australian biology olympiad 2008

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

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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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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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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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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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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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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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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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.


15

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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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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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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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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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.


20

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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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


22


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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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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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.


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)


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.


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.

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


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.


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.


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.

australian biology olympiad 2008

Documents

marks total marks

marks section b

marks section c

answer section c

answer sheet

written answer questions

answer booklet

answer sections