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VCE Biology Units 3&4 Free Online Pack 0

A free online Skill Align units 3&4 end-of-year examination showcase with original Biology Units 3&4 questions, worked solutions, marking guidance and a diagnostic checklist. No PDF download or checkout is provided.

VCE Units 3&4 End-of-Year Examination 2026 Edition - Pack 0 v1.0
Pack 0 is free to read in your browser. It includes the questions, worked solutions, marking guidance and diagnostic checklists below. There is no public checkout or PDF download.

Exam-pack paper structure

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Question and Answer Book

50 questions

120 marks

Estimated duration: Reading time 15 minutes; writing time 2 hours 30 minutes

Reading: 15 minutes · Writing: 2 hours 30 minutes

Read free Pack 0 online

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Skill Align VCE Biology Units 3&4 Practice Examination - 2026 Edition, Pack 0

A complete original free online examination aligned to the current VCE Biology assessment contract.

Paper
Question and Answer Book
Reading
15 minutes
Writing
2 hours 30 minutes
Assessment
120 marks

No calculator or reference material is permitted. Only approved basic stationery may be brought into the examination. Answer every question.

Section A - Multiple-choice questions

Answer all 40 questions. Select the correct or best answer. Each question is worth 1 mark.

Question 1

1 mark
The two strands of a DNA molecule are described as antiparallel because they
  1. run in the same 5-prime to 3-prime direction with complementary bases.
  2. run in opposite 5-prime to 3-prime directions.
  3. alternate backbone direction at each successive nucleotide.
  4. are joined by hydrogen bonds while both backbones keep the same orientation.

Question 2

1 mark
During transcription, RNA polymerase
  1. builds an RNA strand complementary to the DNA template strand.
  2. copies both DNA strands into a single double-stranded RNA product.
  3. joins amino acids in the order specified by the DNA template.
  4. removes introns while it translates the mature transcript.

Question 3

1 mark
A change expected during eukaryotic pre-mRNA processing is
  1. replacement of RNA uracil bases with thymine during transcript maturation.
  2. removal of introns and joining of exons.
  3. translation of retained exons before they are joined.
  4. duplication of the promoter into the mature mRNA.

Question 4

1 mark
At a ribosome, each tRNA contributes to translation by
  1. copying the gene's template strand into pre-mRNA.
  2. joining adjacent mRNA nucleotides through peptide bonds.
  3. binding a DNA promoter to recruit RNA polymerase.
  4. carrying a specific amino acid and pairing its anticodon with a codon.

Question 5

1 mark
In the supplied protein-synthesis pathway, which event produces the labelled mRNA intermediate?
Diagram Preview
DNA template gene P RNA transcript mRNA P polypeptide protein P
  1. Replication of the DNA template before cell division
  2. Translation of the transcript at a ribosome
  3. Post-translational folding of the polypeptide
  4. Transcription of the DNA template

Question 6

1 mark
A base substitution in a protein-coding region does not change the amino acid sequence. The mutation is most accurately described as
  1. missense because one amino acid must be replaced.
  2. nonsense because the substitution creates a stop codon.
  3. silent.
  4. frameshift because one coding base has changed.

Question 7

1 mark
A regulatory protein that prevents RNA polymerase from accessing a promoter will most directly reduce
  1. replication of the chromosome region containing the gene before cell division.
  2. translation of unrelated mature mRNAs in the cytosol.
  3. splicing of a transcript that has already been produced.
  4. transcription of the associated gene.

Question 8

1 mark
A change in a protein's primary structure can alter its function because it may change
  1. the codon assignments used when the altered mRNA is translated.
  2. the gene's phosphodiester bonds after translation finishes.
  3. chromosome number without changing the protein's amino-acid sequence.
  4. interactions that determine folding and the shape of a binding site.

Question 9

1 mark
In polymerase chain reaction, primers are required because they
  1. separate complementary DNA strands during the heating step.
  2. provide defined starting points for DNA polymerase.
  3. cleave template DNA at specific recognition sequences.
  4. determine fragment size by adding different numbers of nucleotides.

Question 10

1 mark
In gel electrophoresis, a shorter DNA fragment usually migrates farther than a longer fragment because it
  1. has a stronger positive charge than the longer fragment.
  2. contains proportionally more negatively charged phosphate groups.
  3. is attracted towards the negative rather than positive electrode.
  4. moves more readily through the gel matrix.

Question 11

1 mark
ATP hydrolysis can support an endergonic cellular reaction when
  1. ATP acts as the enzyme that lowers the coupled reaction's activation energy.
  2. the reactions are coupled so the overall energy change is favourable.
  3. ATP hydrolysis makes the endergonic reaction favourable without molecular coupling.
  4. ADP releases more transferable free energy than ATP hydrolysis.

Question 12

1 mark
Above an enzyme's optimum temperature, activity commonly decreases because
  1. changes to protein structure alter the active site.
  2. substrate molecules move too slowly to collide productively as temperature rises above the optimum.
  3. the reaction no longer requires activation energy above the optimum.
  4. enzyme molecules are consumed and converted into additional substrate.

Question 13

1 mark
The supplied graph shows enzyme activity across temperature. Which conclusion is supported directly by the plotted means?
Graph Preview
102035506560402001020355065temperature (°C)enzyme activity
  1. The exact optimum is 35°C despite untested intermediate temperatures.
  2. Temperature alone caused the differences even if other conditions varied.
  3. Most enzyme molecules were substrate-saturated at 35°C.
  4. Activity is greatest at 35°C among the tested temperatures.

Question 14

1 mark
A competitive inhibitor commonly reduces enzyme activity by
  1. occupying the active site and reducing substrate binding.
  2. binding an allosteric site and lowering functional enzyme concentration.
  3. increasing active-site complementarity for the normal substrate.
  4. binding only the final product without interacting with the enzyme.

Question 15

1 mark
The light-dependent reactions of photosynthesis directly produce
  1. glucose, reduced NADP and carbon dioxide.
  2. ATP, reduced NADP and oxygen.
  3. pyruvate, reduced NAD and molecular oxygen.
  4. ATP and carbohydrate through direct carbon fixation.

Question 16

1 mark
The Calvin cycle depends on the light-dependent reactions because it uses
  1. ATP and reduced NADP.
  2. oxygen and glucose made during photolysis.
  3. pyruvate and reduced NAD made during glycolysis.
  4. water as the carbon source for carbohydrate.

Question 17

1 mark
Most ATP from aerobic cellular respiration is generated during
  1. glycolysis and substrate-level phosphorylation in the cytosol.
  2. lactate fermentation after oxygen accepts the final electrons.
  3. oxidative phosphorylation across the inner mitochondrial membrane.
  4. the citric acid cycle through direct production of most ATP.

Question 18

1 mark
End-product inhibition can stabilise a biochemical pathway because accumulated product
  1. reduces activity of an earlier enzyme in the pathway.
  2. increases activity of each earlier enzyme until more product accumulates.
  3. is converted back to the initial substrate whenever its concentration rises.
  4. irreversibly inactivates an earlier enzyme whenever product concentration rises.

Question 19

1 mark
Which defence is part of the first line of defence against pathogens?
  1. Local inflammation following pathogen entry into tissue
  2. Intact skin limiting pathogen entry
  3. Antibody secretion after plasma-cell differentiation
  4. Cytotoxic T-cell killing of infected host cells

Question 20

1 mark
During a local inflammatory response, increased capillary permeability assists defence by
  1. allowing immune components to move from blood into affected tissue.
  2. preventing plasma proteins and fluid from entering affected tissue.
  3. creating antigen-specific memory before lymphocyte activation.
  4. retaining circulating phagocytes inside nearby capillaries.

Question 21

1 mark
An antigen-presenting cell contributes to adaptive immunity by
  1. secreting antibodies before antigen-specific B-cell activation.
  2. destroying pathogens without any receptor-mediated recognition.
  3. preventing complementary lymphocytes from undergoing clonal selection.
  4. displaying antigen fragments that can activate specific helper T cells.

Question 22

1 mark
Clonal selection begins when
  1. an antigen binds to a lymphocyte with a complementary receptor.
  2. the antigen binds with similar affinity to many lymphocyte receptors.
  3. a memory lymphocyte releases histamine before recognising antigen.
  4. a phagocyte produces antibodies with several unrelated specificities.

Question 23

1 mark
The graph compares antibody levels after first and second exposure to the same antigen. The faster second response is best explained by
Graph Preview
first exposureday 7day 21second exposureday 35day 496040200exposureantibody level
  1. loss of receptor specificity after the first exposure.
  2. slower clonal expansion of the selected lymphocytes.
  3. activation of antigen-specific memory cells.
  4. replacement of adaptive responses by external barriers.

Question 24

1 mark
Vaccination can provide long-term protection primarily by
  1. strengthening non-specific barriers at common pathogen entry sites.
  2. preventing future mutations in the targeted pathogen population.
  3. generating antigen-specific memory cells without requiring the natural disease.
  4. supplying ready-made antibodies that remain at peak concentration for life.

Question 25

1 mark
A monoclonal antibody treatment is specific because its antibody molecules
  1. bind equally to structurally unrelated antigens.
  2. have binding sites complementary to the same target epitope.
  3. are produced by many unrelated lymphocyte clones with different receptors.
  4. lack the variable regions that normally interact with epitopes.

Question 26

1 mark
A fall in reported cases after an intervention is not by itself proof that the intervention caused the fall because
  1. case counts cannot be compared quantitatively across defined time periods.
  2. the timing establishes causation even without a comparison group.
  3. other variables or changes over time may also affect case numbers.
  4. a correlation excludes changes in testing or pathogen transmission.

Question 27

1 mark
The ultimate source of new heritable alleles in a population is
  1. natural selection acting on existing allele differences.
  2. genetic drift changing the frequency of existing alleles.
  3. gene flow transferring existing alleles between populations.
  4. mutation.

Question 28

1 mark
Natural selection will most directly change allele frequencies when individuals differ in
  1. acquired responses that are not transmitted to offspring.
  2. heritable traits that affect reproductive success.
  3. heritable traits with identical effects on reproductive contribution.
  4. non-heritable traits that change only after reproduction.

Question 29

1 mark
Genetic drift is expected to have the strongest proportional effect in
  1. a very large randomly mating population with the same allele frequencies.
  2. a small isolated population.
  3. a population in which the allele under study is absent.
  4. two large populations connected by frequent breeding migration.

Question 30

1 mark
Gene flow between two populations occurs when
  1. both populations experience similar selection without interbreeding.
  2. a new allele arises in a non-reproductive body cell.
  3. individuals migrate but leave no offspring in the recipient population.
  4. individuals or gametes move between populations and reproduce.

Question 31

1 mark
The supplied evidence flow links a shared derived DNA sequence to a biological inference. Which final inference is most defensible?
Diagram Preview
Flow diagram. shared derived DNA sequence; then homologous character; then supported grouping; then recent common ancestor. shared derived DNA sequencehomologous charactersupported groupingrecent common ancestor
  1. The taxa share a more recent common ancestor than taxa lacking that sequence.
  2. The taxa belong to the same species because they share one sequence.
  3. The sequence arose independently because both taxa required it.
  4. The inferred grouping cannot change when additional genes are analysed.

Question 32

1 mark
Allopatric speciation becomes more likely when geographic isolation
  1. increases interbreeding until population allele frequencies converge.
  2. prevents mutation, selection and drift in both isolated groups.
  3. reduces gene flow and populations diverge genetically.
  4. exposes both populations to necessarily identical selection pressures.

Question 33

1 mark
Radiometric dating contributes to evolutionary evidence by estimating
  1. the complete phenotype and behaviour of each ancestral organism.
  2. phylogenetic relatedness without assumptions about the dated material.
  3. the substitution rate of a sampled gene without an external calibration.
  4. the numerical age of suitable rocks or fossils using isotope decay.

Question 34

1 mark
Evidence from hominin fossils and molecular comparisons is best interpreted as showing that human evolution
  1. was a single linear progression in which each species replaced the previous one.
  2. ended after bipedal locomotion first appeared in the fossil record.
  3. was branching, with multiple hominin lineages existing at some times.
  4. can be reconstructed reliably from one skull feature without molecular evidence.

Question 35

1 mark
Which statement is a testable directional hypothesis?
  1. Nitrate concentration will affect mean algal growth under controlled conditions.
  2. If nitrate concentration increases, mean algal growth rate will increase under otherwise controlled conditions.
  3. The aim is to compare algal growth across nitrate concentrations.
  4. Does nitrate concentration alter mean algal growth under controlled conditions?

Question 36

1 mark
In an investigation of a new antibacterial compound dissolved in ethanol, the most appropriate negative control receives
  1. ethanol without the antibacterial compound.
  2. the highest antibacterial-compound concentration without ethanol.
  3. the compound in water without matching the delivery solvent.
  4. neither treatment followed by a different incubation temperature.

Question 37

1 mark
The supplied investigation-design model varies light intensity and measures oxygen production. Which factor should be controlled to strengthen the comparison?
Diagram Preview
light intensity same algalvolume and oxygen producedper minute comparetreatment means
  1. Oxygen production, which is the measured response
  2. Temperature of each algal culture
  3. Light intensity, which is the assigned explanatory variable
  4. Algal strain, with a different strain allocated to each intensity

Question 38

1 mark
Replicate measurements mainly improve an investigation by
  1. eliminating systematic error from an incorrectly calibrated instrument.
  2. guaranteeing that the directional hypothesis will be supported.
  3. allowing variability to be estimated and reducing reliance on one observation.
  4. removing the need to control variables that affect the response.

Question 39

1 mark
Before collecting identifiable human genetic information, a researcher should ensure participants
  1. cannot withdraw their unused sample after consenting to collection.
  2. receive privacy information only after identifiable results are published.
  3. permit broad secondary research uses described only in general terms.
  4. give informed voluntary consent and understand how privacy will be protected.

Question 40

1 mark
Two treatment means differ slightly, but their measurements show substantial overlap and no inferential analysis is supplied. The most defensible conclusion is that
  1. the larger sample mean establishes a treatment effect without uncertainty.
  2. overlap establishes that the population treatment effects are identical.
  3. the overlapping measurements should be excluded before analysis.
  4. the data alone do not establish a reliable treatment effect.

Section B - Short-answer and extended-response questions

Answer all 10 questions. Question 10 is a standalone 10-mark extended response. The number of marks indicates the biological evidence required.

Question 1

8 marks
Stimulus

Researchers compare cells carrying normal allele P with cells carrying variant allele P*. The supplied diagram distinguishes a coding-strand DNA substitution from its transcribed mRNA codon. P* also carries a defined enhancer change that reduces activator binding. Both cell groups receive the same transcription signal.

Use the scenario and supplied visual to answer all parts of this question.
Diagram Preview
normalcoding-strand ...GAA... normal mRNA ...GAA... amino acid Glu variantcoding-strand ...GTA... variant mRNA ...GUA... amino acid Val
(a) 3 marks
Explain why the coding-strand DNA change can alter protein function even though gene length is unchanged.
(b) 3 marks
The P* cells produce much less P mRNA. Explain how the stated enhancer change could cause this result and name one measurement that would test it.
(c) 2 marks
Explain why equal cell number and the same transcription signal are important controls.

Question 2

8 marks
Stimulus

An enzyme converts substrate S to product P. Researchers measured mean enzyme activity at increasing substrate concentrations, first without inhibitor and then with inhibitor I. The supplied graph shows the no-inhibitor treatment; the inhibitor treatment reached the same maximum activity but required more substrate.

Use the scenario and supplied visual to answer all parts of this question.
Graph Preview
1248166040200124816substrate concentrationenzyme activity
(a) 2 marks
Explain why the plotted activity approaches a maximum at high substrate concentration.
(b) 3 marks
Use the inhibitor result to infer a likely inhibition mechanism.
(c) 3 marks
Design one further comparison that tests whether inhibitor I binds reversibly.

Question 3

8 marks
Stimulus

A previously unvaccinated person is exposed to virus R. The supplied model summarises one adaptive immune pathway. Months later, the person is exposed to the same viral antigen again.

Use the scenario and supplied visual to answer all parts of this question.
Diagram Preview
antigen-presentingcell helper T-cellactivation selected B-cellclone plasma cellssecrete antibody memory B cells
(a) 2 marks
Explain how antigen presentation initiates the pathway shown.
(b) 3 marks
Explain how clonal selection leads to antibody production.
(c) 3 marks
Predict and explain the response to the later exposure.

Question 4

8 marks
Stimulus

Two similar towns recorded weekly cases of disease K. Town A introduced a vaccination campaign at week 4; Town B did not. By week 10, Town A fell from 80 to 28 weekly cases and Town B fell from 76 to 55. Vaccination coverage in Town A reached 72%.

Use the scenario to answer all parts of this question.
(a) 3 marks
Describe the comparative evidence without claiming that vaccination was the only cause.
(b) 3 marks
Explain two biological reasons why greater vaccination coverage can reduce transmission.
(c) 2 marks
Identify two additional data items needed to evaluate the campaign more confidently.

Question 5

8 marks
Stimulus

A bacterial population contains rare heritable variants before exposure to antibiotic Z. The supplied flow diagram represents changes during repeated treatment. After several generations, resistant bacteria are common.

Use the scenario and supplied visual to answer all parts of this question.
Diagram Preview
Flow diagram. pre-existing heritable variation; then antibiotic selection pressure; then differential survival; then differential reproduction; then resistance allele becomes common. pre-existing heritable variationantibiotic selection pressuredifferential survivaldifferential reproductionresistance allele becomes common
(a) 3 marks
Explain why the antibiotic did not cause bacteria to mutate because they needed resistance.
(b) 3 marks
Explain the change in frequency of the resistance allele.
(c) 2 marks
Predict one effect of stopping unnecessary use of antibiotic Z and justify the prediction.

Question 6

8 marks
Stimulus

Species J, K and L share homologous gene Q. Pairwise sequence differences are J-K: 4, J-L: 19 and K-L: 18. A fossil assigned to the lineage containing J and K is dated to 3.2 million years ago with an uncertainty of plus or minus 0.2 million years.

Use the scenario to answer all parts of this question.
(a) 2 marks
Infer the closest pair and state the evidence.
(b) 3 marks
Explain how the molecular and fossil evidence make different contributions to the reconstruction.
(c) 3 marks
Explain why the conclusion should remain open to revision.

Question 7

8 marks
Stimulus

Scientists use CRISPR-Cas9 to disrupt a regulatory sequence near gene M in cultured plant cells. Edited cells show 35% of the gene M mRNA and 42% of the protein level measured in sham-edited controls.

Use the scenario to answer all parts of this question.
(a) 3 marks
Explain how CRISPR-Cas9 can produce a targeted change.
(b) 3 marks
Interpret the mRNA and protein results.
(c) 2 marks
State two checks needed before attributing the result specifically to the intended edit.

Question 8

8 marks
Stimulus

Students investigate the effect of light intensity on the rate of photosynthesis in equal algal samples. The supplied design model shows the planned comparison. Dissolved oxygen is measured every two minutes for 20 minutes.

Use the scenario and supplied visual to answer all parts of this question.
Diagram Preview
light intensity equal algae,temperature and change indissolved oxygen comparereplicated
(a) 2 marks
Write a directional hypothesis for the investigation.
(b) 2 marks
Identify the independent variable and the operational dependent variable.
(c) 2 marks
State one safety control and one biological control variable.
(d) 2 marks
Explain why replicate samples and a dark control improve the investigation.

Question 9

6 marks
Stimulus

Students tested catalase at pH 4, 6, 7, 8 and 10. Mean oxygen-production rates were 1.2, 4.8, 6.1, 5.5 and 0.9 mL per minute respectively. Each mean came from three trials, but the students did not record spread or instrument uncertainty.

Use the scenario to answer all parts of this question.
(a) 2 marks
Describe the pattern and identify the supported optimum among the tested values.
(b) 2 marks
Explain one molecular reason for lower activity at extreme pH.
(c) 2 marks
Evaluate the strength of the conclusion and propose one improvement.

Question 10

10 marks
A proposal suggests releasing gene-edited mosquitoes carrying a heritable construct that reduces survival of offspring. The aim is to reduce transmission of a mosquito-borne virus. Write an evidence-based evaluation of the proposal. Integrate molecular biology, inheritance and population change, immunity or disease transmission, investigation design, ethics and the limits of prediction.

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Worked Solutions And Marking Guide

Section A Question 1

Answer: run in opposite 5-prime to 3-prime directions.

The sugar-phosphate backbones have opposite orientations.

Section A Question 2

Answer: builds an RNA strand complementary to the DNA template strand.

RNA polymerase uses one DNA strand as the template for RNA synthesis.

Section A Question 3

Answer: removal of introns and joining of exons.

Splicing removes introns and joins retained exons.

Section A Question 4

Answer: carrying a specific amino acid and pairing its anticodon with a codon.

Complementary codon-anticodon pairing positions the correct amino acid.

Section A Question 5

Answer: Transcription of the DNA template

Transcription produces an RNA copy from a DNA template.

Section A Question 6

Answer: silent.

Degeneracy of the genetic code permits some substitutions to leave the amino acid unchanged.

Section A Question 7

Answer: transcription of the associated gene.

Promoter access controls initiation of transcription for the associated gene.

Section A Question 8

Answer: interactions that determine folding and the shape of a binding site.

The amino-acid sequence influences folding and hence binding-site shape.

Section A Question 9

Answer: provide defined starting points for DNA polymerase.

DNA polymerase extends from primers that flank the target sequence.

Section A Question 10

Answer: moves more readily through the gel matrix.

Smaller fragments experience less resistance while moving towards the positive electrode.

Section A Question 11

Answer: the reactions are coupled so the overall energy change is favourable.

Energy-releasing ATP hydrolysis can be coupled to an energy-requiring process.

Section A Question 12

Answer: changes to protein structure alter the active site.

Excess heat disrupts interactions maintaining enzyme shape.

Section A Question 13

Answer: Activity is greatest at 35°C among the tested temperatures.

The highest plotted mean occurs at 35°C; the other claims require evidence not shown.

Section A Question 14

Answer: occupying the active site and reducing substrate binding.

Competition for the active site lowers the frequency of enzyme-substrate complex formation.

Section A Question 15

Answer: ATP, reduced NADP and oxygen.

Light reactions split water and generate ATP and reduced NADP.

Section A Question 16

Answer: ATP and reduced NADP.

ATP and reducing power drive carbon fixation and carbohydrate production.

Section A Question 17

Answer: oxidative phosphorylation across the inner mitochondrial membrane.

The electron transport chain and chemiosmosis produce most ATP aerobically.

Section A Question 18

Answer: reduces activity of an earlier enzyme in the pathway.

Feedback inhibition limits further product formation when product is abundant.

Section A Question 19

Answer: Intact skin limiting pathogen entry

Physical and chemical barriers act before pathogens enter tissues.

Section A Question 20

Answer: allowing immune components to move from blood into affected tissue.

Greater permeability supports movement of immune cells and proteins into damaged or infected tissue.

Section A Question 21

Answer: displaying antigen fragments that can activate specific helper T cells.

Presented antigen can activate a compatible helper T cell.

Section A Question 22

Answer: an antigen binds to a lymphocyte with a complementary receptor.

Only lymphocytes with complementary receptors are selected to proliferate.

Section A Question 23

Answer: activation of antigen-specific memory cells.

Memory lymphocytes enable a faster and larger secondary response.

Section A Question 24

Answer: generating antigen-specific memory cells without requiring the natural disease.

Vaccination exposes the immune system to antigen safely enough to establish memory.

Section A Question 25

Answer: have binding sites complementary to the same target epitope.

A single clone produces antibodies with the same antigen-binding specificity.

Section A Question 26

Answer: other variables or changes over time may also affect case numbers.

Alternative explanations must be considered before making a causal claim.

Section A Question 27

Answer: mutation.

Mutation creates new DNA sequence variants; other processes change allele frequencies.

Section A Question 28

Answer: heritable traits that affect reproductive success.

Heritable differences linked to reproductive success drive differential allele transmission.

Section A Question 29

Answer: a small isolated population.

Chance sampling effects are proportionally larger in small populations.

Section A Question 30

Answer: individuals or gametes move between populations and reproduce.

Successful migration transfers alleles between populations.

Section A Question 31

Answer: The taxa share a more recent common ancestor than taxa lacking that sequence.

A shared derived character supports relative recency of common ancestry, not identity or certainty.

Section A Question 32

Answer: reduces gene flow and populations diverge genetically.

Restricted gene flow permits divergence through mutation, selection and drift.

Section A Question 33

Answer: the numerical age of suitable rocks or fossils using isotope decay.

Known isotope decay can provide absolute-age estimates for suitable material.

Section A Question 34

Answer: was branching, with multiple hominin lineages existing at some times.

Multiple evidence types support a branching history rather than a linear ladder.

Section A Question 35

Answer: If nitrate concentration increases, mean algal growth rate will increase under otherwise controlled conditions.

It predicts the direction of a measurable relationship between defined variables.

Section A Question 36

Answer: ethanol without the antibacterial compound.

A solvent-only control isolates any effect of the delivery solvent.

Section A Question 37

Answer: Temperature of each algal culture

Temperature can affect photosynthetic rate and should be held constant.

Section A Question 38

Answer: allowing variability to be estimated and reducing reliance on one observation.

Replication reveals spread and makes the mean less dependent on one result.

Section A Question 39

Answer: give informed voluntary consent and understand how privacy will be protected.

Consent, privacy and transparent data-use arrangements are central ethical requirements.

Section A Question 40

Answer: the data alone do not establish a reliable treatment effect.

The small difference must be interpreted with variability and suitable analysis.

Section B Question 1

(a) The coding-strand DNA substitution changes GAA to GTA, which is transcribed as an mRNA change from GAA to GUA. The changed mRNA codon can specify a different amino acid, such as glutamate to valine. The amino-acid substitution can alter folding or a binding site and therefore protein function.

The coding-strand DNA substitution changes GAA to GTA, which is transcribed as an mRNA change from GAA to GUA. The changed mRNA codon can specify a different amino acid, such as glutamate to valine. The amino-acid substitution can alter folding or a binding site and therefore protein function.

(b) Reduced activator binding at the P* enhancer can reduce recruitment of transcription machinery at the promoter. Reduced transcription initiation would lower P mRNA abundance. Quantifying P mRNA in matched cells, normalised to a validated reference gene, would test the prediction.

Reduced activator binding at the P* enhancer can reduce recruitment of transcription machinery at the promoter. Reduced transcription initiation would lower P mRNA abundance. Quantifying P mRNA in matched cells, normalised to a validated reference gene, would test the prediction.

(c) Equal cell number makes mRNA amount comparable between groups. The same signal prevents signal strength from confounding the effect of allele status.

Equal cell number makes mRNA amount comparable between groups. The same signal prevents signal strength from confounding the effect of allele status.

Detailed marking criteria

Part a (1 mark)

The coding-strand DNA substitution changes GAA to GTA, which is transcribed as an mRNA change from GAA to GUA.

Part a (1 mark)

The changed mRNA codon can specify a different amino acid, such as glutamate to valine.

Part a (1 mark)

The amino-acid substitution can alter folding or a binding site and therefore protein function.

Part b (1 mark)

Reduced activator binding at the P* enhancer can reduce recruitment of transcription machinery at the promoter.

Part b (1 mark)

Reduced transcription initiation would lower P mRNA abundance.

Part b (1 mark)

Quantifying P mRNA in matched cells, normalised to a validated reference gene, would test the prediction.

Part c (1 mark)

Equal cell number makes mRNA amount comparable between groups.

Part c (1 mark)

The same signal prevents signal strength from confounding the effect of allele status.

Section B Question 2

(a) More active sites are occupied as substrate concentration increases. At high substrate concentration, most active sites are occupied so enzyme concentration limits the rate.

More active sites are occupied as substrate concentration increases. At high substrate concentration, most active sites are occupied so enzyme concentration limits the rate.

(b) Reaching the same maximum at higher substrate is consistent with competitive inhibition. Inhibitor I competes with substrate S for the active site. High substrate concentration reduces the inhibitor's effect by increasing substrate-active-site encounters.

Reaching the same maximum at higher substrate is consistent with competitive inhibition. Inhibitor I competes with substrate S for the active site. High substrate concentration reduces the inhibitor's effect by increasing substrate-active-site encounters.

(c) Expose enzyme to inhibitor I, then remove or greatly dilute free inhibitor before adding substrate. Include an otherwise identical untreated-enzyme control. Recovery of activity after inhibitor removal would support reversible binding.

Expose enzyme to inhibitor I, then remove or greatly dilute free inhibitor before adding substrate. Include an otherwise identical untreated-enzyme control. Recovery of activity after inhibitor removal would support reversible binding.

Detailed marking criteria

Part a (1 mark)

More active sites are occupied as substrate concentration increases.

Part a (1 mark)

At high substrate concentration, most active sites are occupied so enzyme concentration limits the rate.

Part b (1 mark)

Reaching the same maximum at higher substrate is consistent with competitive inhibition.

Part b (1 mark)

Inhibitor I competes with substrate S for the active site.

Part b (1 mark)

High substrate concentration reduces the inhibitor's effect by increasing substrate-active-site encounters.

Part c (1 mark)

Expose enzyme to inhibitor I, then remove or greatly dilute free inhibitor before adding substrate.

Part c (1 mark)

Include an otherwise identical untreated-enzyme control.

Part c (1 mark)

Recovery of activity after inhibitor removal would support reversible binding.

Section B Question 3

(a) An antigen-presenting cell processes virus R and displays an antigen fragment. A helper T cell with a complementary receptor binds the displayed antigen-MHC complex.

An antigen-presenting cell processes virus R and displays an antigen fragment. A helper T cell with a complementary receptor binds the displayed antigen-MHC complex.

(b) A B cell with a complementary receptor is selected after binding the antigen and receiving activation signals. The selected B cell proliferates to form a clone. Plasma cells from the clone secrete antibodies specific to the viral antigen.

A B cell with a complementary receptor is selected after binding the antigen and receiving activation signals. The selected B cell proliferates to form a clone. Plasma cells from the clone secrete antibodies specific to the viral antigen.

(c) The secondary response will be faster and usually larger than the primary response. Antigen-specific memory cells persist and rapidly undergo clonal expansion on re-exposure. Faster antibody production can clear the virus before severe disease develops.

The secondary response will be faster and usually larger than the primary response. Antigen-specific memory cells persist and rapidly undergo clonal expansion on re-exposure. Faster antibody production can clear the virus before severe disease develops.

Detailed marking criteria

Part a (1 mark)

An antigen-presenting cell processes virus R and displays an antigen fragment.

Part a (1 mark)

A helper T cell with a complementary receptor binds the displayed antigen-MHC complex.

Part b (1 mark)

A B cell with a complementary receptor is selected after binding the antigen and receiving activation signals.

Part b (1 mark)

The selected B cell proliferates to form a clone.

Part b (1 mark)

Plasma cells from the clone secrete antibodies specific to the viral antigen.

Part c (1 mark)

The secondary response will be faster and usually larger than the primary response.

Part c (1 mark)

Antigen-specific memory cells persist and rapidly undergo clonal expansion on re-exposure.

Part c (1 mark)

Faster antibody production can clear the virus before severe disease develops.

Section B Question 4

(a) Town A cases fell by 52, from 80 to 28. Town B cases fell by 21, from 76 to 55. The larger fall in Town A is associated with, but does not alone prove, an effect of the campaign.

Town A cases fell by 52, from 80 to 28. Town B cases fell by 21, from 76 to 55. The larger fall in Town A is associated with, but does not alone prove, an effect of the campaign.

(b) Vaccinated people are less likely to become susceptible hosts or develop transmissible infection. Fewer infectious contacts interrupt transmission chains. Reduced circulation indirectly protects some susceptible people through population-level immunity.

Vaccinated people are less likely to become susceptible hosts or develop transmissible infection. Fewer infectious contacts interrupt transmission chains. Reduced circulation indirectly protects some susceptible people through population-level immunity.

(c) Comparable testing and reporting rates in both towns are needed to exclude surveillance differences. Information about other controls, movement, prior immunity or demographic differences is needed to assess confounding.

Comparable testing and reporting rates in both towns are needed to exclude surveillance differences. Information about other controls, movement, prior immunity or demographic differences is needed to assess confounding.

Detailed marking criteria

Part a (1 mark)

Town A cases fell by 52, from 80 to 28.

Part a (1 mark)

Town B cases fell by 21, from 76 to 55.

Part a (1 mark)

The larger fall in Town A is associated with, but does not alone prove, an effect of the campaign.

Part b (1 mark)

Vaccinated people are less likely to become susceptible hosts or develop transmissible infection.

Part b (1 mark)

Fewer infectious contacts interrupt transmission chains.

Part b (1 mark)

Reduced circulation indirectly protects some susceptible people through population-level immunity.

Part c (1 mark)

Comparable testing and reporting rates in both towns are needed to exclude surveillance differences.

Part c (1 mark)

Information about other controls, movement, prior immunity or demographic differences is needed to assess confounding.

Section B Question 5

(a) Mutations arise independently of whether they will be useful in a later environment. The resistant variant was already present before antibiotic exposure. The antibiotic selected among existing heritable variants rather than directing a needed mutation.

Mutations arise independently of whether they will be useful in a later environment. The resistant variant was already present before antibiotic exposure. The antibiotic selected among existing heritable variants rather than directing a needed mutation.

(b) Antibiotic Z kills or inhibits susceptible bacteria more strongly. Resistant bacteria survive and reproduce at a higher rate under treatment. They pass the resistance allele to descendants, increasing its frequency over generations.

Antibiotic Z kills or inhibits susceptible bacteria more strongly. Resistant bacteria survive and reproduce at a higher rate under treatment. They pass the resistance allele to descendants, increasing its frequency over generations.

(c) Selection favouring resistance would weaken when the antibiotic is absent. If resistance carries a fitness cost, susceptible bacteria may increase relative to resistant bacteria.

Selection favouring resistance would weaken when the antibiotic is absent. If resistance carries a fitness cost, susceptible bacteria may increase relative to resistant bacteria.

Detailed marking criteria

Part a (1 mark)

Mutations arise independently of whether they will be useful in a later environment.

Part a (1 mark)

The resistant variant was already present before antibiotic exposure.

Part a (1 mark)

The antibiotic selected among existing heritable variants rather than directing a needed mutation.

Part b (1 mark)

Antibiotic Z kills or inhibits susceptible bacteria more strongly.

Part b (1 mark)

Resistant bacteria survive and reproduce at a higher rate under treatment.

Part b (1 mark)

They pass the resistance allele to descendants, increasing its frequency over generations.

Part c (1 mark)

Selection favouring resistance would weaken when the antibiotic is absent.

Part c (1 mark)

If resistance carries a fitness cost, susceptible bacteria may increase relative to resistant bacteria.

Section B Question 6

(a) Species J and K are inferred to be the closest pair. Their sequences differ at only 4 positions, fewer than either comparison with L.

Species J and K are inferred to be the closest pair. Their sequences differ at only 4 positions, fewer than either comparison with L.

(b) Sequence similarity provides evidence about relative relatedness. The fossil provides evidence of morphology and occurrence in the past. Radiometric age with uncertainty constrains the timing of the lineage but is not an exact divergence date.

Sequence similarity provides evidence about relative relatedness. The fossil provides evidence of morphology and occurrence in the past. Radiometric age with uncertainty constrains the timing of the lineage but is not an exact divergence date.

(c) The inference uses one gene and may not represent the whole evolutionary history. Fossil assignment and dating have uncertainty. Additional genes or fossils could support or change the proposed relationship.

The inference uses one gene and may not represent the whole evolutionary history. Fossil assignment and dating have uncertainty. Additional genes or fossils could support or change the proposed relationship.

Detailed marking criteria

Part a (1 mark)

Species J and K are inferred to be the closest pair.

Part a (1 mark)

Their sequences differ at only 4 positions, fewer than either comparison with L.

Part b (1 mark)

Sequence similarity provides evidence about relative relatedness.

Part b (1 mark)

The fossil provides evidence of morphology and occurrence in the past.

Part b (1 mark)

Radiometric age with uncertainty constrains the timing of the lineage but is not an exact divergence date.

Part c (1 mark)

The inference uses one gene and may not represent the whole evolutionary history.

Part c (1 mark)

Fossil assignment and dating have uncertainty.

Part c (1 mark)

Additional genes or fossils could support or change the proposed relationship.

Section B Question 7

(a) A guide RNA base-pairs with a complementary target sequence. Cas9 cuts DNA at or near the targeted site. Cellular DNA repair can introduce a sequence change that disrupts the regulatory site.

A guide RNA base-pairs with a complementary target sequence. Cas9 cuts DNA at or near the targeted site. Cellular DNA repair can introduce a sequence change that disrupts the regulatory site.

(b) Disrupting the regulatory sequence is associated with reduced transcription or mRNA abundance. Less mRNA provides fewer templates for translation. The protein reduction is consistent with, but does not prove only, reduced mRNA availability.

Disrupting the regulatory sequence is associated with reduced transcription or mRNA abundance. Less mRNA provides fewer templates for translation. The protein reduction is consistent with, but does not prove only, reduced mRNA availability.

(c) Sequence the target region and assess likely off-target sites to verify edit specificity. Use independent edited lines or a rescue treatment to distinguish the intended edit from clone-specific effects.

Sequence the target region and assess likely off-target sites to verify edit specificity. Use independent edited lines or a rescue treatment to distinguish the intended edit from clone-specific effects.

Detailed marking criteria

Part a (1 mark)

A guide RNA base-pairs with a complementary target sequence.

Part a (1 mark)

Cas9 cuts DNA at or near the targeted site.

Part a (1 mark)

Cellular DNA repair can introduce a sequence change that disrupts the regulatory site.

Part b (1 mark)

Disrupting the regulatory sequence is associated with reduced transcription or mRNA abundance.

Part b (1 mark)

Less mRNA provides fewer templates for translation.

Part b (1 mark)

The protein reduction is consistent with, but does not prove only, reduced mRNA availability.

Part c (1 mark)

Sequence the target region and assess likely off-target sites to verify edit specificity.

Part c (1 mark)

Use independent edited lines or a rescue treatment to distinguish the intended edit from clone-specific effects.

Section B Question 8

(a) Increasing light intensity will increase the mean rate of oxygen production by the algae. The prediction applies while other relevant conditions are controlled.

Increasing light intensity will increase the mean rate of oxygen production by the algae. The prediction applies while other relevant conditions are controlled.

(b) The independent variable is assigned light intensity. The dependent variable is change in dissolved oxygen per unit time.

The independent variable is assigned light intensity. The dependent variable is change in dissolved oxygen per unit time.

(c) Keep water and electrical equipment separated or use a low-voltage sealed light source. Keep algal amount, temperature, solution volume or carbon dioxide availability constant.

Keep water and electrical equipment separated or use a low-voltage sealed light source. Keep algal amount, temperature, solution volume or carbon dioxide availability constant.

(d) Replicates reveal variability and permit a more reliable treatment mean. A dark control estimates oxygen change not caused by light-dependent photosynthesis.

Replicates reveal variability and permit a more reliable treatment mean. A dark control estimates oxygen change not caused by light-dependent photosynthesis.

Detailed marking criteria

Part a (1 mark)

Increasing light intensity will increase the mean rate of oxygen production by the algae.

Part a (1 mark)

The prediction applies while other relevant conditions are controlled.

Part b (1 mark)

The independent variable is assigned light intensity.

Part b (1 mark)

The dependent variable is change in dissolved oxygen per unit time.

Part c (1 mark)

Keep water and electrical equipment separated or use a low-voltage sealed light source.

Part c (1 mark)

Keep algal amount, temperature, solution volume or carbon dioxide availability constant.

Part d (1 mark)

Replicates reveal variability and permit a more reliable treatment mean.

Part d (1 mark)

A dark control estimates oxygen change not caused by light-dependent photosynthesis.

Section B Question 9

(a) Mean rate increases from pH 4 to pH 7 and decreases at higher pH. The highest tested mean is 6.1 mL per minute at pH 7.

Mean rate increases from pH 4 to pH 7 and decreases at higher pH. The highest tested mean is 6.1 mL per minute at pH 7.

(b) Extreme pH can alter charges and interactions maintaining enzyme structure. A changed active-site shape or chemistry reduces productive substrate binding.

Extreme pH can alter charges and interactions maintaining enzyme structure. A changed active-site shape or chemistry reduces productive substrate binding.

(c) Without spread or uncertainty, the reliability of differences between means cannot be judged. Record individual trials and present a suitable spread or uncertainty measure with additional pH values near 7.

Without spread or uncertainty, the reliability of differences between means cannot be judged. Record individual trials and present a suitable spread or uncertainty measure with additional pH values near 7.

Detailed marking criteria

Part a (1 mark)

Mean rate increases from pH 4 to pH 7 and decreases at higher pH.

Part a (1 mark)

The highest tested mean is 6.1 mL per minute at pH 7.

Part b (1 mark)

Extreme pH can alter charges and interactions maintaining enzyme structure.

Part b (1 mark)

A changed active-site shape or chemistry reduces productive substrate binding.

Part c (1 mark)

Without spread or uncertainty, the reliability of differences between means cannot be judged.

Part c (1 mark)

Record individual trials and present a suitable spread or uncertainty measure with additional pH values near 7.

Section B Question 10

Indicative answer: Explains that the edit changes a defined DNA sequence or gene-expression process. Links the molecular change to reduced offspring survival through a plausible protein or regulatory effect. Explains that inheritance is required for the construct to spread beyond released individuals. Applies selection or fitness reasoning to predict possible changes in construct frequency. Explains how reducing the mosquito population or competent vectors could interrupt viral transmission. Recognises that reduced transmission does not directly create antigen-specific immunity in people. Proposes a staged controlled study with comparison populations, replication and defined transmission outcomes. Identifies monitoring for resistance, off-target effects or ecological consequences over multiple generations. Applies an ethical requirement such as community engagement, informed governance, proportional risk control or environmental responsibility. Reaches a qualified conclusion that weighs evidence and uncertainty rather than claiming guaranteed success or zero risk.

Detailed marking criteria

Integrated response (1 mark)

Explains that the edit changes a defined DNA sequence or gene-expression process.

Integrated response (1 mark)

Links the molecular change to reduced offspring survival through a plausible protein or regulatory effect.

Integrated response (1 mark)

Explains that inheritance is required for the construct to spread beyond released individuals.

Integrated response (1 mark)

Applies selection or fitness reasoning to predict possible changes in construct frequency.

Integrated response (1 mark)

Explains how reducing the mosquito population or competent vectors could interrupt viral transmission.

Integrated response (1 mark)

Recognises that reduced transmission does not directly create antigen-specific immunity in people.

Integrated response (1 mark)

Proposes a staged controlled study with comparison populations, replication and defined transmission outcomes.

Integrated response (1 mark)

Identifies monitoring for resistance, off-target effects or ecological consequences over multiple generations.

Integrated response (1 mark)

Applies an ethical requirement such as community engagement, informed governance, proportional risk control or environmental responsibility.

Integrated response (1 mark)

Reaches a qualified conclusion that weighs evidence and uncertainty rather than claiming guaranteed success or zero risk.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Nucleic acids and proteins A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, B1, B7 26 ___ Review DNA regulation, gene expression, protein synthesis, protein structure and biotechnology.
Regulation of biochemical pathways A11, A12, A13, A14, A15, A16, A17, A18, B2 16 ___ Review photosynthesis, cellular respiration, enzymes and pathway regulation.
Responses to pathogens A19, A20, A21, A22, A23, A24, A25, A26, B3, B4 24 ___ Review pathogen entry, innate and adaptive immunity, vaccination and disease control.
Relatedness and biological change A27, A28, A29, A30, A31, A32, A33, A34, B5, B6 24 ___ Review mutation, selection, speciation, evidence for evolution and phylogenetic reasoning.
Scientific inquiry A35, A36, A37, A38, A39, A40, B8, B9, B10 30 ___ Review hypotheses, variables, ethics, data processing, method evaluation and evidence-based conclusions.

What is included

Question and Answer Book questions (120 marks)

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Diagnostic checklist shown online

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