Teacher’s notes: Practical activities
Introduction
The following notes have been prepared to assist teachers in planning and conducting practical activities related to
the prescribed Outcomes in each Area of Study.
•
The practical activities chosen for inclusion in the Heinemann Biology 1 Student Workbook have been
selected because they represent the most relevant and effective activities related to the prescribed Outcomes in
the current VCE Biology course for Units 1 and 2. They provide students with an opportunity to develop skills
of scientific enquiry and method, including experimental design and evaluation, constructing and testing
hypotheses, gathering and analysing data, as well as opportunities to consider technological advances.
•
Practical activities are fully set out in the Heinemann Biology 1 Student Workbook with clear instructions as
well as space for students to answer. This feature means that students have a clear and concise log of relevant
practical activities undertaken during the course bound together in one comprehensive volume.
•
Materials required are listed in the materials list on the opening page for each activity. However, some
materials that warrant special attention are also included here.
•
The comments in the Hints and comments section of the teacher’s notes are arranged approximately in the
order in which they arise in each activity. Numbered steps refer to procedure points in an activity; bold
question numbers refer to circled questions in the activity. Suggested answers are also provided for most
questions to give some guidance as to the direction that the activity is expected to take, as well as some
indication of the depth of discussion expected.
•
Each practical activity has been carefully written so that it can be completed within the suggested timeframes
for this course. Class sessions are based on a time allowance of 50 minutes. However, variations in local
timetabling and other constraints need to be considered to make the best possible use of the time available.
Some activities lend themselves to independent student work, while others will make more efficient use of
time when students are working in groups. Preliminary preparation completed by students before an activity
will also impact on the time taken to complete laboratory work. It is recommended that teachers read these
teacher’s notes in conjunction with the activity in the student workbook in order to form an idea of alternative
strategies for best completing activities given such constraints.
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Unit 1: Unity and diversity
Area of Study 1: Cells in action
Outcome: Design, conduct and report on a practical investigation related to cellular structure, organisation and
processes.
Practical activity 1: Molecules on the move—osmosis and partially permeable
membranes
Text reference: Heinemann Biology 1 Student Workbook, pages 24–25
Estimated time: 50 minutes. This includes about 20 minutes setting-up time and the remainder of the lesson for
discussion and completing the written component of the report after observations have been made.
Scope of activity
The focus of this activity is the importance of biological molecules and how they enter and leave cells by crossing
cell membranes. This activity is designed to demonstrate that partially permeable membranes allow relatively small
molecules such as water and iodine/potassium iodide to pass across while other larger molecules such as starch do
not.
Preparation
•
Dialysis tubing is available from Southern Biological Services.
•
Iodine/potassium iodide solution: dissolve 6 g potassium iodide in 1 L distilled water. Add 3 g iodine crystals
and leave to stand for 24 h to allow the iodine crystals to dissolve. Store in brown bottles.
•
Starch is a 5% solution: add 5 g of soluble starch to 100 mL of cold water, mix thoroughly and then heat to
just boiling.
Hints and comments
Begin with a demonstration: add a couple of drops of iodine/potassium iodide solution to a test tube containing a
little starch solution. Starch solution, which is cloudy white, changes colour to blue–black in the presence of
iodine.
Step 3: It is important to remove all bubbles. This can be a little difficult to do—try tilting the bag and the funnel
on their sides to dislodge any bubble that are present.
1
Students should observe that the solution inside the bag has changed colour from cloudy white to blue–black.
The yellow appearance of the solution in the gas jar remains the same.
2
a
3
The size of the different molecules accounts for their movements—molecules of starch are too large to pass
through the pores in the cellulose tubing, while molecules of iodine/potassium iodine and water are small
enough to pass through.
4
a
The level of solution in the thistle funnel will be a little higher than at the start of the investigation.
b
Starch solution has been retained inside the cellulose tubing, while iodine/potassium iodide and some
free water molecules have moved into the bag, i.e. there has been a net movement of particles into the
bag.
Iodine/potassium iodide molecules have passed from the gas jar through the tiny pores in the cellulose
tubing into the bag. This is apparent because the colour change inside the bag indicates that a reaction has
occurred between the starch and the iodine. No such colour change has occurred in the solution in the gas
jar, indicating that starch molecules have not passed out of the bag.
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Conclusions
1
Osmosis is a special kind of diffusion in which water molecules pass across a partially permeable membrane
from an area of high concentration of free water molecules to an area of low concentration of free water
molecules.
2
Partially permeable membranes are selective in relation to which molecules pass across based on the size of
the molecules. Very small molecules can ‘fit’ through the membrane pores, while relatively large molecules
cannot.
Practical activity 2: Shaping up—body shape vs diffusion
Text reference: Heinemann Biology 1 Student Workbook, pages 26–27
Estimated time: 50 minutes
Scope of activity
This activity uses a simple technique to illustrate a biologically important concept related to the efficient exchange
of materials by the cells and tissues of organisms. Students are guided through the technique and a series of
questions to help them think about adaptations that enable complex multicellular organisms to exchange materials
with their external environment.
Preparation
•
To make agar jelly you will need sodium hydroxide solution and phenolphthalein solution.
•
To make sodium hydroxide solution, add 8.0 g NaOH to 200 mL distilled water and stir to dissolve.
•
To make phenolphthalein solution, dissolve 0.5 g phenolphthalein in 100mL of 95% ethanol (95 mL ethanol
and 5 mL water).
•
To prepare 2 L of agar jelly, add 40.0 g plain agar to 1800 mL of very hot water and stir continuously until the
agar is dissolved. Allow to cool to below 60°C.
•
Stir continuously while adding the sodium hydroxide solution to the agar after the temperature has dropped
below 60°C. Then add the phenolphthalein solution in small amounts until the jelly is a deep pinkish purple
colour—this should take about 60 mL of phenolphthalein.
•
To pour the jelly, straight-sided clear glass or rigid plastic containers are best. Use a plastic ruler as a gauge of
depth.
•
To prepare 0.1 M sulfuric acid, carefully add 5.5 mL concentrated H2SO4 per litre distilled water.
•
All students should wear disposable gloves and safety glasses during this activity.
•
It is recommended to have several containers of sulfuric acid placed around the laboratory to reduce
congestion when students come to collect it.
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Hints and comments
3
Block number
1
2
3
Block dimensions (mm)
20 × 20 × 20
20 × 20 × 10
20 × 20 × 5
2400 mm2 or
1600 mm2 or
1200 mm2 or
24 cm2
16 cm2
12 cm2
8000 mm3 or
4000 mm3 or
2000 mm3 or
8 cm3
4 mm3
2 mm3
3:1
4:1
6:1
Depth of clear layer of block (mm)
Surface area (SA, mm2)
Volume (V, mm3)
Surface area to volume ratio (SA/V)
4
Phenolphthalein is a pink indicator that turns clear in the presence of acid. When the phenolphthalein in the
agar blocks comes into contact with the sulfuric acid, it changes from pink to clear. The time taken for the
central region of the agar blocks to change colour depends on the time it takes for the acid to diffuse.
5
a, b Expect Block 3 to become completely clear first, followed by Block 2, then 1. Block 1 has the least depth
so we would expect the acid to reach the central region of this block first. Block 2 is of medium depth
compared to 1 and 3, and Block 3 has the greatest depth so diffusion will take the longest.
6
a
Block 1
b
Block 1
c
Block 3
7
The surface area to volume ratio of an object increases proportionally as its volume decreases.
Conclusions
8
A large cubic block of agar has a relatively low surface area for its volume compared to a smaller, flatter
block. As the size of the agar blocks decrease, the surface area to volume ratio increases, and this results in
diffusion of the acid through the block more quickly.
9
The microscopic size of cells gives them a relatively large surface area for their volume, i.e. a maximum
surface area to volume ratio. A high surface area to volume ratio is important for the efficient exchange of
materials (nutrients and wastes) across cell membranes.
Further ideas
Cut the following shapes from the agar jelly:
•
a block 30 mm × 20 mm from a 3 mm thick sheet of jelly
•
a block 10 mm × 18 mm from a 10 mm thick sheet of jelly
•
a sphere with a 15 mm diameter.
Each of these shapes has approximately the same volume.
Predict the order in which the blocks will go clear. Place the shapes in beakers containing 0.1 M sulfuric acid to see
if your predictions are correct. Explain the results.
Practical activity 3: Capable catalase—investigating enzyme efficiency
Text reference: Heinemann Biology 1 Student Workbook, pages 28–30
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Estimated time: 50 minutes, including about 20 minutes for student discussion and design, then about 30 minutes
to conduct the activity and gather data.
Scope of activity
This activity provides students with an opportunity to design and conduct a scientific investigation. The task
requires careful planning in order to ensure the activity is as controlled as possible. Students also need to be aware
of safety considerations. Carrying out an activity they have designed themselves should also provide occasion for
evaluation and redesign of steps that clearly involve margins of error, e.g. comparing the reaction rates for different
volumes of liver.
Preparation
•
Use 3% commercial hydrogen peroxide.
•
Eye protection must be worn.
•
Washed sand together with a little distilled water is useful for grinding the liver with a mortar and pestle. The
decanted liquid contains catalase.
Hints and comments
1
Experimental procedure: Students will come up with a range of approaches. Essential elements in
experimental design to investigate catalase in liver and affects of temperature include:
Step 1: Put on disposable gloves, and then collect a chopping board, scalpel and section of liver.
Step 2: Cut liver into six equally sized cubes (about 1 cm3 each). Set aside two blocks. Cut two of the blocks
into smaller pieces using the scalpel—each cut block should be divided into the same number of pieces, and
about the same size pieces. Take the last two blocks and grind separately using a mortar and pestle.
Step 3: Place one uncut cube, one cut cube and one ground cube into separate test tubes filled to about 2 cm
depth with distilled water. Place these test tubes into a large beaker half filled with water and placed over a
Bunsen burner. The Bunsen burner should be set up using heat mat, tripod and gauze mat. Bring to the boil.
Allow to simmer for about 10 minutes. Remove from heat and allow to cool.
Step 4: Set up three test tubes in a test tube rack for the remaining unheated liver pieces. Fill each test tube to
about 2 cm depth of hydrogen peroxide.
Step 5: Add the uncut cube of liver into the first test tube and record your observations.
Step 6: Add the cut liver cube to the second test tube. Record your observations.
Step 7: Add the ground liver cube to the third test tube. Record your observations.
Step 8: Drain the distilled water from each of the heated test tubes. Repeat steps 4–7 using these heated liver
pieces. Compare your observations.
2
Observations: bubbling activity (evolution of gas), test tube becomes warmer.
3
a
Activity varies in length of time, depending on the surface area of the liver that is exposed. The unheated
ground liver will show a greater degree of activity than the cut cube, and the cut cube will show a greater
degree of activity than the uncut cube. The heated cubes should uniformly show no activity.
b
Adding more hydrogen peroxide will lead to continuation of activity.
c
Enzymes in the liver act on the substrate, hydrogen peroxide, breaking it down. When all of the hydrogen
peroxide has been broken down, the activity ceases. The observation that activity continues when more
substrate is added indicates that the enzyme present in the liver continues to be available to catalyse the
reaction, i.e. while it has been involved in the activity, it has not been changed by the activity.
a
2H2O2 → 2H2O + O2
4
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The gas produced in this reaction is oxygen gas.
b
Light a splint with a match. When it is well alight, blow out the flame. Then immediately place the
glowing end of the splint into the mouth of the test tube in which the reaction is occurring. Oxygen gas
produced should reignite the flame.
5
The same volume of liver is used in each test tube. The vigour and time taken for enzyme activity depends on
the surface area of the liver. The more surface area of liver exposed, the greater the amount of the enzyme
catalase available to react with the hydrogen peroxide. When more enzyme is in contact with the substrate, the
reaction proceeds more quickly and more vigorously. Hence, the greatest degree of activity will be observed
in the unheated ground liver, as it has the greatest surface area of liver cells (and hence catalase) exposed to
the substrate.
6
a
No reaction is observed in any of the heated liver samples, regardless of surface area available. (Note: if a
reaction is observed, it indicates that the liver has not been heated enough or for long enough; even if
some reaction is observed, it will be less than for the unheated liver, indicating that heating has affected
the ability of the enzyme to function.)
b
Heating the liver denatures the enzyme. This renders the enzyme unable to function. Denaturation is
irreversible.
Conclusions
7
The addition of more hydrogen peroxide to the liver, after the initial reaction was complete, resulted in further
activity. This indicates that the enzyme present in the liver was still available at the end of the reaction to react
again.
8
In this controlled activity, the same sized liver pieces were subject to heating and then used in the same way as
the unheated liver pieces. When heated liver was added to the hydrogen peroxide substrate, no reaction was
observed. Heating rendered the liver enzyme, catalase, non-functional.
9
Enzyme activity is affected by the amount of enzyme that is available for the reaction, and this is in turn
affected by the surface area of tissue that is in contact with substrate (since the cells contain the enzyme). The
amount of substrate also affects enzyme activity.
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Area of Study 2: Functioning organisms
Outcome: Describe and explain the relationship between features and requirements of functioning organisms and
how these are used to construct taxonomic systems.
Practical activity 4: Food tubes—comparing digestive systems
Text reference: Heinemann Biology 1 Student Workbook, pages 71–72
Estimated time: 50 minutes
Scope of activity
This activity has a comparative focus—students are challenged to apply their knowledge and understanding of
digestion of macromolecules, diet and the role of digestive systems to compare the digestive systems of different
mammals. Consideration of visual representations of different digestive systems will lead to inferences about diet.
Hints and comments
Southern Biological Services has a range of dissection charts that could be useful.
Entering key words into search engines will be useful in answering Conclusion Q6, e.g. ‘vampire bat, teeth, diet’.
1, 2
•
Sheep and pigs have a number of similarities when compared with a bat. The sheep is a herbivore with a diet
high in cellulose from plant cell walls. Its stomach is modified into several fermentation chambers where the
microbial breakdown of cellulose occurs. Sheep are foregut fermenters. The caecum is also well developed.
The small intestine is very long, to maximise digestion and absorption.
•
Pigs are omnivorous. Their digestive systems do not need to handle the same bulk of cellulose and so the
stomach does not have fermentation chambers. The small intestine is not as long as for the sheep.
•
Vampire bats are specialised feeders. They feed on blood that already contains the products of digestion,
absorbed into the bloodstream of the organism that the bat has fed on. There is not a lot of digestion required,
except perhaps for haemoglobin (a protein). The stomach is more a blood storage organ, the small intestine is
much reduced, and there is no caecum since there is no cellulose in the diet
3
Digestion in a sheep would be much less efficient if the gut proportions were similar to that of the vampire
gut. A smaller stomach would mean smaller fermentation chambers, so cellulose would be subject to less
bacteria and less cellulose would be broken down in a given time. This would mean that more undigested
plant material proceeds to the small intestine, and this undigested material will be discarded by the body. A
shorter small intestine also means a reduced surface area, and less time for absorption of digested materials,
resulting in some digested materials being removed from the body in the faeces.
4
A sheep has an exclusively herbivorous diet. All of its nutrients are obtained from plant material. To make use
of the energy-rich carbohydrates contained in the cellulose cell walls that surround plant cells, sheep depend
on a population of fermentation bacteria that are able to digest cellulose. Pigs are omnivores, obtaining their
nutrients from a range of foods—they do not depend on the breakdown of cellulose to obtain energy-rich
nutrients, and so do not need a supply of fermenting bacteria. Digestive systems of omnivores have enzymes
that are efficient at breaking down animal material into simple compounds that can be readily absorbed.
Crushing plant material in mechanical digestion releases the contents of cells. Undigested cellulose is
discarded.
Conclusions
7
Comparison of dentition will be the focus here. Sheep are herbivores—they have incisors and well-developed
premolars and molars to facilitate mechanical digestion of plant material; they have no canines. Pigs have
incisors, canines and molars. While domesticated varieties of pigs have reduced canines, their wild cousins,
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such as boars, retain well-developed canines. Vampire bats have very sharp canine teeth used to pierce the
skin of their prey. Incisors are used to trim the fur at the bite site. Vampire bats also produce a substance in
their saliva that prevents clotting of blood from their victim—a feature that ensures continued flow of food.
Further ideas
Compare the diets and digestive systems of other selected vertebrates as well as some invertebrates, e.g.:
•
Birds (phylum Chordata, sub-phylum Vertebrata, class Aves)
•
Planaria (phylum Platyhelminthes—flatworms)
•
Earthworms (phylum Annelida, class Oligochaeta)
•
Cockroach (phylum Arthropoda, class Insecta)
a
Birds have no teeth for mechanical digestion. Suggest reasons for this. Describe the adaptation evident in
birds that allows for mechanical digestion.
b
In what ways is the digestive system of planaria different from the digestive systems of the other animals
listed above?
Practical activity 5: Tubes for transport—vascular tissue in plants
Text reference: Heinemann Biology 1 Student Workbook, pages 73–76
Estimated time: 100 minutes
Scope of activity
The arrangement and distribution of vascular tissue is the focus of this activity. The gross arrangement and some
cellular detail are examined using celery, and then more detailed observations are made using stained, prepared
slides.
Preparation
•
A range of dye solutions would be suitable, e.g. red food colouring, eosin (0.1 g per 100 mL water). Make
sure that the petioles are cut cleanly before placing them in dye. The time taken for the movement of dye in
the celery is variable. It will certainly be in the leaves if left to stand overnight. A trial with the dye you intend
to use would be advisable before class use.
•
Toluidine blue stain is harmful if swallowed. Eye protection must be worn.
•
Iodine/potassium iodide solution is made by dissolving 4.0 g of KI in 100 mL distilled water, and adding 2.0 g
iodine crystals. Allow to stand for 24 h so that iodine crystals will dissolve. Store in glass dropping bottles,
preferably dark. Iodine does not actually stain the tissues but it provides contrast.
•
Prepared slides of Helianthus should be readily available from biological suppliers such as Southern
Biological Services. The slides required are: TS Helianthus stem (separate bundles), LS Helianthus mature
stem, and TS Helianthus young and old root (for those who want to look at the roots as well). Equivalent
slides of Curcurbita (pumpkin) are also suitable, especially the longitudinal section, which shows sieve tubes
very clearly.
•
Single-edged razor blades for section cutting should be new for best results. After use, they can be kept in
good condition by drying them thoroughly and lightly wiping a smear of very fine oil over the cutting edge.
Many schools have demonstration models of plant stems. These would be most useful for class discussion.
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Hints and comments
1
While some dye may have diffused out of the vascular bundles, the greatest intensity of dye should be clearly
visible in the venation of the plant tissue. This will be most clear in the leaf.
3
The dye will be most intense in tissue of the vascular bundles. However, some of the dye may have diffused
out of the vascular bundles and will be apparent in the surrounding tissue.
5
If a water-soluble dye has been used, it will move with the water molecules. However, if a dye that forms a
fine suspension is used, its particles may be too large to move with the water, and will be confined to the
vascular tissue while the water moves to surrounding tissue.
Step 4: Demonstration of the section cutting technique is recommended here. It needs to be done quickly so that
the sections do not dry out. You may prefer to use toluidine blue stain instead of iodine. It stains lignified tissue a
blue colour and non-lignified tissue a purple colour. It works satisfactorily for the transverse section but is inclined
to be very intense in the longitudinal section.
Step 5: The bundle must be dissected out under a stereoscopic microscope. It is important that only a small piece of
bundle (not the full width) is taken, otherwise there are so many parenchyma cells that not much else can be seen.
If possible, tease out the bundle under the stereoscopic microscope. The more teasing that is done, the better the
observations that will be made. By moving the slide gently from side to side, the cells are spread out before being
squashed.
6
Spiral thickening of the xylem vessels should be clearly seen. (see Figure 1.55a in Heinemann Biology 1
Student Workbook). The spirals may have become dislodged from some cells. If the light is stopped right
down, it should also be possible to see the sieve tube cells and companion cells. The sieve tube cells can be
recognised by V-shaped stained contents, usually just above or below a sieve plate, which can also be seen if
focusing is done carefully (see Figure 1.55b in Heinemann Biology 1 Student Workbook).
7
Thickened walls and spiral thickening give strength to the cell. The spirals can be compared to the wire coils
inside a vacuum cleaner hose, providing structural support.
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8, 9, 10 The following figures show details of Helianthus vascular tissue. You may wish to download and enlarge
these for use on OHP transparencies or data projection.
Conclusions
11 A class discussion using models, if available, would be of value before students attempt this task. A videoflex
would also be helpful.
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Further ideas
1
Cut a small branch from the common garden plant variegated Pittosporum. Use petroleum jelly to coat two
leaves on both the upper and lower epidermis, two leaves on the upper epidermis only, and two leaves on the
lower epidermis only. Stand the stems in coloured dye solution for at least 24 hours. Record your observations
of dye movement and attempt to explain the observations.
2
Obtain a prepared slide of the transverse section of a woody stem where the vascular tissue produced by the
cambium is a complete cylinder. Draw a diagram of the section and label the following: bark (including
phloem), recently formed xylem rings of old xylem (growth rings) and rays. Reference materials may be
required for interpretation.
Practical activity 3: Variations on a theme—flower structure and reproduction
Text reference: Heinemann Biology 1 Student Workbook, pages 77–79
Estimated time: 50 minutes
Scope of activity
This activity utilises flowers that are readily available from household gardens to illustrate typical floral
reproductive structures, with examples of both inferior and superior ovary positions. While the activity requires
students to complete and formally report on the dissection of two flower heads, it is recommended that students be
given the opportunity to examine a range of flowers in order to recognise the same reproductive structures in
different species.
Preparation
Have a variety of flower heads available for students, ensuring that those listed are well represented, since they
represent textbook examples for the study. Include some less familiar examples for students to use as a means of
comparison and for practising identification of the same structures in different species. Encourage students to bring
their own flower heads to class—this will enhance the variety of samples available.
Hints and comments
Students may choose to make drawings of more than one example of inferior and superior ovary flower heads. This
is a useful exercise in crystallising structural patterns. Students who decide to do this are encouraged to fix
additional diagrams into their workbook.
Conclusions
7
Typical structural features of a flower: sepals, petals, stamens, pistil.
9
Typical garden flowers will be colourful and scented, suggesting pollination by an animal agent, usually an
insect such as a bee. Tubular flowers, e.g. Grevillea, may be bird-pollinated.
10 a
The colour and scent of roses are features that attract insect pollinators.
b
Wind-pollinated plants are typically unscented and not colourful, e.g. grasses— tussock grass, crop
plants such as wheat.
c
Grevilleas are pollinated by the honeyeater—it features a long, slender beak that can penetrate to the base
of the flower where it feeds on nectar. While feeding, its head feathers brush against the stamens,
collecting pollen. This pollen is transferred to the stigma of the next Grevillea flower it visits.
d
Some species of orchids are pollinated by moths.
The native Banksia is pollinated by the tiny honey possum, Tarsipes rostratus.
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Further ideas
1
Composite flowers: Use the information and knowledge gained from this activity as a starting point for the
examination of a very different kind of flower—the sunflower.
•
Remove one of the yellow ‘petals’, taking with it its point of attachment to the stalk. With a scalpel, split this
‘petal’ open lengthwise at its base.
•
The ‘petal’ is actually an entire flower; it is called a ray flower. Examine the flower with a hand lens. Look for
the following structures: sepals, petals, stamens, pistil, ovary and ovules.
•
Draw a labelled diagram of the ray flower.
•
Now remove one of the dark centre structures from the sunflower, again making sure that the base point is still
attached.
•
Cut this structure lengthwise using a scalpel. Examine it with a hand lens. This structure is called a tube
flower. Look for the same features as you did for the ray flower.
•
Draw a labelled diagram of the tube flower.
2
i
In what ways are ray and tube flowers similar? How do they differ?
ii
Biologists call this type of flower arrangement a composite. Suggest why this is an appropriate name.
Comparative anatomy—animals and plants: Compare reproduction in flowering plants with reproduction in
mammals. Use the following as the basis for comparison:
• male and female reproductive structures
• fertilisation
• development of the embryo
• methods of ensuring the survival of offspring.
3
Investigate the reproductive structures and strategies of conifers.
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Unit 2: Organisms in their environment
Area of Study 1: Adaptations of organisms
Outcome: Explain and analyse the relationship between environmental factors, and adaptations and distribution of
living things.
Practical activity 7: Ways and means—mangrove adaptations
Text reference: Heinemann Biology 1 Student Workbook, pages 120–123
Estimated time: 50 minutes
Scope of activity
This activity addresses the theme of adaptation, in part by looking at zonation patterns of organisms that make up a
saltmarsh community. The information is provided largely in visual form; students are asked to consider different
kinds of adaptations for the different conditions within the saltmarsh zone. It may be valuable to conduct a
preliminary discussion about structural, functional and behavioural adaptations.
Preparation
Any relevant visual material will be a useful support.
Hints and comments
1
a
Daily changes include the day/night cycle, high and low tides.
b
Seasonal changes include day length, temperature fluctuations, variations in rainfall.
2
This is a hypothetical situation. All other things being equal, one could possibly expect there to be a
movement of the entire pattern towards the open water as a result of increasing sediment build-up at the
landward edge. As well as considering adaptations to extremes conditions, this feature of change in
ecosystems links into the theme of succession considered in Area of Study 2.
3
a
Succulents are plants that feature fleshy stems and/or leaves.
b
The fleshy tissue of succulents is adapted to store water. In the environment of a saltmarsh, the saline
environment tends to cause excess water loss in organisms. The ‘succulent’ stems and leaves of plants
such as the glasswort are adaptations for water storage, allowing the plant to retain water.
4
The boundary between the swamp paperbark and the mixed saltmarsh is at the limit of the very high tides. The
area is not inundated with salt water at each high tide. The salt water dries out and leaves a crust of salt on the
soil. It is too salty for plants to grow.
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5
Guide students in giving careful thought to the idea of adaptations and the survival advantage that each
confers on an organism. Some suggestions are made in the table below.
Description of adaptation
Survival value of adaptation
Type of adaptation
Mangrove plants can tolerate up to
100 times the salt concentration of
other plants
Able to survive in harsh
conditions in which other
plants cannot survive
physiological
Pneumatophores:
structural
grow upwards, protruding above the
water line
access oxygen
have lenticels
allow air to enter
have tissue with air spaces
allow for diffusion of air within
plant tissue
allow fresh water in, but exclude most
salt
reduce salt concentration in
the tissues
Leaves:
are thick & leathery
reduce water loss/ retain water
structural
have salt glands on the upper surface
remove excess salt
structural
are able to ‘pump’ salt out through
these glands
reduce salt concentration
physiological
germinate on the parent plant
overcomes problem of seed
germination in salt water
physiological, behavioural
show rapid seedling growth when
settled in mud
establishes young plant
physiological, behavioural
Seeds:
Many examples of behavioural adaptations of animals are provided. Ensure students do not concentrate on
these alone.
Conclusions
8
The distribution of organisms in the saltmarsh community is dependent on the adaptations they have for
tolerating the kinds of conditions in the different zones. In the high tide zone, beaded glasswort forms the
dominant vegetation—they are adapted to conditions that are exposed for much of the time, with infrequent
inundation by seawater. The shrubby glasswort has its roots in the softer mud that is more regularly subject to
the tides, and can withstand fluctuations in water level and salt concentration. The mangrove community is
further down the shoreline, with its roots permanently embedded in waterlogged mud that is regularly
inundated with seawater. Pneumatophores are specialised aerial roots that allow them to survive in waterlogged conditions. Eel grass, which is salt tolerant, dominates the mudflats that are almost constantly below
the water line.
Students may concentrate on the plant community in this item, since it forms the visual constant—encourage
them to include a discussion of examples of some animal life included in the description of the saltmarsh
community, e.g. acorn barnacles/adaptations to drying out, conniwink.
© Reed International Books Australia
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Practical activity 8: Chemical regulators—investigating hormones
Text reference: Heinemann Biology 1 Student Workbook, pages 124–125
Estimated time: 50 minutes should be sufficient to allow for class discussion followed by written student
responses.
Scope of activity
This second-hand data activity is intended to provoke some thought about the nature of hormones, and provide a
means for students to make inferences about how hormones have their effect compared with the nervous system.
Preparation
Slides or photographs showing the differences between castrated and normal male animals would make an
interesting and useful introduction to this activity.
Hints and comments
Important information is given in the background to this activity, describing the way in which the different birds
were treated. Note especially that with Bird 2, the blood vessels reconnected but the nerves did not.
1
This indicates that the testes cannot regenerate. Some students may believe that they can.
2
The important point in this question is that the blood vessels that rejoined to the circulatory system seem to be
necessary for normal testes function, but the nervous system appears not to be.
3
a
The testes are necessary for the development of secondary sex characteristics.
b
Location of the testes does not appear to be critical, so long as there is a blood supply to them.
4
Nerve tissue is not necessary for the development of secondary sex characteristics, but the presence of testes
is. The target tissues of the body at which secondary sex characteristics develop must respond to a substance
produced by the testes that travels in the blood, rather than responding to nervous stimuli.
5
The testes are necessary for the maintenance of secondary sex characteristics.
6
The testes are the tissue that is vital in determining that a bird remains a cockerel, rather than the combs and
wattles.
7
Hormones.
Conclusions
9
The presence of testes in male birds is vital in the development of secondary sex characteristics. The testes
release a substance or substances that must be carried in the bloodstream to other parts of the body where they
have their effect. These substances are called hormones. The absence of the testes means that these substances
are not present in the body, and this results in the failure of secondary sex characteristics to develop. In this
case, males do not develop into sexually mature adults. Hormones released by the testes are important in the
development of secondary sex characteristics.
Further ideas
1
Find out how castration affects development to sexual maturity in the following animals:
Cattle: bulls → steers
Horses: stallions → geldings
Sheep: rams → wethers
© Reed International Books Australia
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Practical activity 9: Australian ectotherms and endotherms—temperature regulation
Text reference: Heinemann Biology 1 Student Workbook, pages 126–130
Estimated time: 50 minutes should be adequate time for this activity—after 10 minutes of introductory discussion,
students should be able to work through reading, analysis and written component individually.
Scope of activity
This second-hand data activity focuses on a range of Australian animals and the different kinds of adaptations they
possess to regulate body temperature. There is an opportunity to contrast mechanisms used by ectotherms with
those present in endotherms. Students need to consider the nature of adaptations—structural, behavioural and
physiological.
Hints and comments
A: Types of temperature regulation
1
2
a
Endotherms generate their own body heat from the metabolic activities of cells (endo: inside; therm:
temperature). Endotherms typically maintain constant body temperature.
b
Ectotherms are unable to generate enough body heat from metabolic activities; they acquire most of the
heat they need from external sources, e.g. radiant heat from the sun, or heated conducted to their bodies
from warm surfaces, such as rocks and earth (ecto: external). The body temperature of ectotherms
fluctuates in line with environmental temperatures.
a
True endotherms: cat, human and platypus; their body temperatures remain relatively constant despite
changing external temperature, indicating that they have internal mechanisms that control body
temperature. This is a characteristic of endotherms.
b
The lizard is the true ectotherm; its body temperature appears to be strongly affected by the external
temperature. This is a characteristic of ectotherms.
Note that the echidna has a body temperature that appears to be intermediate between the two categories.
3
A warm body temperature provides optimum conditions for efficient cell function. Both endotherms and
ectotherms function best when their internal temperature is relatively warm. They have different mechanisms
for achieving this.
B: Some methods of mammalian temperature regulation
4
Observation
Explanation
Type of
adaptation
An echidna living in cold
regions hibernates during
winter
During winter, food resources are scarce, and so
energy inputs are not sufficient to generate enough
body heat through metabolic processes. During
hibernation, the body’s metabolic rate is greatly
reduced, which in turn reduces body temperature.
Reduced metabolic needs are met from energy
reserves in the body.
behavioural and
physiological
Your skin often looks quite
flushed on a hot day
Blood vessels in the outer layers of the skin dilate,
increasing the volume of blood at the skin, which
increases heat loss to atmosphere.
physiological
Many Australian marsupials
salivate and lick their fur on
hot days
This is a form of evaporative cooling—body heat is
lost as the saliva evaporates.
behavioural
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Observation
Explanation
Type of
adaptation
Whales have a thick layer of
blubber (fat) under their skin
Fat is an effective insulator—it reduces heat loss
from the body at the skin.
structural
You tend to feel cooler on a
hot, dry day than on a humid
day of the same temperature
In dry heat, there is a large concentration gradient
between moisture on the skin surface and moisture
in the air. Dry heat, therefore, allows significant
evaporation to occur, which has a cooling effect. In
high humidity, there is little or no concentration
gradient and so little or no evaporation—therefore,
the cooling effect is reduced.
physiological
A small mammal, with the
same body temperature and
insulating mechanisms as a
larger mammal, loses more
heat than the larger mammal
The small mammal has a greater surface area to
volume ratio, resulting in greater heat loss. The
larger mammal has a smaller surface area for its
volume, and so loses less heat per unit volume.
structural
Dogs pant on a hot day
This is another example of evaporative cooling, and
involves the surface of a dog’s tongue and mouth
cavity.
behavioural
Cats often look ‘fatter’ on a
cold day
Cats have their hair ‘fluffed out’, trapping a layer of
air next to the skin. This reduces the temperature
gradient between the skin and air. Less heat is lost
from the body.
behavioural
C: Examples of Australian endotherms
5
a
Breathing rate in potoroos increases rapidly at about 35°C.
b
The graph suggests that they are able to cope well with temperatures up to 35°C. However, above that
level potoroos display a significantly increased breathing rate, indicative of evaporative cooling
(resulting from panting), which is used as a means of regulating temperature, thereby avoiding heat
stress.
6
Observation
Explanation
Type of
adaptation
The tail lacks hair
Exposed skin can be used for temperature
regulation.
structural
Sweat glands observed in
dense rings around the tail
Allow for heat loss through evaporative cooling, i.e.
heat is lost as water is evaporated.
structural
At high temperatures, the tail
appears to be quite wet
Sweat produced from sweat glands during periods
of high temperature; evaporative cooling.
physiological
At high temperatures,
potoroos continually twitch
their tails from side to side
This action may disperse sweat beads from the tail,
thereby maintaining the temperature and humidity
gradient between the skin and the air. This means
sweating can continue to be effective as a means
of cooling the body.
behavioural
7
The significantly reduced body temperature of the echidna indicates that it has entered a state of torpor.
8
The echidna’s metabolic rate would slow significantly as it entered torpor. Reduced metabolic rate means less
energy is needed, so there would be less oxygen consumed. When an animal is in torpor, most body systems
are in ‘shut down’ mode, so heart rate would be expected to drop.
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9
In cold conditions, entering a state of torpor is an adaptation to survival because it means the echidna will not
be expending a lot of energy searching for a scarce food source (ants and termites). It is more biologically
efficient to slow down body processes by entering torpor, so that energy is conserved.
10 Ectotherms are generally unable to raise their body temperature through heat generated in metabolic
processes, and must rely on heat acquired from the environment. This means that body processes are not
always operating at efficient or optimum levels. They tend to lose heat to the environment quite easily.
Because they are reliant on the environment for body heat, there will be times of the year when the
environment cannot adequately satisfy their temperature requirements. As a result, ectothermic animals are
often inactive during winter.
Conclusion
11 Students can be expected to include all of the key terms used in this activity, in particular endotherm and
ectotherm. Stronger students may also include heterotherm (echidnas are heterotherms). Student answers
should show clear links between different kinds of organisms, and the mechanisms and strategies used to
regulate body temperature in changing external temperature conditions, e.g. metabolic rate, insulation, panting
and sweating, blood flow to skin, hibernation. Students should also be able to identify adaptations as
structural, behavioural or physiological.
© Reed International Books Australia
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Area of Study 2: Dynamic ecosystems
Outcome: Design, conduct and report on a field investigation related to the interactions between living things and
their environment, and explain how ecosystems change over time.
Practical activity 10: Plants in their place—using line transects
Text reference: Heinemann Biology 1 Student Workbook, pages 162–165
Estimated time: About 100 minutes. This activity requires some field work outside the classroom—this is likely to
be somewhere in the school grounds. The time taken will depend on the efficiency with which students are able to
run the transect. 100 minutes is expected to be a maximum.
Scope of activity
This activity is designed to give students practice at using a line transect to describe the vegetation pattern in an
area, and to make students aware of some of the limitations of the technique. Once familiar with the technique,
students should be able to use it where appropriate, in the wider study of an ecosystem.
Preparation
It is intended that this initial activity be conducted near the school or in the school grounds if a there is a suitable
location. Ideally, it should have a range of plant diversity and yet not be too dense. Avoid areas dominated by
grasses or other very small plants, as students tend to try counting every one of them. Do not make the initial
transect line too long.
Clipboards can be kept dry using clear polythene bags. Bags with dimensions of 30 cm × 40 cm allow sufficient
space to write.
For transect lines, use a long tape measure—at least 10 m. It can be made from builder’s line with distances
marked with felt-tipped pens; a metre rule is useful for measuring intermediate distances.
Plant identification
The identification of the main species present in a community is most important. Before the study commences,
teachers should plan to be familiar with the key species in the area(s) chosen for students to study.
A little preparation time spent producing some aids to identification will increase the effectiveness of the activity:
•
Visit the area and either collect a sample of each species with all parts necessary for identification, or take
clear photographs of those parts. Check for necessary permission before removal of specimens.
•
The specimens can be pressed, mounted on stiff card and labelled. Laminating specimens is an ideal way of
preserving them and ensuring they are waterproof, but several layers of clear food wrap is an adequate
alternative. This becomes a field herbarium.
•
The specimens can be used to identify diagrams or photographs in published reference works, and a list of
reference diagrams can be compiled.
•
If the coloured photographs are clear and show the necessary detail, they can be mounted on card, labelled and
waterproofed. Digital cameras and a PC make this a cost-effective and easy-to-manage approach.
The more sets of identification aids there are, the more efficiently each group can work. This also means there is a
reduced demand on students to be collecting and removing specimens from the area.
© Reed International Books Australia
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Hints and comments
Step 3: This is observational and does not yet require the use of the transect line.
Step 4: Suggest that each member of the group counts one species as they move along the transect line—this
avoids confusion. If one student is counting each of the plants along the transect line, then they should count only
one species at a time.
4
If some species are very dense, e.g. grasses, it may be more practical to measure the fraction of the transect
line that the population takes up, rather than counting individual plants. This can be converted into a
percentage.
Steps 5, 6; Q5: The table below represents an example of how Table 2.17 may be completed.
Table 2.17 Ranking species
First
Species
ranking
Tree 1
4
Tree 2
5
Grass 1
Line transect count
Line 1
Line 2
Second
Total
ranking
7
3
1
5
1
28
1
Fern 1
3
15
2
Shrub 1
2
6
4
–
↑
↑
↑
This column
is based on
initial
estimate
Where possible, more than one line
transect should be done and the counts
totalled
This column
is based on
transect
counts
The plant species ranked 1 is the most common in the area.
Conclusions
6–9 Answers to these items are dependent on the area sampled. These questions are designed to encourage
students to think about the validity of data-gathering techniques.
Further ideas
1
As well as using a number system to nominate different plant species, e.g. ‘Tree 1’, you may prefer to identify
some or all of the plants to the level of family, genus and even species using keys and descriptions in
reference books.
2
Instruct students to record their transect data using a profile diagram, as per Figure 2.42. It will be important
to estimate scale when presenting data in this way.
© Reed International Books Australia
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Practical activity 11: The flatweed census—population estimation
Text reference: Heinemann Biology 1 Student Workbook, pages 166–169
Estimated time: 100 minutes, about 50 minutes in the field and about 50 minutes to complete calculations and
remainder of written report
Scope of activity
This activity extends students’ skills from simple transect lines (Activity 10: Plants in their place) to estimation and
calculation of population density in sample two-dimensional areas or quadrats. The activity also challenges
students to think about, discuss and justify some of the procedures used in such an activity.
Preparation
Flatweeds in a lawn or playing field are the best plant species to count because they are easy to identify, and lawns
usually have a fairly defined and measurable shape. If the data from different groups are to be combined, make
sure that all groups know which flatweed species they are counting. If there is a large number of groups, several
flatweed species could be used.
The information on plant identification in the notes for Activity 10: Plants in their place is also relevant here.
Quadrats
These can be made quite cheaply using 16 mm outside diameter electric PVC pipe and right-angle joiners (see
Figure below).
Simple alternatives include 4 × 1 m lengths of dowel or 4 × 1 m rules. Quadrats can also be purchased from
suppliers such as Southern Biological Services.
© Reed International Books Australia
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Hints and comments
Step 2: The shape of most lawn areas, if not rectangular or square, can be divided into rectangles or triangles for
ease of calculation. Marker pegs may be useful.
Step 4: Ecologists use complex mathematical procedures to determine the percentage of an area that should be
sampled to provide meaningful estimations for an area, so there is not a definitive answer to this. However, keep in
mind that sampling techniques are meant to save time, so the number of quadrats used should not be too high.
Step 5: Students may need to re-read the background information at the beginning of Activity 11 in the Student
Workbook.
Step 9: If the quadrat used has sides of 0.25 m, then multiply the number of flatweeds by 16 to give the number per
square metre.
9, 10 Multiply the average flatweed density per square metre by the number of square metres of lawn.
Conclusions
14 The more samples that are taken, the more accurate will be the estimates.
15, 16 Students consider the validity of the sampling techniques used.
Further ideas
1
After making an initial count of weeds using the method described earlier, carry out a weed removal program.
At a later time, carry out a similar population estimate. Write a report on the findings.
2
The quadrat sampling technique can be used to compare the plant populations in two areas that differ in some
way. Before starting, decide on the factor you wish to investigate and also the plant species you will consider.
Try to keep all other factors constant. For example, a mown area and an unmown area could be investigated to
discover the effect of mowing on the dandelion population of a lawn. Other factors that could be investigated
include:
• grazing
• fertilisation
• shade
• angle of the ground (sloping or flat)
• drainage.
This activity provides students with an opportunity to design and carry out an investigation. Students would need
to suggest a hypothesis and design an experiment that takes into account the control of variables.
© Reed International Books Australia
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Practical activity 12: A numbers game—factors affecting population size
Text reference: Heinemann Biology 1 Student Workbook, pages 170–175
Estimated time: 100 minutes if all completed in class time. After initial discussion, students can do some data
analysis during class time and complete the remainder of the activity out of class time. It will be worthwhile
following up this activity with some class discussion.
Scope of activity
This activity addresses the patterns associated with population growth by considering hypothetical population
growth in the first instance, and then graduates to examples of natural population growth. It considers the stages in
a ‘normal’ population growth curve, challenging students to analyse and graph data as well as make predictions
about population growth based on their knowledge of the factors that affect population dynamics.
Hints and comments
A: Hypothetical population growth
1
Time
Generation
number
Number of
bacteria
6.30 am
1
1
7.00 am
2
2
7.30 am
3
4
8.00 am
4
8
8.30 am
5
16
9.00 am
6
32
9.30 am
7
64
10.00 am
8
128
10.30 am
9
256
11.00 am
10
512
11.30 am
11
1024
12.00 am
12
2048
12.30 am
13
4096
1.00 pm
14
8192
1.30 pm
15
16384
2.00 pm
16
32768
3
It is very difficult to plot the data in a conventional way. Early values are very small, later values are very
large.
4
The characteristic curve is a ‘J’ curve.
5
The population size is doubling. Numbers become very large very quickly.
6
Exponential growth.
B: Real population growth
9
a
Both curves feature the initial ‘J’ shape.
© Reed International Books Australia
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b
The bee population numbers plateau or level out, while the deer population numbers crash.
10 Stable populations usually show small fluctuations over time. Wide fluctuations do not occur in stable
populations. The bee population curve is expected to be the most common.
11 The bee population is stable, indicating that it is probably in balance with the available resources. Minor
fluctuations are likely to be the result of minor shifts in resources and predator populations.
12 The dramatic decline in the deer population suggests some significant environmental impact, probably a lack
of resources such as food, or possibly disease.
13 a
b
14 a
b
Similar ‘J’ phase of the graph.
Different: population continues to increase.
Availability of adequate food or space may contribute to environmental resistance in the bee population.
A lack of food resources or the introduction of a disease may contribute to environmental resistance in
the deer population.
Conclusions
15 Correct order of terms: equilibrium, exponential, negative acceleration, positive acceleration
16
© Reed International Books Australia
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17 Migratory animals; the population of animals in a particular location will vary greatly depending on the time
of the year. Immigration and emigration also affect human populations.
18 a
b
Grasshopper populations generally reach plague proportions due to an abundance of food. Population
numbers will fall again when food resources decline.
The starfish population increases as a result of the lack of competition from other species and the lack of
predators. Eventually, the population may crash due to overuse of resources, or it may reach equilibrium.
19 Given abundant resources and a balance of environmental factors, such as predators, a population becoming
established in a particular area is likely to show a pattern of growth that follows the ‘S’-shaped curve, with
relatively slow growth initially, followed by exponential growth and then a levelling off. At this point the
population becomes stable. It is likely to remain stable if there is a balance of environmental factors, such as
food and other resources, as well as predators. A lack of food resources or the introduction of predators or
disease can lead to dramatic declines in populations. Immigration and emigration also impact on population
sizes.
Further ideas
The human population
The human ecosystem is effectively the entire world. Humans are atypical animals, in that they are able to survive
in almost every type of environment because of the ability to modify our environment. Human demands on the
environment are large in terms of both resources and energy.
With an understanding of some of the ideas about hypothetical and natural population growth, we can consider the
growth of human populations.
There is a vast amount of current data about human population growth available on the Internet. Enter key words,
e.g. human population, into a search engine to find relevant statistics. The data in the following table is taken from
a US web page (a billion is 1000 million, so 0.3 billion is 300 million).
World population growth from 0 AD to 1998
Year
Population
(billions)
0
0.30
1000
0.31
1250
0.40
1500
0.50
1750
0.79
1800
0.98
1850
1.26
1900
1.65
1910
1.75
1920
1.86
1930
2.07
1940
2.30
1950
2.52
1960
3.02
1970
3.70
1980
4.44
© Reed International Books Australia
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Year
Population
(billions)
1990
5.27
1998
5.90
1
Plot the data in the table on a graph. Plot time since 0 AD on the horizontal axis.
2
To which of the graphs you have already drawn for this activity is the human population graph most similar?
Does the assumption that was stated for that population hold for the human population? Explain.
3
Discuss in detail some of the implications of this comparison. Search the Internet for current information on
population growth trends to help you. Consider the following factors:
• overall population growth rate
• life expectancy
• infant mortality rate per 1000 live births
• child mortality rate per 1000 live births
• fertility rate
• average life span.
4
During the 1990s, the United Nations looked at predictions of population growth beyond 2000 (see following
table). Find out what the world’s human population is in 2005. How does this compare with the predictions
made in this table? Describe the pattern of human population growth. Find out what the current predictions for
world populations are. Enter these into the table.
Predicted population growth beyond 2000
Year
Population (billions)
1990s predictions
5
2000
6.06
2010
6.79
2020
7.50
2030
8.11
2040
8.58
2050
8.91
Current predictions
Find out the carrying capacity of our planet in terms of our human population. Discuss the implications of this
compared with the population trends you considered in Q4 above. Outline any measures that could be used to
address this issue.
© Reed International Books Australia
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