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WACE Biology ATAR — Free Online Pack 0

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WACE ATAR course written examination 2026 Edition - Pack 0 v2.0
Updated 10 Oct 2026

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WACE Biology ATAR

39 questions

100 marks attempted (210 authored marks offered)

Reading: 10 minutes · Writing: 3 hoursFree online paper · no PDF download or checkout

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Covered in this pack

  • WACE Biology ATAR Year 12 Unit 3 - Continuity of life on Earth: Gene-pool change through mutation, selection, drift and gene flow 22 marks · 3 questions
  • WACE Biology ATAR Year 12 Unit 3 - Continuity of life on Earth: Microevolution, allopatric isolation and speciation 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 3 - Continuity of life on Earth: Mutation, selection pressure, differential reproduction and allele-frequency change 4 marks · 4 questions
  • WACE Biology ATAR Year 12 Unit 3 - Continuity of life on Earth: Phylogenetic trees based on protein, genomic or anatomical information 21 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 3 - Continuity of life on Earth: Reduced genetic diversity and increased extinction risk 20 marks · 1 question
  • WACE Biology ATAR Year 12 Unit 3 - Heredity: DNA sequencing, profiling, PCR, gel electrophoresis and recombinant DNA applications 4 marks · 4 questions
  • WACE Biology ATAR Year 12 Unit 3 - Heredity: Dominance, co-dominance, incomplete dominance, sex linkage, multiple alleles and polygenic inheritance 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 3 - Heredity: Gene and chromosome mutation caused by replication, division or environmental damage 21 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 3 - Heredity: Triplet genetic code, transcription, translation and protein function 2 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 3 - Heredity: Variation arising through meiosis, fertilisation and mutation 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 3 - Science as a Human Endeavour: Biotechnology used to monitor species, assess breeding gene pools and support quarantine 20 marks · 1 question
  • WACE Biology ATAR Year 12 Unit 3 - Science as a Human Endeavour: Genetic-diversity and environmental risks of transgenic organisms, including non-target effects 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Homeostasis: Nitrogenous waste strategies of vertebrates in relation to water availability 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Homeostasis: Stimulus-response models and negative feedback 2 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 4 - Homeostasis: Structural, behavioural and physiological thermoregulation in endotherms and ectotherms 42 marks · 4 questions
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Pathogen growth, host density, transmission mode and organism movement 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Pathogen invasion and transmission between hosts 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Prescribed bacterial, fungal, protist and viral disease examples 21 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Quarantine, herd immunity, life-cycle disruption and physical prevention 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Structural distinctions among bacterial, fungal, protist and viral pathogens 21 marks · 2 questions
  • WACE Biology ATAR Year 12 Unit 4 - Infectious disease: Zoonotic transmission between vertebrate species 1 mark · 1 question
  • WACE Biology ATAR Year 12 Unit 4 - Science Inquiry Skills: Investigation, data analysis, evidence evaluation and communication for homeostasis and disease 1 mark · 1 question

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Skill Align WACE Biology ATAR Year 12 - Free Online Pack 0

A complete independently authored WACE Biology ATAR practice examination with worked answers and marking guidance.

Paper
Question and Response Book
Reading
10 minutes
Writing
3 hours
Assessment
100 marks attempted (210 authored marks offered)

Up to three calculators may be used. Calculators must not contain programs or text that could compromise the examination.

Section One: Multiple-choice

Answer all 30 questions. Select the best answer for each question. This section is worth 30% of the examination.

Question 1

1 mark
Why is DNA replication described as semi-conservative?
  1. Each daughter molecule contains one old and one new DNA strand
  2. Both parental strands remain together in one daughter molecule
  3. Only half of each chromosome is copied before cell division
  4. New DNA molecules contain alternating old and new nucleotides

Question 2

1 mark
Which event directly produces independent assortment during meiosis?
  1. Exchange of segments between non-sister chromatids
  2. Random orientation of homologous chromosome pairs at metaphase I
  3. Replication of homologous chromosomes before meiosis I
  4. Random fusion of gametes following meiosis

Question 3

1 mark
A single nucleotide is inserted near the start of a protein-coding sequence. What is the most likely consequence?
  1. The reading frame changes for codons downstream of the insertion
  2. One amino acid is replaced while the original downstream reading frame is retained
  3. One complete codon is added while downstream triplet boundaries are retained
  4. The coding sequence is shortened by one nucleotide and its reading frame is restored

Question 4

1 mark
What is produced directly by transcription of a protein-coding gene?
  1. A complementary DNA strand
  2. A chain of amino acids
  3. A complementary RNA transcript
  4. A DNA-RNA primer for replication

Question 5

1 mark
During PCR, what is the main role of primers?
  1. They separate the target strands by breaking hydrogen bonds
  2. They define the target and start DNA synthesis
  3. They supply the enzyme that extends the new DNA strands
  4. They join the amplified fragments into a single DNA molecule

Question 6

1 mark
In agarose gel electrophoresis, which DNA fragments usually migrate furthest from the wells in a fixed time?
  1. The shortest fragments
  2. The longest fragments
  3. Fragments with the most genes
  4. Fragments containing the most adenine

Question 7

1 mark
A cyclone reduces an island population to six randomly surviving individuals. Which process is most likely to cause immediate allele-frequency change?
  1. Directional selection favouring the most resistant phenotype
  2. Gene flow from a newly arriving migrant population
  3. A founder effect after colonisation of a new island
  4. Genetic drift through a population bottleneck

Question 8

1 mark
Which condition is necessary for natural selection to change a population over generations?
  1. Environmental differences produce non-heritable variation
  2. All phenotypes have similar reproductive success
  3. Heritable variation affects reproductive success
  4. Mating reshuffles existing alleles without a fitness difference

Question 9

1 mark
Two populations occupy the same region but flower in different seasons. What directly reduces gene flow between them?
  1. Geographic isolation
  2. Hybrid sterility
  3. Temporal reproductive isolation
  4. Gamete incompatibility

Question 10

1 mark
On a phylogenetic tree, which pair of taxa is interpreted as most closely related?
  1. The pair drawn closest together at the branch tips
  2. The pair with the most similar present-day body shapes
  3. The pair occupying the most similar present-day habitats
  4. The pair sharing the most recent common ancestor

Question 11

1 mark
Why can antibiotic use rapidly increase the frequency of a resistance allele in a bacterial population?
  1. Antibiotic exposure causes susceptible cells to acquire the required resistance allele
  2. Susceptible bacteria are removed while resistant bacteria survive and reproduce
  3. Resistance rises by migration even when the treated population is closed
  4. Random allele sampling alone explains the repeated survival advantage under treatment

Question 12

1 mark
A roan calf has separate red hairs and white hairs, while each parent has only one of those hair colours. Which inheritance pattern best explains the calf's coat?
  1. Incomplete dominance, because the hairs combine into one intermediate pigment
  2. Complete dominance, because one parental hair colour is absent
  3. Polygenic inheritance, because coat colour must be controlled by many loci
  4. Co-dominance, because both allele products are expressed

Question 13

1 mark
Two body cells contain the same genome but produce different proteins. What best explains this?
  1. Their ribosomes assign different amino acids to the same codon
  2. Their genomes contain different protein-coding genes despite having the same DNA
  3. Their proteins differ because transcription copies amino-acid sequences into RNA
  4. Different sets of genes are expressed in the two cell types

Question 14

1 mark
What distinguishes negative feedback in homeostasis?
  1. The response reinforces the original deviation from the reference range
  2. The response opposes the initial deviation from a set range
  3. The effector's response is independent of the detected internal condition
  4. The regulated variable tracks the external environment without correction

Question 15

1 mark
Which response increases heat loss when human core temperature rises?
  1. Vasoconstriction of skin arterioles
  2. Vasodilation of skin arterioles
  3. Rapid skeletal-muscle shivering
  4. Reduced sweat secretion

Question 16

1 mark
A pest-resistant crop reduces caterpillar damage, but larvae of a native butterfly that feed on nearby weeds also have lower survival. Which risk is shown most directly?
  1. The butterfly population is evolving resistance because its larval survival is lower
  2. A non-target species may be harmed by exposure to the introduced product
  3. Pollen-mediated gene flow is established by the decline in butterfly larval survival
  4. The introduced product affects target pests without any evidence of ecological cost

Question 17

1 mark
A forensic sample contains too little DNA at a selected locus for direct comparison. Why is PCR used before electrophoresis?
  1. To separate the DNA fragments into bands by size
  2. To identify the sample's donor without comparing DNA
  3. To make many copies of the selected DNA region
  4. To remove contaminating DNA while retaining the target

Question 18

1 mark
An endotherm's core temperature rises during exercise on a hot day. Which paired response most directly increases heat loss?
  1. Reduced skin blood flow together with shivering
  2. Greater skin blood flow together with evaporative cooling
  3. Fur erection together with increased metabolic heat production
  4. Movement into sunlight together with reduced evaporation

Question 19

1 mark
A freshwater fish releases ammonia directly across its gills. Which environmental feature makes this strategy viable?
  1. Low ammonia solubility allows nitrogen to leave as a concentrated precipitate
  2. Continuous access to water rapidly dilutes the toxic ammonia
  3. Fresh water lowers ammonia toxicity by converting it into urea outside the gills
  4. The surrounding water supplies the energy needed to synthesise uric acid

Question 20

1 mark
A drought-tolerant leaf has a thick waxy cuticle. How does this feature conserve water most directly?
  1. It reduces the leaf's exposed surface area
  2. It traps humid air within sunken stomatal pits
  3. It reduces the concentration of dissolved salts in the roots
  4. It reduces evaporation through the leaf surface

Question 21

1 mark
Tuberculosis is caused by a cellular pathogen that reproduces by binary fission and can be treated with appropriate antibiotics. Which major group is it?
  1. A virus
  2. A fungus
  3. A bacterium
  4. A protist

Question 22

1 mark
Two cattle yards use the same testing method. The crowded yard has more close contacts per animal and a larger fraction of new respiratory infections. Which mechanism is most plausible?
  1. Crowding gives each animal fewer opportunities to encounter infectious droplets
  2. Host density increases the mutation rate of the pathogen rather than host contacts
  3. The fraction infected rises solely because the crowded yard contains more animals
  4. Higher host density increases transmission opportunities

Question 23

1 mark
A nursery stops plants and soil from an affected site being moved into disease-free districts. Which transmission opportunity is reduced?
  1. Airborne spore production within the affected nursery
  2. Movement of infected hosts or contaminated material
  3. Pathogen reproduction in plants that remain at the site
  4. Local insect transmission between plants in the nursery

Question 24

1 mark
Why are antibiotics generally ineffective against viral infections?
  1. Rapid viral mutation is the main reason standard bacterial targets cannot be inhibited
  2. Viral protein coats prevent antibiotic entry into infected hosts' tissues
  3. Viruses lack the antibiotic targets found in bacteria
  4. Antibiotics must activate host recognition before acting on a pathogen

Question 25

1 mark
How can high vaccination coverage protect some susceptible people?
  1. It reduces transmission opportunities through the population
  2. Vaccinated people transfer their protection to susceptible neighbours during contact
  3. Coverage lowers disease severity without changing the chance of onward transmission
  4. Exposure to vaccinated people makes the pathogen reproduce more slowly in others

Question 26

1 mark
Which change can increase the risk of zoonotic spillover?
  1. More separation of wildlife and livestock
  2. Reduced handling of wildlife by people
  3. More wildlife-livestock-human contact
  4. Less shared use of water by wildlife and livestock

Question 27

1 mark
A plant flowers earlier in warmer years, but its specialist pollinator emergence time changes little. What is the most direct risk?
  1. Earlier flowers necessarily receive more pollinator visits
  2. Pollinator emergence must advance by the same amount
  3. The plants become reproductively isolated from each other
  4. A timing mismatch reduces pollination success

Question 28

1 mark
A marine species is increasingly recorded at higher latitudes as ocean temperatures rise. Which response is this?
  1. A seasonal change in reproductive timing
  2. An individual acclimatisation response
  3. A distributional range shift
  4. A change in local allele frequency

Question 29

1 mark
Why can greater genetic diversity increase a population's resilience to environmental change?
  1. It makes useful mutations more likely to arise in response to need
  2. It reduces variation so more individuals match the present environment
  3. It increases individual survival independently of selection pressures
  4. It increases the chance of advantageous heritable variants

Question 30

1 mark
Students compare pathogen growth at two temperatures. Which action most directly improves the validity of the comparison?
  1. Use a different growth medium at each temperature
  2. Keep inoculum size, medium and incubation time constant
  3. Measure one plate at the low temperature and ten at the high temperature
  4. Round all measurements to the nearest whole number before recording

Section Two: Short answer

Answer all five questions. Use the supplied data and biological evidence. This section is worth 50% of the examination.

Question 31

20 marks
Stimulus

A conservation team compares a 420-base-pair marker in three quokka populations. The reference ladder is in lane 1. Lane 2 contains two bands at 420 bp and 300 bp, lane 3 contains one band at 420 bp, and lane 4 contains one band at 300 bp. The 300 bp allele contains a deletion. Two lane-2 parents produce 80 offspring: 19 show only the 420 bp band, 42 show both bands and 19 show only the 300 bp band.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Diagram Preview
Lane 1Ladder Lane 2Parent Lane 3A Lane 4B wells migrationdistance
(a) 3 marks
With reference to the 420 bp and 300 bp fragments in the gel, explain why they migrate to different positions.
(b) 4 marks
Using the 420 bp and 300 bp bands in lanes 2-4 and the 19:42:19 offspring counts, identify the genotypes represented by lanes 2, 3 and 4, using L for the 420 bp allele and d for the 300 bp allele. Justify lane 2.
(c) 4 marks
Using the 420 bp and 300 bp bands in lanes 2-4 and the 19:42:19 offspring counts, calculate the expected numbers for a 1:2:1 genotype ratio among 80 offspring and compare the observations with that expectation.
(d) 4 marks
With reference to the 420 bp and 300 bp bands in lanes 2-4 and the 19:42:19 offspring counts, explain two controls or procedural checks needed before comparing allele frequencies among populations.
(e) 5 marks
For the 420 bp and 300 bp bands in lanes 2-4 and the 19:42:19 offspring counts, evaluate one benefit and one limitation of using this marker to guide conservation breeding.

Question 32

20 marks
Stimulus

Four island skink populations descend from a mainland ancestor. Molecular comparisons show that populations A and B share 98.8% sequence identity at the sampled loci, C shares 95.1% with A, and D shares 92.4% with A. A and B live on neighbouring islands but breed in different months. The tree summarises the inferred branching order.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Diagram Preview
ABCDshared ancestor
(a) 4 marks
Using the 98.8%, 95.1% and 92.4% sequence comparisons, island branching order and different breeding months, use the tree and molecular evidence to identify the closest relatives and explain the inference.
(b) 4 marks
With reference to the 98.8%, 95.1% and 92.4% sequence comparisons, island branching order and different breeding months, explain how isolation, mutation and genetic drift could produce divergence after colonisation of separate islands.
(c) 5 marks
With reference to the 98.8%, 95.1% and 92.4% sequence comparisons, island branching order and different breeding months, explain how different breeding months could contribute to speciation even if A and B later occupy the same island.
(d) 7 marks
For the 98.8%, 95.1% and 92.4% sequence comparisons, island branching order and different breeding months, propose and evaluate evidence, including an appropriate control comparison, to test whether A and B should be treated as separate biological species. Include one explained limitation of the biological species concept and explain how the proposed results would inform the species-status decision.

Question 33

20 marks
Stimulus

Lizards begin at 19 degrees C. In sun they reach 31 degrees C in 24 minutes; in shade they reach 24 degrees C. Each treatment uses twelve animals, and surface temperature is held constant within treatment.

Use the stimulus and supplied visual to answer all parts of this question.
Diagram Preview
sun or shade matched start body temperatureafter 24 minutes compare temperaturechanges
(a) 4 marks
With reference to the named stimulus, calculate and compare the temperature changes in sun and shade.
(b) 5 marks
With reference to the named stimulus, explain how shade choice contributes to behavioural thermoregulation.
(c) 5 marks
Use the temperature results to explain one benefit of shade selection. Explain one limitation of this response, identifying the condition in which that limitation matters.
(d) 6 marks
With reference to the named stimulus, improve the investigation so its conclusion is reliable and evidence-bounded.

Question 34

20 marks
Stimulus

A remote community records daily new influenza cases during an outbreak. Influenza is caused by a virus that enters respiratory epithelial cells and uses those cells to make new viral particles. Infection damages respiratory tissue, and virus-containing respiratory particles can reach another person during close contact. A rapid-testing and isolation program begins after day 4. New cases are 8, 12, 18, 25, 23, 17, 11 and 7 on days 1 to 8. Testing access also increases after day 4, and no untreated comparison community is available.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Graph Preview
day 1day 2day 3day 4day 5day 6day 7day 83020100timecases
(a) 4 marks
Using the day 1-8 case counts, the day-4 testing and isolation change and the absence of a comparison community, describe the trend before and after the program begins.
(b) 6 marks
With reference to the day 1-8 case counts, the day-4 testing and isolation change and the absence of a comparison community, explain how rapid testing and isolation could reduce transmission.
(c) 4 marks
With reference to the named influenza context, explain how the virus enters and affects a host and how it can be transmitted to another host.
(d) 6 marks
For the day 1-8 case counts, the day-4 testing and isolation change and the absence of a comparison community, design a stronger ethical evaluation using surveillance data.

Question 35

20 marks
Stimulus

A shrub occurs in fragmented populations along a 700 km rainfall gradient. Southern populations have high seedling survival during dry years but low genetic diversity. Northern populations have greater diversity but lower dry-year survival. Managers propose moving pollen from southern populations into northern populations before rainfall declines further.

Use the supplied stimulus to answer all parts of this question.
(a) 5 marks
Using the southern and northern survival and genetic-diversity results, explain the supported population-genetic conclusion and the additional evidence needed to demonstrate evolutionary adaptation.
(b) 5 marks
With reference to the 700 km rainfall gradient, contrasting survival and diversity, and the proposed south-to-north pollen transfer, explain how assisted gene flow could increase adaptive capacity.
(c) 4 marks
Using the northern and southern evidence, evaluate four distinct biological risks involving local gene combinations, offspring fitness, biosecurity and performance across rainfall conditions.
(d) 6 marks
For the 700 km rainfall gradient, contrasting survival and diversity, and the proposed south-to-north pollen transfer, design a staged test that could support a management decision.

Section Three: Extended answer

Answer two questions: one of Questions 36 or 37 for Unit 3, and one of Questions 38 or 39 for Unit 4. This section is worth 20% of the examination.

Question 36

20 marks
A sequencing study finds a one-base substitution that changes an mRNA codon and one amino acid in an enzyme. Activity is 62% of the reference form in replicated assays.
(a) 10 marks
Explain the pathway from this DNA substitution to the altered enzyme activity. Include transcription, translation, protein structure and active-site behaviour, and interpret the reported activity.
(b) 10 marks
Evaluate whether the assays establish that this substitution causes the whole-organism phenotype. Consider linked variants, assay conditions and gene-environment effects, then propose a controlled follow-up that distinguishes these explanations.

Question 37

20 marks
In insect fields containing resistance alleles R1 and R2, survival at a standard pesticide dose rises from 3% to 64% over five years. Nearby unsprayed refuges retain susceptible insects.
(a) 10 marks
Explain how new variation, natural selection, gene flow and genetic drift could change resistance frequencies. Distinguish the origin of variation from the effect of pesticide exposure.
(b) 10 marks
Assess how susceptible refuges could slow the spread of resistance. Develop a monitoring plan that distinguishes movement of resistance alleles from independent origins in different fields.

Question 38

20 marks
During a desert morning, air temperature rises from 20 to 32 degrees C. Lizards move between sun and shade; mammals show changes in skin blood flow and evaporation.
(a) 10 marks
Distinguish ectothermy from endothermy and explain how the lizards' behaviour can regulate temperature during this 12-degree rise. Include heat exchange, tolerance and the availability of suitable sites.
(b) 10 marks
Explain the mammals' physiological responses and their water or energy costs. Propose an ethical comparison using direct measurements to test the effectiveness of thermoregulation under these conditions.

Question 39

20 marks
Tuberculosis cases in a crowded settlement number 6, 11, 19 and 24 in successive months. Access to diagnosis varies, and residents move between neighbourhoods. Tuberculosis is caused by bacteria that infect the respiratory system.
(a) 10 marks
Explain the transmission and host effects of tuberculosis. Relate crowding, ventilation, delayed diagnosis and movement to the rising case counts, and state what the counts alone cannot establish.
(b) 10 marks
Justify a combined program of case detection, antibiotics and ventilation. Explain practical limitations and design a fair evaluation that accounts for access to care and population movement.

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Worked Solutions And Marking Guide

Section One: Multiple-choice Question 1

Answer: Each daughter molecule contains one old and one new DNA strand

Complementary base pairing uses each parental strand as a template, so each product conserves one original strand.

Section One: Multiple-choice Question 2

Answer: Random orientation of homologous chromosome pairs at metaphase I

Each homologous pair can orient independently at metaphase I, producing different combinations of maternal and paternal chromosomes in gametes.

Section One: Multiple-choice Question 3

Answer: The reading frame changes for codons downstream of the insertion

An insertion not divisible by three shifts the codon reading frame and can alter many downstream amino acids.

Section One: Multiple-choice Question 4

Answer: A complementary RNA transcript

RNA polymerase uses a DNA template to synthesise a complementary RNA transcript; translation produces the polypeptide later.

Section One: Multiple-choice Question 5

Answer: They define the target and start DNA synthesis

Short primers anneal to sequences flanking the target and provide a free end from which thermostable DNA polymerase extends.

Section One: Multiple-choice Question 6

Answer: The shortest fragments

Shorter DNA fragments move more readily through the gel matrix and therefore travel further.

Section One: Multiple-choice Question 7

Answer: Genetic drift through a population bottleneck

A small random survivor group may carry allele frequencies unlike the original population, producing bottleneck-driven genetic drift.

Section One: Multiple-choice Question 8

Answer: Heritable variation affects reproductive success

Selection changes allele frequencies when heritable phenotypic differences lead to consistent differences in reproductive success.

Section One: Multiple-choice Question 9

Answer: Temporal reproductive isolation

Different breeding or flowering times prevent mating or cross-pollination even without geographic separation.

Section One: Multiple-choice Question 10

Answer: The pair sharing the most recent common ancestor

Relatedness is inferred from branching order and the recency of the shared common ancestor, not page distance.

Section One: Multiple-choice Question 11

Answer: Susceptible bacteria are removed while resistant bacteria survive and reproduce

The antibiotic is a selection pressure: resistant variants have higher survival and reproductive success under treatment.

Section One: Multiple-choice Question 12

Answer: Co-dominance, because both allele products are expressed

Separate expression of both allele products is the defining evidence for co-dominance.

Section One: Multiple-choice Question 13

Answer: Different sets of genes are expressed in the two cell types

Cell differentiation depends largely on regulated gene expression, so different genes are transcribed and translated in different cell types.

Section One: Multiple-choice Question 14

Answer: The response opposes the initial deviation from a set range

Negative feedback reduces a deviation and returns the regulated variable towards its normal range.

Section One: Multiple-choice Question 15

Answer: Vasodilation of skin arterioles

Vasodilation increases warm blood flow near the skin surface, increasing heat transfer to the environment.

Section One: Multiple-choice Question 16

Answer: A non-target species may be harmed by exposure to the introduced product

The native butterfly is not the intended pest, so reduced survival is a non-target effect.

Section One: Multiple-choice Question 17

Answer: To make many copies of the selected DNA region

PCR amplifies a targeted region so that enough DNA is available for analysis.

Section One: Multiple-choice Question 18

Answer: Greater skin blood flow together with evaporative cooling

Vasodilation increases heat transfer to the body surface and evaporation removes heat.

Section One: Multiple-choice Question 19

Answer: Continuous access to water rapidly dilutes the toxic ammonia

Ammonia is highly toxic and requires substantial water for safe excretion.

Section One: Multiple-choice Question 20

Answer: It reduces evaporation through the leaf surface

A waxy cuticle provides a barrier to water diffusion from epidermal surfaces.

Section One: Multiple-choice Question 21

Answer: A bacterium

The tuberculosis pathogen is a bacterium.

Section One: Multiple-choice Question 22

Answer: Higher host density increases transmission opportunities

More close contacts can increase transmission opportunities for a respiratory pathogen.

Section One: Multiple-choice Question 23

Answer: Movement of infected hosts or contaminated material

Quarantine reduces geographic transfer; it does not guarantee eradication at the affected site.

Section One: Multiple-choice Question 24

Answer: Viruses lack the antibiotic targets found in bacteria

Common antibiotics target bacterial processes such as cell-wall synthesis or bacterial ribosomes, which viruses do not possess.

Section One: Multiple-choice Question 25

Answer: It reduces transmission opportunities through the population

When many people are immune, infectious chains are less likely to reach susceptible individuals, although protection is not absolute.

Section One: Multiple-choice Question 26

Answer: More wildlife-livestock-human contact

Habitat disruption and intensified interfaces can increase cross-species contact and opportunities for a pathogen to enter a new host population.

Section One: Multiple-choice Question 27

Answer: A timing mismatch reduces pollination success

Differing phenological responses can reduce overlap between interacting species and lower reproductive success.

Section One: Multiple-choice Question 28

Answer: A distributional range shift

A geographic change in occurrence that tracks environmental conditions is a range shift.

Section One: Multiple-choice Question 29

Answer: It increases the chance of advantageous heritable variants

More heritable variation provides more potential phenotypes on which selection can act when conditions change.

Section One: Multiple-choice Question 30

Answer: Keep inoculum size, medium and incubation time constant

Controlling other plausible causes isolates temperature as the independent variable and improves causal validity.

Section Two: Short answer Question 31

(a) Both DNA fragments are negatively charged and move towards the positive electrode. The shorter 300 bp fragment passes through the gel matrix more readily and therefore migrates farther than the 420 bp fragment.

Both DNA fragments are negatively charged and move towards the positive electrode. The shorter 300 bp fragment passes through the gel matrix more readily and therefore migrates farther than the 420 bp fragment.

(b) Lane 2 is Ld, lane 3 is LL and lane 4 is dd. Lane 2 has both allele-sized bands, so the individual is heterozygous.

Lane 2 is Ld, lane 3 is LL and lane 4 is dd. Lane 2 has both allele-sized bands, so the individual is heterozygous.

(c) Expected numbers are 20 LL, 40 Ld and 20 dd. Observed values 19, 42 and 19 are close to that expectation, with deviations of -1, +2 and -1.

Expected numbers are 20 LL, 40 Ld and 20 dd. Observed values 19, 42 and 19 are close to that expectation, with deviations of -1, +2 and -1.

(d) Any two valid checks with explanations earn full credit. For example, use the same primers and PCR conditions so amplification differences do not mimic allele-frequency differences, and include a no-template control so contaminating DNA is detected. Equivalent checks include a positive control, equal-quality DNA or standardised gel conditions with a reference ladder.

Any two valid checks with explanations earn full credit. For example, use the same primers and PCR conditions so amplification differences do not mimic allele-frequency differences, and include a no-template control so contaminating DNA is detected. Equivalent checks include a positive control, equal-quality DNA or standardised gel conditions with a reference ladder.

(e) The distinguishable 420 bp and 300 bp bands let breeders identify carriers and choose parents that retain both marker alleles. One limitation is that this single DNA region samples only a small part of the genome, so retaining its two alleles does not establish high genome-wide diversity. Alternatively, explain that the marker has no demonstrated relationship to fitness, or that it cannot establish reproductive compatibility. Use the marker alongside broader genetic or breeding evidence appropriate to the chosen limitation; only one explained limitation is required.

The distinguishable 420 bp and 300 bp bands let breeders identify carriers and choose parents that retain both marker alleles. One limitation is that this single DNA region samples only a small part of the genome, so retaining its two alleles does not establish high genome-wide diversity. Alternatively, explain that the marker has no demonstrated relationship to fitness, or that it cannot establish reproductive compatibility. Use the marker alongside broader genetic or breeding evidence appropriate to the chosen limitation; only one explained limitation is required.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.
  • For part (e), credit one benefit and its application, one limitation and its explanation, and a justified evaluation. Genome-wide diversity, fitness and reproductive compatibility are alternative limitations, not cumulative requirements.

Detailed marking criteria

Part a.1 (1 mark)

Negatively charged DNA moves towards the positive electrode.

Part a.2 (1 mark)

The 300 bp fragment is shorter than the 420 bp fragment.

Part a.3 (1 mark)

The shorter 300 bp fragment moves farther through the gel matrix.

Part b.1 (1 mark)

Lane 2 has genotype Ld.

Part b.2 (1 mark)

Lane 3 has genotype LL.

Part b.3 (1 mark)

Lane 4 has genotype dd.

Part b.4 (1 mark)

Both allele-sized bands in lane 2 show that the individual is heterozygous.

Part c.1 (1 mark)

The expected number with genotype LL is 20.

Part c.2 (1 mark)

The expected number with genotype Ld is 40.

Part c.3 (1 mark)

The expected number with genotype dd is 20.

Part c.4 (1 mark)

The observed 19, 42 and 19 are close to the expected 20, 40 and 20, or differ by -1, +2 and -1.

Part d.1 (1 mark)

Identifies a first valid control or procedural check for the DNA comparison.

Part d.2 (1 mark)

Links the first check to the particular error it detects or prevents.

Part d.3 (1 mark)

Identifies a second distinct valid control or procedural check.

Part d.4 (1 mark)

Justifies the second check by explaining its contribution to a valid comparison between populations.

Part e.1 (1 mark)

Identifies a valid marker benefit, such as retaining both marker alleles in the breeding population.

Part e.2 (1 mark)

Explains how the distinguishable bands or identified genotypes inform that breeding decision.

Part e.3 (1 mark)

Identifies one valid limitation, such as limited genomic coverage, unknown fitness effects or unknown reproductive compatibility.

Part e.4 (1 mark)

Explains why the chosen limitation restricts the conservation conclusion or breeding decision.

Part e.5 (1 mark)

Justifies combining the marker with additional evidence that addresses the chosen limitation.

Section Two: Short answer Question 32

(a) A and B are closest relatives because they share the most recent common ancestor on the tree and have the highest sampled sequence identity. The conclusion applies to the sampled loci and taxa.

A and B are closest relatives because they share the most recent common ancestor on the tree and have the highest sampled sequence identity. The conclusion applies to the sampled loci and taxa.

(b) Restricted gene flow separates allele pools. New mutations introduce variants, and random sampling in small founder populations changes allele frequencies through drift. Independent changes accumulate, increasing genetic divergence.

Restricted gene flow separates allele pools. New mutations introduce variants, and random sampling in small founder populations changes allele frequencies through drift. Independent changes accumulate, increasing genetic divergence.

(c) Different breeding months prevent many matings between A and B even if they occupy the same island. This temporal pre-fertilisation isolation reduces allele transfer between the gene pools. Selection and drift can then increase divergence; separate species form only if effective reproductive isolation becomes sufficiently strong.

Different breeding months prevent many matings between A and B even if they occupy the same island. This temporal pre-fertilisation isolation reduces allele transfer between the gene pools. Selection and drift can then increase divergence; separate species form only if effective reproductive isolation becomes sufficiently strong.

(d) Measure natural mating, fertilisation and fertile-offspring production where contact is possible, and combine this with behavioural, genomic and ecological evidence. A controlled breeding study could test compatibility but may not represent mate choice in nature. Include within-population crosses under the same conditions as between-population crosses to check that the method permits successful breeding. One limitation is that geographically isolated populations cannot be observed interbreeding naturally, so separation alone does not establish reproductive incompatibility. Alternatively, explain why interbreeding cannot assess asexual reproduction or be tested in fossils; only one explained limitation is required. Reproductive isolation despite opportunities to mate would support separate species, whereas natural interbreeding with fertile offspring would support a shared biological species.

Measure natural mating, fertilisation and fertile-offspring production where contact is possible, and combine this with behavioural, genomic and ecological evidence. A controlled breeding study could test compatibility but may not represent mate choice in nature. Include within-population crosses under the same conditions as between-population crosses to check that the method permits successful breeding. One limitation is that geographically isolated populations cannot be observed interbreeding naturally, so separation alone does not establish reproductive incompatibility. Alternatively, explain why interbreeding cannot assess asexual reproduction or be tested in fossils; only one explained limitation is required. Reproductive isolation despite opportunities to mate would support separate species, whereas natural interbreeding with fertile offspring would support a shared biological species.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.
  • For part (d), award the limitation mark for any one correctly explained limitation of the biological species concept. Additional limitations do not replace the separate marks for a justified control comparison and linking proposed evidence to the species-status decision.

Detailed marking criteria

Part a.1 (1 mark)

Populations A and B share 98.8% sequence identity at the sampled loci.

Part a.2 (1 mark)

The tree places populations A and B at the most recent shared branching point.

Part a.3 (1 mark)

The molecular and branching evidence therefore identify A and B as the closest sampled relatives.

Part a.4 (1 mark)

The relationship conclusion is limited to the sampled loci and taxa.

Part b.1 (1 mark)

Island separation restricts gene flow between the populations.

Part b.2 (1 mark)

New mutations introduce heritable variants into each isolated gene pool.

Part b.3 (1 mark)

Founder effects and genetic drift change allele frequencies by random sampling.

Part b.4 (1 mark)

Independent allele-frequency changes accumulate and increase genetic divergence.

Part c.1 (1 mark)

Different breeding months prevent many mating opportunities between populations A and B.

Part c.2 (1 mark)

The breeding-time difference is temporal pre-fertilisation reproductive isolation.

Part c.3 (1 mark)

Fewer matings reduce the transfer of alleles between the A and B gene pools.

Part c.4 (1 mark)

Selection and genetic drift can increase allele-frequency differences while gene flow remains low.

Part c.5 (1 mark)

Separate species can form only if effective reproductive isolation becomes sufficiently strong.

Part d.1 (1 mark)

Natural mating observations test whether populations A and B mate where contact occurs.

Part d.2 (1 mark)

Fertilisation and fertile-offspring measurements test reproductive compatibility.

Part d.3 (1 mark)

Behavioural, genomic and ecological evidence can corroborate the mating evidence.

Part d.4 (1 mark)

Controlled crosses may not represent natural mate choice.

Part d.5 (1 mark)

One valid limitation is explained: natural interbreeding cannot be observed between isolated populations, interbreeding does not apply to asexual organisms, or breeding cannot be tested in fossils; accept an equivalent explained limitation.

Part d.6 (1 mark)

A within-population control cross under the same conditions checks that the breeding method permits fertilisation and fertile offspring; accept another appropriate justified control comparison.

Part d.7 (1 mark)

A proposed result is linked to the decision: reproductive isolation despite mating opportunities supports separate species, or natural interbreeding with fertile offspring supports a shared biological species.

Section Two: Short answer Question 33

(a) The lizards warm by 12 degrees C in sun and by 5 degrees C in shade. The final temperature in sun is 7 degrees C higher than in shade. These results describe the treatments but do not alone prove that shade is the only cause of the difference.

The lizards warm by 12 degrees C in sun and by 5 degrees C in shade. The final temperature in sun is 7 degrees C higher than in shade. These results describe the treatments but do not alone prove that shade is the only cause of the difference.

(b) Moving into or remaining in shade is behavioural thermoregulation. Shade reduces radiant heat absorbed from direct sunlight, lowers net heat gain and limits the rise in body temperature. Moving back into sunlight when cooler could increase heat gain.

Moving into or remaining in shade is behavioural thermoregulation. Shade reduces radiant heat absorbed from direct sunlight, lowers net heat gain and limits the rise in body temperature. Moving back into sunlight when cooler could increase heat gain.

(c) The lower final temperature in shade, 24 rather than 31 degrees C after 24 minutes, supports a benefit of limiting overheating. Maintaining a viable temperature range supports enzyme and cellular function. One possible limitation is reduced feeding opportunity: if food occurs mainly in exposed areas, remaining in shade reduces access and can lower energy intake. Alternatively, explain inadequate protection when shaded sites are still too hot, or the inability to choose shade where it is absent. Only one limitation, its relevant condition and its consequence are required.

The lower final temperature in shade, 24 rather than 31 degrees C after 24 minutes, supports a benefit of limiting overheating. Maintaining a viable temperature range supports enzyme and cellular function. One possible limitation is reduced feeding opportunity: if food occurs mainly in exposed areas, remaining in shade reduces access and can lower energy intake. Alternatively, explain inadequate protection when shaded sites are still too hot, or the inability to choose shade where it is absent. Only one limitation, its relevant condition and its consequence are required.

(d) Randomly allocate similar lizards to replicated sun and shade enclosures, standardise starting temperature, surface, airflow and measurement time, and record body temperature repeatedly with calibrated equipment. Report variation among animals and restrict the conclusion to the tested species, temperatures and 24-minute period.

Randomly allocate similar lizards to replicated sun and shade enclosures, standardise starting temperature, surface, airflow and measurement time, and record body temperature repeatedly with calibrated equipment. Report variation among animals and restrict the conclusion to the tested species, temperatures and 24-minute period.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.
  • For part (c), award the three limitation-related marks for one coherent limitation, its condition and its consequence. Do not require food access, hot shade and absent shade together.

Detailed marking criteria

Part a.1 (1 mark)

Calculates a 12 degrees C rise in sun from 19 to 31 degrees C.

Part a.2 (1 mark)

Calculates a 5 degrees C rise in shade from 19 to 24 degrees C.

Part a.3 (1 mark)

Calculates a 7 degrees C final sun-to-shade difference.

Part a.4 (1 mark)

Limits the comparison to the two experimental treatments.

Part b.1 (1 mark)

Identifies movement into or remaining in shade as a behavioural response.

Part b.2 (1 mark)

Classifies shade choice as behavioural thermoregulation.

Part b.3 (1 mark)

Shade reduces radiant heat absorbed from direct sunlight.

Part b.4 (1 mark)

Links reduced radiant heat gain to a slower rise in lizard body temperature.

Part b.5 (1 mark)

Returning to sunlight can increase heat gain when the lizard is cooler.

Part c.1 (1 mark)

Links the lower temperature in shade, or the smaller temperature rise, to reduced overheating risk.

Part c.2 (1 mark)

Avoiding excessive temperature helps maintain enzyme or cellular function.

Part c.3 (1 mark)

Identifies one valid limitation of shade selection.

Part c.4 (1 mark)

Identifies a specific condition under which the chosen limitation matters.

Part c.5 (1 mark)

Explains the biological consequence of that limitation under the stated condition.

Part d.1 (1 mark)

Uses independent replicated groups of lizards in both sun and shade treatments.

Part d.2 (1 mark)

Randomly allocates similar lizards to treatments.

Part d.3 (1 mark)

Standardises starting temperature, surface conditions, airflow and exposure time.

Part d.4 (1 mark)

Records lizard body temperature repeatedly with calibrated equipment.

Part d.5 (1 mark)

Reports variation among the twelve animals in each treatment.

Part d.6 (1 mark)

Limits the conclusion to the tested species, temperatures and 24-minute exposure.

Section Two: Short answer Question 34

(a) The plotted series starts at 8 in day 1, reaches a maximum of 25 in day 4, and finishes at 7 in day 8. The final value is lower than the initial value.

The plotted series starts at 8 in day 1, reaches a maximum of 25 in day 4, and finishes at 7 in day 8. The final value is lower than the initial value.

(b) Testing identifies infected people sooner. Isolation reduces their effective contacts during the infectious period, decreasing opportunities for pathogen transfer and potentially lowering the effective reproduction number below one.

Testing identifies infected people sooner. Isolation reduces their effective contacts during the infectious period, decreasing opportunities for pathogen transfer and potentially lowering the effective reproduction number below one.

(c) Influenza virus reaches respiratory surfaces in virus-containing particles and enters susceptible respiratory epithelial cells. It uses host-cell machinery to produce new viral particles. Infection damages respiratory tissue and can impair respiratory function. New particles leave the infected person in respiratory particles and can reach another susceptible person's respiratory surfaces during close contact.

Influenza virus reaches respiratory surfaces in virus-containing particles and enters susceptible respiratory epithelial cells. It uses host-cell machinery to produce new viral particles. Infection damages respiratory tissue and can impair respiratory function. New particles leave the infected person in respiratory particles and can reach another susceptible person's respiratory surfaces during close contact.

(d) Predefine new confirmed cases per population and test positivity, use consistent case definitions, record testing volume and intervention timing, and compare matched communities or interrupted time series while adjusting for vaccination, mobility and earlier confirmed case counts. Protect privacy through de-identification and community governance. Replicate over sufficient time and report uncertainty.

Predefine new confirmed cases per population and test positivity, use consistent case definitions, record testing volume and intervention timing, and compare matched communities or interrupted time series while adjusting for vaccination, mobility and earlier confirmed case counts. Protect privacy through de-identification and community governance. Replicate over sufficient time and report uncertainty.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.

Detailed marking criteria

Part a.1 (1 mark)

Identifies initial value 8.

Part a.2 (1 mark)

Identifies maximum value 25.

Part a.3 (1 mark)

Maximum occurs at day 4.

Part a.4 (1 mark)

Identifies final value 7.

Part b.1 (1 mark)

Testing identifies infected people earlier in their infectious period.

Part b.2 (1 mark)

Earlier identification permits isolation to begin sooner.

Part b.3 (1 mark)

Isolation reduces the infected person's effective contacts.

Part b.4 (1 mark)

Fewer effective contacts reduce opportunities for pathogen transfer.

Part b.5 (1 mark)

Interrupted transmission chains lower the expected number of secondary cases.

Part b.6 (1 mark)

An effective reproduction number below one produces a declining case count.

Part c.1 (1 mark)

Influenza virus in inhaled particles enters susceptible respiratory epithelial cells.

Part c.2 (1 mark)

Infected host cells are used to produce new viral particles.

Part c.3 (1 mark)

Respiratory-cell damage can impair respiratory function.

Part c.4 (1 mark)

Respiratory particles can carry new virus to another susceptible host.

Part d.1 (1 mark)

New confirmed cases per population and test positivity are predefined as outcomes.

Part d.2 (1 mark)

Matched communities or an interrupted time series provide comparison evidence.

Part d.3 (1 mark)

Vaccination, mobility and earlier confirmed case counts are measured as potential confounders.

Part d.4 (1 mark)

The same case definition and testing rule are applied throughout surveillance.

Part d.5 (1 mark)

De-identification and community governance protect participants' privacy.

Part d.6 (1 mark)

Repeated observations over sufficient time are reported with uncertainty.

Section Two: Short answer Question 35

(a) Southern and northern populations differ in dry-year survival and genetic diversity, but the observations are population comparisons rather than reversible responses measured within individuals. An evolutionary-adaptation claim requires evidence that survival differences are inherited and cause different reproductive success across generations. Common-garden breeding evidence could test those requirements.

Southern and northern populations differ in dry-year survival and genetic diversity, but the observations are population comparisons rather than reversible responses measured within individuals. An evolutionary-adaptation claim requires evidence that survival differences are inherited and cause different reproductive success across generations. Common-garden breeding evidence could test those requirements.

(b) Pollen-mediated fertilisation can introduce southern alleles into northern offspring. Sexual reproduction and recombination place those alleles into new genetic combinations, increasing heritable drought-response variation. If the introduced variants improve reproductive success as conditions dry, natural selection can increase their frequency across generations.

Pollen-mediated fertilisation can introduce southern alleles into northern offspring. Sexual reproduction and recombination place those alleles into new genetic combinations, increasing heritable drought-response variation. If the introduced variants improve reproductive success as conditions dry, natural selection can increase their frequency across generations.

(c) Introduced alleles could disrupt locally adapted northern gene combinations, and crosses between divergent populations could reduce fertility or survival. Pollen transfer could also move pathogens, while drought-associated traits could reduce fitness under wetter conditions.

Introduced alleles could disrupt locally adapted northern gene combinations, and crosses between divergent populations could reduce fertility or survival. Pollen transfer could also move pathogens, while drought-associated traits could reduce fitness under wetter conditions.

(d) Use replicated northern controls, southern pollen sources and several crossing proportions in common gardens and reciprocal sites. Randomise blocks and standardise seed age and soil treatment. Measure germination, survival and lifetime seed production across dry and normal years, genotype offspring to confirm pollen-mediated gene flow, and apply biosecurity screening and stopping rules before broader release.

Use replicated northern controls, southern pollen sources and several crossing proportions in common gardens and reciprocal sites. Randomise blocks and standardise seed age and soil treatment. Measure germination, survival and lifetime seed production across dry and normal years, genotype offspring to confirm pollen-mediated gene flow, and apply biosecurity screening and stopping rules before broader release.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.

Detailed marking criteria

Part a.1 (1 mark)

Southern and northern populations differ in dry-year seedling survival.

Part a.2 (1 mark)

The populations also differ in measured genetic diversity.

Part a.3 (1 mark)

The population comparisons do not show whether the survival difference is inherited.

Part a.4 (1 mark)

Evolutionary adaptation requires inherited differences that affect reproductive success across generations.

Part a.5 (1 mark)

Common-garden breeding across generations could test inheritance and fitness under dry conditions.

Part b.1 (1 mark)

Pollen-mediated fertilisation introduces southern alleles into northern offspring.

Part b.2 (1 mark)

Sexual reproduction combines southern and northern alleles in the offspring.

Part b.3 (1 mark)

Recombination produces new combinations of drought-response alleles.

Part b.4 (1 mark)

The new allele combinations increase heritable variation in drought response.

Part b.5 (1 mark)

Natural selection can increase introduced variants that improve reproductive success under drying conditions.

Part c.1 (1 mark)

Introduced alleles could disrupt locally adapted northern gene combinations.

Part c.2 (1 mark)

Crosses between divergent populations could reduce offspring fertility or survival.

Part c.3 (1 mark)

Transferred pollen could introduce a plant pathogen into northern populations.

Part c.4 (1 mark)

Traits associated with dry-year survival could reduce fitness under wetter conditions.

Part d.1 (1 mark)

Replicated northern controls are compared with several southern-pollen crossing proportions.

Part d.2 (1 mark)

Blocks are randomised and seed age and soil treatment are standardised.

Part d.3 (1 mark)

Germination, survival and lifetime seed production provide distinct fitness measures.

Part d.4 (1 mark)

Reciprocal sites and dry and normal years test environmental dependence.

Part d.5 (1 mark)

Offspring genotyping confirms whether southern pollen produced the measured descendants.

Part d.6 (1 mark)

Biosecurity screening and predefined stopping rules precede any broader release.

Section Three: Extended answer Question 36

(a) The substituted DNA base changes the sequence transcribed into mRNA and hence a codon read by a ribosome. During translation, that codon specifies a different amino acid. The changed amino-acid sequence can alter the protein's three-dimensional structure and active-site shape or behaviour, affecting substrate binding or catalysis. Activity at 62% of the reference indicates partial function, not absence of the enzyme. A named R-group interaction is not needed to explain these links.

The substituted DNA base changes the sequence transcribed into mRNA and hence a codon read by a ribosome. During translation, that codon specifies a different amino acid. The changed amino-acid sequence can alter the protein's three-dimensional structure and active-site shape or behaviour, affecting substrate binding or catalysis. Activity at 62% of the reference indicates partial function, not absence of the enzyme. A named R-group interaction is not needed to explain these links.

(b) Replication strengthens the activity comparison but association alone does not establish causation for the organism. Linked variants, assay conditions and interactions between genotype and environment could contribute. Compare the sequence variants in a shared genetic background under matched conditions, including a reference form and procedural controls. Repeat the activity assays and measure the relevant whole-organism phenotype. Conclusions should remain limited to the tested genotypes and environments.

Replication strengthens the activity comparison but association alone does not establish causation for the organism. Linked variants, assay conditions and interactions between genotype and environment could contribute. Compare the sequence variants in a shared genetic background under matched conditions, including a reference form and procedural controls. Repeat the activity assays and measure the relevant whole-organism phenotype. Conclusions should remain limited to the tested genotypes and environments.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.
  • For part (a), sequence, protein structure, active-site behaviour and activity are the required links. R-group terminology or a named molecular interaction is not compulsory.

Detailed marking criteria

Part a.1 (1 mark)

Identifies the one-base DNA substitution.

Part a.2 (1 mark)

Links the substitution to transcription of altered mRNA.

Part a.3 (1 mark)

Identifies the changed mRNA codon.

Part a.4 (1 mark)

States that ribosomes read codons during translation.

Part a.5 (1 mark)

Links the changed codon to incorporation of a different amino acid.

Part a.6 (1 mark)

Links the changed amino-acid sequence to a possible change in the protein's three-dimensional structure.

Part a.7 (1 mark)

Explains how altered protein structure can change the active site's shape or behaviour.

Part a.8 (1 mark)

Links changed substrate binding or catalysis at the active site to altered enzyme activity.

Part a.9 (1 mark)

Interprets 62% activity as partial function.

Part a.10 (1 mark)

Avoids claiming that the enzyme is absent.

Part b.1 (1 mark)

Uses the replicated assays as supporting evidence.

Part b.2 (1 mark)

Distinguishes association from proof of causation.

Part b.3 (1 mark)

Identifies linked variants as an alternative.

Part b.4 (1 mark)

Identifies assay conditions as an alternative.

Part b.5 (1 mark)

Identifies gene-environment interaction as an alternative.

Part b.6 (1 mark)

Uses a shared genetic background.

Part b.7 (1 mark)

Matches environmental conditions.

Part b.8 (1 mark)

Specifies reference and procedural controls.

Part b.9 (1 mark)

Specifies repeated enzyme-activity measurements.

Part b.10 (1 mark)

Measures whole-organism phenotype before drawing a bounded conclusion.

Section Three: Extended answer Question 37

(a) Mutation can introduce heritable R1 or R2 variants, while recombination forms new allele combinations. These processes do not occur because insects need resistance. Pesticide exposure favours resistant survivors, whose reproduction increases resistance alleles in the gene pool. Migration transfers alleles between fields and chance sampling causes drift, particularly in small populations. These processes can operate together; the survival rise alone does not identify their separate contributions.

Mutation can introduce heritable R1 or R2 variants, while recombination forms new allele combinations. These processes do not occur because insects need resistance. Pesticide exposure favours resistant survivors, whose reproduction increases resistance alleles in the gene pool. Migration transfers alleles between fields and chance sampling causes drift, particularly in small populations. These processes can operate together; the survival rise alone does not identify their separate contributions.

(b) Refuges retain susceptible breeders that can interbreed with resistant insects, reducing the relative transmission of resistance alleles. Success depends on refuge size, connectivity and movement. Sample multiple fields and refuges repeatedly across generations, comparing resistance alleles and linked markers. Record pesticide dose and insect movement alongside survival. Shared resistance and surrounding marker patterns may support spread, whereas different genetic backgrounds may suggest separate origins, subject to sampling uncertainty.

Refuges retain susceptible breeders that can interbreed with resistant insects, reducing the relative transmission of resistance alleles. Success depends on refuge size, connectivity and movement. Sample multiple fields and refuges repeatedly across generations, comparing resistance alleles and linked markers. Record pesticide dose and insect movement alongside survival. Shared resistance and surrounding marker patterns may support spread, whereas different genetic backgrounds may suggest separate origins, subject to sampling uncertainty.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.

Detailed marking criteria

Part a.1 (1 mark)

Identifies mutation as a source of a new resistance allele.

Part a.2 (1 mark)

Identifies R1 and R2 as heritable resistance alleles.

Part a.3 (1 mark)

Recombination can reshuffle resistance alleles.

Part a.4 (1 mark)

States that treatment does not direct the required mutation.

Part a.5 (1 mark)

Links heritable variation to the population gene pool.

Part a.6 (1 mark)

Links pesticide exposure to differential survival.

Part a.7 (1 mark)

Links survivor reproduction to increased resistance-allele frequency.

Part a.8 (1 mark)

Explains movement of resistance alleles by gene flow.

Part a.9 (1 mark)

Random allele-frequency change by genetic drift.

Part a.10 (1 mark)

States that selection, gene flow and drift can act together.

Part b.1 (1 mark)

States that refuges retain susceptible alleles.

Part b.2 (1 mark)

Explains mating between susceptible and resistant individuals.

Part b.3 (1 mark)

Predicts dilution of resistance alleles among offspring.

Part b.4 (1 mark)

Links movement between refuge and treated areas to the outcome.

Part b.5 (1 mark)

Qualifies the prediction using refuge size or connectivity.

Part b.6 (1 mark)

Uses replicated sampling through time.

Part b.7 (1 mark)

Samples multiple treated and refuge locations.

Part b.8 (1 mark)

Compares haplotypes or linked markers around R1 and R2.

Part b.9 (1 mark)

Records treatment history and movement between sites.

Part b.10 (1 mark)

Tests shared spread against independent mutation as alternative explanations.

Section Three: Extended answer Question 38

(a) Ectotherms obtain most heat externally, whereas endotherms generate substantial metabolic heat. Both function within tolerable body-temperature ranges. As air temperature rises by 12 degrees C, basking increases a lizard's radiant heat gain, while shade reduces that input. Moving between sites can keep body temperature within a functional range, but its effectiveness depends on access to cooler and warmer microhabitats and cannot guarantee a fixed temperature.

Ectotherms obtain most heat externally, whereas endotherms generate substantial metabolic heat. Both function within tolerable body-temperature ranges. As air temperature rises by 12 degrees C, basking increases a lizard's radiant heat gain, while shade reduces that input. Moving between sites can keep body temperature within a functional range, but its effectiveness depends on access to cooler and warmer microhabitats and cannot guarantee a fixed temperature.

(b) Greater skin blood flow can transfer internal heat to the surface, and evaporation removes heat while consuming water. Metabolic heat production also has an energy cost; hot or humid conditions can limit heat loss. Use replicated animals under safe standardised heat, airflow and hydration conditions for matched durations. Record body temperature and behaviour, water loss and metabolic activity, with welfare stopping rules. Report variation and restrict conclusions to the conditions and species tested.

Greater skin blood flow can transfer internal heat to the surface, and evaporation removes heat while consuming water. Metabolic heat production also has an energy cost; hot or humid conditions can limit heat loss. Use replicated animals under safe standardised heat, airflow and hydration conditions for matched durations. Record body temperature and behaviour, water loss and metabolic activity, with welfare stopping rules. Report variation and restrict conclusions to the conditions and species tested.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.

Detailed marking criteria

Part a.1 (1 mark)

Calculates the 12 degrees C rise in air temperature.

Part a.2 (1 mark)

Identifies body temperature as the regulated condition.

Part a.3 (1 mark)

Distinguishes the lizard as an ectotherm from the mammal as an endotherm.

Part a.4 (1 mark)

Explains that both organisms exchange heat with the environment.

Part a.5 (1 mark)

Limits claims to temperatures within each organism's tolerance.

Part a.6 (1 mark)

Identifies movement between sun and shade as behavioural thermoregulation by the lizard.

Part a.7 (1 mark)

Sunlight increases radiant heat gain.

Part a.8 (1 mark)

Shade reduces radiant heat gain.

Part a.9 (1 mark)

Links changed heat gain to lizard body temperature.

Part a.10 (1 mark)

Lack of suitable shade as a limit on the behavioural response.

Part b.1 (1 mark)

Explains that increased skin blood flow transfers mammalian core heat towards the surface.

Part b.2 (1 mark)

Evaporation removes heat from the mammal.

Part b.3 (1 mark)

Identifies body-water loss as a cost of evaporative cooling.

Part b.4 (1 mark)

Identifies metabolic heat production as an endotherm energy cost.

Part b.5 (1 mark)

Hot humid conditions can limit heat loss.

Part b.6 (1 mark)

Uses replicated lizards and mammals under ethically safe temperature treatments.

Part b.7 (1 mark)

Standardises radiant heat, airflow, hydration and observation time.

Part b.8 (1 mark)

Measures body temperature and behavioural position repeatedly.

Part b.9 (1 mark)

Measures mammal water loss and metabolic rate.

Part b.10 (1 mark)

Reports variation and restricts conclusions to the tested species and 20-to-32-degree range.

Section Three: Extended answer Question 39

(a) Bacteria leave an infectious person's respiratory tract in airborne material and enter another person's respiratory tract. Infection can damage lung tissue and impair gas exchange despite host defences. Crowding increases contacts, poor ventilation allows infectious material to accumulate, delayed diagnosis prolongs opportunities for exposure and movement connects neighbourhoods. The rising counts are consistent with spread but cannot establish which factor caused it or separate transmission changes from changes in detection.

Bacteria leave an infectious person's respiratory tract in airborne material and enter another person's respiratory tract. Infection can damage lung tissue and impair gas exchange despite host defences. Crowding increases contacts, poor ventilation allows infectious material to accumulate, delayed diagnosis prolongs opportunities for exposure and movement connects neighbourhoods. The rising counts are consistent with spread but cannot establish which factor caused it or separate transmission changes from changes in detection.

(b) Early case detection enables treatment; effective antibiotics reduce viable bacteria, and ventilation reduces airborne exposure. Combining them addresses infectious sources and transmission, but access, treatment completion and resistance may limit success. Compare matched areas or a staged program using the same case definition and testing effort. Track ventilation, treatment completion and new cases alongside occupancy, movement and access. Protect privacy, involve the community and maintain care while reporting uncertainty.

Early case detection enables treatment; effective antibiotics reduce viable bacteria, and ventilation reduces airborne exposure. Combining them addresses infectious sources and transmission, but access, treatment completion and resistance may limit success. Compare matched areas or a staged program using the same case definition and testing effort. Track ventilation, treatment completion and new cases alongside occupancy, movement and access. Protect privacy, involve the community and maintain care while reporting uncertainty.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent biological terminology and logically equivalent calculations or interpretations.
  • Do not double-penalise a consequential error when later reasoning is internally consistent unless a later criterion independently tests the same concept.
  • Apply only the stated analytic criteria when a response is ambiguous; do not infer unsupported biological meaning.

Detailed marking criteria

Part a.1 (1 mark)

Classifies the tuberculosis pathogen as a bacterium.

Part a.2 (1 mark)

Describes infectious droplets leaving an infected person's respiratory tract.

Part a.3 (1 mark)

Describes inhaled droplets entering another person's respiratory tract.

Part a.4 (1 mark)

Relates bacterial growth in lung tissue to impaired gas exchange.

Part a.5 (1 mark)

Relates a local defensive response to restriction of bacterial growth.

Part a.6 (1 mark)

Links crowded housing to more frequent close respiratory contacts.

Part a.7 (1 mark)

Links poor ventilation to accumulation of airborne droplets.

Part a.8 (1 mark)

Links delayed diagnosis to a longer opportunity for transmission.

Part a.9 (1 mark)

Movement can connect infected and susceptible groups.

Part a.10 (1 mark)

Rejects attribution of the rising monthly counts to one factor without comparison evidence.

Part b.1 (1 mark)

Case detection can reduce time spent unknowingly exposing contacts.

Part b.2 (1 mark)

An appropriate completed antibiotic course can reduce viable bacteria.

Part b.3 (1 mark)

Ventilation can dilute or remove infectious droplets.

Part b.4 (1 mark)

Justifies combining strategies that act at different stages of transmission.

Part b.5 (1 mark)

Treatment access or antibiotic resistance as a limit on program effectiveness.

Part b.6 (1 mark)

Uses a matched comparison community or a staged program introduction.

Part b.7 (1 mark)

Applies the same case definition and testing schedule throughout the study.

Part b.8 (1 mark)

Measures ventilation performance, treatment completion and new-case counts separately.

Part b.9 (1 mark)

Records housing occupancy, movement and clinic-access changes during follow-up.

Part b.10 (1 mark)

Uses community governance, confidential data and guaranteed clinical care.

Diagnostic Checklist

Review focusSectionReflection and next action
Unit 3 - Continuity of life on Earth 7, 8, 9, 10, 11, 27, 28, 29, 32, 35, 37 Review evolutionary evidence, allele-frequency change, speciation, selective breeding and genetic-diversity consequences.
Unit 3 - Heredity 1, 2, 3, 4, 5, 6, 12, 13, 17, 36 Review DNA, cell division, gene expression, inheritance patterns, pedigrees and prescribed DNA technologies.
Unit 3 - Science as a Human Endeavour 16, 31 Review transgenic-organism benefits and risks, conservation biotechnology and viable-gene-pool planning.
Unit 4 - Homeostasis 14, 15, 18, 19, 20, 33, 38 Review negative feedback, thermoregulation, nitrogenous waste, animal water/salt balance, xerophytes and halophytes.
Unit 4 - Infectious disease 21, 22, 23, 24, 25, 26, 34, 39 Review pathogen groups and examples, transmission, spread, climate effects, rapid evolution and management strategies.
Unit 4 - Science Inquiry Skills 30 Practise valid investigations, data analysis, uncertainty, evidence evaluation and biological communication in Unit 4 contexts.

What is included

Question and Response Book questions (100 marks)

Worked solutions and marking guidance shown online

Diagnostic checklist shown online

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How to use this pack

  1. Attempt: Open the complete paper online and work under the timing and conditions shown on this page.
  2. Mark: Use the supplied worked solutions or response support and marking guidance to check answers and method.
  3. Review: Use the supplied diagnostic or review support to identify the next areas for revision.

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  • 2026 edition · Pack 0 · v2.0
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2026 exam timetable reference View SCSA 2026 timetable
  • Written examination Mon 9 Nov 2026 2:00 pm Upcoming7 days left

Times are Western Australian local time. Check your personalised timetable for your venue and arrangements.