An increase in the humidity of the leaf can cause the stoma to open.D) Exposure to blue light can stimulate the opening of the stoma.


DNA contains nitrogenous bases.

A) Glucose is a monosaccharide that does not contain any nitrogen.B) Cellulose is polysaccharide that does not contain any nitrogen.C) Oils are found in certain seeds do not contain any nitrogen.

Graphic Campbell

OBJ: LO 2.8 | LO 2.9 NAT: SP 4.1 | SP 1.1 | SP 1.4

ID: B

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29. ANS: BThe surface area of the leaves found in desert plants is reduced. In certain succulents the leaves present as spines. A. Many desert plants may have their stomata open at night for gas exchange and keep them closed during the day to minimize evaporation thus conserving water. C. Many desert plants may have a fibrous root system spreading over a large area to collect as much water as possible.D. Mycorrhizae function in providing certain minerals to the roots.

OBJ: LO 1.2 NAT: SP 2.2 | SP 5.3 30. ANS: C

Many angiosperms are pollinated by animals.

A) Angiosperms ovules are enclosed in a cellular jacket.B) Angiosperms have sperm cells that are delivered by the pollen tube.D) Angiosperms do not produce cones.


Water molecules are held together by cohesion of hydrogen bonds so the movement of one molecule causes a pull on them all.

A) Transpiration cause a negative pressure on the continuous column of water in the plant’s xylem.C) Gas bubbles in the xylem would disrupt the pull of the column of water that depends on the hydrogen bonding of the water molecules.D) Water molecules on the margins of leaves are caused by root pressure, not transpiration pull.

OBJ: LO 2.8 NAT: SP 4.1 32. ANS: A

The correct answer is A as increases in the length of a plant arises from the apical meristem. The height of the nail in the tree will still be at 1.5 meters above the ground. The height of the nail would not be changed.





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3. A student was asked to determine the number of stomata per cm2 on the lower leaf surface of Chlorophytumcomosum more commonly known as the airplane plant. The student measured the diameter of the field of view

of the microscope at 100× as shown below on the left and determined it to be 1.5 mm. The student then carefullyremoved the lower epidermis of the leaf and viewed the tissue under the microscope at a total magnification of




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2. An experiment is performed where plants are exposed to a mixture of gases for several hours and then analyzed.One of the gases included in the mixture is radioactive CO2 that contains the C-14 isotope of carbon rather than theC-12 isotope. Which of the following structures would contain the highest concentration of organic moleculescontaining the C-14 isotope?


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3. While studying transpiration, a scientist used a dendrometer to record the small daily changes in the diameter of atree trunk at two different heights (2 meters and 3 meters) above the ground at the same time. The diameterdecreased in the daytime and first occurred at the 3 m location.


A) An overcast day without much wind.B) A cool eveningC) Increased transpirationD) A humid afternoon

4. A certain short-day plant flowers only when days are less than 12 hours long. Which of the following would causeit to flower?


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predictable.



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of the seedlings.

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A) The closing of stomata at night to prevent evaporative water loss.B) The reduction in surface area of leaves to reduce water loss. C) The formation of a fibrous root system spreading over a small area to prevent

dehydration.D) The association of mycorrhizae with the root system to prevent water loss.

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direction.



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A) They require extensive periods of dormancy during which the embryo develops. B) The developing embryo within is dependent upon the gametophyte for nutrition.C) They are surrounded by brightly colored, sweet fruit. D) They germinate very soon after being released from the ovary.

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A) The presence of low levels of O2 in the leaf. B) Exposure of the leaf to sunlight.C) A decrease in humidity in the leaf’s

environment.D) Exposure of the leaf to green light.











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A) Water flows from the root tissue into the sucrose solution.B) Water flows from the sucrose solution into the root tissue. C) Water flow will be in both directions thus the concentrations would remain unchanged.D) Water flow occurs only as ATP is hydrolyzed in the tissue.







germinate.




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AP* Biology: Botany


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direction.

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A) 4B) 8C) 16D) 24


A) 4B) 8C) 16D) 24




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predictable.



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of the seedlings.







germinate.

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20.




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A) The presence of low levels of O2 in the leaf. B) Exposure of the leaf to sunlight.C) A decrease in humidity in the leaf’s

environment.D) Exposure of the leaf to green light.



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A) A 9-hour night and 15-hour day with 1 minute of darkness after 7 hours

B) An 8-hour day and 16-hour night with a flash of white light after 8 hours

C) A 13-hour night and 11-hour day with 1 minute of darkness after 6 hours

D) A 12-hour day and 12-hour night flash of red light after 6 hours








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ExamView - AP Biology Botany MC - Weeblyrhsscience.weebly.com/uploads/1/2/4/9/12490351/ap...Version A 5 9. An experiment is performed where plants are exposed to a mixture of gases