Ch 30 The adaptation that made possible the colonization of dry land environments by seed plants is most likely the result of the evolution of _____. pollen heterospory sporophylls ovules cones

pollen *In bryophytes and seedless vascular plants, flagellated sperm must swim through a film of water to reach the egg cells. In seed plants, the use of airborne pollen to bring gametes together is a terrestrial adaptation.

Pine Life Cycle In pines, the female gametophyte contains _____, each of which contains a(n) _____. archegonia ... sperm cell antheridia ... egg antheridia ... sperm cell microsporangia ... egg cell archegonia ... egg

archegonia ... egg *In pines, a megaspore repeatedly grows and divides, giving rise to a female gametophyte. The female gametophyte is the site in which egg-bearing gametophytes develop.

Pine Life Cycle In pines, an embryo is a(n) _____. seed immature sporophyte immature female gametophyte immature male gametophyte food reserve for the immature sporophyte

immature sporophyte *The diploid embryo will develop into a seedling and then into a mature pine tree.

Pine Life Cycle In pine trees, pollen grains get to the ovule via the _____. eggs integument pollen cone megaspore micropyle

micropyle *Pollen grains gain entry into the ovule via the micropyle.

Pine Life Cycle Which of these statements is true about the gametophyte tissue that surrounds the pine embryo? It functions as a triploid food reserve. It functions as a haploid food reserve. It functions as a diploid food reserve. It develops from the fusion of a microspore and a megaspore. It is the remnant of the pollen tube.

It functions as a haploid food reserve. *This gametophyte tissue is a source of nourishment for the embryo.

Pine Life Cycle Of the four haploid cells produced by a pine cone's megasporocyte (megaspore mother cell), _____ survive(s). three four one two integuments

one

Pine Life Cycle In the pine, microsporangia form _____ microspores by _____. haploid ... mitosis triploid ... fertilization diploid ... mitosis diploid ... meiosis haploid ... meiosis

haploid ... meiosis *Diploid microsporangia form haploid microspores by meiosis.

Ch 30 What is true of stamens, sepals, petals, carpels, and pinecone scales? They are modified leaves. None are capable of photosynthesis. They are found on angiosperms. They are found on flowers. They are female reproductive parts.

They are modified leaves.

Ch 30 The question refers to the following description. The cycads, a mostly tropical phylum of gymnosperms, evolved about 300 million years ago and were dominant forms during the Age of the Dinosaurs. Though their sperm are flagellated, their ovules are pollinated by beetles. These beetles get nutrition (they eat pollen) and shelter from the microsporophylls. Upon visiting megasporophylls, the beetles transfer pollen to the exposed ovules. In cycads, pollen cones and seed cones are borne on different plants. Cycads synthesize neurotoxins, especially in the seeds, that are effective against most animals, including humans. Which feature of cycads distinguishes them from most other gymnosperms? 1. They have exposed ovules. 2. They have flagellated sperm. 3. They are pollinated by animals. 1 only 1, 2, and 3 2 and 3 2 only 3 only

2 and 3

Ch 30 Harold and Kumar are pre-med and pre-pharmacy students, respectively. They complain to their biology professor that they should not have to study about plants because plants have little relevance to their chosen professions. From reading their biology textbook, what would Harold and Kumar discover? Prescription drugs that enter the water table are responsible for the extinction of many plants. All rain forest plants contain at least one chemical useful as a medicine. About one-quarter of all prescription drugs come from seed plants. Much of what was once rain forest has been replanted with fields of medicinally valuable plants.

About one-quarter of all prescription drugs come from seed plants.

Ch 30 Unlike most angiosperms, grasses are pollinated by wind. As a consequence, some unnecessary parts of grass flowers have almost disappeared. Which of the following parts would you expect to be most reduced in a grass flower? anthers petals stamens ovaries carpels

petals *Petals play a role in attracting pollinators.

Ch 30 What adaptations should one expect of the seed coats of angiosperm species whose seeds are dispersed by frugivorous (fruit-eating) animals, as opposed to angiosperm species whose seeds are dispersed by other means? 1. The exterior of the seed coat should have barbs or hooks. 2. The seed coat should contain secondary compounds that irritate the lining of the animal's mouth. 3. The seed coat should be able to withstand low pH's. 4. The seed coat, upon its complete digestion, should provide vitamins or nutrients to animals. 5. The seed coat should be resistant to the animals' digestive enzymes. 4 only 2 and 3 3, 4, and 5 1 and 2 3 and 5

3 and 5

Primary and Secondary Growth Secondary growth NEVER occurs in _____. stems roots stems and leaves leaves roots and leaves

leaves *Secondary growth never occurs in leaves.

Primary and Secondary Growth _____ provides cells for secondary growth. Vascular cambium Apical meristem Secondary xylem Secondary phloem The root

Vascular cambium *Vascular cambium is lateral meristem that provides cells for secondary growth.

Primary and Secondary Growth Vascular cambium forms wood toward the stem's _____ and secondary phloem toward the stem's _____. center ... center surface ... center center ... surface surface ... surface top ... bottom

center ... surface *Wood, or secondary xylem, is formed toward the stem's center, and secondary phloem is formed toward the stem's surface.

Primary and Secondary Growth What is the function of cork? regulating the opening and closing of stomata providing cells for primary growth providing cells for secondary growth providing a site for photosynthesis insulation and waterproofing

insulation and waterproofing *Cork insulates and waterproofs roots and stems.

Primary and Secondary Growth How is the supply of vascular cambium maintained? by the differentiation of secondary xylem by the division of its cells by the differentiation of cork by the differentiation of secondary phloem by the differentiation of apical meristem

by the division of its cells *When a vascular cambium cell divides, one cell differentiates and the other cell remains meristematic.

Ch 35 Compared to most animals, the growth of most plants is best described as primary. derivative. weedy. perennial. indeterminate.

indeterminate.

Ch 35 Which part of a plant absorbs most of the water and minerals taken up from the soil? sections of the root that have secondary xylem storage roots root hairs root cap the thick parts of the roots near the base of the stem

root hairs

Ch 35 Most of the growth of a plant body is the result of cell differentiation. cell elongation. cell division. reproduction. morphogenesis.

cell elongation.

Ch 35 Which of the following would not be seen in a cross-section through the woody part of a root? vessel elements parenchyma cells root hairs sieve-tube elements sclerenchyma cells

root hairs

Interpreting Data: Effects of Auxin on Cell Elongation For Graph: Assignment #8- Question #8 What does the x-axis of this graph represent? the concentration of auxin in the roots the extent that elongation is promoted or inhibited the concentration of auxin in liters per gram the concentration of auxin in grams per liter

the concentration of auxin in grams per liter

Interpreting Data: Effects of Auxin on Cell Elongation What does the distance between two white vertical lines on this graph represent? a 100-fold change in auxin concentration a doubling of auxin concentration a change in auxin concentration of 1 g/L an increase in elongation

a 100-fold change in auxin concentration

Interpreting Data: Effects of Auxin on Cell Elongation What does the point on the far right end of the blue line indicate? A very high concentration of auxin greatly promotes elongation of the roots. A very high concentration of auxin greatly inhibits elongation of the roots. A very high concentration of auxin greatly promotes elongation of the stems. A very high concentration of auxin greatly inhibits elongation of the stems.

A very high concentration of auxin greatly inhibits elongation of the stems.

Interpreting Data: Effects of Auxin on Cell Elongation As labeled on the graph, the blue line crosses the dashed zero line at an auxin concentration of 0.9 g/L. What does this indicate? At this auxin concentration, elongation of stems is neither promoted nor inhibited. At this auxin concentration, elongation of stems is promoted, but elongation of roots is inhibited. At this auxin concentration, elongation of roots is neither promoted nor inhibited. At this point on the graph, the auxin concentration is 0 g/L.

At this auxin concentration, elongation of stems is neither promoted nor inhibited.

Interpreting Data: Effects of Auxin on Cell Elongation Is this statement supported or not supported by the graph? Very low concentrations of auxin promote elongation of the roots. supported not supported cannot be determined from the graph

supported

Interpreting Data: Effects of Auxin on Cell Elongation Is this statement supported or not supported by the graph? Auxin concentrations below 0.9 g/L always promote elongation. supported not supported cannot be determined from the graph

not supported

Interpreting Data: Effects of Auxin on Cell Elongation Is this statement supported or not supported by the graph? Any given concentration of auxin either promotes stem elongation or promotes root elongation, but never both. supported not supported cannot be determined from the graph

not supported

Interpreting Data: Effects of Auxin on Cell Elongation Is this statement supported or not supported by the graph? There are concentrations of auxin that promote stem elongation while simultaneously inhibiting root elongation. supported not supported cannot be determined from the graph

supported

Ch 39 A plant mutant that shows normal gravitropic bending but does not store starch in its plastids would require a reevaluation of the role of ____________ in gravitropism. statoliths light differential growth calcium auxin

statoliths

Ch 39 How may a plant respond to severe heat stress? by reorienting leaves to increase evaporative cooling by increasing the proportion of unsaturated fatty acids in cell membranes, reducing their fluidity by creating air tubes for ventilation by producing heat-shock proteins, which may protect the plant's proteins from denaturing by initiating a systemic acquired resistance response

by producing heat-shock proteins, which may protect the plant's proteins from denaturing

Ch 39 In order for a plant to initiate chemical responses to herbivory, phytoalexins must be released. volatile "signal" compounds must be perceived. it must be past a certain developmental age. the plant must be directly attacked by an herbivore. gene-for-gene recognition must occur.

volatile "signal" compounds must be perceived.

Flowering Lab For a short-day plant, the _____ is critical in determining if flowering will occur. maximum number of hours of light minimum number of hour of light maximum number of hours of darkness None of these is correct, since short-day plants are also known as day-neutral plants. minimum number of hours of darkness

minimum number of hours of darkness *Short-day plants are actually long-night plants.

Flowering Lab For a long-day plant, the _____ is critical in determining if flowering will occur. maximum number of hours of light minimum number of hours of light maximum number of hours of darkness minimum number of hours of darkness None of these is correct, since short-day plants are also known as day-neutral plants.

maximum number of hours of darkness *Long-day plants are short-night plants.

Flowering Lab Day-neutral plants flower regardless of _____. photoperiod day length or night length day length, night length, or photoperiod night length day length

day length, night length, or photoperiod *Photoperiod refers to seasonal changes in the relative lengths of day and night.

Flowering Lab Which of these would inhibit flowering in a short-day plant with a critical night length of 12 hours? 12 hours of light followed by 12 hours of dark 12 hours of light, 6 hours of dark, a flash of red light, a flash of far-red light, 6 hours of dark All of these regimens would inhibit flowering in a short-day plant with a critical night length of 12 hours. 12 hours of light, 6 hours of dark, a flash of red light, a flash of far-red light, a flash of red light, a flash of far-red light, 6 hours of dark 12 hours of light, 6 hours of dark, a flash of red light, 6 hours of dark

12 hours of light, 6 hours of dark, a flash of red light, 6 hours of dark *The flash of red light shortens the length of the dark period, so flowering would not occur.

Flowering Lab Which of these would stimulate flowering in a long-day plant with a critical night length of 12 hours? 8 hours of light and 16 hours of dark 12 hours of light, 6 hours of dark, a flash of red light, a flash of far-red light, a flash of red light, a flash of far-red light, 6 hours of dark 12 hours of light, 6 hours of dark, a flash of red light, 6 hours of dark 12 hours of light followed by 12 hours of dark 12 hours of light, 6 hours of dark, a flash of red light, a flash of far-red light, 6 hours of dark

12 hours of light, 6 hours of dark, a flash of red light, 6 hours of dark *The flash of red light shortens the length of the dark period and thus flowering would not occur.

Water Transport in Plants: Transpiration (BioFlix tutorial) Transpiration provides the driving force for the movement of water from the soil to the highest leaves of plants. Transpiration is also linked to other processes in plants, including photosynthesis and the transport of mineral nutrients. Which of the following statements correctly describe(s) a relationship between transpiration and other processes in plants? Select all that apply. Water is pulled from the roots to the leaves by transpiration, whereas mineral nutrients diffuse from the roots to the leaves. The large surface area exposed to air inside the leaf maximizes the plant’s ability to absorb CO2 while minimizing water loss through transpiration. Transpiration is important in cooling leaves on warm, sunny days. In most plants, the highest rate of transpiration occurs when the rate of photosynthesis is also highest. To minimize water loss during dry conditions, most plants must also restrict their ability to carry out photosynthesis. Open stomata provide a low-resistance pathway for CO2 to enter and for water to exit the leaf.

Transpiration is important in cooling leaves on warm, sunny days. In most plants, the highest rate of transpiration occurs when the rate of photosynthesis is also highest. To minimize water loss during dry conditions, most plants must also restrict their ability to carry out photosynthesis. Open stomata provide a low-resistance pathway for CO2 to enter and for water to exit the leaf. *Both water and CO2 diffuse as gases through stomata. Therefore, the opening of stomata to permit CO2 to rapidly enter the leaf also promotes water loss through transpiration. Conversely, when the stomata close to limit water loss under dry conditions, the entry of CO2 into the leaf is restricted and photosynthesis slows. Transpiration pulls both water and mineral nutrients from the roots to the leaves. It also cools the leaves as water evaporates from cell surfaces inside the leaves.

Ch 36 All of the following normally enter the plant through the roots except calcium. carbon dioxide. potassium. nitrogen. water.

carbon dioxide

Ch 36 Which of the following experimental procedures would most likely reduce transpiration while allowing the normal growth of a plant? putting the plant in drier soil decreasing the relative humidity around the plant subjecting the leaves of the plant to a partial vacuum injecting potassium ions into the guard cells of the plant increasing the level of carbon dioxide around the plant

increasing the level of carbon dioxide around the plant

Ch 36 Water flows into the source end of a sieve tube because the companion cell of a sieve tube actively pumps in water. sucrose has been transported out of the sieve tube by active transport. sucrose has diffused into the sieve tube, making it hypotonic. sucrose has been actively transported into the sieve tube, making it hypertonic. water pressure outside the sieve tube forces in water.

sucrose has been actively transported into the sieve tube, making it hypertonic.

Ch 36 Select the correct mechanism of stomatal opening and closing. The radial orientation of cellulose microfibrils in the cell walls of guard cells causes them to bow outward and open the stomatal pore when turgid. The cell walls of guard cells are thickened on the side of the stomatal opening and the thinner walls bow outward when the guard cells become turgid, to close the stomata. Light stimulates proton pumps in the plasma membrane of the guard cells, causing them to lose K+ and become flaccid.

The radial orientation of cellulose microfibrils in the cell walls of guard cells causes them to bow outward and open the stomatal pore when turgid.

Activity: Soil Formation and Nutrient Uptake What are the largest particles formed from the breakdown of rock? Sand Silt Gravel Clay

Gravel *Gravel particles are the largest particles formed from the breakdown of rock.

Activity: Soil Formation and Nutrient Uptake Which of the following is an elemental ion? K+ HCO3- NO3- K

K+ *K+ is an elemental ion.

Activity: Soil Formation and Nutrient Uptake True or false? Soil texture affects the amount of water available to plants; water is held best by clay and sand particles. True False

False *Soil texture does indeed affect the amount of water available to plants, but water is held best by the smallest soil particles, clay and silt.

Activity: Soil Formation and Nutrient Uptake Which of the following steps occurs first during soil formation? Mosses grow on the rock surface. Organic material is added to the rock surface. Lichens grow on the rock surface. Weathering of solid rock occurs.

Weathering of solid rock occurs. *Weathering is the process by which solid rock is broken down into smaller pieces and is the first step in soil formation.

Activity: Soil Formation and Nutrient Uptake Why is the decomposition of dead organisms important for soil formation? The process makes negatively charged ions more easily absorbed by plant roots. The process produces acidic compounds that dissolve the rock surface. The process adds organic matter to the soil, which is necessary to support the growth of larger plants. The process increases the amount of silt in the soil.

The process adds organic matter to the soil, which is necessary to support the growth of larger plants. *Decomposition of dead organisms adds organic matter, or humus, to the soil; humus is necessary to support plant growth.

Activity: Soil Formation and Nutrient Uptake Which of the following statements about nutrient uptake by plants is true? Plants can easily absorb mineral ions from soil with large amounts of organic matter. Plants require other elements besides carbon, hydrogen, and oxygen to grow, and they can obtain these in soil. Root hairs increase the volume of roots for more efficient absorption of water and nutrients. Positively charged ions remain dissolved in water and are easily absorbed by plant roots.

Plants require other elements besides carbon, hydrogen, and oxygen to grow, and they can obtain these in soil. *This statement is true; plants also require elements such as phosphorus, sulfur, and magnesium.

Activity: Soil Formation and Nutrient Uptake True or false? Plants use both active and passive transport processes to transport ions against their concentration gradients. True False

False *Plants use passive transport, which requires no energy expenditure, to transport ions down their concentration gradients, but they can only use active transport, which requires energy in the form of ATP, to transport ions against their concentration gradients.

Ch 37 The sundew plant has to digest insects because _____. it has lost the ability to perform photosynthesis it obtains nitrogen from their bodies that it cannot get from the soil it's a method of self-cleaning to rid the plants of insects that get stuck in the plant it lives in a dry environment and uses moisture from the insects' bodies its flowers are fertilized by pollen in its digestive tract

it obtains nitrogen from their bodies that it cannot get from the soil *The sundew lives in nitrogen-poor soil and obtains its nitrogen from the digestion of insects.

Ch 37 The relationship between legumes and Rhizobium is _____. parasitic commensalistic predatory mutualistic competitive

mutualistic *Both the legume and the Rhizobium benefit from this relationship. The legume gains a supply of fixed nitrogen from Rhizobium, and Rhizobium gains organic nutrients from the legume.

Ch 37 Nitrogen-fixing bacteria in the soil _____. convert nitrates to N2 use nitrates to make amino acids that plants can use change ammonium into nitrates convert nitrate to ammonium convert atmospheric nitrogen to ammonia

convert atmospheric nitrogen to ammonia *Nitrogen fixation is the conversion of N2 (atmospheric nitrogen) to NH3.

Ch 37 Identify the components of the rhizosphere, the soil layer that surrounds plant roots. Select all that apply. Fungi Dirt Bacteria

Fungi Dirt Bacteria

Ch 37 How do plants obtain organic molecules? Plants take up organic molecules through their roots. Plants synthesize their own organic molecules. Plants take in organic molecules through their stomata.

Plants synthesize their own organic molecules. * As autotrophs, plants synthesize their own organic molecules. They incorporate inorganic nutrients taken up from soil into components of these organic molecules.

Ch 37 If a wide-spectrum fungicide that kills all fungal species were used extensively in a forest, what effects would you expect the treatment to have on the forest vegetation? It would harm the few plants that benefit from mutualistic relationships with mycorrhizal fungi. It would greatly reduce the ability of most plants to take up water and minerals from soil. It would benefit plants by killing off fungal pathogens.

It would greatly reduce the ability of most plants to take up water and minerals from soil. *Most plants rely on mycorrhizal fungi to take up water and minerals from soil. If these fungi were killed, it would greatly reduce the ability of most plants to take up water and minerals from soil.

LOOK AT ON MASTERING BIOLOGY ASSIGNMENT #7= 2A,2B,3A,&6A ASSIGNMENT #8= 1A,1B,1C,2A,2B,2E, 7A,7B,&7C ASSIGNMENT #9= 1A,1B&1C