1
External Examination 2014
2014 BIOLOGY
FOR OFFICE
USE ONLY
SUPERVISOR
CHECK
QUESTION
BOOKLET
1
ATTACH SACE REGISTRATION NUMBER LABEL
TO THIS BOX
9 pages, 6 questions
RE-MARKED
Monday 10 November: 9 a.m.
Time: 3 hours
Section B: Short-answer Questions
Part 1
Examination material: Question Booklet 1 (9 pages)
Question Booklet 2 (11 pages)
Question Booklet 3 (8 pages)
Multiple-choice Question Booklet (14 pages)
one blue multiple-choice answer sheet
one SACE registration number label
Approved dictionaries and calculators may be used.
Instructions to Students
1.
You will have 10 minutes to read the paper. You must not write in your question booklets, or on your blue
multiple-choice answer sheet, or use a calculator during this reading time but you may make notes on the
scribbling paper provided.
2.
This paper is in three sections: Section A is in the Multiple-choice Question Booklet; Part 1 of Section B is in
Question Booklet 1; Part 2 of Section B is in Question Booklet 2; and Section C is in Question Booklet 3.
Section A: Multiple-choice Questions (Questions 1 to 25)
Answer Section A on the separate blue multiple-choice answer sheet, using black or blue pen.
Answer all questions in Section A.
Section B: Short-answer Questions (Questions 26 to 36)
Answer all Part 1 of Section B (Questions 26 to 31) in the spaces provided in Question Booklet 1.
Write on page 9 of Question Booklet 1 if you need more space.
Answer all Part 2 of Section B (Questions 32 to 36) in the spaces provided in Question Booklet 2.
Write on page 11 of Question Booklet 2 if you need more space.
Section C: Extended-response Questions (Questions 37 and 38)
Answer both questions in Section C in Question Booklet 3.
Write on page 8 of Question Booklet 3 if you need more space.
3.
In Section B there is no need to fill all the space provided; clear, well-expressed answers are required. If you
delete part or all of an answer you should clearly indicate your final answer and label it with the appropriate
question number.
4.
The allocation of marks and suggested allotment of time are as follows:
Section A
50 marks
40 minutes
Section B
120 marks
110 minutes
Section C
30 marks
30 minutes
Total
200 marks
180 minutes
5.
Attach your SACE registration number label to the box at the top of this page. Copy the information from your
SACE registration number label into the boxes on your blue multiple-choice answer sheet and on the front covers
of Question Booklets 2 and 3.
6.
At the end of the examination, place Question Booklets 2 and 3, and your blue multiple-choice answer sheet,
inside the back cover of Question Booklet 1.
STUDENT’S DECLARATION ON THE USE OF
CALCULATORS
By signing the examination attendance roll I declare that:
• my calculators have been cleared of all memory
• no external storage media are in use on these calculators.
I understand that if I do not comply with the above conditions
for the use of calculators I will:
• be in breach of the rules
• have my results for the examination cancelled or amended
• be liable to such further penalty, whether by exclusion from
future examinations or otherwise, as the SACE Board of
South Australia determines.
2
SECTION B: SHORT-ANSWER QUESTIONS (Questions 26 to 36)
(120 marks)
You should spend about 110 minutes on this section. Answers may be in note form. The allocation
of marks is shown in brackets at the end of each part of each question. Answer all questions in the
spaces provided.
PART 1 (Questions 26 to 31)
(60 marks)
26. It is possible to genetically modify an animal by adding copies of a particular gene to the nucleus
of a fertilised egg cell. If the gene is incorporated in an appropriate position on a chromosome, it
can be successfully expressed.
(a) State one function of a gene.
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marks)
(b) Name one method that could be used in the laboratory to produce multiple copies of a
particular gene.
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marks)
(c) Name one method that could be used to transfer a gene to the nucleus of a fertilised
egg cell.
____________________________________________________________________________________________ (2
marks)
(d) Explain why it is necessary to add copies of a particular gene to a fertilised egg cell,
rather than to a cell of an adult animal, in order to genetically modify all of the animal’s
cells.
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3
marks)
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27. Refer to the following schematic diagram, which represents a process that involves DNA and
occurs in prokaryotic cells before cell division:
(a) State the location in the cell where this process occurs.
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marks)
(b) Explain why it is essential that this process occurs before the cell divides.
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marks)
(c) Explain how the diagram above shows that this process is semi-conservative.
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4
marks)
28. Oestrogens are hormones that are involved in the reproductive cycles of vertebrates and some
insects. They are able to bind to oestrogen receptor proteins on the surface of cell membranes.
(a) State the subunits that make up an oestrogen receptor protein.
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marks)
(b) Explain why the order of these subunits is critical to the function of the oestrogen
receptor proteins.
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marks)
(c) Explain the role of tRNA in ensuring the correct order of the subunits in oestrogen
receptor proteins.
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5
marks)
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29. Refer to the following photomicrograph, which shows an organelle, X:
X
1 μm
Source: Department of Anatomical Science, University of Adelaide
Organelle X contains DNA, part of which codes for an enzyme, ATP synthase. This enzyme is
involved in one part of a series of reactions that produce ATP. Without the ATP synthase, ATP is
not formed by this metabolic pathway.
(a) State the name of organelle X.
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marks)
(b) Explain why the reactions in metabolic pathways often involve a series of several small
steps.
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marks)
(c) Use one piece of evidence to explain the evolutionary link between prokaryotic cells and
organelle X.
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6
marks)
30. The well-researched p53 protein is associated with DNA in normal human cells. It is able to
regulate the cell cycle, and control the timing of cell division. High concentrations of p53 protein
restrict cell division. If the base sequence of a cell’s DNA changes or is damaged, the p53 protein
can stop the cell from dividing and allow it to repair the damaged DNA. If such a repair is
unsuccessful, p53 may trigger the death of the cell.
Very low concentrations of p53 protein may have serious consequences for the cell.
(a) State the term used to describe a permanent change in the base sequence of DNA.
____________________________________________________________________________________________ (2
marks)
(b) State one environmental factor, other than a chemical, that may cause a permanent
change in the base sequence of DNA.
____________________________________________________________________________________________ (2
marks)
(c) Explain the effect on a person if one of his or her cells is unable to produce the
p53 protein.
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marks)
(d) Explain one benefit to society that may result from continued research into the cellular
role of the p53 protein.
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7
marks)
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31. The use of antibiotics to prevent bacterial infections in farmed animals and to treat undiagnosed
fever in human beings has increased substantially in recent years. Scientific research shows a
direct link between the overuse of antibiotics and the evolution of antibiotic-resistant bacteria
that are now prevalent in hospitals, where they cause infections that are difficult to treat.
The continuous presence of antibiotics in the environment ensures that the bacteria retain the
genes for antibiotic resistance.
Antibiotic-resistant bacteria are at a disadvantage in an environment that is free of antibiotics.
A recent study of various animal farms where the use of antibiotics is prohibited has shown a
decrease in the proportion of antibiotic-resistant bacteria present.
(a) Explain how the prohibition of antibiotics could lead to an increase in the proportion of
bacteria that are not resistant to antibiotics.
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marks)
(b) Discuss how the use of prescription drugs can affect the well-being of individuals.
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8
marks)
You may write on this page if you need more space to finish your answers to any questions in
Part 1 of Section B. Make sure to label each answer carefully (e.g. 29(b) continued).
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© SACE Board of South Australia 2014
9
PLEASE TURN OVER
External Examination 2014
2014 BIOLOGY
FOR OFFICE
USE ONLY
SUPERVISOR
CHECK
SACE REGISTRATION NUMBER
FIGURES
QUESTION
BOOKLET
CHECK
LETTER
SEQ
BIN
BIOLOGY
RE-MARKED
Monday 10 November: 9 a.m.
Section B: Short-answer Questions
2
Part 2
Write your answers to Part 2 of Section B in this question booklet.
2
11 pages, 5 questions
2
SECTION B: SHORT-ANSWER QUESTIONS
PART 2 (Questions 32 to 36)
(60 marks)
Answer all questions in the spaces provided.
32. Glucose transporters (GLUTs) are proteins that are important in regulating the movement of
glucose through the cell membrane. Some types of GLUTs are active in muscle cells where they
facilitate the movement of glucose through the membrane and into a cell that already contains a
higher concentration of glucose than is contained in the tissue fluid that surrounds it.
(a) Draw a labelled diagram to represent a section of cell membrane that shows how GLUT
proteins are likely to be arranged.
(4 marks)
(b) State the name of the process by which a substance (such as glucose) moves through
the cell membrane and into a cell that already contains a higher concentration of the
substance than is contained in the tissue fluid that surrounds it.
____________________________________________________________________________________________ (2
3
marks)
PLEASE TURN OVER
33. The Marathon des Sables is an ultramarathon held in the Sahara desert. Competitors run about
250 kilometres over 5 to 6 days in extremely hot, dry conditions.
(a) Write the chemical equation for the aerobic heat-producing process that occurs in the
muscle cells of the competitors.
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marks)
(b) State one response involved in the control of body temperature, and explain how it would
increase the competitors’ heat loss.
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marks)
Rhabdomyolysis is a condition that results in rapid muscle meltdown. The muscles overheat as
a result of increased heat production and reduced heat loss. The muscle cells die and release
chemicals into the bloodstream. One of these chemicals is a protein called myoglobin, which
affects membrane transport proteins on the tubules of the kidney. Myoglobin can enter the
nephron, and cause the tubules to break down.
(c) State one structural feature of an exchange surface such as the nephron.
____________________________________________________________________________________________ (2
marks)
(d) People with kidney damage resulting from rhabdomyolysis have dark red or brown urine,
containing proteins and red blood cells.
Explain why the urine of a healthy person does not contain proteins or red blood cells.
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4
marks)
(e) Rhabdomyolysis can be treated early with intravenous hydration to prevent kidney failure.
The increased levels of fluid in the blood dilute the myoglobin, and the patient produces
an increased volume of urine.
Explain why increased levels of fluid in the blood result in reduced reabsorption of water from
the nephron.
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5
marks)
PLEASE TURN OVER
34. In an article published in 2010 in the Molecular Biology and Evolution journal, scientists
hypothesised that sexual reproduction must be beneficial to organisms as it is more common than
asexual reproduction.
Source: © iStockphoto.com|Vitalii Hulai
The scientists had conducted an investigation in which they sequenced mitochondrial genomes
of both sexually and asexually reproducing varieties of a New Zealand freshwater snail,
Potamopyrgus antipodarum. It was found that the sexually reproducing snails had accumulated
harmful DNA mutations at about half the rate of the asexually reproducing snails.
(a) State the type of cell division in the asexually reproducing snails.
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marks)
(b) Compare the offspring of the asexually reproducing snails with the offspring of the
sexually reproducing snails with reference to the similarity of each group to its parents.
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marks)
(c) State the independent variable in this investigation.
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marks)
(d) State why it would be important for scientists to repeat this investigation using other
species before reaching a conclusion about the differences between asexual reproduction
and sexual reproduction.
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6
marks)
35. Native birds are natural predators of pests that damage coffee plants on Mount Kilimanjaro,
Tanzania.
Coffee plants are pollinated by bees and other insects.
Explain why retaining native habitat around coffee plantations is the best way to maximise the
yield of coffee bean crops.
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7
marks)
PLEASE TURN OVER
36. The Amazon river basin contains half of the world’s tropical rainforests and plays a pivotal role
in Earth’s climate system. In recent years scientists have been recording the amounts of carbon
absorbed by and released from the Amazon forest.
(a) State the name of the process that produces glucose and is responsible for the
absorption of carbon from the atmosphere.
____________________________________________________________________________________________ (2
marks)
(b) State the function of starch in the tissues of the trees in the Amazon forest.
____________________________________________________________________________________________ (2
marks)
The net amounts of carbon released during wet and dry years were recorded.
In the dry years of 2000, 2005, and 2010 the net amounts of carbon released to the atmosphere
were 460, 540, and 480 million tonnes respectively.
In the wet years of 2001 and 2006 the net amount of carbon released by the trees in the Amazon
forest was zero.
However, in the wet year of 2011 the forest absorbed 250 million tonnes of carbon.
The cyclic nature of the wet and dry years is critical for the storage of carbon and the role of
the Amazon forest as a ‘carbon sink’. One of the factors that contribute to the amount of carbon
released during a dry year is the occurrence of fires in the forest. If no fires occur, the amount of
carbon that the forest can absorb is the same as, or greater than, the amount released.
(c) Using the data above, construct a table that shows the year, the net amount of carbon
released, and whether the year was wet or dry.
(4 marks)
8
(d) Using the data on page 8, state the year in which the Amazon was most affected by fire,
and explain your answer.
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marks)
(e) Explain why the net amount of carbon released by the trees in the Amazon forest was
zero in 2001 and 2006.
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marks)
The fires that occur in the Amazon forest destroy parts of the community. Research has shown
that most of the large trees do not survive and that the region affected by the fire regenerates
slowly, if at all. Scientists suggest that, if the fires in dry years continue to occur, large regions of
the Amazon ecosystem will be permanently changed.
(f ) Define the term ‘community’.
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marks)
(g) State the name of the process that leads over time to changes in the mix of species.
____________________________________________________________________________________________ (2
marks)
Question 36 continues on page 10.
9
PLEASE TURN OVER
(h) Describe how community change, such as in the Amazon ecosystem after a fire, leads
over time to changes in the mix of species.
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10
marks)
You may write on this page if you need more space to finish your answers to any questions in
Part 2 of Section B. Make sure to label each answer carefully (e.g. 34(b) continued).
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© SACE Board of South Australia 2014
11
PLEASE TURN OVER
External Examination 2014
2014 BIOLOGY
SACE REGISTRATION NUMBER
FIGURES
QUESTION
BOOKLET
CHECK
LETTER
SEQ
BIN
BIOLOGY
Monday 10 November: 9 a.m.
Section C
Write your answers to Section C in this question booklet.
3
3
8 pages, 2 questions
2
SECTION C: EXTENDED-RESPONSE QUESTIONS (Questions 37 and 38)
(30 marks)
Answer both questions in this section.
Write your answers in this question booklet:
• Question 37, on pages 4 and 5, is worth 15 marks.
• Question 38, on pages 6 and 7, is worth 15 marks.
You should spend about 30 minutes on this section,
5 to 10 minutes planning and 20 to 25 minutes writing.
Credit will be given for clear, well-expressed answers that
are well organised and relevant to the questions.
3
PLEASE TURN OVER
37. The islands in the Caribbean Sea are home to more than 150 species of lizard, all belonging to
the genus Anolis. It has been hypothesised that all these species are the descendants of two
original populations of lizard. Each species has unique features that enable it to live in its habitat.
The lizards are found in mountain ranges, woodlands, and rainforests. Many Caribbean islands
have only one species of Anolis lizard.
Source: © Andy Rhodes | Dreamstime.com
Source: © Linda Johnsonbaugh | Dreamstime.com
Source: © Henner Damke | Dreamstime.com
Source: © Philip Delos | Dreamstime.com
Source: © Jason P Ross | Dreamstime.com
Describe:
• two important events that occur during gamete production, increasing the chances of
survival and reproduction of Anolis lizards
• how natural selection resulted in the evolution of more than 150 species of Anolis in
the Caribbean islands.
(15 marks)
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PLEASE TURN OVER
38. Refer to the following diagram, which shows the apparatus for an experiment in which students
studied the effect of light intensity on the rate of photosynthesis in pondweed:
test tube containing dilute
solution of NaHCO3
bubbles
clamp stand
light source
pondweed
beaker containing
dilute solution of
NaHCO3
distance between light
source and beaker
The students placed a small piece of pondweed in a solution of sodium hydrogen carbonate
(NaHCO3), which provided a source of carbon dioxide. The number of bubbles of gas
released by the pondweed in 60 seconds was counted.
The distance between the light source and the edge of the beaker was changed and the
light intensity was measured and recorded at each distance. Each time the distance between
the light source and the beaker was changed, the students waited for 2 minutes before next
counting the bubbles.
The results are shown in the following table:
Light intensity
(arbitrary units)
Number of bubbles of gas
released in 60 seconds
8
9
32
30
128
103
512
107
The students were concerned about errors in their experiment.
• Describe the pattern of the students’ results, and write a valid conclusion for their
experiment.
• Explain how the students could have reduced the effect of experimental errors.
(15 marks)
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You may write on this page if you need more space to finish your answers to Questions 37
and 38. Make sure to label each answer carefully (e.g. 37 continued).
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© SACE Board of South Australia 2014
8
MULTIPLE-CHOICE QUESTION BOOKLET
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External Examination 2014
2014 BIOLOGY
Monday 10 November: 9 a.m.
Multiple-choice Question Booklet
Write your answers to Section A: Multiple-choice Questions
on the separate blue multiple-choice answer sheet.
Pages: 14
Questions: 25
SECTION A: MULTIPLE-CHOICE QUESTIONS (Questions 1 to 25)
(50 marks)
Answer all questions in this section.
Each of the twenty-five multiple-choice questions in Section A involves choosing from four
alternative answers. Read each question carefully. Then indicate the one alternative that you
consider best answers the question by shading the bubble by the appropriate letter alongside
the question number on the blue multiple-choice answer sheet. Use black or blue pen. It is
in your interest to give an answer to every question in this section of the paper, as no marks
are deducted for incorrect answers. Each question is worth 2 marks. You should spend about
40 minutes on this section.
1.
Cellulose is a macromolecule made up of
J.
simple sugars.
K.
amino acids.
L.
glycogen.
M. nucleotides.
2.
Refer to the following sequence of nucleotide bases which represents a section of an mRNA
molecule:
UAC CCG AAU UAG.
Which one of the following sequences of nucleotide bases would be found on the strand of DNA
that was transcribed to produce this section of an mRNA molecule?
J.
TAC CCG AAT TAG.
K.
UAC CCG AAU UAG.
L.
AUG GGC UUA AUC.
M. ATG GGC TTA ATC.
3.
A probe used to locate a gene can be
J.
radioactively labelled to enable it to bind to a complementary sequence of nucleotide
bases.
K.
used to cut double-stranded DNA into smaller fragments.
L.
made from a short RNA molecule.
M. made from a sequence of amino acids.
2
4.
The human PHOX2B gene is on chromosome 4. This gene is essential for development of the
nervous system in human embryos.
Which one of the following statements is correct?
J.
There will be one copy of the PHOX2B gene in each cell of a healthy human being.
K.
A mutation in the PHOX2B gene will prevent the production of a functional protein.
L.
As a result of crossing over, the PHOX2B gene will be on a chromosome other than
chromosome 4.
M. The base sequence of the PHOX2B gene may vary between individuals.
Refer to the following graph, which shows the effect of temperature on the rate of an
enzyme-controlled chemical reaction:
rate of reaction (arbitrary units)
5.
2
1
0
10
20
30
temperature (°C)
40
50
When the temperature
J.
falls below 10°C the enzyme is denatured.
K.
rises above 37°C the enzyme–substrate complex forms at a faster rate.
L.
falls below 37°C the enzyme–substrate complex forms at a slower rate.
M. rises above 50°C there is insufficient energy to activate the enzyme.
3
PLEASE TURN OVER
6.
A series of enzymes is necessary for the production of folic acid from 4-aminobenzoic acid.
The chemical sulfanilamide has a structure similar to that of 4-aminobenzoic acid, as shown in the
diagram below:
NH2
O
S
HO
O
O
C
NH2
NH2
sulfanilamide
4-aminobenzoic acid
When a small amount of sulfanilamide is added to cells, it is most likely to
J.
replace 4-aminobenzoic acid as the substrate of the first enzyme in the series.
K.
reduce the rate of production of folic acid.
L.
bind to the 4-aminobenzoic acid, distorting the shape of the active site.
M. denature the folic acid that is produced.
7.
Refer to the following table, which identifies cellular structures A, B, C, D, and E:
Cellular structure
A
cell membrane
B
ribosome
C
chloroplast
D
cell wall
E
mitochondrion
Which one of the following combinations of these structures may be found in prokaryotic cells
and in eukaryotic cells?
J.
A, B, and D.
K.
A, B, and C.
L.
A, D, and E.
M. B, D, and E.
8.
After a cell divides, each resulting daughter cell has a
J.
surface area-to-volume ratio the same as that of the parent cell.
K.
reduced volume, but the same surface area as that of the parent cell.
L.
greater surface area-to-volume ratio than that of the parent cell.
M. reduced volume, but a greater surface area than that of the parent cell.
4
9.
As part of the process of bacterial cell reproduction
J.
DNA replicates and the resulting chromosomes separate during binary fission.
K.
the pair of homologous chromosomes separate during binary fission.
L.
DNA replicates and the resulting chromosomes separate during mitosis.
M. the pair of homologous chromosomes separate during mitosis.
10. Which one of the following statements about energy use in cells is correct?
J.
The formation of ATP from ADP and phosphate releases energy for use in cells.
K.
The movement of sodium ions with the concentration gradient does not require cells to
provide energy.
L.
The transportation of water from the soil into a plant root cell requires cells to provide
energy.
M. The maintenance of a stable internal cellular environment does not require cells to
provide energy.
11. Which one of the following diagrams represents the chromosomes contained in a cell that has a
diploid number of 4 and has just completed mitotic division?
J.
K.
L.
M.
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12. A sodium chloride solution with a concentration of 0.9% is described as ‘isotonic’. If normal human
red blood cells were placed in this solution, there would be no net movement of water into or out
of the cells.
If the same red blood cells were placed in a sodium chloride solution with a concentration of 0.2%,
they would
J.
burst as water moved into them from the solution.
K.
gain water but their cell walls would prevent them from bursting.
L.
lose water and shrivel.
M. pump sodium chloride in to achieve equilibrium.
13. Fertilisation is necessary to
J.
ensure genetic variation in offspring.
K.
restore the diploid number.
L.
increase the chance of survival of offspring.
M. produce gametes.
14. Which one of the following combinations correctly identifies the product(s) of fermentation and the
amount of energy released relative to aerobic respiration in plant, animal, or yeast organisms?
Organism
Product(s) of fermentation
Amount of energy released
relative to aerobic respiration
J.
plant
lactic acid
low
K.
animal
lactic acid
low
L.
yeast
ethanol and carbon dioxide
high
M.
animal
ethanol and carbon dioxide
high
15. One role of lymph capillaries is to
J.
maintain a concentration gradient of gases at the alveoli.
K.
transport glucose to respiring muscle cells.
L.
transport hormones to target organs.
M. transport lipids from villi.
6
16. Refer to the following graph, which shows the effect that the concentration of carbon dioxide in the
lungs has on the activity of the enzyme Na,K-ATPase. This enzyme increases the reabsorption of
alveolar fluid in human beings.
Na,K-ATPase activity (%)
100
50
0
low
high
relative concentration of CO2
Patients with acute lung failure have impaired gas exchange as a result of the accumulation of
alveolar fluid. This condition is treated with mechanical ventilation, which has a side effect of
causing carbon dioxide to accumulate in the lungs.
From the information above, it would be reasonable to conclude that mechanical ventilation
J.
results in an increased concentration of carbon dioxide in the lungs and increases the
rate of diffusion of oxygen from the lungs to the blood capillaries.
K.
increases the amount of alveolar fluid, which increases the surface area of the alveoli,
resulting in increased gas exchange.
L.
reduces Na,K-ATPase activity, increasing the efficiency of both oxygen and carbon
dioxide exchange.
M. decreases the rate of diffusion of carbon dioxide from the blood capillaries to the alveoli.
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17. In 2001 at Henry Doorly Zoo and Aquarium in Nebraska, USA, a female hammerhead shark
(Sphyrna tiburo) gave birth to a baby shark. Genetic tests revealed that the baby shark had
genetic material only from the mother, but was not genetically identical to her. The female
hammerhead shark and the baby shark had the same number of chromosomes.
If the only male sharks in the aquarium were tiger sharks (Carcharias taurus), the best explanation
for this occurrence would be that the
J.
female hammerhead shark and a male tiger shark produced gametes by meiosis and
mated, producing the baby shark.
K.
female hammerhead shark produced gametes by meiosis, one of which divided by
mitosis, producing the baby shark.
L.
female hammerhead shark produced gametes by meiosis, two of which fused, producing
the baby shark.
M. female hammerhead shark and a male tiger shark produced gametes by mitosis and
mated, producing the baby shark.
8
18. Scientists at Washington University have recently discovered that odour receptors are found not
only in the nose, but also in the lungs.
When odour receptors in the nose detect chemicals such as cigarette smoke, they send a
nerve impulse to the brain and the smell of the cigarette smoke is then sensed.
Odour receptors in the lungs are called pulmonary neuroendocrine cells (PNECs). When PNECs
detect chemicals such as cigarette smoke, they release hormones that cause the airways to
constrict. In some people constriction of the airways results in an asthma attack. It is thought that
people with asthma may have hypersensitive PNECs.
Which one of the following statements is not consistent with this information?
J.
The airways of people with asthma would be constricted before they sensed the smell of
cigarette smoke.
K.
Hormones released by PNECs would travel through the bloodstream.
L.
Drugs that block hormone receptors in the airways could be used to treat asthma.
M. The constricted airways will not return to normal until the smell can no longer be sensed.
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19. Refer to the following diagram. Adult stem cells in bone marrow can differentiate into a variety of
cell types found in different organs:
brain
skin
adult stem cells in
bone marrow can
differentiate into
lungs
heart
liver
muscle
Source: © Gaby Kooijman |Dreamstime.com (adapted)
Stem cells that are transplanted from a healthy donor into a patient (recipient) will divide and
differentiate into cells of like form and function to the surrounding cells.
The adult stem cells from the bone marrow of the donor would be genetically identical to the cells
of the
J.
recipient, and divide by mitosis.
K.
donor, and divide by mitosis.
L.
recipient, and divide by meiosis.
M. donor, and divide by meiosis.
10
20. Most of the energy that flows through the producers in a community is
J.
captured in the bonds of chemicals.
K.
released during cellular respiration.
L.
transformed into glucose.
M. lost as heat.
21. The peafowl and the turkey are able to interbreed and produce offspring.
peafowl
turkey
Source: © Iakov Filimonov |Dreamstime.com
Source: © Irina Khomenko |Dreamstime.com
For a biologist to conclude that the peafowl and the turkey are the same species, the resulting
offspring would need to
J.
be physically similar to both parent organisms.
K.
be able to produce offspring with another peafowl or turkey.
L.
be unable to produce gametes.
M. inherit half of their chromosomes from the turkey.
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22. Refer to the following diagram, which shows organisms on an r – K reproductive strategy
continuum:
r
K
fly
crab
toad
rat
lion
elephant
Sources: fly © Andrei Mihalcea | Dreamstime Stock Photos; crab © Zakidesign | Dreamstime Stock Photos; toad © Isonphoto | Dreamstime.com;
rat © iStockphoto | whiteway; lion © Andres Rodriguez | Dreamstime Stock Photos; elephant © Kellers | Dreamstime.com
Which one of the following statements is not consistent with this r – K continuum?
J.
Rats could be found in stable communities, and provide some parental care for their
offspring.
K.
Flies would provide no parental care for their offspring.
L.
Elephants would provide extensive parental care for their offspring, which would probably
be slow to mature sexually.
M. Crabs have low reproductive effort, and a long lifespan.
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23. The productivity of a community is determined primarily by the
J.
increase in biomass of the consumers.
K.
presence of fertiliser.
L.
availability of water and light energy.
M. amount of oxygen available for producers.
24. Refer to the following diagram, which shows part of the phosphorus cycle:
geological uplift and
weathering by water
plants
animals
phosphates dissolved in water
sedimentation,
new rock formed
mycorrhizal
fungi in roots
phosphates in soil
dead and
decaying
matter
detritus
decomposers
Which one of the following statements is consistent with the diagram above?
J.
Mycorrhizal fungi are the only source of phosphates for animals.
K.
Decomposers enable phosphates from organisms to be recycled into the abiotic
environment.
L.
Phosphates are an energy source for decomposers.
M. Geological uplift and weathering by water are the only processes that make phosphates
available to organisms.
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25. Refer to the following data from an experiment in which the effect of temperature on the
germination of sunflower seeds was investigated in four trials. Ten sunflower seeds were used at
each temperature in each trial:
Temperature
(°C)
Number of sunflower seeds germinated
Average
number
of seeds
germinated
Trial 1
Trial 2
Trial 3
Trial 4
0
0
0
0
0
0
5
0
0
0
0
0
10
2
4
2
4
3
15
8
8
10
6
8
20
10
10
10
10
10
25
9
7
8
8
8
30
2
1
1
4
2
35
0
0
0
0
0
40
0
0
0
0
0
Which one of the following statements is consistent with the data from this experiment?
J.
Temperature was deliberately kept constant.
K.
The data at 20°C were the most accurate.
L.
The dependent variable was the number of sunflower seeds that germinated.
M. The data at 15°C and 25°C show the same precision.
© SACE Board of South Australia 2014
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