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BIOLOGY 2008 National Qualifying Examination
Time Allowed: Reading Time: 15 minutes
Examination Time: 120 minutes
INSTRUCTIONS
! Attempt all questions
! Permitted materials: Non-programmable, NON-GRAPHICAL
calculator, pens, pencils, erasers and a ruler.
! Answer SECTIONS A and B on the ANSWER SHEET PROVIDED.
! Answer SECTION C in the answer booklet provided. Write in pen
and use pencil only for graphs.
! Do not write on this question paper. It will not be
marked.
! Particular attention should be paid to giving clear diagrams
and explanations.
! All numerical answers must have correct units.
! Marks will not be deducted for incorrect answers.
MARKS
SECTION A 47 multiple choice questions 47 marks
SECTION B 17 written answer questions 17 marks
SECTION C 5 written answer questions 36 marks
Total marks for the paper 100 marks
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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SECTION A: MULTIPLE CHOICE
USE THE ANSWER SHEET PROVIDED
1. Which of the following is the BEST reason for including
protein in the diet?
a. Energy for the body. b. Fibre for digestion. c. Raw materials
for cell growth and repair. d. Vitamins for fighting disease.
2. Which of the following organisms are used to convert milk
into yoghurt?
a. Bacteria. b. Protozoa. c. Viruses. d. Algae. e. Fungi.
3. The growth of some plants can be improved by spreading bone
meal (ground-up bones)
around their roots. What does bone meal supply to plants that
improves growth?
a. Energy. b. Minerals. c. Vitamins. d. Carbon dioxide.
4. Tissues are found in living things. What is the definition of
a tissue?
a. A group of cells with similar structure and function. b. A
group of cells with different structure and function working
together. c. A group of organelles contained inside a cell. d. The
substances that constitute the walls of a cell.
5. Which one of the following characteristics is most likely to
be found in mammals that are
subject to predation by other mammals?
a. Eyes on the sides of head. b. Teeth that are long and
pointed. c. Claws on the feet. d. Ears that cannot move.
6. Which of the following is NOT a function of the blood?
a. Digestion of food. b. Protection against disease. c.
Transport of waste materials away from the tissues. d. Transport of
oxygen to different parts of the body.
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7. Malaria is a disease in humans caused by the protozoan
parasite Plasmodium, which is
transmitted by mosquitoes. The drug chloroquine has been widely
used to treat malaria for
several decades. Recently, the number of malaria cases not
responding to chloroquine has
increased.
What of the following is the most likely explanation?
a. Malaria sufferers are now excreting chloroquine before it can
kill the parasite. b. Mutations conferring chloroquine resistance
now arise more frequently in
Plasmodium.
c. Chloroquine favours the survival and reproduction of
resistant parasites. d. A new species of Plasmodium has
emerged.
8. Which of the following features of the platypus is
characteristic of mammals?
a. Predation of other animals. b. Production of milk. c. Nesting
and laying of eggs. d. Webbed feet.
9. Eukaryotic genes tend to consist of coding regions (exons)
and non-coding regions
(introns). The figure shows how such a gene leads to the
production of a protein.
Which of the following statements is true?
a. Thymine content of (1) and (2) is approximately equal. b. The
process occurring between (2) and (3) takes place in the cytosol.
c. (4) can hybridise with (2). d. The number of amino acid residues
in (5) must equal the number of nucleotide
residues in (2).
e. All processes occurring between (3) and (5) take place in the
nuclei.
1.
2.
3.
4.
5.
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10. Seeds develop from which part of the plant?
a. Flower. b. Leaf. c. Root. d. Stem.
11. When an active muscle cell experiences a shortage of oxygen
the pH changes due to the
build up of certain by-products of energy metabolism. Which of
the following correctly lists
the nature of the pH change and the major substance
responsible?
pH change substance
a. decrease carbon dioxide
b. decrease lactic acid
c. increase carbon dioxide
d. increase lactic acid
e. decrease pyruvate
12. The graphs below show the effects of temperature and pH on
enzyme activity.
Which statement explains the enzyme activity at the point
shown?
a. At P, hydrogen bonds are formed between enzyme and substrate.
b. At Q, the kinetic energy of enzyme and substrate is highest. c.
At R, peptide bonds in the enzyme begin to break. d. At S, the
substrate is completely denatured.
rate of
reaction
temperature pH
S
Q
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13. The following graph shows the effect of pH on the structure
of a protein that consists
entirely of repeating residues of a single amino acid.
Which of the following statements is true?
a. At pH2 the protein has lost its secondary structure. b. At
pH2 the protein has lost its tertiary structure. c. At pH10 the
protein has lost its primary structure. d. At pH10 the protein has
lost its secondary structure.
14. The primary reason scientists repeat the measurements they
take during experiments is
so that they can:
a. check that the equipment is working. b. list all the results
in a table. c. estimate experimental error. d. change the
experimental conditions.
15. The body can reduce local blood flow by constricting blood
vessels. This is particularly
important in
a. thermal regulation. b. preventing capillary rupture. c.
absorbing the correct amount of waste carbon dioxide. d.
lengthening the lifespan of red blood cells. e. keeping the walls
of thicker blood vessels elastic, in case of damage.
16. Which of the following characteristics of water makes life
on Earth possible?
a. It has a low specific heat capacity. b. It has a low heat of
vaporisation. c. It has a low relative surface tension. d. It is
found in all three states in the natural environment. e. The liquid
form is denser than the solid.
increasing
symmetry
pH
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17. Identical animal cells were placed in solutions of differing
water potentials. The diagram
shows the volume of the cells at the start and the end of the
experiment. Which cell was
placed in the solution with the lowest (most negative) water
potential?
A B C D
18. Aldosterone is secreted by the adrenal cortex. It is
regulated by ACTH, a hormone
secreted by the anterior pituitary gland. Normally, negative
feedback occurs where
aldosterone inhibits the secretion of ACTH.
Addisons disease occurs when the aldosterone secreting cells of
the adrenal cortex are
impaired, resulting in lowered aldosterone secretion. Which of
the following test results best
confirms the presence of Addisons disease?
a. The injection of ACTH fails to increase the secretion of
aldosterone. b. The ACTH level in the blood is low. c. The level of
aldosterone in the blood is low. d. The administration of
aldosterone alleviates symptoms. e. The removal of the adrenal
cortex worsens symptoms.
19. Insulin is an important protein hormone in the regulation of
blood glucose levels.
Insulin-dependent diabetics are unable to synthesise their own
insulin and must rely on the
biotechnology industry to produce the insulin they cannot.
Mature insulin consists of 2
polypeptide chains linked by several disulfide bonds. To
synthesise correctly assembled
insulin, the two amino acid chains are generated in separate
strains of E. coli and purified.
The purified chains are then combined under conditions favouring
disulfide formation.
Which of the following aspects of insulin structure remain
unchanged throughout this
process?
a. Primary structure. b. Tertiary structure. c. Quaternary
structure. d. All aspects of structure are altered during this
process. e. All aspects of structure remain unchanged during this
process but differ from the
structure of insulin produced in healthy individuals.
volume at start
end result
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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20. The primary production of an ecosystem is the amount of
biomass added through
photosynthesis, measured in g/m2/year.
Average Primary
Production (g/m2/yr)
Percentage of Total Primary
Production of Earth
Open ocean 125 24
Tropical rainforest 2200 22
Algal beds and reefs 2500 2
What is the most likely reason for the disparity between the
average primary production and
the percentage of the total primary production of the earth of
the different ecosystems?
a. Cyanobacteria in the ocean can fix carbon into biomass at
very high rates. b. Algae fix carbon very slowly. c. Algae fix
carbon at the highest rates. d. Algal beds make up a much lower
proportion of the total surface area of the earth
than the open ocean.
e. Phytoplankton in the ocean have a very high turnover
rate.
21. The proportion of adenine bases in a sample of DNA was found
to be 12%. Which of the
following statements is true? The proportion of:
a. uracil bases in the sample is 12%. b. uracil bases in the
sample is 88%. c. thyroxine bases in the sample is 12%. d. cytosine
bases in the sample is 38%. e. cytosine bases in the sample is
12%.
22. Consider the following diagrams:
Which of the specimens shown above is most closely related to
specimen E?
a. Specimen A. b. Specimen B. c. Specimen C. d. Specimen D.
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23. An investigator measured the amount of oxygen used by brine
shrimps living in 3% and
15% salt solutions. The results are presented in the graph
below.
The investigator also made the following observations:
Brine-shrimp living in 15% salt solution were less active than
those in 3% salt solution.
The animals grew less rapidly in the 15% salt solution. Females
living in 3% salt solution produced more eggs.
What hypothesis was being investigated?
a. Brine-shrimp have the ability to maintain a uniform internal
concentration of salt. b. The cells of the brine-shrimp performing
the work of active transport require extra
oxygen.
c. Brine-shrimp living in a 15% salt solution pump water instead
of pumping salt. d. A 15% salt solution contains less dissolved
oxygen than 3% salt solution.
24. In the 1960s the drug cholesterlower, after the usual period
of carefully monitored
clinical trials, was declared by its manufacturers to be safe
and very effective at lowering
cholesterol levels in the blood. The World Health Organisation
carried out exactly the same
kind of trial on the drug but for a much longer period than the
usual five years. The results in
1980 showed that the mortality rate from all causes was 25%
higher for those on
cholesterlower than for those who, though similar in other
respects, had not taken the drug.
Which of the following is a conclusion that can be drawn from
the above passage?
a. The five-year trial period may not be sufficient to identify
all drug side effects. b. Taking cholesterlower reduces life
expectancy by 25%. c. Cholesterlower is less effective at reducing
cholesterol levels than was at first
thought.
d. After the original trials, the manufacturers concealed the
side effects of cholesterlower.
e. The monitoring programme instigated by the World Health
Organisation was carried out efficiently.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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25. A study investigating the factors contributing to
evolutionary changes in animal size was
conducted. The mean size of all mammals in a region of North
America was calculated (using
data from the fossil record) for a period of 10 million years.
The temperature in the region
was also calculated and graphs representing the findings are
shown below.
Which of the following is the most likely explanation of the
trend evident in the graphs?
a. Larger animals can more easily avoid predation, thus having a
selective advantage during cold periods when food is scarce.
b. Larger animals produce less CO2 through respiration, which
contributes to low atmospheric CO2 levels and decreases global
temperature.
c. The cold climate stimulates animals to grow larger, a
characteristic they then pass on to their offspring.
d. Larger animals use energy to heat themselves more efficiently
and have a selective advantage during cold periods.
e. The increased surface area to volume ratio of larger mammals
gives them a selective advantage during cold periods.
26. What is the primary function of large leaves found on
seedlings growing on the forest
floor?
a. Provision of shade for their root systems. b. Elimination of
excess water that is entering via the roots. c. To allow for leaf
damage by insects. d. Acquisition of as much light as possible for
photosynthesis.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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Use the following information to answer questions 27-29.
A microbiologist was testing the effect of antibiotics on one
strain of pathogenic bacteria. She
plated out the bacteria on a suitable agar medium and placed
small discs soaked in antibiotic
solutions of equal concentration on the agar. She then incubated
the plates under matched
conditions and measured the region of no growth surrounding the
discs. The following
results were obtained.
Diameter of zone with no bacterial growth (mm)
Antibiotic
Experiment Positive control Negative control
1 8 22 2
2 19 23 0
3 15 21 1
4 9 24 3
5 10 23 2
27. Which antibiotic appears to be the most effective?
a. 1 b. 2 c. 3 d. 4 e. 5
28. What would be a suitable negative control for this
experiment?
a. Antibiotic discs but no bacteria. b. Antibiotic discs and a
different, antibiotic-resistant strain of bacteria. c. Discs soaked
in a harmless solution and the same strain of bacteria. d. Discs
soaked in a known toxic solution and the same strain of bacteria.
e. Antibiotic discs and yeast.
29. There is at least one variable that the microbiologist has
not controlled in this experiment.
What is it?
a. The amount of antibiotic applied to the agar. b. The strain
of bacteria used. c. The temperature at which the plates were
incubated. d. The rate at which the antibiotic solutions diffuse
through the agar. e. The amount of moisture available to the
bacteria.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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30. Consider the following diagram and key:
The following key has been devised for a group of toadstool
fungi.
1a Gills present under cap 2
1b Gills not present under cap 5
2a Basal cup absent 3
2b Basal cup present 4
3a Gills pink, turning brown Psalliota campestris
3b Gills not as above Lepiota gracilenta
4a Cap red with white flecks Amanita muscaria
4b Cap white Rozites australiensis
5a Annulus present Boletus elegans
5b Annulus absent Hydnum repandum
Using the above key the following specimen may be identified
as:
a. Lepiota gracilenta. b. Amanita muscaria. c. Rozites
australiensis. d. Boletus elegans. e. Hydnum repandum.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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31. The diagram shows a red blood cell and the concentrations of
ions, in mmol dm!3
, in the
plasma and the cell.
Which ions are actively transported into and out of the
cell?
into cell out of cell
a. Cl- K+
b. K+ Na+
c. Na+ Cl-
d. Na+ K+
32. Use the pedigree below showing the inheritance of a
recessive characteristic in a family.
Which one of the lists given in the answer key below contains
individuals in this pedigree who
are definitely heterozygous for the recessive
characteristic?
a. 1, 2 and 7. b. 3, 6 and 7. c. 1, 3 and 6. d. 1, 5 and 6.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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Use the following information to answer the questions 33-34.
The diagram below shows a food web. Arrows represent the
direction of energy flow and a
different letter represents each species.
Answer
key
Type of organism
a. Primary consumer
b. Primary producer
c. Tertiary consumer
d. Herbivore
e. Decomposer
33. Which answer most correctly describes the role of organisms
D ? Select your response
from the above key.
34. Which answer most correctly describes the role of organisms
Z? Select your response
from the above key.
35. When pure breeding black Andalusian chickens are crossed
with pure breeding white
Andalusian chickens the first generation offspring are all grey
in colour (known as blue
Andalusians). What is the expected phenotypic ratio when two of
these blue Andalusians
are mated and produce offspring?
a. 25% grey, 50% black, 25% white. b. 50% black, 50% white. c.
25% black, 50% grey, 25% white. d. 50% black, 50% grey.
A B
D
C
E
Z
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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36. A few shoots from the water plant, Elodea, were placed
upside down in water and
illuminated. Bubbles of constant size that were emitted from the
leaves and stem were
counted and the rate of bubble formation calculated at various
light intensities. The results are
summarised in the following figure.
The above experiment was repeated but a strong bicarbonate
solution was added to provide
an excess of CO2. The results are shown in the following
figure.
What does experiment 2 tell us about experiment 1?
a. CO2 is the limiting factor at X in experiment 1.
b. Light can be made to be limiting by decreasing other factors.
c. Temperature is limiting at Y in experiment 2. d. CO
2 is limiting at X in experiment 2.
37. A scientific study seeks to establish the optimum water
temperature for growing trout
under farming conditions. Which of the following factors is
likely to be an unavoidable
source of experimental error?
a. Length of study being curtailed by trout lifespan. b. Death
of trout at extremely high or low temperatures. c. Cost of keeping
trout tanks at different temperatures. d. Variation of optimum
temperature between individuals. e. Obtaining accurate measurements
of trout growth.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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Use the following information to answer the questions 38-39
The gag reflex is a reflex contraction of the muscles of the
throat, which stops material from
entering the throat (except in swallowing) and helps to prevent
choking. The sensory nerve in
this reflex is the glossopharyngeal nerve, and the motor nerve
to the throat muscles is the
vagus nerve. For questions 38-39, match the correct label to the
components of the gag reflex.
Each may be used once, more than once or not at all.
a. Afferent limb
b. Efferent limb
c. Integrator
d. Signal
e. Receptor
38. Glossopharyngeal nerve.
39. Pressure sensors in the throat.
40. Periodically, the sun develops relatively cool dark areas
known as sunspots. Scientists
have found that periods of high sunspot activity coincide with
stormy periods on Earth. Hence
sunspots cause storms on Earth. Which of the following is the
best statement of the flaw in
the argument above?
a. It disputes the fact that storms are the result of
low-pressure systems in the Earth's atmosphere.
b. It ignores the influence of periods of low sunspot activity
on Earth's weather systems. c. It assumes that because sunspots and
storms occur at the same time, sunspots cause
storms.
d. It overlooks the fact that there is always a storm somewhere
on Earth. e. It ignores the fact that there are stormy periods in
some areas but not in others while
there is sunspot activity.
41. A DNA segment has this nucleotide sequence:
A A G C T C T T A C G A A T A T T C
Which mRNA sequence is complementary to this sequence?
a. A A G C T C T T A C G A A T A T T C b. T T C G A G A A T G C
T T A T A A G c. A A G C U C U U A C G A A U A U U C d. U U C G A G
A A U G C U U A U A A G
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Australian Science Olympiads 2008 Biology National Qualifying
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42. The diagram below represents the relationships between
organisms in a remote pond
ecosystem.
From this information which of the following is the most likely
to be correct?
a. DDT introduced to the ecosystem would be in highest
concentration in the tissues of Detritivore 1.
b. The introduction of consumer 4 individuals from an external
population would lead to a temporary increase in numbers of
producer 2.
c. Disease in the producer 1 population would lead to an
increase in the producer 3 population.
d. Extermination of consumer 3 would cause a sustained increase
in the population of consumer 2.
e. Consumer 1 is more adaptable with regards to its food source
than consumer 3.
43. In humans, the hormone insulin is secreted by the pancreas
in response to increased blood
glucose levels. Insulin helps cells take up glucose from the
bloodstream. Type II diabetes is a
condition where body cells lose their sensitivity to
insulin.
How will a person with type II diabetes respond to a rise in
blood glucose?
Blood insulin levels
a. No change
b. Increase
c. Decrease
Producer 1 1
Consumer 1
Producer 2
Detritivore 1
Consumer 3
Consumer 2
Consumer 4
Producer 3
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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44. Different types of cells from the same organism can be
distinguished by the different
proteins they produce. Scientists often use these differences to
identify subtly different cell
types that look morphologically identical. The figure below
shows two graphs representing
the presence of particular proteins on the surface of a
heterogenous cell population isolated
from the blood. For example the graph on the left depicts three
distinct cell subsets within the
population; cells lacking both protein A and protein B
(AnegBneg), cells with protein A but not
protein B (AposBneg) and cells with both protein A and protein B
(AposBpos). The plot on the
right depicts the same cell population but is looking at the
protein markers A and C.
From the information provided which of the following is true of
this cell population?
a. There is a cell subset which is AnegBnegCneg. b. All Bneg
cells are Cpos. c. There is a cell subset positive for all three
proteins. d. All Bpos cells are Cneg. e. There are more Bneg cells
than Bpos cells.
45. Consider the following key:
1a Wings present 2
1b Wings absent Order Apterygota
2a With one pair of wings Order Diptera
2b With two pairs of wings 3
3a Front wings of coarser texture than hind wings 4
3b All wings membranous. May be hair or scale covered 8
4a Basal two-thirds of front wing thickened, remainder
membranous Order Hemiptera
4b Whole of front wing of same texture 5
5a Front wings hard and horny Order Coleoptera
5b Front wings slightly thickened with distinct veins 6
6a Mouthparts of piercing type Order Hemiptera
6b Mouthparts of biting type 7
7a Hind legs much longer than other legs Order Orthoptera
7b All legs more or less equal in length Order Blattodea
8a Wings and body completely covered by fine scales or hairs
Order Lepidoptera
8b Wings without scales or hairs 9
9a Hind and front wings linked by a row of hooks. Front of
abdomen
narrowed to form a 'waist' Order Hymenoptera
9b Wings not joined. No 'waist' Order Odonata
Aneg
Apos
Aneg
Apos
Bneg
Bpos
Cneg
Cpos
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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Lepidoptera would have the following characteristics:
a. two pairs of membranous wings covered in fine scales. b. one
pair of membranous wings lacking scales or hairs. c. two pairs of
wings, the front pair being of coarse texture. d. two pairs of
membranous wings which are not hooked together and lack hairs
or
scales.
46. The enzyme subtilisin is a protease, originally found in the
bacteria Bacillus subtilis, that
is produced in vast quantities annually for use in laundry
powders (Have you ever wondered
about the mysterious enzymes mentioned in all those laundry
powder ads?). In order to
generate washing powders that work over a range of different
conditions, subtilisin has been
the focus of many protein engineers attempting (and succeeding)
to create subtilisin variants
that have optimal activity in conditions other than those
naturally occurring in B.subtilis. The
graphs below show the activities of engineered subtilisin
variants over a range of
temperatures and pH.
Which variant would be most appropriate for use in a cold water
laundry detergent at 20oC?
a. A. b. B. c. C. d. D.
Temperature (C)
Rel
ativ
e A
ctiv
ity
pH
Rel
ativ
e A
ctiv
ity
Variant A Variant A
Variant B
Variant B
Variant C
Variant D Variant C
Variant D
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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47. When an elephant jumps off a very tall platform, it
initially gains speed rapidly. Its
acceleration decreases due to air resistance until it reaches
terminal velocity. The elephant
then falls at this velocity until it opens its parachute. The
parachute slows down the elephant
until it reaches a new steady speed, which it maintains until it
reaches the ground.
Which of these graphs shows this?
END OF SECTION A
speed
time
speed
time
speed
time
speed
time
speed
time
a. b.
c. d.
e.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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SECTION B: SHORT ANSWER QUESTIONS
USE THE ANSWER SHEET PROVIDED TO WRITE YOUR ANSWER
DO NOT USE THE ANSWER BOOKLET
48. (1 mark) What percentage of offspring from an AaBB x aaBb
cross will be either AaBb
OR aaBB? Write your answer as a percentage on the section B
answer sheet.
49. (1 mark) Penetrance refers to the proportion of individuals
of a given genotype who
display the associated phenotype. For example in a particular
plant seed coat colour can be
either brown or red. The brown phenotype is seen in all
individuals with genotypes RR or Rr
and 40% of individuals with genotype rr. The remaining 60% of
individuals with genotype rr
have the red phenotype. Red seed coat is therefore said to be
60% penetrant.
What proportion of the seeds resulting from the crossing of
genotype Rr with Rr would have
brown seed coats? Write your answer as a percentage on the
section B answer sheet.
50. (1 mark) Tinkerbell has 2 pigs. She estimates the weight of
one as being 85 kg and the
other as about 72 kg. She has ready access to a cheap source of
potatoes but wishes to feed
her pigs a 50/50 nutritional mix of potatoes and meal. Two
tables from her feed book are
reproduced below.
How much meal should Tinkerbell feed her pigs each day?
Calculate an amount in kg that
would be sufficient to feed both pigs. Write your answer on the
section B answer sheet.
State of pig Amount of
meal to feed each day(for feeding meal alone)
80 kg bacon weight
2.5 kg
60 kg -80 kg 2.0 kg
50 kg -60 kg 1.8 kg
40 kg -50 kg 1.6 kg
30 kg -40 kg 1.4 kg
20 kg -30 kg 1.0 kg
Weaners 8 10 weeks
Up to 1.0 kg
Weaners 6 weeks to weaning
250 g -350 g
Nutritional equivalents to 1 kg of meal
Carrots 8 kg
Comfrey 5 kg
Grass 5 kg
Kale 7 kg
Potatoes 5 kg
Skim milk 6 litres
Swedes 8 kg
Whey 9 -10 kg
ccccc
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
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51. (1 mark) An organism that reproduces sexually has the
genotype AaBbCcDd. How
many unique haploid gametes can this organism generate? Write
your answer as a whole
number on the section B answer sheet.
52. (1 mark) Farmer Pan is planning to use a 10 acre field to
supply winter silage for his
cattle and hay for his sheep. He reckons to get 100 bales of hay
from each acre. Before cutting
the hay he will first cut an acre round the edge of the field as
silage. This allows him to
manoeuvre his hay making machinery. After making his hay in June
he will cut the whole
field as second cut silage in August and again as third cut
silage in September. His farm
manual provides the following information.
How many bales of silage will Farmer Pan have at the end of his
harvest? Write your answer
as a whole number on the section B answer sheet.
53. (1 mark) To measure the amount of air in soil, four steps
were followed as
illustrated in the diagram below, using identical beakers
throughout.
Calculate the % (by volume) of air in the soil sample. Write
your answer as a
percentage on the section B answer sheet.
As silage As hay
First cut 7 bales per acre 100 small bales per acre
Second cut 5 bales per acre Not applicable
Third cut 4 bales per acre Not applicable
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
22
Use the following information to answer the questions 54-59.
An experiment is set up to test if chemical X is mutagenic.
Alanine is an amino acid that is essential for bacterial growth.
Many bacteria can synthesise
alanine from pyruvate as shown by the reaction below:
A particular strain of Escherichia coli has a point mutation in
the gene that codes for enzyme
Y, which makes the enzyme non-functional. This strain of E. coli
was inoculated onto agar
plates and incubated for 24 hours. Additional chemicals were
added as shown below.
Plate A Alanine and pyruvate
Plate B Pyruvate only
Plate C Pyruvate and a known mutagen
Plate D Pyruvate and chemical X
After incubation the plates were removed and inspected. The
results are represented in the
diagram below:
54. (1 mark) Which of the plates serve(s) as a positive control?
Write your answer
on the section B answer sheet.
Decide whether the following statements are true (T) or false
(F). Mark the answer sheet
either T or F.
55. (1 mark) The growth on plate A indicates that alanine is a
potent mutagen.
56. (1 mark) On plates C and D, colonies grow only when a
mutation has occurred in the
faulty gene for enzyme Y.
57. (1 mark) The difference in number of colonies on plates C
and D must be due to
differing concentrations of the mutagens.
58. (1 mark) Plate B is unnecessary for the purposes of this
experiment.
59. (1 mark) Repetition would improve the accuracy of the
results.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
23
Use the following information to answer the questions 60-64.
Below is a pedigree showing members of a particular family
affected by a disease that can
lead to kidney failure.
By examining the pedigree, determine whether the following
statements are true or false.
For questions 60-62, mark the answer sheet either T or F.
60. (1 mark) The lack of affected individuals in generations I
and III indicates the disorder
arises from a spontaneous mutation.
61. (1 mark) There is sufficient information to conclude that
the trait is X-linked.
62. (1 mark) Individual III3 is heterozygous for the trait
63. (1 mark) Assuming that the trait is X-linked, determine the
probability that individual
IV1 (sex unknown) is affected by the disorder. Write your answer
as a decimal fraction on
section B answer sheet.
64. (1 mark) Assuming that the trait is X-linked, determine the
probability that individual
IV1 (sex unknown) is a carrier of the disorder. Write your
answer as a decimal fraction on
the section B answer sheet.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
24
SECTION C: SHORT ANSWER QUESTIONS
USE THE ANSWER BOOKLET PROVIDED TO WRITE YOUR ANSWER
65. (19 marks total) During photosynthesis, plants use light
energy to incorporate carbon,
sourced from CO2, into glucose molecules. Photosynthesis can be
studied by providing
radioactive CO2 (labelled with carbon-14) to plants and
determining the amount of
radioactive carbon present in plant tissue.
A healthy plant was kept for 24 hours in an enclosed environment
at constant temperature and
humidity, with a constant supply of oxygen and radioactive CO2.
The apparatus was set up
near a window so the plant would be exposed to natural
light.
Leaf samples were taken at 3 hour intervals and the amount of
radioactivity present
determined. The results are shown below.
Time point (hours) Radioactivity in leaf tissue
(counts per minute)
0 67
3 88
6 124
9 153
12 161
15 159
18 157
21 155
24 181
a. (5 marks) Plot the data in the table above on the axes
provided.
b. (1 mark) Approximately what range of time points corresponds
to night time?
c. (3 marks) Describe and explain the shape of your graph during
the time period in part b.
d. (2 marks) The experiment is repeated using a similar plant,
the leaves of which have been
smeared with oil before the experiment starts. On your graph,
draw and label a line to show
the level of radioactivity in leaf tissue under these conditions
for the same time period.
e. (4 marks) In another experiment, radioactive water (where the
oxygen atoms were
radioactively labelled) was added to plants. No radioactivity
was detected in leaves. How
might you account for this?
In another experiment, Farmer Loo investigated the effects of
two different strains (707 and
313) of Rhizobium on the growth of a leguminous plant. The
results are given in the table
below.
Fresh mass per plant (g) Fertiliser added
(kg/ha) No Rhizobium Strain 707 Strain 313
0 7.0 12.4 9.7
20 8.4 7.7 7.7
40 9.4 6.6 8.0
f. (2 marks) Draw a bar chart to represent the data given.
g. (2 marks) Summarise the trend shown by the data.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
25
66. (6 marks total) The graph shows two methods by which
immunity can be brought about.
days since injection given
level o
f
imm
un
ity
(arb
itrary
un
its)
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80 90 100
passive
active
Safe level of immunity
Unsafe level of immunity
A safe level of immunity is given at 60 arbitrary units.
a. (2 marks) How long does it take for each method to give a
safe level of immunity?
I. Active
II. Passive
b. (2 marks) Which method becomes ineffective first and after
how long?
c. (2 marks) With reference to the data in the graph compare the
active and passive immune responses.
Lev
el o
f im
mu
nit
y
(arb
itra
ry u
nit
s)
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
26
67. (5 marks total) Flowering in flowering plants is regulated
by photoperiodism, which is
the relative lengths of daylight to darkness. Plants may be long
day plants, short day plants or
neutral to the effects of the daylength.
Consider the following information.
There is a plant pigment that exists in two forms, P660 and
P730.
P660 has a maximum light absorption around 660 nm (red light)
and P730
has a maximum light
absorption around 730nm (far red light).
Red light is absorbed by P660 which converts it to P730. Far red
light is absorbed by P730 which converts it to P660.
P730 in the dark slowly converts to P660 and it is this slow
conversion that is the clock
by which the plant measures night length.
Flowering in long day plants is stimulated only if the level of
P730
stays above a critical value.
Flowering in short day plants is stimulated only if the level of
P730
drops below a critical
value.
a. (2 marks) Draw a simple flow chart to illustrate the
interconversion between P660 and P730.
A number of Poinsettia plants were subjected to three different
patterns of illumination (blank
spaces) and darkness (black spaces). The following results were
obtained.
No flowering
Flowering
No flowering
0 12 24
Time (hrs)
b. (3 marks) Using the information above, deduce whether
Poinsettias are long day plants, short day plants or day neutral
plants. Explain your answer.
68. (4 marks) Draw up a table comparing the causes and effects
of global warming and the
depletion of the ozone layer.
69. (2 marks) Sam the scholar was doing a thought experiment one
day, idly weighing out
20mL of water for an osmosis experiment on a balance. He put his
thumb into the beaker of
water, thereby displacing the water. Never mind, he reasoned,
its not the volume thats
important; its the mass Im after. Would the mass of the water
increase, decrease or stay the
same? Explain.
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Australian Science Olympiads 2008 Biology National Qualifying
Examination
27
You have finished. Yay!
Acknowledgements:
The examiners gratefully acknowledge contributions from a
variety of sources including the
following:
Chris Maslanka, writer of the Pyrgic Puzzles column, the
Guardian newspaper for the germ
of an idea in question 69.
Integrity of the Competition
To ensure the integrity of the competition and to identify
outstanding students the
competition organisers reserve the right to re-examine or
disqualify any student or
group of students before determining a mark or award where there
is evidence of
collusion or other academic dishonesty.
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2008 National Qualifying Exam Biology
Solutions
Section A & B (Multiple Choice)
Section C We will always set a graphing question and expect
students to determine the most appropriate format, size and choice
(bar, line etc) of graph. Too few students write an appropriate
title: radioactivity vs time is not a meaningful title. The title
should reflect the content of the graph properly, describing the
relationship between the variables. 65. a). Marks are awarded for
title, correct orientation of axes, labelling of axes with units,
scale, plotting and drawing a line graph. Mean score for the whole
question: 7.9
b). Approx 12-21. Allow 10-22 c). Describe (1) level of
radioactivity increases at first, levels off for about 12 hours,
then rises again. Explain (2) relate the increased levels of
radioactivity to periods of carbon fixation and photosynthesis: the
levelling off, slight decrease to periods of no photosynthesis,
even some loss as a result of respiration.
Question Answer Question Answer Question Answer Question Answer
Q1 C Q18 A Q35 C Q52 97 Q2 A Q19 A Q36 A Q53 25 Q3 B Q20 D Q37 D
Q54 A and C Q4 A Q21 D Q38 A Q55 F Q5 A Q22 A Q39 E Q56 T Q6 A Q23
B Q40 C Q57 F Q7 C Q24 A Q41 D Q58 F Q8 B Q25 D Q42 B Q59 T Q9 C
Q26 D Q43 B Q60 F Q10 A Q27 B Q44 D Q61 F Q11 B Q28 C Q45 A Q62 T
Q12 A Q29 D Q46 D Q63 0.25 Q13 D Q30 D Q47 A Q64 0.25 Q14 C Q31 B
Q48 50 Q15 A Q32 D Q49 85 Q16 E Q33 C Q50 2.25kg Q17 A Q34 E Q51
16
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2008 National Qualifying Exam Biology
Solutions d). Drawn (1) and labelled (1) on the graph: we expect
to see no rise in radioactivity in this sample from the same
starting point. e). This was most creatively answered and suggests
that students are not well informed about the nature of
radioactivity: some of the following were common responses: it will
dissipate on its way to the leaf; radioactivity is left behind in
the soil; the Caspian strip will stop it coming into the plant;
insufficient carbon dioxide / sunlight / temperature humidity;
oxygen is not required by the plant in any form; radioactivity is
taken up by roots and not by the leaves; oxygen was already in the
plant so it didnt take up any radioactive oxygen; Examiners were
looking for a response to the information (data) given. We are not
expecting students to be well versed in the complexities of
biochemistry: we want to test understanding and reasoning. This
question does that well: it was weighted for marks for suggesting
that water is split (2) during photosynthesis and the oxygen is
released (2) as a waste product. We prefer to test reasoning and
understanding rather than knowledge directly and this is a trend
reflected in the International Biology Olympiad. f. Marks were
awarded for layout and labelling of the graph, which asks students
to process data: this is not beyond students in year 10. Graphing
seems to be an issue for many students and we would encourage some
direct teaching about the principles of graphing. For a
user-friendly guide to graphing, look at:
http://www.lmpc.edu.au/resources/Science/research_projects/graphs/graphs.htm.nsw.edu.au
g). We would expect to see some of the following: the mass produced
to increase in the absence of Rhizobium when fertiliser is added
and that when there is no Rhizobium present that adding fertiliser
will result in increased mass of plant produced. In future we will
increase the mark allocation for these types of questions in part f
and g and expect to tease more out of the data. It was surprising
how many students wrote about radioactivity in the Rhizobium
infected plants. 66. The answer booklet and the question paper were
not set out to help or mislead students and we regret any confusion
because of the structuring of the answer booklet. Examiners awarded
marks for answers in any order in part a. to minimise effects of
any confusion. Mean score: 4.3
a. active: 24 days, passive 6 days. We allowed 1 day either side
for 0.5 mark each. b. Passive after 45 days: this is a graph
reading skill c. Compare means you need to refer to both active and
passive in your answer: you can
get full marks from looking carefully at the data in the graph.
Students should be encouraged to develop this thinking skill.
Passive is fast acting compared to active; passive lasts a short
time, whereas active lasts longer and is safer for longer or words
to that effect.
No prior knowledge of immune response was required. There were
some elaborate responses to include antibiotics, antibodies (seems
to be a commonly held misconception that the 2 are almost the same
thing) secondary immune response, helper T cells, etc. This is
knowledge that we would not expect year 11 students to have.
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2008 National Qualifying Exam Biology
Solutions 67. a). Students should be able to draw a simple flow
chart given the information presented such as: red light / day
P660 dark(slow) P730
far red light b). short day (1) night must be long enough or no
flowering when short nights (1) P730 has to drop below a critical
level (1) This question tests reasoning and logic and demands no
prior knowledge. Mean score: 2.25 68. Environmental issues such as
the ozone hole and global warming are covered in lower school
science lessons across Australia. Tabulating information / data
should be within students abilities. An ideal student response
would be tabulated such as:
Causes Effects Ozone depletion results from chemical reactions
high in the upper atmosphere such as CFCs used in the manufacture
of coolants whereas global warming (or the enhanced greenhouse
effect) results from excess carbon compounds such as methane,
carbon dioxide being released in the atmosphere. Largely due to
industrialisation, combustion of fossil fuel, cattle farming, waste
products cause heat to be re-radiated back to the earth.
Increase in uv rays which allow increases in mutation / DNA
damage / eg. skin cancer whereas global warming results in
widespread environmental changes such as polar ice caps melting /
sea temperatures / floods / droughts etc
Examiners gave credit where due. The concepts were not well
understood, with a mean score of 1.7. We expected a comparison to
be made and factually correct responses to be described. We were
challenged by some students to verify the claims made (and did so)
and we will continue in the future to examine closely all student
responses. It was very disappointing that so many students are not
well schooled in these issues, particularly given the topical
debate on the environment and the urgency to promote scientific
literacy in science lessons.
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2008 National Qualifying Exam Biology
Solutions 69. We expect the level of the scale to rise, though
the mass of the water would stay the same. Look to Archimedes for
an explanation. In summary: We expect students to be able to
demonstrate the facility to list, identify, perform simple
numerical calculations, etc. through to more cognitively demanding
tasks such as explain, compare and evaluate. Some guided
preparation for students might be appropriate so they become
accustomed to thinking about and responding to a variety of
questions. We will test directly knowledge that is specified by the
curriculum bodies. This will include simple Mendelian genetics
(which should be covered by all students in year 10), scientific
processes and skills, environmental issues, nomenclature of
biological organisms, simple calculations and comprehension of
lower school science topics, many at considerable depth. The
genetics questions in the multiple choice part of the paper (32
[mean of 0.47] and 35 [mean of 0.86] suggests reasonable grasp of
genetics and the performance in questions 60-64 was highly
variable. There will inevitably be questions on topics that
teachers and students may not have covered in school lessons but
the range of marks on this paper (less than 10% to 92%) suggests
that many students are very well equipped with strong biological
knowledge and the confidence to tackle difficult questions. At the
other end of the scale, many other students experience a
challenging paper for which they seem unprepared. We strongly
discourage students in year 9 from entering and would discourage
entering whole classes. The NQEs are aimed at the most able science
students in Australia, and should not be regarded as an opportunity
to grade a whole class of students. We recommend that only the most
able students in the school attempt the NQE for marking: once the
paper has been marked, schools are free to use this and our
suggested answers in school for class teaching purposes. The
evidence of collusion (identical scripts) in some papers from some
schools suggests that some care needs to be taken when supervising
students to maintain the integrity of the paper and competition. We
welcome suggestions, your input and feedback from teachers. We are
always keen to have new question writers and teacher input to
devise fair, challenging and testing NQE papers. We owe teachers a
huge debt of thanks in managing the process at school level: in
identifying students who can benefit from the challenge of the NQE
and in supervising and supporting those most able in our
schools.