Download Notes to Instructors

EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
213
EQA
Answers
?
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
214
Soil
surface
Generic plant
215
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
f
e
d
c
b
a
pith
Root
Stem
Petiole
Leaf
dorsal
dorsal
Apical meristem
endodermis
xylem
endodermis
phloem
xylem
xylem
xylem
phloem
phloem
Same as Column 1
phloem
xylem
Same as Column 1
Column 2
Cross sections if plant
is a herbaceous dicot
phloem
phloem
xylem
phloem
xylem
Column 1
Cross sections if
plant is a monocot
endodermis
2º xylem
2º phloem
2º xylem
cambium
2º phloem
Same as Column 1
Same as Column 2
Same as Column 1
Column 3
Cross sections if plant
is a woody dicot
EQA
Use the information in your drawings to answer the questions on the next pages.
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
216
EQA
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
Monocot
Leaf
Stem
Root
Apical meristem
shoot vs root
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)
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
217
EQA
Herbaceous dicot Vascular traces
(veins) in the
leaf are arranged
in a netlike
pattern.
Vascular bundles
are arranged in a
ring between a
thin outer cortex
layer and a
central region of
In a cross section
pith.
of the leaf you
would not expect
to see the
vascular bundles
evenly spaced. In
addition, the
bundles would
likely be of
varying size.
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 meristems of the
roots are generally 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.
A ring of
vascular
cambium forms
between the
central xylem
and phloem and
lays down secondary xylem to
the inside and
secondary
phloem to the
outside.
As in monocots,
the apical meristems of the
roots are generally 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.
Vascular
cambium is
continuous and
produces secondary xylem to
the inside and
In a cross section
secondary
of the leaf you
phloem to the
would not expect
outside. Variato see the
tions in water
vascular bundles
availability (wet
evenly spaced. In
vs dry seasons)
addition, the
can, result in
bundles would
growth rings.
likely be of
varying size.
New root growth
is similar in
cross section to
that of herbaceous dicots.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
218
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
219
EQA
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
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
220
EQA
(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
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
221
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
222
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
223
EQA
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).
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
224
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
225
EQA
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.
Answers
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
226
EQA
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 c. Which of the
need to maintain? List what
requirements in parts
components you would need
a and b could be
to maintain in the atmosphere
recycled? What could
and why each would be
not be recycled? Explain.
necessary.
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-tocarbon 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.)
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
227
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.
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
228
EQA
Answers
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.
(continued on next page)
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
229
EQA
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.
4. What is pollination?
How does it differ from
fertilization?
Pollination occurs when
pollen lands on the stigma
of a genetically compatible
plant.
A complete flower includes
sepals, petals, stamens, and
carpel(s).
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.
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 micropyle, 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.
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 underground 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.
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
230
EQA
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.
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
231
EQA
Answers
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 crosslinks 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 concentrations 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.
(continued on next page)
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
232
233
Copyright © Pearson Education, Inc., publishing as Benjamin Cummings
d. What design modifications would you need to make to support plants with different photoperiods—for example,
long-day versus short-day plants?
If the different plant species you were growing had different photo-periods, you would need to growth them in separate
cham-bers. You could then set up different timer controls in each chamber to regulate the number of hours of light
(versus dark) per day (24 hours).
A hydroponic growth system should be set up to include a meshwork mat to support seeds or seedlings in early growth
and to sup-port the stems and maintain the roots below the mat (in the hydroponic solution) in later growth.
c. What kind of physical environment would you need to maintain appropriate phototropic responses among plants?
In space or in the lab, you would need to set up banks of overhead lights. Light bulbs that are rich in red and blue
wavelengths should be used.
b. What light characteristics would you use to maximize plant growth per unit time?
In photosynthesis, the chlorophyll molecules in photo-systems I and II respond to or absorb light in the red and blue
wavelengths of the visible spectrum. As a result, if nothing else is limiting, the more red and blue light available, the
faster the photosynthetic rate per unit time.
a. How do phototropism and photoperiodism differ?
Phototropism is the growth of a plant toward or away from light. In general, shoots grow toward light (are positively
photo-tropic) and roots grow away from light (are negatively phototropic). Again, this response appears to be the result
of assymetric distribution of auxin. (Refer to Figures 39.5 and 39.6.)
Photoperiodism is a physiological response to changes in day length. For example, the physi-ological responses that
cause flowering in some species of plants are triggered by changes in day length.
2. Another problem with growing plants in space relates to a plant’s light requirements and phototropic responses versus
the photoperiods required for the plant to flower and produce fruit.
EQA