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- One old and one new DNA strand in each daughter molecule
- Both parental strands remain together in one daughter molecule.
- Only half of each chromosome is copied before cell division.
- New DNA molecules contain alternating old and new nucleotides.
Question 2
1 mark- Separation of sister chromatids during mitosis
- Replication of DNA before meiosis I
- Random orientation of homologous chromosome pairs at metaphase I
- Pairing of identical alleles during fertilisation
Question 3
1 mark- Only one amino acid can change and all later codons remain unchanged.
- The chromosome number of the cell doubles.
- Every intron is translated into the protein.
- The reading frame changes for codons downstream of the insertion.
Question 4
1 mark- A chain of amino acids
- A duplicated chromosome
- A complementary RNA transcript
- A phospholipid bilayer
Question 5
1 mark- They separate DNA fragments according to length.
- They join amino acids during translation.
- Primers define the target and start DNA synthesis
- They cut DNA at specific recognition sites.
Question 6
1 mark- The longest fragments
- The shortest fragments
- Fragments with the most genes
- Fragments containing the most adenine
Question 7
1 mark- Genetic drift through a population bottleneck
- Directional selection caused by identical survival fitness
- Gene flow from the extinct individuals
- Artificial selection by the surviving organisms
Question 8
1 mark- Every individual acquires the same useful trait.
- Heritable variation affects reproductive success.
- Mutations occur because organisms need them.
- All phenotypes leave equal numbers of offspring.
Question 9
1 mark- Increased mutation rate
- Identical selection pressure
- Temporal reproductive isolation
- Polymerase chain reaction
Question 10
1 mark- The pair sharing the most recent common ancestor
- The pair drawn closest together on the page
- The pair with the longest terminal labels
- The pair living in the most similar habitat
Question 11
1 mark- Susceptible bacteria are removed while resistant bacteria survive and reproduce.
- The antibiotic directs every bacterium to make the same resistance mutation.
- Resistance alleles disappear when competition decreases.
- Bacteria stop reproducing whenever an antibiotic is present.
Question 12
1 mark- Incomplete dominance, because the hairs combine into one intermediate pigment
- Co-dominance with both allele products expressed
- Complete dominance, because one parental hair colour is absent
- Polygenic inheritance, because coat colour must be controlled by many loci
Question 13
1 mark- Each cell type permanently loses all unused genes.
- The genetic code assigns different amino acids in each tissue.
- DNA uses uracil instead of thymine in one cell type.
- Different sets of genes are expressed in the two cell types.
Question 14
1 mark- The response amplifies every change until a structure fails.
- The response opposes the initial deviation from a set range.
- The receptor and effector are always the same cell.
- The regulated variable is kept at one perfectly constant value.
Question 15
1 mark- Vasoconstriction of skin arterioles
- Rapid skeletal-muscle shivering
- Reduced sweat secretion
- Vasodilation of skin arterioles
Question 16
1 mark- The crop has lost the inserted gene because caterpillar damage fell
- Every nearby species must acquire the transgene
- The observation proves that the butterfly and caterpillar are the same species
- A non-target species may be harmed by exposure to the introduced product
Question 17
1 mark- To separate fragments by their movement through a gel
- To translate the DNA directly into a protein
- To make many copies of the selected DNA region
- To count chromosomes in every cell of the sample
Question 18
1 mark- Greater skin blood flow together with evaporative cooling
- Reduced skin blood flow together with shivering
- Fur erection together with increased metabolic heat production
- Movement into sunlight together with reduced evaporation
Question 19
1 mark- Ammonia is non-toxic and can be stored indefinitely
- The fish converts ammonia into solid uric acid at the gill surface
- Fresh water prevents nitrogen-containing molecules from diffusing
- Continuous access to water rapidly dilutes the toxic ammonia
Question 20
1 mark- It permanently closes every stomatal pore
- It increases the exposed wet surface of the leaf
- It pumps water vapour from the air into the xylem
- It reduces evaporation through the leaf surface
Question 21
1 mark- A virus
- A bacterium
- A fungus
- A protist
Question 22
1 mark- More transmission opportunities at higher host density
- Crowding changed the pathogen from a bacterium into a virus
- Every animal in the crowded yard must have been infected initially
- Host density can affect only genetic disease
Question 23
1 mark- Mutation of every pathogen already in the district
- Movement of infected hosts or contaminated material
- Inheritance of resistance alleles from parent plants
- Heat loss from plant leaves during transport
Question 24
1 mark- Viruses are always larger than bacterial cells.
- Antibiotics can act only at temperatures below body temperature.
- Viruses lack the antibiotic targets found in bacteria
- All viruses are protected by human antibodies before infection.
Question 25
1 mark- It reduces transmission opportunities through the population.
- It guarantees that no vaccinated person can carry a pathogen.
- It changes every pathogen into a harmless species.
- It replaces innate immunity in unvaccinated people.
Question 26
1 mark- Complete separation of host populations
- Elimination of all pathogen genetic variation
- More wildlife-livestock-human contact
- Identical immune histories in every host
Question 27
1 mark- A timing mismatch reduces pollination success.
- Every plant immediately becomes a new species.
- Gene expression stops in both organisms.
- The pollinator becomes an abiotic factor.
Question 28
1 mark- Primary succession on bare rock
- A distributional range shift
- Artificial gene cloning
- Loss of all homeostatic regulation
Question 29
1 mark- It ensures every individual has the same phenotype.
- A greater chance that advantageous heritable variants are present
- It prevents all mutations from occurring.
- It removes the need for reproduction.
Question 30
1 mark- Use a different growth medium at each temperature.
- Measure one plate at the low temperature and ten at the high temperature.
- Round all measurements to the nearest whole number before recording.
- Keep inoculum size, medium and incubation time constant.
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 marksA 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.
Question 32
20 marksFour 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.
Question 33
20 marksLizards 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.
Question 34
20 marksA 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.
Question 35
20 marksA 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.
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 marksUnit 3 alternative - DNA sequencing and protein function
Question 37
20 marksUnit 3 alternative B - Rapid evolution in a fragmented landscape
Question 38
20 marksUnit 4 alternative - desert thermoregulation
Question 39
20 marksA remote community records 6, 11, 19 and 24 active tuberculosis cases over four monthly surveys. Tuberculosis is caused by a bacterium that can be carried in airborne respiratory droplets. Crowded housing and limited clinic access affect exposure and the speed of diagnosis.
Worked Solutions And Marking Guide
Section One: Multiple-choice Question 1
Answer: One old and one new DNA strand in each daughter molecule
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: Primers 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 with both allele products 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: More transmission opportunities at higher host density
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: A greater chance that advantageous heritable variants are present
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) Use equal-quality DNA and the same PCR primers, reagent concentrations, cycle conditions, gel concentration, voltage, run time and reference ladder. Include positive and no-template controls so amplification failure or contamination is not mistaken for a population difference. Any two explained checks earn credit.
Use equal-quality DNA and the same PCR primers, reagent concentrations, cycle conditions, gel concentration, voltage, run time and reference ladder. Include positive and no-template controls so amplification failure or contamination is not mistaken for a population difference. Any two explained checks earn credit.
(e) The marker can identify carriers and help retain both alleles or reduce mating among close genetic profiles. However, one marker may not represent genome-wide diversity or fitness, and prioritising it could reduce other variation. Use it with multiple markers, pedigree and ecological evidence rather than as a sole decision rule.
The marker can identify carriers and help retain both alleles or reduce mating among close genetic profiles. However, one marker may not represent genome-wide diversity or fitness, and prioritising it could reduce other variation. Use it with multiple markers, pedigree and ecological evidence rather than as a sole decision rule.
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)
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)
Standardises DNA quality and PCR primers, reagents or cycle conditions.
Part d.2 (1 mark)
Standardised PCR prevents amplification differences being mistaken for allele-frequency differences.
Part d.3 (1 mark)
Standardises gel concentration, voltage, run time or the reference ladder.
Part d.4 (1 mark)
Standardised gel conditions permit valid comparison of band positions among populations.
Part e.1 (1 mark)
The marker can retain both alleles or reduce mating between similar profiles.
Part e.2 (1 mark)
Links the marker result to a specific conservation breeding decision.
Part e.3 (1 mark)
One marker does not measure genome-wide diversity.
Part e.4 (1 mark)
The marker does not measure fitness or reproductive compatibility.
Part e.5 (1 mark)
Concludes that the marker should be combined with multiple markers, pedigree and ecological evidence.
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. The biological species concept is difficult to apply to geographically isolated, asexual or fossil organisms.
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. The biological species concept is difficult to apply to geographically isolated, asexual or fossil organisms.
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)
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)
The biological species concept is difficult to apply to geographically isolated populations.
Part d.6 (1 mark)
The biological species concept cannot test asexual or fossil organisms by interbreeding.
Part d.7 (1 mark)
Converging natural and controlled evidence is required before assigning separate 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) Shade choice can limit overheating and keep body temperature within a viable range. Remaining shaded may reduce access to food, mates or basking sites, and it cannot cool the lizard below available environmental temperatures. Regulation is limited if suitable shade is absent.
Shade choice can limit overheating and keep body temperature within a viable range. Remaining shaded may reduce access to food, mates or basking sites, and it cannot cool the lizard below available environmental temperatures. Regulation is limited if suitable shade is absent.
(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.
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)
Shade choice reduces the risk of overheating.
Part c.2 (1 mark)
Links a viable body-temperature range to continued enzyme and cellular function.
Part c.3 (1 mark)
Identifies reduced access to food, mates or basking sites as a behavioural cost.
Part c.4 (1 mark)
Explains that an ectotherm cannot cool below the temperatures available in its habitat.
Part c.5 (1 mark)
Absence of suitable shade as an environmental limit.
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
Indicative answer: Identifies the one-base DNA substitution; links the substitution to transcription of altered mRNA; identifies the changed mRNA codon; states that ribosomes read codons during translation; links the changed codon to incorporation of a different amino acid. Links amino-acid sequence to R-group interactions; links R-group interactions to protein folding; explains how folding can alter an active site; interprets 62% activity as partial function; avoids claiming that the enzyme is absent. Uses the replicated assays as supporting evidence; distinguishes association from proof of causation; identifies linked variants as an alternative; identifies assay conditions as an alternative; identifies gene-environment interaction as an alternative. Uses a shared genetic background; matches environmental conditions; includes reference and procedural controls; repeats enzyme-activity measurements; measures whole-organism phenotype before drawing a bounded conclusion.
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
Integrated response (1 mark)
Identifies the one-base DNA substitution.
Integrated response (1 mark)
Links the substitution to transcription of altered mRNA.
Integrated response (1 mark)
Identifies the changed mRNA codon.
Integrated response (1 mark)
States that ribosomes read codons during translation.
Integrated response (1 mark)
Links the changed codon to incorporation of a different amino acid.
Integrated response (1 mark)
Links amino-acid sequence to R-group interactions.
Integrated response (1 mark)
Links R-group interactions to protein folding.
Integrated response (1 mark)
How folding can alter an active site.
Integrated response (1 mark)
Interprets 62% activity as partial function.
Integrated response (1 mark)
Avoids claiming that the enzyme is absent.
Integrated response (1 mark)
Uses the replicated assays as supporting evidence.
Integrated response (1 mark)
Distinguishes association from proof of causation.
Integrated response (1 mark)
Identifies linked variants as an alternative.
Integrated response (1 mark)
Identifies assay conditions as an alternative.
Integrated response (1 mark)
Identifies gene-environment interaction as an alternative.
Integrated response (1 mark)
Uses a shared genetic background.
Integrated response (1 mark)
Matches environmental conditions.
Integrated response (1 mark)
Specifies reference and procedural controls.
Integrated response (1 mark)
Specifies repeats enzyme-activity measurements.
Integrated response (1 mark)
Measures whole-organism phenotype before drawing a bounded conclusion.
Section Three: Extended answer Question 37
Indicative answer: A high-quality response treats R1 and R2 as pre-existing or independently arising heritable variants rather than directed responses. Insecticide removes susceptible insects and gives resistant carriers higher reproductive success, increasing resistance-allele frequency. Gene flow can spread alleles among fields, while small local populations may also drift; recombination changes associations with other loci. Connected untreated refuges can maintain susceptible alleles and mating, potentially diluting resistance, although they may also serve as movement corridors. A defensible study samples sites and years before and after treatment, genotypes R1, R2 and linked neutral markers, records dose and movement, uses replicated refuge-connectivity treatments or landscape comparisons, and interprets shared haplotypes as spread while distinct local backgrounds support repeated origins, with uncertainty and alternative explanations retained.
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
Integrated response (1 mark)
Identifies mutation as a source of a new resistance allele.
Integrated response (1 mark)
Identifies R1 and R2 as heritable resistance alleles.
Integrated response (1 mark)
Recombination can reshuffle resistance alleles.
Integrated response (1 mark)
States that treatment does not direct the required mutation.
Integrated response (1 mark)
Links heritable variation to the population gene pool.
Integrated response (1 mark)
Links pesticide exposure to differential survival.
Integrated response (1 mark)
Links survivor reproduction to increased resistance-allele frequency.
Integrated response (1 mark)
Explains movement of resistance alleles by gene flow.
Integrated response (1 mark)
Random allele-frequency change by genetic drift.
Integrated response (1 mark)
States that selection, gene flow and drift can act together.
Integrated response (1 mark)
States that refuges retain susceptible alleles.
Integrated response (1 mark)
Explains mating between susceptible and resistant individuals.
Integrated response (1 mark)
Predicts dilution of resistance alleles among offspring.
Integrated response (1 mark)
Links movement between refuge and treated areas to the outcome.
Integrated response (1 mark)
Qualifies the prediction using refuge size or connectivity.
Integrated response (1 mark)
Uses replicated sampling through time.
Integrated response (1 mark)
Samples multiple treated and refuge locations.
Integrated response (1 mark)
Compares haplotypes or linked markers around R1 and R2.
Integrated response (1 mark)
Records treatment history and movement between sites.
Integrated response (1 mark)
Tests shared spread against independent mutation as alternative explanations.
Section Three: Extended answer Question 38
Indicative answer: Calculates the 12 degrees C rise in air temperature; identifies body temperature as the regulated condition; distinguishes the lizard as an ectotherm from the mammal as an endotherm; explains that both organisms exchange heat with the environment; limits claims to temperatures within each organism's tolerance Identifies movement between sun and shade as behavioural thermoregulation by the lizard; explains that sunlight increases radiant heat gain; explains that shade reduces radiant heat gain; links changed heat gain to lizard body temperature; identifies lack of suitable shade as a limit on the behavioural response Explains that increased skin blood flow transfers mammalian core heat towards the surface; explains that evaporation removes heat from the mammal; identifies body-water loss as a cost of evaporative cooling; identifies metabolic heat production as an endotherm energy cost; explains that hot humid conditions can limit heat loss Uses replicated lizards and mammals under ethically safe temperature treatments; standardises radiant heat, airflow, hydration and observation time; measures body temperature and behavioural position repeatedly; measures mammal water loss and metabolic rate; reports variation and restricts conclusions to the tested species and 20-to-32-degree range
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
Integrated response (1 mark)
Calculates the 12 degrees C rise in air temperature.
Integrated response (1 mark)
Identifies body temperature as the regulated condition.
Integrated response (1 mark)
Distinguishes the lizard as an ectotherm from the mammal as an endotherm.
Integrated response (1 mark)
Explains that both organisms exchange heat with the environment.
Integrated response (1 mark)
Limits claims to temperatures within each organism's tolerance.
Integrated response (1 mark)
Identifies movement between sun and shade as behavioural thermoregulation by the lizard.
Integrated response (1 mark)
Sunlight increases radiant heat gain.
Integrated response (1 mark)
Shade reduces radiant heat gain.
Integrated response (1 mark)
Links changed heat gain to lizard body temperature.
Integrated response (1 mark)
Lack of suitable shade as a limit on the behavioural response.
Integrated response (1 mark)
Explains that increased skin blood flow transfers mammalian core heat towards the surface.
Integrated response (1 mark)
Evaporation removes heat from the mammal.
Integrated response (1 mark)
Identifies body-water loss as a cost of evaporative cooling.
Integrated response (1 mark)
Identifies metabolic heat production as an endotherm energy cost.
Integrated response (1 mark)
Hot humid conditions can limit heat loss.
Integrated response (1 mark)
Uses replicated lizards and mammals under ethically safe temperature treatments.
Integrated response (1 mark)
Standardises radiant heat, airflow, hydration and observation time.
Integrated response (1 mark)
Measures body temperature and behavioural position repeatedly.
Integrated response (1 mark)
Measures mammal water loss and metabolic rate.
Integrated response (1 mark)
Reports variation and restricts conclusions to the tested species and 20-to-32-degree range.
Section Three: Extended answer Question 39
Indicative answer: Tuberculosis is bacterial. Airborne droplets enable respiratory entry, bacterial growth damages lung tissue, barriers reduce entry, and a defensive response can restrict bacterial growth. Crowding increases close contacts, poor ventilation permits droplet accumulation, delayed diagnosis extends exposure, movement connects groups, and the monthly counts alone do not identify the dominant cause. Case detection shortens unnoticed exposure, antibiotics reduce viable bacteria when completed correctly, ventilation lowers droplet concentration, combined action interrupts different transmission stages, and resistance or access barriers limit success. A valid evaluation compares matched communities or time periods, applies consistent testing, measures mechanisms and case outcomes, records competing changes, and protects community rights.
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
Integrated response (1 mark)
Classifies the tuberculosis pathogen as a bacterium.
Integrated response (1 mark)
Describes infectious droplets leaving an infected person's respiratory tract.
Integrated response (1 mark)
Describes inhaled droplets entering another person's respiratory tract.
Integrated response (1 mark)
Relates bacterial growth in lung tissue to impaired gas exchange.
Integrated response (1 mark)
Relates a local defensive response to restriction of bacterial growth.
Integrated response (1 mark)
Links crowded housing to more frequent close respiratory contacts.
Integrated response (1 mark)
Links poor ventilation to accumulation of airborne droplets.
Integrated response (1 mark)
Links delayed diagnosis to a longer opportunity for transmission.
Integrated response (1 mark)
Movement can connect infected and susceptible groups.
Integrated response (1 mark)
Rejects attribution of the rising monthly counts to one factor without comparison evidence.
Integrated response (1 mark)
Case detection can reduce time spent unknowingly exposing contacts.
Integrated response (1 mark)
An appropriate completed antibiotic course can reduce viable bacteria.
Integrated response (1 mark)
Ventilation can dilute or remove infectious droplets.
Integrated response (1 mark)
Justifies combining strategies that act at different stages of transmission.
Integrated response (1 mark)
Treatment access or antibiotic resistance as a limit on program effectiveness.
Integrated response (1 mark)
Uses a matched comparison community or a staged program introduction.
Integrated response (1 mark)
Applies the same case definition and testing schedule throughout the study.
Integrated response (1 mark)
Measures ventilation performance, treatment completion and new-case counts separately.
Integrated response (1 mark)
Records housing occupancy, movement and clinic-access changes during follow-up.
Integrated response (1 mark)
Uses community governance, confidential data and guaranteed clinical care.
Diagnostic Checklist
| Review focus | Section | Reflection 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. |