Test Bank for Botany An Introduction to Plant Biology, Seventh Edition James D. Mauseth

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 1—Introduction to Plants and Botany

Multiple Choice

1. Rose plants often get black spots on their leaves. This is an: A) observation.

B) hypothesis.

C) theory.

D) law.

E) explanation. <Answer: A> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

2. Scientists can use the scientific method to study all of the following except: A) how a plant produces sugars.

B) the interpretation of a painting.

C) the chemical composition of paint.

D) how an ecosystem functions.

E) what makes chili peppers hot. <Answer: B> <A-head: Areas Where the Scientific Method Is Inappropriate> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

3. Which of the following is not part of the scientific method?

A) Explaining phenomena controlled by unknowable forces B) Testing of all proposed explanations

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

C) Controlled experiments D) Explaining observable phenomena E) Carefully documented observations <Answer: A> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

4. Which of the following is the probable sequence in which organisms evolved?

A) Prokaryotic bacteria → eukaryotic algae → cyanobacteria → land plants B) Eukaryotic bacteria → cyanobacteria → eukaryotic algae → land plants C) Cyanobacteria → prokaryotic bacteria → eukaryotic algae → land plants D) Cyanobacteria → eukaryotic algae → prokaryotic bacteria → land plants E) Prokaryotic bacteria → cyanobacteria → eukaryotic algae → land plants <Answer: E> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

5. The first organisms to evolve have been dated at about: A) 3.5 billion years and were much like present-day bacteria.

B) 2.8 billion years and were photosynthetic like present-day cyanobacteria.

C) 3.5 billion years and had a nucleus.

D) 1.5 billion years and were terrestrial.

E) 3.5 billion years and possessed organelles which had specialized functions. <Answer: A> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

6. All of the following contribute to carbon dioxide additions to the atmosphere except: A) animals, fungi, and bacteria, which produce carbon dioxide as a by-product of respiration.

B) burning wood, coal, natural gas, and other fossil fuels.

C) volcanic eruptions.

D) formation of calcium carbonate algal shells.

E) magma eruptions at mid-ocean ridges. <Answer: D> <A-head: Concepts> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

7. Which of the following adaptations enable evergreen species to survive the physiological drought that occurs during freezing winter months?

A) The aboveground shoot system dies while maintaining a viable subterranean system.

B) The leaves are shed.

C) The leaves have extra thick cuticles.

D) The subterranean root system goes dormant.

E) The bark is shed. <Answer: C> <A-head: Diversity of Plant Adaptations> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

8. The population of grass plants growing in an open field: A) share identical genetic information.

B) carry out a metabolism that is uniquely biological and unrelated to the laws of physics and chemistry.

C) are adapted to grow and reproduce in the conditions of the open field.

D) have well protected fibrous leaves that are no longer capable of photosynthesis.

E) lack a means of storing and using information. <Answer: C> <A-head: Using Concepts to Understand Plants>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Difficult> <Subject: Chapter 1> <Taxonomy: Analysis>

9. Which of the following is a logical question to ask concerning the biological phenomena that roots absorb water?

A) Do plants have sources of energy other than sunlight?

B) Do plants create carbohydrate reserves in their root systems?

C) Do plants have alternative pathways by which water can be absorbed?

D) Are roots fibrous and tough in order to be unpalatable to insects?

E) Are some leaf modifications intended to trap insects as a source of nutrition? <Answer: C> <A-head: Plants versus the Study of Plants> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

10. All plants have a scientific name consisting of: A) a minimum of three descriptive adjectives.

B) the family name and genus name.

C) the phylum name and a specific epithet.

D) the genus name and a specific epithet.

E) the species name and subspecies name. <Answer: D> <A-head: Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

11. Derived features that have evolved more recently from ancestral features are known as: A) plesiomorphic B) apomorphic C) eukaryotic

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

D) prokaryotic <Answer: B> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

12. Which of the following phrases is the correct description of a scientific theory?

A) A hypothesis that has been proven over and over again and becomes generally accepted as truth B) An educated statement that makes a prediction that can be tested C) logical explanations for simple observations without verification D) the principle of never being certain of a conclusion and always being willing to consider new evidence.

E) assuming that processes or structures have a purpose <Answer: A> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

True/False

13. The scientific method can, ultimately, explain any concept. <Answer: False> <A-head: Areas Where the Scientific Method Is Inappropriate> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

14. At one time, fungi were classified as plants.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: True> <A-head: Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

15. The idea that mitochondria and chloroplasts originated as independent, prokaryotic organisms is now a law. <Answer: False> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

16. Eukaryotes with mitochondria and chloroplasts evolved into algae and plants. <Answer: True> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

17. The principles of chemistry and physics apply only to nonliving things, not to plants. <Answer: False> <A-head: Using Concepts to Understand Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

Fill-in-the-Blank

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

18. When new features arise through mutation, is the process by which the new features are eliminated from the population or passed on to future generations. <Answer: natural selection> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Analysis>

19. In contrast to most observations, which are reasonably accurate and trustworthy, are human constructs based on observations, intuition, and expectations. <Answer: interpretations> <A-head: Plants versus the Study of Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

20. Assuming that plants have human characters such as decision-making capabilities is called . <Answer: anthropomorphism> <A-head: Using Concepts to Understand Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

21. Numerous scientific studies have shown that all plants are composed of cells. This statement is an example of a well-supported . <Answer: theory> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

22. A key evolutionary step in the history of life occurred when the DNA was packaged within its own nucleus, creating a group of organisms called . <Answer: eukaryotes> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Easy> <Taxonomy: Recall>

Matching

23. Match the following terms with their examples.

A) Logical extension of an observation without verification <Answer: Speculative philosophy> B) Creation of species by God <Answer: Religious system> C) Complex reactions of photosynthesis were developed by a wise creator <Answer: Intelligent design> D) Investigates a testable hypothesis <Answer: Scientific method> E) Willingness to consider new evidence that is not in support of an established theory <Answer: Skepticism> [F] Mountains erode over time but the law of gravity is invariable <Answer: Constancy and universality> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

24. Match the following terms with their definitions or examples.

A) Three groups into which all organisms are classified. <Answer: Domains> B) Bacteria, cyanobacteria, and archaeans <Answer: Prokaryotes>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

C) Cells with nuclei <Answer: Eukaryotes> D) Protozoans, fungi, and animals <Answer: Eukaryotic cells without chloroplasts> E) Protists <Answer: Eukaryotic cells with chloroplasts> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

Essay

25. Explain why many scientists oppose the addition of intelligent design to science curricula. <Answer: The fundamental premise of intelligent design is that only an intelligent being could create the elaborate structures and complex phenomena of life. This premise does not allow for the testing of alternate hypotheses, the planning of future experiments, and developing a more detailed understanding of the process to be studied. In contrast, the scientific method allows scientists to develop hypotheses that may be tested and then supported or not based on observations.> <A-head: Scientific Method> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

26. Microtubules are long, tubular structures found in virtually all eukaryotic cells. They are composed of a protein called tubulin, which is remarkably similar in all of the eukaryotic cells studied so far, whether they are fungal, plant, animal, or algal cells. Is tubulin an apomorphic or plesiomorphic feature? Explain your Answer. <Answer: Tubulin is likely a plesiomorphic feature, as it likely evolved before the fungal, plant, animal, and algal lineages diverged. Features that appear relatively unchanged over evolutionary time are plesiomorphic features. Apomorphic features evolved more recently and are less likely to have evolved independently among the four lineages described above.> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Difficult> <Taxonomy: Analysis>

27. Consider the statement “Leaves absorb sunlight.” Use the fundamental concepts related to the study of plants to explain how this relationship is based on the metabolic requirements of the plant and not a decision-making capacity of the plant. <Answer: Some of the information in the plant’s genes causes the plant to produce leaves. The structure and metabolism of leaves results in the absorption of carbon dioxide and sunlight to create organic compounds used by the plant. Plants have leaves because they inherited leaf genes from their ancestors, not in order to absorb carbon dioxide and sunlight.> <A-head: Using Concepts to Understand Plants> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

28. Describe two human–plant interactions that cause global warming. Describe two human– plant interactions that could reverse global warming. <Answer: <Answers will vary but may include conserving existing trees, growing more trees, and engineering long-lived plants with superior carbon dioxide absorption properties.> <A-head: Concepts> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

29. Explain some of the evidence scientists use to show that not all organisms evolve at the same rate. <Answer: Relictual features are those that remain relatively unchanged over evolutionary history. In contrast, derived features are those that have evolved more recently. Some plants have more relictual features such as ferns, whereas others, like angiosperms, have more derived features. Angiosperms and ferns evolved at different rates.> <A-head: Origin and Evolution of Plants> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

30. Consider the diversity of physical gradients presented by a mountain. Explain some of the stresses found at different elevations and how specific plant adaptions enable them to survive in different elevational zones. <Answer: The higher elevations are cold and windy; therefore, we would expect short-statured plants that can withstand extreme cold to be close to the boundary layer of the earth. The steep slopes tend to be dry, so we might expect conifers that are adapted to dry conditions. Water accumulates in the flat valley bottoms, so sedges and rushes could thrive, as these plants have special root systems that allow them to persist in water-logged environments.> <A-head: Diversity of Plant Adaptations> <Subject: Chapter 1> <Complexity: Difficult> <Taxonomy: Analysis>

31. Which are more people likely to agree on, observations about a particular plant or explanations of the observations? <Answer: Observations are reasonably accurate and trustworthy if the observer was careful and reported truthfully. Interpretations are based on human constructs involving observations, intuition, previous experience, calculations, and expectations. Observations tend to be more objective and thus several people can make similar observations. Interpretations are <Subjective and can vary from one individual to the next.> <A-head: Plants Versus the Study of Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

32. Explain why there is difficulty in creating a singular definition of a plant. <Answer: Botany is the scientific study of plants, but plants have many types and variations.

Most plants have leaves, stems, and roots but not all plants have flowers, although many do. A definition that is too inclusive is not useful but one that is too exclusive may be too complicated.

Confusion exists within the scientific community, as scientists have not always agreed on whether fungi are plants and whether algae should be classified within the plant kingdom.> <A-head: Plants> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Application>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

33. List the five absolute characteristics of life. Give an example of a non-living system that shares one of these characteristics. <Answer: 1. Metabolism involving the exchange of energy and matter with the environment. 2. Nonrandom organization. 3. Growth. 4. A system of heredity and reproduction. 5. A capacity to respond to the environment such that metabolism is not adversely affected. Examples of non-living systems that shares one of the characteristics of life: 1. Rivers absorb water from creeks, mix it with mud and boulders, and then “excrete” it into oceans. 2. Crystals have an orderly arrangement as do many cloud patterns, weather patterns, and ripple patterns in flowing streams. 3. Mountains and crystals grow. 4. Fires reproduce but are not alive. 5. Mountains respond to the environment by growing as geological forces push them upward and by becoming smaller as erosion wears them away.> <A-head: The Characteristics of Life> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Recall>

34. Explain why coccolithophorids are so effective at sequestering carbon in their bodies.. <Answer: Carbon dioxide is a greenhouse gas that contributes to global warming. Carbon can be sequestered in the bodies of plants and animals only for as long as the animal is alive, then it returns to the atmosphere through the carbon cycle. Coccolithophorids are microscopic algae that secrete a shell composed of calcium carbonate. After death, the dense calcium carbonate shells of coccolithophorids sink to the cold depths of the ocean where low temperatures slow their decay.> <A-head: Algae and Global Warming> <Subject: Chapter 1> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 2—Overview of Plant Life

Multiple Choice

1. The primary vegetative organs of plants are: A) stems, leaves, nodes, and axillary buds.

B) sepals, petals, stamens, and carpels.

C) sporangia, seeds cones, and pollen cones.

D) roots, stems, and leaves.

E) aboveground stems, underground roots, leaves, and flowers. <Answer: D> <A-head: Overview of Plant Structure> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

2. Which of the following statements best accounts for the fact that a single plant stem may have as many as 20 vascular bundles?

A) Vascular bundles serve a similar purpose in plants as arteries do in animals.

B) This feature provides the plant with safety through redundancy.

C) If a small mammal removes the entire top half of the plant, the vascular system will not be damaged.

D) No plant puts all of its sugars, waters, and minerals in 30 vascular bundles.

E) A single large vascular bundle would confer higher plants greater fitness. <Answer: B> <A-head: Overview of Plant Structure> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Analysis>

3. The process of respiration breaks down molecules of glucose and transfers its energy to:

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

A) carbohydrate.

B) starch.

C) water.

D) ATP. E) chlorophyll. <Answer: D> <A-head: Overview of Plant Metabolism> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

4. When leaves are shed in the autumn, this process is likely a direct result of which of the following forms of plant communication?

A) Nutrient requirements of developing fruits B) Large amounts of hormones that indicate leaves require water C) Decreased amounts of hormones that indicate leaves require water D) Root growth to increase water and nutrient absorption E) Tendrils bending toward touch <Answer: C> <A-head: Overview of Information in Plants> <Subject: Chapter 2> <Complexity: Difficult> <Taxonomy: Analysis>

5. A botanist finds an unknown plant and upon close examination finds that it is spore bearing, diploid, contains a vascular system, and does not form seeds. This plant is likely an example of a(n): A) moss or liverwort.

B) angiosperm.

C) fern or lycophyte.

D) hornwort or fern.

E) gnetophyte or cycad. <Answer: C> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Moderate> <Taxonomy: Application>

6. Which of the following groups of plants have a dominant haploid generation?

A) Ferns, lycophytes, and cycads B) Mosses, lycophytes, and ferns C) Bryophytes, gymnosperms, and angiosperms D) Mosses, liverworts, and hornworts E) Hornworts, grasses, liverworts <Answer: D> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Application>

7. Current botanical evidence suggests that early flowering plants diverged into: A) lycophytes and euphyllophytes.

B) monocots and dicots.

C) embryophytes and spermatophytes.

D) ferns, lycophytes, and eudicots.

E) basal angiosperms, monocots, and eudicots. <Answer: E> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Application>

8. Seasonal changes on the earth are largely driven by: A) energy from Earth’s hot interior.

B) the energy available in sunlight.

C) erupting volcanoes.

D) continental drift forming Central America.

E) the breakdown of limestone.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: B> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

9. Plants directly affect the temperature of the Earth’s surface by: A) absorbing carbon dioxide from the atmosphere.

B) releasing free oxygen into the atmosphere.

C) releasing carbon dioxide into the atmosphere as they respire their food.

D) releasing photosynthetic waste products into the atmosphere.

E) releasing photosynthetic waste products into the biosphere. <Answer: A> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Application>

10. Some fig (Ficus spp.) trees are only pollinated by a single species of fig wasp, and that same fig wasp requires that species of fig tree to complete its life cycle. This is an example of: A) a co-evolutionary relationship.

B) neutralism.

C) commensalism.

D) a predatory relationship.

E) a competitive relationship. <Answer: A> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Analysis>

True/False

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

11. The fundamental organization of flowers includes sepals, petals, stamens, and carpels. <Answer: True> <A-head: Overview of Plant Structure> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

12. Plants synthesize all the organic compounds needed for their structure and metabolism from glucose. <Answer: True> <A-head: Overview of Plant Metabolism> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

13. All plants produce both spores and seeds. <Answer: False> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

14. DNA is stored only in the nucleus of plant cells. <Answer: False> <A-head: Overview of Information in Plants> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

15. The gigantic continent Pangaea split into Laurasia and Gondwana before plants evolved from algae. <Answer: False> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. Energy provided by sunlight is captured by the process of

.

<Answer: photosynthesis> <A-head: Overview of Plant Metabolism> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

17. The sugar produced in the leaves through photosynthesis is carried in the , where it is transported downward to the roots and upward to fruits and flowers. <Answer: phloem> <A-head: Overview of Plant Structure> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Application>

18. Three sources from which plants receive information include DNA, other plants parts, and the . <Answer: environment> <A-head: Overview of Information in Plants> <Subject: Chapter 2> <Complexity: Easy>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Taxonomy: Recall>

19. A

is a group of plants that evolved from a common ancestor.

<Answer: clade> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

20. Life on Earth occurs at the Earth’s surface in the <Answer: biosphere> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

Matching

21. Correctly match each group of plants with its corresponding example.

A) Bryophyte <Answer: hornwort> B) Vascular cryptogams <Answer: lycophyte> C) Cone-bearing seed plant <Answer: gingko> D) Monocot <Answer: lilies> E) Eudicot <Answer: lupine> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Difficult>

.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Taxonomy: Analysis>

22. Match each term with its correct definition.

A) Neither organism benefits or is harmed <Answer: neutralism> B) Both organisms benefit <Answer: mutualism> C) One organism benefits whereas the other is neither harmed nor helped <Answer: commensalism> D) One organism harms another without receiving any benefit itself <Answer: amensalism> E) Both organisms harm each other <Answer: competition> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Easy> <Taxonomy: Recall>

Essay

23. Ash (Fraxinus) flowers are small, green and hardly apparent to the human eye. In contrast, the flowers of Datura wrightii are large and showy with fused petals forming a tube. Explain the adaptive advantages of each of these types of flowers. <Answer: Ash flowers are probably wind-pollinated. Wind-pollinated flowers often are not showy or they do not possess petals, as petal production would be a waste of energetic resources.

In contrast, Datura wrightii likely has a showy flower in order to attract a specific pollinator that will ensure pollination of other plants of the same species. The fusing of the petals further specifies the type of pollinator, allowing only those with a complementary morphology to access the nectar (if present) and thus the pollen.> <A-head: Overview of Plant Structure> <Subject: Chapter 2> <Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

24. Explain the role of glucose in the processes of photosynthesis and respiration. <Answer: Glucose stores energy and mass. It is the end product of photosynthesis that combines carbon dioxide and energy from the sun to create glucose. Glucose is the starting point of respiration that breaks down glucose into carbon dioxide transferring energy to ATP.> <A-head: Overview of Plant Metabolism> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

25. What kind of information is used by deeply buried bulbs in order for the shoots to grow upward and the roots to grow downward? <Answer: Deeply buried bulbs are responding to information in their environment. The bulbs detect gravity, which informs the roots to grow downward and the shoots to grow upward.> <A-head: Overview of Information in Plants> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

26. Explain why botanists no longer use the term gymnosperm. <Answer: Recent evidence shows that both angiosperms and gymnosperms share a common ancestor and should therefore be grouped together. Scientists suspect that seeds evolved only once and all seed plants alive today are members of the same clade. Dividing gymnosperm from angiosperms artificially divides a single clade, so botanists no longer support that division and they no longer use the term gymnosperm.> <A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Difficult> <Taxonomy: Analysis>

27. Describe three ways in which bryophytes differ from angiosperms. <Answer: Bryophytes lack vascular tissue, seeds, and flowers whereas angiosperms possess all of those features. Bryophytes are also composed of a haploid body, which contrasts with the diploid body of angiosperms.>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<A-head: Overview of Plant Diversity and Evolution> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

28. Explain how heterotrophs and autotrophs mutually benefit from the existence of the other. <Answer: Some autotrophs such as plants release oxygen as a waste product of photosynthesis.

The free oxygen in turn is used in the cellular respiration of heterotrophs to release energy from their food. The carbon dioxide released from heterotrophic aerobic respiration can be used by plants in photosynthesis.> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Recall>

29. Provide two examples of how the physical process of continental drift has affected the biosphere. <Answer: Continental drift has caused North and South America to move toward each other, forming Central America. The connection of the two continents has enabled the migration of species across the two landmasses. The creation of Central America has also changed ocean circulation patterns such that warm water was forced to flow toward the poles. As a result, plants and animals in the northern parts of the northern hemisphere have experienced global warming.> <A-head: Overview of Plant Ecology> <Subject: Chapter 2> <Complexity: Moderate> <Taxonomy: Application>

30. In Scotland, farmers have noted a higher abundance of thorny thistle in pastures with relatively high sheep densities. Explain how this scenario could be the result of predation. <Answer: Herbivores like sheep eat plants, thereby harming the plants. Some plants evolved defenses against the herbivores. The sheep are grazing on the grasses and avoiding the thistle due to their thorny defenses. Eventually due to the low grass numbers, the thistles outcompete the grasses and become the dominant plant in the fields.> <A-head: Overview of Plant Ecology> <Subject: Chapter 2>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 3—Ethnobotany

Multiple Choice

1. The first crop plant was likely to have been: A) rice.

B) wheat.

C) lentils.

D) millet.

E) potatoes. <Answer: E> <A-head: Concepts> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

2. Malted barley has been roasted in order to: A) “polish” the grain.

B) convert starch to the glucose required by yeast to produce ethyl alcohol.

C) sterilize the germ.

D) distinguish barley used for livestock feed from barley meant for human consumption.

E) produce silage for cattle. <Answer: B> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

3. Sucrose, glucose, and fructose are commonly harvested from: A) cassava, sugar cane, and sugar beets.

B) sugar cane only.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

C) sugar beets, maple trees, and sweet potatoes.

D) sugar beets, sugar cane, and maple trees.

E) sugar beets only. <Answer: D> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Analysis>

4. Linoleic and alpha-linoleic acids contain fatty acids considered essential for animal diets because they: A) are not trans fatty acids.

B) are digested slowly and evenly into the bloodstream.

C) are abundant in antioxidants.

D) contain omega-3 and omega-6 fatty acids.

E) come in the form of extracted oils. <Answer: D> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Analysis>

5. The mature fruits of black pepper owe their pungent flavor to: A) piperine.

B) eugenol.

C) capsaicin.

D) sinigrin.

E) menthol. <Answer: A> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

6. Devil’s trumpet contains atropine, which: A) disrupts movement of chloride ions in the nerve cells of the brain and spinal cord.

B) interferes with sodium metabolism in the heart.

C) causes hallucinations, seizures, and sometimes death.

D) increases the risk of cancer of the lungs, throat, pancreas, and mouth.

E) stabilizes irregular heartbeats. <Answer: C> <A-head: Plants that Provide Drugs> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Recall>

7. A drug that was originally made from willow bark and is now synthesized artificially is: A) ipecac syrup.

B) salicylic acid.

C) morphine.

D) digitoxin.

E) quinine. <Answer: B> <A-head: Plants that Provide Drugs> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

8. Fibrovascular bundles are characterized by: A) vascular bundles with phloem fiber caps.

B) fiber cells surrounding vascular bundles.

C) fiber bundles located just to the interior of the epidermis.

D) fibrous trichomes.

E) xylary fibers found in the xylem of hardwood trees. <Answer: B> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Moderate> <Taxonomy: Analysis>

9. Resins are used for: A) artificial flavoring, aromas, and rosin.

B) the production of synthetic sealers.

C) carnauba and candelilla waxes.

D) waterproofing canvas, rosin, and drying oils.

E) forming thin polymeric, protective water-resistant films. <Answer: A> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Analysis>

10. Old-growth forests are unique from other forest types in all of the following ways except: A) they have never been logged.

B) they are home to insects that inhabit large, dead trees.

C) they are comprised only of old trees.

D) mature dead trees are left on site rather than being removed.

E) they have had little mineral removal. <Answer: C> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

11. Scientists have genetically modified the plant Cucurbito pepo to be rich in beta-carotene, a precursor of vitamin A.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: False> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Recall>

12. In terms of grams of food consumed per day, the greatest part of our diet is usually comprised of carbohydrates. <Answer: True> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

13. Natural selection resulted in the evolution of plant chemicals that interact with animal nerve cells. <Answer: True> <A-head: Plants that Provide Drugs> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Application>

14. Burlap is one of the strongest plant fibers known, and its strength only increases when wet. <Answer: False> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

15. Conifer wood is predominately used for furniture, baseball bats, and tool handles. <Answer: False> <A-head: Plants that Provide Fibers, Wood, and Chemicals>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. Wheat and lentils were domesticated in an area stretching from southeastern Turkey to western Afghanistan known as the . <Answer: Fertile Crescent> <A-head: Concepts> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

17. In a process called parts behind.

, fungi attack soft plant parts, leaving the hard

<Answer: retting> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

18. The final step of tea processing is when we make a(n) hot water over the tea leaves to extract their flavor.

by pouring

<Answer: infusion> <A-head: Plants that Provide Drugs> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Application>

19.

fibers are those that occur anywhere except in hardwood xylem.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: Extraxylary> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

20. woods are appreciated for their beauty and used to create small objects of beauty like boxes and musical instruments. <Answer: Artisan> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each plant with its active compound.

A) Kola tree <Answer: Caffeine B) Marijuana <Answer: Tetrahydrocannabinol C) Common periwinkle <Answer: Vinblastine D) Foxglove <Answer: Digitoxin E) Erythroxylum <Answer: Cocaine F) Papaver <Answer: Opium <A-head: Plants that Provide Drugs> <Subject: Chapter 3 <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

22. Match each type of fiber with its example.

A) Xylary fiber <Answer: Pine tracheids B) Bast fiber <Answer: Jute C) Hard fiber <Answer: Sisal D) Seed fiber <Answer: Cotton E) Whole leaves <Answer: Longleaf pine needles <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Explain some of the conditions that led to the development of early agriculture. <Answer: At least 125,000 years ago, human ancestors learned to use fire and even cook food.

About 12,000 years ago, the earth began a period of climatic warming known as the Holocene Epoch, an interglacial period in which the glaciers retreated. Early humans began saving seeds and planting them the following growing season and practicing artificial selection on their crop plants. > <A-head: Concepts> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Analysis>

24. Why does whole wheat flour tend to go rancid more quickly than white flour? <Answer: To obtain whole wheat flour, the seed coat and fruit wall or the “bran” of the grain is ground with the embryo. The embryo contains the enzymes necessary to digest the endosperm

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

and thus the digestive enzymes end up mixed through the flour, causing self-digestion, and the flour becomes rancid. > <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Recall>

25. Why are sweet potatoes considered to be healthier than white potatoes? <Answer: Sweet potatoes contain beta-carotene, the precursor to vitamin A. They also have complex polysaccharides that are digested slowly, releasing sugar into the bloodstream gradually. They also have a lower glycemic index value than white potatoes.> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Easy> <Taxonomy: Recall>

26. What is “baking chocolate” and why is it so bitter? <Answer: Baking chocolate is chocolate liquor that has been mixed with fats. The chocolate liquor comes from roasted and ground cacao nibs. It is bitter because the milk solids and sugars characteristic of confectionary chocolate have not been added.> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Analysis>

27. Why did British soldiers stationed in India take to the habit of drinking gin and tonics in the afternoon? <Answer: Quinine, from the Cinchona officinalis tree, kills the protozoans that cause malaria.

The effective dose of quinine is bitter and distasteful. Mixing the quinine with gin makes it more palatable. British soldiers and medical workers drank gin and tonics in India as a prophylactic measure against getting malaria.> <A-head: Plants that Provide Drugs> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Application>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

28. Why is burlap often used to protect the roots of trees and shrubs used in landscaping projects? <Answer: Burlap is a geotextile that allows moisture to escape while oxygen diffuses inward. It is also biodegradable. This coarse cloth makes it easier to transport trees while protecting the root system. Trees can be planted with burlap surrounding their root ball. The roots can penetrate the burlap and after 1 to 2 years the burlap rots away.> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Moderate> <Taxonomy: Application>

29. How does engineered wood differ from sawn wood? In what kinds of products is it used? <Answer: Engineered woods are produced by gluing small pieces of wood together, whereas sawn wood is shaped and cut wood. Plywood, particleboard, and oriented strand wood are examples of engineered wood. These woods can be used in furniture, veneers, and laminated beams.> <A-head: Plants that Provide Fibers, Wood, and Chemicals> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Application>

30. Distinguish between unsaturated, saturated, and trans fats. Which is the unhealthiest and why? <Answer: In a saturated fat, each carbon is attached to two others by single bonds. Each carbon is attached to two hydrogen atoms. In an unsaturated fat, two adjacent carbons are attached by a double bond. Each carbon is only attached to one hydrogen. Trans fats are not made by plants and involve processing that causes the two parts of the carbon backbone to lie on opposite sides.

Trans fats in the diet increase the risk of heart disease, raise the level of LDL (bad) cholesterol, and reduce HDL (good) cholesterol.> <A-head: Food Plants> <Subject: Chapter 3> <Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 4—Cell Structure

Multiple Choice

1. Cell membranes are most permeable to: A) hydrophobic substances.

B) positively charged substances.

C) negatively charged substances.

D) potassium.

E) calcium. <Answer: A> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Analysis>

2. Organelles each possess a unique selectively permeable membrane that ensures ordered metabolic reactions. This is an example of the importance of A) vesicles.

B) facilitated diffusion.

C) impermeable membranes.

D) freely permeable membranes.

E) compartmentalization. <Answer: E> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

3. Which of the following is an example of a prokaryotic organism?

A) Plant

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

B) Animal C) Bacterium D) Fungus E) Protist <Answer: C> <A-head: Basic Cell Types> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

4. Oily seeds, such as peanuts, that store energy in fat molecules can convert that stored fat to the carbohydrates needed for growth in: A) nuclei.

B) chloroplasts.

C) dictyosomes.

D) glyoxysomes.

E) peroxisomes. <Answer: D> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

5. If you wanted to investigate membrane structure, you might isolate and study a plant cell’s: A) hyaloplasm.

B) tonoplast.

C) ribosomes.

D) mitotic spindle.

E) microfilaments. <Answer: B> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

6. The red color of beets is due to a water-soluble pigment called anthocyanin, which is most likely found in a beet cell’s: A) vacuole.

B) chromoplasts.

C) cytoplasm.

D) nucleus.

E) chloroplasts <Answer: A> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

7. Cells on the surface of the leaves of the carnivorous plant sundew (Drosophyllum) secrete enzymes to digest insects. These enzymes are processed by which organelle?

A) vacuole, B) nucleus.

C) dictyosomes.

D) cytosol.

E) chloroplasts. <Answer: C> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Application>

8. If the mitochondria were removed from a plant cell, what process would immediately stop in the cell?

A) Respiration B) Photosynthesis C) Lipid synthesis D) Protein synthesis E) Starch synthesis

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: A> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

9. The sperm cells are the only cells present in plants which possess: A) flagella B) chromoplasts C) amyloplasts.

D) cell walls <Answer: A> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Application>

10. The term protoplasm excludes a cell’s: A) nucleus.

B) endoplasmic reticulum.

C) cell wall.

D) mitochondria.

E) microtubules. <Answer: C> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Easy> <Taxonomy: Recall>

True/False

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

11. An absolute requirement for a cell to exist is the presence of at least one membrane. <Answer: True> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

12. Fungal cells contain most types of plastids but do not contain chloroplasts. <Answer: False> <A-head: Fungal Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

13. Once an ion from the soil crosses a plasma membrane into a root cell, that ion could move to the top of the plant without crossing another plasma membrane. <Answer: True> <A-head: Associations of Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

14. When information to build a plant cell is transcribed from DNA, the copies are in the form of messenger RNA. <Answer: True> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Easy> <Taxonomy: Recall>

15. You would expect to find the same basic types of cells in flowers and roots.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: True> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

Fill-in-the-Blank

16. Nuclei, mitochondria, and vacuoles are all examples of within the protoplasm of plant cells.

found

<Answer: organelles> <A-head: Concepts> <Subject: Chapter 4> <Complexity: Easy> <Taxonomy: Recall>

17. The lipid layer of a biological membrane doubles over to make a in which the fatty acids are away from water and the phosphate groups make contact with the water. <Answer: bilayer> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Easy> <Taxonomy: Recall>

18. A organelles. <Answer: eukaryotic> <A-head: Basic Cell Types> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

cell is one that has a membrane-bound nucleus and multiple

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

19. Plant cell walls are composed of cellulose, whereas fungal cell walls contain . <Answer: chitin> <A-head: Fungal Cells> <Subject: Chapter 4> <Complexity: Easy> <Taxonomy: Recall>

20. Collectively, intercellular spaces and cell walls comprise the the plant.

of

<Answer: apoplast> <A-head: Associations of Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

Matching

21. Match the following terms with their definitions.

A) Nothing passes through <Answer: impermeable membrane> B) Virtually anything can pass through <Answer: freely permeable membrane> C) Certain substances pass through quickly, while others pass through slowly <Answer: selectively permeable> D) Hydrophilic, charged molecules diffuse through the membrane <Answer: facilitated diffusion> E) Intrinsic membrane proteins bind a molecule and force it through the membrane <Answer: active transport> F) A vesicle fuses with the membrane, releasing its contents to the exterior of the cell <Answer: exocytosis> G) Invagination of the cell membrane forms a vesicle that carries external material into the cell <Answer: endocytosis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<A-head: Membranes> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each structure with its corresponding component.

A) Mictrotubule <Answer: tubulin> B) Chloroplast <Answer: chlorophyll> C) Ribosome <Answer: rRNA> D) Amyloplast <Answer: starch> E) Cell wall <Answer: cellulose> [F] Microfilament <Answer: actin> [G] Middle lamella <Answer: pectic substances> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Analysis>

Essay

23. What is the major advantage of cell specialization for an organism? Are there any disadvantages? <Answer: Cell specialization allows for a division of labor that makes the entire organism more efficient. A disadvantage of cell specialization is that it causes each cell to depend more on others such that damage to one part of the organism may result in the death of all cells, not just those that were directly damaged.> <A-head: Concepts>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Recall>

24. Describe the three major properties of biological membranes and how each property is a critical part of membrane function. <Answer: Membranes can grow, allowing the transport of materials. Membranes are selectively permeable, which allows for compartmentalization. Lastly, membranes are dynamic, which allows cells to change their functions over time.> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Recall>

25. Cell membranes have a discrete structure, yet are also referred to as a fluid mosaic membrane. Explain membrane structure and how it can also form a “fluid mosaic.” <Answer: All biological membranes are composed of proteins and two layers of phospholipid molecules. The hydrophilic heads of the phospholipids form hydrogen bonds with water molecules and the hydrophobic tails are buried within the interior of the membrane. Despite the clear structure, the membrane is actually a heterogeneous liquid containing a number of proteins.

Some intrinsic proteins can diffuse laterally throughout the membrane. Others interact with adjacent proteins, forming domains that are different from surrounding regions of the membrane, creating patches of differentiation.> <A-head: Membranes> <Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Application>

26. In phloem tissue, the metabolism of enucleate sieve tube elements is controlled by adjacent companion cells. How can these cells communicate? <Answer: Companion cells and sieve tube elements can communicate through cytoplasmic connections that occur through the plasmodesmata. Cells can also secrete chemical compounds that inform nearby cells of metabolic and developmental activities.> <A-head: Associations of Cells> <Subject: Chapter 4> <Complexity: Difficult>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Taxonomy: Application>

27. If pigments from spinach leaves are separated using a technique called chromatography, five pigments are isolated: 2 green chlorophylls, 2 yellow xanthophylls, and an orange carotene. Why do spinach leaves look green? <Answer: Pigments may be present in cells as discrete droplets called plastoglobuli. Lipid pigments are present in leaf cells in small amounts contained in plastoglobuli, but they are masked by the abundance of green chlorophyll, giving the leaf a green color.> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Difficult> <Taxonomy: Recall>

28. One theory suggests that mitochondria began as independent prokaryotic cells that were incorporated into a host cell and gradually evolved into present-day mitochondria. What are two pieces of evidence that support this idea? <Answer: Mitochondria have their own DNA and ribosomes. They can also divide. These features would allow simple single-celled organisms such as an early prokaryote to survive. <A-head: Plant Cells <Subject: Chapter 4 <Complexity: Difficult <Taxonomy: Analysis

29. Sometimes the outer portion of a white potato turns green. Explain what is happening in the potato cells. <Answer: As a plant organ changes, so might its plastids by extensively altering their membranes and proteins. Starchy potatoes contain large numbers of amyloplasts, which are starch-storing plastids. Amyloplasts can be converted to chlorophyll, which contain chloroplasts when exposed to light, thus turning those cells green.> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

30. Why can most plant cells increase in size more rapidly than animal cells? <Answer: Growing plant cells contain large central vacuoles that can expand rapidly, causing the cell to grow rapidly as well. Animal cells must synthesize complete protoplasm to grow, but plant cells only need to take more water into their vacuoles.> <A-head: Plant Cells> <Subject: Chapter 4> <Complexity: Moderate> <Taxonomy: Application>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 5—Growth and Division of the Cell

Multiple Choice

1. The pool of free nucleotides necessary for DNA replication is mainly synthesized during: A) G1 of interphase.

B) G1 of prophase.

C) S of interphase D) S of prophase.

E) G2 of interphase. <Answer: A> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

2. If you examine a root tip, you would find most of the cells in: A) anaphase.

B) telophase.

C) metaphase.

D) interphase.

E) prophase. <Answer: D> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

3. Which of the following is not present in interphase?

A) G3 B) S C) G1

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

D) G2 <Answer: A> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Easy> <Taxonomy: Recall>

4. If a zygote contains 12 chromosomes, how many chromosomes did the original egg cell contain?

A) 3 B) 4 C) 6 D) 12 E) 24 <Answer: C> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Analysis>

5. During meiosis, DNA replication occurs during: A) G1.

B) S. C) G2.

D) interkinesis.

E) cytokinesis. <Answer: B> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Easy> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

6. If a normal body cell of a plant contains 5 picograms of DNA, then that cell at the end of prophase of mitosis contains: A) 2.5 picograms.

B) 5 picograms.

C) 10 picograms.

D) 15 picograms.

E) 20 picograms. <Answer: C> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

7. Which of the following is the order of events in one cell cycle?

A) Interphase → metaphase → anaphase → prophase → telophase B) Prophase → metaphase → anaphase → telophase C) Interphase → prophase → metaphase → anaphase → telophase D) Prophase → metaphase → anaphase → interphase → telophase E) Interphase → prophase → anaphase → telophase <Answer: C> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

8. In a microscope, you observe a cell with condensed chromosomes in a spherical group, no nucleolus, and no nuclear envelope. Each chromosome consists of two chromatids. This cell is most likely in: A) interphase.

B) prophase.

C) anaphase.

D) telophase.

E) metaphase. <Answer: B>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Application>

9. What is the direct source of new membrane in a dividing cell’s cell plate?

A) Plasma membrane B) Nuclear envelope C) Mitochondria D) Tonoplast E) Dictyosome <Answer: E> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

10. Which of the following would tell you that a cell is in metaphase I of meiosis?

A) Diploid number of chromosomes; homologous chromosomes paired B) Haploid number of chromosomes C) Haploid number of chromosomes; chromosomes paired D) Diploid number of chromosomes; two groups of homologous chromosomes; each chromosome composed of two chromatids E) Haploid number of chromosomes; two groups of chromosomes; each chromosome composed of one chromatid <Answer: A> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

11. In general, the shortest phase in the cell cycle is G1. <Answer: False> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

12. DNA replication occurs during S phase. <Answer: True> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Easy> <Taxonomy: Recall>

13. Meiosis is a part of cytokinesis. <Answer: False> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Easy> <Taxonomy: Recall>

14. A cell cannot divide unless centrioles are present in the cell. <Answer: False> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Easy> <Taxonomy: Recall>

15. In some eukaryotic cells, the nuclear envelope does not break down during prophase of mitosis.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: True> <A-head: Cell Division in Algae> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. When the nuclei of some root cells are much larger than those of adjacent root cells, this likely due to . <Answer: endoreduplication> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

17. Following duplication, the protein at the centromere.

connects the sister chromatids

<Answer: cohesion> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

18. When chromosomes are arranged in the synaptonemal complex, occurs and pieces of DNA are exchanged between homologous chromosomes. <Answer: crossing-over> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

19. The chromosomes.

is the point of attachment of spindle microtubules to

<Answer: kinetochore> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

20. Not all green algae form a phragmoplast during cytokinesis; some algae form a , a group of microtubules oriented parallel to the plane of division. <Answer: phycoplast> <A-head: Cell Division in Algae> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Application>

Matching

21. Match each stage of prophase with its distinguishing characteristic.

A) Zygotene <Answer: Chromosomes pair> B) Pachytene <Answer: Crossing-over occurs> C) Leptotene <Answer: Chromosomes begin to condense> D) Diakinesis <Answer: Homologous chromosomes become untangled> E) Diplotene <Answer: Homologous chromosomes begin to move away from one another> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Recall>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

22. Match each event with its corresponding phase.

A) Chiasmata form <Answer: prophase I> B) Centromere division <Answer: metaphase II> C) Separation of homologs <Answer: anaphase I> D) Reappearance of nuclear envelope <Answer: telophase II> E) Separation of chromatids <Answer: anaphase II> [F] Homologous chromosomes align at center of cell <Answer: metaphase I> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

Essay

23. Why are genes packaged in chromosomes rather than existing as individual genes? <Answer: It would be difficult to organize and correctly copy 30,000 to 60,000 pieces of DNA. Chromosomes organize genes, making the duplication process more efficient. The histone proteins that complex with DNA to form chromosomes also provide protection and structure.> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Moderate> <Taxonomy: Application>

24. Is interphase of the cell cycle really a resting phase? Why or why not?

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: Interphase was originally named a “resting phase” because researchers were focused on the division phases of the cell cycle. The cell is actually metabolically active during interphase because it is synthesizing nucleotides, duplicating DNA, making tubulin, and preparing for cell division.> <A-head: Growth Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Application>

25. Tetraploid cells have twice the normal number of chromosomes. This condition is often accomplished in a lab by exposing cells to colchicine. Why does this work? <Answer: This drug disrupts microtubules and thus prevents the formation of a spindle.

Consequently, the duplicated chromosomes fail to separate in mitosis. Onion cells exposed to colchicine for several days may have over 1,000 chromosomes inside.> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

26. Which division process is more like mitosis—meiosis I or II? Provide at least two ways in which they are different. <Answer: Mitosis and meiosis II are most similar to each other in that both produce daughter cells fairly similar to the mother cell. However, the final product of meiosis II is four haploid daughter cells, whereas the final product of mitosis is two diploid daughter cells. Prophase II of meiosis II is not always necessary, unlike the requirement of prophase to proceed metaphase in mitosis.> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

27. How could you tell whether an onion root cell is in metaphase of mitosis or metaphase I of meiosis? In metaphase of mitosis or metaphase II of meiosis?

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: In metaphase of mitosis, the homologous chromosomes are not paired as they are in metaphase I of meiosis. In metaphase II of meiosis, there will be half the number of chromosomes present as there are in metaphase of mitosis.> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

28. Describe two ways in which prokaryotic cell division differs from that of eukaryotic cell division. <Answer: There is no nucleus in prokaryotic cells and thus no breakdown of the nuclear envelope in the prokaryotic cell division as there is in eukaryotic cell division. In prokaryotes, cytokinesis occurs through infurrowing of the plasma membrane. The cytokinesis of eukaryotes may occur through infurrowing, formation of a phycoplast, formation of a phragmoplast, or some combination of these methods.> <A-head: Cell Division of Prokaryotes> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

29. Why do some scientists believe that dinoflagellates originated earlier in the evolution of eukaryotes than plants? <Answer: Dinoflagellates have unusual nuclear characteristics that may represent evolutionary lineages that originated earlier in the history of eukaryotes than did plants. The nuclei of dinoflagellates do not possess histones and their chromosomes are always condensed, yet histone presence and nucleosome structure is constant in all other eukaryotes. Also, in contrast with plants, there are gaps in the nuclear envelope of dinoflagellates that allow microtubules to pass through the nucleus.> <A-head: Cell Division in Algae> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Application>

30. A farmer notices a white spot occurring on a developing tomato leaf. What is a reasonable explanation for this phenomenon?

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: Plastids contain the DNA necessary for their growth and functioning. All cells must receive at least one plastid during division in order to produce this organelle. A cell missing a plastid cannot contain chloroplasts and there will not be green. When these cells reproduce, their daughter cells will also lack plastids, appear white, and cause a white spot to develop on the leaf.> <A-head: Division of Chloroplasts and Mitochondria> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Application>

31. The amount of DNA per cell at the beginning of G1 is 0.40 picograms. Draw a graph plotting the amount of DNA per cell vs. phase of the mitotic cell cycle. <Answer: The graph should plot the amount of DNA (picograms/cell) on the y-axis and mitotic phase (prophase, metaphase, anaphase, and telophase). The graph will show a flat line of 8 picograms of DNA per cell for all four mitotic phases.> <A-head: Division Phase of the Cell Cycle> <Subject: Chapter 5> <Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

CHAPTER 6—Tissues and the Primary Growth of Stems

Multiple Choice

1. In the coastal deserts near Lima, Peru, fog is frequent, but rain never falls. Which of the following adaptations would be advantageous for a plant living in this type of habitat to possess?

A) leaves with extensive surface area.

B) leaves that absorb moisture from the air.

C) a mass of green photosynthetic roots connected to a tiny portion of stem.

D) a shoot that becomes active only when flowers are to be produced.

E) production of defensive spines to deter herbivory. <Answer: B> <A-head: Concepts> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

2. A peach pit consists of the inner layer of fruit surrounding a seed, giving the seed protection.

The fruit inner layer is most likely composed of: A) parenchyma cells with thick primary walls.

B) sclereids.

C) fibers.

D) collenchyma.

E) tracheids. <Answer: B> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

3. If each stem node has two leaves and each leaf pair is at a right angle to the pair below it, the phyllotaxy is said to be:

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

A) alternate.

B) alternate and distichous.

C) opposite and distichous.

D) opposite and decussate.

E) spiral. <Answer: D> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

4. On a per unit basis, which of the following would be the most expensive to build energetically?

A) A moss plant that is entirely parenchyma B) A herbaceous plant that consists mainly of parenchyma, collenchyma, and vascular tissue C) A cactus that consists mainly of parenchyma but also contains vascular tissue and sclerenchymatous spines D) A woody plant composed mainly of xylem, plus some parenchyma and phloem E) An aquatic plant that contains mainly aerenchyma, in addition to vascular tissue <Answer: D> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

5. The vessel elements of an aquatic plant would most likely have what type of secondary wall?

A) Scalariform thickenings B) Annular thickenings C) Solid with circular bordered pits D) Reticulate thickenings E) All of these are correct. <Answer: B> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Moderate> <Taxonomy: Analysis>

6. The vascular system of a plant differs from the vascular system of an animal. One way they differ is that in plants: A) all transport tubes are composed of living cells.

B) cells move through the system without encountering any barriers.

C) solutions move across semi-permeable barriers constantly.

D) a hole in the system could let all of the solution leak out.

E) there is only one type of transport tissue. <Answer: C> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

7. Mature tracheary elements are unique in a plant because they: A) have an elongated shape.

B) are alive and transport mainly sugars.

C) are dead and transport mainly water and minerals.

D) have a primary wall only and are dead.

E) are alive and lack a nucleus. <Answer: C> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

8. The subapical meristem produces young cells of the pith and cortex called: A) protoderm B) provascular tissue.

C) ground meristem.

D) metaxylem

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

E) protoderm and ground meristem. <Answer: C> <A-head: Stem Growth and Differentiation> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

9. In a plant, the least amount of resistance occurs as water moves between: A) tracheids through pit-pairs.

B) tracheids through perforations.

C) tracheids through plasmodesmata.

D) vessel elements through pit-pairs.

E) vessel elements through perforations. <Answer: E> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

10. Which of the following is wrong about stem vascular bundles?

A) Can be scattered or located in a ring B) Cells of primary xylem or larger than cells of primary phloem C) Relative amount of cell types with the bundles can vary D) Phloem is found to the inside and xylem to the outside E) Can be interconnected with other vascular bundles <Answer: D> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

True/False

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

11. Shoots are not important in energy storage; their primary purpose is to provide support for the leaves. <Answer: False> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

12. Because herbaceous plants are composed of only the primary plant body, they are only capable of becoming smaller plants with primarily annual life cycles. <Answer: False> <A-head: Concepts> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

13. Pits are areas in a cell wall where there is no primary wall. <Answer: False> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

14. All vascular bundles contain both xylem and phloem and are said to be collateral. <Answer: True> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

15. Protoxylem cells have annular secondary walls that enable them to stretch as cells around them grow. <Answer: False> <A-head: Stem Growth and Differentiation> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

Fill-in-the-Blank

16. There are low depressions in the secondary cell wall as it is deposited along the inner surface of the primary wall; these areas become in the secondary wall that allow lateral transfer of water and sugars between adjacent cells. <Answer: pits> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

17. Vines, with their elongated regions have the ability to explore their immediate surroundings and in this way often find sunnier sites. <Answer: internode> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

18. The

cells are responsible for nuclear control of the sieve cells.

<Answer: albuminous> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Difficult>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Taxonomy: Recall>

19. Mature epidermal cells develop from meristem.

found in the subapical

<Answer: protoderm> <A-head: Stem Growth and Differentiation> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

20.

cells are pigmented parenchyma cells involved in photosynthesis.

<Answer: Chlorenchyma> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

Matching

21. Match each structure with its definition.

A) Elongated horizontal stems with no expanded leaves <Answer: stolon B) Short shoots with thick, fleshy leaves <Answer: bulbs C) Swollen underground stems that grow for a relatively short period <Answer: tubers D) Vertical, thick stems with thin, papery leaves <Answer: corms E) Fleshy, horizontal, underground stems <Answer: rhizomes <A-head: External Organization of Stems> <Subject: Chapter 6>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Complexity: Easy> <Taxonomy: Recall>

22. Match each process with its complementary structure.

A) Water transport <Answer: xylem> B) Strong support <Answer: fibers> C) Photosynthesis <Answer: chlorenchyma> D) Flexible support <Answer: collenchyma> E) Sugar transport <Answer: phloem> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

Essay

23. How might a reduced shoot system evolve in land plants? <Answer: Nearly “shootless” plants may have had ancestors with roots that were more resistant to water stress. This could happen through mutations in stems and leaves that caused them to be more water conserving. Alternatively, mutations could occur that enhanced the ability of the root to perform photosynthesis.> <A-head: Concepts> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

24. Starting with the outer surface of a plant and moving inward, compare the sequence of tissues you would encounter in a mature eudicot stem and its subapical meristem.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: In a mature eudicot stem, one would expect to first encounter the cuticle when starting at the outer surface. The tissue order following the cuticle would be epidermis, cortex, phloem, xylem, cortex, and finally the pith. In the subapical meristem, one would encounter cells that are beginning to differentiate into the tissues described above. One would first encounter the protoderm followed by the protophloem, metaphloem, metaxylem, and protoxylem.> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

25. On a walk in the country, you observe two trees of the same species, one in a pasture, the other in the woods. Would you expect them to have the same shape? Why or why not? <Answer: No, I would not expect the trees growing in the different environments to have the same shape due to the growth of the axillary buds. Some axillary buds develop into branches, whereas others remain dormant. The tree growing in the shady, wooded environment is likely to have more dormant axillary buds so that resources may be concentrated on the growth of the trunk, allowing the tree to reach brighter light at the top of the forest canopy. Such a tree may have a long and narrow crown. In contrast, it may be more advantageous for most of the axillary buds of the open-grown pasture tree to grow and form leaves with access to sunlight. We might expect the crown of this tree to be round and full.> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

26. Why might a cactus be more resistant to pathogens and insects than an aquatic plant? <Answer: One might expect a cactus to possess a thick cuticle to protect it from drying out in harsh desert conditions. The cutin and wax of the cuticle provide defense against pathogens. Due to its easy access to water, the aquatic plant has less of a need for a thick cuticle and thus does not have the same level of protection against pathogens.> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

27. Why are stomata often called compromise structures? <Answer: The epidermis and cuticle protect the underlying tissues from desiccation and pathogen infection. Stomata are actual holes in the epidermis that allow for gas exchange.

Therefore the stomata “compromise” the integrity of the epidermis but they are required for carbon dioxide to enter and oxygen to leave the plant.> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

28. Many perennial plants in an area such as Pennsylvania have to survive a cold period. What structural adaptations might permit them to survive this unfavorable time of year? <Answer: Underground storage shoots such as rhizomes, tubers, corms, and bulbs allow perennial plants to go through a dormant period. To survive cold periods, many plants shed their leaves and enter dormancy. The production of new leaves, roots, and stems under favorable environmental conditions requires reserved carbohydrates, which can be stored in the subterranean shoots. Bud scales can also protect overwintering tissues.> <A-head: External Organization of Stems> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

29. Describe how collenchyma and parenchyma interact to provide support for an elongating shoot tip. <Answer: Elongating shoot tips must be long and flexible and resistant to breaking. In a shoot tip, primary walled parenchyma cells are surrounded by collenchyma cells with uneven cell wall thickenings. The turgid parenchyma exert pressure against the collenchyma, creating support for the elongating shoot.> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

30. Compare the energy and materials requirements needed to build cell walls of the three basic types of plant cells.

Test Bank for Botany: An Introduction to Plant Biology , Seventh Edition James D. Mauseth

<Answer: Parenchyma have only primary cell walls that remain thin and thus require the least energy and material investment to build. Collenchyma have a primary cell wall that remains thin in some places and thicker in others. Collenchyma require more glucose for their production than parenchyma. Sclerenchyma has both a primary and a thick, lignified secondary cell wall. These cells require the highest energy and materials requirements to build.> <A-head: Basic Types of Cells and Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

31. Compare the pathway of water movement through tracheids vs. vessel elements. Does one have an advantage over the other? If yes, why? <Answer: Tracheids obtain their water from other tracheids below them. The water moves through pit-pairs and the pit membrane, made up of the set of primary walls and middle lamella between them. Vessel elements obtain their water from adjacent vessel elements through pitpairs and also through the vessel elements below through perforations, areas where the primary wall has been digested. Water moves more quickly through vessel elements than through tracheids, which have more friction. Narrow tracheids are better at resisting the inward tension placed on them in dry environments and are thus advantageous in this situation. In humid, moist environments, there is less inward traction and vessel elements are more advantageous.> <A-head: Internal Organization of Stems: Arrangement of Primary Tissues> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 7—Leaves

Multiple Choice

1. Which of the following types of leaves would most likely minimize leaf damage due to insects?

A) Simple and petiolate B) Simple and sessile C) Compound and petiolate D) Compound and sessile E) All leaves are equally vulnerable to insect damage. <Answer: C> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

2. The major type of tissue found in foliage leaves is: A) parenchyma.

B) xylem.

C) phloem.

D) epidermis.

E) collenchyma. <Answer: A> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Application>

3. The leaf of poison ivy has three leaflets attached to the end of the petiole. This leaf is: A) simple.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) parallel C) pinnately compound.

D) palmately compound.

E) practically compound. <Answer: D> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

4. In a typical foliage leaf, the majority of photosynthesis occurs in the: A) upper epidermis.

B) palisade parenchyma.

C) bundle sheath parenchyma.

D) spongy mesophyll.

E) lower epidermis. <Answer: B> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

5. If a plant produces leaves that are oriented horizontally, which of the following might you expect to find in greater abundance on the upper side relative to the lower side?

A) Stomata B) Cuticle C) Spongy mesophyll D) Epidermal cells E) Veins <Answer: B> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. Leaves are quite vulnerable to damage by fungi and chewing insects. A leaf adaptation that would not be a deterrent to those pests would be: A) a thick cuticle on the surface of the epidermis.

B) a bundle sheath around leaf vascular bundles.

C) stomata in the epidermis.

D) trichomes on the surface.

E) a blade divided into leaflets. <Answer: C> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application >

7. If a plant leaf is exposed to radioactively labeled carbon dioxide, the radioactivity first appears in palisade and spongy mesophyll cells. The radioactivity would most likely move next to cells of the: A) midrib.

B) lateral veins.

C) minor veins.

D) midrib and lateral veins.

E) midrib and minor veins. <Answer: C> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

8. In most cases, bud scales: A) have a long petiole.

B) have a very thin cuticle.

C) are compound.

D) often have a thin layer of corky bark.

E) have extensive chlorenchyma.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: D> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

9. Leaves that are flat and stay on a plant for numerous years, have abundant fibers, have thickwalled epidermal cells, and have a thick cuticle are: A) bud scales.

B) sclerophyllous.

C) spines.

D) leaves with Kranz anatomy.

E) succulent. <Answer: B> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

10. Which of the following adaptations of succulent leaves unfortunately results in a reduced capacity for carbon dioxide uptake?: A) Krantz anatomy.

B) transparent mesophyll.

C) reduced surface-to-volume ratio.

D) subterranean chlorenchyma.

E) numerous stomata. <Answer: C> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

Copyright © 2017 by Jones & Bartlett Learning, LLC, an Ascend Learning Company

Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. A petiole is a specialized stem. <Answer: False> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

12. In a pinnately compound leaf, the leaflets are in pairs along the rachis. <Answer: True> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

13. In the midrib section of the leaf, primary xylem is located on the lower side and primary phloem is located on the upper side. <Answer: False> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

14. Conifer leaves are usually sclerophyllous. <Answer: True> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 15. Monocot leaves typically have a sheathing leaf base. <Answer: True> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. In basal angiosperms and eudicots, the veins appear in a netted pattern known as venation. <Answer: reticulate> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Recall>

17. The South American shrub Azara lanceolata appears to have two different kinds of leaves, when in actuality this plant has large true leaves and smaller, round, photosynthetic that are enlarged offshoots of the petiole. <Answer: stipules> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

18. All leaves are initiated at the shoot apical meristem with the formation of the . <Answer: leaf primordium> <A-head: Initiation and Development of Leaves> <Subject: Chapter 6> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

19. leaves are thick and fleshy with low surface-area-to-volume ratios that enhance water conservation. <Answer: Succulent> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

20. Needle-shaped cactus help protect the stem cortex.

are modified leaves of the axillary buds that

<Answer: spines> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Easy> <Taxonomy: Application>

Matching

21. Match each function with its complementary anatomical part.

A) Water distribution <Answer: vein> B) Leaflet support <Answer: rachis> C) Blade attachment to the stem <Answer: petiole> D) Photosynthesis <Answer: lamina> E) Leaf detachment <Answer: abscission zone> <A-head: External Structure of Foliage Leaves>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

22. Match each modified leaf with its primary function.

A) Foliage leaf <Answer: Photosynthesis> B) Bud scale <Answer: Protection of dormant shoot apical meristem> C) Trap leaves <Answer: Nitrogen procurement> D) Tendril <Answer: Support> E) Kranz anatomy <Answer: C4 photosynthesis> F) Spine <Answer: Protection of active photosynthetic tissue> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Describe three functions of leaves. <Answer: Leaves are responsible for photosynthesis, bud protection (bud scales), support (tendrils), nutrient storage (fleshy leaves of bulbs), and nitrogen procurement (carnivorous plant leaves).> <A-head: Concepts> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 24. Are the veins of a leaf connected to (continuous with) the vascular bundles of a stem? Why or why not? <Answer: The vascular bundles of the leaf are continuous with those of the stem. The vascular bundles distribute water from the stem into the leaf and at the same time collect sugars produced by photosynthesis in the leaf and carry them to the stem for use or storage elsewhere in the plant.> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

25. Why are a majority of foliage leaves flat and thin? <Answer: Most leaves are flat and thin and therefore have a large surface area for maximal absorption of light and CO2. Thin and flat leaves also minimize internal self-shading. Such leaves are also easy to support on fine branches, thus requiring less energy investment in support structures.> <A-head: External Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

26. Explain why cells are packed differently in the palisade mesophyll versus spongy mesophyll of a leaf. <Answer: The cells of the palisade mesophyll are packed in rows just inside the upper epidermis to provide efficient access to light. However, there are spaces surrounding these cells to provide maximum surface area exposure to carbon dioxide dissolved in the cytoplasm. The spongy mesophyll is below the palisade parenchyma and its open, loose arrangement allows carbon dioxide to diffuse rapidly from the stomata into the leaf interior.> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Application>

27. In oleander, the stomata are mainly located in crypts in the lower leaf epidermis. This suggests that oleander normally lives in what type of environment? Why?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: Crypts in the lower epidermis are epidermal depressions in the leaf that create small regions of nonmoving air. The crypts often contain trichomes. The crypts and trichomes trap water molecules and diffuse out of the open stomata and cause them to be reabsorbed by the leaf.

This feature is adaptive to dry environments where water conservation traits are advantageous. <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Analysis>

28. Why can trees seem to develop leaves overnight in the spring? <Answer: In many perennial plants, leaves are initiated during the growing season prior to the maturation of the leaf. After the initiation of the leaf, it enters a dormant stage that lasts over the winter. When the spring arrives and the conditions are conducive for growing, the leaf bud emerges from its dormant state, only requiring water absorption for growth through the swelling of the vacuole. The leaf seems to develop overnight because it was “preformed” the previous growing season.> <A-head: Initiation and Development of Leaves> <Subject: Chapter 6> <Complexity: Difficult> <Taxonomy: Analysis>

29. Explain how an insect walking on the surface of a Venus flytrap leaf can get caught. What happens to the insect once it is caught? <Answer: As insects walk across the leaf of the Venus flytrap, it brushes against trigger hairs.

When two trigger hairs are stimulated, the motor cells of the midrib quickly lose water, causing the two halves of the trap to close. The teeth along the leaf margins trap the insect, and glands secrete digestive fluid that digests the insect. The trap reopens after digestion and absorption as the midrib motor cells again fill with water.> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

30. If shifting sands in a South African desert bury all but the leaf tips of Lithops (stone plants), the plants can still survive. Why?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: These plants survive due to their unique translucent underground leaves that act as optical fibers. The majority of the leaf persists in a relatively moist and cool subterranean environment. The exposed leaf tips allow enough sunlight to be absorbed by buried chlorenchyma cells to enable photosynthesis to occur.> <A-head: Morphology and Anatomy of Other Leaf Types> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Recall>

31. What are the advantages and disadvantages of trichomes on a leaf surface? <Answer: Some of the advantages of trichomes include providing shade on the upper surface of a leaf, preventing water loss from stomata on the lower surface, deterring herbivory and pathogen infection, and preventing the blockage of stomata by films of water or dust. The major disadvantage of trichomes is their production costs.> <A-head: Internal Structure of Foliage Leaves> <Subject: Chapter 6> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 8—Roots

Multiple Choice

1. You would expect most roots to do all of the following except: A) absorb water.

B) absorb minerals.

C) photosynthesis.

D) produce and export hormones to the shoot.

E) anchor a plant in one place. <Answer: C> <A-head: Concepts> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Application>

2. Palm trees, which are monocots, can increase in diameter every year due to the production of additional: A) wood.

B) lateral roots.

C) taproots.

D) adventitious roots.

E) Monocots cannot increase in diameter. <Answer: D> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Application>

3. Which mitotically inactive portion of the root tip acts as a reserve of healthy cells in case the apical meristem becomes damaged?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) root cap.

B) zone of cell division.

C) zone of elongation.

D) quiescent center.

E) zone of maturation. <Answer: D> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

4. A root is not uniform along its length but has distinct zones. Starting at the tip, these zones proceed as which of the following?

A) Apical meristem → root cap → zone of elongation → zone of maturation/root hairs B) Root cap → zone of elongation → apical meristem → zone of maturation/root hairs C) Root cap → apical meristem → zone of elongation → zone of maturation/root hairs D) Root cap → apical meristem → zone of maturation/root hairs → zone of elongation <Answer: C> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

5. The main function of a root cap is: A) absorption of water and minerals from the soil.

B) root anchorage.

C) protection of the root apical meristem.

D) hormone production.

E) plant vegetative reproduction. <Answer: C> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. If a water molecule entered a mature root at the root surface, it would cross what tissues, in their correct order, on its way to the center?

A) Epidermis, endodermis, cortex, phloem, xylem B) Epidermis, cortex, pericycle, endodermis, phloem, xylem C) Cortex, endodermis, phloem, pericycle, xylem D) Epidermis, cortex, endodermis, pericycle, phloem, xylem E) Endodermis, epidermis, cortex, pericycle, phloem, xylem <Answer: D> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

7. In dicot root xylem, metaxylem cells would be the: A) innermost, narrow cells in vascular bundles.

B) innermost, wide cells in vascular bundles.

C) innermost, wide cells in a solid mass of xylem.

D) outermost, wide cells in a solid mass of xylem.

E) outermost, narrow cells in vascular bundles. <Answer: C> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Application>

8. Minerals entering the xylem must first go through endodermal cells because suberin and lignin are found in the cells’: A) side walls.

B) top and bottom walls.

C) tangential and side walls.

D) side, top, and bottom walls.

E) side, top, bottom, and tangential walls.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: D> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Recall>

9. The downward direction of root growth in response to gravity would be most adversely affected by an inhibition of: A) protein synthesis.

B) lipid synthesis.

C) starch synthesis.

D) DNA synthesis.

E) cellulose synthesis. <Answer: C> <Subject: Chapter 8> <A-head: Internal Structure of Roots> <Complexity: Moderate> <Taxonomy: Application>

10. The rapid diffusion of oxygen to submerged portions of roots of some marsh plants occurs through: A) chlorenchyma.

B) aerenchyma.

C) parenchyma.

D) xylem.

E) phloem. <Answer: B> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. The taproot develops from the radicle. <Answer: True> <A-head: Concepts> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Recall>

12. Like stem trichomes, root hairs may be multicellular. <Answer: False> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

13. Plasmodesmata connect cells of the root. <Answer: True> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Recall>

14. Lateral root formation destroys cells of the epidermis and cortex, thus severing conductive ties with the parent root. <Answer: False> <Subject: Chapter 8> <A-head: Origin and Development of Lateral Roots> <Complexity: Difficult> <Taxonomy: Analysis>

15. Not only do plants make use of food stored in roots, but people do as well.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: True> <Subject: Chapter 8> <A-head: Other Types of Roots and Root Modifications> <Complexity: Moderate> <Taxonomy: Analysis>

Fill-in-the-Blank

16. Dictyosomes in the root cap secrete a complex polysaccharide rich in carbohydrates and amino acids known as . <Answer: mucigel> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Recall>

17. The mitotically inactive reserve of healthy cells near the root apical meristem is known as the . <Answer: quiescent center> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Recall>

18. The root epidermis of orchids is called barrier preventing excess water from leaving the roots.

and acts as a waterproof

<Answer: velamen> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

19. Some tropical trees, like Ficus macrophylla, have tall, platelike that brace the trunk, preventing it from blowing over in the wind. <Answer: buttress roots> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Recall>

20. Most monocots have a mass of similarly sized roots constituting a system. <Answer: fibrous root> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Recall>

Matching

21. Match each root structure and its major function.

A) Root cap <Answer: Protection> B) Root hair <Answer: Increased surface area> C) Apical meristem <Answer: Cell production> D) Parenchyma cells <Answer: Food storage> E) Endodermis <Answer: Control of mineral absorption> [F] Pericycle <Answer: Lateral root origin> [G] Xylem <Answer: Mineral transport>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Subject: Chapter 8> <A-head: Internal Structure of Roots> <Complexity: Moderate> <Taxonomy: Analysis>

22. Match each structure with its major function.

A) Mycorrhizal root <Answer: Increased phosphorus absorption> B) Prop root <Answer: Stabilization and support> C) Storage root <Answer: Accumulation of carbohydrate> D) Contractile root <Answer: Pull stem downward> E) Haustorial root <Answer: Absorption from host> [F] Root nodule <Answer: Nitrogen fixation> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8> <Complexity: Easy> <Taxonomy: Recall>

Essay

23. Relate the overall shape of a root to its function and environment. <Answer: Water and minerals are distributed on all sides of a root, thus making the conical shape advantageous for nutrient and water absorption on all sides. Roots also have a high surface-tovolume ratio, further enhancing nutrient absorption. The cylindrical shape also improves root penetration through soil. Branched root systems and deep taproots improve plant stability.> <A-head: Concepts> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

24. Distinguish between radicle, taproot, lateral root, and adventitious root. Would you expect the internal distribution of tissues to be the same or different in these roots? Why? <Answer: Fleshy taproots develop from the embryonic root, the radicle. In systems with taproots, these roots become the largest root in the system and are often used in carbohydrate storage. Lateral roots branch off the taproot and are smaller than the taproot. Adventitious roots arise from stems. Although the major internal tissues are present in all of these roots, the proportions of these tissues will differ. The radicle is the youngest root and therefore the least differentiated.> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Application>

25. Would grass or carrots be most effective for control of erosion? Why? <Answer: Grasses have greater soil-holding capacity and are therefore likely to be more effective in controlling erosion. Grasses are monocots and have fibrous root systems. The horizontal shoots of grasses can also produce additional adventitious roots. Such systems support numerous roots that explore and hold large volumes of soil. In contrast, the fleshy taproots of carrots extend downward and support more limited lateral branching and thus in general inhabit a smaller volume of soil.> <A-head: External Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Application>

26. List and discuss the functions of the root cap. <Answer: The primary roles of the root cap are to protect the apical meristem and detect gravity.

The cells in the layer closest to the root meristem are also meristematic. Root cap cells are constantly being sloughed off and regenerated. The cells also detect gravity because their dense starch grains settle to the lower side of the cell. Lastly the root cap secretes complex polysaccharides known as mucigel that lubricate the passage of the root through the soil.> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

27. Relate the development and structure of the endodermis to its function. <Answer: The tangential walls of the endodermis are close to the xylem or cortex and have thin primary walls allowing the passage of minerals from the soil to the xylem. The radial walls, those touching other endodermal cells, are encrusted with lignin and suberin, forming a waterproof casparian strip. This forces any minerals that are moving through the apoplast of the plant to either be accepted by the plasma membrane of the epidermis and thereby move to the xylem or be stopped from entering the vascular tissue. In this way harmful minerals are excluded by the epidermis.> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Application>

28. Lateral roots originate in the pericycle of a root. What is the advantage to a plant of this place of origin? <Answer: Lateral roots are formed deep within the mature portion of the root. One advantage of this is that the developing root primordium is protected from breaking off while its growth is initiated. As the root primordium grows through the cortex, it is protected by a thin covering of torn endodermis cells. By the time the lateral root emerges, it has formed a protective root cap and the protoxylem and protophloem have differentiated to the point where it has established a critical conductive connection with the vascular tissues of the parent root.> <Subject: Chapter 8> <A-head: Origin and Development of Lateral Roots> <Complexity: Difficult> <Taxonomy: Analysis>

29. Why do farmers often rotate legumes and non-legumes from year to year in a field? <Answer: Legume plant species form mutualistic relationships with specialized bacteria that fix nitrogen. This relationship enhances soil fertility by adding nitrogen to the soil. To decrease agricultural pests from year to year, it is beneficial to rotate crops. Rotating between legumes and non-legumes adds nitrogen to the soil and reduces the need for nitrogen amendments.> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Analysis>

30. Why would it be economically beneficial to farmers if a variety of corn was developed that produced root nodules? <Answer: Root nodules are the result of a mutualistic relationship between plants and specialized bacteria that fix nitrogen. This relationship enhances soil fertility by adding nitrogen to the soil.

To decrease agricultural pests from year to year, it is beneficial to rotate crops. Rotating between legumes and non-legumes adds nitrogen to the soil and reduces the need for nitrogen amendments.> <A-head: Other Types of Roots and Root Modifications> <Subject: Chapter 8> <Complexity: Difficult> <Taxonomy: Analysis>

31. Discuss the possible routes a mineral ion might take from the soil to the xylem in the zone of maturation. <Answer: Minerals from the soil may be accepted by the plasma membrane of a root hair or other epidermal cell and then travel symplastically to the xylem. A mineral may also travel through the apoplast until it reaches the endodermis, where it can be accepted by the plasma membrane of an endodermal cell and then travel through the symplast to the xylem. Lastly a mineral may be accepted by the plasma membrane of a cortex cell and then travel through the symplast to the xylem.> <A-head: Internal Structure of Roots> <Subject: Chapter 8> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 9—Structure of Woody Plants

Multiple Choice

1. The cells in primary tissues were produced by cell divisions in a(n): A) vascular cambium.

B) cork cambium.

C) apical meristem.

D) vascular cambium and an apical meristem.

E) vascular cambium, cork cambium, and apical meristem. <Answer: C> <A-head: Concepts> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

2. Anticlinal divisions of fusiform initials may produce: A) vessel elements.

B) new fusiform initials.

C) ray initials.

D) sieve tube members.

E) storage parenchyma cells. <Answer: B> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

3. The cells that transport sugars from the leaves to the roots of a woody flowering plant are: A) sieve tube members produced by fusiform initials of the vascular cambium.

B) sieve tube members produced by fusiform initials of the cork cambium.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) tracheids produced by fusiform initials of the vascular cambium.

D) companion cells produced by ray initials of the vascular cambium.

E) fibers produced by fusiform initials of the cork cambium. <Answer: A> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

4. Which of the following is shown by the accompanying diagram?

A) Storied cambium with uniseriate rays B) Storied cambium with biseriate rays C) Storied cambium with multiseriate rays D) Nonstoried cambium with uniseriate rays E) Nonstoried cambium with biseriate rays <Answer: A> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Analysis>

5. Many people buy furniture made of oak. The wood of this tree is particularly good for furniture construction because it contains large amounts of: A) fibers.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) parenchyma.

C) sclereids.

D) tracheids.

E) vessel elements. <Answer: A> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Application>

6. The maple sap tapped in the spring to make maple syrup comes from the xylem of sugar maple trees. The sugar in this sap was stored in: A) root cortex cells.

B) phloem parenchyma cells.

C) xylem axial parenchyma cells.

D) xylem ray upright cells.

E) xylem ray procumbent cells. <Answer: D> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Application>

7. One function of a woody stem is support. In gymnosperms, this is mainly a function of: A) tracheids.

B) fibers.

C) vascular cambium.

D) sieve cells.

E) All of these are correct. <Answer: A> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

8. Although variable from plant to plant, cork cambia are not initiated in a stem’s: A) epidermis.

B) cortex.

C) primary phloem.

D) secondary phloem.

E) vascular cambium <Answer: E> <A-head: Outer Bark> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Application>

9. The first cork cambium in a dicot root usually arises from: A) primary phloem cells.

B) epidermal cells.

C) primary xylem cells.

D) pericycle cells.

E) primary phloem, parenchyma, and pericycle cells. <Answer: D> <A-head: Secondary Growth in Roots> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

10. One way herbaceous plants can increase their conduction capacity is to produce: A) additional leaves.

B) storage roots.

C) secondary tissues.

D) adventitious roots.

E) contractile roots. <Answer: D> <A-head: Concepts>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Recall>

True/False

11. A mature woody plant is composed of secondary tissues, such as secondary xylem and phloem, but no primary tissues. <Answer: False> <A-head: Concepts> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Recall>

12. It is possible for fusiform initials to become ray initials and vice versa. <Answer: True> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Recall>

13. In plants that produce annual rings in secondary xylem, annual rings will also be seen in secondary phloem. <Answer: True> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Recall>

14. In a cross section of a tree, the oldest xylem is found near the center and the oldest phloem is found next to the xylem.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: False> <A-head: Outer Bark> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

15. You can distinguish a cross section of a woody root from a cross section of a woody stem because the secondary xylem of the root is star-shaped. <Answer: False> <A-head: Secondary Growth in Roots> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

Fill-in-the-Blank

16. In woody species, , such as wood and bark, are produced in the stem and root from meristems other than apical meristems. <Answer: secondary tissues> <A-head: Concepts> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Recall>

17. In woody plant species, the cells located between the metaxylem and metaphloem never undergo cell cycle arrest and constitute the . <Answer: fascicular cambium> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

18. is derived from the vascular cambium and may contain cellular components such as sieve elements, fibers, nonconducting parenchyma, companion cells, and albuminous cells. <Answer: Secondary phloem> <A-head: Secondary Phloem> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

19. Aerenchymatous cork cells called into the stem.

allow oxygen to diffuse

<Answer: lenticels> <A-head: Outer Bark> <Subject: Chapter 9> <Complexity: Easy> <Taxonomy: Recall>

20. Palm trees increase their girth through growth in which adventitious roots are added to the base.

, a form of primary

<Answer: establishment growth> <A-head: Anomalous Forms of Growth> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each structure with its complementary term.

A) Annual ring <Answer: Spring wood + summer wood>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) Nonconducting xylem <Answer: Heartwood> C) Conducting xylem <Answer: Sapwood> D) Wide growth rings on upper side of branch <Answer: Tension wood> E) Xylem composed mainly of tracheids <Answer: Softwood> F) Xylem containing a large number of fibers <Answer: Hardwood> G) Wide growth rings on underside of branch <Answer: Compression wood> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each form of secondary growth with its corresponding example.

A) Secondary vascular bundles <Answer: dragon tree> B) Included phloem <Answer: Bougainvillea> C) Production of parenchymatous xylem and phloem <Answer: sweet potatoes> D) Unequal activity of the vascular cambium <Answer: Bauhinia> E) Vascular cambia create secondary xylem and secondary phloem <Answer: basal angiosperms> <A-head: Anomalous Forms of Growth> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Recall>

Essay

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 23. Why are woody plants more likely to survive adverse environmental conditions than nonwoody plants? <Answer: Woody plants have a greater capacity to move water, minerals, and carbohydrates throughout the plant body, and their capacity increases on an annual basis rather than being predetermined as it is for herbs. As conductive capacity increases, so does its ability to form leaves, grow roots, produce seeds, photosynthesize, and form defensive chemicals. Due to the defensive chemicals and the production of wood and bark, the woody species are better defended than herbaceous species. All of these traits are advantageous for survival under adverse environmental conditions.> <A-head: Concepts> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

24. Could a woody plant survive if the vascular cambium had no fusiform initials? Ray initials?

Why? <Answer: Yes, as both cell types may be formed from the other. In side-by-side fusiform initials, a central cell may undergo transverse divisions and be transformed into ray initials. Similarly, in a broad group of ray initials, the central cells may elongate and convert to fusiform initials.> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

25. Would you expect the length of the sieve tube members to be related to the length of the fibers and vessel elements of the same tree? Why? <Answer: Yes, sieve tube members, fibers, and vessel elements are all produced from fusiform initials. The lengths of the fusiform initials likely influence the length of the further differentiated cells. Typical lengths of fusiform initials vary from 140 to 8,700 μm, and length varies by species.> <A-head: Vascular Cambium> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 26. When viewing the cross sections of two different woods, how can you tell which wood is from a tropical tree and which is from a temperate tree? <Answer: Temperate climates produce annual rings due to the construction of early wood and late wood. Early wood is produced early in the growing season when developing leaves lack a thick cuticle and are therefore prone to water loss. Trees need a high conduction capacity at this time and thus a high proportion of wide vessels. Later in the growing season, the tree is bigger and heavier, requiring strong wood composed of thick fibers and/or narrow, thick-walled tracheids. The last cells to develop are often just heavy fibers with thick secondary walls, making a clear distinction between the end of one year’s growth and the beginning of the next year’s growth. These same patterns are not usually found in tropical trees. In mild, tropical climates, the cambium remains almost continually active, making it difficult to distinguish one year of growth from another.> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

27. Often in storms with high winds, trees fall over, revealing a hollow center. Why could the tree function normally before it was blown over? <Answer: The center of a tree contains heartwood with dead cells that no longer conduct water.

Oftentimes these cells are impregnated with phenolic compounds, and lignin contain tyloses to inhibit bacterial and fungal growth. The outer xylem rings, or the sapwood, are the conductive layers. If the sapwood remains intact, a tree can perform its metabolic functions even if it is hollow in the center.> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

28. What is the function of xylem rays? Phloem rays usually connect to xylem rays. Why? <Answer: Ray parenchyma store carbohydrates and other nutrients during dormant periods and conduct material over short distances in wood. The size, shape, and number of phloem rays match that of xylem rays because both are produced by the same ray initials.> <A-head: Secondary Xylem> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

29. Why do different species of trees have different patterns of outer bark? <Answer: In some species, only a small amount of cork is formed, so bark is thin and consists mainly of dead secondary phloem. In other species, cork is produced in large quantities and becomes quite thick. Sometimes the cork cambia form close together and each is sheet-like, causing bark to peel off in sheets. In some species, as bark patches peel away, the freshly exposed tissues become oxidized and change color. Some bark has different ratios of cork, sclereids, secondary phloem, and phelloderm, each affecting the bark morphology. If the first cork cambium arises by reactivation of the epidermal cells, the first outer bark contains only periderm and cuticle and is therefore smooth. If the first cork cambium arises in the cortex, the first bark is still smooth; however, if later cork cambia arise in the secondary phloem, the outer bark will only consist of cork and phloem.> <A-head: Outer Bark> <Subject: Chapter 9> <Complexity: Difficult> <Taxonomy: Analysis>

30. A slow way to kill a tree is to girdle the trunk: the outer and inner bark is removed in a strip around the trunk’s entire circumference. Why does that kill the tree? <Answer: Removing the inner bark removes the secondary phloem, which is responsible for conduction of carbohydrates and minerals up and down the stem. When the inner bark is girdled, the root systems starve the sugars made by photosynthesis in the leaves and are prevented from reaching the roots. Removing the outer bark removes the protective barrier that protects the stem from animal and microbial attack. The tree becomes more vulnerable to pathogens and other disturbances.> <A-head: Secondary Phloem> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Analysis>

31. Why doesn’t a 100-year-old tree have just an epidermis on its surface? <Answer: A long-lived tree has to battle insects, fungi, and environmental harshness many times longer than annual plants. An epidermis is a living layer of cells with metabolic maintenance costs that are not as well protected as the cells in the periderm. An epidermis also is only a few layers thick at most. Outer bark is made up of multiple layers of cells, including lignified

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank sclereids for strength. Many of the cells in the outer bark are encrusted with suberin and lignin and are dead. There is very little nutritious tissue, which helps deter animals and microorganisms.> <A-head: Outer Bark> <Subject: Chapter 9> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 10—Flowers and Reproduction

Multiple Choice

1. Under what conditions would it be likely for asexual reproduction to be more advantageous to a species than sexual reproduction?

A) A rare late spring frost B) An unusually severe summer drought C) A stable climate D) Global warming E) Introduction of a fungal disease <Answer: C> <A-head: Concepts> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Analysis>

2. A grove of aspen trees in Utah is actually considered to be a single plant because: A) each aspen tree was formed from a plantlet from the leaf margins of the mother tree.

B) birds dispersed and planted seeds so closely to each other that the root systems joined.

C) each tree developed from an adventitious shoot bud from the same root system.

D) the leading edge of the stand survives independently of the mother tree.

E) seeds of the mother tree were equally adapted to the Utah environment. <Answer: C> <A-head: Asexual Reproduction> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

3. The two halves of a lima bean or pea are the: A) cotyledons.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) epicotyl.

C) embryo.

D) radicle.

E) hypocotyl. <Answer: A> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Recall>

4. Which of the following is the order in which flower parts are attached to the receptacle, from lowest or highest?

A) Petals → sepals → stamens → carpels B) Sepals → petals → stamens → carpels C) Sepals → stamens → petals → carpels D) Sepals → petals → carpels → stamens <Answer: B> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Recall>

5. Many organisms are oogamous and produce: A) small, motile sperm cells and large, nonmotile egg cells.

B) large, motile sperm cells and small, nonmotile egg cells.

C) sperm and egg cells that are the same size and are both motile.

D) small, motile sperm cells and large, motile egg cells.

E) small, nonmotile sperm cells and large, nonmotile egg cells. <Answer: A> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. Concerning angiosperm seeds, the: A) endosperm in young eudicot seeds is haploid and surrounds the embryo in a mature seed.

B) endosperm in young eudicot seeds is triploid and food is usually stored in the cotyledons in a mature seed.

C) endosperm in a young monocot seed is diploid and is usually stored around the embryo in a mature seed.

D) endosperm in a young monocot seed is triploid and food is usually stored in the one cotyledon in a mature seed.

E) chromosome complement of endosperm is the same in all young seeds and the site of food storage is the same in all mature seeds. <Answer: B> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

7. Which of the following is shown by the accompanying diagram?

A) Epigynous flower B) Apogynous flower C) Perigynous flower D) Hypogynous flower

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank E) Hypergynous flower <Answer: D> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Recall>

8. It is usually advantageous to a species if the offspring of a sporophyte are genetically different than the parent sporophyte. Which of the following would be the least likely to ensure genetically different offspring?

A) Pistil and stamens within a flower mature at different times.

B) The pollen produced on a plant cannot germinate on stigmas on the same plant.

C) The species is dioecious.

D) The species is monoecious.

E) The plant is pollinated by bees. <Answer: D> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

9. Wind-pollinated flowers, in contrast to insect pollinated flowers, are usually characterized by: A) a greatly reduced stigma and style.

B) reduced or no petals and sepals.

C) producing few pollen grains.

D) brightly colored sepals or stamens.

E) zygomorphy. <Answer: B> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 10. Which type of adaptation is likely to be beneficial to a seed that is dispersed by animals?

A) presence of sails or parachutes.

B) presence of hooks or stickers.

C) buoyancy.

D) mildew resistance.

E) soft edible seed coat. <Answer: B> <A-head: Fruit Types and Seed Dispersal> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Recall>

True/False

11. Cell division by meiosis always produces sperm and egg cells. <Answer: False> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Recall>

12. If flower color is controlled by a gene in the plastids of a plant, then the offspring of two plants of different color will be the same color as the maternal parent. <Answer: True> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

13. Zygomorphic flowers are adapted to animal pollination. <Answer: True>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Recall>

14. An incomplete flower can also be a perfect flower. <Answer: True> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

15. Many flowers last a long time because they produce secondary tissues and become woody. <Answer: False> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Recall>

16. For most sexually reproducing plants, the potential advantages of sexual reproduction outweigh the potential disadvantages. <Answer: True> <A-head: Concepts> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Application>

Fill-in-the-Blank

17. As the dog cholla (Opuntia schotti) spreads through the process of , it branches and spreads from a central point, forming a characteristic

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank ring of living growth that continues to survive and surround the older portion of the plant that eventually dies. <Answer: fragmentation> <A-head: Asexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Application>

18. A(n) typical of eudicots.

seed has cotyledons rich in starch, oil, or proteins and is

<Answer: exalbuminous> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

19. Within the anther, the microspore mother cells are adjacent to a layer of nurse cells known as . These cells contribute to microspore development and maturation. <Answer: tapetum> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Recall>

20. The two sperm cells in the macrogametophyte are formed from a small, lens-shaped . <Answer: generative cell> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 21. The pollination of a carpel by the pollen of a different individual is and helps ensure genetic diversity in the resulting offspring. <Answer: cross-pollination> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Moderate> <Taxonomy: Application>

Matching

22. Match each function with its corresponding structure.

A) Usually attracts pollinators <Answer: Corolla> B) Produces pollen <Answer: Androecium> C) Point of attachment for floral parts <Answer: Receptacle> D) Protects the flower bud <Answer: Calyx> E) Contains the ovules <Answer: Gynoecium> F) Supports the flower <Answer: Pedicel> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Easy> <Taxonomy: Recall>

23. Match each type of fruit with its definition.

A) Berry <Answer: Fleshy fruit in which as layers are soft> B) Caryposis <Answer: Simple and small with a testa that becomes fused to the fruit wall during maturation> C) Pome

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Fleshy fruit with a papery or leathery endocarp> D) Legume <Answer: Dehiscent fruit breaks open along both sides> E) Drupe <Answer: Fleshy fruit with hard, sclerenchymatous endocarp> F) Aggregate fruit <Answer: Carpels of flower not fused but grow together during fruit maturation> G) Multiple fruit <Answer: Fruits of inflorescence grow together during fruit maturation> <A-head: Fruit Types and Seed Dispersal> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Recall>

Essay

24. What are the advantages and disadvantages of a strawberry plant reproducing by stolons? By seeds? <Answer: Through asexual reproduction, such as the production of stolons, plants produce identical offspring that are as adapted as the parent but are not better adapted to the environment.

In the case of a changing environment, all offspring could be equally adversely affected by the change. However, stolons also allow for rapid colonization of new environments and allow even isolated individuals to reproduce. When strawberries reproduce by seed through sexual reproduction, the offspring are genetically different from the parent and some offspring are more adapted to the environment, have the ability to colonize different sites, and are potentially adaptable to new habitat conditions. The disadvantages are that some offspring are less adapted than their parents, some offspring colonize more slowly, and isolated offspring cannot reproduce.> <A-head: Concepts> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

25. Draw and label a megagametophyte of a flowering plant and compare it to the microgametophyte.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: The megagametophyte should resemble Figure 9-15D. It is more complex and larger than a microgametophyte. Megagametophytes vary by species but they are generally comprised of a central cell with polar nuclei. Three cells at the top of the central cell are called antipodals and two of the three cells at the bottom are called synergids. Each synergid flanks the egg cell.

The microgametophyte is sometimes referred to as pollen and contains vegetative cells and two sperm cells.> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

26. Why are kernels of field corn much harder than the kernels of sweet corn? <Answer: Field corn is harvested after the endosperm has become cellular, starch-filled, and dry.

Sweet corn is harvested just as the endosperm begins to convert from a milky coenocytic form of endosperm to the starchy, hard form.> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

27. All seeds (with a few exceptions) are composed of three parts: an embryo, a seed coat, and stored food. From what structures in the ovule did each of these develop? How does the structure of a mature eudicot seed usually differ from the structure of a mature monocot seed? <Answer: The embryo and the stored food developed from the nucellus, whereas the seed coat developed from the integuments. A mature eudicot seeds the cotyledons, becoming filled with starch, oil, or protein while the endosperm shrinks. Such a seed is exalbuminous. In the mature monocot seed, most of the endosperm remains an albuminous seed.> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

28. In most flowers, the sepals do not attract pollinators and cannot produce pollen or ovules. So isn’t the production of sepals a waste of energy? Why or why not?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Sepals are modified leaves that surround and enclose the other flower parts as they mature. These other flower parts include the ovary containing the egg and the anthers containing the pollen and sperm. The plants require these protected parts to complete their life cycle.

Therefore the production of sepals is not a waste of energy but an investment in reproduction.> <A-head: Sexual Reproduction> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

29. In many flowers that have a superior ovary, the petals fuse to one another to form a tube.

What is the advantage of this to the plant? <Answer: A superior ovary is above the other flower parts. The petals often fuse to form a tube to better protect the ovary from damage or predation. The fused petals may also ensure pollinator specificity by better fitting the body of the pollinator.> <A-head: Flower Structure and Cross-Pollination> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Application>

30. Outline the relative selective advantages and disadvantages of flower size, number of ovules per flower, and flower grouping. <Answer: Big flowers have higher production or energy costs. The advantage of a large flower is that it is easily seen by pollinators; however, disadvantages are that it is a large energy investment and one large flower can easily by damaged or removed without ever setting seed. It is advantageous to maximize the number of ovules per flower and make smaller flowers so that if one or a few are lost, there are still others available for pollination and seed set. It can also be advantageous to group small flowers on an inflorescence to realize the advantages of small flowers while making the impact of one large flower. Inflorescences also provide control over the initiation, maturation, and opening of the flowers so the plant can be flowering over a longer period of time.> <A-head: Inflorescences and Pollination> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

31. Why are immature fleshy fruits unpalatable?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: There is a division of labor in fruits. Some parts are protective, some are attractive, and others allow germination. If animals are to disperse seeds, part of the fruit must be edible or attractive while other parts are protective. The seeds of edible fruits must not be destroyed by their animal dispersers. The hard seed coats or endocarps protect seeds as they pass through the animal digestive tract. Hard, bitter, unripe seeds deter frugivores until the protective structures around the seed have developed.> <A-head: Fruit Types and Seed Dispersal> <Subject: Chapter 10> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 11- Energy Metabolism: Photosynthesis

Multiple Choice

1. Which of the following organisms is not a photoautotroph?

A) A white pine tree.

B) A geranium.

C) Anabaena, a cyanobacterium.

D) Agaricus, the common field mushroom.

E) Spirogyra, a green alga. <Answer: D> <A-head: Concepts> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Recall>

2. ATP is synthesized from ADP and phosphate in: A) the cytoplasm using the oxidation of organic molecules as the energy source.

B) mitochondria using light energy as the energy source.

C) chloroplasts using light energy as the energy source.

D) chloroplasts using the oxidation of organic molecules as the energy source. <Answer: C> <A-head: Energy and Reducing Power> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Analysis>

3. To promote plant growth, lights called grow lights should produce light enriched in what wavelengths?

A) Red and green.

B) Red and blue.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) Yellow and blue.

D) Blue and violet.

E) Orange and violet . <Answer: B> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Application>

4. Photosystem I and photosystem II must be connected for the formation of: A) ATP. B) NADPH. C) NADH. D) H2O. E) Reduced ferredoxin. <Answer: B> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Recall>

5. The strongest evidence supporting the hypothesis that chlorophyll is the essential pigment for photosynthesis is that: A) all plants are green.

B) it is present in all plants.

C) its absorption spectrum matches the action spectrum of photosynthesis.

D) its absorption spectrum does not match the action spectrum of photosynthesis.

E) chlorophyll is green in color. <Answer: C> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. The major function of carotenoid pigments is to: A) absorb light wavelengths different than chlorophyll and pass most of that energy to chlorophyll a.

B) absorb the same light wavelengths as chlorophyll a and pass most of that energy to chlorophyll a.

C) pass absorbed light energy to chlorophyll by fluorescence.

D) reflect excessive light, protecting chlorophyll molecules from intense light.

E) absorb excessive light, protecting chlorophyll molecules from intense light. <Answer: E> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Analysis>

7. Which of the following is not a protective adaptation against intense sunlight?

A) Vertical leaves.

B) Accessory pigments.

C) Abundant chlorophyll.

D) A heavy coating of wax on the leaves.

E) Hairy leaves. <Answer: C> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

8. The main advantage of Kranz anatomy is that: A) RuBP carboxylase is found only in bundle sheath chloroplasts.

B) RuBP carboxylase is found only in palisade parenchyma chloroplasts.

C) RuBP carboxylase is found only in spongy mesophyll chloroplasts.

D) RuBP carboxylase is found only in mesophyll chloroplasts.

E) PEP carboxylase is found only in bundle sheath chloroplasts. <Answer: A>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

9. If you identified a desert plant as belonging to the family Crassulaceae, it would very likely: A) have thin leaves that were dropped when water was scarce.

B) exhibit C3 photosynthesis.

C) open its stomates at night and close them during the day.

D) photosynthesize very efficiently and grow rapidly.

E) exhibit a low rate of photorespiration at all times. <Answer: C> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Analysis>

10. The rate of photosynthesis in a plant is measured at X ppm CO2 at a specific light intensity. If the amount of CO2 is increased to Y ppm and the light intensity remains the same, the rate of photosynthesis increases. These results suggest that at X ppm CO2: A) light is limiting the rate of photosynthesis.

B) the amount of water available is limiting the rate of photosynthesis.

C) the amount of CO2 available is limiting the rate of photosynthesis.

D) the plant is damaged by the light intensity used.

E) no photosynthesis is occurring. <Answer: C> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. Roots are usually heterotrophic. <Answer: True> <A-head: Concepts> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Recall>

12. The carbon in carbon dioxide is more reduced than the carbon in a carbohydrate. <Answer: False> <A-head: Energy and Reducing Power> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

13. A major function of photosystem II is to reduce P700, replenishing electrons lost in photosystem I. <Answer: True> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Recall>

14. Rapidly growing plants commonly exhibit Crassulacean acid metabolism. <Answer: False> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

15. The light compensation point is the level of light at which the rate of respiration exceeds that of photosynthesis.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: True> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. Organisms that gather energy from the sun and use it to assimilate their own tissues are known as . <Answer: photoautotrophs> <A-head: Concepts> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Recall>

17. Plastoquinones and are examples of electron carriers that transport electrons over short distances within thylakoid membrane during photosynthesis. <Answer: cytochromes> <A-head: Energy and Reducing Power> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

18. Following the production of PGAL in the Calvin/Benson cycle, glucose is anabolically synthesized through . <Answer: gluconeogenesis> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

19. Up to 30% of all ATP and NADPH produced by a chloroplast can be lost through , a wasteful process that occurs when RuBP binds with oxygen. <Answer: photorespiration> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

20. In C4 metabolism, mesophyll cells contain the enzyme a high affinity for carbon dioxide.

, which has

<Answer: PEP carboxylase> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each structure with its function.

A) Chlorophyll a <Answer: Component of P700 reaction center> B) CF0–CF1 complex <Answer: Chemiosmotic phosphorylation> C) Cyclic electron transport <Answer: Production of extra ATP without making NADPH> D) Chlorophyll b <Answer: Accessory pigment(s)> E) Noncyclic electron transport <Answer: Electron flow from water to NADPH> F) Carotenoids <Answer: Absorption of excess light> G) RuBp <Answer: Substrate for carbon dioxide>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank H) PGAL <Answer: Reduced carbon compound> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

22. Match each compound with its function.

A) PEP carboxylase <Answer: Enzyme with a high affinity for carbon dioxide> B) Oxaloacetate <Answer: First stable carbon compound of the C4 pathway> C) Malate <Answer: Compound that decarboxylates to yield CO2 for use in the C3 pathway> D) Phosphoglycolate <Answer: Carbon compound formed during photorespiration> E) Phosphoenolpyruvate <Answer: Substrate for CO2> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

Essay

23. The second law of thermodynamics says that the entropy of the universe is constantly increasing. Are organisms an exception to this law? <Answer: No, organisms are not an exception to this law. The second law refers to the state of entropy in the universe. As organisms use energy inputs from the sun to create ordered, living systems, atomic reactions generating sunlight cause greater disorder on the sun than sunlight causes order in living systems. The entire system, the sun plus life, becomes more disordered.

Also, once living systems die, their molecules become disordered as entropy increases.> <A-head: Concepts> <Subject: Chapter 11>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Application>

24. What is phosphorylation? Distinguish between photophosphorylation, substrate-level phosphorylation, and oxidative phosphorylation. <Answer: Phosphorylation is a method by which a phosphate group is added to molecules such as adding phosphate to ADP to form ATP. Photophosphorylation involves sunlight as the energy sources and takes place in chloroplasts and thus occurs only in organs and organisms that contain chloroplasts. Substrate-level phosphorylation occurs in the cytosol and does not involve oxygen.

In the last stages of respiration, oxidative phosphorylation in the mitochondria in the presence of oxygen.> <A-head: Energy and Reducing Power> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Recall>

25. Why is NADPH said to have reducing power, and why is reducing power important to photosynthesis? <Answer: Reducing power is the ability to force electrons onto compounds. Reducing power is important in plants because they take in carbon dioxide and water, highly oxidized forms of those compounds, and convert them to reduced compounds like carbohydrates and fats. NADPH is an important electron carrier that can reduce other compounds by giving up an electron and forcing it onto another compound.> <A-head: Energy and Reducing Power> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Application>

26. Discuss the advantages and disadvantages of a plant cell storing energy in each of the following molecules: ATP, NADPH, glucose, sucrose, starch. <Answer: ATP and NADPH are short-term storage and energy and reducing power because they are highly reactive and short-lived. They can both be reused multiple times in cellular metabolism. Glucose and sucrose are stable enough to move throughout the plant and last for weeks to months. In large quantities they cause cells to absorb water through osmosis. Starch is a

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank stable, long-term energy storage molecule. It may last for years but it too large to be easily transported.> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Recall>

27. Why would growing plants under only green light basically be like growing the plants in the dark? <Answer: Plants appear green because they reflect green wavelengths and thus require other wavelengths of light to power photosynthesis. Growing under only green lights would not provide enough energy to run the photosynthetic machinery, and thus carbohydrates would not be formed for use in plant metabolism.> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Easy> <Taxonomy: Application>

28. What does it mean if a protein is conserved evolutionarily? What does this tell you about the connection between the protein’s structure and function? <Answer: An evolutionarily conserved protein, like RUBISCO, means that the amino acid sequence for this protein is nearly identical in all plants. Since the genetic sequence dictates both structure and function, this tells us that function is closely related to structure. In the case of RUBISCO, the proper tertiary and quaternary structure must be present in order to form a functional active site.> <A-head: Photosynthesis> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Analysis>

29. If a plant is to grow larger, which must occur in a 24-hour period: photosynthesis exceeds respiration or respiration exceeds photosynthesis? Explain your choice. <Answer: Photosynthesis must exceed respiration. Photosynthesis is the fixing of carbon that yields biomass. Respiration is the breakdown of carbohydrate to access energy and fixed carbon is lost from the system as carbon dioxide.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Difficult> <Taxonomy: Application>

30. In temperate zones, why do many forest wildflowers carry out most of their annual growth and reproduction in the early spring before the trees produce leaves? <Answer: The quality, quantity, and duration of light reaching the forest floor is highest in the early spring than in the summer after the leaves of the forest canopy have fully developed. Many wildflowers take advantage of the abundant light reaching the forest floor before they are shaded by the canopy trees.> <A-head: Environmental and Internal Factors> <Subject: Chapter 11> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 12—Energy Metabolism: Respiration

Multiple Choice

1. Which of the following are characteristics of H2O and O2 that make them selectively advantageous for use in the metabolic reactions of plants and other organisms?

A) They function as waste products of photosynthesis and respiration.

B) They are mostly nontoxic but can become toxic in large quantities.

C) They are abundant, nontoxic, and inexpensive to synthesize.

D) They were equally abundant in atmospheres of prehistoric and modern earth.

E) They are immune to mutations. <Answer: C> <A-head: Concepts> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Analysis>

2. Plants as a whole are: A) obligate aerobes.

B) obligate anaerobes.

C) facultative aerobes.

D) facultative anaerobes. <Answer: A> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Easy> <Taxonomy: Recall>

3. Which of the following is not a difference between aerobic respiration and photosynthesis?

A) Site of the process B) Function of the electron transport chain

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) Energy source for phosphorylation D) General type of redox process E) Role of O2 <Answer: B> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Analysis>

4. The process of photorespiration can salvage some of the carbon in glycolate, converting it to a useful form. The organelles involved in photorespiration are: A) chloroplasts, endoplasmic reticulum, and dictyosomes.

B) mitochondria and peroxisomes.

C) chloroplasts, mitochondria, and peroxisomes.

D) chloroplasts and peroxisomes.

E) chloroplasts, mitochondria, and glyoxysomes. <Answer: C> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Recall>

5. As electrons are passed from one compound to another in the electron transport chain: A) electrons are pumped from the matrix out of the mitochondrion.

B) NADH is pumped from the cytosol into the matrix.

C) protons are pumped from the matrix into the cristae lumen.

D) total potential energy of the compounds is increased.

E) only oxidation reactions occur. <Answer: C> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 6. If plant roots stand in water until all of the dissolved oxygen has been used, the cells switch to anaerobic respiration. The main reason that the pyruvate produced in glycolysis is further metabolized is to: A) regenerate NAD necessary for glycolysis to occur so ATP can be synthesized.

B) produce ATP using the energy in pyruvate.

C) produce lactic acid, which can be further metabolized by the root cells.

D) produce ethanol and CO2, which are used in photosynthesis. <Answer: A> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Recall>

7. A mature parenchyma cell redifferentiates into a sclereid. Which of the following best describes what happens metabolically in that cell?

A) Glycolysis dominates the pentose phosphate pathway.

B) Glycolysis occurs; the pentose phosphate pathway shifts in favor of ribose production.

C) Glycolysis occurs; the pentose phosphate pathway shifts in favor of erythrose production.

D) No glycolysis; the pentose phosphate pathway shifts in favor of ribose production.

E) No glycolysis; the pentose phosphate pathway sifts in favor of erythrose production. <Answer: C> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Analysis>

8. Which of the following is false about thermogenic respiration?

A) The last electron carrier is cytochrome oxidase.

B) It generates considerable amounts of heat.

C) It is insensitive to cyanide.

D) Electron carriers do not react with carbon monoxide. <Answer: A> <A-head: Types of Respiration> <Subject: Chapter 12>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Easy> <Taxonomy: Recall>

9. The synthesis of proteins from amino acids is a very endergonic process. In most organisms, the energy to drive this synthesis comes from all of the following except: A) glycolysis.

B) photorespiration.

C) the citric acid cycle.

D) the electron transport chain and chemiosmotic oxidative phosphorylation.

E) respiration from lipids. <Answer: B> <A-head: Total Energy Yield of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

True/False

10. Cellular respiration occurs only in animals; photosynthesis occurs only in plants. <Answer: False> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Easy> <Taxonomy: Recall>

11. Molecular oxygen is required for cyanide-resistant respiration. <Answer: True> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

12. Plants construct all 20 amino acids required for protein synthesis from intermediates of respiration; no external supply of amino acids is needed. <Answer: True> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Recall>

13. In general, a plant can live longer without oxygen than an animal can. <Answer: True> <A-head: Environmental and Internal Factors> <Subject: Chapter 12> <Complexity: Easy> <Taxonomy: Application>

14. The net ATP gain to a cell from glycolysis is two molecules. <Answer: True> <A-head: Total Energy Yield of Respiration> <Subject: Chapter 12> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

15. In the final step of the citric acid cycle, the electron acceptor molecule is formed and then reused in the earlier steps of this cycle when it receives an acetyl group from acetyl CoA. <Answer: oxaloacetate> <A-head: Types of Respiration> <Subject: Chapter 12>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Application>

16. The first electron acceptor molecule of the mitochondrial electron transport chain is . <Answer: flavin mononucleotide> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Recall>

17. Rice plants are considered flooded conditions and in drier, well-drained soils.

, anaerobes as they can germinate under

<Answer: facultative> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

18. In flooded environments, in carbohydrates.

allows roots to access the energy stored

<Answer: anaerobic respiration> <A-head: Environmental and Internal Factors> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

19. The for glucose is equal to 1 and represents the ratio of carbon dioxide that is liberated per unit of oxygen consumed. <Answer: respiratory quotient> <A-head: Respiratory Quotient> <Subject: Chapter 12>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Easy> <Taxonomy: Recall>

Multiple Response 20. If the mitochondria were removed from a cell, which of the following would immediately stop?

A) Glycolysis B) The citric acid cycle C) The pentose phosphate pathway D) Oxidative phosphorylation in the electron transport chain of respiration E) Malate-aspartate shuttle F) Thermogenic respiration <Answer: B, D, E, F> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Analysis>

Matching

21. Match each process with the biochemical pathway with which it is primarily associated.

A) Pyruvate as the final electron acceptor <Answer: Glycolysis> B) Movement of malate to carry reducing power across the inner mitochondrial membrane <Answer: NADH shuttle> C) Thermogenic respiration <Answer: Mitochondrial conversion of NADH to heat> D) Breakdown of pyruvate to form NADH and CO2 <Answer: Acetyl CoA formation> E) Establishment of electrochemical gradient through proton buildup in crista lumen <Answer: Mitochondrial electron transport chain> F) Phosphorylation of glucose-6-phosphate to form basis for nucleotides <Answer: Pentose phosphate pathway>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Analysis>

22. Match each respiratory pathway with its corresponding energy yield A) Heat-generating respiration <Answer: none> B) Anaerobic respiration <Answer: 2 ATPs> C) Citric acid cycle <Answer: 24 ATPs> D) Lipid respiration <Answer: variable> E) Pentose phosphate pathway <Answer: 0 to 12 ATPs> F) Sum of three parts of aerobic respiration <Answer: 36 to 38 ATPs> <A-head: Total Energy Yield of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Why is it important for cellular respiration to occur in plant cells? <Answer: Photosynthesis produces excess energy and reducing power that are stored as starch and glucose. Plant cells need to access that energy and reduced carbon. Respiration is the process that breaks down carbon compounds into simpler molecules while simultaneously generating ATP, which is used to power other metabolic processes.> <A-head: Concepts> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

24. Why is it important for mitochondria to have two membranes around the matrix? <Answer: The outer mitochondrial membrane helps to make the mitochondria a discrete organelle by regulating the passage of compounds moving in and out. The inner membrane is convoluted and folded into cristae as a means of increasing surface area for intrinsic proteins such as electron carriers of electron transport chains. This unique configuration also allows for the establishment of an electrochemical gradient that is strong enough to power chemiosmotic phosphorylation of ADT to form ATP.> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

25. What makes the citric acid cycle a metabolic cycle? <Answer: The citric acid cycle breaks down pyruvate to generate ATP. However, it also forms NADH and FADH2 to drive synthesis of ATP via the mitochondrial electron transport chain.

This is a cycle because in one of the final steps, fumarate reacts with water to form malate. The malate transfers a final set of electrons to NAD+ and is thus transformed into the original acceptor molecule, oxaloacetate, which may be used to begin the cycle again.> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Analysis>

26. Many people like to eat apple cores, seeds and all. In moderation, there is no danger from this. But if consumed in great quantity over a short period of time, this could be dangerous because apple seeds contain cyanide. Why is it dangerous? Be as specific as possible. <Answer: When certain toxins such as cyanide or azide are present in cells, electrons in the mitochondrial electron transport chain cannot pass from cytochrome c to oxygen. This prevents the formation of ATP and cell dies.> <A-head: Types of Respiration> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

27. What effect will global warming have on the rate of cellular respiration in plants? What effect might this have on overall growth rate? <Answer: In most tissues, a 10° increase in temperature will double the respiration rate, in the temperature range of 5°C and 25°C. Most global warming models predict an increased temperature change that will speed up respiration rates but will not be so great as to exceed temperature thresholds and cause enzyme or other cellular damage. Increased respiration rates will provide plants with greater access to energy and carbon required to synthesize biomass.

However, ultimately growth rates are controlled by the balance of respiration and photosynthesis so both processes must be taken into account when predicting growth rates.> <A-head: Environmental and Internal Factors> <Subject: Chapter 12> <Complexity: Difficult> <Taxonomy: Application>

28. Why is it important to grow houseplants in containers that have a hole in the bottom? <Answer: As the final electron acceptor of the mitochondrial electron transport chain, oxygen is required for aerobic cellular respiration to occur. In waterlogged soils, there is less oxygen available to the root systems. Although anaerobic respiration will allow roots to survive over short periods, it is not sufficient for root growth and the maintenance of the healthy metabolism.

When continually flooded, the roots of many tree species die, ultimately resulting in the death of the plant due to the inability to absorb water and minerals. Holes in the bottoms of pots allow excess water to drain out, thereby maintaining aerobic conditions in the soil.> <A-head: Environmental and Internal Factors> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

29. The energy released in the aerobic oxidation of glucose is 686 kcal/mole; the hydrolysis of ATP yields 7.3 kcal/mole. If 36 molecules of ATP are produced per glucose, what is the efficiency of this process? If your answer is less than 100%, what happened to the energy that does not end up in ATP? <Answer: Assuming that 36-36 molecules of ATP are generated during aerobic respiration, the efficiency is 38% to 40%. Approximately 60% of the energy is lost as heat.> <A-head: Total Energy Yield of Respiration>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Analysis>

30. Why do most wines contain no more than 12% to 14% ethanol? <Answer: Ethanol is produced by the fermentation (anaerobic respiration) of glucose by yeasts.

Ethanol production kills the yeast if it builds up to a concentration of 18% to 20%. Most U.S. wines contain 12% to 14% ethanol and still have some sugar left, so the fermentation is stopped artificially.> <A-head: Fermentation of Alcoholic Beverages> <Subject: Chapter 12> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 13—Transport Processes

Multiple Choice

1. Carbon dioxide molecules diffuse from the atmosphere into plant cells. This movement is due to: A) a molecule pump powered by ATP produced in respiration.

B) a molecule pump powered by ATP produced in photosynthesis.

C) the random movement of CO2 molecules.

D) CO2 molecules being carried along with a stream of water.

E) wind currents. <Answer: C> <A-head: Diffusion, Osmosis, and Active Transport> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Application>

2. If the water potential of mature mesophyll leaf cells is –0.1 MPa, then water will diffuse: A) to root cells.

B) to developing fruit cells.

C) to expanding leaves at the stem tip.

D) into cells of the phloem.

E) It is impossible to predict with this information. <Answer: E> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Application>

3. If a cell has a water potential of –0.3 MPa, water will diffuse from that cell toward a cell whose water potential is:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) –0.4 MPa. B) −0.3 MPa. C) −0.2 MPa. D) −0.1 MPa. E) 0 MPa. <Answer: A> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Application>

4. When a plant cell is placed in deionized (pure) water: A) water diffuses into the cell because the water potential of the cell is less negative than the water potential of the deionized water.

B) water diffuses into the cell because the water potential of the cell is more negative than the water potential of the deionized water.

C) the cell and water are at equilibrium, so there is no net movement of water into or out of the cell.

D) water diffuses out of the cell because the water potential of the cell is less negative than the water potential of the deionized water.

E) water diffuses out of the cell because the water potential of the cell is more negative than the water potential of the deionized water. <Answer: B> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Analysis>

5. Just after the sun rises, the initial event that leads to the opening of stomatal pores is the: A) diffusion of water into guard cells from adjacent cells.

B) diffusion of water out of guard cells into adjacent cells.

C) active transport of K+ into guard cells from adjacent cells, lowering the water potential.

D) diffusion of K+ into guard cells from adjacent cells, lowering the water potential.

E) synthesis of glucose, lowering the water potential in the guard cells. <Answer: C>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Short Distance Intercellular Transport> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Application>

6. To move sucrose molecules from leaves to roots, the sucrose molecules must enter sieve elements. This involves: A) moving the molecules through plasmodesmata.

B) the use of ATP to pump sucrose molecules with their concentration gradient across the plasma membrane.

C) the use of ATP to pump sucrose molecules against their concentration gradient across the plasma membrane.

D) the pumping of sucrose molecules across the plasma membrane with their concentration gradient without the use of ATP. E) the pumping of sucrose molecules across the plasma membrane against their concentration gradient without the use of ATP. <Answer: C> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Analysis>

7. Which of the following about phloem translocation is false?

A) It is explained by pressure flow hypothesis.

B) At sources, sugars are actively transported to sieve elements.

C) Water moves out from sieve elements at sources.

D) At sinks, sugars are actively transported from sieve elements to sink cells. <Answer: C> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Analysis>

8. The source of energy driving the transport of water through the xylem is:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

A) the sun.

B) ATP generated by respiration.

C) ATP produced by photosynthesis.

D) NADPH produced by photosynthesis.

E) transpiration. <Answer: A> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Analysis>

9. Which of the following statements is true?

A) The contents of the phloem are under pressure and the contents of the xylem are under tension.

B) Sieve tube members are empty, dead cells.

C) Vessel element protoplasm is unique because the vacuolar membrane disintegrates, allowing vacuolar water to mix with the cytosol.

D) Phloem sap movement is driven by the atmospheric water potential.

E) Phloem sap moves only short distances, whereas xylem sap moves long distances. <Answer: A> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Application>

10. Water moves from the soil into the xylem of a root via: A) transpiration.

B) diffusion and osmosis.

C) active transport.

D) membrane vesicles.

E) translocation. <Answer: B> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Application>

True/False

11. Membranes are effective barriers against the diffusion of water and can prevent water from entering or leaving a cell. <Answer: False> <A-head: Diffusion, Osmosis, and Active Transport> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Application>

12. Transfer cells are involved in rapid short-distance transport. <Answer: True> <A-head: Short-Distance Intercellular Transport> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Recall>

13. Sources of sugars that are transported in the phloem are always leaves. <Answer: False> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Application>

14. Although water normally diffuses into a root, it is possible, under the right conditions, for water to diffuse out of a root. <Answer: True>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Application>

15. If a plant is exposed to blue light and abscisic acid simultaneously, the stomata will remain open. <Answer: False> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Analysis>

Fill-in-the-Blank

16. The speed by which water can move through membranes can be increased by the presence of , which act as protein channels through the membrane. <Answer: aquaporins> <A-head: Diffusion, Osmosis, and Active Transport> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Recall>

17. If a plant cell continues to lose water beyond the point at which the protoplast pulls completely away from the walls, it is said to be . <Answer: plasmolyzed> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 18. Most small molecules easily move through the consists of both the cell wall and intercellular spaces.

of the plant, which

<Answer: apoplast> <A-head: Short-Distance Intercellular Transport> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Recall>

19. The rate at which sugar and nutrients are transported through the phloem is known as . <Answer: mass transfer> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Recall>

20. In order for water to move up a stem, on average the leaf water potential must be 0.2 MPa more negative than root water potential in order to account for and gravity. <Answer: friction> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each tissue with its most plausible water potential value.

A) Leaves of plants in well-watered soil with good leaf growth <Answer: −0.5 MPa> B) Leaves at full turgor <Answer: 0.0 MPa>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) Most roots in dry soil <Answer: −0.2 MPa> D) Leaves of plants in very dry soil <Answer: −3.0 MPa> E) Dry seeds capable of germination <Answer: −20 MPa> F) Leaves of desert shrub in extremely dry soil <Answer: −6.0 MPa> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Easy> <Taxonomy: Recall>

22. Match each feature of long-distance transport in xylem with its corresponding description.

A) Embolism <Answer: Air bubble> B) Cavitation <Answer: Breaking of water columns under extreme tension> C) Adhesive properties of water <Answer: Wall-induced reinforcement of water column> D) Cohesive properties of water <Answer: Loss of water from the tracheary element through the transstomatal transpiration> E) Xylem transport in pendant epiphytes <Answer: Water moves in direction of gravity> F) Xylem transport in stolons <Answer: Water movement unimpeded by gravity> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Analysis>

Essay

23. Why do most land plants have xylem and phloem?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Vascular tissues such as xylem and phloem are selectively advantageous for shoots to grow upright, thereby elevating leaves into sunlight above competing plants. The vascular system allows sugars synthesized by photosynthesis in the leaves to travel downward to other plant parts. The elevation of photosynthetic tissues also enabled plants to elevate their reproductive structures, thus making them more visible to pollinators and more accessible to wind currents.> <A-head: Concepts> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Application>

24. Why does a cell store carbohydrate as starch rather than glucose? <Answer: Osmotic potential is related to the number of particles present in solution. A solution composed of 2 g of glucose in 100 mL of water has an osmotic potential twice as negative as a solution containing only 1 g. If a molecule of starch contains 1,000 glucose units, then a cell storing carbohydrate as starch will have a more positive osmotic potential and therefore water potential than one storing an equivalent amount of glucose. The cell containing 1,000 glucose units would have a much more negative water potential, and this will affect water movement through the plant body.> <A-head: Water Potential> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Application>

25. Most plants must have an adequate supply of 16 essential elements to survive. Why is potassium one of those elements? <Answer: Potassium is an essential element in plants because it is a critical component of shortdistance transport systems. K+ is actively transported into guard cells, lowering their water potential and causing them to open as they swell with water moving into the cells by osmosis.

Potassium is also used in the trigger hairs of Venus flytraps. The trap opens as potassium accumulates in the motor cells located along the midrib of the modified leaf that serves as the trap. As water diffuses in the motor cells, they become turgid and the trap opens. When the trap closes, the membrane suddenly becomes freely permeable and the potassium and water rush out of the cells. As the motor cells collapse, the trap closes.> <A-head: Short-Distance Intercellular Transport> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

26. Both the inner membrane of a mitochondrion and the plasma membrane of a transfer cell have numerous folds. Why is this advantageous in both cases? <Answer: Transfer cell walls have numerous fingerlike outgrowths on the interior surface along which the plasma membrane adheres. The folding of the plasma membranes of transfer cells and mitochondrion drastically increase the surface area to hold many molecular pumps, allowing high-volume transport of materials.> <A-head: Short-Distance Intercellular Transport> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Recall>

27. What happens in the sieve elements of a vascular bundle if an insect takes a bite out of a stem, severing the vascular bundle? <Answer: P-protein is found as a fine network adjacent to the plasma membrane. When a sieve element is injured, the phloem sap surges toward the break, sweeping the P-protein into the center of the cell where it becomes a tangled mess. It is too large to pass through the sieve area or sieve plate and forms a P-protein plug. A second polymer, callose, precipitates out of the protoplasm due to pressure drop and contributes to the plug formed by the P-protein. Together these compounds work to seal the rupture and prevent further leaking.> <A-head: Long-Distance Transport: Phloem> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Recall>

28. What limits the height of trees? Why? <Answer: Due to the adhesive qualities of water, water molecules adhere to the cell walls of the tracheary elements, providing them with extra strength. It is this feature that allows trees like redwoods to reach heights of 100 meters. Water is pulled upward through the trees but 100 m is the limit for xylem, as the water column cannot support chains of water greater than 100 m, even with the reinforcement provided by the cell walls.> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

29. Would it be possible for a net movement of water out of roots into the soil to occur? Justify your answer. <Answer: Water moves in the direction of the more negative water potential. In some cases as soils dry, their water potentials decline and may become lower than those of the root tissues. In this case the tension on the water column increases and the hydrogen bonding breaks, causing a break in the water column, called cavitation. Molecules above the cavitation point are drawn upward and those below rush downward with some of that water moving out of the root and into the soil.> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Difficult> <Taxonomy: Analysis>

30. Is transpiration always detrimental to a plant? Why or why not? <Answer: When the soil water supply is adequate, transpiration is actually beneficial to a plant.

The upward movement of water is the primary means by which minerals are transported through the plant. Transpiration is also responsible for evaporative cooling, which can protect leaf tissues from damage in high-heat situations. However, if soil water is in limited supply, transpiration can be lethal for the plant.> <A-head: Long-Distance Transport: Xylem> <Subject: Chapter 13> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 14—Soils and Mineral Nutrition

Multiple Choice

1. Organisms capable of nitrogen fixation include which of the following?

A) Plants and cyanobacteria B) Fungi and bacteria C) Plants and fungi D) Bacteria and cyanobacteria E) Plants and bacteria <Answer: D> <A-head: Concepts> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

2. Which is a true statement regarding element essentiality?

A) Fruits can develop without zinc; therefore, it is not essential.

B) Rubidium, an element similar to potassium, can substitute for the essential element potassium in a plant’s growth and development.

C) The function of boron is unknown, but growth will not occur without it; therefore, it is essential.

D) Plants usually contain aluminum; therefore, that element is essential for plant growth.

E) Plants grow much better in a nutrient solution containing EDTA, a compound that keeps iron in solution; therefore, EDTA is essential for plant growth. <Answer: C> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Analysis>

3. Which of the following is false about criteria for essentiality for a mineral?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

A) The element must be necessary for complete plant development in its entire life cycle.

B) The element cannot be substituted to be effective and must be necessary by itself.

C) The element can be substituted to be effective.

D) The element must be acting inside the plant and not outside of it. <Answer: C> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Analysis>

4. Which of the following plants would be most likely to develop deficiency symptoms?

A) A plant growing in a tropical rain forest with associated mycorrhizae B) Cacti growing in the desert C) A crop plant growing in a field that has been cultivated for many years D) An aquatic plant growing in lake sediments E) A plant growing naturally on serpentine soil <Answer: C> <A-head: Mineral Deficiency Diseases> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

5. If a plant is deficient in phosphorus, a typical deficiency symptom would be: A) older leaves that are purple.

B) young leaves that are purple.

C) older leaves that are chlorotic.

D) young leaves that are necrotic.

E) dead apical meristems. <Answer: A> <A-head: Mineral Deficiency Diseases> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. Essential elements would not be made available to plants by: A) alternate freezing and thawing of rock.

B) runoff from rainstorms and avalanches.

C) acids released by an organism called a lichen.

D) the release of CO2 by roots.

E) being incorporated into the crystalline matrix of rocks. <Answer: E> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Analysis>

7. In a tropical rain forest, where there is abundant vegetation and rainfall, the soil contains: A) abundant cations due to soil acidity but few anions due to leaching.

B) abundant cations and anions due to soil acidity.

C) abundant anions and few cations due to soil acidity and leaching.

D) few cations or anions due to soil acidity, leaching, and negatively charged soil particles.

E) high concentrations of hydroxyl ions, making the soil alkaline. <Answer: D> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Analysis>

8. If cubic meters of different soils are compared, in which is the total surface area of all particles greatest?

A) Coarse sand B) Fine sand C) Silt D) Clay E) All would have the same total surface area. <Answer: D>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Analysis>

9. The conversion of N2 into ammonia is called: A) nitrogen reduction.

B) nitrate reduction.

C) nitrogen assimilation.

D) denitrification.

E) nitrogen fixation. <Answer: E> <A-head: Nitrogen Metabolism> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

10. If populations of free-living bacteria in the soil make nitrogen from the air available to organisms and that nitrogen is eventually absorbed and incorporated into plants, what is the most likely sequence of events in the soil and then in the plants?

A) Nitrogen fixation → nitrate reduction → ammonia oxidation → nitrite reduction B) Nitrogen fixation → ammonia oxidation → nitrite reduction → nitrate reduction C) Nitrogen fixation → ammonia oxidation → nitrate reduction → nitrite reduction D) Nitrate reduction → nitrite reduction → ammonia oxidation → nitrogen fixation E) Ammonia oxidation → nitrogen fixation → nitrate reduction → nitrite reduction <Answer: E> <A-head: Nitrogen Metabolism> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. The majority of elements essential for animal growth are obtained directly from the soil. <Answer: False> <A-head: Concepts> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

12. Strontium, which is chemically similar to calcium, can substitute for that element in most plants. <Answer: False> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

13. In some environments, plants cannot grow because excessive minerals in the soil cause osmotic drought. <Answer: True> <A-head: Mineral Deficiency Diseases> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

14. The process of respiration in root cells can make cations in the soil available for absorption into a root. <Answer: True> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 15. Plant tissues only need small amounts of minerals compared with animals. <Answer: True> <A-head: Storage of Minerals Within Plants> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. In research regarding the mineral nutrition of plants, a can be substituted for soil, enabling the researcher exact control of the chemical composition of the growing medium. <Answer: hydroponic solution> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

17. One symptom of mineral deficiency is chlorophyll, turn yellowish, and become brittle or papery.

, when leaves lack

<Answer: chlorosis> <A-head: Mineral Deficiency Diseases> <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Recall>

18. Cations can be liberated from soil micelles and dissolve in the soil solution through the process of . <Answer: cation exchange> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Application>

19. The stems of epiphytic contain hollow chambers that ants use as living spaces and the plants use as a source of nitrogenous compounds. <Answer: ant-plants> <A-head: Nitrogen Metabolism> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

20. The mineral holding substance to store amino acids and essential elements.

permits the protein body of a seed

<Answer: phytin> <A-head: Storage of Minerals Within Plants> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Application>

Matching

21. Match each element with its corresponding function.

A) Sulfur <Answer: Component of amino acids> B) Copper <Answer: Component of plastocyanin> C) Potassium <Answer: Stomatal opening and closing> D) Magnesium <Answer: Component of chlorophyll> E) Manganese <Answer: Chlorophyll synthesis> F) Calcium <Answer: Integrity of membranes>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank G) Molybdenum <Answer: Nitrogen reductase> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each term with its function.

A) Nitrogenase <Answer: An enzyme that uses N2 as a substrate> B) Allantoic acid <Answer: Common transport form of nitrogen in legumes> C) Nitrate reductase <Answer: Enzyme that carries electrons by means of a molybdenum atom> D) Glutamate <Answer: Initial acceptor molecule of nitrogen assimilation> E) Nitrification <Answer: Oxidation of nitrogen> F) Nitrogen fixation <Answer: Reduction of nitrogen> <A-head: Nitrogen Metabolism > <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Imagine that an expedition has returned from Mars, where they found a plantlike organism.

Chemical analysis of the organism reveals that it contains 27 different chemical elements.

Describe the experiments you would perform to determine which elements are essential for the organism. What three things would you have to prove for each element? <Answer: Julius von Sachs’s hydroponic experiments could be replicated, in which one begins by growing the plant in the 27 different elements. Each successive trial would remove one element to determine if that one was essential for growth. Essentiality for growth would be based

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank on three premises: (1) the element is necessary for the plant to complete its life cycle, (2) the element cannot be substituted with another element, and (3) the element must be acting within the plant. If an element was determined not to be essential, it would continue to be removed from the solution used for further trials.> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Application>

24. Why has it been so difficult to determine that elements such as molybdenum and chlorine are essential for plants? <Answer: Chlorine is required in such minute amounts that plants are provided with enough chlorine by the touch of a human finger. Experimenters needed to wear gloves to determine if chlorine was an essential element.> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

25. Explain why a plant root might contain nonessential elements. <Answer: The endodermis of the root cannot exclude all minerals completely. A living plant usually contains at least trace quantities of every element in the soil whether it is necessary for the plant or not.> <A-head: Essential Elements> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Recall>

26. The formation of potholes in the streets of New York City is an example of what type of weathering? Describe the process. <Answer: Potholes are an example of physical weathering. This weathering is due to the breakdown of the asphalt by the physical processes of freeze-thaw cycles and driving over the streets. Wind and rain are other physical processes that contribute to the breakup of asphalt into smaller pieces, forming potholes.> <A-head: Soils and Mineral Availability>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 14> <Complexity: Easy> <Taxonomy: Application>

27. Acid precipitation can alter soil and kill the small roots of a plant. Discuss one reason why acid precipitation might kill roots. <Answer: Acid precipitation leads to increased concentrations of protons and a decline in soil pH. In acidic soils, aluminum and manganese can become so soluble as to reach toxic levels. The buildup of protons in the soil also causes leaching of base cations, leading to infertile soils that could result in root death.> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

28. Many houses in the Philadelphia, Pennsylvania, area have azalea bushes in the yard that bloom profusely in the spring, whereas very few houses in Columbus, Ohio, do. What does this suggest might be a difference in the soil in these two cities? <Answer: Azalea bushes require acidic soils to survive. When planted in alkaline soils, azaleas show stress symptoms immediately, often due to a lack of iron. The Philadelphia soils are likely to be more acidic than those in Columbus.> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Difficult> <Taxonomy: Analysis>

29. Why do some farmers often rotate crops, such as corn or wheat, with legumes, such as alfalfa or clover? <Answer: Alfalfa and clover form a symbiotic relationship with nitrogen fixing bacteria that fix atmospheric nitrogen into forms of nitrogen that can be used by plants. These crops do not need costly nitrogenous fertilizers added and add plant usable forms of nitrogen to the soil.

Alternating the nitrogen-fixing crops with non-nitrogen fixing crops improves the nitrogen availability in the soil.> <A-head: Nitrogen Metabolism> <Subject: Chapter 14>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Analysis>

30. Although acid rain is usually considered detrimental, it can also be beneficial to a plant.

Why? <Answer: In acidic soils, the high concentration of protons causes more cations to be released from soil micelles. The released cations are then available for plant uptake.> <A-head: Soils and Mineral Availability> <Subject: Chapter 14> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 15—Development and Morphogenesis

Multiple Choice

1. If the site of perception of an environmental stimulus is a leaf, then the site of transduction is a: A) root.

B) stem.

C) leaf.

D) bud.

E) flower. <Answer: C> <A-head: Environmental Complexity> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Recall>

2. The most common long-term response of a plant part to environmental stimuli is: A) rapid movement.

B) growth.

C) hormone production.

D) turgor changes.

E) no response. <Answer: B> <A-head: Responding to Environmental Stimuli> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

3. Flowers of night-blooming cereus are open at night and closed during the day. This is an example of a:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) positive phototropic response.

B) negative phototropic response.

C) photonastic response.

D) photomorphogenic response.

E) positive phototaxis response. <Answer: C> <A-head: Responding to Environmental Stimuli> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Application>

4. Which of the following was the first plant hormone discovered?

A) Gibberellin B) Auxin C) Abscisic acid D) Ethylene E) Cytokinin <Answer: B> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Recall>

5. If a plant shoot grows too large, the root system cannot adequately supply it with water and minerals. This happens because the: A) leaves produce cytokinins.

B) leaves produce gibberellins.

C) stems produce cytokinins.

D) roots produce cytokinins.

E) roots produce gibberellins. <Answer: D> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. A piece of mica is inserted in the tip of a grass coleoptile and the seedling is exposed to unilateral light. In response, the coleoptile: A) bends toward the light because of a redistribution of auxin.

B) bends away from the light because of a redistribution of auxin.

C) bends toward the light because of destruction of auxin on the lighted side.

D) bends toward the light because extra auxin is synthesized on the darker side.

E) does not bend because auxin cannot be redistributed. <Answer: E> <A-head: Hormones as Signals of Environmental Factors> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Analysis>

7. If all of the fruits (seeds) on the outside of a strawberry are removed when it is young, it will not develop. If the strawberry is sprayed with a hormone solution, normal development occurs.

The hormone most likely to cause this response is: A) cytokinin.

B) gibberellin.

C) ethylene.

D) auxin.

E) abscisic acid. <Answer: D> <A-head: Hormones as Signals of Environmental Factors> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Analysis>

8. If plants are kept in the dark, they have elongated internodes, making them spindly. This effect is due to: A) an overproduction of auxin.

B) phytochrome in the Pr form.

C) phytochrome in the Pfr form.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank D) an overproduction of gibberellins.

E) a lower auxin/cytokinin ratio in the stems. <Answer: B> <A-head: Flowering> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Application>

9. Endogenous rhythms: A) cycle between three states.

B) are dependent on exogenous temperature.

C) are dependent on exogenous light.

D) equip a negative feedback loop. <Answer: D> <A-head: Flowering> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Application>

10. If you want to produce a bushier shrub with more branches, you should: A) water the plant with a gibberellin solution.

B) spray the plant with cytokinin.

C) spray the plant with auxin.

D) pinch off the stem tips.

E) pinch off the root tips. <Answer: D> <A-head: Activation and Inhibition of Shoots by Auxin> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Application>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. The different parts of a terrestrial plant live in different microenvironments. <Answer: True> <A-head: Environmental Complexity> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Recall>

12. Vines exhibit a positive thigmotropic response if they encounter a fence post. <Answer: True> <A-head: Responding to Environmental Stimuli> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Application>

13. In most cases, hormones are synthesized at their place of effect. <Answer: False> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

14. The development of spine clusters in prickly pear cacti can be controlled by cytokinins, which stimulate pad growth, or gibberellins, which stimulate the development of additional spines. <Answer: True> <A-head: Interactions of Hormones in Shoots> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 15. According to the ABC model of flower part organization, a combination of B and C genes produces stamens. <Answer: True> <A-head: Flowering> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. In order for a plant to respond to its environment, it must perceive the environmental change and the information must be before it can elicit a response. <Answer: transduced> <A-head: Environmental Complexity> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Application>

17. The ability of sperm cells to swim toward egg cells is due to response in which a cells swims away or toward a stimulus.

,a

<Answer: taxis> <A-head: Responding to Environmental Stimuli> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

18. In order to cause perception and response of a stimulus, in some cases, a minimum must be present during the presentation time. <Answer: threshold> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Application>

19. There are specialized cells in root caps that contain response to gravity, thereby informing the roots to grow down.

that sink in

<Answer: statoliths> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Easy> <Taxonomy: Application>

20. The action spectrum of phototropism matches that of a pigment called . <Answer: phototropin> <A-head: Hormones as Signals of Environmental Factors> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each hormone with a response it elicits.

A) Salicyclic acid <Answer: Resistance to pathogens> B) Ethylene <Answer: Fruit abscission> C) Auxin <Answer: Apical dominance> D) Cytokinin <Answer: Prevention of fruit abscission> E) Brassinosteroids <Answer: Leaf morphogenesis> F) Abscisic acid <Answer: Stomatal closure>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank G) Gibberellin <Answer: Release of seeds from dormancy> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each photoreceptor with an example of its common effects.

A) Zeitlupe <Answer: Circadian rhythms and photoperiod-controlled flowering> B) Phytochromes <Answer: Seed germination and seedling development> C) Phototropins <Answer: Bending relative to light> D) Cryptochromes <Answer: Photomorphogenesis and circadian rhythms> E) UVR8 <Answer: Initiates stress responses> <A-head: Flowering> <Subject: Chapter 15> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Distinguish between morphogenesis and differentiation. <Answer: Morphogenesis describes how a plant develops to have a characteristic shape and structure. Differentiation describes the process by which individual cells become specialized and therefore increase the overall complexity of the plant.> <A-head: Concepts> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

24. What is the advantage to a biennial plant of vernalization? <Answer: Biennial plants that require vernalization spend their first year in a vegetative phase and can only enter a reproductive phase after receiving a cold winter. In their second year, the short nights of early summer stimulate flowering but in their first year they don’t respond to that stimulus. The advantage of vernalization is that in the second year of growth, the plants flower relatively early in the summer, when the environmental conditions are suitable for successful growth and the reproductive structures have long exposure to pollinator activity. These plants would not have the energetic resources to become reproductive so early in their first year of growth.> <A-head: Environmental Complexity> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Analysis>

25. Explain why gardeners do not have to worry about planting seeds upside down. <Answer: The first structure to emerge from a germinating seed is the embryonic root, which is followed by the embryonic shoot. Most roots are positively gravitropic, growing downward in response to gravity. Shoots are negatively gravitropic, growing upward in response to gravity.

Thus the newly germinated seedling orients itself regardless of seed orientation.> <A-head: Responding to Environmental Stimuli> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Application>

26. Explain the expression “One rotten apple spoils the barrel.” <Answer: Climacteric fruits such as apples ripen slowly, but once they are in the final stages, developmental changes occur rapidly, such as softening of cell walls, color changes, conversion of starch to simpler sugars, and flavor development. Ethylene, a gaseous hormone, stimulates these changes. At first so little ethylene is present that the changes occur slowly; however, one of its effects is that once it is produced, it stimulates further production through a continuous positive feedback system. When one apple produces ethylene, it stimulates surrounding apples to do the same.> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Difficult>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Analysis>

27. Salicylic acid, from which aspirin is derived, is produced in the inner bark of willow (Salix) trees. Willow trees do not get headaches, so what good is salicylic acid to the tree? <Answer: Salicylic acid is involved in the resistance to pathogens. This hormone activates disease-resistant genes throughout the plant, a process known as systemic acquired resistance.> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Analysis>

28. If a hormone is transported from its site of synthesis to all parts of a plant, why don’t all parts of the plant respond to the hormone? <Answer: Hormones are released into general circulation and are not carried to specific target cells. Many regions do not respond to the hormone, however, because they don’t have the proper receptor molecules. Hormone receptors are often embedded in membranes and cause a metabolic response once they bind with the hormone, creating a hormone-receptor complex.> <A-head: Communication Within the Plant> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Application>

29. A row of intact oat coleoptiles receives unilateral light that has first passed through a prism.

Describe the appearance after 2 hours of the coleoptiles receiving each component color of white light: violet, blue, green, yellow, orange, and red. <Answer: Light is the stimulus of phototropisms and blue is the most effective wavelength. After 2 hours of blue light, the oat coleoptile will grow toward the light as the darker side of the coleoptile receives more auxin and thus grows more rapidly, bending toward the light. The other colors of light would not be expected to impact the coleoptile growth.> <A-head: Hormones as Signals of Environmental Factors> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

30. Poinsettias are short-day plants. How could you get a plant to bloom at the equator? <Answer: Short-day plants are really long-night plants that bloom at critical night lengths and longer. Experiments have shown that if plants such as poinsettias are given artificially long nights, they can be induced to flower.> <A-head: Flowering> <Subject: Chapter 15> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank Chapter 16

Multiple Choice

1. In the development of a fiber cell, which of the following genes would you expect to be turned on in the developing fiber cell but not in a developing parenchyma cell?

A) Genes for production of cellulose B) Genes for production of hemicelluloses C) Genes for production of lignin D) Genes for production of rRNA E) Genes for production of tRNA <Answer: C> <A-head: Concepts> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

2. A cell cannot survive without: A) a nucleus.

B) mitochondria.

C) ribosomes.

D) chloroplasts.

E) a nucleolus. <Answer: C> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Analysis>

3. Histone proteins: A) act as enzymes.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) are the basis for nucleosomes in chloroplasts.

C) protect nuclear DNA and act as the structural basis of nucleosomes.

D) are the main regulator of transcription. <Answer: C> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Analysis>

4. If a polypeptide chain contains 144 amino acids, what is the minimum number of nucleotides in the mRNA coding for this chain?

A) 72 B) 144 C) 288 D) 432 E) 576 <Answer: D> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

5. Which of the following is the most energy-efficient level of control of metabolism?

A) Processing of hnRNA into mRNA B) Rate of translation C) Rate of transcription D) Activation/inactivation of an enzyme E) Rate of mRNA transport from nucleus to hyaloplasm <Answer: C> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. If a segment of DNA has the base sequence ATTGCA, the RNA transcribed from it would be: A) TAACGT. B) UAACGU. C) GCCAUG. D) GCCATG. E) CGGTAC. <Answer: B> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Application>

7. If a DNA segment has the base sequence TGACTCAAGCTT, then the anticodons of the tRNA molecules that bind to the mRNA transcribed from that sequence would be: A) ACU, GAG, UUC, GAA. B) TGA, CTC, AAG, CTT. C) UGA, CUC, AAG, CUU. D) UG, AC, UC, AA, GC, UU. E) UGAC, UCAA, GCUU. <Answer: C> <A-head: Protein Synthesis> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

8. During translation, the mRNA binds first to: A) the DNA molecule and RNA polymerase II. B) RNA polymerase II only.

C) the nuclear envelope and initiation factors.

D) the large subunit of a ribosome, complexed with initiation factors and an initiation tRNA. E) the small subunit of a ribosome, complexed with initiation factors and an initiation tRNA. <Answer: E> <A-head: Protein Synthesis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

9. If you isolated a particular chromosome from the cells of a plant, isolated the DNA, melted it, then found that DNA hybridization occurred quite rapidly, that chromosome would probably contain: A) multiple copies of rRNA and/or tRNA genes.

B) multiple copies of chloroplast protein genes.

C) a few copies of rRNA genes.

D) a few copies of tRNA genes.

E) multiple copies of glycolysis enzyme genes. <Answer: A> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

10. If you were to run a DNA hybridization test on the following plants, in comparison to a daisy, which of the following would you expect to have the least amount of DNA annealed to the DNA of daisy? Plants tested: daisy, fern, brown alga, liverwort, pine.

A) Liverwort B) Pine C) Fern D) Brown alga E) Daisy <Answer: D> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Application>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. All genes contain the information to make a particular protein or polypeptide. <Answer: False> <A-head: Concepts> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

12. Exon sequences are responsible for protein expression. <Answer: True> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Recall>

13. The transcription of all RNA molecules is catalyzed by the same type of RNA polymerase; therefore, transcription is easier to control. <Answer: False> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

14. In all proteins, the first amino acid is methionine. <Answer: True> <A-head: Protein Synthesis> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall>

15. DNA microarrays are used to study gene expression.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: True> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. The genetic code is said to be most amino acids.

because multiple codons exist for

<Answer: degenerate> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Easy> <Taxonomy: Recall>

17. When nucleotides are read in the wrong sets of three, this is called a , and the result is often useless proteins. <Answer: frameshift error> <A-head: Protein Synthesis> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall>

18. Proteins called enhancer regions.

control gene activity by binding to promoter or

<Answer: transcription factors> <A-head: Control of Protein Levels> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

19. Plastid DNA can be cut with restriction endonucleases and separated by gel electrophoresis and made visible by staining to create a of the DNA. <Answer: restriction map> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall>

20. A commonly used vector for inserting bacterial genes into plants is a from the bacterium Agrobacterium tumesfasciens. <Answer: ti plasmid> <A-head: Genetic Engineering of Plants> <Subject: Chapter 16> <Complexity: Moderate > <Taxonomy: Recall>

Matching

21. Match each process with its mediating enzyme.

A) Translocation of ribosome along mRNA <Answer: Large ribosomal activating enzymes> B) Preparation of small ribosomal unit to bind mRNA <Answer: Eukaryotic initiation factors> C) DNA digesting enzymes <Answer: DNases> D) Transcription of hnRNA <Answer: RNA Polymerase II> E) Attachment of amino acid to tRNA <Answer: Amino acid activating enzyme> F) Transcription of most rRNA <Answer: RNA polymerase I> <A-head: Storing Genetic Information>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

22. Match each statement with a corresponding structure.

A) Used as vector in plant engineering <Answer: Ti plasmid> B) Excised from hnRNA <Answer: Intron> C) Necessary for attachments of NA polymerase II <Answer: TATA box> D) Controls transcription <Answer: Promoter> E) Synthesized using reverse transcriptase <Answer: cDNA> F) Contains the complete information to make a polypeptide <Answer: Structural gene> [G] Contains genes for rRNA <Answer: Nucleolar DNA> [H] Codes for part of mature mRNA <Answer: Exon> [I] Stops transcription <Answer: Multiple adenine sequence> <A-head: Protein Synthesis> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

Essay

23. If a nucleic acid is single-stranded, how could you determine whether it is RNA or DNA? <Answer: A single-stranded nucleic acid of DNA contains thymine and no uracil, whereas RNA contains uracil in place of thymine. DNA structure is synthesized with deoxyribose, whereas RNA uses ribose.> <A-head: Concepts>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Application>

24. What kind of experiment(s) would you perform to determine how closely related two species of plants are? <Answer: DNA sequence analysis enables species identification to be determined. When comparing sequences for different species, one can examine the number of differences that occur. Closely related species will have fewer differences than distantly related species.> <A-head: Storing Genetic Information> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

25. Why is protein synthesis such a complicated process? <Answer: Because structure and function are inextricably linked, completion of organism’s’ life cycles require that proteins function correctly. The complicated process of protein synthesis ensures that the correct copy of the instructions is made (transcription), that the instructions are correctly translated to new amino acid sequences (translation), and that ultimately the correct protein is synthesized. The process must be efficient, controlled, and precise.> <A-head: Protein Synthesis> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

26. Outline the steps of initiation, elongation, and termination of mRNA translation. <Answer: Protein synthesis begins with a complex initiation process that prepares the ribosomal subunits for reading and aligning with the mRNA. During elongation, the mRNA lies within a channel between the ribosome subunits, and tRNA molecules carry amino acids to the ribosomemRNA complex. Enzymes located in the large ribosomes subunits attach the amino groups in a repetitive process, beginning with methionine and its start codon and ending when a stop codon is reached. When the protein is completed, it releases and the ribosome subunits detach from each other.> <A-head: Protein Synthesis> <Subject: Chapter 16>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Recall>

27. Briefly describe how a piece of DNA can be cloned using bacteria. Why would it be important to use the same type of restriction endonuclease in the harvest phase as in the preparation phase? <Answer: The plant DNA can be cut into fragments using restriction endonucleases and mixed with bacteria. The bacteria take up some of the DNA and are cultured on petri dishes. The cultured bacteria are tested for the protein of interest. The DNA of interest can be isolated and treated with the same restriction endonucleases used to obtain the original plant fragments. At least one fragment of the bacterial digest should match that of the plant digest; the matching fragment contains the gene of interest.> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Analysis>

28. How might a scientist use recombinant DNA techniques to make a tomato plant that is resistant to insect attacks? <Answer: Bacillus thuringiensis is a bacterium with a gene that codes for a protein that is toxic to caterpillars. The researchers can use restriction endonucleases to isolate the gene of interest and introduce it into the plant genome using a ti plasmid. Ideally the gene should be isolated as well as its promoter for the best probability of success.> <A-head: Genetic Engineering of Plants> <Subject: Chapter 16> <Complexity: Difficult> <Taxonomy: Application>

29. Explain how restriction fragment length polymorphisms can be used to evaluate how closely related two species are? <Answer: (Will vary) Restriction endonucleases always cleave DNA at a specific sequence called a palindrome. After the restriction endonucleases have acted, these fragments can be used to study heredity and the evolution of DNA. If two species are closely related, then they will have restriction fragments that are similar in size. If two species are not closely related, then their

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank restriction fragments will be of different sizes, there is restriction fragment length polymorphism.> <A-head: Analysis of Genes and Recombinant DNA Techniques> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall> 30. Explain how CRISPR-Cas9 might be used to introduce a desired gene into a DNA sequence. <Answer: (Will vary) CRISPR-Cas9 is a protein complex that binds to a short piece of guide RNA. That guide RNA is complementary to a particular region of DNA so CRISPR-Cas9 binds only to that specific site. CRISPR-Cas9 then cleaves both strands of DNA and can also insert a new sequence of DNA into the site.> <A-head: CRISPR-Cas9> <Subject: Chapter 16> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 17—Genetics

Multiple Choice

1. Which statement about DNA replication is correct?

A) DNA polymerase adds nucleotides at the 5' end of the growing nucleotide chain.

B) Short strands of DNA act as primers.

C) A replicon is an enzyme for DNA replication.

D) DNA replication is semiconservative.

E) Chromatin compacts as DNA replication begins. <Answer: D> <A-head: Replication of DNA> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Analysis>

2. Which of the following mutations is the most likely to be harmful to a plant?

A) A point mutation in an rRNA gene B) An insertion in an intron of an enzyme gene C) A deletion in spacer DNA D) An inversion in a tRNA gene E) A point mutation in the region of DNA coding for a start codon <Answer: E> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

3. Navel oranges contain no seeds, so they must be propagated vegetatively. If this seedless condition could be traced back to a single branch on a normal, seed-bearing tree, the cause of this alteration was probably a:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) somatic mutation.

B) reproductive mutation.

C) transposition.

D) mutation in one seed produced by the tree.

E) None of these are correct. <Answer: A> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

4. In Jimson weed, the allele for spiny pods is dominant to the allele for smooth pods. If two plants heterozygous for this trait are crossed, their offspring would be expected to produce pods in which of the following ratios?

A) 3 spiny pods: 1 smooth pod B) 3 smooth pods: 1 spiny pod C) 2 spiny pods: 1 smooth pod D) 4 spiny pods: 1 smooth pod E) 1 spiny pod: 2 intermediate pods: 1 smooth pod <Answer: A> <A-head: Monohybrid Crosses> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

5. Plants that “breed true” or are “pure bred” for a trait controlled by a single pair of alleles: A) are heterozygous.

B) are homozygous dominant.

C) are homozygous recessive.

D) are either homozygous dominant or homozygous recessive.

E) cannot be genotypically determined. <Answer: D> <A-head: Monohybrid Crosses> <Subject: Chapter 17> <Complexity: Easy>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

6. A particular species of plant produces flowers of two different sizes and colors: large flowers (L) are dominant to small (I), and red flowers (R) are dominant to white (r). A plant heterozygous for the two traits is selfed, producing an F1 population of 80 plants. How many plants in the F1 generation will produce large, red flowers?

A) 60 B) 45 C) 40 D) 15 E) 5 <Answer: B> <A-head: Dihybrid Crosses> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

7. Flower color in sweet peas is controlled by two genes, C/c and E/e. C is dominant to c and E is dominant to e. A plant produces purple flowers only if it contains at least one dominant allele for each gene; otherwise it produces white flowers. If two plants heterozygous for both genes are crossed, what will be the phenotypic ratio for purple: white flowers?

A) 3:1 B) 4:1 C) 9:7 D) 5:3 E) 15:1 <Answer: C> <A-head: Dihybrid Crosses> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 8. Tallness (T) is dominant to dwarfness (t) and broad leaves (B) are dominant to narrow leaves (b). To determine the genotype of a tall, broad-leaved plant, it would be best to cross it with a plant that is: A) tall with broad leaves.

B) tall with narrow leaves.

C) dwarf with broad leaves.

D) dwarf with narrow leaves. <Answer: D> <A-head: Dihybrid Crosses> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

9. The protein portion of phytochrome is misformed when plant genes are exposed to UV light.

The result is that seed germination slows and all stem growth becomes etiolated. This is an example of: A) epinasty.

B) epistasis.

C) gene linkage.

D) codominance.

E) pleiotropy. <Answer: E> <A-head: Multiple Genes for One Character> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

10. The mitochondria in some strains of Paramecium aurelia carry a gene that codes for a poison that kills other strains. If during sexual reproduction a poison-producing cell is the mother and a nonproducing cell is the father, then the F1 progeny would most likely be: A) all poison producers.

B) all nonpoison producers.

C) 50% poison producers and 50% nonpoison producers.

D) 75% poison producers and 25% nonpoison producers.

E) 25% poison producers and 75% nonpoison producers.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: A> <A-head: Other Aspects of Inheritance> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

11. In eukaryotic cells, DNA chains are replicated in one continuous piece from the 5' of the 3' end. <Answer: False> <A-head: Replication of DNA> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Recall>

12. One probable result of destruction of the ozone layer is increased mutations in plant cells, most of which will be harmful. <Answer: True> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Easy> <Taxonomy: Application>

13. An individual who is homozygous for a particular gene has two different alleles of that gene. <Answer: False> <A-head: Monohybrid Crosses> <Subject: Chapter 17> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 14. Two genes that produce 5% recombinant F1s are farther apart than two genes that produce 15% recombinant F1s. <Answer: False> <A-head: Dihybrid Crosses> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

15. Higher plants are the evolutionary product of mutations in junk DNA. <Answer: True> <A-head: Other Aspects of Inheritance> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

Fill-in-the-Blank

16. When DNA is replicated, a strand of the DNA double helix is cut; as the DNA uncoils and then separates, it forms a at each end of the replicon. <Answer: replication fork> <A-head: Replication of DNA> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

17. UV light, X-rays, and radiation are all examples of changes to DNA sequences.

that cause

<Answer: mutagens> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Applications>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

18. In some red-flowered plant species, the heterozygote condition produces less pigment than the homozygous dominant plant and therefore the flowers appear pink. In this case the pair of alleles exhibit . <Answer: incomplete dominance> <A-head: Monohybrid Crosses> <Subject: Chapter 17> <Complexity: Moderate> <Taxonomy: Application>

19. Complex traits such as leaf shape and cold hardiness are the result of many genes, a situation known as . <Answer: epistasis> <A-head: Multiple Genes for One Character> <Subject: Chapter 17> <Complexity: Easy> <Taxonomy: Recall>

20. During meiosis II, the two chromatids may remain together during anaphase II in an event known as , which causes a change in the chromosome number per nucleus. <Answer: nondisjunction> <A-head: Other Aspects of Inheritance> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Recall>

Matching

21. Match each base sequence change with the term that best describes what has happened.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) ATCTGAG → ATGTGAG <Answer: Point mutation> B) GCAGTCT → GCACT <Answer: Deletion> C) TGACTGC → TGAGCCTGC <Answer: Insertion> D) ACGATGA -→ ACTAGGA <Answer: Inversion> E) ATCTGAG → ATGTGAG or GCAGTCT → GCACTCT <Answer: Mutation> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Analysis>

22. In a species of grain, herbicide resistance (H) is dominant to herbicide susceptibility (h) and large kernel size (K) is dominant to small kernels (k). Using the accompanying information, match the crosses with their correct ratios.

A) Kk x KK <Answer: All phenotypically dominant> B) Kk x Kk <Answer: 3:1 phenotypic ratio and 1:2:1 genotypic ratio> C) kk x kk <Answer: All phenotypically recessive> D) HhKk x HhKk <Answer: 9:3:3:1 phenotypic ratio> E) Hh x hh <Answer: 1:1 genotypic ratio> <A-head: Dihybrid Crosses> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

Essay

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 23. In many large fields, the plants are exactly the same variety of crop plant. What is the potential danger in this practice? <Answer: The same variety of crop plant means that they will all be equally adapted to the environment; none will be more adapted than their parents or siblings. This situation is similar to the results of asexual reproduction. If drought or an infestation of pathogenic microorganisms affects the crop, then all the plants will be equally affected, leaving little to no yield.> <A-head: Concepts> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

24. Considering the process of DNA replication, why isn’t each molecule of DNA actually a mixture of DNA and RNA? <Answer: As DNA replication begins, the double helix opens and free nucleotides diffuse to the single-stranded DNA and pair with its pieces. These are ribonucleotides that are polymerized into short pieces of primer RNA. The primer RNAs act as substrates for DNA polymerase. As the Okazaki fragments grow, the RNAs are depolymerized and the Okazaki fragments are ligated together without any intervening pieces of RNA.> <A-head: Replication of DNA> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

25. Describe the process of DNA replication. Why does each chromosome have thousands of replication start sites instead of just one? What are Okazaki fragments? <Answer: As DNA replication begins, the double helix opens, forming a replicon. Free nucleotides diffuse to the single-stranded DNA and pair with its pieces. These are ribonucleotides that are polymerized into short pieces of primer RNA. The primer RNAs act as substrates for DNA polymerase to add deoxyribonucleotides to the 3’ end of the growing nucleic acid. Each strand of DNA is used as a template for making a complementary strand. This is semiconservative replication because each new double helix contains one new molecule and one conserved old one. As the DNA uncoils, large numbers of initiation sites allow numerous DNA polymerases to work simultaneously and thus each replicon can be replicated in as little as 1 to 3 hours. As new pieces of DNA grow at the initiation sites, these fragments are called Okazaki fragments. As the Okazaki fragments grow, the RNAs are depolymerized and the Okazaki fragments are ligated together without any intervening pieces of RNA.> <A-head: Replication of DNA>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

26. Describe how a particular plant gene could be located using transposons. <Answer: Transposons are being sequenced and engineered to contain markers. The mutants of interested are located and restriction endonucleases are used to cut the DNA. The DNA is denatured and combined with radioactive DNA complementary to the transposon. When the piece is found, it can be sequenced and the transposon sequence is known. The DNA on either side of the transposon is the gene of interest.> <A-head: Mutations> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

27. You are given a tall mustard plant. Assume complete dominance and that tall (T) is the dominant trait over short (t). How would you go about determining its genotype? <Answer: The genotype can be revealed with a test cross. The plant with the unknown genotype can be crossed with a homozygous recessive (tt) plant. If the plant being tested is homozygous dominant, then 100% of the progeny will be tall. If the plant being tested is heterozygous, 50% of the progeny will be tall and the other half will be short.> <A-head: Monohybrid Crosses> <Subject: Chapter 17> <Complexity: Easy> <Taxonomy: Application>

28. What are the roles of independent assortment and crossing over in generating variety in a population? <Answer: Independent assortment occurs if the alleles of two genes move independently of the other. During anaphase I, there is no way to predict which poles the chromosomes will migrate.

At the conclusion of meiosis II, the sister chromatids separate, resulting in four haploid cells with unique allele combinations. If two genes are located far apart on the same chromosome, crossing over may occur during prophase I and then independent assortment may occurs as described above.> <A-head: Dihybrid Crosses>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

29. Explain how a sporophyte could be triploid, tetraploid, or pentaploid. <Answer: Occasionally cells of sporophytes fail to undergo mitosis after S phase DNA replication and the cell becomes tetraploid. If they produce a part of the sporophyte that initiates flowers, pollen, or eggs, diploid cells are produced that could become triploid zygotes if fertilized by haploid gametes. This process can repeat to form pentaploid zygotes.> <A-head: Other Aspects of Inheritance> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

30. How could a plant produce striped or spotted leaves? <Answer: Plant plastids are inherited from the mother plant. During cell division, plastids move into each of the daughter cells but occasionally it receives no plastids or mutant plastids. If this occurs in the leaf primordium, the cells continue to grow and divide, resulting in a patch of cells that do not contain chlorophyll and create white or other colored spots in the leaf.> <A-head: Other Aspects of Inheritance> <Subject: Chapter 17> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 18—Population Genetics and Evolution

Multiple Choice

1. The genetic recombination that occurs during sexual reproduction is important only if: A) The two sexual partners have different genotypes.

B) The two sexual partners have different phenotypes.

C) The two sexual partners have the same genotype.

D) The two sexual partners have the same phenotype.

E) The two sexual partners are genetically identical. <Answer: A> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

2. Which of the following is not capable of evolving?

A) A population of fruit flies B) Your professor C) A culture of bacteria D) The collective cats of a city E) A particular species of plant <Answer: B> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Application>

3. Natural selection is least likely to occur in which of the following situations?

A) A population of mature oak trees with a great deal of genetic diversity produces a large number of acorns over many years in an undisturbed forest.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) A seed of a self-pollinating weed species is carried by a bird into an open, plowed field and establishes a population of that species of weed.

C) A deadly fungal disease attacks the weeds in a cornfield each year over several years.

D) Insects preferentially pollinate the individuals of a species of woodland flower that have the largest amount of nectar. <Answer: B> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Analysis>

4. All of the following are foundations of the type concept of species except: A) God created all types of organisms.

B) characters of species exhibit a range of values.

C) all aspects of earth are imperfect manifestations of a perfect ideal.

D) generalizations consider the average as representative.

E) one ideal specimen should represent a species. <Answer: B> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

5. An example of a postzygotic isolating mechanism between closely related species would be which of the following?

A) One species flowers in April, the other in August.

B) The two species are pollinated by different species of bees.

C) A zygote produced by syngamy of sperm and egg of the two species contains an uneven number of chromosomes.

D) The pollen from one species will not germinate on the stigma of the other.

E) One species is terrestrial, the other totally aquatic. <Answer: C> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Difficult>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Analysis>

6. If two species exhibit convergent evolution, the result will be: A) one species.

B) two distinct species that resemble one another very closely.

C) a variety of species produced over a short period of time.

D) two distinct species that appear very different from each other.

E) It is impossible to predict what might happen. <Answer: B> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Application>

7. In a large population of plants, which of the following is least likely to cause a change in allele frequencies within the population?

A) A forest fire that destroys part of the population.

B) Radioactive fallout from an accident at a nuclear power plant.

C) Microhabitats within the range of the population where certain phenotypes have a better chance of surviving.

D) The preference of a pollinator for a certain flower color.

E) Wind pollination of the flowers. <Answer: E> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

8. Two organisms are considered to be a distinct species if they: A) look markedly different.

B) have some metabolic difference.

C) cannot interbreed and produce fertile offspring.

D) are found in different habitats.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank E) have different mating rituals. <Answer: C> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Application>

9. The second atmosphere around Earth lacked: A) hydrogen.

B) water.

C) ammonia.

D) hydrogen sulfide.

E) molecular oxygen. <Answer: E> <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Analysis>

10. Energy was available on the early second earth atmosphere for chemosynthesis from all of the following except: A) UV and gamma radiation from the sun.

B) lightning.

C) heat from coalescence of gas and dust and from radioactivity.

D) cellular respiration. <Answer: D> <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Analysis>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. Evolution by natural selection occurs as the types and abundances of alleles present in a population change over time. <Answer: True> <A-head: Concepts> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Recall>

12. Natural selection causes differential survival within a population.. <Answer: True> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Recall>

13. Rates of evolution are the same in all species, regardless of the environment in which the species occur or the effect of a mutation on the individual within a species. <Answer: False> <A-head: Rates of Evolution> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Recall>

14. The only outcome of divergent speciation is the development of one subpopulation into a new species while the other subpopulation remains the original species. <Answer: False> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

15. Evolution of bacteriochlorophyll led to the conditions suitable for the evolution of aerobic respiration. <Answer: False> <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Application>

Fill-in-the-Blank

16. Rice has been cultivated for thousands of years; thus its almost entirely of alleles that have been artificially selected for.

is made

<Answer: gene pool> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

17. Mimulus lewisii produces nectar ideal for bumblebees and grows in the same habitat as M. cardinalis, which produces flowers with a different morphology and nectar suited toward hummingbirds, providing an example of speciation. <Answer: sympatric> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

18. If gene flow does not keep a species homogenous throughout its range, then the different parts of it range are said to be isolated. <Answer: reproductively> <A-head: Speciation>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

19. If a single seed is responsible for the establishment of a new population, that seed would be considered the individual. <Answer: founder> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Application>

20. Current scientific studies suggest that of storing hereditary information.

was the molecules capable

<Answer: RNA> <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Easy> <Taxonomy: Recall>

Matching

21. Match each type of evolutionary process with its best example.

A) Hawaiian silverswords <Answer: Adaptive radiation> B) Cactus and euphorb spines <Answer: Convergent evolution> C) Embryo dies in early developmental stages <Answer: Hybrid inviability> D) Homologous chromosomes fail to form synaptonemal complexes <Answer: Hybrid sterility> E) Herbs such as Caltha growing in high alpine habitat patches <Answer: Allopatric speciation>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank F) Coconut rafts to a new continent and becomes a unique species over time <Answer: Phyletic speciation> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

22. Match each scientist with his best recognized contribution to advancing scientific theory.

A) Darwin <Answer: Natural selection> B) Galileo <Answer: The sun is the center of the solar system.> C) Mendel <Answer: Genetic factors do not change during sexual reproduction.> D) S. Miller <Answer: Evidence of life synthesized from inorganic compounds.> E) A. Oparin <Answer: Postulation of chemosynthetic theory> <A-head: Concepts> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

Essay

23. What two conditions are necessary for natural selection to occur? Why must they occur? <Answer: First, the population must produce more offspring than can possibly grow and survive to maturity in that habitat. Secondly, the progeny must differ from each other in their types of alleles. These conditions must occur because natural selection can only operate on existing alleles and the variety of alleles provides the raw material for change in populations. In order for some alleles to be expressed in higher proportions, there must be differential survival in the population.> <A-head: Population Genetics> <Subject: Chapter 18>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Application>

24. How can cabbage, broccoli, and cauliflower look very different from one another, yet all belong to the same species? <Answer: Through artificial selection, crop breeders have increased the frequency of certain alleles while eliminating or minimizing the frequency of others. When plant breeders find traits they are interested in, they focus their breeding efforts on producing varieties that exhibit the desired traits.> <A-head: Population Genetics> <Subject: Chapter 18> <Complexity: Moderate> <Taxonomy: Application>

25. The early ancestors of cacti were large, woody trees. Gymnocalycium is a modern cactus with a succulent growth form with a short stature that remains close to the ground. How might such drastic phenotypical changes occur in plants with shared phylogenetic ancestry? <Answer: Evolutionary changes that result in a loss of structure or metabolism can come about quickly. In desert conditions, mutations that disrupted lamina formation were advantageous and cacti lost their leaves over a relatively short period of time, as little as 10 million years. Other mutations also occurred so that the genes involved in stem elongation began producing shorter stems and the genes for succulents began developing. Mutations that prevented for formation of opaque bark were also favorable in the desert conditions.> <A-head: Rates of Evolution> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Application>

26. One solution to the problem of species extinction is captive breeding in zoos or gardens.

What are some of the problems associated with this solution? <Answer: The gene pool of captive populations is so small that rapid fluctuations and genetic drift can cause them to evolve rapidly. As the captive population diverges rapidly from its wild relatives, the gene pool becomes enriched in deleterious mutations. Artificial selection pressures in captive populations also tend to select for unusual or rare traits (especially in the case of plants) that could prove disadvantageous in the wild.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Speciation> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Analysis>

27. Distinguish between phyletic and divergent speciation and the causes of each. <Answer: Phyletic speciation occurs gradually, in which one species becomes so changed it must be considered a new species. Phyletic speciation occurs as a result of gene flow through such things as pollen transfer, seed dispersal, and vegetative propagation. Divergent speciation results in a population that splits, such that one evolves into a new species and the other remains unchanged or evolves into a third species. If gene flow does not maintain a level of homogeneity within the entire range of the population, then divergent speciation might occur. This situation can occur when populations are reproductively isolated through abiological or biological reproductive barriers.> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Recall>

28. Explain why, in the absence of living organisms, moon rocks may contain amino acids. <Answer: The chemosynthetic theory postulates that amino acids can be synthesized from inorganic precursors. According to the theory, if the moon possesses the right inorganic chemicals, appropriate energy sources, a great deal of time, and an absence of oxygen, then it is possible to create amino acids.> <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Application>

29. Discuss the reasons why the evolution of photosynthesis was so important to life on Earth. <Answer: Photosynthesis resulted in the buildup of free oxygen in the atmosphere, transforming the early reducing atmosphere to an oxidizing one. Aerobic respiration became possible when abundant atmospheric oxygen was available to drive the mitochondrial electron transport chain.

Free oxygen also allows for the formation of ozone, which shields the earth from harmful UV radiation.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Evolution and the Origin of Life> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Application>

30. Which type of speciation, phyletic or divergent, do you think occurs more commonly? Justify your answer. <Answer: Divergent evolution is likely to be more common. As biological and abiological selection pressure changes over time, some individuals will be better adapted to those changes and therefore have a greater contribution to the gene pool under certain circumstances.

Eventually this will lead to the creation of new species while the original species may remain or evolve into yet a third species.> <A-head: Speciation> <Subject: Chapter 18> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 19—Classification and Systematics

Multiple Choice

1. Which of the following is most likely to be a phylogenetic classification system?

A) Trees, shrubs, vines, and herbs B) Aquatic and terrestrial plants C) Bee-pollinated, hummingbird-pollinated, beetle-pollinated, and wind-pollinated plants D) Plants producing bilateral flowers with 5 sepals, 5 petals, 10 stamens, and one hypogynous gynoecium; plants producing radial flowers with 3 sepals, 3 petals, 3 stamens, and one epigynous gynoecium composed of 3 fused carpels E) Bilateral or radial flowers <Answer: D> <A-head: Concepts> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Analysis>

2. Which of the following is the correct way to write the species name of corn?

A) Zea mays B) Zea mays C) Zea Mays D) Mays E) mays <Answer: B> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Application>

3. The taxonomic level between family and class is:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) genus.

B) species.

C) division (phylum).

D) order.

E) kingdom. <Answer: D> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

4. Organisms are given scientific names because: A) Latin is the language of scholars.

B) it frustrates amateur botanists.

C) it gives taxonomists something to do.

D) it makes communication about organisms easier.

E) all species must be named. <Answer: D> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Analysis>

5. If you were a scientist trying to study the taxonomy of a group of plants, you might study all of the following except the plants’: A) phytochemistry.

B) plastid DNA sequences.

C) microscopic anatomy using a scanning electron microscope.

D) rRNA base sequences.

E) human uses. <Answer: E> <A-head: Cladistics > <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. In which of the following situations would a particular plant species be more likely to evolve into a large family?

A) An environment where only a few phenotypes are fit enough to survive and almost all mutations are selectively disadvantageous B) An environment where numerous phenotypes are successful and many mutations are advantageous C) An environment that is very stable D) A species pollinated by one species of insect E) A species living in only one particular type of habitat <Answer: B> <A-head: Cladistics > <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Application>

7. Betalains are red pigments found in beets and other members of the order Caryophyllales (cacti, carnations, amaranths, and most carnivorous plants). This pigment is not found in other groups, and the Caryophyllales are derived from a common ancestor. The presence of betalain pigment in members of this group is a: A) homoplasy.

B) synapomorphy.

C) clear indication that the group is polyphyletic.

D) All of these are correct. <Answer: B> <A-head: Cladistics > <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Application>

8. A common ancestor with dark green leaves has given rise to four species, G, H, I, and J. G, H, and I have dark leaves; J has mottled leaves. A classification scheme has included G, H, and I but omitted J. This classification scheme would be considered to be:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) polyphyletic.

B) monophyletic.

C) paraphyletic.

D) semiphyletic. <Answer: C> <A-head: Cladistics > <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Application>

9. Which of the following organisms is not categorized under Eukarya?

A) Animalia B) Protista C) Myceteae D) Cyanobacteria E) Plantae <Answer: D> <A-head: The Major Lines of Evolution> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

10. The scientist who contributed most to plant classification in the B.C. era was: A) Pliny the Elder.

B) Dioscorides.

C) Theophrastus.

D) Carolus Linnaeus.

E) Gaspard Bauhin. <Answer: C> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

True/False

11. Plants are still being described today. <Answer: True> <A-head: Concepts> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

12. The taxonomy of many organisms depends on the opinions of the taxonomer. <Answer: True> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Application>

13. The rbcL sequence is used for phylogentic studies. <Answer: True> <A-head: Cladistics> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Recall>

14. The discovery of a new species of plant, while interesting in its own right, usually has very little effect on our knowledge of plants already described. <Answer: False> <A-head: Taxonomic Studies> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

15. The concept of the genus was first established by Carolus Linnaeus. <Answer: False> <A-head: The Major Lines of Evolution> <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. In an unnatural,

group, members do not share common ancestry.

<Answer: polyphyletic> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Application>

17. Each point at which cladogram branches is called a the divergence of one taxon into two.

and represents

<Answer: node> <A-head: Cladistics> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

18. A classification system for fossils that combines features of both artificial and natural systems uses to establish groupings. <Answer: form genera> <A-head: Other Types of Classification Systems> <Subject: Chapter 19> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

19. Although each plant has only one type of specimen, it is also represented by several , which are distributed among herbaria around the world. <Answer: isotypes> <A-head: Taxonomic Studies> <Subject: Chapter 19> <Complexity: Easy> <Taxonomy: Recall>

20. The old classification of Protista was based on the observation that these organisms were placed together because they had a low level of evolutionary advancement. This is an example of a . <Answer: grade classification> <A-head: The Major Lines of Evolution > <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each term with its definition.

A) Synapomorphy <Answer: Shared derived characters> B) Homoplasy <Answer: Result of convergent evolution> C) Plesiomorphy <Answer: Relictual characters> D) Symplesiomorphy <Answer: Ancestral characters shared by two or more groups> E) Apomorphy <Answer: Derived characters> F) Autoapomorphy

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Unique derived characters that are only present in one group> <A-head: Cladistics> <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each characteristic with its corresponding kingdom.

A) Cell walls contain chitin <Answer: Fungi> B) No true nucleus <Answer: Bacteria> C) Photosynthesis <Answer: Plantae, bacteria, and protista> D) Produce tissues of very specialized cells; no photosynthesis <Answer: Animalia> E) Very simple eukaryotic organisms <Answer: Protista> F) Produce tissue of very specialized cells; photosynthesis <A-head: The Major Lines of Evolution> <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. Imagine that scientists have discovered a novel life form on Mars that does not fit within our existing classification system of life. Describe the best approach for assigning a name to this new species. <Answer: The same goals we use in our current classification system would apply to the Martian species. The scientist should assign a name based on developing a natural system of classification that groups closely related organisms. The name should reflect phylogenetic relationships.> <A-head: Concepts> <Subject: Chapter 19>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Application>

24. What criteria would you use to establish that a particular plant species is a new species? <Answer: Members of the same species can interbreed with each other successfully but cannot interbreed with members of another species. If the members of the two populations cannot breed successfully in nature, then they can be considered separate species. If the occasional crosspollination results in a fertile adult, the two groups might be considered a subspecies and an evolutionary stepping stone to becoming distinct species.> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Application>

25. Why might a species be reclassified and thus renamed? <Answer: Taxonomists believe that all species included in a genus must be descendants of a common ancestor, forming a monophyletic group. When systematists discover they have accidently created a polyphyletic group, they begin work on a more accurate classification, which sometimes involves a name change.> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Application>

26. How would you determine whether two similar structures are homologous or analogous? <Answer: Systematists study multiple features to determine homology or analogy. They make macroscopic observations, scanning electron microscope observations, study plastids in the phloem, study aspects of metabolism, and sequence DNA. In determining the phylogenetic relationships, the principle of parsimony is applied. We prefer the simplest hypothesis possible to explain evolutionary relationships. Combining parsimonious theory with all the data sources, scientists determine how the characters evolved.> <A-head: Cladistics> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

27. Create a simple cladogram using hypothetical taxa, correctly labeling all parts, showing an apomorphy, a synapomorphy, and a symplesiomorphy. <Answer: The cladogram may resemble that in Figure 18-5.> <A-head: Cladistics> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Recall>

28. Compare and contrast artificial and natural classification systems. <Answer: Natural systems of classification reflect evolutionary relationships and are preferred by botanists. Artificial systems of classification are often based on easily observed characters.

These are described as artificial because plants that are not closely related can be grouped together. Artificial systems are sometimes used as an adjunct to natural systems. Artificial systems can also be useful for people who are interested in various functions of plants, such as their suitability for woodworking.> <A-head: Other Types of Classification Systems> <Subject: Chapter 19> <Complexity: Moderate> <Taxonomy: Application>

29. Outline the process for naming a new plant species. <Answer: Taxonomists have created the International Code of Botanical Nomenclature, which describes the naming of new species. A name must be chosen that has not previously been used, and it must have a detailed plant description written in multiple languages. The name and description must be published in a peer-reviewed journal. A single preserved plant is identified as the type specimen and stored in an herbarium.> <A-head: Taxonomic Studies> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 30. Give your students a set of objects, such as different types of nails, pins, or clothes fasteners, and have them write a taxonomic key to identify them. <Answer: The key should be constructed of couplets. The first choice should be broadest and easier to discern. The couplets should become increasingly narrow as the key proceeds.

Ultimately, a key user should end with a “species” identification.> <A-head: Levels of Taxonomic Categories> <Subject: Chapter 19> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 20—Algae and the Origin of Eukaryotic Cells

Multiple Choice

1. Some heterokonts (stramenophiles) are photosynthetic and gained their chloroplasts by: A) primary endosymbiosis with a photosynthetic cyanobacterium.

B) primary endosymbiosis with a photosynthetic glaucophyte chloroplast.

C) secondary endosymbiosis with a green alga.

D) secondary endosymbiosis with a red alga.

E) secondary endosymbiosis with a dinoflagellate. <Answer: D> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Application>

2. Eukaryotic DNA is: A) located in the cytoplasm.

B) enclosed in double membranes.

C) simple and circular.

D) carrying approximately 3,000 genes. <Answer: B> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Recall>

3. Which of the following matches the overall algal body plan to its description?

A) Nonmotile colony—cells held tightly by a middle lamella and divide transversely B) Coencytic—cytokinesis does not occur, resulting in giant multinucleate cells C) Parenchymatous—cells held tightly by a middle lamella and divide in three planes

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank D) Membranous—cells adhere loosely, not an individual organism <Answer: B> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Analysis>

4. The algal group with the greatest diversity of body plans is the: A) green algae.

B) red algae.

C) euglenoids.

D) brown algae.

E) dinoflagellates. <Answer: A> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Recall>

5. The pigments of the red algae most closely resemble the pigments of: A) true plants.

B) cyanobacteria.

C) pyrrophyta.

D) phaeophyta E) None of these is correct. <Answer: B> <A-head: Red Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

6. In structure and function, trumpet cells in brown algae most closely resemble:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) mesophyll cells in leaves.

B) stomata.

C) sieve elements in phloem.

D) endodermal cells in a root.

E) tendrils on the leaf of a pea plant. <Answer: C> <A-head: Brown Algae and Their Relatives: The Heterokonts> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Recall>

7. The major characteristic used to identify species of diatoms is: A) their shape.

B) the pattern of ridges, depressions, and pores in the cell wall.

C) their photosynthetic pigments.

D) their type of cell division.

E) the number of positions of their flagella. <Answer: B> <A-head: Brown Algae and Their Relatives: The Heterokonts> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Application>

8. A characteristic of the Chrysophyta unifying its three classes is the presence of: A) two anterior, unequal flagella.

B) silica in the cell wall.

C) chlorophylls a and c.

D) laminarin as a food storage product. <Answer: C> <A-head: Brown Algae and Their Relatives: The Heterokonts> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

9. Which of the following correctly pairs the algal group with its flagellar apparatus?

A) Dinoflagellates—one trailing, one in a transverse groove B) Heterokonts—numerous flagella located in a ring C) Euglenoids—a short whiplash and a long tinsel D) Red algae—four whiplash E) Brown algae <Answer: A> <A-head: Dinoflagellates> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Analysis>

10. Red tides are dangerous because the: A) dinoflagellates that cause them produce a strong neurotoxin.

B) euglenoids that cause them produce a toxin that causes severe flulike symptoms.

C) green algae that cause them produce a severe allergic reaction in most people.

D) diatoms that cause them produce severe gastric distress.

E) red algae that cause them can disrupt shipping by clogging propellers. <Answer: A> <A-head: Dinoflagellates> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Analysis>

True/False

11. More DNA is found per cell in eukaryotes than in prokaryotes. <Answer: True> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

12. All spores in green algae are produced by mitosis. <Answer: False> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

13. Cytokinesis in red algae resembles most other plants except in the direction of cell wall growth between the new daughter cells. <Answer: True> <A-head: Red Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

14. The chromosomes of dinoflagellates retain some prokaryotic characteristics, such as a lack of histones. <Answer: True> <A-head: Dinoflagellates> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

15. Euglenoids are characterized by their lack of a cell wall and the presence of a pellicle. <Answer: True> <A-head: Euglenoids> <Subject: Chapter 20> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

Fill-in-the-Blank 16. Chloroplasts originated through a event where an early eukaryote engulfed but did not digest a prokaryote, giving rise to a clade containing red algae, green algae, glaucophytes. <Answer: primary endosymbiosis> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

17. The informal name has been used to describe the monophyletic clade that includes charophytes and embryophytes. <Answer: streptophyte> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

18. algae cells.

is a branched glucose polymer that occurs in the cytoplasm of red

<Answer: Floridean starch> <A-head: Red Algae> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Recall>

19. Many brown algae grow at the pneumatocyst and the blade.

located at the junction between the

<Answer: intercalary meristem> <A-head: Brown Algae and Their Relatives: The Heterokonts> <Subject: Chapter 20>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Recall> 20. are heterokont protists that were once thought to be fungi because they have no chloroplasts and consist of coenocytic tubes with no cross walls. <Answer: Oomycetes> <A-head: Dinoflagellates> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Application>

Matching

21. Match each algal division with an example of its characteristic trait.

A) Chrysophyta <Answer: Members share similar biochemistry with the brown algae> B) Rhodophyta <Answer: Never possess flagella> C) Euglenophyta <Answer: Nonpigmented members are classified as protozoans by zoologists> D) Pyrrhophyta <Answer: Lack histones and undergo intranuclear mitosis> E) Phaeophyta <Answer: Commonly possess large, complex bodies found on rocky coasts> F) Chlorophyta <Answer: Ancestors of true plants> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each type of algal body construction with its representative genus.

A) Unicellular species

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Chlamydomonas> B) Filamentous <Answer: Ulothrix> C) Parenchymatous <Answer: Chara> D) Coenocytic <Answer: Derbesia> E) Motile colony <Answer: Eudorina> F) Laminar <Answer: Ulva> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Moderate> <Taxonomy: Recall>

Essay

23. What is the major characteristic used to separate true algae from true plants? Why is this characteristic significant? Would any alga be classified as a plant on this basis? <Answer: Reproductive structures are critical in distinguishing plants from algae. Algae are photosynthetic organisms whose reproductive structures are completely converted to spores or gametes that, when released, leave nothing but empty walls. The reproductive structures of plants are always complex and multicellular, only some of the inner cells become reproductive. These reproductive structures involve two tissues, reproductive and sterile, unlike the one reproductive tissue typical of algae. This reflects a more complex level of organization in plants. Using this criteria, the green algae Chara could be classified as a plant.> <A-head: Concepts> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Analysis>

24. Discuss why the kingdom Protista is now considered to be an informal grade, “protista.”

What has become, taxonomically, of the groups formerly classified in the kingdom Protista?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Protists were originally grouped together because they are mainly unicellular and possess many ancestral characters. As such, they were considered to be a low level or low grade of evolution. They lacked the derived characters of plants, animals, and fungi. It is now recognized that protists represent numerous clades and the group we call algae are not monophyletic. The ancestor that is common to algae is also common to animals, fungi, and plants. Now we use algae as an informal name for these varied groups. Within the algae, six major divisions have been identified, including the euglenoids, dinoflagellates, chrysophyta, green algae, brown algae, and red algae.> <A-head: Concepts> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Analysis>

25. Is it more likely that chloroplasts had a single origin or multiple origins? Defend your answer. <Answer: DNA sequences, electron microscopy, and biochemical evidence suggest that it is most likely that a single primary endosymbiosis was followed by multiple secondary endosymbiosis. Red algae, green algae, and glaucophytes all share features with cyanobacteria and likely belong to a clade that developed after the primary endosymbiosis. Several events of secondary endosymbiosis produced other algal lines such as euglenoids from engulfing green algae and heterokonts from engulfing red algae.> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Analysis>

26. List 10 characteristics of prokaryotes and then compare each to the situation in eukaryotes. <Answer: See Table 19-1.> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Recall>

27. Distinguish between the autogenous and endosymbiont theories. Include evidence to support one of those theories.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: The autogenous theory postulates that eukaryotes evolved from prokaryotes by increasing in complexity over time. It is suggested that the endomembrane system specialized to become mitochondria and plastids. The endosymbiont theory postulates that a prokaryotic cell evolved to the point of having some eukaryotic features such as 80S ribosomes and a nuclear envelope. One of these early eukaryotic cells engulfed a prokaryote but did not digest it, but rather formed an association with it and eventually the engulfed prokaryote evolved into mitochondria. Plastids arose in a similar fashion to mitochondria. Plastids and mitochondria share many features with prokaryotes, such as possession of their own DNA and ribosomes, division through cleavage, lack of microtubules, circular DNA, and ribosomes, which are sensitive to the same antibiotics as prokaryotic ribosomes.> <A-head: Origin of Eukaryotic Cells> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Application>

28. Using labeled diagrams, show the life cycle of Derbesia; indicate where mitosis, meiosis, and fertilization occur. <Answer: The diagram should resemble Figure 19-21.> <A-head: Green Algae> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Application>

29. Although there is no typical red algae life cycle, explain one feature that is unique to them. <Answer: The carposporophyte generation occurs only in red algae life cycles. The carposporophyte produces diploid spores through mitosis. Carpospores germinate and grow into diploid individuals called tetrasporophytes that are similar to regular sporophytes. These have sporangia in which the cells undergo meiosis, producing haploid tetraspores that grow into gametophytes.> <A-head: Red Algae> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Recall>

30. Explain how the reproductive structures of Fucus are adapted to the littoral zone.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: The littoral zone exists between the high and low tide marks. In this region, seaweeds are periodically exposed to air and full sunlight. The ends of the Fucus branches are called receptacles that are swollen with hydrophilic compounds. Scattered over the receptacles are tiny openings called conceptacles that lead to small cavities. The conceptacle cells produce either eggs or sperm. At low tide, the seaweed is exposed to air and the conceptacles contract, squeezing out the gametes. When the tide comes in, the gametes are washed free and fertilization occurs in the water.> <A-head: Brown Algae and Their Relatives: The Heterokonts> <Subject: Chapter 20> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank Chapter 21

Multiple Choice

1. Which of the following is a true moss?

A) Reindeer moss B) Spanish moss C) Club moss D) Pond moss E) Tortula <Answer: E> <A-head: Characters of Nonvascular Plants> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall>

2. The structures labeled number 2 are:

A) spores.

B) calyptera cells.

C) part of the columella.

D) sperm cells.

E) egg cells. <Answer: A> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall> Copyright © 2021 by Jones & Bartlett Learning, LLC, an Ascend Learning Company

Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

3. The cells labeled number 2 were produced by:

A) mitosis.

B) meiosis.

C) binary fission.

D) mitosis or meiosis. <Answer: A> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Recall>

4. Moss spores are released from a capsule when: A) cells of the seta respond to humidity, twisting and turning to shake the spores out of the capsule.

B) cells of the operculum respond to humidity by changing shape, opening holes.

C) cells of the capsule wall respond to humidity by changing shape and splitting open the capsule wall.

D) cells of the peristome respond to humidity, bending the peristome teeth outward.

E) elaters uncoil, pushing the spores out. <Answer: D> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

5. If you cut a cross section through the shoot apex of a moss, you might find: Copyright © 2021 by Jones & Bartlett Learning, LLC, an Ascend Learning Company

Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

A) an apical cell shaped like an inverted pyramid.

B) microgametangia.

C) megagametangia.

D) protoxylem and protophloem.

E) several layers of substantially differentiated tissues. <Answer: A> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Application>

6. Rhizoids of mosses: A) provide anchorage to the gametophore.

B) absorb water for the rest of the plant.

C) absorb minerals for the rest of the plant.

D) have chloroplasts.

E) conduct water and minerals for the rest of the plant. <Answer: A> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Easy> <Taxonomy: Recall>

7. On some soil samples, you find filaments that contain chloroplasts. How would you know if they are algae or the protonema of a moss?

A) A protonema has numerous small chloroplasts in each cell; algae have one or two large chloroplasts.

B) A protonema is branched; algae are always unbranched.

C) A protonema will grow on soil; algae will not grow on soil.

D) A protonema is perennial. <Answer: A> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Application>

8. The gametophyte of leafy liverworts resembles the gametophyte of a moss because both have: A) a stem with three rows of leaves.

B) pointed leaves with a midrib.

C) leaves with conducting tissue.

D) a stem with thick-walled cells on the outside.

E) a reduced row of leaves that grow flat against the substrate. <Answer: A> <A-head: Division Hepatophyta: Liverworts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Analysis>

9. Gas exchange in the gametophyte of larger thallose liverworts such as Marchantia occurs principally by: A) diffusion through epidermal cells lacking a cuticle.

B) stomata on the upper surface.

C) stomata on the lower surface.

D) gasses dissolved in water that flows over the surface of the body of the liverwort.

E) permanently open large pores on the upper surface that open into internal chambers. <Answer: E> <A-head: Division Hepatophyta: Liverworts> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Analysis>

10. In the hornwort sporophyte: A) transfer cells carry nutrients from the gametophyte to the sporophyte.

B) basal meristem cells continuously produce new spore-generating tissue.

C) the foot develops rhizoids and the sporophyte becomes independent from the gametophyte.

D) spores are produced only by mitosis.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: D> <A-head: Division Anthocerotophyta: Hornworts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Recall>

True/False

11. In all nonvascular plants, the sporophyte must obtain minerals and sugars from the gametophyte. <Answer: True> <A-head: Characters of Nonvascular Plants> <Subject: Chapter 21> <Complexity: Easy> <Taxonomy: Recall>

12. In mosses, the sporophyte and gametophyte are structurally similar. <Answer: False> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Easy> <Taxonomy: Application>

13. Moss sperm usually locate mature eggs by swimming randomly. <Answer: False> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall>

14. Liverwort gametophores are always bisexual.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: False> <A-head: Division Hepatophyta: Liverworts> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Application>

15. Hornworts have a single large chloroplast in each cell. <Answer: True> <A-head: Division Anthocerotophyta: Hornworts> <Subject: Chapter 21> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. Although originally all placed within the division Bryophyta as three separate classes, the Bryophyta, Hepatophyta, and now each form their own division. <Answer: Anthocerotophyta> <A-head: Classification of Nonvascular Plants> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Application>

17. The leafy stems of mosses are called

.

<Answer: gametophores> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 18. The microgametangia of mosses are called short stalk, sterile cells, and sperm cells.

and they consist of a

<Answer: antheridia> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall>

19. Some cells within a liverwort sporangium do not undergo meiosis and instead differentiate into that function to push spores out when the sporangium opens. <Answer: elaters> <A-head: Division Hepatophyta: Liverworts> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall>

20. Two clear differences between hornworts and liverworts are that hornworts do not have seta or discrete . <Answer: sporangium> <A-head: Division Anthocerotophyta: Hornworts> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Application>

Matching

21. Match each group of organisms with their common structures.

A) No seeds, no vascular tissue <Answer: Nonvascular plants> B) Vascular tissue and seeds <Answer: Spermatophytes> C) No seeds, no vascular tissues, lack of sterile cells surrounding gametangia

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Charophytes> D) Vascular tissue, no seeds <Answer: Vascular cryptogams> <A-head: Concepts> <Subject: Chapter 21> <Complexity: Moderate> <Taxonomy: Recall>

22. Match each characteristic with a group or groups of plants.

A) Sporophyte dependent on gametophyte <Answer: Mosses, liverworts, hornworts> B) Leaves with two rounded lobes and no midrib <Answer: Liverworts> C) Sporophyte has a basal meristem <Answer: Hornworts> D) May contain leptoids <Answer: Mosses> E) Sporangium contains elaters <Answer: Liverworts and hornworts> F) Sporangium has columella <Answer: Mosses and hornworts> G) Capsule often covered by calyptra <Answer: Mosses and liverworts> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Analysis>

Essay

23. Why have nonvascular plants stayed fairly small in size? <Answer: Nonvascular plants use turgid parenchyma cells in their bodies and thus cannot grow very tall. Vascular plants possess xylem and phloem, which confer the stem strength to grow tall.

Also the lack of the vascular tissues in bryophytes prohibits the construction of root and leaf systems that support and maintain large body mass. The small, delicate gametophytes further

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank limit these plants to mild environments that can be found at a micro-scale in biomes across the planet.> <A-head: Concepts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

24. List and discuss four ways that plants cope with dry periods. <Answer: See Table 20-2. Dormant, drought-resistant spores and the ability of the entire plants to go dormant, such as may occurs in many mosses, are useful strategies for tolerating long, dry periods. Brief dry periods may be buffered through conserving water such as a multicellular body with a low surface-to-volume ratio. A waterproof cuticle on leaves is extremely advantageous in dry environments. The grouping of sporangia and gametangia into protective structures further protects these necessary tissues from drying out.> <A-head: Concepts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

25. Most mosses do not retain water well. What adaptations have evolved to permit them to inhabit terrestrial areas? <Answer: Leaves and stems are so small as to form narrow spaces to act as capillary channels that allow water movement. Some mosses have leptoids and hydroids to transport water. Their small size allows them to use these various water transport mechanisms over the required short distances. Their small size also enables them to thrive in permanently moist microhabitats. Some mosses are tolerant of desiccation and remain dormant but alive when conditions are very dry.> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

26. In what ways do mosses enable vascular plants to become established in an area? <Answer: Mosses can colonize bare rock and dissolve the rock with acids that leak from their cells. Gametophore leaves and stems catch and hold dust particles that create small pockets of soil that can catch spores and seeds, thereby leading to the establishment of vascular plants.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank Mosses can also hold water and further improve the microenvironment for seedling establishment.> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

27. How do water and minerals move in most mosses? How fast does it occur? <Answer: The majority of mosses conduct water along the exterior of their stems through capillary action. Sugars are moved between adjacent parenchyma cells by slow transport. Some mosses, such as those in the family Polytrichaceae, possess hydroids and leptoids. Hydroid are elongated cells that lose their cytoplasm when mature and conduct water and minerals. Leptoids conduct mainly sugars and resemble the sieve cells of vascular plants.> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Analysis>

28. Why do you think that the fingerlike cap of a mature Marchantia archegoniophore is at the top of a long stalk? <Answer: The fingerlike cap is an archegoniophore and the archegonia are located on the underside. Antheridia are located on the upper side of the antheridiophore. When a raindrop hits the antheridiophore, the sperm are captured in the raindrops and splash upward. The long stalk of the archegoniophore is able to catch the splashing raindrops. The sperm then swim into the archegonium neck and fertilize the egg.> <A-head: Division Hepatophyta: Liverworts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Analysis>

29. When would be the best time of year to find hornworts? Why? <Answer: Hornworts tends to be small, inconspicuous thalloid plants that grow in moist soils, often in the shade. They produce characteristic sporophytes that resemble “horns” that grow continuously from a basal meristem. In temperate climates, hornworts typically live for less than a year. They grow as winter annuals, appearing in the cool, moist autumn months, growing

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank during winter, producing sporophytes in the spring, and dying before summer. One of the best times to find them is early spring when the conditions tend to be moist and sporophytes are present.> <A-head: Division Anthocerotophyta: Hornworts> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Application>

30. You are walking through the forest and encounter a small mosslike plant. You begin studying the plant with your hand lens. What characters might you look for to determine whether the plant is a moss, liverwort, or hornwort? <Answer: The gametophyte of a moss has leaves aligned in three rows and most are pointed with a midrib. Leafy liverworts typically have two rounded lobes with no midrib, whereas thallose liverworts are flat, ribbonlike, and suppressed. The leafy liverworts also grow from two apical points and have a basal pouch. Hornwort gametophytes resemble those of thallose liverworts but are often quite small and brittle. The sporophytes of liverworts tend to be more delicate than those of mosses, whereas the sporophytes of hornworts look like “horns” and lack a seta or distinct sporangium.> <A-head: Division Bryophyta: Mosses> <Subject: Chapter 21> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 22—Vascular Plants Without Seeds

Multiple Choice

1. The major difference between Sciadophyton and Rhynia is the: A) absence of leaves in Rhynia.

B) presence of a cuticle in Sciadophyton.

C) presence of a pith in Rhynia.

D) presence of terminal sporangia in Rhynia.

E) presence of vascular tissue in Sciadophyton. <Answer: D> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

2. Which of the following is false about early vascular plants’ xylem structure?

A) It is a protostele.

B) There is no pith in the center.

C) It is an endarch or exarch protostele.

D) It is a siphonostele.

E) The center is a solid mass of xylem. <Answer: D> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Application>

3. Features present in living groups of plants that are similar to very early species are said to be: A) primitive.

B) advanced.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) original.

D) plesiomorphic.

E) apomorphic. <Answer: D> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Application>

4. Early lycophytes are thought to have evolved from: A) psilotum-like plants.

B) Rhynia-like plants.

C) Zosterophyllum-like plants.

D) Cooksonia-like plants.

E) The evolutionary origin is unknown. <Answer: C> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Recall>

5. The now-extinct lycophytes Lepidostrobus and Lepidophloios share many characteristics with seed plants except: A) heterospory.

B) megaspore development within the megaspore wall.

C) megaspore retention within the sporangia.

D) indehiscence of the sporangia.

E) sporangium protected by thick-walled cells. <Answer: D> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. Psilotum is unique among living vascular plants because the: A) gametophyte produces rhizoids.

B) gametophyte is heterotrophic.

C) sporophyte has an endarch siphonostele.

D) sporophyte lacks stomates.

E) gametophyte contains vascular tissue. <Answer: E> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Recall>

7. A sorus is a: A) group of sporangia on the upper side of a fern leaf.

B) group of sporangia on the underside of a fern leaf.

C) group of sporangia along the edge of a fern stem.

D) single sporangium on the underside of a fern leaf.

E) group of fern spores. <Answer: B> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Recall>

8. Which group of characteristics best describes sporophytes of the Arthrophyta?

A) Jointed, herbaceous, aerial stems; whorled reduced megaphylls; subterranean rhizomes; roots B) Jointed, woody, aerial stems; whorled reduced microphylls; subterranean rhizomes; roots C) Dichotomous, herbaceous, aerial stems; alternate reduced megaphylls; roots D) Herbaceous, aerial stems; opposite reduced megaphylls; subterranean rhizomes; rhizoids E) Woody, aerial stems; whorled reduced megaphylls; subterranean rhizomes; rhizoids <Answer: A> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Analysis>

9. Megaphylls evolved from: A) microphylls that became more elaborate due to hormonal interactions.

B) enations that became vascularized.

C) spines that became vascularized.

D) three-dimensional branching systems that became planated and webbed.

E) the elaboration of ligules. <Answer: D> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Recall>

10. Which of the following is false about euphyllophytes?

A) They form a monophyletic clade.

B) Their roots have an exarch xylem.

C) They have a 30-kb inversion in the large single-copy region of their plastid DNA. D) They have megaphylls.

E) Their roots have an endarch xylem. <Answer: E> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

True/False

11. Zosterophyllophytes had lateral sporangia. <Answer: True>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Recall>

12. Some living lycophytes have roots, leaves, and a small amount of secondary growth. <Answer: True> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Recall>

13. Microphylls are always small, as their name implies. <Answer: False> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Recall>

14. All ferns are eusporangiate. <Answer: False> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Recall>

15. The division of seedless vascular plants that contains the largest number of living species is the Pteridophyta. <Answer: True> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Easy>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

Fill-in-the-Blank

16. An important step in plant evolution was the appearance of the cycle, in which there is both a multicellular gametophyte and multicellular sporophyte.

life

<Answer: dibiontic> <A-head: Concepts> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Recall>

17. In Asteroxylon, the

were photosynthetic and contained stomata.

<Answer: enations> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Application>

18. A distinguishing feature between Selaginella and lycopodium is the presence of a on the upper surface of the Selaginella leaves. <Answer: ligule> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Application>

19. A unique feature of quillworts in the genus roots to absorb CO2 from mud.

is the ability of their

<Answer: Isoetes>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Application>

20. The reproductive structures of Equisetum consist of umbrella-shaped grouped onto strobili. <Answer: sporangiophores> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Moderate> <Taxonomy: Recall>

Matching

21. Match each group of plants with its root and/or leaf type.

A) Equisetophytes <Answer: Endarch siphonostele and megaphyll> B) Rhyniophytes <Answer: Endarch protostele> C) Lycophytes <Answer: Exarch siphonostele and microphyll> D) Zosterophyllophytes <Answer: Exarch protostele> E) Psilotum <Answer: Protostele and megaphyll> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

22. Match each plant division with its prominent characteristic.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) Equal dichotomous branching <Answer: Rhyniophytes> B) Vascularized gametophytes <Answer: Psilophyta> C) Have microphylls <Answer: Lycophyta> D) Exhibit circinate venation <Answer: Pteridophyta> E) Have silica in their epidermis <Answer: Arthrophyta> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

Essay

23. Distinguish between the transformation hypothesis and the interpolation hypothesis for the origin of life cycles in early terrestrial plants. <Answer: The interpolation hypothesis postulates that a small sporophyte came into existence when a zygote germinated mitotically. Over time the elaboration of the sporophyte allowed it to increase in size and complexity while the gametophyte generation remained small. In this way the sporophyte generation was interpolated into a monobiontic life cycle. An alternative hypothesis is the transformation theory that describes an elaboration of both the gametophyte and sporophyte generations after a dibiontic life cycle originated. This theory postulates that the early ancestors divided into two clades with the first clade being nonvascular plants that possess a simplified sporophyte dependent on the gametophyte. The second clade were the vascular plants in which the sporophyte became increasingly complex and the gametophytes became simpler.> <A-head: Concepts> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Application>

24. How might the sporophytes of ancient plants be distinguished from gametophytes?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Fossils of Rhynia and Aglaophyton sporophytes show dichotomous branching patterns with terminal sporangium. In the same rocks where these fossils were found, other plants, Lyonophyton and Sciadophyton, were also found. The flattened ends of these latter plants contained gametangia but not sporangia. Because these plants were found in the same area, it is plausible that they might be alternate generations of the same species.> <A-head: Early Vascular Plants> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

25. Why have no fossils ever been found with wood more than 10 cm thick? <Answer: Some ancient plants such as extinct lycophytes had a vascular cambium and secondary growth. However, the cells in the vascular cambium could not undergo radial division and new fusiform initials could not be produced. Older cells became stretched circumferentially until they were no longer functional. This limited the circumference of early plants with secondary growth.> <A-head: The Microphyll Line of Evolution: Lycophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Application>

26. Knowing that you have taken a botany course, a friend asks for advice about a “sick” fern that has brown spots all over the undersides of its leaves. What would be your advice? <Answer: This fern is not sick. Most ferns have sori, or clusters or sporangia, on the undersides of their leaves where meiosis occurs to create spores. Often these sori appear as brown dots.> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Application>

27. Antheridia and archegonia produced by an individual fern gametophyte do not mature at the same time. What are the advantages of this timing? Disadvantages? <Answer: Asynchronous timing of antheridia and archegonia maturation helps to prevent selfing and thus increase genetic diversity in the population. This is an advantage that can help maintain long-term survival of species in changing environments. The primary disadvantage is that if there

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank are no compatible antheridia or archegonia in the nearby environment, then the plant may not be able to successfully reproduce.> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Analysis>

28. How did megaphylls evolve? <Answer: Megaphyll evolution is described by the telome theory. Initially dichotomous branching of stems was replaced with overtopping where some branches grew longer than others.

The last dichotomy is known as a telome. The lateral branches grew only in one plane and parenchymatous, photosynthetic tissues developed between the branches. This could lead to the evolution of leaves.> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Recall>

29. What is the difference between a genus and a form genus? <Answer: A genus is a level of taxonomic organization of organisms. It is located between family and species. Form genera are used for the classification of fossils. Many fossils represent fragments of organisms, and if a new fossil of a leaf is found that doesn’t look like any known plants, it is classified as a new form genus of a leaf. Both genera and form genera may change over time if enough evidence accumulates to warrant the change.> <A-head: The Megaphyll Line of Evolution: Euphyllophytes> <Subject: Chapter 22> <Complexity: Easy> <Taxonomy: Analysis>

30. Why is the term vascular cryptogam only used informally by botanists? <Answer: Vascular cryptogams refer to a grade or level of evolutionary advancement, and current biological organization is based on phylogenetic relationships. If we were to form a clade called vascular cryptogams, we would be creating a group that contains an ancestral group but not all of its descendants. In this case, leaving out the lignophytes would create a paraphyletic group, which does not accurately represent known phylogenies.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: The Term “Vascular Cryptogams”> <Subject: Chapter 22> <Complexity: Difficult> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 23—Seed Plants I: Gymnosperms

Multiple Choice

1. It is difficult to fully analyze the evolutionary relationships between extant and extinct seed plants because: A) many of the extinct groups are hypothetical and there is no evidence of them.

B) from groups known only from fossils, no DNA evidence is available for comparison with living groups.

C) the fossils provide no structural information about early groups of seed plants.

D) older classifications are often incorrect. <Answer: B> <A-head: Concepts> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

2. Fossil material of progymnosperms, a now extinct group, never contains: A) megaphylls.

B) two kinds of spores.

C) seeds.

D) secondary xylem and phloem.

E) cork cambia. <Answer: C> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23> <Complexity: Easy> <Taxonomy: Recall>

3. Place the following the order, from left to right, in which they evolved: I. Seeds

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank II. Trimerophytes III. Vascular cambium A) I, II, III B) III, II, I C) II, III, I D) III, I, II E) I, III, II <Answer: C> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

4. Seed fern wood was softer than progymnosperm wood because of all the following except: A) its rays were not uniseriate.

B) its tracheids were longer and wider than progymnosperm tracheids.

C) it contained a lot of axial parenchyma.

D) it was manoxylic. <Answer: C> <A-head: Division Pteridospermophyta: Seed Ferns> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

5. A common ornamental shrub is yew (Taxus), belonging to the order Taxales. The principal way this order differs from the Coniferales is the: A) absence of secondary growth.

B) structure of the leaves.

C) lack of seed cones.

D) xylem structure.

E) megaspore retention. <Answer: C> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Recall>

6. The accompanying diagram shows the parts of a conifer cone. Which of the following correctly pairs the name of the structure with the letter labeling it?

A) X-bract B) Y-seed C) Y-ovuliferous scale D) Z-bract E) W-ovuliferous scale <Answer: E> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

7. The ovuliferous scale of a conifer is homologous to a: A) branching system.

B) leaf.

C) bract.

D) leaf bud.

E) leaf and bract <Answer: A> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> Copyright © 2021 by Jones & Bartlett Learning, LLC, an Ascend Learning Company

Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Difficult> <Taxonomy: Recall>

8. Most support in cycads is provided by: A) collenchyma.

B) pycnoxylic wood.

C) manoxylic wood.

D) turgor pressure in the abundant parenchyma.

E) persistent leaf bases. <Answer: E> <A-head: Division Cycadophyta: Cycads> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

9. The ancestors of the ginkgos were one of the: A) Taxales.

B) Coniferales.

C) Pteridospermophyta.

D) Gnetophyta.

E) The origin is completely unknown. <Answer: C> <A-head: Division Ginkgophyta: Maidenhair Tree> <Subject: Chapter 23> <Complexity: Easy> <Taxonomy: Recall>

10. You have just discovered a new living plant and upon examination it has the following characteristics: secondary growth, simple leaves, compound seed cones, and compound pollen cones. This plant should be classified in the division: A) Cycadophyta.

B) Gnetophyta.

C) Pteridospermophyta.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank D) Ginkgophyta.

E) Coniferophyta. <Answer: B> <A-head: Division Gnetophyta> <Subject: Chapter 23> <Complexity: Easy> <Taxonomy: Application>

True/False

11. Seed plants are grouped under spermatophytes. <Answer: True> <A-head: Concepts> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

12. Progymnosperms were not capable of producing true wood and therefore their diameters were limited to a maximum of 10 cm. <Answer: False> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

13. Adaptations of pine needles suggest that pines may have evolved in a habitat short of water all or part of the year. <Answer: True> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

14. The stem of a cycad contains a protostele. <Answer: False> <A-head: Division Cycadophyta: Cycads> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

15. Cycadeoids are commonly used as ornamentals in states such as Florida. <Answer: False> <A-head: Division Cycadeoidophyta: Cycadeoids> <Subject: Chapter 23> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. Cycads and their relative produce small amounts of is soft and spongy.

wood, which

<Answer: manoxylic> <A-head: Concepts> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

17. The integument of Chauleria possessed a cells to swim to the egg.

that permitted sperm

<Answer: micropyle> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Application>

18. The of cycads are not fused structures as are the ovuliferous scales of conifers; instead they are simple and often resemble a foliage leaf. <Answer: megasporophylls> <A-head: Division Cycadophyta: Cycads> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Recall>

19. Pollen cones are microsporophylls.

with a single short unbranched axis that bears

<Answer: simple cones> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Application>

20. In planted landscapes such as parks, microsporangiate Ginkgo trees are preferred over the megasporangiate ones as their seeds produce the putrid-smelling . <Answer: butyric acid> <A-head: Division Ginkgophyta: Maidenhair Tree> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Recall>

Matching

21. Match each characteristic with a group of plants.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) Vessels <Answer: Gnetophyta> B) Simple, dichotomously veined leaves <Answer: Ginkgophyta> C) Lack seeds <Answer: Progymnospermophyta> D) Resin canals <Answer: Coniferophyta> E) Bisexual cones <Answer: Cycadeoidophyta> F) Seeds produce on the foliage leaves <Answer: Pteridospermophyta> <A-head: Division Pteridospermophyta: Seed Ferns> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Analysis> 22. Match each part of a coniferous cone with its function.

A) Microspore <Answer: Develops into a small gametophyte> B) Air bladders <Answer: Increase buoyancy> C) Generative cell <Answer: Becomes two nonmotile sperm cells> D) Tube cell <Answer: Becomes the pollen tube> E) First body cells produced <Answer: Degenerate> F) Ovuliferous scale <Answer: Megasporophyll> G) Bracts <Answer: True leaves of cone axis> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

Essay

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

23. Compare the advantages and disadvantages of two life cycles: one in which alternate generations are completely independent of the preceding generation and one where they are not. <Answer: When one generation is dependent on the other, then they can share resources. For example, if the megametophyte is retained by the sporophyte, then the developing embryo may take advantage of the absorptive capacities of the sporophyte roots and stems. A disadvantage may be that the retained megagametophyte cannot be easily reached by swimming sperm. A further disadvantage of this system is that if the sporophyte perishes, so too will the dependent gametophyte. An advantage of independent generations is that gametophytes could produce eggs and sperms that could easily find each other.> <A-head: Concepts> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

24. The description of aneurophytes and trimerophytes are very similar. Why are they placed in separate divisions? <Answer: Both trimerophytes and aneurophytes share the traits of little webbing between their ultimate branches and their primary xylem was arranged in a protostele. However, aneurophytes have a vascular cambium and are capable of secondary growth.> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

25. How could you distinguish between fossil microspores and pollen grains? <Answer: As fossils developed into microspores, changes occurred in the nature of their cell wall, in their method of germination, and in the nature of the microgametophyte they produced.> <A-head: Division Progymnospermophyta: Progymnosperms> <Subject: Chapter 23> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank 26. How do cycads differ from seed ferns? How are they similar? <Answer: Unlike seed ferns, cycad foliage leaves do not bear ovules. Cycads are always dioecious and produce seed cones and pollen cones on separate plants. All seed ferns are extinct, whereas some cycads have remained extant. Both groups produce manoxylic woods with a thick cortex and tough outer layers formed by persistent leaf bases.> <A-head: Division Pteridospermophyta: Seed Ferns> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

27. In what ways do cycads differ from cycadeoids? <Answer: The two groups differ in the differentiation of stomatal complexes and in leaf trace organization. Individual cones of cycadeoids contain both microsporophylls and megasporophylls. In contrast, cycads are always dioecious, possessing either pollen cones or seed cones. Cycadeoids are extinct, whereas cycads are extant.> <A-head: Division Cycadophyta: Cycads> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

28. Both ginkgos and taxads produce single seeds that are not in a seed cone. How could you tell the difference between them? <Answer: Taxads produce single seeds enclosed in a red, fleshy aril protecting the seed. The arils are produced on stems also containing needlelike leaves (sporophyll). Ginkgos produce paired ovules at the end of short stalks. The ovule is exposed and unprotected at maturity.> <A-head: Division Coniferophyta: Conifers> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

29. You’re walking in the park and pass several female ginkgo trees covered with mature reproductive structures. Your friend remarks that the ginkgo trees have produced a lot of fruit.

Would you agree or disagree with your friend’s statement? Defend your answer.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Fruits are mature carpels found in angiosperms that enclose ovules. Ginkgos are a type of gymnosperm, which means “naked ovules”; the ovules are often located on flat sporophylls like cones or on short stalks as in ginkgos. The ovules of ginkgos are large and develop a threelayered seed coat, but they are not fruits in a botanical sense.> <A-head: Division Ginkgophyta: Maidenhair Tree> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

30. The Gnetophyta exhibit several features that are characteristic of angiosperms rather than gymnosperms. What are those characteristics? Why are gnetophytes considered “gymnosperms”? <Answer: Shared features between the Gnetophyta and the angiosperms include broad leaves and a stout trunk with many branches. The wood itself lacks vessels and axial parenchyma and thus resembles that of conifers. Gingko reproduction is dioecious and has large ovules that resemble those of pteridosperms and cycads.> <A-head: Division Gnetophyta> <Subject: Chapter 23> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank CHAPTER 24—Seed Plants II: Angiosperms

Multiple Choice

1. The most recently evolved group of plants are the: A) ferns.

B) conifers.

C) gnetophytes.

D) flowering plants.

E) cycads. <Answer: D> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Recall>

2. What is the most probable order of evolution?

A) Vessel elements → sieve tubes → herbaceousness → double fertilization → bisexual flowers B) Bisexual flowers → vessel elements → double fertilization → sieve tubes → herbaceousness C) Double fertilization → bisexual flowers → vessel elements → sieve tubes → herbaceousness D) Herbaceousness → double fertilization → bisexual flowers → vessel elements → sieve tubes E) Double fertilization → sieve tubes → bisexual flowers →vessel elements → herbaceousness <Answer: C> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Analysis>

3. Which of the following is more likely to be a derived feature in angiosperms?

A) Deciduous leaves B) Being a small tree Copyright © 2021 by Jones & Bartlett Learning, LLC, an Ascend Learning Company

Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank C) Little axial parenchyma D) Tracheary elements E) All rays narrow and tall <Answer: A> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Application>

4. A synapomorphy that unites the Magnoliophyta is the: A) presence of wood.

B) absence of cones.

C) presence of flowers.

D) interactions with fungi.

E) leaf shape and size. <Answer: C> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Recall>

5. The earliest definite angiosperm leaf fossils are from the: A) Eocene Epoch and are common.

B) Pleistocene and are rare.

C) Lower Cretaceous and are rare.

D) Carboniferous Period and are rare.

E) Devonian Period and are common. <Answer: C> <A-head: Changing Concepts About Early Angiosperms> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank 6. Eudicots are recognized by: A) straplike leaves.

B) flower parts generally in threes.

C) parallel leaf venation.

D) vascular bundles in the stem arranged in one ring.

E) no members having secondary growth. <Answer: D> <A-head: Classification of Flowering Plants> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Application>

7. A flower characteristic that is quite derived and that is a product of considerable evolutionary modification is: A) petals fused into a tube.

B) stamens fused to the petals.

C) sepals very small.

D) an inferior gynoecium.

E) a compound gynoecium. <Answer: D> <A-head: Monocots> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Analysis>

8. About 50% of our food calories come from which family?

A) Rosaceae B) Fabaceae C) Poaceae D) Asteraceae E) Vitaceae <Answer: C> <A-head: Monocots> <Subject: Chapter 24>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank <Complexity: Easy> <Taxonomy: Recall>

9. Which of the following plants is a member of the most derived clade of eudicots?

A) Sunflower B) Amborella C) Peanut D) Venus flytrap E) Grape <Answer: A> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Recall>

10. A pigment uniquely characteristic of the Caryophyllaceae is: A) anthocyanin.

B) chlorophyll b.

C) betalain.

D) B-carotene.

E) xanthophyll. <Answer: C> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Recall>

True/False

11. Classifying Nymphaeales as monocots creates a monophyletic group. <Answer: False>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank <A-head: Classification of Flowering Plants> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Application>

12. The carpels of Magnolia flowers are not fused together into a pistil. <Answer: True> <A-head: Basal Angiosperms> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Recall>

13. Monocots lack ordinary secondary growth. <Answer: True> <A-head: Monocots> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Application>

14. The major pigment in flowers of Caryophyllales is anthocyanin. <Answer: False> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Recall>

15. Asterids are the most derived large clade of eudicots. <Answer: True> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Easy> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank

Fill-in-the-Blank

16. The 257,000 species of angiosperms are classified in one division called the . <Answer: Magnoliophyta> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Recall>

17. C. E. Bessey developed the hypothesis of the flower, which described the Magnolia-type flower as one that was generalized and relictual. <Answer: ranalean> <A-head: Changing Concepts About Early Angiosperms> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Application>

18. The pollen grains of basal angiosperms and monocots are all having only a single germination pore.

,

<Answer: uniaperturate> <A-head: Basal Angiosperms> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Application>

19. Collectively, the Commelinales, Arecales, Poales, and Zingiberales are known as the monocots. <Answer: commelinoid>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank <A-head: Monocots> <Subject: Chapter 24> <Complexity: Moderate> <Taxonomy: Recall>

20. All eudicots share the distinctive feature of derivation of that form.

pollen grains or a

<Answer: tricolpate> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Application>

Matching

21. Match each plant or plant product with a family.

A) Onion <Answer: Alliaceae> B) Asparagus <Answer: Asparagaceae> C) Sugar cane <Answer: Poaceae> D) Pineapple <Answer: Bromeliaceae> E) Banana <Answer: Musaceae> F) Ginger <Answer: Zingiberaceae> G) Cattails <Answer: Typhaceae> H) Aloe vera <Answer: Aspholodelaceae> <A-head: Monocots> <Subject: Chapter 24> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank <Taxonomy: Recall>

22. Match each plant or plant product with a family from the list below.

A) Beets <Answer: Amaranthaceae> B) Venus flytrap <Answer: Droseraceae> C) Grapes <Answer: Vitaceae> D) Apples <Answer: Rosaceae> E) Peanuts <Answer: Fabaceae> F) Milkweed <Answer: Asclepiadaceae> G) Lettuce <Answer: Asteraceae> H) Coffee <Answer: Rubiacaea> I) Celery <Answer: Apiaceae> J) Mistletoe <Answer: Sanatalaceae> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Recall>

Essay

23. What characteristics did early flowers most likely have? In what ways have they evolved since then? <Answer: Early flowers were complete and had many parts, spiral arrangement, radial symmetry, superior ovary position, no fused parts, and mostly winder pollination and seed dispersal. Derived features include being imperfect with fewer numbers of parts arranged in

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank whorls. Other derived features include bilateral symmetry, inferior ovaries, fusion of parts, and many types of pollination and seed dispersal syndromes.> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Application>

24. Describe how the closed carpel may have evolved from a leaflike megasporophyll. <Answer: Carpels in the basal angiosperms are leaflike, resembling young leaves that failed to open. The ovary edges have rows of secretory hairs that secrete a liquid that seals the seam and functions as a stigma. If such carpels produced ovules only at their base and the upper part elongate, it would begin to resemble a close carpel. In closed carpels, the sporophyll primordia crowd against each other and grow shut.> <A-head: Concepts> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Application>

25. Describe the floral characteristics of wind-pollinated angiosperms. Why are these adaptations important in the reproduction of these plants? <Answer: Wind-pollinated flowers do not need to attract pollinators and thus do not need to be large and colorful. Key adaptations for such plants to effect successful pollination include the production of large amounts of pollen and a stigma with a large and potentially convoluted surface area. Both of these features maximize the probability of the correct pollen finding the stigma of the same species.> <A-head: Changing Concepts About Early Angiosperms> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Application>

26. Discuss how the basal angiosperms differ from the other groups of angiosperms. <Answer: The basal angiosperms became reproductively separate from the other angiosperms very early. Depending on the lineage, some have more in common with eudicots while others have more in common with monocots. In the basal angiosperms, the carpals are not fused

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank together into a pistil, as are those of the other angiosperms. Unlike eudicots, the basal angiosperms are uniaperturate.> <A-head: Basal Angiosperms> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Analysis>

27. In some members of the Alismatidae, flowers lack a perianth completely. Under what conditions does this occur most often? What is the advantage to the plant? Why can the flowers function without a perianth? <Answer: It is necessary for a species to be adapted to the most adverse conditions that it frequently encounters. Flowers that often lack a perianth are those species in which the flowers emerge from the water and pollination occurs below the surface. The main purpose of the perianth is to attract pollinators. Species pollinated underwater don’t need to attract pollinators, as pollen simply moves through the water column. In this case the perianth is unnecessary and therefore energetically inefficient to produce.> <A-head: Monocots> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Analysis>

28. Explain how parallel venation in monocot leaves might have evolved. <Answer: One theory suggests that the monocot ancestors had broad leaves and lived as aquatic plants. Over time the leaves evolved to a reduced simple form lacking a blade. Some plants moved into drier habitats where broad and long lamina were more adaptive. Mutations resulted in a basal meristem and these were selected for, resulting in strap-shaped leaves. In some lineages a marginal meristem evolved.> <A-head: Monocots> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Analysis>

29. Name four drugs produced by members of the Asteridae and explain how each is used medicinally.

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Botany: An Introduction to Plant Biology, Sixth Edition James D. Mauseth Test Bank <Answer: Vinca contains vinblastine and vincristine used to treat cancer. Cinchona produces the antimalarial quinine. Cephaelis provides us with ipecac, an emetic that induces vomiting in the case of poisoning. Cardiac glucosides that induce the heart muscles to beat more slowly and improve circulation are extracted from Digitalis.> <A-head: Eudicots> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Recall>

30. Discuss the role of germplasm banks in maintaining genetic diversity of plants. <Answer: Germplasm banks are seed-storage facilities that maintain many varieties of seeds in cool, dry conditions. These facilities maintain plant genetic diversity. At certain time intervals, the seeds are planted and grown into mature plants so that fresh seeds may be collected. Seeds may also be kept viable for long periods of time at very low temperatures or in liquid nitrogen. In the future, attempts may be made to simply keep the DNA molecules in solution or in cloning vectors.> <A-head: Classification of Flowering Plants> <Subject: Chapter 24> <Complexity: Difficult> <Taxonomy: Analysis>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 25—Populations and Ecosystems

Multiple Choice

1. Which one of the following is not a negative impact of an isolated individual of a population?

A) Disease susceptibility increases B) Prone to natural disasters such as fires C) Safe from predators and pathogens D) Isolated individual cannot support pollinator population <Answer: C> <A-head: Concepts> <Subject: Chapter 25> <Complexity: Easy> <Taxonomy: Analysis>

2. The habitat of a plant species would include all of the following except: A) a person who picks some of the species’ flowers once.

B) nearby plants of other species.

C) the composition of the soil.

D) insects that lay their eggs in the plant’s leaves.

E) length of the growing season. <Answer: A> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Easy> <Taxonomy: Application>

3. If two species, A and B, occupy the same ecological niche in an area, all of the following outcomes are possible except: A) species A is better adapted and species B is ultimately excluded from the ecosystem.

B) both species diverge from their original niches, reducing competition.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank C) species B diverges from its original niche, reducing competition.

D) over time no species level changes should occur. <Answer: D> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Analysis>

4. Throughout its range, a species of fern is locally adapted to wetter conditions in some areas. In other areas, it is locally adapted to drier conditions. Other populations are adapted to higher light intensities; some populations are adapted to lower light intensities. These locally adapted populations are referred to as: A) populons.

B) ecotones.

C) ecotypes.

D) edaphonomes.

E) competitors. <Answer: C> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Analysis>

5. In a soil profile clay.

is rich in nutrients and contains both humus and

A) A horizon B) B horizon C) C horizon D) D horizon <Answer: B> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

6. Eastern red cedar grows best on less acidic soils, which are commonly found on limestone outcrops in its range. The type of geographic distribution of this plant would be: A) sparse.

B) clumped.

C) random.

D) uniform.

E) allelopathic. <Answer: B> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Easy> <Taxonomy: Application>

7. r conditions would not be produced by: A) a high wind blowing down a large tree in a forest.

B) a drought.

C) the abandonment of a field by a farmer.

D) the closing of the tree canopy in late spring or early summer.

E) a volcanic eruption that produces a new island. <Answer: D> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Analysis>

8. If a few seeds of a species blow into an area, the population will at first increase exponentially and then increase further, but not exponentially, until it reaches its carrying capacity. What will be the relationship between birth rate (germination) and death rate as time progresses?

A) Birth rate >>> death rate → birth rate > death rate → birth rate = death rate B) Birth rate = death rate → birth rate >>> death rate → birth rate > death rate → birth rate = death rate C) Birth rate > death rate → birth rate >>> death rate → birth rate = death rate D) Birth rate >>> death rate indefinitely

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank E) There is no way to predict how birth rate and death rate will compare. <Answer: A> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

9. A cow eats 2,500 kg of grass. About how much of the energy in the grass will be respired?

A) 2,500 kg B) 250 kg C) 1,250 kg D) 2,250 kg E) 1,875 kg <Answer: D> <A-head: The Structure of Ecosystems> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Application>

10. In general, areas with mild climates and rich soils have many species but intense competition.

The competition has resulted in natural selection for: A) niche partitioning, with many species in very narrow niches.

B) limiting factors being the same for almost all species.

C) niche overlap tolerance, with many species occupying the same niche.

D) temporal patterns being the same for almost all species.

E) generation time being the same for almost all species. <Answer: A> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. Living component along with the nonliving component is considered as the community. <Answer: False> <A-head: Concepts> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

12. A plant’s habitat would not include mycorrhizal fungi in the soil. <Answer: False> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Easy> <Taxonomy: Application>

13. If a plant needs 1 unit or iron, 3 units of phosphorus, and 5 units of calcium to survive, then the growth of a population of that plant in an area that contains 20 units of iron, 50 units of phosphorus, and 100 units of calcium would be limited by the amount of iron available. <Answer: False> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

14. Plants in temperate regions have a higher tolerance range than a plant from a tropical region of the world. <Answer: True> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

15. Climates of the high mountaintops are similar to high latitude places on earth. <Answer: True> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Easy> <Taxonomy: Recall>

Fill-in-the-Blank

16. Ants protect acacia trees and in exchange receive food from the acacias in the form of that are filled with glycogen. <Answer: Beltian bodies> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

17. The aspects of the habitat that definitely affect plant fitness are known as the plant’s habitat. <Answer: operational> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Recall>

18. At any given time and locality, the determines the health of the plant.

is the one factor that

<Answer: limiting factor> <A-head: The Structure of Populations> <Subject: Chapter 25>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Moderate> <Taxonomy: Application>

19. Terpenes given off by plants can inhibit the growth of other plants in the laboratory in an example of . <Answer: allelopathy> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

20. The that coexist in the ecosystem.

of an ecosystem describes the number and diversity of species

<Answer: species composition> <A-head: The Structure of Ecosystems> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

Matching

21. Match each example with a type of ecological relationship.

A) Indian pipe and a mycorrhizal fungus that has partnered with a chlorophyllous plant <Answer: Parasitism> B) Light blight, a fundal disease, of potato <Answer: Pathogenic> C) Epiphytic bromeliads on a tree <Answer: Commensalism> D) Mealybugs that suck sap out of the phloem of the plant <Answer: Predation> E) Two species of plants that produce flowers of the same size, shape, and color at the same time of year <Answer: Competition>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank F) Yucca moths lay their eggs in the gynoecium of the yucca flowers and also pollinate them <Answer: Mutualism> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

22. Match each Raunkiaer life-form with an example.

A) Phanerophyte <Answer: Sugar maple tree> B) Hemicryptophyte <Answer: Agave plant growing as a rosette> C) Therophyte <Answer: Annual weeds> D) Geophyte <Answer: Daffodil bulb> E) Chamaephyte <Answer: Rose shrub> <A-head: The Structure of Ecosystems> <Subject: Chapter 25> <Complexity: Moderate> <Taxonomy: Application>

Essay

23. Relate what you know about the effects of fire on ecosystems and the debate about putting out fires in Yellowstone National Park. <Answer: Fire is an inherent component of the ecosystems in Yellowstone National Park. Some species, such as those with serotinous cones, require fire to release seeds and prepare the seed bed for germination. The ash resulting from fire adds minerals to the soil to further stimulate growth. Periodic fire helps to eliminate the buildup of excessive fuel loads that result in catastrophic fires that can kill trees and lead to ecosystem damage. Given that the ecosystems in Yellowstone have involved under the adaptive pressure associated with periodic fire, the best management strategy should include periodic burning.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

24. One element often lost from grassland ecosystems during fires is nitrogen. What group of flowering plants (besides grasses) would you expect to find in grasslands? Why? <Answer: Although fire is an inherent component of many ecosystems, it can result in nitrogen losses from the burned site. Plants that form a relationship with nitrogen-fixing bacteria can afford to inhabit soils low in nitrogen; thus, this group of plants would likely be found in firedominated ecosystems.> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

25. If two species occupy exactly the same ecological niche in an ecosystem, what are the possible outcomes of this situation? Why? <Answer: A niche is the particular set of conditions an organism needs for survival. The competitive exclusion theory postulates that whichever species is less adapted is excluded from the ecosystem by superior competitors. An alternate theory suggests that when two species require the same niche conditions, each one will be pushed to the parts of the habitat that are not used by the other species. In the region where they overlap, usage is determined by competition.

A long-term result of competition would be species modification through natural selection.

Eventually divergent speciation could occur.> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

26. Sycamores and oaks often grow in the same general area, but oaks tend to be restricted to well-drained soils on hills, while sycamores are mainly found on wetter soils on floodplains.

Each type of tree is capable of growing on either well-drained or wet soil. Why do these two species exhibit this distribution pattern?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Sycamores and oaks probably have similar ecological niches. A niche is the particular set of conditions an organism needs for survival. An alternate theory to the competitive exclusion theory suggests that when two species require the same niche conditions, each one will be pushed to the parts of the habitat that are not used by the other species. In the region where they overlap, usage is determined by competition. In this case the sycamore are better able to grow and survive in the wetter soils, whereas the oak is the better competitor on the drier soils.> <A-head: Plants in Relationship to Their Habitats> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Analysis>

27. Distinguish between generation time and biotic potential. Describe how each affects population growth. <Answer: Generation time is the length of time from the birth of one individual until the birth of its first offspring. It affects the rapidity of population growth. The biotic potential is the number of offspring produced by an individual that actually live long enough to reproduce under ideal conditions. Since some seeds do not germinate and other seedlings die prior to reproduction, the biotic potential does not equal the numbers of seeds produced. This is the potential of plant reproduction over their lifetimes.> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Application>

28. Diagram the typical growth curve exhibited by most species when they invade a new habitat.

Explain what is happening in each part of the growth curve. <Answer: See Figure 25-23B. Initially, the population increases quickly. As the growth curve approaches the carrying capacity, K, and resources become limiting, the growth rate begin to slow. When the population size equals K, the population increase stops and the death rate equals the birth rate.> <A-head: The Structure of Populations> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Application>

29. Draw a pyramid of energy and explain why it has that particular shape.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: The pyramid should resemble Figure 25-36. Energy enters the biosphere through autotrophic metabolism such as the photosynthesis of plants. As energy moves through each trophic level, some is lost as heat; therefore, there is less energy at the higher trophic levels than there is at the base.> <A-head: The Structure of Ecosystems> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Application>

30. How might global warming affect the temporal structure of an ecosystem? <Answer: The changes that an ecosystem undergoes with time are its temporal structure. Plants change dramatically with the seasons. If temperatures warm, this could cause spring to come earlier and winter to arrive later. These changes in the timing of the seasons could have ecosystem impacts through changes in the timing of events such as leaf out, flowering, pollination, and photoperiod.> <A-head: The Structure of Ecosystems> <Subject: Chapter 25> <Complexity: Difficult> <Taxonomy: Application >

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 26—Community Ecology

Multiple Choice

1. Which of the following is an example of a community?

A) Two populations of salamanders living on two different mountaintops B) The organisms in Crescent Lake C) All the species and their habitats in North America D) The Green Ash trees on the outskirts of a beaver pond E) The beech trees and oak trees found throughout the world <Answer: B> <A-head: Concepts> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Application>

2. Which of the following is an example of a study of communities conducted at a regional scale?

A) Butterflies in a backyard B) Butterflies in a city park C) Butterflies in the northern hardwood forest D) Butterflies in the temperate forest of the United States E) Butterflies in temperate forests across the globe <Answer: B> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Analysis>

3. All of the following statements about communities may be true except: A) several species may be abundant and most species have just a few individuals or are rare.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank B) it may have a few species with only a few individuals.

C) the least abundant species within a community is the most frequently observed.

D) no community will have equal numbers of individuals of every species.

E) if many rare species are found, then the community may not have been searched enough. <Answer: C> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Analysis>

4. Which of the following is an example of the paradox of enrichment?

A) An invasive species is removed from the environment, providing more space for a native shrub; herbivore populations explode on the native shrub, causing both populations to go extinct.

B) If there are just a few deer, grasses will flourish and increase in abundance.

C) As the population of native plants grows, so does a predatory herbivore species, but at some point the herbivore overconsumes and the native plant declines, leading to a decline of the herbivore.

D) The population of an invasive vine species increases dramatically until it equals carrying capacity.

E) Both deer and grass populations remain steady, neither increasing nor decreasing. <Answer: A> <A-head: Predator–Prey Interactions> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

5. Harvesting permits for wild ginseng, Panax quinquefolius, limit collection to 3 pounds of wet weight per harvester over a 2-week period. Which harvesting method is this an example of?

A) Maximum sustained yield B) Fixed effort harvesting C) Fixed quota harvesting D) Fixed effort and fixed quota harvesting E) Maximum sustained yield, fixed effort, and fixed quota harvesting <Answer: D> <A-head: Predator–Prey Interactions>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Application>

6. Which of the following is not an example of a nurse plant?

A) A patch of moss provides a moist seedbed for seedlings B) Forest trees create shady conditions for shade-loving plants like ferns C) Lupine harboring Rhizobium D) Selaginella pilifera catching organic matter in which Echinocereus pectinatus becomes established E) Milkweed vine growing on mesquite <Answer: C> <A-head: Beneficial Interactions Between Species> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Application>

7. Some patches of alpine habitat are found underneath rock overhangs and therefore provide less light than required for many alpine wildflowers; however, wildflowers persist in these patches because: A) unoccupied patches maintain healthy populations.

B) such low-quality patches are required for ecological balance.

C) some patches are occupied by the sun-loving species and some are not.

D) by definition alpine environments are patchy and heterogeneous.

E) new members migrate in from high-quality patches rapidly enough to offset the losses of existing members. <Answer: E> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Analysis>

8. Metapopulations are best defined as:

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank A) all individuals of a species.

B) populations that maintain distinctive attributes through geographic isolation.

C) local populations that are interconnected by migration and gene flow between patches.

D) several small populations that collectively equal one large population.

E) populations capable of surviving in both and high- and low-quality patches. <Answer: C> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Analysis>

9. A food web is created by: A) choosing one predator and identifying its main prey and its supporting plant species.

B) identifying the keystone species in the ecosystem.

C) tracing all the prey of the top carnivores and then following the food sources of each of the prey species.

D) calculating the biomass of the dominant species as a measure of all energy inputs.

E) identifying the communities in each of the three trophic levels. <Answer: C> <A-head: Interconnectedness of Species: Food Chains and Food Webs> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Recall>

10. Migration corridors that interconnect patches are beneficial for community conservation for all of the following reasons except: A) they help to mitigate the effects of habitat fragmentation.

B) they can provide for the colonization of new habitat patches.

C) they have the potential to aid in the poleward migration of species as the earth’s climate changes.

D) they enable the migration of exotic species.

E) they promote gene flow. <Answer: D> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Complexity: Easy> <Taxonomy: Application>

True/False

11. Nurse plants are an example of a mutualism in which the plants below the nurse plants benefit from the favorable habitat conditions and the nurse plants themselves are supplied with greater access to nitrogen. <Answer: False> <A-head: Beneficial Interactions Between Species> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Application>

12. Scientists studying communities focus on the growth, interbreeding, and survival of a single species. <Answer: False> <A-head: Concepts> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Recall>

13. A species checklist measures species richness. <Answer: True> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Application>

14. When population size equals carrying capacity, the death rate equals the germination rate.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: True> <A-head: Predator–Prey Interactions> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Application>

15. A fugitive species is one that colonizes an area quickly to become the dominant species that is not easily displaced. <Answer: False> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Application>

Fill-in-the-Blank

16. The more-or-less predictable sequence of changes in community patterns over time is . <Answer: succession> <A-head: Concepts> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Recall>

17. Ecologists comparing species diversity across logged and unlogged plots in different temperate forests within the Pacific Northwest are assessing diversity. <Answer: beta> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

18. If a predator’s functional response is determine the predator’s feeding rate and handling time.

, then prey density will

<Answer: prey dependent> <A-head: Predator–Prey Interactions> <Subject: Chapter 26> <Complexity: Moderate> <Taxonomy: Application>

19. is an example of a beneficial interaction between species where one organism helps another without receiving any benefit in return. <Answer: Facilitation> <A-head: Beneficial Interactions Between Species> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Recall>

20. Humans practice dispersal when they collect seeds from one patch and plant them in a new patch with the appropriate ecological conditions. <Answer: assisted> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

Matching

21. Match each type of species interaction description with its example.

A) Optimal foraging theory <Answer: In one out of five plant visits, an herbivore finds a plant that lacks spines and possesses soft, edible berries.> B) Intraspecies competition

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Several bracken fern offspring compete for germination space.> C) Exploitation competition <Answer: The vegetative sprouts of American beech trees colonize an area following a logging event so quickly that few other species are able to establish.> D) Interference competition <Answer: In the winter, dead bracken fern leaves accumulate on shrubs, preventing their access to sunlight.> E) Apparent competition <Answer: An invasive plant is eliminated from an area, resulting in increased native plant densities; then population densities of an associated herbivore that prefers one species over the others also increase.> F) Prey-dependent response <Answer: Deer feeding rates on grass increase as the grass species become more abundant.> G) Zero growth isocline <Answer: Both aphid and hibiscus plant populations are neither increasing nor decreasing.> <Answer: A> <A-head: Predator–Prey Interactions> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Analysis>

22. Match each ecological interaction with its example.

A) Mutualism <Answer: Plants and Rhizobium> B) Facilitation <Answer: Beaver ponds> C) Primary succession <Answer: Lichens on bare rock> D) Nurse plant <Answer: Spineless living star cactus growing underneath hedgehog cactus> E) Climax community <Answer: Spruce-fir forest> F) Community restoration <Answer: Reintroduction of wolves into Yellowstone National Park> <A-head: Beneficial Interactions Between Species> <Subject: Chapter 26> <Complexity: Easy> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

Essay

23. Explain two threats to successful community restoration projects and how they may be mitigated. <Answer: Community restoration projects seek to minimize human impacts on native communities by encouraging the development of native communities. Two major threats to these projects are habitat loss and habitat fragmentation. To mitigate the effects of habitat loss, many project attempts are made to conserve lands and reconstruct habitats required by native flora and fauna. Scientists and land managers also attempt to connect smaller parcels of land through the construction of protected corridors. These attempts can ameliorate the problems caused by both the loss of large habitats and its fragmentation.> <A-head: Concepts> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

24. Provide one ecological and one historical explanation for the observation that some of the highest levels of community diversity exist in tropical climates. <Answer: Many species can thrive in equatorial regions, which tend to have stable, warm climates with few temperature extremes. From an ecological perspective, there is lots of energy available to the primary producers and thus more species at higher trophic levels can also be supported. About 50 million years ago, tropical conditions covered most of the earth’s surface.

After that point the Earth began to cool and tropical conditions were limited to equatorial regions. Plants and animals have had more to time to adapt to tropical conditions than temperate conditions. When considering the Earth’s past climate history, tropical regions may support more diversity because that has been more time for niche-partitioning and coevolution.> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

25. How might community ecologists measure and define the community diversity of an area?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: Rather than measuring species richness and completing checklists, community ecologists often focus on several organisms rather than all of them and may use factors other than species as the objective of diversity studies. Some examples of those other factors include the diversity of growth forms, herbivore feeding methods, diversity at different trophic levels, and the diversity of alleles.> <A-head: Diversity> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Analysis>

26. How does the Lotka-Volterra modeling the net rate of change in prey numbers differ from that which describes the net rate of change of predator numbers? <Answer: The formula dN/dt denotes how prey numbers change in response to rN – aNP. Where r is the intrinsic rate of increase for the prey species, N is the number of individuals of prey species, a is predators per capita attack rate, and P is the number of predator individuals present.

The rate of net change in predator populations is described by dP/dt, which is determined by faNp – qP. The latter expression has been modified from that of the prey equation to include f, a constant that indicates the predator’s efficiency at converting the prey it has eaten into new predators. The predator equation also considers q, the predator’s per capita mortality rate.> <A-head: Predator–Prey Interactions> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Analysis>

27. Explain how “cheating” could evolve in plant–pollinator relationships. <Answer: Natural selection favors organisms that reduce their costs. Cheating refers to obtaining benefits without paying costs. Nectar robbing bees bite through petals and obtain nectar without struggling around the stamens and becoming dusted with pollen. The bees fly from flower to flower, gaining their sugary rewards without doing any pollinating. These bees are cheating the plant-pollinator system.> <A-head: Beneficial Interactions Between Species> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

28. Explain the four assumptions used in common models of metapopulations.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: Metapopulation models assume some part of the environment is composed of discrete patches in which the species can live. They also assume that some patches are occupied, whereas other suitable patches are not. Empty patches are colonized by migration from occupied patches.

Populations within patches have a probability of going extinct in that patch.> <A-head: Metapopulations in Patchy Environments> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Recall>

29. Explain how a dominant species differs from a keystone species. <Answer: A dominant species is easily recognized and will have a strong impact on the structure of the community, but that impact is related to its abundance. The presence or absence of a keystone species has dramatic consequences for community structure. A keystone species had an impact that is out of proportion to its size or the number of individuals present.> <A-head: Interconnectedness of Species: Food Chains and Food Webs> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

30. In what ways has the Convention on International Trade in Endangered Species (CITES) actually hindered botanical conservation? <Answer: CITES is meant to prevent the illegal trade of endangered plants and animals; however, it does not take into account that plants are different from animals, particularly regarding the modularity of plants. Seeds, cuttings, and other plant parts are banned from trade when they could be used to propagate new plants in botanical gardens or other safe areas. Some parts of the world have diverse plant communities that have not been accurately inventoried.

CITES makes it harder to get scientific collection permits and thus limits potentially fruitful collaborations between scientists and local people with plant knowledge. Inspection agents are not trained in identification of all plant species and therefore confuse endangered and nonendangered species, thereby limiting trade of plants that aren’t endangered.> <A-head: Interconnectedness of Species: Food Chains and Food Webs> <Subject: Chapter 26> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank CHAPTER 27—Biomes

Multiple Choice

1. What is the main similarity between temperate and tropical rain forests?

A) The yearly temperature pattern B) The dominant trees C) Species diversity D) Daily fog E) A large amount of precipitation each year <Answer: E> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Easy> <Taxonomy: Analysis>

2. The entire United States would have a uniform distribution of rainfall if it: A) had mountains in the west and the rest was flat.

B) had mountains in the east and in the west.

C) had mountains in the center but none on the coasts.

D) was fairly flat. <Answer: D> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Analysis>

3. The climate of the earth is due to which of the following factors?

I. The presence of atmosphere and oceans II. The changes in the distance of the earth from the sun over the course of a year III. The movement of cold water from the deep ocean to the surface

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank IV. The earth’s tilted axis of rotation A) II and III are correct.

B) III and IV are correct.

C) I and II are correct.

D) I and IV are correct.

E) I, II, III and IV are correct. <Answer: D> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Analysis>

4. Which one of the following is not a result of oceanic heat distribution?

A) Reduction of temperature gradient between tropics and poles B) Increase in humidity at the poles C) Decrease in humidity at the poles D) Increased humidity and large temperature extremes at the poles <Answer: C> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Analysis>

5. As North America collided with Eurasia, forming the Appalachian Mountains, what group of plants was first evolving?

A) Aneurophytales B) Trimerophytophyta C) Rhyniophytes D) Archaeopteridales E) Medullosales <Answer: C> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Moderate>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Taxonomy: Recall>

6. Conifers evolved: A) as Gondwanaland formed and the climate became cooler and drier.

B) as Gondwanaland collided with Laurasia and climate became hotter and drier.

C) as Pangaea formed with its diverse climate.

D) as North America and Eurasia moved northward and climate became cooler and moister.

E) during the Triassic, when the climate was dry. <Answer: D> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

7. During the time of Pangaea, if you were a glaciologist (a person who studies glaciers), you would most likely be: A) doing research at the Laurasian portion of Pangaea.

B) doing research at the Gondwanaland portion of Pangaea.

C) anywhere in Pangaea because the entire continent was covered in glaciers.

D) out of luck, as there were virtually no glaciers at the time of Pangaea. <Answer: B> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

8. What type of biome most often supports herds of large herbivores?

A) Tropical rain forest B) Shrubland C) Temperate deciduous forest D) Savanna E) Boreal coniferous forest

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: D> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Easy> <Taxonomy: Application>

9. If you were a botanist interested in the study of lichens, which biome would be the best source of those organisms?

A) Tropical rain forest B) Arctic tundra C) Temperate deciduous forest D) Desert E) Grassland <Answer: B> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Application>

10. If you find a deposit of bauxite in an area that currently has a climate that is cool and dry, you could conclude that the climate at the time that the bauxite formed was: A) similar to what it is now.

B) actually much colder and wetter than it is now.

C) actually much warmer and wetter than it is now.

D) warmer but dryer than it is now.

E) cool but much wetter than it is now. <Answer: C> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Analysis> True/False

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

11. It is the dominant plants that most strongly characterize a biome. <Answer: True> <A-head: Concepts> <Subject: Chapter 27> <Complexity: Easy> <Taxonomy: Recall>

12. In the Northern Hemisphere, the prevailing westerlies occur farther south than the northeast trade winds. <Answer: False> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

13. The latest major event in continental drift was the collision of South America and North America. <Answer: True> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

14. A temperate deciduous forest has a thick layer of organic matter on the forest floor. <Answer: False> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

15. Chaparral in California has a mild, warm summer every year.

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

<Answer: False> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

Fill-in-the-Blank

16. Biomes are driven in part by continental position, which can shift over time due to . <Answer: continental drift> <A-head: Concepts> <Subject: Chapter 27> <Complexity: Easy> <Taxonomy: Recall>

17. During the directly as possible toward the sun.

, on December 21 or 22, the South Pole points as

<Answer: winter solstice> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

18. Just after the Period, the world’s climate warmed and many new plant groups evolved, such as the Cycadophyta and Ginkgophyta. <Answer: Permian> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Recall>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

19. In the northeastern United States, contain balsam fir.

can extend down to 760 m and

<Answer: subalpine forest> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Recall>

20. In the summer, arctic tundra soil can be very marshy due to drainage prevention of . <Answer: permafrost> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Easy> <Taxonomy: Application>

Matching

21. Match each continental position with a geological time period.

A) South America separated from Africa <Answer: Cretaceous Period> B) Collision of South America and North America <Answer: 5 to 13 million years ago> C) Gondwanaland located in the Southern Hemisphere <Answer: Cambrian Period> D) Pangaea was completely formed <Answer: Devonian Period> E) Laurasia formed <Answer: Silurian Period> F) Pangaea began to break up <Answer: Jurassic Period> <A-head: Continental Drift>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Recall>

22. Match each characteristic with a biome.

A) Greatest species diversity <Answer: Tropical rain forest> B) Richest soil in essential elements <Answer: Temperate grasslands> C) Majority of plant biomass may be underground <Answer: Arctic tundra> D) May be dominated by 1 to 2 species <Answer: Taiga> E) Soil thin and rocky, in some cases reduced to pebbles <Answer: Desert> F) Transitional biome often between grasslands and forests <Answer: Woodlands> G) Often found in the rain shadow on the landward side of mountain ranges <Answer: Montane forest> H) Termites can be primary grazer <Answer: Tropical savanna> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Moderate> <Taxonomy: Analysis>

Essay

23. Explain how the western part of Washington State can support lush vegetation while the eastern part is grassland. <Answer: The prevailing westerlies bring moist air from the Pacific Ocean onto land in Washington, where it is forced to rise by coastal mountain ranges. The rising air rains over the western slopes. After the air passes the summit, it descends and warms, and the rains stop. The decreased rain on the landward side of the mountains is called a rain shadow. Eastern Washington is in a rain shadow, a condition in which grasslands thrive.>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

24. For their latitude, which is basically the same as Canada, Ireland, and England are warmer than they should be. Why? <Answer: Oceanic currents are the cause of the warmer water in Europe and cooler water in Canada. Trade winds blow across the tropics from east to west, pushing surface waters into equatorial currents. The Atlantic Equatorial current is warmed as it absorbs huge amounts of energy and is deflected northward at the tip of Brazil up to Florida. The warm water continues to move along the Gulf Stream, and as it nears New Jersey and New York, it is deflected to Europe and does not travel northward to Canada. In fact, at the turning point, cold polar water moves south along Canada’s east coast.> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

25. Explain in terms of meteorology why tropical rain forests receive so much rainfall and why the Sahara Desert is so hot and dry. <Answer: Evaporation driven by solar radiation in equatorial regions causes the air to warm, expand, and rise high into the atmosphere. The expanding air cools, decreasing its ability to hold moisture, and its water vapor condenses into rain over the tropical rain forests. The air is pushed northward and southward and by the time it reaches the horse latitudes, it has cooled, contracted, and become dense enough to sink. The descending air masses are extremely dry, resulting in some of the world’s driest desert biomes like the Sahara Desert.> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

26. Describe the origin of the northeast trade and prevailing westerly winds. How does each influence climate in the United States?

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank <Answer: The Earth’s rotation causes air moving toward the equator to be deflected northward; this becomes the northeast trade wind. Air spreading toward the poles from the horse latitudes is deflected eastward and blows as a prevailing westerly wind. During the winter, the United States is under the influence of the prevailing westerlies such that winter weather primarily comes from the Pacific and Arctic Oceans. During the summer, the northeast trade winds bring summer rainstorms westward from the Atlantic onto the East Coast.> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

27. Explain why there are so few native species of plants in the Himalayan Mountains. <Answer: During the Mesozoic Era, India and Africa separated. At the time of separation, India was at the edge of the south temperate zone and its flora was adapted to that climate. It migrated rapidly, in less than 70 million years, into the north temperate zone. The migration caused mass extinctions. When it collided with the Eurasian continent, the Himalayan Mountains were formed. At this time, flora from Eurasia that were well adapted to the north temperate zone invaded the Indian continent and outcompeted the native Indian flora, resulting in few native species in the Himalayans in modern times.> <A-head: Continental Drift> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

28. Many tropical rain forests are being cut down or burned by people to grow crops. Knowing what you do about soils, predict their success. Justify your prediction. <Answer: Tropical soils are not ideal for agricultural purposes. Soils transform rapidly in tropical rain forests due to high temperatures and moisture. Many elements are leached from the soil, leaving behind aluminum and iron oxides. Humus decays rapidly and there is little development of soil horizons. Almost all of the available essential elements are sequestered in biomass and thereby removed when the forest trees are burned or otherwise harvested and removed from the site. Following forest removal, the remaining thin, nutrient-poor soils are not ideal for agricultural success.> <A-head: World Climate> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

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Botany: An Introduction to Plant Biology, Seventh Edition James D. Mauseth Test Bank

29. Why are the only two continents that have temperate rain forests North and South America? <Answer: The west coasts of North and South America both have coastal mountain ranges. The prevailing westerlies bring moist air from the Pacific Ocean onto land, where it is forced to rise by coastal mountain ranges. The rising air reliably rains over the western slopes throughout most of the year, with the exception of a brief period of summer dryness. After the air passes the summit, it descends and warms, and the rains stop. The climate patterns are the same on the two continents but the species composition is entirely different.> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

30. Explain how the age of a rock formation can be shown by paleomagnetism. <Answer: Paleomagnetism allows the discovery of past continental position by examining the magnetic materials of rocks. When rocks solidify, the magnetic minerals are fixed in alignment with the rock position and the Earth’s magnetic field. If a rock is found with a magnetic orientation that does not point to the north magnetic pole, then that rock has been moved since it was formed. Examining blocks of rock of known ages reveals the position of the rock at the time of formation.> <A-head: The Current World Biomes> <Subject: Chapter 27> <Complexity: Difficult> <Taxonomy: Application>

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