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WACE Human 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 Human Biology ATAR

40 questions

100 marks attempted (219 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 Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-biotechnology-and-society 42 marks · 2 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-defence-and-immunity 24 marks · 5 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-endocrine-coordination 2 marks · 2 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-homeostasis 6 marks · 6 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-infection-and-transmission 2 marks · 2 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-nervous-coordination 26 marks · 3 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u3-science-inquiry-skills 18 marks · 1 question
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u4-biotechnology-and-evolutionary-evidence 29 marks · 6 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u4-evidence-and-dating 22 marks · 3 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u4-evolutionary-processes 4 marks · 4 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u4-hominin-evolution 22 marks · 3 questions
  • WACE Human Biology ATAR Year 12 Units 3 and 4 - wace-human-biology-u4-human-variation 22 marks · 3 questions

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

A complete independently authored practice examination with worked answers, analytic marking guidance and qualitative diagnostics.

Paper
Question and Response Book
Reading
10 minutes
Writing
3 hours
Assessment
179 raw marks attempted (219 raw marks offered, including optional questions). Results are weighted to a total of 100%.

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
During a reflex test at Karri Coast Clinical Research Centre, Amina withdraws a hand immediately after a painful stimulus. Which pathway correctly orders the structures involved?
  1. receptor, motor neuron, interneuron, sensory neuron, effector
  2. effector, sensory neuron, spinal cord, motor neuron, receptor
  3. receptor, sensory neuron, interneuron, motor neuron, effector
  4. receptor, sensory neuron, motor neuron, brain cortex, effector

Question 2

1 mark
A nerve-conduction study involving Amina at Karri Coast Clinical Research Centre compares equally wide myelinated and unmyelinated axons. Why is conduction usually faster in the myelinated axon?
  1. Depolarisation is regenerated mainly at nodes of Ranvier, producing saltatory conduction.
  2. Neurotransmitter diffuses along the full length of the myelinated axon.
  3. Myelin causes every part of the axon membrane to depolarise simultaneously.
  4. The myelinated axon contains more synapses between each pair of nodes.

Question 3

1 mark
A hormone is present in blood samples from many tissues collected at Karri Coast Clinical Research Centre, but only some tissues respond. What best explains this specificity?
  1. Only target cells receive blood directly from the endocrine gland.
  2. The hormone enters every cell but is destroyed in non-target cells before binding.
  3. Target cells produce the hormone and therefore recognise it more rapidly.
  4. Only target cells possess receptors with a complementary shape for that hormone.

Question 4

1 mark
Researchers at Karri Coast Clinical Research Centre apply a water-soluble peptide hormone to cultured target cells. Which event is expected first?
  1. The hormone passes through the phospholipid bilayer and binds directly to DNA.
  2. The hormone binds to a receptor in the cell-surface membrane.
  3. The hormone is converted into a steroid before entering the nucleus.
  4. The hormone binds to ribosomes and immediately changes translation.

Question 5

1 mark
During a five-day desert ecology survey, Amina's core temperature rises above its normal range. Which coordinated response would increase heat loss?
  1. cutaneous vasoconstriction and shivering
  2. cutaneous vasodilation and increased sweating
  3. decreased sweating and increased thyroxine release only
  4. reduced skin blood flow and piloerection

Question 6

1 mark
After Amina consumes a carbohydrate-rich meal during a study at Karri Coast Clinical Research Centre, blood glucose rises. Which response contributes directly to restoring the normal range?
  1. Alpha cells secrete glucagon, increasing hepatic glycogen breakdown and glucose release.
  2. Beta cells reduce insulin secretion, increasing hepatic glucose output.
  3. Beta cells secrete insulin, increasing glucose uptake and glycogen formation.
  4. Alpha cells secrete glucagon, increasing glucose uptake and glycogen formation.

Question 7

1 mark
Before breakfast, Amina's blood glucose is below the regulated range in a fasting trial at Karri Coast Clinical Research Centre. Which response is most appropriate?
  1. Pancreatic alpha cells release glucagon, promoting hepatic glycogenolysis and gluconeogenesis.
  2. Pancreatic beta cells release insulin, promoting glucose uptake and glycogenesis.
  3. Pancreatic alpha cells release glucagon, promoting hepatic glycogenesis and glucose uptake.
  4. Pancreatic beta cells reduce insulin while skeletal muscle releases glycogen-derived glucose into blood.

Question 8

1 mark
After prolonged sweating during a five-day desert ecology survey, Amina's plasma osmolarity increases. What response is expected?
  1. Less ADH is released, increasing collecting-duct permeability and concentrating urine.
  2. More aldosterone directly inserts aquaporins without changing sodium reabsorption.
  3. More ADH is released, decreasing collecting-duct permeability and increasing urine volume.
  4. More ADH is released, increasing water reabsorption in kidney collecting ducts.

Question 9

1 mark
A fluid-balance investigation at Karri Coast Clinical Research Centre records reduced blood volume in Amina. Which action of aldosterone would support recovery?
  1. It decreases distal-nephron sodium reabsorption, so more water returns to blood.
  2. It increases glomerular filtration, so sodium and water are retained in plasma.
  3. It increases sodium reabsorption in the distal nephron, so water is retained osmotically.
  4. It inserts aquaporins into collecting ducts without affecting sodium handling.

Question 10

1 mark
During a graded exercise test involving Amina, arterial carbon dioxide concentration rises. Which response helps restore the regulated range?
  1. Chemoreceptor input decreases ventilation, retaining more carbon dioxide in blood.
  2. Chemoreceptor input increases ventilation rate and depth, removing more carbon dioxide.
  3. Respiratory centres increase pulmonary blood flow without increasing alveolar ventilation.
  4. Respiratory centres increase breathing rate but reduce depth enough to lower alveolar ventilation.

Question 11

1 mark
The fictional respiratory virus Kappa is usually acquired through inhaled droplets. Which defence acts as a first-line barrier in the respiratory tract?
  1. Mucus traps particles while cilia move them deeper towards the alveoli.
  2. Respiratory epithelial cells secrete antigen-specific antibodies before exposure.
  3. Cilia destroy inhaled virus directly without moving contaminated mucus.
  4. Mucus traps particles and cilia move the mucus towards the throat.

Question 12

1 mark
A small skin abrasion sustained by Amina becomes red, warm and swollen. Which process best accounts for these signs?
  1. Histamine promotes local vasodilation and increased capillary permeability.
  2. Antibodies cause local vasoconstriction and prevent tissue-fluid movement.
  3. Memory cells lower local blood flow until a specific pathogen is identified.
  4. Interferon closes capillaries and stops phagocytes reaching the tissue.

Question 13

1 mark
During infection with respiratory virus Kappa, pyrogens reset Amina's hypothalamic temperature set point upward. Which response is expected while body temperature remains below the new set point?
  1. Shivering and cutaneous vasoconstriction
  2. Sweating and cutaneous vasodilation
  3. Shivering and cutaneous vasodilation
  4. Sweating and cutaneous vasoconstriction

Question 14

1 mark
Laboratory testing identifies respiratory virus Kappa as an RNA virus. Why would an antibiotic that blocks bacterial cell-wall synthesis not treat this infection?
  1. The viral capsid contains peptidoglycan that becomes inaccessible inside host cells.
  2. The virus synthesises peptidoglycan only while leaving the host cell.
  3. Viruses lack bacterial cell walls and reproduce using host-cell machinery.
  4. The viral envelope has the same peptidoglycan target but prevents antibiotic binding.

Question 15

1 mark
At Karri Coast Clinical Research Centre, one group receives a vaccine against respiratory virus Kappa and another receives prepared antibodies after exposure. Which comparison is correct?
  1. Vaccination provides passive immunity, whereas prepared antibodies cause long-term memory-cell formation.
  2. Vaccination induces active immunity and memory, whereas prepared antibodies provide temporary passive immunity.
  3. Both treatments rely on the recipient first producing the administered antibodies.
  4. Both treatments create the same long-lived population of memory B and T cells.

Question 16

1 mark
Months after vaccination at Karri Coast Clinical Research Centre, Amina encounters the same antigen again. Why is the secondary response usually faster?
  1. Long-lived circulating antibody alone accounts for every secondary response after antibody concentration falls.
  2. Naive lymphocytes become non-specific after first exposure and therefore bind the antigen faster.
  3. Innate phagocytes retain antigen-specific receptors and replace lymphocyte clonal expansion.
  4. Memory lymphocytes specific to the antigen can rapidly undergo clonal expansion.

Question 17

1 mark
Genetic testing at Karri Coast Clinical Research Centre detects a new mutation in one of Amina's skin cells. Under which circumstance could a new mutation be inherited by offspring?
  1. It occurs in any somatic cell and spreads through the bloodstream to the gonads.
  2. It occurs in a germline cell that contributes to a gamete involved in fertilisation.
  3. It changes a skin-cell protein after translation but leaves DNA unchanged.
  4. It is beneficial to the individual, regardless of which cell contains it.

Question 18

1 mark
Allele R is dominant to r. In a population survey, 84% of people express the dominant phenotype. Without genotype data or further assumptions, what can be concluded about the frequency of R?
  1. It cannot be determined exactly because the dominant phenotype includes both RR and Rr genotypes.
  2. It must be 0.84 because every person with the dominant phenotype has two R copies.
  3. It must exceed 0.50 because dominance makes an allele more common.
  4. It must be 0.42 because phenotype frequency is always twice allele frequency.

Question 19

1 mark
In the fictional Lake Nuru district, malaria is common and people heterozygous for a haemoglobin allele have greater survival than either homozygote. What long-term outcome is most likely if other factors remain similar?
  1. Selection must fix the protective allele because heterozygotes have the highest fitness.
  2. Selection must eliminate the allele because one homozygous genotype has lower survival.
  3. Selection can maintain both alleles because the heterozygote has the highest reproductive success.
  4. Genetic drift maintains both alleles specifically because heterozygotes leave more offspring.

Question 20

1 mark
The Beacon Island founders began when a small, non-representative group left the mainland source population. Which process best explains an initially different allele-frequency distribution?
  1. Directional mutation caused by the new environment
  2. Gene flow caused by continuous interbreeding with the source
  3. Stabilising selection that guarantees the source frequencies
  4. Founder effect, a form of genetic drift

Question 21

1 mark
Several adults move from the mainland source population to Beacon Island founders and reproduce there. Which evolutionary process is demonstrated most directly?
  1. Genetic drift, because movement itself transfers alleles before reproduction
  2. Mutation, because reproduction by migrants creates new alleles in the receiving population
  3. Natural selection, because only advantageous migrant alleles can enter the receiving gene pool
  4. Gene flow, because alleles move between populations through migration and reproduction

Question 22

1 mark
A geological barrier separates Beacon Island founders from the mainland source population for many generations, and accumulated differences eventually prevent fertile offspring. What has occurred?
  1. Geographic isolation without speciation because fertile offspring are still possible
  2. Speciation following reproductive isolation
  3. Gene flow maintaining one interbreeding population
  4. Temporary phenotypic acclimatisation with unchanged reproductive compatibility

Question 23

1 mark
Researchers comparing DNA from Beacon Island founders and the mainland source population use PCR before sequencing. What is PCR used to do?
  1. Produce many copies of a selected DNA region
  2. Separate DNA fragments according to size
  3. Determine the nucleotide order of the selected region directly
  4. Cut the selected region at restriction-enzyme recognition sites

Question 24

1 mark
After PCR at Karri Coast Clinical Research Centre, negatively charged DNA fragments are separated by gel electrophoresis. Which fragments generally migrate furthest through the gel?
  1. The largest fragments
  2. The fragments with the greatest DNA concentration
  3. The smallest fragments
  4. The fragments with the most restriction sites regardless of length

Question 25

1 mark
A bioinformatics comparison finds fewer nucleotide differences between Beacon Island founders and the mainland source population than between either group and a distant population. Which inference is best supported?
  1. The founder and source populations share a more recent common ancestry, assuming comparable sequences and mutation patterns.
  2. The founder population is biologically more advanced than the distant population.
  3. The compared sequence proves the exact migration date without any calibration.
  4. All individuals in the founder and source populations must have identical genomes.

Question 26

1 mark
At Rift Site K, an undisturbed sediment layer containing a hominin fossil lies below a layer containing a well-characterised index fossil. What can stratigraphy establish most directly?
  1. The hominin fossil is exactly 1.4 million years old.
  2. The two layers have the same relative age because both contain fossils.
  3. The lower layer formed after the upper layer because deeper sediment is always younger.
  4. The hominin-bearing layer is relatively older than the layer above it.

Question 27

1 mark
Volcanic ash bracketing a fossil at Rift Site K is estimated to be about 1.4 million years old. Which method is most suitable for obtaining an absolute age for the ash?
  1. Radiocarbon dating of the mineral crystals
  2. Comparison of tooth shape only
  3. Potassium-argon dating
  4. Stratigraphic superposition without isotope measurement

Question 28

1 mark
A pelvis and femur recovered from Rift Site K are being assessed for habitual bipedalism. Which combination provides the strongest support?
  1. A long narrow pelvis and a divergent grasping big toe
  2. A short broad pelvis and a femur angled inward towards the knee
  3. Very long forelimbs and curved finger bones only
  4. A foramen magnum at the rear of the skull and no lumbar curve

Question 29

1 mark
Archaeologists at Rift Site K identify an Acheulian tool assemblage. Which hominin is most strongly associated with this culture in the Year 12 Human Biology course?
  1. Homo habilis
  2. Australopithecus afarensis
  3. Homo erectus
  4. Homo neanderthalensis

Question 30

1 mark
Researchers reconstructing a maternal migration pathway from the mainland source population to Beacon Island founders choose mitochondrial DNA. Which property makes it useful for this purpose?
  1. It is inherited mainly through the maternal line and normally does not undergo meiotic recombination.
  2. It is inherited equally from both parents and recombines like an autosomal chromosome.
  3. It follows only the paternal line because sperm contribute most embryonic mitochondria.
  4. It contains the complete nuclear genome, allowing every ancestral lineage to be traced as one unit.

Section Two: Short answer

Answer all six questions. Show working where a calculation is required and use the supplied biological evidence. This section is worth 50% of the examination.

Question 31

18 marks
Stimulus

A classroom investigation models a contact route for a fictional virus V that can infect nasal mucosa after viable particles are deposited there. A harmless fluorescent tracer is used instead of virus. Each trial starts with 80 tracer units on a door handle. A clean gloved hand touches the handle and then an artificial mucosal target; no person touches their face. The handle is cleaned after tracer application, and the gloved hand is washed after touching the handle, only in the assigned conditions. Contact time and pressure are kept constant. Each condition has 12 independently reset trials with fresh materials.

Candidate data (condition: mean tracer units recovered from the target): neither intervention: 40; hand wash only: 10; handle cleaning only: 20; both interventions: 4. The graph displays these means. No viable-virus counts or human infection outcomes are measured.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Graph Preview
40Neither10Hand wash20Handle clean4Bothmean tracer reaching target (units)
(a) 4 marks
Calculate the percentage of the initial tracer that reaches the target under each of the four conditions. Show working.
(b) 4 marks
Explain the possible infection route represented by the model, including why material on intact skin does not by itself establish infection.
(c) 5 marks
Compare the intervention results and evaluate the claim that combining the interventions eliminates infection and reduces infection risk by exactly 90%.
(d) 5 marks
Propose a follow-up that tests whether the interventions reduce transfer of viable microorganisms. Explain how it would provide a fair, reliable comparison without deliberately exposing people to infection.

Question 32

22 marks
Stimulus

A team at Karri Coast Clinical Research Centre is comparing injected recombinant insulin with an experimental somatic gene therapy for adults whose pancreatic beta cells produce insufficient insulin. The gene therapy uses a modified, non-replicating vector to deliver a functional insulin-related coding sequence to selected somatic cells. The diagram is a generic negative-feedback scaffold; labels A, B and C must be interpreted in the low-blood-glucose context specified in Part (b), not as a diagram of the gene therapy.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Diagram Preview
A: stimulus B: controlresponse C: outcome opposes change
(a) 2 marks
Define a target cell and explain why insulin does not produce the same response in every cell.
(b) 5 marks
Use the scaffold to explain the negative-feedback response to low blood glucose, identifying an appropriate stimulus, control response and outcome.
(c) 4 marks
Contrast the action of water-soluble and lipid-soluble hormones on target cells.
(d) 7 marks
Outline how recombinant DNA technology can be used to produce human insulin in bacteria.
(e) 4 marks
Evaluate one potential benefit and one limitation of the proposed somatic gene therapy compared with repeated insulin injections.

Question 33

20 marks
Stimulus

Two fictional sites contain older and younger sediment deposits. At each site the older deposit lies below the younger one and the layers are undisturbed. The same sieving and identification procedure counts distinct hominin skeletal specimens; the count is not a count of individuals.

Candidate data (deposit: excavated volume; specimens recovered): A older: 10 m³; 20 specimens. A younger: 1 m³; 4 specimens. B older: 6 m³; 12 specimens. B younger: 2 m³; 8 specimens. The graph shows the specimen counts. Older deposits show rapid burial in low-oxygen conditions; younger deposits show weathering and greater bone fragmentation.

The older deposit at each site also contains index-fossil species I. Independent reference records show that I was widespread and existed from 2.10 to 2.00 million years ago. These index fossils have not been reworked. A report claims that fewer specimens in each younger deposit demonstrate a decline in the living hominin population.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Graph Preview
20A older4A younger12B older8B youngeridentified hominin specimens recovered
(a) 4 marks
Calculate the number of recovered hominin specimens per cubic metre for each deposit. Show working and units.
(b) 6 marks
Evaluate the population-decline claim using your calculations, sampling volumes and the supplied preservation evidence.
(c) 5 marks
Use the supplied index-fossil information to assess the proposed correlation of the two older deposits. Explain what it establishes about age and what it cannot establish about precise timing or evolutionary ancestry.
(d) 5 marks
Design a field comparison to test whether the recovery-density change reflects a biological population change. Justify the sampling and evidence needed.

Question 34

24 marks
Stimulus

A fictional neurological assessment compares a control participant with P, who has localised damage to a motor area of the cerebrum, and Q, who has localised cerebellar damage. All understand the instructions, recognise touch and pain, and have intact peripheral sensory and motor nerves. Direct motor-nerve stimulation produces normal skeletal-muscle contraction. Each attempts ten instructed reaches to a visible target; success means ending within 1 cm of it within three seconds.

Candidate data (participant: successful reaches out of ten): control: 9; P: 0; Q: 3. P fails to initiate the instructed movement, while Q initiates movement but frequently overshoots the target. P still withdraws the hand in a spinal reflex after a suitable painful stimulus and consciously reports the pain.

A separate simplified signalling model delivers a cue only to the left cerebral hemisphere. That hemisphere supplies motor commands to the right hand, while the right hemisphere supplies commands to the left hand. The corpus callosum is the only route represented for cue transfer between hemispheres. Both hands respond normally when their controlling hemisphere receives a cue directly.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Graph Preview
9Control0P3Qsuccessful reaches (out of ten)
(a) 6 marks
Use the graph and behavioural observations to explain how P's and Q's movement impairments differ from the control and from each other.
(b) 6 marks
Explain why P's spinal withdrawal and conscious pain report can persist despite loss of the instructed movement. Predict withdrawal and awareness if only the ascending pain pathway to the cerebrum is blocked and all local spinal connections remain intact.
(c) 6 marks
Identify three structures or materials that protect the brain and explain one protective contribution of each.
(d) 6 marks
In the supplied signalling model, predict and explain the two hand responses if the corpus-callosum route is blocked. Explain why directly cueing each hemisphere is useful for interpreting the result.

Question 35

13 marks
Stimulus

A sample of 100 adults from Beacon Island founders contained 36 HH, 48 Hh and 16 hh genotypes at one marker locus. Researchers amplified the locus with PCR and separated restriction fragments by gel electrophoresis. Lane 1 is a size reference; Lanes 2, 3 and 4 contain pooled profiles from the mainland source population, Beacon Island founders and a distant comparison population respectively. Matching band positions represent fragments of the same measured length, but this single marker does not represent the complete genome.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Diagram Preview
L1 size ref L2 source L3 founders L4 comparison wells migrationdistance
(a) 3 marks
Calculate the frequency of allele h in the sampled gene pool. Show working.
(b) 4 marks
Explain how mutation, natural selection, gene flow and genetic drift can each affect variation or allele frequency in a human population.
(c) 3 marks
Explain why the marker locus was amplified before gel electrophoresis and how the PCR product makes the displayed comparison possible.
(d) 3 marks
Interpret the gel comparison between the mainland source population and Beacon Island founders, and state one caution.

Question 36

12 marks
Stimulus

Illustrative coding-strand DNA sequences from one homologous nine-base region are shown below. They are teaching data, not measurements of complete primate genomes. Read every coding strand 5′ to 3′, with the reading frame starting at the first base. The supplied codon table gives the corresponding mRNA codons and amino acids.

Human: `GAA TTT GGC`

Chimpanzee: `GAG TTC GGT`

Gorilla: `GAG TTT GGA`

No mutation rate or dating calibration is supplied. A report claims that identical amino-acid sequences prove identical genomes, and that the DNA differences alone establish an exact divergence age of ten million years.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
Diagram Preview
mRNA codon reference (read5' → 3')GAAGluGAGGluUUUPheUUCPheGGCGlyGGUGlyGGAGly
(a) 3 marks
Compare each non-human DNA sequence with the human sequence. Identify the differing positions and state which is more similar within this excerpt.
(b) 4 marks
Use the codon table to determine the amino-acid sequence for each species. Explain why protein comparison would not reveal the DNA differences in this excerpt.
(c) 5 marks
Evaluate both claims in the report and explain what additional molecular and chronological evidence would be needed for broader evolutionary conclusions.

Section Three: Extended answer

Answer two questions: one of Questions 37 or 38 for Unit 3, and one of Questions 39 or 40 for Unit 4. This section is worth 20% of the examination.

Question 37

20 marks
Stimulus

Unit 3 alternative A — Parkinson's disease involves loss of dopamine-producing neurons and can impair movement. Researchers investigate replacing these cells. All results below are fictional. Stem cells can divide and can be induced to differentiate into specialised cells; an undifferentiated stem cell is not yet a dopamine-producing neuron.

In a simplified laboratory model, replacement neurons receive an identical stimulus that generates action potentials. They release dopamine near receiver neurons with dopamine receptors. Receiver activity is measured as a change from baseline in relative units; it is not a measurement of human movement. Independent matched cultures contain equal numbers of living replacement and receiver neurons. The two blockers act only at their stated targets and do not change cell survival.

Candidate data (condition: dopamine released, receiver activity): no blocker: 80, 60; dopamine-vesicle release blocked: 4, 3; receiver dopamine receptors blocked: 79, 4. Differences of one unit are within measurement uncertainty; the larger differences are reproducible.

A separate pilot study gives eight volunteers replacement cells. Their average movement-task performance improves after three months. There is no comparison group; medication doses and rehabilitation are also increased during this period.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
(a) 7 marks
Explain chemical communication from the stimulated replacement neurons to the receiver neurons, and use both blocker conditions to distinguish neurotransmitter release from receptor action.
(b) 6 marks
Explain why cell replacement requires more than transplanting living stem cells. Include differentiation, function at the target tissue and two distinct biological safety concerns.
(c) 7 marks
Evaluate the claim that the pilot study proves a permanent cure. Propose three justified design features for a stronger clinical investigation and one ethical safeguard for participants.

Question 38

20 marks
Stimulus

Unit 3 alternative B — A fictional virus has original form O and variant V. V has a changed cell-attachment region but retains several other surface regions. Antibodies can bind different surface regions; in this model, only binding that obstructs attachment prevents entry into cells. No specialist assay knowledge is required.

Researchers test the same purified antibody preparation from donors vaccinated against O at the same concentration against each form. A binding test counts virus particles with any antibody attached. A separate entry test counts cells entered by virus after a fixed time. Cell type, starting viable-virus numbers and conditions are matched. Results are reproducible across independently prepared cultures.

Candidate data (virus form: particles with antibody attached out of 100; cells entered without antibody; cells entered with antibody): O: 90; 200; 20. V: 85; 200; 160.

An updated vaccine would contain a non-infectious version of V's attachment-region protein antigen. Its clinical safety, duration of protection and effect on disease incidence have not yet been measured.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
(a) 6 marks
Calculate the percentage reduction in cell entry for each form, using its no-antibody control. Explain why the binding results alone would give a misleading comparison of protection against O and V.
(b) 6 marks
Explain how the updated antigen could induce production of antibodies that obstruct V attachment and establish immune memory. Do not assume every antibody produced will prevent entry.
(c) 8 marks
Design a laboratory comparison of antibodies induced by the original and updated vaccines that tests entry prevention against V. Explain the controls and replication needed. Then identify the additional clinical evidence and one practical access consideration needed, and give a conditional recommendation about a community vaccine change.

Question 39

20 marks
Stimulus

Unit 4 alternative A — Researchers investigate how a fictional later-hominin community organised tool production. No hominin remains are found. A composite tool combines a shaped stone point, a wooden shaft and plant-fibre binding with adhesive. Stone provides a hard cutting tip, wood transmits force, and the binding and adhesive secure the join. The point is shaped before mounting; its base and the shaft socket must fit. Several recovered examples repeat this construction.

At Quarry Q there are local stone blocks, partly shaped points and large waste flakes. At Camp C, 12 km away, there are mounted points and small flakes from edge maintenance. The point stone matches Q's distinctive source. Two large flakes from Q physically refit onto missing surfaces of two points found at C, demonstrating that they came from the same original pieces. For this model, natural movement between sites has been excluded. There is no evidence identifying which individuals carried or manufactured the objects.

A museum claims that mounting a point always makes a better cutting tool and that these finds prove the makers belonged to one particular hominin species. Researchers can safely test modern replicas using an enclosed mechanical apparatus; no human hunting experiment is needed.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
(a) 7 marks
Explain how manufacturing this composite tool combines materials and coordinated actions. Use the supplied construction evidence to infer practical skills and planning, and state a limit on what it reveals about cognition.
(b) 6 marks
Evaluate the proposal that all stages of point manufacture occurred at Camp C. Explain what the two sites jointly support about production and movement, and assess whether they identify the tool makers' species.
(c) 7 marks
Propose a fair replica investigation of the claim that mounting a point always produces a better cutting tool. Specify both cutting performance and a separate cost or failure measure, explain controls and replication, and evaluate how far the results could support conclusions about past tool use.

Question 40

20 marks
Stimulus

Unit 4 alternative B — A fictional laboratory investigates chromosome changes in cultured human body cells. A normal cell has 46 chromosomes. Letters A to G label consecutive large segments of one chromosome, not single bases or alleles. Its reference order is A-B-C-D-E-F-G. Cell X still has 46 chromosomes, but one copy has order A-B-E-D-C-F-G, with every labelled segment retained once. In a separate observed mitotic division of a 46-chromosome cell, both sister chromatids of one chromosome move to the same pole; after division the daughter cells have 47 and 45 chromosomes.

A separate screening experiment assigns cultures at random to no chemical, low dose or high dose of a suspected mutagen. Each starts with 2000 living cells. Cells do not divide during this screening interval. All other conditions are matched. At 48 hours, researchers count surviving cells and identify those with a detectable chromosome alteration. They do not count dead cells or observe when alterations first arise. The following is one illustrative culture per condition; independent repeats have not yet been completed.

Candidate data (condition: surviving cells; survivors with an alteration): no chemical: 1000; 8. Low dose: 1000; 24. High dose: 200; 10. The detection method is equally sensitive in all three conditions. A report claims that high dose is safest because it leaves fewer altered cells than low dose. For a follow-up, a validated method is available to record chromosome status in the same tracked cells over time; its technical procedure is not assessed.

Use the stimulus and supplied visual, where present, to answer all parts of this question.
(a) 7 marks
Distinguish the chromosome change in Cell X from the segregation error in the observed division. Explain why a chromosome count alone cannot show that a cell has no mutation, and whether X's biological effect can be established from segment order alone.
(b) 7 marks
Calculate the percentage of surviving cells with an alteration in each condition. Use these percentages and the survival data to evaluate the report's safety claim, and explain why these are not direct measurements of the rate at which new mutations arise.
(c) 6 marks
Propose a stronger investigation that distinguishes production of new chromosome alterations from preferential survival of cells already carrying alterations. Explain the starting-cell evidence, comparisons, observations and replication needed, without exposing people to the chemical.

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Worked Solutions And Marking Guide

General marking principles

  • Award a mark only for a distinct, scientifically accurate element that addresses the stated part.
  • Accept accurate equivalent 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.
  • Keep conclusions within the limits of the supplied data; association alone does not establish causation.

Section One: Multiple-choice Question 1

Answer: receptor, sensory neuron, interneuron, motor neuron, effector

A withdrawal reflex begins at a receptor, travels along a sensory neuron to an interneuron in the central nervous system, then leaves by a motor neuron to an effector.

Section One: Multiple-choice Question 2

Answer: Depolarisation is regenerated mainly at nodes of Ranvier, producing saltatory conduction.

Myelin insulates internodes, so action potentials are regenerated at nodes of Ranvier and appear to jump between nodes.

Section One: Multiple-choice Question 3

Answer: Only target cells possess receptors with a complementary shape for that hormone.

Hormones circulate widely, but a response requires the appropriate receptor on or inside a target cell.

Section One: Multiple-choice Question 4

Answer: The hormone binds to a receptor in the cell-surface membrane.

Water-soluble hormones cannot freely cross the lipid bilayer, so they bind membrane receptors and activate intracellular signalling.

Section One: Multiple-choice Question 5

Answer: cutaneous vasodilation and increased sweating

Vasodilation increases heat transfer to the skin and evaporation of sweat removes heat.

Section One: Multiple-choice Question 6

Answer: Beta cells secrete insulin, increasing glucose uptake and glycogen formation.

Raised blood glucose stimulates pancreatic beta cells to release insulin, promoting cellular uptake and glycogenesis.

Section One: Multiple-choice Question 7

Answer: Pancreatic alpha cells release glucagon, promoting hepatic glycogenolysis and gluconeogenesis.

Low blood glucose stimulates alpha cells to secrete glucagon, which raises blood glucose mainly through liver glycogenolysis and gluconeogenesis.

Section One: Multiple-choice Question 8

Answer: More ADH is released, increasing water reabsorption in kidney collecting ducts.

Osmoreceptors detect increased plasma osmolarity. More ADH makes collecting ducts more permeable to water, concentrating urine and conserving water.

Section One: Multiple-choice Question 9

Answer: It increases sodium reabsorption in the distal nephron, so water is retained osmotically.

Aldosterone promotes sodium reabsorption in the distal nephron; water follows by osmosis, helping restore blood volume.

Section One: Multiple-choice Question 10

Answer: Chemoreceptor input increases ventilation rate and depth, removing more carbon dioxide.

Chemoreceptors detect changes associated with elevated carbon dioxide and stimulate increased ventilation, which removes more carbon dioxide.

Section One: Multiple-choice Question 11

Answer: Mucus traps particles and cilia move the mucus towards the throat.

Respiratory mucus traps pathogens and cilia move contaminated mucus away from the lungs as a non-specific barrier.

Section One: Multiple-choice Question 12

Answer: Histamine promotes local vasodilation and increased capillary permeability.

Inflammatory mediators such as histamine increase local blood flow and capillary permeability, producing redness, heat and swelling.

Section One: Multiple-choice Question 13

Answer: Shivering and cutaneous vasoconstriction

During the onset of fever, the body is below the raised set point and responds as if cold: shivering produces heat and vasoconstriction reduces heat loss.

Section One: Multiple-choice Question 14

Answer: Viruses lack bacterial cell walls and reproduce using host-cell machinery.

A cell-wall-targeting antibiotic acts on a bacterial structure that viruses do not possess; viruses replicate within host cells.

Section One: Multiple-choice Question 15

Answer: Vaccination induces active immunity and memory, whereas prepared antibodies provide temporary passive immunity.

A vaccine stimulates the recipient's adaptive response and memory cells. Transferred antibodies act immediately but are temporary and do not normally create immune memory.

Section One: Multiple-choice Question 16

Answer: Memory lymphocytes specific to the antigen can rapidly undergo clonal expansion.

Antigen-specific memory B and T cells persist and respond rapidly through clonal expansion and differentiation.

Section One: Multiple-choice Question 17

Answer: It occurs in a germline cell that contributes to a gamete involved in fertilisation.

Only mutations present in germline cells and passed through a fertilising gamete can enter the offspring's genome.

Section One: Multiple-choice Question 18

Answer: It cannot be determined exactly because the dominant phenotype includes both RR and Rr genotypes.

Dominance describes expression in heterozygotes, not allele abundance. The survey does not separate RR from Rr, so the number of R copies is unknown.

Section One: Multiple-choice Question 19

Answer: Selection can maintain both alleles because the heterozygote has the highest reproductive success.

Heterozygote advantage can maintain both alleles because heterozygous individuals leave more surviving offspring in that environment.

Section One: Multiple-choice Question 20

Answer: Founder effect, a form of genetic drift

A small founding sample may carry allele frequencies that differ by chance from the source population, reducing or changing genetic variation.

Section One: Multiple-choice Question 21

Answer: Gene flow, because alleles move between populations through migration and reproduction

Migration followed by reproduction transfers alleles between gene pools and is therefore gene flow.

Section One: Multiple-choice Question 22

Answer: Speciation following reproductive isolation

Restricted gene flow allows populations to diverge. When reproductive isolation prevents fertile offspring, they are separate species under the biological species concept.

Section One: Multiple-choice Question 23

Answer: Produce many copies of a selected DNA region

PCR amplifies a selected DNA region so that enough target DNA is available for detection, comparison or sequencing.

Section One: Multiple-choice Question 24

Answer: The smallest fragments

Smaller DNA fragments pass more easily through the gel matrix and therefore migrate further under the electric field.

Section One: Multiple-choice Question 25

Answer: The founder and source populations share a more recent common ancestry, assuming comparable sequences and mutation patterns.

Greater sequence similarity generally supports more recent common ancestry, but the inference depends on suitable comparable DNA and does not imply progress or complete identity.

Section One: Multiple-choice Question 26

Answer: The hominin-bearing layer is relatively older than the layer above it.

In undisturbed strata, lower layers are relatively older. Stratigraphy provides relative order, not an exact numerical age.

Section One: Multiple-choice Question 27

Answer: Potassium-argon dating

Potassium-argon dating is appropriate for old volcanic material. Radiocarbon dating is limited to much younger once-living material.

Section One: Multiple-choice Question 28

Answer: A short broad pelvis and a femur angled inward towards the knee

A short broad pelvis supports trunk stability and an inward femoral angle places the knees under the body during bipedal walking.

Section One: Multiple-choice Question 29

Answer: Homo erectus

Acheulian tools are conventionally associated in the course with Homo erectus.

Section One: Multiple-choice Question 30

Answer: It is inherited mainly through the maternal line and normally does not undergo meiotic recombination.

Maternal inheritance and the absence of ordinary meiotic recombination allow mitochondrial lineages to be followed through populations, using accumulated mutations as markers.

Section Two: Short answer Question 31

(a) Divide each target value by the initial 80 units and multiply by 100: neither = 40/80 × 100 = 50%; hand wash = 10/80 × 100 = 12.5%; handle cleaning = 20/80 × 100 = 25%; both = 4/80 × 100 = 5%.

Divide each target value by the initial 80 units and multiply by 100: neither = 40/80 × 100 = 50%; hand wash = 10/80 × 100 = 12.5%; handle cleaning = 20/80 × 100 = 25%; both = 4/80 × 100 = 5%.

(b) An inanimate contaminated handle can act as a fomite, transferring material to a hand. Subsequent hand contact could deposit viable V particles on nasal mucosa, its supplied portal of entry. Intact skin is a physical barrier, so material on its surface is not evidence that the barrier has been crossed. Even deposition at the appropriate entry site does not establish infection without viable virus successfully infecting host cells.

An inanimate contaminated handle can act as a fomite, transferring material to a hand. Subsequent hand contact could deposit viable V particles on nasal mucosa, its supplied portal of entry. Intact skin is a physical barrier, so material on its surface is not evidence that the barrier has been crossed. Even deposition at the appropriate entry site does not establish infection without viable virus successfully infecting host cells.

(c) Both interventions leave the least tracer: 4 units compared with 10 after hand washing, 20 after handle cleaning and 40 with neither. Hand washing alone transfers less tracer than handle cleaning alone. Four units are still detected, so transfer has not been eliminated. Although target tracer falls by 90% relative to 40 units, fluorescence measures material transfer, not viral viability. No infection outcomes were measured, so an exact infection-risk reduction cannot be inferred.

Both interventions leave the least tracer: 4 units compared with 10 after hand washing, 20 after handle cleaning and 40 with neither. Hand washing alone transfers less tracer than handle cleaning alone. Four units are still detected, so transfer has not been eliminated. Although target tracer falls by 90% relative to 40 units, fluorescence measures material transfer, not viral viability. No infection outcomes were measured, so an exact infection-risk reduction cannot be inferred.

(d) Use a suitable harmless microorganism in a contained bench model and quantify viable organisms recovered from each artificial target, for example by a validated culture assay. Compare the four intervention conditions using the same starting viable load, contact procedure and materials. Repeat independent trials with fresh materials, randomise condition allocation and blind outcome measurement. Apply appropriate containment and hygiene; deliberately infecting people is unnecessary, and viable-transfer results would still require separate evidence before claiming a clinical risk reduction.

Use a suitable harmless microorganism in a contained bench model and quantify viable organisms recovered from each artificial target, for example by a validated culture assay. Compare the four intervention conditions using the same starting viable load, contact procedure and materials. Repeat independent trials with fresh materials, randomise condition allocation and blind outcome measurement. Apply appropriate containment and hygiene; deliberately infecting people is unnecessary, and viable-transfer results would still require separate evidence before claiming a clinical risk reduction.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (4 marks)

  • 1 mark: Calculates 40/80 × 100 = 50% for neither intervention.
  • 1 mark: Calculates 10/80 × 100 = 12.5% for hand washing alone.
  • 1 mark: Calculates 20/80 × 100 = 25% for handle cleaning alone.
  • 1 mark: Calculates 4/80 × 100 = 5% for both interventions.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (4 marks)

  • 1 mark: Identifies the handle as a fomite that can transfer material to a hand.
  • 1 mark: Traces hand-to-nasal-mucosa transfer to the supplied portal of entry.
  • 1 mark: Explains that intact skin forms a physical barrier rather than treating surface contamination as established infection.
  • 1 mark: Distinguishes deposition at an entry site from successful infection by viable virus.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (5 marks)

  • 1 mark: Uses 4 versus 10, 20 and 40 units to identify the combined condition as having the lowest mean transfer.
  • 1 mark: Uses 10 versus 20 units to compare the two single interventions.
  • 1 mark: Uses the remaining 4 units to reject complete elimination of material transfer.
  • 1 mark: Distinguishes tracer transfer from the presence or quantity of viable virus.
  • 1 mark: Rejects an exact infection-risk reduction because the investigation measured no infection outcomes.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Part d (5 marks)

  • 1 mark: Proposes a suitable harmless microbial surrogate and a valid measurement of viability rather than fluorescence alone.
  • 1 mark: Retains intervention comparisons with matched starting viable load and standardised contacts and materials.
  • 1 mark: Uses independent repeated trials with fresh or decontaminated materials to assess reproducibility.
  • 1 mark: Reduces allocation or measurement bias through randomisation or blinded outcome assessment, with a correct explanation.
  • 1 mark: Uses a contained, appropriately safe model without deliberate human infection and limits clinical inference.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part d checkpoint. Accept another justified safe design and validated viability assay when it meets the comparison, replication and evidence-limit checkpoints.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part d checkpoints.

Section Two: Short answer Question 32

(a) A target cell possesses a receptor complementary to a particular hormone. Cells lacking a functional insulin receptor do not form the hormone-receptor complex needed to initiate the insulin response.

A target cell possesses a receptor complementary to a particular hormone. Cells lacking a functional insulin receptor do not form the hormone-receptor complex needed to initiate the insulin response.

(b) Low blood glucose is detected by pancreatic alpha cells. They secrete glucagon into the blood. Glucagon binds receptors on target cells, especially liver cells, promoting glycogenolysis and gluconeogenesis. Glucose is released into blood, raising the concentration towards its normal range; the reduced deviation then lowers the glucagon stimulus.

Low blood glucose is detected by pancreatic alpha cells. They secrete glucagon into the blood. Glucagon binds receptors on target cells, especially liver cells, promoting glycogenolysis and gluconeogenesis. Glucose is released into blood, raising the concentration towards its normal range; the reduced deviation then lowers the glucagon stimulus.

(c) Water-soluble hormones bind receptors in the plasma membrane and activate intracellular second-messenger or enzyme pathways, commonly producing a rapid, shorter response. Lipid-soluble hormones cross the membrane, bind intracellular receptors and alter gene transcription, commonly producing a slower but longer-lasting response.

Water-soluble hormones bind receptors in the plasma membrane and activate intracellular second-messenger or enzyme pathways, commonly producing a rapid, shorter response. Lipid-soluble hormones cross the membrane, bind intracellular receptors and alter gene transcription, commonly producing a slower but longer-lasting response.

(d) The human insulin coding sequence is obtained and cut or prepared with a restriction enzyme. A bacterial plasmid is removed and cut with the same enzyme to create compatible ends. DNA ligase joins the insulin sequence into the plasmid. The recombinant plasmid is introduced into a bacterium, transformed cells are selected and cultured, the inserted sequence is expressed, and the insulin product is harvested and purified.

The human insulin coding sequence is obtained and cut or prepared with a restriction enzyme. A bacterial plasmid is removed and cut with the same enzyme to create compatible ends. DNA ligase joins the insulin sequence into the plasmid. The recombinant plasmid is introduced into a bacterium, transformed cells are selected and cultured, the inserted sequence is expressed, and the insulin product is harvested and purified.

(e) Successful somatic gene therapy could provide more sustained production and reduce repeated injections, whereas injected insulin permits dose adjustment and can be stopped. Gene delivery may be incomplete, expression may be poorly regulated and immune or off-target effects may occur. A trial would need long-term expression, glucose-control and safety evidence before replacing established treatment. An additional valid observation is that the change is not expected to be inherited because somatic cells rather than gametes are targeted.

Successful somatic gene therapy could provide more sustained production and reduce repeated injections, whereas injected insulin permits dose adjustment and can be stopped. Gene delivery may be incomplete, expression may be poorly regulated and immune or off-target effects may occur. A trial would need long-term expression, glucose-control and safety evidence before replacing established treatment. An additional valid observation is that the change is not expected to be inherited because somatic cells rather than gametes are targeted.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (2 marks)

  • 1 mark: Defines a target cell as a cell with a receptor specific or complementary to the hormone.
  • 1 mark: Explains that cells without a functional insulin receptor cannot initiate the insulin signalling response.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (5 marks)

  • 1 mark: Identifies low blood glucose as the detected stimulus.
  • 1 mark: Identifies pancreatic alpha cells as the responding endocrine cells and glucagon as the hormone released.
  • 1 mark: States that glucagon travels in blood and binds to receptors on target cells, especially liver cells.
  • 1 mark: Links glucagon to glycogenolysis and/or gluconeogenesis that raises blood glucose.
  • 1 mark: Explains that restoring blood glucose towards the normal range reduces the original stimulus by negative feedback.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (4 marks)

  • 1 mark: States that water-soluble hormones bind to plasma-membrane receptors.
  • 1 mark: Links water-soluble hormone binding to second-messenger or intracellular enzyme pathways.
  • 1 mark: States that lipid-soluble hormones cross the membrane and bind intracellular receptors.
  • 1 mark: Links lipid-soluble hormone-receptor complexes to altered transcription and a generally slower, longer-lasting response.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Part d (7 marks)

  • 1 mark: Obtains the human insulin coding sequence.
  • 1 mark: Cuts or prepares the insulin sequence with a restriction enzyme to create compatible ends.
  • 1 mark: Removes a bacterial plasmid and cuts it with the same restriction enzyme.
  • 1 mark: Uses DNA ligase to join the insulin sequence into the plasmid.
  • 1 mark: Introduces the recombinant plasmid into bacteria and selects transformed cells.
  • 1 mark: Cultures the transformed bacteria so they multiply and express the inserted sequence.
  • 1 mark: Harvests and purifies the produced insulin.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part d checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part d checkpoints.

Part e (4 marks)

  • 1 mark: Explains a potential sustained-treatment or reduced-injection benefit of successful somatic gene delivery.
  • 1 mark: Contrasts this with the controllable, adjustable and reversible dosing of injected insulin.
  • 1 mark: Identifies a relevant gene-therapy limitation such as incomplete delivery, poorly regulated expression, immune response or off-target effect.
  • 1 mark: Reaches a justified evaluation of effectiveness and safety that recognises limitations in the available evidence.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part e checkpoint. Somatic non-inheritance is a valid additional observation, not a compulsory checkpoint or a substitute for the requested comparison of treatments.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part e checkpoints.

Section Two: Short answer Question 33

(a) A older: 20/10 = 2 specimens m⁻³. A younger: 4/1 = 4 specimens m⁻³. B older: 12/6 = 2 specimens m⁻³. B younger: 8/2 = 4 specimens m⁻³.

A older: 20/10 = 2 specimens m⁻³. A younger: 4/1 = 4 specimens m⁻³. B older: 12/6 = 2 specimens m⁻³. B younger: 8/2 = 4 specimens m⁻³.

(b) Less sediment was excavated from the younger deposit at each site, so raw counts are not directly comparable. Recovery density is higher in both younger deposits, rising from 2 to 4 specimens m⁻³. The raw counts therefore do not demonstrate the claimed population decline. Specimen density is still not living-population size: rapid low-oxygen burial can improve older preservation, whereas weathering can destroy younger remains or fragmentation can change what is recoverable. Several specimens can come from one individual, and deposition or recovery conditions can alter densities without population change.

Less sediment was excavated from the younger deposit at each site, so raw counts are not directly comparable. Recovery density is higher in both younger deposits, rising from 2 to 4 specimens m⁻³. The raw counts therefore do not demonstrate the claimed population decline. Specimen density is still not living-population size: rapid low-oxygen burial can improve older preservation, whereas weathering can destroy younger remains or fragmentation can change what is recoverable. Several specimens can come from one individual, and deposition or recovery conditions can alter densities without population change.

(c) The widespread, short-lived index species provides a useful correlation between distant deposits. Because unreworked I occurs in both older deposits, each is consistent with deposition within its 2.10–2.00-million-year range. This supports overlapping age ranges, rather than proving identical deposition dates. It does not identify an exact year within that range or show that a hominin at one site was an ancestor of a hominin at the other; ancestry needs separate anatomical or molecular evidence.

The widespread, short-lived index species provides a useful correlation between distant deposits. Because unreworked I occurs in both older deposits, each is consistent with deposition within its 2.10–2.00-million-year range. This supports overlapping age ranges, rather than proving identical deposition dates. It does not identify an exact year within that range or show that a hominin at one site was an ancestor of a hominin at the other; ancestry needs separate anatomical or molecular evidence.

(d) Sample multiple equivalent older and younger deposits rather than one small excavation at each age. Excavate matched volumes, or use measured volumes to standardise recovery density. Use the same sieving, identification and counting rules. Record burial conditions, weathering and fragmentation so differences in preservation can be assessed. Verify the ages independently and report variation among samples before interpreting a repeatable density trend as possible biological evidence.

Sample multiple equivalent older and younger deposits rather than one small excavation at each age. Excavate matched volumes, or use measured volumes to standardise recovery density. Use the same sieving, identification and counting rules. Record burial conditions, weathering and fragmentation so differences in preservation can be assessed. Verify the ages independently and report variation among samples before interpreting a repeatable density trend as possible biological evidence.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (4 marks)

  • 1 mark: Calculates 2 specimens m⁻³ for A older from 20/10.
  • 1 mark: Calculates 4 specimens m⁻³ for A younger from 4/1.
  • 1 mark: Calculates 2 specimens m⁻³ for B older from 12/6.
  • 1 mark: Calculates 4 specimens m⁻³ for B younger from 8/2.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (6 marks)

  • 1 mark: Explains that unequal excavation volumes confound direct comparison of the raw counts.
  • 1 mark: Interprets the calculated density change as an increase in both younger deposits; accept consistent follow-through from Part (a).
  • 1 mark: Rejects the report's population-decline conclusion as unestablished by the raw counts.
  • 1 mark: Links rapid burial or low oxygen in the older deposits to reduced destruction or decay and better preservation.
  • 1 mark: Links younger weathering or fragmentation to a change in preservation or recoverability.
  • 1 mark: Distinguishes recovered skeletal-specimen density from living-population size, including that specimens need not represent separate individuals.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint. Apply consequential marking to a consistent interpretation of Part (a); do not penalise its arithmetic error again. Accept another valid preservation or deposition explanation tied to the supplied evidence.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (5 marks)

  • 1 mark: Explains why a widespread species with a short known duration is useful for correlating distant deposits.
  • 1 mark: Uses the unreworked I specimens to place both older deposits within the supplied 2.10–2.00-million-year interval.
  • 1 mark: Distinguishes overlapping supported age ranges from proof that the two deposits formed simultaneously.
  • 1 mark: States that the index fossil does not supply an exact numerical date within its duration.
  • 1 mark: Explains that chronological correlation alone cannot establish direct hominin ancestry.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Part d (5 marks)

  • 1 mark: Samples multiple representative deposits or sites at each age to test whether the pattern is reproducible.
  • 1 mark: Matches excavated volume or normalises by measured volume to make recovery rates comparable.
  • 1 mark: Standardises recovery, taxonomic identification and specimen-counting methods.
  • 1 mark: Records and assesses preservation or depositional conditions as alternative causes of the density pattern.
  • 1 mark: Uses independently supported ages and variation among samples to make a qualified temporal comparison.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part d checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part d checkpoints.

Section Two: Short answer Question 34

(a) The control succeeds in 9/10 reaches, P in 0/10 and Q in 3/10. P's failure to initiate movement is consistent with the supplied damage to a cerebral motor area that initiates voluntary motor output. Q can initiate movement, but overshooting is consistent with impaired cerebellar coordination of its timing and precision. Normal directly evoked contractions support preserved peripheral motor and muscle function rather than inability of either muscle to contract at all.

The control succeeds in 9/10 reaches, P in 0/10 and Q in 3/10. P's failure to initiate movement is consistent with the supplied damage to a cerebral motor area that initiates voluntary motor output. Q can initiate movement, but overshooting is consistent with impaired cerebellar coordination of its timing and precision. Normal directly evoked contractions support preserved peripheral motor and muscle function rather than inability of either muscle to contract at all.

(b) Sensory input can reach spinal circuits despite the localised cerebral motor injury. Interneurons in the spinal cord can activate a motor neuron to skeletal muscle without requiring the damaged cerebral motor area to initiate the reflex. Separate ascending signals can reach sensory-processing regions of the cerebrum, allowing awareness of pain. Blocking only that ascending pain pathway would remove conscious awareness of this stimulus while preserving the intact local withdrawal circuit.

Sensory input can reach spinal circuits despite the localised cerebral motor injury. Interneurons in the spinal cord can activate a motor neuron to skeletal muscle without requiring the damaged cerebral motor area to initiate the reflex. Separate ascending signals can reach sensory-processing regions of the cerebrum, allowing awareness of pain. Blocking only that ascending pain pathway would remove conscious awareness of this stimulus while preserving the intact local withdrawal circuit.

(c) The skull provides a hard bony enclosure that resists mechanical injury. The meninges form protective tissue coverings around the central nervous system. Cerebrospinal fluid cushions and supports the brain, reducing mechanical stress and movement against the enclosure. These protections reduce risk but do not guarantee prevention of every injury.

The skull provides a hard bony enclosure that resists mechanical injury. The meninges form protective tissue coverings around the central nervous system. Cerebrospinal fluid cushions and supports the brain, reducing mechanical stress and movement against the enclosure. These protections reduce risk but do not guarantee prevention of every injury.

(d) The left hemisphere still receives the cue and can send commands to the right hand, so that response is retained. The cue cannot pass along the represented corpus-callosum route to the right hemisphere, so the left hand does not respond to this left-only cue. Directly cueing the right hemisphere should restore the left-hand response under the stated intact-pathway conditions. These controls demonstrate that both downstream motor pathways work and help distinguish failure of interhemispheric transfer from failure of a hand or its motor output.

The left hemisphere still receives the cue and can send commands to the right hand, so that response is retained. The cue cannot pass along the represented corpus-callosum route to the right hemisphere, so the left hand does not respond to this left-only cue. Directly cueing the right hemisphere should restore the left-hand response under the stated intact-pathway conditions. These controls demonstrate that both downstream motor pathways work and help distinguish failure of interhemispheric transfer from failure of a hand or its motor output.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (6 marks)

  • 1 mark: Uses the control's 9/10 and P's 0/10 successes to describe the measured contrast.
  • 1 mark: Links P's failure to initiate instructed movement to the supplied cerebral motor-area damage and voluntary motor output.
  • 1 mark: Uses Q's 3/10 successes versus the control's 9/10 to describe Q's reduced accuracy.
  • 1 mark: Links Q's overshooting to impaired cerebellar coordination or precision.
  • 1 mark: Distinguishes P's initiation problem from Q's preserved initiation but impaired coordination.
  • 1 mark: Uses normal directly evoked contractions to distinguish the central deficits from complete peripheral motor or muscle failure.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (6 marks)

  • 1 mark: Explains that the intact peripheral sensory pathway can still carry the stimulus to the spinal cord.
  • 1 mark: Links spinal interneurons to motor-neuron activation and skeletal-muscle withdrawal.
  • 1 mark: Explains that initiating the spinal reflex does not require initiation by the damaged cerebral motor area.
  • 1 mark: Links P's conscious pain report to an intact ascending route and cerebral sensory processing distinct from the damaged motor area.
  • 1 mark: Predicts retained withdrawal when only the ascending pain pathway is blocked because the local spinal circuit remains intact.
  • 1 mark: Predicts loss of conscious awareness of that stimulus when its ascending pain pathway to the cerebrum is blocked.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (6 marks)

  • 1 mark: Identifies the skull or cranium.
  • 1 mark: Links its hard bony enclosure to protection against mechanical injury.
  • 1 mark: Identifies the meninges.
  • 1 mark: Links the meningeal tissue coverings to enclosure, support or protection of the brain.
  • 1 mark: Identifies cerebrospinal fluid.
  • 1 mark: Explains cushioning, buoyancy or reduced mechanical stress as a protective contribution of cerebrospinal fluid.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint. Award one identification mark and one correctly explained protective-contribution mark for each of three distinct valid structures or materials. Accept the blood-brain barrier with an accurate explanation of selective restriction of harmful substances in blood, in place of one listed example; do not count the same structure twice.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Part d (6 marks)

  • 1 mark: Predicts a retained right-hand response to the left-hemisphere cue.
  • 1 mark: Explains that the represented left-to-right-hand motor path does not require transfer across the corpus callosum.
  • 1 mark: Predicts absence of the left-hand response to that left-only cue.
  • 1 mark: Explains that blocked interhemispheric transfer prevents that cue reaching the right hemisphere controlling the left hand.
  • 1 mark: Predicts restoration of the left-hand response when the right hemisphere is cued directly under the supplied conditions.
  • 1 mark: Uses the directly cued controls to distinguish transfer failure from a downstream motor-pathway or hand defect.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part d checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part d checkpoints.

Section Two: Short answer Question 35

(a) Allele h count = 2(16) + 48 = 80. Total allele count = 2(100) = 200. Frequency = 80/200 = 0.40.

Allele h count = 2(16) + 48 = 80. Total allele count = 2(100) = 200. Frequency = 80/200 = 0.40.

(b) Mutation creates new alleles. Natural selection changes allele frequencies when inherited variants cause differences in survival and reproduction in a particular environment. Gene flow transfers alleles when migrants reproduce in another population. Genetic drift changes allele frequencies by chance, with stronger effects in small populations such as founder groups.

Mutation creates new alleles. Natural selection changes allele frequencies when inherited variants cause differences in survival and reproduction in a particular environment. Gene flow transfers alleles when migrants reproduce in another population. Genetic drift changes allele frequencies by chance, with stronger effects in small populations such as founder groups.

(c) The original samples contain too little of the selected marker for a clear comparison. Primers direct repeated amplification of the same homologous locus in each sample, producing enough copies for restriction treatment and visible gel bands. Using the same target region means aligned bands can be compared as fragments from equivalent loci.

The original samples contain too little of the selected marker for a clear comparison. Primers direct repeated amplification of the same homologous locus in each sample, producing enough copies for restriction treatment and visible gel bands. Using the same target region means aligned bands can be compared as fragments from equivalent loci.

(d) The lanes labelled the mainland source population and Beacon Island founders share two visible bands aligned at the same relative heights, supporting two shared measured fragment lengths at this marker. Each lane also contains one visible band that does not align with the other, so the displayed pooled profiles are not identical. Because only one pooled marker is shown, the gel cannot establish complete genomic identity, precise ancestry or migration direction.

The lanes labelled the mainland source population and Beacon Island founders share two visible bands aligned at the same relative heights, supporting two shared measured fragment lengths at this marker. Each lane also contains one visible band that does not align with the other, so the displayed pooled profiles are not identical. Because only one pooled marker is shown, the gel cannot establish complete genomic identity, precise ancestry or migration direction.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (3 marks)

  • 1 mark: Counts two h alleles for every hh individual and one for every Hh individual.
  • 1 mark: Uses 200 as the total number of alleles in 100 diploid individuals.
  • 1 mark: Reports an allele-h frequency of 0.40.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (4 marks)

  • 1 mark: Explains that mutation creates a new allele or DNA variant.
  • 1 mark: Explains that natural selection changes allele frequency through differential reproductive success of inherited variants.
  • 1 mark: Explains that gene flow transfers alleles between populations through migration and reproduction.
  • 1 mark: Explains that genetic drift is chance allele-frequency change, particularly influential in small populations.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (3 marks)

  • 1 mark: Explains that amplification produces enough copies of the selected marker for detection or analysis.
  • 1 mark: Explains that target-specific primers cause the same homologous locus to be amplified in each sample.
  • 1 mark: Links the comparable PCR products to restriction fragments that can be separated and compared as visible gel bands.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Part d (3 marks)

  • 1 mark: Identifies two bands aligned at the same visible heights in the source and founder lanes as shared fragment lengths.
  • 1 mark: Identifies one additional non-aligned band in each lane, so the displayed profiles are not identical.
  • 1 mark: Limits the conclusion because one pooled marker cannot establish complete genomic identity, precise ancestry or migration direction.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part d checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part d checkpoints.

Section Two: Short answer Question 36

(a) Numbering bases from 1 to 9, chimpanzee differs at 3, 6 and 9, while gorilla differs at 3 and 9. Gorilla therefore has fewer differences from the human sequence in this short excerpt. This comparison is restricted to the shown region.

Numbering bases from 1 to 9, chimpanzee differs at 3, 6 and 9, while gorilla differs at 3 and 9. Gorilla therefore has fewer differences from the human sequence in this short excerpt. This comparison is restricted to the shown region.

(b) Replacing coding-strand T with U gives mRNA codons: human GAA UUU GGC; chimpanzee GAG UUC GGU; gorilla GAG UUU GGA. All translate to Glu–Phe–Gly. The supplied codons show genetic-code redundancy: different codons can encode the same amino acid. The shown substitutions are synonymous or silent at the protein-sequence level, so protein comparison alone conceals these nucleotide differences.

Replacing coding-strand T with U gives mRNA codons: human GAA UUU GGC; chimpanzee GAG UUC GGU; gorilla GAG UUU GGA. All translate to Glu–Phe–Gly. The supplied codons show genetic-code redundancy: different codons can encode the same amino acid. The shown substitutions are synonymous or silent at the protein-sequence level, so protein comparison alone conceals these nucleotide differences.

(c) The displayed DNA already differs despite identical amino-acid sequences, contradicting genome identity. Nine bases from one region cannot represent the whole genome or establish the full species relationship. An exact divergence age cannot be calculated without a justified rate and dating calibration. Broader inference needs reliably aligned homologous regions across representative samples, and independent fossil chronology can help test or calibrate the timing hypothesis.

The displayed DNA already differs despite identical amino-acid sequences, contradicting genome identity. Nine bases from one region cannot represent the whole genome or establish the full species relationship. An exact divergence age cannot be calculated without a justified rate and dating calibration. Broader inference needs reliably aligned homologous regions across representative samples, and independent fossil chronology can help test or calibrate the timing hypothesis.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (3 marks)

  • 1 mark: Identifies chimpanzee differences at positions 3, 6 and 9.
  • 1 mark: Identifies gorilla differences at positions 3 and 9.
  • 1 mark: Identifies greater human-gorilla similarity in this excerpt from two differences versus three, without claiming a genome-wide result.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (4 marks)

  • 1 mark: Uses the supplied mRNA codons to obtain Glu–Phe–Gly for all three species.
  • 1 mark: Explains genetic-code redundancy using a supplied pair such as GAA/GAG or UUU/UUC.
  • 1 mark: Classifies the shown substitutions as synonymous or silent because amino-acid identity is unchanged.
  • 1 mark: Explains that amino-acid comparison therefore cannot detect these observed coding-DNA differences.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (5 marks)

  • 1 mark: Rejects identical genomes because the supplied DNA differs even though its encoded amino acids are identical.
  • 1 mark: Limits the relatedness inference because one nine-base region does not represent the full genomes or species relationship.
  • 1 mark: Rejects the exact ten-million-year date because no mutation rate or dating calibration is supplied.
  • 1 mark: Requires multiple reliably compared homologous genomic regions and representative samples for a broader relationship inference.
  • 1 mark: Uses independently dated fossil evidence to test or calibrate a chronological hypothesis rather than assigning an age from uncalibrated differences.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Section Three: Extended answer Question 37

(a) An action potential depolarises the replacement neuron's terminal. Voltage-gated calcium channels open and calcium enters. Calcium promotes vesicle fusion and dopamine release by exocytosis. Dopamine diffuses to complementary receptors on receiver neurons, changing their intracellular signalling and activity. Blocking release reduces dopamine from 80 to 4 units and receiver activity from 60 to 3, supporting dependence on released transmitter. Receptor blockade leaves release effectively unchanged at 79 versus 80 but reduces receiver activity to 4; released dopamine cannot produce the usual response without effective receptor action. Release therefore precedes receptor-mediated signalling, and the two manipulations interrupt different steps.

An action potential depolarises the replacement neuron's terminal. Voltage-gated calcium channels open and calcium enters. Calcium promotes vesicle fusion and dopamine release by exocytosis. Dopamine diffuses to complementary receptors on receiver neurons, changing their intracellular signalling and activity. Blocking release reduces dopamine from 80 to 4 units and receiver activity from 60 to 3, supporting dependence on released transmitter. Receptor blockade leaves release effectively unchanged at 79 versus 80 but reduces receiver activity to 4; released dopamine cannot produce the usual response without effective receptor action. Release therefore precedes receptor-mediated signalling, and the two manipulations interrupt different steps.

(b) Stem cells must differentiate into the required neuronal type rather than merely survive. Suitable cells must synthesise and release dopamine and deliver it where surviving receiver neurons can respond. Release must be regulated appropriately; the laboratory response alone does not establish useful integration into a person's movement-control network. Uncontrolled division of residual undifferentiated cells could produce an abnormal growth. Immune rejection could damage transplanted cells and reduce their survival. Other justified biological concerns, such as inappropriate transmitter release or disruption of existing neural activity, are acceptable as distinct safety concerns.

Stem cells must differentiate into the required neuronal type rather than merely survive. Suitable cells must synthesise and release dopamine and deliver it where surviving receiver neurons can respond. Release must be regulated appropriately; the laboratory response alone does not establish useful integration into a person's movement-control network. Uncontrolled division of residual undifferentiated cells could produce an abnormal growth. Immune rejection could damage transplanted cells and reduce their survival. Other justified biological concerns, such as inappropriate transmitter release or disruption of existing neural activity, are acceptable as distinct safety concerns.

(c) Improvement after transplantation is an association; increased medication and rehabilitation could contribute, and the uncontrolled study cannot isolate the cells' effect. Three months cannot establish permanent benefit or late safety. A stronger investigation could randomly allocate eligible consenting participants to cell replacement plus standard care or standard care alone, keep background medication and rehabilitation comparable where clinically appropriate, and use blinded assessors with the same objective movement task. Long-term repeated assessment is another valid design feature. Participants need informed consent that explains the experimental status, uncertainty and material risks, with an opportunity to withdraw from follow-up without losing ordinary care. Each justified feature earns credit on its own merits; sham brain surgery is not required.

Improvement after transplantation is an association; increased medication and rehabilitation could contribute, and the uncontrolled study cannot isolate the cells' effect. Three months cannot establish permanent benefit or late safety. A stronger investigation could randomly allocate eligible consenting participants to cell replacement plus standard care or standard care alone, keep background medication and rehabilitation comparable where clinically appropriate, and use blinded assessors with the same objective movement task. Long-term repeated assessment is another valid design feature. Participants need informed consent that explains the experimental status, uncertainty and material risks, with an opportunity to withdraw from follow-up without losing ordinary care. Each justified feature earns credit on its own merits; sham brain surgery is not required.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (7 marks)

  • 1 mark: Links the arriving action potential to depolarisation of the replacement neuron's terminal.
  • 1 mark: Links terminal depolarisation to opening calcium channels and calcium entry.
  • 1 mark: Links calcium entry to vesicle fusion and dopamine exocytosis.
  • 1 mark: Explains diffusion to complementary receptors and a change in receiver-cell signalling or activity.
  • 1 mark: Uses 80 to 4 dopamine units and 60 to 3 activity units to support dependence on transmitter release.
  • 1 mark: Uses 79 versus 80 dopamine units with activity falling to 4 to locate the receptor block downstream of release.
  • 1 mark: Distinguishes release and receptor action as successive necessary steps in this model rather than treating either blocker as evidence of cell death.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (6 marks)

  • 1 mark: Explains differentiation into the required specialised neuronal type rather than equating a living stem cell with a functional neuron.
  • 1 mark: Identifies synthesis and release of dopamine as required replacement-cell functions.
  • 1 mark: Links the location of released dopamine to surviving receiver cells with functional receptors.
  • 1 mark: Explains the need for appropriately regulated activity within the host neural network, beyond a response in culture.
  • 1 mark: Explains one relevant biological safety concern, such as uncontrolled cell division causing abnormal growth.
  • 1 mark: Explains a second distinct biological safety concern, such as immune rejection damaging the graft.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint. Award the two safety marks for any two distinct, correctly explained biological risks; the indicative examples are not compulsory.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (7 marks)

  • 1 mark: Distinguishes observed improvement after transplantation from proof that the cells caused it.
  • 1 mark: Identifies changed medication or rehabilitation as a specific alternative explanation in the uncontrolled pilot.
  • 1 mark: Explains why three months cannot establish permanent benefit or long-term safety.
  • 1 mark: Proposes and justifies one design feature that improves causal inference, measurement or follow-up.
  • 1 mark: Proposes and justifies a second distinct design feature rather than repeating the first.
  • 1 mark: Proposes and justifies a third distinct design feature rather than repeating either previous feature.
  • 1 mark: Explains a relevant participant safeguard, such as informed voluntary consent acknowledging experimental risks and uncertainty.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint. Accept any three justified, distinct improvements, including appropriate comparison/allocation, comparable co-treatments, blinded standardised assessment, representative recruitment or longer repeated follow-up. Do not require invasive sham surgery.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Section Three: Extended answer Question 38

(a) For O, reduction is (200 - 20)/200 × 100 = 90%; for V it is (200 - 160)/200 × 100 = 20%. The matched no-antibody controls establish the entry baseline for each form, so the comparison is not caused by different measured baseline entry. Binding remains common for V at 85 of 100 particles, but that test counts antibody attached at any surface region. Antibodies recognising retained regions can bind without obstructing the altered attachment region. Consequently, similar overall binding does not imply similar entry prevention; neither laboratory assay directly measures protection from disease in people.

For O, reduction is (200 - 20)/200 × 100 = 90%; for V it is (200 - 160)/200 × 100 = 20%. The matched no-antibody controls establish the entry baseline for each form, so the comparison is not caused by different measured baseline entry. Binding remains common for V at 85 of 100 particles, but that test counts antibody attached at any surface region. Antibodies recognising retained regions can bind without obstructing the altered attachment region. Consequently, similar overall binding does not imply similar entry prevention; neither laboratory assay directly measures protection from disease in people.

(b) B lymphocytes with receptors complementary to the updated antigen can be selected. With appropriate helper T-cell stimulation, selected B cells divide by clonal expansion and differentiate into plasma cells. Plasma cells secrete antibodies with that specificity; antibodies binding an appropriate part of the attachment region can obstruct entry. Some selected cells become long-lived memory cells, permitting a faster response on later exposure. Binding must still be tested functionally because not every antibody to the antigen will block attachment.

B lymphocytes with receptors complementary to the updated antigen can be selected. With appropriate helper T-cell stimulation, selected B cells divide by clonal expansion and differentiate into plasma cells. Plasma cells secrete antibodies with that specificity; antibodies binding an appropriate part of the attachment region can obstruct entry. Some selected cells become long-lived memory cells, permitting a faster response on later exposure. Binding must still be tested functionally because not every antibody to the antigen will block attachment.

(c) Compare preparations from original-vaccine and updated-vaccine recipients in the same V entry assay. Standardise antibody concentration and sampling time after vaccination, as well as cells and viable-virus input, so a difference is not simply due to a different amount or test condition. Include a V no-antibody control to calculate entry reduction. Use independent donor samples and repeat cultures, reporting variation rather than only a pooled result. A community recommendation additionally needs comparative disease-incidence or severity data and safety data, since laboratory entry reduction is not clinical effectiveness. Cost or distribution requirements could affect equitable access and achievable uptake. Recommend a change only if the combined clinical benefit, safety and access evidence supports it; high binding alone is insufficient.

Compare preparations from original-vaccine and updated-vaccine recipients in the same V entry assay. Standardise antibody concentration and sampling time after vaccination, as well as cells and viable-virus input, so a difference is not simply due to a different amount or test condition. Include a V no-antibody control to calculate entry reduction. Use independent donor samples and repeat cultures, reporting variation rather than only a pooled result. A community recommendation additionally needs comparative disease-incidence or severity data and safety data, since laboratory entry reduction is not clinical effectiveness. Cost or distribution requirements could affect equitable access and achievable uptake. Recommend a change only if the combined clinical benefit, safety and access evidence supports it; high binding alone is insufficient.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (6 marks)

  • 1 mark: Calculates (200 - 20)/200 × 100 = 90% reduction for O.
  • 1 mark: Calculates (200 - 160)/200 × 100 = 20% reduction for V.
  • 1 mark: Explains that the matched no-antibody controls provide each form's baseline entry for the comparison.
  • 1 mark: Interprets 85 of 100 V particles as frequent binding at some surface region, not necessarily at the attachment region.
  • 1 mark: Explains how antibodies binding retained non-blocking regions could give high binding but weak entry prevention.
  • 1 mark: Rejects equivalent protection from similar binding and limits the result to the measured laboratory outcomes.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (6 marks)

  • 1 mark: Explains selection of B lymphocytes with receptors complementary to the updated antigen.
  • 1 mark: Identifies helper T-cell stimulation in an appropriate response to this protein antigen.
  • 1 mark: Explains clonal expansion of selected B lymphocytes.
  • 1 mark: Links differentiation into plasma cells to secretion of antigen-specific antibodies.
  • 1 mark: Links appropriate antibody binding at the attachment region to obstruction of entry without assuming all binding is protective.
  • 1 mark: Links formation of persistent memory cells to a faster response on later exposure.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (8 marks)

  • 1 mark: Compares original-vaccine and updated-vaccine antibody preparations using V cell entry as the functional outcome.
  • 1 mark: Standardises antibody concentration and relevant sampling or assay conditions to make the comparison interpretable.
  • 1 mark: Includes a matched V no-antibody control and relates it to calculation of entry reduction.
  • 1 mark: Uses independent donor samples and repeated cultures to assess variability and reproducibility.
  • 1 mark: Identifies comparative disease incidence or severity as required clinical-effectiveness evidence.
  • 1 mark: Identifies clinical safety or adverse-effect evidence as an additional requirement.
  • 1 mark: Explains how one practical factor, such as cost or distribution, could affect equitable access or uptake.
  • 1 mark: Makes a conditional recommendation based on the combined evidence rather than treating the laboratory result as sufficient.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint. Accept another controlled functional comparison with appropriate independent replication; a named specialist assay or vaccine-manufacturing procedure is not required.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Section Three: Extended answer Question 39

(a) The stone contributes a hard cutting tip, the wood transmits force and the binding with adhesive secures the connection. Manufacture requires shaping the point before mounting it and matching the point base to the shaft socket. Coordinating these dependent actions supports planning of a finished object made from several components. Repeated fitting and shaping are consistent with learned technical skill and manual control. The artefacts do not measure intelligence, prove language or reveal the makers' precise thoughts; the inference concerns the capacities needed for this manufacture.

The stone contributes a hard cutting tip, the wood transmits force and the binding with adhesive secures the connection. Manufacture requires shaping the point before mounting it and matching the point base to the shaft socket. Coordinating these dependent actions supports planning of a finished object made from several components. Repeated fitting and shaping are consistent with learned technical skill and manual control. The artefacts do not measure intelligence, prove language or reveal the makers' precise thoughts; the inference concerns the capacities needed for this manufacture.

(b) Partly shaped points and large waste flakes support early shaping at Q, so exclusive manufacture at C is not supported. Mounted points and small maintenance flakes support use or maintenance at C. The physical refits link specific camp points to waste flakes found at Q more strongly than a general resemblance in stone type would. This favours early shaping at Q, but movement of waste means that deposition alone does not prove where it was produced. With natural movement excluded, people moved the relevant objects between sites, although exchange or transport by different people remains possible. These data support production organised across locations, not a particular individual's itinerary. No human remains identify a species, and technology alone is not an exclusive species marker.

Partly shaped points and large waste flakes support early shaping at Q, so exclusive manufacture at C is not supported. Mounted points and small maintenance flakes support use or maintenance at C. The physical refits link specific camp points to waste flakes found at Q more strongly than a general resemblance in stone type would. This favours early shaping at Q, but movement of waste means that deposition alone does not prove where it was produced. With natural movement excluded, people moved the relevant objects between sites, although exchange or transport by different people remains possible. These data support production organised across locations, not a particular individual's itinerary. No human remains identify a species, and technology alone is not an exclusive species marker.

(c) Compare mounted and unmounted replicas made with matched stone, point shape and edge sharpness. Apply a standard force and stroke using the apparatus to cut standardised test material. Measure cutting performance, such as depth per stroke, and a separate cost or failure measure, such as preparation time, breakage or join failure. Use multiple independently made replicas and repeated trials, randomising test order and allowing for wear. Test more than one relevant material or task because superiority in one condition does not justify 'always'. The results would demonstrate what replicas can do under the test conditions; they would not prove that the ancient tools were used for that task.

Compare mounted and unmounted replicas made with matched stone, point shape and edge sharpness. Apply a standard force and stroke using the apparatus to cut standardised test material. Measure cutting performance, such as depth per stroke, and a separate cost or failure measure, such as preparation time, breakage or join failure. Use multiple independently made replicas and repeated trials, randomising test order and allowing for wear. Test more than one relevant material or task because superiority in one condition does not justify 'always'. The results would demonstrate what replicas can do under the test conditions; they would not prove that the ancient tools were used for that task.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (7 marks)

  • 1 mark: Links the stone material to a hard cutting tip.
  • 1 mark: Links the wooden shaft to transmission of force.
  • 1 mark: Links binding and adhesive to securing the separate components.
  • 1 mark: Explains the dependence of mounting on prior shaping and compatible point-base and socket dimensions.
  • 1 mark: Uses coordination of components and dependent actions to support planning of the finished tool.
  • 1 mark: Uses repeated fitting or shaping to infer learned technical skill or manual control.
  • 1 mark: Limits the cognitive inference rather than claiming a measured intelligence level, definite language or access to the makers' thoughts.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (6 marks)

  • 1 mark: Uses partly shaped points and large waste flakes to favour early shaping at Q over an unsupported claim of manufacture exclusively at C; proof of the production location is not required.
  • 1 mark: Uses mounted points and small maintenance flakes to support later use or maintenance at C.
  • 1 mark: Explains that physical refitting links particular objects across sites rather than only a shared raw-material type.
  • 1 mark: Infers human movement of the relevant objects because natural transport is excluded.
  • 1 mark: Retains uncertainty about who transported or made them, including exchange between people as a possibility.
  • 1 mark: Rejects a definite species identification because there are no hominin remains and the technology is not an exclusive species marker.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (7 marks)

  • 1 mark: Compares mounted and unmounted points with matched stone, geometry and initial edge condition.
  • 1 mark: Controls applied force or motion and the material being cut so mounting is the intended difference.
  • 1 mark: Defines a measurable cutting-performance outcome, such as depth per standard stroke.
  • 1 mark: Defines a distinct cost or failure outcome relevant to the judgement of 'better'.
  • 1 mark: Uses independent replicas and repeated trials with a justified control for order or wear.
  • 1 mark: Explains why different tasks or materials must be considered before claiming universal superiority.
  • 1 mark: Distinguishes demonstrated replica capability from proof of the ancient objects' actual use.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint. Accept other safe, measurable cutting tasks and justified controls that isolate mounting while evaluating both performance and a distinct cost or failure outcome.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Section Three: Extended answer Question 40

(a) In X, segment C-D-E has reversed to E-D-C: this is an inversion, a structural chromosome mutation. The specified rearrangement retains all labelled segments and need not change the chromosome count. In the observed division, sister chromatids fail to separate normally during mitosis, a nondisjunction event. One daughter receives an extra chromosome and the other one fewer, giving 47 and 45. A count of 46 would miss X's structural change and could also miss a change within a gene. The effect of X cannot be established from order alone: rearrangement may disrupt a gene or its regulation, but no affected gene or functional measurement is supplied.

In X, segment C-D-E has reversed to E-D-C: this is an inversion, a structural chromosome mutation. The specified rearrangement retains all labelled segments and need not change the chromosome count. In the observed division, sister chromatids fail to separate normally during mitosis, a nondisjunction event. One daughter receives an extra chromosome and the other one fewer, giving 47 and 45. A count of 46 would miss X's structural change and could also miss a change within a gene. The effect of X cannot be established from order alone: rearrangement may disrupt a gene or its regulation, but no affected gene or functional measurement is supplied.

(b) The percentages among survivors are 8/1000 × 100 = 0.8%, 24/1000 × 100 = 2.4% and 10/200 × 100 = 5.0%. Thus high dose leaves fewer altered cells in absolute terms than low dose, but a larger proportion of its survivors are altered. With no division during screening, survival is also much lower: 200/2000 = 10% at high dose versus 1000/2000 = 50% at low dose. Neither finding justifies calling high dose safest. This single-culture comparison has no estimate of between-culture variation. The counts are a snapshot of survivors, not observations of new mutation events; differential death of altered and unaltered cells could change their proportions, and the time or origin of an alteration was not recorded.

The percentages among survivors are 8/1000 × 100 = 0.8%, 24/1000 × 100 = 2.4% and 10/200 × 100 = 5.0%. Thus high dose leaves fewer altered cells in absolute terms than low dose, but a larger proportion of its survivors are altered. With no division during screening, survival is also much lower: 200/2000 = 10% at high dose versus 1000/2000 = 50% at low dose. Neither finding justifies calling high dose safest. This single-culture comparison has no estimate of between-culture variation. The counts are a snapshot of survivors, not observations of new mutation events; differential death of altered and unaltered cells could change their proportions, and the time or origin of an alteration was not recorded.

(c) Use contained cell cultures and establish the detectable chromosome status of starting cell lines, rather than assuming all cells begin without alterations. Randomly assign matched cultures to chemical and no-chemical conditions. Track the same cells from before exposure through the non-dividing screening interval, recording when an alteration first becomes detectable rather than counting only the final survivors. Record deaths of altered and unaltered cells separately to test differential survival. Repeat with independent cultures and report variation in newly detected events among tracked cells over the same exposure period, using a consistent detection method. Use an approved contained laboratory system with safe handling and disposal, not chemical exposure of volunteers.

Use contained cell cultures and establish the detectable chromosome status of starting cell lines, rather than assuming all cells begin without alterations. Randomly assign matched cultures to chemical and no-chemical conditions. Track the same cells from before exposure through the non-dividing screening interval, recording when an alteration first becomes detectable rather than counting only the final survivors. Record deaths of altered and unaltered cells separately to test differential survival. Repeat with independent cultures and report variation in newly detected events among tracked cells over the same exposure period, using a consistent detection method. Use an approved contained laboratory system with safe handling and disposal, not chemical exposure of volunteers.

Mark allocation

  • Award marks for independently observable evidence; do not require the indicative-answer wording or sequence.
  • Accept accurate equivalent Human Biology terminology and logically equivalent calculations or interpretations.
  • Do not infer an unsupported mechanism or conclusion from a vague statement.

Detailed marking criteria

Part a (7 marks)

  • 1 mark: Identifies reversal of C-D-E to E-D-C as an inversion.
  • 1 mark: Explains that every labelled segment is retained once, so the rearrangement changes order without loss or duplication of the labelled material.
  • 1 mark: Identifies failure of sister-chromatid separation during mitosis as nondisjunction.
  • 1 mark: Explains that unequal distribution produces the observed 47- and 45-chromosome daughter cells.
  • 1 mark: Explains why a normal chromosome count does not exclude a structural mutation such as X's.
  • 1 mark: Explains why a chromosome count would also miss a gene-level sequence mutation.
  • 1 mark: Limits claims about X's biological effect because gene disruption or altered regulation requires additional functional or sequence evidence.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part a checkpoint.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part a checkpoints.

Part b (7 marks)

  • 1 mark: Calculates 8/1000 × 100 = 0.8% among no-chemical survivors.
  • 1 mark: Calculates 24/1000 × 100 = 2.4% among low-dose survivors.
  • 1 mark: Calculates 10/200 × 100 = 5.0% among high-dose survivors.
  • 1 mark: Rejects the count-only safety claim using the higher altered proportion and poorer survival at high dose.
  • 1 mark: Recognises that one culture per condition provides no measure of independent between-culture variation.
  • 1 mark: Explains that differential survival can alter the composition of the cells remaining at 48 hours.
  • 1 mark: Distinguishes an endpoint prevalence of alterations from the rate of new mutation events, whose occurrence was not observed.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part b checkpoint. Accept logically consistent arithmetic follow-through after an isolated numerical error. A rate of new mutation requires events and an appropriate opportunity or time denominator; endpoint proportions alone are insufficient.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part b checkpoints.

Part c (6 marks)

  • 1 mark: Establishes detectable starting chromosome status so pre-existing alterations can be distinguished from newly observed changes.
  • 1 mark: Uses matched randomly allocated chemical and no-chemical cultures as the controlled comparison.
  • 1 mark: Observes the same cells through time and records newly detected alteration events over a defined exposure period rather than only an endpoint proportion.
  • 1 mark: Separately tracks deaths of altered and unaltered cells to test preferential survival.
  • 1 mark: Repeats independent cultures using a consistent detection method and observation period, and assesses variation.
  • 1 mark: Uses a contained laboratory model with appropriate handling and disposal rather than deliberately exposing people.

Acceptable alternatives: Accept equivalent terminology only when it preserves the biological relationship or data interpretation specified in a listed Part c checkpoint. Accept a justified design using matched verified cell lines and serial observations that separates new detectable alterations from differential survival. Naming specialised imaging or sequencing technology is not required.

Do not credit by itself: Do not credit a general statement, unlabelled fact or repeated idea unless it independently satisfies one of the listed Part c checkpoints.

Diagnostic Checklist

Review focusSectionReflection and next action
wace-human-biology-u3-biotechnology-and-society Q32, Q37 Explain recombinant DNA, synthetic hormones, vaccination or gene-therapy applications with evidence limits and social considerations.
wace-human-biology-u3-defence-and-immunity Q12, Q13, Q15, Q16, Q38 Sequence non-specific and specific defences, including inflammation, fever, lymphocytes, antibodies and memory cells.
wace-human-biology-u3-endocrine-coordination Q3, Q4 Link hormone release, transport, target-cell receptors and negative feedback to the stated regulated variable.
wace-human-biology-u3-homeostasis Q10, Q5, Q6, Q7, Q8, Q9 Apply thermoregulation, blood-glucose, fluid-balance and gas-concentration mechanisms to changing conditions and data.
wace-human-biology-u3-infection-and-transmission Q11, Q14 Distinguish pathogen types and transmission routes, then connect prevention or treatment to the relevant mechanism.
wace-human-biology-u3-nervous-coordination Q1, Q2, Q34 Trace nervous-system pathways, impulse transmission, synapses and autonomic or somatic control using precise structures and sequences.
wace-human-biology-u3-science-inquiry-skills Q31 Strengthen hypotheses, variables, controls, reliability, data interpretation and conclusions that remain within the evidence.
wace-human-biology-u4-biotechnology-and-evolutionary-evidence Q23, Q24, Q25, Q30, Q35, Q36 Interpret PCR, gel electrophoresis, DNA sequencing, protein comparison and bioinformatics evidence for relatedness and migration.
wace-human-biology-u4-evidence-and-dating Q26, Q27, Q33 Use fossil, stratigraphic, index-fossil and radiometric evidence with the correct limits of relative and absolute dating.
wace-human-biology-u4-evolutionary-processes Q19, Q20, Q21, Q22 Explain allele-frequency change using mutation, selection, genetic drift, founder effect, gene flow and reproductive isolation.
wace-human-biology-u4-hominin-evolution Q28, Q29, Q39 Compare locomotion, skull, dentition, brain and tool-culture evidence across specified hominin groups.
wace-human-biology-u4-human-variation Q17, Q18, Q40 Calculate allele frequencies and distinguish mutations, gene pools and inherited variation from individual change.

What is included

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  • WACE Human Biology ATAR
  • 2026 edition · Pack 0 · v2.0
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Official examination resources

Open these current official resources alongside the Skill Align paper. They remain hosted and controlled by the assessment authority and are not republished by Skill Align.

SCSA Human Biology course page Official course, syllabus and support-material landing page.
Human Biology ATAR Year 12 syllabus Current Year 12 syllabus used for content scope and examination design.
2026 Human Biology ATAR examination front cover Official 2026 question counts, raw marks, section weights, timing and choice structure.
2025 Human Biology ATAR examination Previous examination used for question-style and quality calibration; the 2026 front cover defines the current raw-mark allocation.
2025 Human Biology ATAR marking key Official marking key used to confirm analytic marking granularity.
2025 Human Biology ATAR examination report Official public report used to target common reasoning and content weaknesses.
2026 Homo erectus enamel-protein study Primary study used to ground the Pack 7 Question 36 palaeoproteomic stimulus and avoid unsupported ancient-DNA claims.

Questions about this exam pack

What is included in Human Biology ATAR Free Online - Pack 0?

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2026 exam timetable reference View SCSA 2026 timetable
  • Written examination Thu 5 Nov 2026 9:20 am Upcoming10 days left

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