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ACT BSSS Biology T — Free Online Pack 0

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ACT BSSS integrated Biology T practice assessment 2026 Edition - Pack 0 v2.0
Updated 12 Oct 2026

2026 Edition

Course alignment: ACT BSSS Biology A/T course — from 2023

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Course alignment: Biology A/T course — from 2023.

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ACT BSSS Biology T

20 questions

100 marks

Reading: Teacher-set · Writing: Suggested 120 minutesFree online paper · no PDF download or checkout

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Covered in this pack

  • Current ACT BSSS Biology A/T course document - Biotechnology and molecular evidence 2 marks · 2 questions
  • Current ACT BSSS Biology A/T course document - DNA, gene expression and continuity 13 marks · 4 questions
  • Current ACT BSSS Biology A/T course document - Evolution and population change 3 marks · 3 questions
  • Current ACT BSSS Biology A/T course document - Homeostasis and response 22 marks · 3 questions
  • Current ACT BSSS Biology A/T course document - Infectious disease and immunity 27 marks · 3 questions
  • Current ACT BSSS Biology A/T course document - Inheritance and variation 12 marks · 3 questions
  • Current ACT BSSS Biology A/T course document - Science inquiry skills 21 marks · 2 questions

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Skill Align ACT BSSS Biology T - Free Online Pack 0

An independently authored, adaptable school-style assessment with worked answers and question-specific marking guidance.

Paper
Integrated practice assessment
Reading
Teacher-set
Writing
Suggested 120 minutes
Assessment
100 marks

No centrally prescribed ACT BSSS Biology examination reference sheet is claimed. All sequences, datasets and statistical information required to answer a question are supplied with it.

Part A - Multiple choice

Questions 1 to 15 (15 marks). Select the best answer for each question.

Question 1

1 mark
In the Molonglo riparian beetle study, a short Pgm coding-strand segment is 5'-ATGGAACTT-3'. Which sequence shows its complementary DNA template strand with correct antiparallel orientation? Write the template with its 3-prime end on the left.
  1. 5'-TACCTTGAA-3'
  2. 3'-TACCTTGAA-5'
  3. 5'-AAGTTCCAT-3'
  4. 5'-AUGGAACUU-3'

Question 2

1 mark
The displayed Pgm sequence is the coding DNA strand 5'-ATGGAACTT-3'. Ignoring RNA processing, which mRNA segment is transcribed?
  1. 3'-AUGGAACUU-5'
  2. 5'-UACCUUGAA-3'
  3. 5'-ATGGAACTT-3'
  4. 5'-AUGGAACUU-3'

Question 3

1 mark
The diagram shows equal-length coding sequences. The supplied code assigns GAA and GAG to Glu. How should the displayed change be classified?
Diagram Preview
Pgm reference coding DNA 5'-ATGGAACTT-3' Pgm variant coding DNA 5'-ATGGAGCTT-3' observed molecular one or more codonsdiffer
  1. a synonymous base substitution
  2. a one-base frameshift
  3. an in-frame codon deletion
  4. a nonsense base substitution

Question 4

1 mark
A diploid cell from a synthetic Molonglo riparian beetle population contains 12 chromosomes before DNA replication. How many chromosomes should one normal gamete contain after meiosis?
  1. 12
  2. 24
  3. 6
  4. 10

Question 5

1 mark
At a separate experimental pigment locus, all beetles have the same assayed Bb genotype and are randomly allocated to cool or warm rearing. The graph shows the proportion developing pigment. Which interpretation fits the evidence?
Graph Preview
80cool20warmrearing environmentpigmented individuals (%)
  1. the warm group must have inherited more recessive alleles
  2. the two groups demonstrate different segregation ratios at conception
  3. environment can affect phenotype expression within one genotype
  4. the pigment phenotype uniquely identifies the bb genotype

Question 6

1 mark
A DNA profile for the Pgm study contains fragments of 180, 420 and 710 base pairs. Under the same gel conditions, which fragment should migrate furthest from the wells?
  1. the 420 base-pair fragment
  2. the 180 base-pair fragment
  3. the 710 base-pair fragment
  4. all fragments should migrate the same distance

Question 7

1 mark
A valid Pgm PCR model begins with one target molecule and doubles it in each of 4 ideal cycles. How many target molecules are predicted?
  1. 16
  2. 5
  3. 12
  4. 8

Question 8

1 mark
In a separate beetle population, a focal allele rises from 24% to 39%. What is the increase in percentage points?
  1. 62.5 percentage points
  2. 39 percentage points
  3. 24 percentage points
  4. 15 percentage points

Question 9

1 mark
In a synthetic Molonglo riparian beetle population, thirty marked beetles enter from a neighbouring reach carrying the P allele. Which mechanism most directly describes the resulting allele-frequency influence?
  1. gene flow
  2. individual organisms evolving because they need to
  3. homeostasis changing the DNA sequence in every cell
  4. translation producing an identical new allele in all offspring

Question 10

1 mark
Two separated beetle populations meet again and produce many fertile offspring in controlled crosses. What does this show?
  1. the populations must use different genetic codes
  2. geographic separation always makes new species
  3. complete reproductive isolation has not been demonstrated
  4. fertility proves their allele frequencies are identical

Question 11

1 mark
In the displayed pathway, a sensor fails to detect rising trehalose although effectors remain intact. What is the most direct limitation?
Diagram Preview
trehalose rises metabolism adjusts regulated range negative feedback
  1. the pathway becomes DNA replication
  2. the original disturbance must disappear
  3. all effectors must permanently increase activity
  4. the corrective response may not be initiated

Question 12

1 mark
A beetle is gaining heat in sunlight while the shaded ground is cooler than its body. Which immediate response would reduce net heat gain?
  1. increase flight-muscle activity while remaining in sunlight
  2. move onto the shaded ground
  3. move onto warmer exposed ground
  4. reduce evaporative water loss while remaining in sunlight

Question 13

1 mark
In a controlled influenza exposure study, a barrier blocks direct spray of large droplets and shared surfaces are excluded. Transmission still occurs across shared air. Which route best explains this result?
  1. inhalation of infectious particles remaining in shared air
  2. direct impact of large droplets blocked by the barrier
  3. transfer from the excluded shared surfaces to the mouth
  4. ingestion of contaminated food that was not shared

Question 14

1 mark
A wound becomes inflamed before a new antigen-specific antibody response develops. Which process belongs to this early innate response?
  1. memory B cells change the pathogen's genome
  2. every lymphocyte makes the same specific antibody
  3. phagocytes engulf invading material
  4. antibodies encode new chromosomes

Question 15

1 mark
Beetle colour is scored from photographs. What most directly reduces observer expectation bias?
  1. use one expected-looking photograph
  2. conceal treatment identities from the scorer
  3. discard unexpected colours
  4. tell the scorer which result is wanted

Part B - Biological concepts and models

Questions 16 and 17 (20 marks). Apply biological mechanisms, models and molecular evidence.

Question 16

10 marks

A separate beetle enzyme locus has coding DNA 5'-ATGTTTCAA-3'; its variant is 5'-ATGTTCCAA-3'. Read from the first base. The supplied code is AUG = Met, UUU/UUC = Phe, CAA = Gln. Enzyme activity is 10 units in reference cells and 4 in variant cells, but protein abundance was not measured.

Diagram Preview
reference coding DNA 5'-ATGTTTCAA-3' variant coding DNA 5'-ATGTTCCAA-3' reading frame read from first base
(a) 5 marks
Determine the two peptide segments using the supplied code and explain the roles of ribosomes and tRNA.
(b) 5 marks
Assess the claim that the variant changes the enzyme's amino-acid sequence and thereby causes the activity decrease; propose a discriminating test.

Question 17

10 marks
Stimulus

A simplified model treats A as the dominant allele for dark elytra and a as recessive. Two heterozygous parents produce 120 offspring. PCR produces a larger A fragment and a smaller a fragment. The gel shows both parental lanes and one offspring lane; smaller fragments migrate further.

Use the supplied context to answer all parts of this question.
Diagram Preview
ladder parent 1 parent 2 offspring wells migrationdistance
(a) 5 marks
Use a probability model to predict genotype proportions and the expected number of offspring with the recessive phenotype.
(b) 5 marks
Interpret the offspring lane and explain two method controls needed before using the gel as genotype evidence.

Part C - Data analysis

Question 18 (20 marks). Analyse the supplied population evidence and justify conclusions.

Question 18

20 marks
Stimulus

A synthetic monitoring program follows a synthetic Molonglo riparian beetle population for five generations during a shift from pale gravel to dark leaf litter. The plotted allele-frequency values are original teaching data. Source data (focal allele frequency): G0 = 18%; G1 = 25%; G2 = 33%; G3 = 41%; G4 = 49%. The phenotype of interest is dark elytra that reduce visual detection. Field notes also record that thirty marked beetles enter from a neighbouring reach carrying the P allele; migration rate and effective population size were not measured directly.

Use the supplied context to answer all parts of this question.
Graph Preview
G0G1G2G3G46040200generationfocal allele frequency (%)
(a) 5 marks
Describe the allele-frequency pattern quantitatively without claiming a cause.
(b) 5 marks
Explain how natural selection could produce the observed change.
(c) 5 marks
Explain why gene flow or genetic drift remains a competing or interacting explanation.
(d) 5 marks
Design a stronger monitoring strategy that could distinguish selection from the alternatives.

Part D - Practical investigation

Question 19 (20 marks). Apply science inquiry skills to the biological investigation.

Question 19

20 marks
Stimulus

Students investigate haemolymph trehalose concentration using a synthetic Molonglo riparian beetle population. The stated disturbance for the feedback model is that haemolymph trehalose concentration rises. The control, moderate-change and large-change diets contain 2%, 6% and 10% digestible carbohydrate respectively. Four organisms are assigned to each condition for the same exposure time. Mean regulated-response indices are 20, 29 and 37; individual values and spread are not reported. The proposed conclusion is that a larger disturbance always produces a stronger successful homeostatic response.

Use the supplied context to answer all parts of this question.
Diagram Preview
diet carbohydrate controlled variables(duration, stage,method) response index compare means andspread
(a) 5 marks
Explain the relevant negative-feedback mechanism using stimulus, detection, coordination and effector response.
(b) 5 marks
Evaluate the design and specify a safer, more valid and ethical procedure.
(c) 5 marks
Interpret the reported means and explain why they do not establish the proposed conclusion.
(d) 5 marks
Propose evidence that would test repeatability and whether the response maintained the internal variable within tolerance limits.

Part E - Extended response

Question 20 (25 marks). Integrate biology, evidence, context, risk and justified evaluation.

Question 20

25 marks
Stimulus

A synthetic community study of influenza A virus compares a setting using improved ventilation plus vaccination with a comparison setting. In this question, the stated transmission route is droplet and aerosol transmission: respiratory particles spread during crowded indoor rehearsals. The graph reports weekly notified cases per 1,000 registered people for four equivalent observation periods; each rate uses that setting's weekly population denominator and the same testing eligibility. Source data (weekly notified cases per 1,000 people): Comparison: week 1 = 12, week 2 = 29, week 3 = 48, week 4 = 63; Intervention: week 1 = 11, week 2 = 20, week 3 = 27, week 4 = 31. However, the intervention population is smaller and participation differs between rooms. No confidence intervals or pathogen genomic data are supplied.

Use the supplied context to answer all parts of this question.
Graph Preview
week 1week 2week 3week 4806040200 Comparison Interventiontimeweekly cases per 1,000 people
(a) 5 marks
Explain how the pathogen and stated transmission route can produce the observed outbreak pattern.
(b) 5 marks
Compare innate and adaptive immune responses relevant to infection and later exposure.
(c) 5 marks
Compare the population-standardised series quantitatively and state a bounded epidemiological conclusion.
(d) 5 marks
Evaluate the evidence, risk and contextual factors that should qualify a decision.
(e) 5 marks
Recommend a staged action and specify evidence that would determine continuation, modification or stopping.

ACT BSSS Biology T assessment is school-based. This independent Skill Align practice assessment is not an official ACT BSSS examination and may be adapted by a college or teacher. Skill Align is not affiliated with, authorised by, sponsored by, approved by, or endorsed by ACT BSSS or the ACT Government.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank. Every question, dataset, answer and marking description in this pack is original Skill Align content.

Worked Solutions And Marking Guide

General marking principles

  • Award each one-mark criterion independently for the specific biological evidence described.
  • Accept equivalent scientifically accurate terminology, representations, calculations or reasoning applied to the supplied context.
  • Carry forward an earlier numerical error when later biology is internally consistent, unless a later criterion independently assesses that step.
  • Use the task-specific evidence analytically across Concepts, Models and Applications, Contexts, and Inquiry Skills.

Question 1

Answer: 3'-TACCTTGAA-5'

Complementary DNA bases pair A-T and C-G, and the two strands run antiparallel.

Question 2

Answer: 5'-AUGGAACUU-3'

An mRNA sequence matches the coding strand in the 5' to 3' direction except uracil replaces thymine.

Question 3

Answer: a synonymous base substitution

One base is replaced, without changing the encoded Glu or shifting the reading frame. GAA -> GAG is synonymous (Glu -> Glu). The displayed products are Met-Glu-Leu and Met-Glu-Leu

Question 4

Answer: 6

Meiosis halves the diploid chromosome number, so a normal gamete contains 12/2 = 6 chromosomes.

Question 5

Answer: environment can affect phenotype expression within one genotype

The assayed genotype is held constant; phenotype differences between allocated environments concern expression, rather than an offspring segregation probability.

Question 6

Answer: the 180 base-pair fragment

Smaller DNA fragments move more readily through the gel matrix and therefore travel further from the wells.

Question 7

Answer: 16

Ideal PCR doubling gives 2^4 = 16 target molecules from one starting molecule.

Question 8

Answer: 15 percentage points

Subtract frequencies expressed as percentages: 39 - 24 = 15 percentage points. The relative percentage increase, a different quantity, is 62.5%.

Question 9

Answer: gene flow

The scenario directly describes gene flow; its population-level effect must be distinguished from change within an individual.

Question 10

Answer: complete reproductive isolation has not been demonstrated

Fertile offspring demonstrate reproductive compatibility under the tested conditions; other barriers remain possible.

Question 11

Answer: the corrective response may not be initiated

Detection is required to initiate the shown corrective pathway.

Question 12

Answer: move onto the shaded ground

Shade reduces radiant heat input and cooler ground permits heat loss; muscle activity, warmer surroundings or less evaporative cooling do not produce the same heat-budget change.

Question 13

Answer: inhalation of infectious particles remaining in shared air

The exclusions leave shared-air inhalation as the supported route. The result does not quantify particle size or establish that other routes never occur.

Question 14

Answer: phagocytes engulf invading material

Inflammation recruits innate cells, including phagocytes, before a primary adaptive response develops.

Question 15

Answer: conceal treatment identities from the scorer

Blinding reduces systematic scoring differences based on expected treatment effects.

Question 16

(a) Both segments encode Met-Phe-Gln because UUU and UUC both encode Phe. Ribosomes read successive mRNA codons, tRNAs pair anticodons with codons and deliver amino acids, and peptide bonds join the amino acids.

(b) The displayed substitution is synonymous and leaves this peptide segment unchanged, so the proposed amino-acid explanation is unsupported. Activity falls by 6 units, or 60%, but abundance, other sequence regions or experimental conditions could differ. Measure activity per equal amount of purified protein from matched replicated genotypes to separate abundance from activity per molecule.

Detailed marking criteria

Part a (1 mark)

Determines the reference peptide as Met-Phe-Gln.

Part a (1 mark)

Determines the variant peptide as Met-Phe-Gln.

Part a (1 mark)

Explains that UUU and UUC specify the same amino acid.

Part a (1 mark)

Links ribosomes with codon reading and peptide-bond formation.

Part a (1 mark)

Links tRNA anticodon pairing with amino-acid delivery.

Part b (1 mark)

Rejects a changed amino-acid sequence for the supplied segment.

Part b (1 mark)

Quantifies the activity decrease as 6 units or 60%.

Part b (1 mark)

Identifies protein abundance or an unmeasured sequence/condition as an alternative.

Part b (1 mark)

Proposes an assay using equal protein amounts under matched conditions.

Part b (1 mark)

Explains how replication and activity per protein distinguish the alternatives.

Question 17

(a) Aa × Aa gives genotype proportions 1/4 AA, 1/2 Aa and 1/4 aa, with phenotype proportions 3/4 dominant and 1/4 recessive. The expected recessive count is 120 × 1/4 = 30. Actual counts may differ through random fertilisation and survival.

(b) The offspring lane contains an upper larger A fragment and a lower smaller a fragment, supporting Aa under the model. This is consistent with heterozygosity, conditional on reliable amplification and allele assignment. Explain any two suitable controls: known-genotype samples verify allele assignment; a ladder checks size; a no-template tube tests contamination; matched extraction/amplification conditions test comparability. Two well-explained controls suffice.

Detailed marking criteria

Part a (1 mark)

States the 1 AA : 2 Aa : 1 aa genotype ratio.

Part a (1 mark)

States the 3 dominant : 1 recessive phenotype ratio.

Part a (1 mark)

Identifies aa as the recessive phenotype genotype.

Part a (1 mark)

Calculates the expected recessive count as 30.

Part a (1 mark)

Distinguishes an expected probability from an exact required offspring count.

Part b (1 mark)

Identifies both the upper A-sized and lower a-sized products.

Part b (1 mark)

Infers heterozygous genotype Aa under the supplied model.

Part b (1 mark)

Explains the purpose of one valid method control.

Part b (1 mark)

Explains the purpose of a second distinct valid method control.

Part b (1 mark)

Qualifies the genotype inference as dependent on reliable amplification or correct allele assignment.

Question 18

(a) Frequency rises in every interval from 18% at G0 to 49% at G4, a net increase of 31 percentage points. The plotted pattern is consistent and positive, but the graph alone does not identify its mechanism.

(b) If the phenotype is heritable and improves survival or reproductive success under a shift from pale gravel to dark leaf litter because it involves dark elytra that reduce visual detection, carriers contribute more offspring. The associated allele is then represented more often in the next generation, increasing its frequency over time.

(c) The supplied context states that thirty marked beetles enter from a neighbouring reach carrying the P allele, which is evidence for gene flow. Gene flow can add alleles through migrants, whereas drift changes frequencies by chance most strongly in small populations. Either process can act with selection, so migrant genotypes and effective population size are required.

(d) Track replicated populations and reference sites; genotype random samples each generation; measure the candidate phenotype, survival and reproductive output; estimate effective population size and migration; compare change at the candidate locus with neutral loci and report sampling uncertainty.

Detailed marking criteria

Part a (1 mark)

States that frequency rises in every plotted interval.

Part a (1 mark)

States the initial value of 18%.

Part a (1 mark)

States the final value of 49%.

Part a (1 mark)

Calculates a net increase of 31 percentage points.

Part a (1 mark)

Keeps the description separate from a causal claim.

Part b (1 mark)

Identifies heritable phenotypic variation in the population.

Part b (1 mark)

Identifies a shift from pale gravel to dark leaf litter as the relevant selection pressure.

Part b (1 mark)

Links the phenotype of interest, dark elytra that reduce visual detection, with a plausible advantage.

Part b (1 mark)

Links the advantage with differential survival or reproduction.

Part b (1 mark)

Links differential reproduction with the observed allele-frequency increase.

Part c (1 mark)

Uses the supplied context to identify gene flow.

Part c (1 mark)

Explains gene flow as allele movement between populations.

Part c (1 mark)

Explains drift as chance sampling of alleles.

Part c (1 mark)

Links drift strength with small effective population size.

Part c (1 mark)

Explains that population mechanisms can interact rather than being mutually exclusive.

Part d (1 mark)

Requires replicated populations or suitable reference sites.

Part d (1 mark)

Requires random repeated genotyping across generations.

Part d (1 mark)

Requires phenotype-linked survival or reproductive evidence.

Part d (1 mark)

Requires effective population-size and migration estimates.

Part d (1 mark)

Requires neutral-locus comparison and quantified uncertainty.

Question 19

(a) The stimulus is that haemolymph trehalose concentration rises. An appropriate sensor detects the deviation. In this organism, metabolic or hormonal signals following carbohydrate sensing coordinate tissue responses that reduce circulating trehalose; any biologically defensible pathway is acceptable and specific hormone names are not required. The resulting response is that trehalose synthesis, transport and use are adjusted to lower the circulating concentration. This opposes the original disturbance so haemolymph trehalose returns towards its regulated range; the response is limited by the organism's tolerance range.

(b) Use a justified non-harmful treatment range from pilot evidence; randomly allocate comparable study units; standardise life stage, acclimation, exposure time and measurement method; use adequate independent replication; and apply organism-appropriate care and biosecurity that uses non-harmful diets, minimises handling and specifies recovery and housing conditions.

(c) The mean response index increases from 20 to 29 and 37 as the disturbance increases. Means from four organisms without individual values, spread or measurement uncertainty cannot show reliability or overlap. The data also do not show whether regulation succeeded or exceeded tolerance limits.

(d) Repeat independent trials with more organisms, record the regulated internal variable and effector response over time, report individual data and confidence intervals, include recovery measurements and predefine the tolerance range. Replicate the study at relevant environmental conditions before generalising.

Detailed marking criteria

Part a (1 mark)

Identifies the stimulus: haemolymph trehalose concentration rises.

Part a (1 mark)

Identifies detection by a biologically appropriate receptor or sensor.

Part a (1 mark)

Explains biologically defensible metabolic or hormonal coordination that lowers circulating trehalose; specific hormone names are not required.

Part a (1 mark)

Identifies the effector response: trehalose synthesis, transport and use are adjusted to lower the circulating concentration.

Part a (1 mark)

Explains how the response means haemolymph trehalose returns towards its regulated range.

Part b (1 mark)

Requires a justified treatment range that avoids foreseeable harm.

Part b (1 mark)

Requires random allocation of comparable study units.

Part b (1 mark)

Controls acclimation, exposure duration and measurement method.

Part b (1 mark)

Requires adequate independent replication rather than four convenient observations.

Part b (1 mark)

Specifies organism-appropriate care or biosecurity that uses non-harmful diets, minimises handling and specifies recovery and housing conditions.

Part c (1 mark)

States the ordered increase across all three reported means.

Part c (1 mark)

Calculates the control-to-large-condition increase as 17 index units.

Part c (1 mark)

Identifies the small sample of four organisms per condition.

Part c (1 mark)

Identifies missing individual values, spread or uncertainty.

Part c (1 mark)

Explains that response magnitude is not proof of successful regulation within tolerance limits.

Part d (1 mark)

Requires independent repeat trials with increased sample size.

Part d (1 mark)

Measures the regulated internal variable as well as the response index.

Part d (1 mark)

Uses time-series and recovery measurements.

Part d (1 mark)

Reports individual data and uncertainty such as confidence intervals.

Part d (1 mark)

Defines tolerance limits and bounds generalisation to tested conditions.

Question 20

(a) Influenza A virus is the infectious agent. The stated route is droplet and aerosol transmission: infectious respiratory particles are inhaled and the virus replicates in susceptible respiratory cells; crowding increases effective contacts while ventilation lowers particle exposure. This mechanism is consistent with the supplied context that respiratory particles spread during crowded indoor rehearsals. Infected people shed the newly replicated virus, becoming new sources of infection at rehearsals. This can occur before symptoms develop and cases are recognised. Successive exposure and infection events, followed by incubation and reporting delays, can produce the rising weekly pattern.

(b) Innate barriers, inflammation, complement and phagocytes respond rapidly and non-specifically. Adaptive B lymphocytes produce antigen-specific antibodies, while T lymphocytes provide cell-mediated functions. Clonal expansion takes time during first exposure, and antigen-specific memory B and T cells enable a faster, stronger response to subsequent exposure to the same antigen.

(c) By week 4 the comparison setting has 63 notified cases per 1,000 people and the intervention setting has 31 per 1,000, a rate difference of 32 per 1,000. Both series rise, but the intervention rate grows more slowly with improved ventilation plus vaccination. The denominator-standardised association is consistent with lower transmission risk, not proof that the intervention alone caused it.

(d) The supplied per-1,000 denominator improves comparability, but missing confidence intervals, genomic linkage and individual exposure data limit causal attribution. The intervention population is smaller and participation differs between rooms. Feasibility, equity, privacy, participation, cost and unintended harms should be considered alongside outbreak severity and transmission biology. A decision can be precautionary while evidence is updated.

(e) Continue improved ventilation plus vaccination as a monitored, time-limited measure because the denominator-standardised series is consistent with benefit and the biological mechanism is plausible. Use matched settings or phased introduction, standardised testing, population and person-time denominators, coverage and contact data, pathogen typing where appropriate, adverse-effect and cost monitoring, and confidence intervals. Predefine thresholds for continuation, modification and withdrawal.

Detailed marking criteria

Part a (1 mark)

Identifies influenza A virus as the infectious agent in the scenario.

Part a (1 mark)

Identifies droplet and aerosol transmission.

Part a (1 mark)

Explains that crowding increases effective contacts and ventilation reduces exposure to infectious respiratory particles.

Part a (1 mark)

Explains that influenza virus replicates in susceptible respiratory cells and is shed by infected people, making them new sources of infection.

Part a (1 mark)

Links successive infections, incubation and reporting delays with the weekly case pattern.

Part b (1 mark)

Identifies a rapid non-specific innate response.

Part b (1 mark)

Explains a role for inflammation, complement or phagocytes.

Part b (1 mark)

Explains antigen-specific antibody production by B lymphocytes.

Part b (1 mark)

Explains a relevant T-lymphocyte cell-mediated role.

Part b (1 mark)

Links antigen-specific memory cells with a faster response to subsequent exposure to the same antigen.

Part c (1 mark)

States that both plotted notification-rate series increase over time.

Part c (1 mark)

States that the intervention rate grows more slowly.

Part c (1 mark)

Uses the week-4 rates 63 and 31 per 1,000 people.

Part c (1 mark)

Calculates the week-4 rate difference as 32 per 1,000 people.

Part c (1 mark)

States a denominator-standardised association while withholding an unsupported causal claim.

Part d (1 mark)

Recognises that the supplied population denominator improves comparability but does not remove uncertainty.

Part d (1 mark)

Identifies missing linkage, exposure, confidence-interval or genomic evidence.

Part d (1 mark)

Uses the stated limitation that the intervention population is smaller and participation differs between rooms.

Part d (1 mark)

Identifies at least one relevant social, ethical or economic consideration.

Part d (1 mark)

Explains why a conditional precautionary decision can be justified under uncertainty.

Part e (1 mark)

Makes an explicit staged recommendation rather than an unconditional claim.

Part e (1 mark)

Uses both the plotted association and a plausible transmission mechanism.

Part e (1 mark)

Requires a matched comparison or phased design with standardised testing.

Part e (1 mark)

Requires uncertainty, coverage, harm and cost monitoring.

Part e (1 mark)

Defines decision thresholds for continuing, modifying or stopping the action.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Biotechnology and molecular evidence Q6, Q7, Q16(b), Q17(b) 12 ___ Review sequencing, profiling, PCR, gel electrophoresis, vectors and molecular-evidence limits.
DNA, gene expression and continuity Q1, Q2, Q3, Q16(a) 8 ___ Review DNA replication, transcription, translation, gene expression and mutation effects.
Evolution and population change Q8, Q9, Q10, Q18(b), Q18(c) 13 ___ Review evolutionary evidence, selection, allele-frequency mechanisms, isolation, speciation and diversity.
Homeostasis and response Q11, Q12, Q19(a) 7 ___ Review negative feedback, signalling, thermoregulation, water balance and tolerance limits.
Infectious disease and immunity Q13, Q14, Q20(a), Q20(b), Q20(c) 17 ___ Review pathogen entry, transmission, innate and adaptive immunity, outbreak factors and control.
Inheritance and variation Q4, Q5, Q17(a) 7 ___ Review meiosis, fertilisation, probability models, inheritance patterns and phenotype influences.
Science as a Human Endeavour Q20(d), Q20(e) 10 ___ Review technology, evidence limits, risk, context, collaboration and justified biological decisions.
Science inquiry skills Q15, Q18(a), Q18(d), Q19(b), Q19(c), Q19(d) 26 ___ Review safe and ethical investigation design, data analysis, uncertainty, validity and reliability.

What is included

Integrated practice assessment questions (100 marks)

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Worked solutions and marking guidance shown online

Diagnostic checklist shown online

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How to use this pack

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

Is this pack right for you?

Designed for

Students, parents, teachers and tutors who want to inspect a complete online ACT BSSS Biology T example before choosing a released PDF pack.

Product at a glance

  • ACT BSSS Biology T
  • 2026 edition · Pack 0 · v2.0
  • 1 online paper
  • Free online view · no checkout or PDF download
  • Free to view online
  • No monthly Online Practice subscription required

Compare before buying

Pack 0 is the free online showcase for this course. Other pack numbers identify separate original products; they do not indicate difficulty or a required order.

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.

Current ACT BSSS Biology A/T course document Official curriculum reference, not a prescribed examination specification. Opens on the ACT BSSS website.
Current ACT BSSS Science Course Framework Official assessment and achievement-standard framework. Opens on the ACT BSSS website.
ACT BSSS annotated Biology examination exemplar Official Year 11 assessment example used as design evidence; it is not a prescribed Year 12 examination. Opens on the ACT BSSS website.

Questions about this exam pack

What is included in Biology Free Online - Pack 0?

Pack 0 includes one full-length showcase paper, worked solutions, marking guidance and diagnostic checklists, all shown online.

Is Pack 0 really free?

Yes. Pack 0 can be read online without checkout or a monthly subscription.

What does Pack 0 mean?

Pack 0 identifies the free online showcase for this course. Other pack numbers identify separate original products; they do not indicate difficulty or a required completion order.

Can I download Pack 0 as a PDF?

No. Pack 0 is intentionally online-only and no downloadable PDF is provided.

Are these official assessment authority examination questions?

No. The questions are original Skill Align material. Skill Align is independent and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by any state assessment authority.

Independent practice resource

ACT BSSS courses and senior secondary assessment are administered by the ACT Board of Senior Secondary Studies. Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by ACT BSSS or the ACT Government. ACT BSSS packs are original Skill Align practice assessment resources, not official ACT BSSS subject examinations.

Each exam pack is listed with a pack label so buyers can distinguish separate original products in the same course series. Pack numbers do not indicate difficulty or a required completion order.

2026 exam timetable reference View ACT BSSS assessment system

ACT BSSS has no external subject-based examinations. Your college sets its own course assessment times; the ACT Scaling Test is separate from these practice packs.