2013 Senior External Examination Biology Paper Two ... · PDF file2013 Senior External...
Transcript of 2013 Senior External Examination Biology Paper Two ... · PDF file2013 Senior External...
2013 Senior External Examination
Biology Thursday 31 October 2013
Paper Two — Question and response book 1 pm to 3:10 pm
Time allowed
• Perusal time: 10 minutes
• Working time: 2 hours
Examination materials provided
• Paper Two — Question and response book
• Notepaper
Equipment/materials allowed
• QSA-approved equipment
• non-programmable calculator
• one A4 sheet of EBI question topic research notes
Directions
Do not write in this book during perusal time.
Paper Two has two parts:
• Part A — Continuity of life
• Part B — Evolution and diversity
Attempt all questions.
Suggested time allocation
• Part A: 70 minutes
• Part B: 50 minutes
Assessment
Paper Two assesses the following assessment criteria:
• Understanding biology (UB)
• Investigating biology (IB)
• Evaluating biological issues (EBI)
Assessment standards are at the end of this book.
After the examination session
The supervisor will collect this book and your research notes when you leave.
Supervisor use only
QSA use only
Supervisor’s initials
Marker number
––
Candidate use
Print your candidate number here
Attach barcode here
Number of books used
1 3
Part A — Continuity of life
Questions 1–9 assess Understanding biology (UB) and Investigating biology (IB).
Question 10 assesses Evaluating biological issues (EBI).
Attempt all questions. Write your responses in the spaces provided.
Suggested time allocation: 70 minutes.
Question 1 (UB)
State one advantage and one disadvantage of asexual reproduction.
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Question 2 (UB)
The diagram below shows a cell undergoing meiotic division.
a. Describe what is happening at this stage of the division.
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b. What is the purpose of this type of cell division (meiosis) and why is it important?
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2013 Biology — Paper Two — Question and response book 1
Question 3 (UB)
a. In mice, black-coat colour (B) is dominant to white coat-colour (b). If a heterozygous black-coat mouse is crossed with a white-coat mouse, determine the probability of producing a white-coat mouse.
b. In snapdragons, red flower colour and white flower colour show incomplete dominance.
The flower colour of a heterozygous plant is pink. What is the chance of producing a plant with red flowers when a pink-flowered plant is crossed with another pink-flowered plant?
2 2013 Biology — Paper Two — Question and response book
Question 4 (UB)
The hair of Himalayan rabbits is white except on the extremities (i.e. ears, nose, tail and feet). The Himalayan rabbit used in the experiment illustrated was homozygous for a gene controlling the synthesis of the black pigment melanin, which affects the colour of the extremities.
The hair on the back is normally white but by shaving the skin (I) and applying an ice pack during the period in which the hair regrew (II), it was found that the new hair was black (III).
Explain the role of the environment on the production of melanin in the Himalayan rabbit.
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I
skin shaved ice pack
II III
2013 Biology — Paper Two — Question and response book 3
Question 5 (UB)
The following diagram shows the process of protein synthesis.
a. Name structures 1, 2 and 3.
b. Explain the difference between transcription and translation.
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c. Label on the diagram above where transcription and translation occur.
Structure 1
Structure 2
Structure 3
1
2
3
nuclear membrane
ribosome
amino acid
tRNA
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4 2013 Biology — Paper Two — Question and response book
Question 6 (UB)
A species of small mammal has a highly visible row of hair that stands upright along the back of the animal. This characteristic, referred to as ridgeback, is determined by a recessive gene and is found in a small proportion of the natural population. In an effort to increase the frequency of this characteristic, an artificial breeding program is set up in a veterinary clinic.
Three distinct matings were set up as part of the program:
i. a heterozygous normal male with a ridgeback female
ii. a heterozygous normal male with a heterozygous normal female
iii. a pure-breeding normal male with a pure-breeding ridgeback female.
Which of the above three matings is likely to increase the number of ridgebacks in the population most quickly? Explain your response in detail, showing all working.
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2013 Biology — Paper Two — Question and response book 5
Question 7 (UB)
Phenylketonuria (PKU) is inherited as an autosomal recessive trait. It is a metabolic disorder and is characterised by the inability to produce the enzyme phenylalanine hydroxylase. The gene controlling the production of the enzyme has two allelic forms:
• enzyme produced (E)
• lack of enzyme (e).
The enzyme phenylalanine hydroxylase catalyses the following reaction:
When the enzyme is absent, very high levels of phenylalanine accumulate in the blood and other tissues.
Children born with PKU will be mentally and physically impaired unless they are placed on a special diet soon after birth. They are kept on this diet until they are about 10 years of age, after which time high blood levels of phenylalanine generally have no effect.
a. John and Betty are both able to produce the phenylalanine hydroxylase enzyme. Their son, Thomas, has PKU. Thomas was placed on a special diet shortly after birth and developed without impairment. He married Sally, who produces the enzyme and has a PKU father.
What is the chance that a child of Thomas and Sally will have PKU? Show all reasoning and working in full.
b. A female with PKU can be taken off the special diet at about 10 years of age. However, her children will be mentally impaired unless she goes back on the diet before and during pregnancy. This happens irrespective of the genotype of the child.
Explain this observation.
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phenylalanine(an amino acid)
tyrosine(an amino acid)
enzyme
6 2013 Biology — Paper Two — Question and response book
Question 8 (UB)
In wild dogs, dark coat is dominant over albino and short hair is dominant over long hair. If two independently assorting genes cause these phenotypes, give the genotypes of the parents in each of the crosses given below. Show working.
(Use symbols D and d for the dark and albino coat alleles and S and s for the short hair and long hair alleles.)
Parental phenotypesNumber of offspring
Dark, short Dark, long Albino, short Albino, long
a. Dark, short x dark, short 89 31 29 11
b. Dark, long x dark, long 0 32 0 10
2013 Biology — Paper Two — Question and response book 7
Question 9 (IB)
Outbreaks of plague locusts can be predicted by studying the effect of various physical factors on the hatching rate of their eggs.
The results of an experiment investigating one of these factors are shown below.
a. Analyse the data to draw a conclusion from this experiment. Justify your response.
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No. of eggs
Temp.(°C)
Days incubated
Soil moisture
(%)
No. of eggs hatched
20 25 10 95 5
20 25 10 90 6
35 25 10 80 19
25 25 10 70 21
20 25 10 60 19
25 25 10 50 18
43 25 10 40 22
50 25 10 20 14
60 25 10 15 11
30 25 10 0 4
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b. A second experiment was conducted to examine the effect of temperature on the hatching rate with the results shown below.
Evaluate the design of the second experiment and propose refinements.
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No. of eggsSoil moisture
(%)Temp.(°C)
No. of eggs hatched
25 95 10 2
25 95 20 4
25 95 30 3
25 95 40 5
25 95 50 4
2013 Biology — Paper Two — Question and response book 9
Question 10 assesses Evaluating biological issues (EBI). Write an extended response below.
Your response MUST refer to biological principles, concepts and ideas.
Plan your response carefully. If you write more than one draft, indicate which is to be marked.
Question 10 (EBI)
Genetically modified foods are derived from organisms that have had specific changes introduced into their DNA by genetic engineering techniques.
Identify the potential risks and benefits of genetically modified foods. Decide whether genetically modified foods should be produced in Australia. Justify your decision.
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10 2013 Biology — Paper Two — Question and response book
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End of Part A
2013 Biology — Paper Two — Question and response book 11
Part B — Evolution and diversity
Questions 1–7 assess Understanding biology (UB).
Attempt all questions. Write your responses in the spaces provided.
Suggested time allocation: 50 minutes.
Question 1 (UB)
The following table shows the common and scientific names of three organisms.
Which two are most closely related? Why?
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Question 2 (UB)
One Australian marsupial is known by all the following common names: pygmy flying possum, flying mouse, feathertail, pygmy flying phalanger, pygmy glider, feathertail glider. The same marsupial is known by the scientific name Acrobates pygmaeus.
Explain why scientific names are important.
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Common name Scientific name
Cougar Puma concolor
Tiger Panthera tigris
Leopard Panthera pardus
12 2013 Biology — Paper Two — Question and response book
Question 3 (UB)
Shown below are four homologous structures. How can they be used as evidence to support the theory of evolution?
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Question 4 (UB)
Fossils of the fish shown below have been found in rocks near the surface at three different locations. The diagram shows a cross-section of these locations.
Are these fossils about the same age or is one significantly older? Explain your response.
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Human Seal Bat Dog
Triassic
Permian
A
BC
2013 Biology — Paper Two — Question and response book 13
Question 5 (UB)
Two of Darwin’s statements are outlined briefly below.
Statement 1
Organisms reproduce at a rate that could more than double the number of offspring each generation. In natural populations, however, numbers remain fairly constant.
Statement 2
There is great variation within a species. No two individuals are exactly alike.
a. Explain Statement 1 in terms of natural selection and “survival of the fittest”.
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b. How does the variation referred to in Statement 2 occur? What is the significance of this variation in terms of the survival of a species?
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Question 6 (UB)
After many unsuccessful attempts to control the rabbit population, the Australian Government introduced a viral disease, myxomatosis, which was deadly to rabbits but not to humans. When first used, the virus killed 95% of rabbits infected. Since the mid-1950s, however, rabbit populations have increased and myxomatosis is no longer effective as a population control measure. Explain why this has happened.
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14 2013 Biology — Paper Two — Question and response book
Question 7 (UB)
The two different species pictured are burrowing animals and live most of their life underground. They are not closely related.
a. Name this type of evolution.
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b. Explain how they came to resemble each other.
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End of Part B
End of Paper Two
Australian marsupial mole(Notoryctes caurinus)
European placental mole(Talpa europaea)
2013 Biology — Paper Two — Question and response book 15
Additional pages for responses (if required)
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Part Question
16 2013 Biology — Paper Two — Question and response book
2013 Biology — Paper Two — Question and response book 17
Additional pages for responses (if required)
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Part Question
Ass
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2013 Biology — Paper Two — Question and response book 21
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