Notes to Instructors - San Dieguito Union High School Districtteachers.sduhsd.net/ahaas/AP Biology/Plant Physiology... · Notes to Instructors ... To understand transport in plants,

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Notes to Instructors What is the overall focus of activities associated with Chapters 35 to 49? Activities from earlier chapters and from the chapters on the diversity of organisms allowed students to develop an understanding of these topics:

• The events that occurred in the evolution of multicellular organisms in general and of plants and animals in particular

• The effects that surface area-to-volume (SA/V) ratios, rates of osmosis and diffusion, and metabolic requirements per unit volume per unit time can have on the evolution of the form and function of organisms

• The effects that different sets of selective forces (aquatic versus terrestrial) can have on the evolution of the form and function of organisms

The activities in Chapters 35−49, which cover the units “Plant Form and Function” and “Animal Form and Function” in Biology, 7th edition, are designed with these objectives:

• These activities build on the understanding of how the structure of each system is related to (or in a sense, determines) its function.

• In addition, they are designed to help students develop a good understanding of both the basic function(s) of each system and the interrelationships among systems.

Chapter 35 Plant Structure, Growth, and Development What is the focus of this activity? In this activity, students are asked to compare the basic form of monocots and dicots to determine key similarities and differences in their structure and function. What is this particular activity designed to do? Activity 35.1 How Does Plant Structure Differ among Monocots, Herbaceous Dicots, and Woody Dicots? The questions in this activity are designed to help students review and understand the basic form and function of the angiosperm plant body as well as how that basic form is modified in monocots, herbaceous dicots, and woody dicots.

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Answers

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Activity 35.1 How Does Plant Structure Differ among Monocots, Herbaceous Dicots, and Woody Dicots?

The chart on the next page shows a drawing of a generic plant. The arrows indicate various plant parts or organs. In the columns to the right of the plant, draw cross sections of the plant at the points indicated by the arrows. To help you visualize the differences in structure among different types of plants, in Column I draw the cross sections assuming the plant is a monocot. In Column II draw them assuming the plant is a herbaceous dicot. Finally, in Column III draw the cross sections as if the plant is a woody dicot. Be sure to label your drawings.

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Use the information in your drawings to answer the questions on the next pages.

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1. In Column I, connect the cross sections of the monocot by drawing a line (color 1) from the water transport tissue in one section to the water transport tissue in the next, and so on. Draw another line (color 2) to connect the food transport tissue from one section to the next. Do the same for the herbaceous and the woody dicot cross sections.

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2. Use the information in the cross sections to fill in the chart below. Note: A distinguishing feature is one that is found in only the given type of organism. For example, a distinguishing feature of mammals is the presence of hair: Hair is a characteristic of all mammals and is not found in any other animals.

Distinguishing feature(s) of

Leaf Stem Root Apical meristem shoot vs root

Monocot

Vascular traces (veins) in the leaf are arranged in parallel. In a cross section of the leaf you would expect to see the vascular bundles evenly spaced and of approximately equal size across the section.

Vascular bundles are scattered throughout the cross section.

Vascular bundles form a ringlike pattern around a central pith. (Vascular bundles are surrounded by a pericycle, which is surrounded by an endodermal layer.)

The apical meristem of the root is covered by a protective root cap. In addition to the apical meristem which give rise to the leaves, many monocots have intercalary meristems, which lie near the base of the stem. For example, grass blades or leaves continue to grow after mowing or grazing. This growth is from their intercalary meristems.

(Continued on next page)

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Herbaceous dicot

Vascular traces (veins) in the leaf are arranged in a netlike pattern.

In a cross section of the leaf you would not expect to see the vascular bundles evenly spaced. In addition, the bundles would likely be of varying size.

Vascular bundles are arranged in a ring between a thin outer cortex layer and a central region of pith.

There is no central pith. Xylem forms a crosslike pattern in the center of the root. Phloem sits in the arms of the cross. As in monocots, this central region is surrounded by a pericycle, which is surrounded by an endodermal layer.

As in monocots, the apical meri-stems of the roots are gener-ally protected by a root cap. There is gener-ally no protec-tive tissue layer over the apical meristems of the aerial portions of the plant.

Woody dicot

Vascular traces (veins) in the leaf s are arranged in a netlike pattern.

In a cross section of the leaf you would not expect to see the vascular bundles evenly spaced. In addition, the bundles would likely be of varying size.

Vascular cambium is continuous and produces sec-ondary xylem to the inside and secondary phloem to the outside. Varia-tions in water availability (wet vs dry seasons) can, result in growth rings.

A ring of vascular cambium forms between the central xylem and phloem and lays down secon-dary xylem to the inside and secondary phloem to the outside.

New root growth is similar in cross section to that of herba-ceous dicots.

As in monocots, the apical meri-stems of the roots are gener-ally protected by a root cap. There is gener-ally no protec-tive tissue layer over the apical meristems of the aerial portions of the plant.

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3. If cross sections were not available, what other characteristics of the plants as a whole could you use to determine whether each was a monocot, a herbaceous dicot, or a woody dicot? Monocot leaves have parallel veins (vascular bundles); dicot leaves have netlike veins. Monocot floral parts are generally in multiples of three; dicot floral parts are in multiples of four or five. Monocots tend to have fibrous root systems; dicots usually have taproot systems. Older woody dicots could obviously be distinguished from herbaceous dicots by the presence of woody lateral growth, including bark. Very young woody dicots would be difficult to distinguish from herbaceous dicots. 4. A cartoon shows a man going to sleep in a hammock suspended between two relatively short trees. The second frame of the cartoon shows the man waking 20 years later and finding his hammock 15 feet higher off the ground. Critique this drawing in terms of what you know about the growth pattern of trees. If the tree is a dicot, then its growth in height occurs by the addition of cells at the apical meristems. As a result, the position of the hammock should not change over time.

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Notes to Instructors Chapter 36 Transport in Vascular Plants

What is the focus of this activity? In this activity, students will examine the basic structure and function of transport systems in higher plants. What is this particular activity designed to do?

Activity 36.1 How Are Water and Food Transported in Plants? This activity is designed to help students review and understand

• how water and specific dissolved minerals can enter and be transported in xylem, • how sugars are transported in phloem, and • how the functions of these two transport systems are interrelated.

What misconceptions or difficulties can this activity reveal?

Activity 36.1 To understand transport in plants, students must have a good understanding of the properties of water and of osmotic relationships (Chapter 3) and the properties of cell membranes (Chapter 7). It is helpful to remind students of these topics and refer them to the appropriate text chapters and activities.

To help students understand how water can be pulled up a tree by transpirational loss, you can use a juice box analogy. When you put negative partial pressure on the straw in a juice box (by sucking on it), you pull juice up the straw. If the negative partial pressure is great enough, you will even collapse the sides of the juice box. This analogy also helps students understand how a tree can decrease in diameter during periods of high transpiration (see question 9).

Many students don’t understand how water in the xylem can move laterally into the phloem. They may assume that xylem vessels and tracheids are laterally impermeable. Figure 35.9, Water-Conducting Cells of the Xylem, in Biology, 7th edition, demonstrates that both tracheids and xylem vessels contain pits in their lateral walls. These pits allow for the lateral movement of water.

Explaining lateral movement sometimes further confuses students, who wonder: If this is true, how can water move up the xylem? Aren’t the lateral pits like holes in the side of a straw, which make it much more difficult to draw water up the straw? You can demonstrate the situation in class. Give each student two straws, and have each of them cut a hole into the side of one straw. Have different students cut different-sized holes from pin holes to up to 1/8 inch across. Now have them test how much suction they need to exert to pull water up both straws. Students will find that it is still possible (though harder) to pull water up the straw with the hole. Next, to simulate the environment of the xylem inside a plant, have students wrap the cut straw with wet paper toweling

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(analogous to the water-filled cell walls of a plant). They will discover that the amount of energy they need to exert to pull water up in this straw is similar to the exertion for the uncut straw. In addition, this demonstration should help students understand how a difference in water potential between phloem (wet toweling) and xylem can lead to movement of some of the water laterally in addition to vertically.

Some students have difficulty understanding how water could initially be drawn up into a tree that is many meters tall. They may assume that the tree started out 50 meters tall with empty xylem. It helps to remind them that even the tallest trees started out as small seedlings. As the trees grew, the xylem and the water columns inside the xylem grew along with them.

Answers

Activity 36.1 How Are Water and Food Transported in Plants? Refer back to the cross sections of the root you developed in Activity 35.1 and trace the pathway of a water molecule from the soil until it leaves the leaf at the top of a tree. Then use playdough, cutout pieces of paper, chalk, or some other material to create a model of transport. Be sure to include all the terms and concepts in the list below.

• Model or demonstrate how water is transported from the soil to the xylem. • Model or demonstrate the transport of K+ ions from the soil to the xylem.

TERMS:

endodermis

oxidative phosphorylation

Casparian strip

oxygen

plasmodesmata

respiration

stele

channels

root hairs

ADP

symplastic transport

ATP

apoplastic transport

epidermis

symplast

lateral transport

apoplast

absorption

xylem

–ψ versus +ψ

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Use your model to answer the questions. 1. How do water’s properties of adhesion and cohesion help maintain the flow of water in the xylem of a plant? Water molecules cohere to one another and adhere to the sides of the xylem vessels or tracheids. As negative pressure is applied to pull the water up in the xylem, each water molecule therefore pulls on the next in the column. As a result, the entire water column rises.

2. a. If water flows from a region of more positive (higher) water potential to a region of more negative (lower) water potential, how does the water potential in the root compare to that in the soil outside the root? The water potential in the soil must be higher than it is in the root.

b. How does the water potential in the air compare to that in the leaf of a plant undergoing transpiration?

If transpiration is occurring, the air must have a lower water potential than the leaf.

3. A student uses a AUB tube for a series of experiments. Sides A and B of the tube are separated by a membrane that is permeable to water but not to sugar or starch. What results would you expect under the experimental conditions given below? Explain your answers in terms of osmotic potential, water potential, and the equation

ψ = ψP + ψS

(Hint: Solute pressure is always negative and a 0.1 M solution of any substance has ψS = −0.23. Therefore, a 0.2 M solution would have ψS = −0.23 × 2 = −0.46.)

Experiment a: A solution of 10 g of sucrose in 1,000 g of water (the molecular weight of sucrose is 342) is added to side A. An equal volume of pure water is added to side B. What will happen to the concentrations of water and sugar in the two sides over time? Explain. A solution with 10 g of sucrose per 1,000 g of water is equivalent to 10 g of sucrose divided by 342 g per mole, or 0.03 mole of sucrose/1,000 g of water. If 0.1 mole of a solution has ψS = −0.23, then 0.03 mole has ψS = −0.23 x 0.3 = −0.069:

ψ = ψP + ψS

ψ = 0 + (−0.069) = −0.069

Because the ψ of pure water (on the other side of the U tube) is zero, water will move from this region of higher water potential to the side with the sucrose, which has a lower water potential.

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Experiment b: A solution of 10 g of soluble starch in 1,000 g of water is added to side A. Assume the molecular weight of soluble starch is about 63,000. An equal volume of pure water is added to side B. What will happen to the concentrations of water and starch in the two sides over time. Explain how this compares with the results in Experiment a. A solution with 10 g of starch per 1,000 g of water is equivalent to 10 g of starch divided by 63,000 g per mole, or 0.00016 mole of starch/1,000 g of water. If 0.1 mole of a solution has ψS = −0.23, then 0.0016 mole has ψS = −0.23 x 0.0016 = −0.0004. Because the ψ of pure water (on the other side of the U tube) is zero, water will move from this region of higher water potential to the side with the starch, which has a lower water potential. This movement will be much slower, however. In fact, it will be 0.069/0.0004 = 172.5 times slower. 4. Fertilizer generally contains nitrogen and phosphorus compounds required by plants. The nitrogen is often in the form of nitrates, and the phosphorus is in the form of phosphates. Based on what you know about chemistry and water potential, why would overfertilizing lead to the death of plants? Adding nitrates and phosphates to the soil will decrease the water potential in the soil. If the water potential in the soil equals the water potential in the root or is less than that in the root, water will not flow into the root (when potentials are equal) and will actually leave the root (when the water potential in the soil is lower than it is in the root). 5. a. One of the most common ways of killing a plant is overwatering. Why does overwatering kill a plant? Roots, like all other living tissues of the plant, require oxygen for cellular respiration. If you overwater a plant, you fill all the air spaces in the soil with water. Root tissue cannot survive without oxygen. b. If overwatering kills plants, why can you sprout roots from cuttings of stems in water? Oxygen from the atmosphere can diffuse into the water in a jar or vase. As a result, the water will contain enough oxygen to allow cellular respiration to continue in the roots. (If the jar is clear and in the sun, green portions of the root underwater will also photosynthesize and add additional oxygen to the water.) However, air spaces in soil are narrow and penetrate deep into the soil. When these are filled with water, the surface area available for the diffusion of oxygen into the water is greatly reduced. 6. Xylem cells are dead when functional. Why must phloem cells be alive when functional? Phloem cells must be capable of loading sugars against a concentration gradient. To do this, they must have a semipermeable membrane and must expend ATP. As a result, they must be alive.

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7. What forces bring about: How are these forces generated in the plant? a. xylem conduction? The major force is the negative partial pressure generated in the xylem.

Transpirational loss of water at the leaf surface (and elsewhere) on the plant generates this negative partial pressure.

b. phloem conduction? The major force a positive hydrostatic pressure in these cells.

This force is generated by the difference in water potential in phloem versus xylem. Sucrose is actively transported into phloem (to levels as high as 25%). This potential difference pulls water out of the xylem and into the phloem, generating the hydrostatic pressure that moves substances in phloem.

8. Refer to your diagram of the cross sectional structure of a typical angiosperm leaf from Activity 35.1. Correlate this structure (that is, the type and placement of cells, and so on) with the activities of the leaf as they relate to photosynthesis, water conservation, and food and water transport. When the stomata are open in the leaf, carbon dioxide can enter and oxygen and water vapor can exit. It is this water loss through the stomata that generates the transpirational pull on the xylem. Closing the stomata reduces water loss; however, it also reduces the amount of carbon dioxide available to the plant (C3 plant). Xylem vessels and phloem sieve cells lie in close contact. This facilitates the movement of water from xylem to phloem to produce the hydrostatic pressure required for the movement of substances in phloem. 9. Scientists have measured the circumference of trees at 2 A.M. and at 2 P.M. If they collect measurements when the ground has adequate moisture and the days are sunny and dry, they find that the circumference (and therefore the diameter) of the tree trunk is smaller at 2 P.M. than at 2 A.M. From your knowledge of the mechanisms of water transport, suggest the reasons for this decrease in circumference. On bright sunny days, the transpirational pull generated by evaporation from the leaves can be so great that it partially collapses (pulls in) the walls of the xylem vessels. This is analogous to what happens when you drink out of a juice box. You can put so much suction on the straw that you collapse (pull inward) the sides of the box. In a tree, this pull can be measured as an actual decrease in circumference (and as a result in diameter).

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10. Outline an experiment that would allow you to determine (a) how fast a substance is carried in the xylem, (b) in what direction the substance flows in the xylem, and (c) what percentages of solutes are in the xylem at various distances away from the leaves or roots. There are a number of different ways to design an experiment. Xylem is dead when functional. As a result, you can insert taps and/or measuring devices into it. For an example of an experiment, place taps and/or measuring devices into the xylem along one side of a branch. Place these at 10-cm intervals in a straight line from near the leaf end to the opposite end of the branch. In one now-classic experiment, Bruno Huber inserted temperature-recording devices at intervals along a tree branch. He heated the water in the xylem at various locations along the branch and then recorded how (if at all) the temperature of the water changed at the other recording devices. This allowed him to determine that the water in the xylem moved vertically. By measuring the time it took for the heated water to move from one recorder to the next, he could also determine the speed of transport in xylem.

If you wanted to determine the amount of solutes present in xylem at various distances away from roots or leaves, you could put taps into the xylem at 10-cm (or other) intervals. Open the taps periodically and remove some of the fluid in the xylem. Analyze the fluid from the different taps for contents (for example, for the percentages of various minerals).

11. When researchers have tried to tap into phloem cells during experiments, they find that the disrupted phloem immediately stops functioning. However, aphids can pierce through plant tissues with their mouth parts and locate individual phloem cells. Once inside a phloem cell, the aphids are essentially force-fed phloem sap. If the aphid body is removed from the mouth parts, phloem sap will continue to flow and can be collected. What kinds of information can be derived from an analysis of phloem sap? Different kinds of information can be derived. For example, you could sample phloem sap at different distances from the leaves and determine the concentration of sugar present. If you added a tracer, you could determine the rate and direction of flow in phloem. For example, you could cover the leaves of a plant with a plastic bag and inject radioactive carbon dioxide (14CO2) into the bag. The carbon will be incorporated into sugar during photosynthesis. Over time, you could sample the sap from each of the aphid mouthparts and determine

• when and where the radioactivity first appears, • how quickly the radioactivity moves from that first to the other aphid mouthpart

taps, and • which direction the sap moves.

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Notes to Instructors Chapter 37 Plant Nutrition

What is the focus of this activity? This activity focuses student attention on the fact that “plant foods” are usually inorganic; that is, unlike animals, most plants are capable of making all of their organic compounds from inorganic precursors. What is this particular activity designed to do? Activity 37.1 What Do You Need to Consider in Order to Grow Plants in Space (or Anywhere Else for That Matter)? This activity is designed to help students understand the basic requirements for plant growth and how knowledge of these requirements can help them solve problems like how to grow plants in space.

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Activity 37.1 What Do You Need to Consider in Order to Grow Plants in Space (or Anywhere Else for That Matter)?

Long-range, manned space travel and manned space stations may someday become a reality. Before this can occur, however, we will have to develop sustainable methods of agriculture suitable for use in space. One of the key methods being investigated is hydroponics, growing plants in water supplemented with nutrients. You are assigned to a team working on the design of plant growth systems for use in a space station. 1. What types of plants would you choose to grow? Explain the reasoning behind your choices. There are many possible answers to this question. All answers should include how much room the proposed crop plant requires for growth, and how much of the crop plant produced can be consumed or otherwise used by humans (or other organisms on the space station). For example, it makes sense to grow plants that produce a high ratio of edible to inedbile matter per plant. On one hand, a wheat plant produces a fairly low ratio of edible to inedible matter per plant. On the other hand, a potato plant produces a much higher ratio of edible to inedible matter.

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2. When you set up the growth system: Note: Minimal answers to the following questions are provided. a. What nutrients

would you need to add to the water? List what you would need, and why each would be necessary.

b. What atmosphere would you need to maintain? List what components you would need to maintain in the atmosphere and why each would be necessary.

c. Which of the requirements in parts a and b could be recycled? What could not be recycled? Explain.

Refer to Table 37.1 of Biology, 7th edition, for the complete list of both macronutrients and micronutrients required for plant growth. This table also indicates the major functions of each nutrient.

The atmosphere in the space station would have to be similar to the atmosphere on Earth. This would be a requirement for both human life and successful plant growth. Plants require carbon dioxide for photosyn-thesis and oxygen for cellular respiration. If the oxygen-to-carbon dioxide ratio became too high, then C3 plants would undergo photorespiration instead of photosynthesis. This would reduce plant growth and crop yield. You would obviously have to balance the number of plants versus animals (and other carbon dioxide producers, for example microbes) on the station to control the oxygen-to-carbon dioxide ratios in the atmosphere over time. (If you chose to grow legumes, symbionts associated with plant roots—for example, Rhizobium in legume roots—could fix atmospheric nitrogen, which would reduce the need for additional fertilizer.)

Oxygen and carbon dioxide could be recycled between the plants and animals on the station. Water in the atmosphere (and in the urine of animals) could be recycled. Many of the minerals in the plants could be recycled by composting inedible parts of the plants and animal feces. These could be converted to inorganic nutrients for plants by the actions of a variety of microbes. Without the microbes, large quantities of organic waste would build up and, as a result, most compounds other than oxygen, carbon dioxide, and water could not be recycled.

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Notes to Instructors Chapter 38 Angiosperm Reproduction and Biotechnology

What is the focus of this activity? All plants display an alternation of generations between a multicellular haploid or gametophyte generation and a multicellular diploid or sporophyte generation. Evolutionarily there has been a general trend in the plant kingdom toward reduction of the haploid generation. This reduction is maximum in the angiosperms, where the female gametophyte is reduced to the embryo sac (seven cells, eight nuclei). The male gametophyte is reduced to even fewer cells: the germinated pollen grain, which contains a generative nucleus, and two sperm nuclei. What is this particular activity designed to do? Activity 38.1 How Can Plant Reproduction Be Modified Using Biotechnology? This activity is designed to help students understand the basic structure and function of male and female gametophytes in angiosperms. They will also learn how biotechnology differs from artificial selection and cloning of plants.

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Activity 38.1 How Can Plant Reproduction Be Modified Using Biotechnology?

Answer questions 1–7. Then in question 8 you will develop a concept map that links the information from all seven answers. 1. Draw a general diagram of the life cycle of a seed plant. Indicate which steps are haploid and which are diploid. Refer to Figure 13.5, part B on page 242 of Biology, 7th edition.

2. Define microsporogenesis and megasporogenesis. In what portion(s) of the flower does each of these processes occur? What is the end product of each process? Microsporogenesis occurs in the sporangia of the anther in flowers. Microsporocytes (2n) undergo meiosis to produce haploid microspores; each then develops into a pollen grain. Megasporogenesis occurs in the sporanium of the ovule in flowers. The megasporocyte (2n) undergoes meiosis to form a haploid megaspore (plus three other haploid cells that do not survive). The megaspore undergoes mitosis to produce the embryo sac, which contains the egg, synergids, antipodal cells, and polar nuclei.

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4. What is pollination? How does it differ from fertilization? Pollination occurs when pollen lands on the stigma of a genetically compatible plant. Fertilization occurs when a sperm nucleus from the pollen tube joins with the egg nucleus in the embryo sac. In double fertilization, the other sperm nucleus (from the pollen tube) joins with the polar nuclei to form the endosperm nucleus.

6. What stages of the life cycle are eliminated or bypassed when plants are cloned naturally? When plants are cloned on the farm or in the laboratory? The gametophyte generation is bypassed. Natural cloning occurs when, for example, plants send out runners or new sprouts arise from under-ground roots or rhizomes. Similarly, cuttings or small pieces of plants can be grown into complete plants on the farm or in the lab. In all of these cases, the genetics of the new plants are identical to those of the parent. (Note: Some variations can arise due to spontaneous mutations.) Biology, 7th edition, includes other examples as well.

3. Draw and label all parts of a complete flower. Indicate the functions of the major parts. A complete flower is diagrammed in Figure 38.2 of Biology, 7th edition. A complete flower includes sepals, petals, stamens, and carpel(s). Sepals enclose and protect the developing flower. Petals attract the attention of pollinators. Anthers contain the male microsporangium. Carpel(s) contain the female megasporangium.

5. Draw and label a mature ovule. Include the micro-pyle, integuments, nucellus, synergids, polar nuclei, egg, and anti-podals. Indicate the functions of each of these structures. See Figure 38.4 of Biology, 7th edition.

7. What does the science of plant biotechnology do that artificial selection and/or cloning practices don’t do? Biotechnology can add genes from other organisms to plants. For example, a gene from Bacillus thuringiensis has been added to some crop plants. It produces the Bt toxin, which acts as an insecticide.

8. Construct a concept map that links all of the information in questions 1–7. You can do this in the space provided.

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Notes to Instructors Chapter 39 Plant Responses to Internal and External Signals

What is the focus of this activity? In Activity 37.1, students looked at some of the nutrient conditions required to develop sustainable agriculture in space. In this activity, students are asked to refine their ideas to include the effects of gravitropism, phototropism, and photoperiodism. What is this particular activity designed to do? Activity 39.1 How Do Gravity and Light Affect Plant Growth Responses? Students apply what they have learned about gravitropism, phototropism, and photoperiodism to propose methods for improving systems for growing plants in space. What misconceptions or difficulties can this activity reveal? Activity 39.1 Many students have difficulty understanding how higher concentrations of auxin can affect stems or shoots differently than they affect roots. It helps to remind students that, although stems and roots are continuous structures, they are distinctly different plant organs. As such, they can respond differently to a given chemical signal. This situation is analogous to the mammalian digestive tract. The esophagus, stomach, and small and large intestines are all continuous; however, each has a different structure and function. In addition, each can respond differently to a single hormone.

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Activity 39.1 How do gravity and light affect plant growth responses?

Review Chapter 39 of Biology, 7th edition, and your answers to Activity 37.1. Then answer the questions. 1. One of the problems associated with growing plants in space is lack of gravity. a. How does gravity affect

the normal growth of a plant’s roots, stems, and other parts? Explain the mechanisms involved.

Under the influence of gravity, auxin accumulates on the lower side of the root and stem. Higher auxin concentrations stimulate cell elongation in the stem but inhibit cell elongation in the root. In the shoot region, a concentration of auxin between 10–8 and 10–4 stimulates cell elongation. According to the acid growth hypothesis, auxin concentrations in this range stimulate proton pumps, which lower the pH in the cell wall. This activates enzymes that break cross-links between cellulose molecules and allow the cell to elongate.

b. How would a lack of gravity affect normal growth?

Normal seed germination and seedling growth begin underground in the absence of light. Under these conditions, the seedling relies on the gravitropic responses of the shoot and root to orient these above ground and underground, respectively. In the absence of gravity, auxin concen-trations in the root and shoot would not vary, and growth of the shoot away from gravity and growth of the root toward gravity would not occur.

c. Propose mechanisms to overcome the problems associated with a lack of gravity.

Plant orientation is also affected by light. (See question 2.) The response to light can help counteract the lack of gravity. In addition, if plants are grown hydroponically, a dense meshwork mat can be used to prevent roots from growing above water.

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t to

flow

er a

nd p

rodu

ce f

ruit.

a. H

ow d

o ph

otot

ropi

sm a

nd p

hoto

peri

odis

m d

iffe

r?

Phot

otro

pism

is th

e gr

owth

of

a pl

ant t

owar

d or

aw

ay f

rom

ligh

t. In

gen

eral

, sho

ots

grow

tow

ard

light

(ar

e po

sitiv

ely

phot

o-tr

opic

) an

d ro

ots

grow

aw

ay f

rom

ligh

t (ar

e ne

gativ

ely

phot

otro

pic)

. Aga

in, t

his

resp

onse

app

ears

to b

e th

e re

sult

of a

ssym

etri

c di

stri

butio

n of

aux

in. (

Ref

er to

Fig

ures

39.

5 an

d 39

.6.)

Ph

otop

erio

dism

is a

phy

siol

ogic

al r

espo

nse

to c

hang

es in

day

leng

th. F

or e

xam

ple,

the

phys

i-ol

ogic

al r

espo

nses

that

ca

use

flow

erin

g in

som

e sp

ecie

s of

pla

nts

are

trig

gere

d by

cha

nges

in d

ay le

ngth

.

b. W

hat l

ight

cha

ract

eris

tics

wou

ld y

ou u

se to

max

imiz

e pl

ant g

row

th p

er u

nit t

ime?

In

pho

tosy

nthe

sis,

the

chlo

roph

yll m

olec

ules

in p

hoto

-sys

tem

s I

and

II r

espo

nd to

or

abso

rb li

ght i

n th

e re

d an

d bl

ue

wav

elen

gths

of

the

visi

ble

spec

trum

. As

a re

sult,

if n

othi

ng e

lse

is li

miti

ng, t

he m

ore

red

and

blue

ligh

t ava

ilabl

e, th

e fa

ster

the

phot

osyn

thet

ic r

ate

per

unit

time.

c. W

hat k

ind

of p

hysi

cal e

nvir

onm

ent w

ould

you

nee

d to

mai

ntai

n ap

prop

riat

e ph

otot

ropi

c re

spon

ses

amon

g pl

ants

? In

spa

ce o

r in

the

lab,

you

wou

ld n

eed

to s

et u

p ba

nks

of o

verh

ead

light

s. L

ight

bul

bs th

at a

re r

ich

in r

ed a

nd b

lue

wav

elen

gths

sho

uld

be u

sed.

A

hyd

ropo

nic

grow

th s

yste

m s

houl

d be

set

up

to in

clud

e a

mes

hwor

k m

at to

sup

port

see

ds o

r se

edlin

gs in

ear

ly g

row

th

and

to s

up-p

ort t

he s

tem

s an

d m

aint

ain

the

root

s be

low

the

mat

(in

the

hydr

opon

ic s

olut

ion)

in la

ter

grow

th.

d. W

hat d

esig

n m

odif

icat

ions

wou

ld y

ou n

eed

to m

ake

to s

uppo

rt p

lant

s w

ith d

iffe

rent

pho

tope

riod

s—fo

r ex

ampl

e,

long

-day

ver

sus

shor

t-da

y pl

ants

? If

the

diff

eren

t pla

nt s

peci

es y

ou w

ere

grow

ing

had

diff

eren

t pho

to-p

erio

ds, y

ou w

ould

nee

d to

gro

wth

them

in s

epar

ate

cham

-ber

s. Y

ou c

ould

then

set

up

diff

eren

t tim

er c

ontr

ols

in e

ach

cham

ber

to r

egul

ate

the

num

ber

of h

ours

of

light

(v

ersu

s da

rk)

per

day

(24

hour

s).

EQA

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Notes to Instructors - San Dieguito Union High School Districtteachers.sduhsd.net/ahaas/AP Biology/Plant Physiology... · Notes to Instructors ... To understand transport in plants,