HSC Biology 2025
Worked solutions to every question in the 2025 HSC Biology exam. Multiple-choice answers with a one-line reason, and a 'Show worked solution' model answer for each Section II question, aligned to the official NESA marking guidelines.
- Marks
- 100
- Time
- 180 min
- Authority
- NESA
- Updated
Every question from the 2025 HSC Biology exam, with a worked answer. Section II solutions are tucked behind a Show worked solution toggle, so you can attempt a question first and reveal the model answer when you are ready.
How to use this page
- Questions are from the 2025 HSC Biology exam, copyright NSW Education Standards Authority (NESA). Open the official PDF (button above) for the original stimulus diagrams and graphs.
- Answers are original model responses by ExamExplained (Claude Opus 4.8), written to the official marking guidelines, not copied from NESA's sample answers.
- Each Section II solution shows the mark split and a short Marker's note from the notes from the marking centre.
Structure and timing
100 marks in 180 minutes is about 1.8 minutes per mark.
- Section I (20 marks): 20 multiple-choice. Allow about 35 minutes.
- Section II (80 marks): Questions 21 to 34, short and extended response. Allow about 2 hours and 25 minutes, in proportion to the marks. Plan the two long answers (Question 30(b), 7 marks, and Question 32, 7 marks) before you write.
Section I - Multiple choice
- Q1
- Amoeba reproduce by binary fission (nucleus divides, cytoplasm divides, two daughter cells). Which is a characteristic of the daughter cells? A. Genetically different to parent B. Genetically identical to parent C. More organelles D. Fewer chromosomes than the parent
Answer: B - asexual division copies the genome exactly, so the daughters are genetically identical to the parent. - Q2
- What type of mutagen is UV light? A. Biochemical B. Biological C. Chemical D. Electromagnetic
Answer: D - UV is a form of electromagnetic radiation. - Q3
- Which row correctly identifies the role of phagocytes and lymphocytes? Answer: A - phagocytes engulf bacteria; lymphocytes produce antibodies.
- Q4
- A storm randomly kills 80% of a frog population and allele frequencies change notably. What process caused this? A. Gene flow B. Genetic drift C. Natural selection D. Survival of the fittest
Answer: B - a random change in allele frequency in a small population is genetic drift (a bottleneck), not selection. - Q5
- A disease spreads by direct contact. Under which conditions does it spread most rapidly? Answer: B - high population density (more contact) with hosts rarely dying quickly, so they have longer to infect others.
- Q6
- A rabies-bitten person is injected with ready-made antibodies. What immunity is this? A. Innate active B. Natural passive C. Acquired active D. Acquired passive
Answer: D - receiving antibodies rather than making your own is acquired passive immunity. - Q7
- Body temperature stays near 41 degrees C while air temperature varies. Type of animal and why? Answer: C - endotherm; body temperature is relatively constant despite changes in air temperature.
- Q8
- Red x white camellia gives offspring with both red and white petals. Reason? Answer: D - both alleles are expressed together, a co-dominance pattern of inheritance.
- Q9
- Dialysis composition of blood before and after? Answer: A - dialysis removes urea but not glucose, so blood goes from high urea, high glucose to low urea, high glucose.
- Q10
- Measles deaths before and after the 1975 vaccine in the NIP. Which is a trend? Answer: C - deaths fell to near zero after the measles vaccine was added to the NIP (a trend read directly from the data, not an unsupported claim).
- Q11
- Benefit of a 95% childhood vaccination rate? Answer: C - herd immunity protects the unvaccinated remaining 5% by reducing pathogen circulation.
- Q12
- A karyotype shows trisomy 21 with X and Y. Conclusion? Answer: D - a female with a chromosomal mutation (an extra chromosome 21 is a chromosomal, not deletion, mutation).
- Q13
- Prevalence constant for over 10 years; a new treatment prolongs life without curing. Effect on incidence and prevalence? Answer: A - incidence is unchanged (no effect on new cases) but prevalence increases (people live longer with the disease).
- Q14
- Which antibody matches the antigen shape of the pathogen shown? Answer: A - antibodies are specific, so the correct one has a binding site complementary to the antigen.
- Q15
- Five steps of a cochlear implant in order? Answer: C - sound picked up by microphone, turned into electrical impulses, transmitted across skin to the implant, sent to electrodes in the cochlea, auditory nerve sends impulses to the brain (3,1,5,2,4).
- Q16
- Identical offspring from separating embryo cells are clones of? Answer: A - they are clones of each other (artificial twinning makes genetically identical embryos, not of the parents or surrogates).
- Q17
- Correct sequence of the cell-division phases shown? Answer: C - S, T, R, Q for the order of the stages depicted.
- Q18
- Bar-eye (dominant, X-linked) male x normal female. Percentage of male and female offspring with bar-eye? Answer: B - the father passes bar-eye on his single X only to daughters, so 0% of males and 100% of females have bar-eye.
- Q19
- Genome of 120 million base pairs, 42% C and G. Number of thymine nucleotides per cell? Answer: C - 58% of bases are A or T, so thymine is 29% of the 240 million bases across both strands, which is 69.6 million.
- Q20
- Best explanation of the cause of T-ALL from the DNA diagram? Answer: D - the regulating proteins can activate oncogene LMO2, switching it on and causing the condition.
Section II - Short and extended response
Question 21 (2 marks)
The diagram shows components of the innate immune system in humans (nasal hair, tear glands, mucus lining, skin, stomach acid, urinary tract).
State the role of TWO components that protect against infection.
Show worked solution
[2 marks]. Two named components with how each protects:
| Component | How it protects against infection |
|---|---|
| Skin | A continuous physical barrier that stops pathogens entering the body. |
| Stomach acid | Strongly acidic, so it kills or inhibits the growth of pathogens that are swallowed. |
(Nasal hair trapping inhaled pathogens, or mucus lining trapping microbes, would also each earn a mark.)
Marker's note. Use the components shown in the stimulus and say how each one works, for example "nasal hairs trap pathogens inhaled in the air", rather than just naming a barrier.
Question 22 (4 marks)
(a) Outline a process used by fungi for reproduction. (2 marks)
(b) Outline an adaptation in a pathogen that facilitates transmission between hosts. (2 marks)
Show worked solution
(a) [2 marks]. Many fungi reproduce asexually by producing spores. The parent fungus releases large numbers of spores; when a spore lands in a favourable environment it germinates and grows into a new, genetically identical fungal organism. (Budding, as in yeast, is also acceptable.)
(b) [2 marks]. Cholera bacteria produce a toxin that causes watery diarrhoea. This flushes large numbers of bacteria out of the host into water sources where sanitation is poor, so other people who drink the water are infected. The adaptation increases transmission between hosts, not just survival inside one host.
Marker's note. In (a) link the process to its outcome (spores carried to a new site grow into a new organism). In (b) the adaptation must help the pathogen move between hosts, not simply benefit it inside a single host.
Question 23 (3 marks)
Compare the processes of artificial insemination and artificial pollination.
Show worked solution
[3 marks].
| Artificial insemination | Artificial pollination | |
|---|---|---|
| Organism group | Animals only | Flowering plants only |
| Male gamete used | Sperm placed into the female reproductive tract | Pollen placed onto the stigma |
Similarity: both are reproductive technologies in which a human deliberately transfers the male gametes to the female part of a chosen individual to control which parents reproduce.
Difference: artificial insemination is carried out in animals (sperm into the female tract), while artificial pollination is carried out in flowering plants (pollen onto the stigma).
Marker's note. Write a genuine comparison (a stated similarity and difference), not two separate descriptions. Do not confuse artificial insemination with IVF, and if you use a table, give it clear headings for "similarity" and "difference".
Question 24 (4 marks)
The flow chart represents the control of body temperature in humans, with the control centre returning temperature to the normal range using mechanism A or mechanism B.
(a) Complete the flow chart to give an example of mechanism A and an example of mechanism B. (2 marks)
(b) Outline how mechanism B maintains homeostasis. (2 marks)
Show worked solution
(a) [2 marks]. Mechanism A (cooling, when temperature rises): sweating. Mechanism B (warming, when temperature falls): shivering.
(b) [2 marks]. When body temperature drops below the normal range, the control centre triggers shivering, the rapid involuntary contraction of skeletal muscles. These contractions release heat, which raises body temperature back into the normal range, returning the body to homeostasis.
Marker's note. Read the flow chart so mechanism A increases heat loss and mechanism B generates heat. In (b) link shivering specifically to heat generation and the return to the normal range.
Question 25 (3 marks)
The graph shows changes in UV level over a single day. The Cancer Council suggests sun protection is needed whenever the UV level is 3 or above. A sunscreen product states sunscreen should be used between 10 am and 4 pm.
Using the graph, evaluate the information provided on the sunscreen product with regard to the Cancer Council suggestion.
Show worked solution
[3 marks]. The product statement is not adequate. From the graph, the UV level is at or above 3 from roughly 8 am to 6 pm, which is wider than the 10 am to 4 pm window the product recommends. So for about two hours in the morning (8 am to 10 am) and about two hours in the late afternoon (4 pm to 6 pm) the UV level already exceeds 3, yet a person following only the product advice would be unprotected. The product therefore underestimates the period needing protection and does not fully match the Cancer Council guidance.
Marker's note. Tie the data in the graph directly to both the Cancer Council level (3) and the product times, and reach a clear, evidence-based judgement rather than a general comment.
Question 26 (4 marks)
The diagram shows the steps in LASIK (Laser-Assisted In Situ Keratomileusis) surgery.
Compare the LASIK technology shown with ONE other technology that can be used to treat a named visual disorder.
Show worked solution
[4 marks]. Visual disorder: myopia (short-sightedness), where light focuses in front of the retina.
- How each treats it: LASIK uses a laser to reshape the cornea so it refracts incoming light less strongly, moving the focal point back onto the retina. Prescription glasses with concave (diverging) lenses also reduce the convergence of light before it reaches the eye, so the image again falls on the retina. Similarity: both correct myopia by adjusting how light is refracted so it focuses on the retina.
- Differences: LASIK is a surgical, permanent and more expensive procedure with a healing period; glasses are non-invasive, removable, cheaper and immediately reversible.
Marker's note. Keep the focus on treatment of a named disorder and how each technology refracts light to fix it, not just a general comparison of cost or healing time. Bring in a technology beyond the stimulus.
Question 27 (6 marks)
The table shows eggs produced by females in two groups. Group A: monotremes 1 to 3, snakes 1 to 100, birds 1 to 17. Group B: crabs 1000 to 2000, sea urchins 100 000 to 2 million, squid 2000 to 3000.
Compare the types of fertilisation that occur in group A and group B animals with reference to the data provided.
Show worked solution
[6 marks]. Group A uses internal fertilisation; Group B uses external fertilisation.
- Similarity: in both groups fertilisation is the fusion of a male and a female gamete to form a zygote.
- Group A (internal): the gametes meet inside the female body, so the chance of an egg being fertilised is high and the gametes are protected from drying out. Far fewer eggs are needed, which matches the data (1 to 100 in snakes, 1 to 3 in monotremes). Internal fertilisation also allows reproduction on land.
- Group B (external): gametes are released into water and fertilise outside the body, so the gametes are spread through a large volume and the probability of any one being fertilised is low. To compensate, huge numbers of eggs are produced, which matches the data (100 000 to 2 million in sea urchins). External fertilisation must occur in water so the gametes do not dry out.
So the number of eggs is far higher in Group B because external fertilisation has a much lower success rate per egg than the protected, higher-probability internal fertilisation of Group A.
Marker's note. Identify the fertilisation type in each group, compare internal and external fertilisation (not fertilisation versus parental care, or external fertilisation versus asexual reproduction), and quote the egg-number data to support the comparison.
Question 28 (9 marks)
Alpha-gal syndrome is a tick-borne allergy to red meat. The stimulus shows how a tick bite transfers alpha-gal, the body makes antibodies, and later meat exposure triggers a reaction.
(a) The flow chart shows antibody production: Alpha-gal to Macrophage to X to Y to Z (then memory cells and antibodies). Describe the role of X, Y and Z in the process of antibody production. (4 marks)
(b) An allergic reaction to alpha-gal is similar to a secondary immune response. The graph shows antibody concentration over time (tick bite, then meat eaten). Describe the features of antibody production shown in the graph. (2 marks)
(c) Explain the role of memory cells in the immune response. (3 marks)
Show worked solution
(a) [4 marks]. A macrophage engulfs the alpha-gal antigen and presents it.
- X (helper T cell): recognises the presented antigen and releases cytokines that activate the matching B lymphocytes.
- Y (B lymphocyte): the specific B cell, once activated, proliferates and differentiates.
- Z (plasma cell): the differentiated B cell that secretes large quantities of antibodies against alpha-gal (some activated B cells instead become memory cells).
(b) [2 marks]. On the first exposure (the tick bite) antibody production is slow and reaches only a low concentration. On the second exposure (eating meat) antibodies are produced much faster and reach a far higher concentration, the pattern of a secondary response.
(c) [3 marks]. Memory cells are long-lived B and T cells made during the first (primary) exposure to a specific antigen. They remain in the body, for example in the lymph nodes, after the infection clears. On re-exposure to the same antigen they recognise it immediately and rapidly divide and differentiate into plasma cells (and cytotoxic T cells), producing antibodies faster and in greater amounts. This rapid secondary response usually neutralises the pathogen before disease develops.
Marker's note. In (a) label or name X, Y and Z explicitly and show the progression from antigen presentation to antibody production. In (b) describe distinct features of each exposure from the graph (speed and level), not a single broad trend. In (c) explain both how memory cells are made and their role on re-exposure.
Question 29 (6 marks)
The Varroa mite is an external parasite of European honey bees.
(a) Why is Varroa mite infection considered to be an infectious disease? (2 marks)
(b) In June 2022, the Varroa mite was detected in Australia at the Port of Newcastle and then spread. Explain TWO procedures that could have been employed to prevent the spread of the Varroa mite in honey bees. (4 marks)
Show worked solution
(a) [2 marks]. The Varroa mite is a parasitic pathogen that lives on and harms honey bees, and it is transmitted from infected bees and hives to healthy ones through direct contact. Because it is caused by a pathogen and spreads between organisms, the infestation is classed as an infectious disease.
(b) [4 marks].
- Early identification and isolation. Regular monitoring of colonies for mites allows infested hives to be detected early and quarantined, stopping bees from those hives mixing with healthy hives and carrying mites to them.
- Destroying infected hives. Euthanising the bees in confirmed infested hives kills the mites along with their hosts, removing the source of infection so it cannot spread to surrounding healthy hives. (Washing and disinfecting shared equipment is also acceptable.)
Marker's note. In (a) tie the answer to "caused by a pathogen" and "transmitted between organisms". In (b) give procedures specific to the Varroa scenario and link each one to how it stops spread through the wider bee population, rather than generic biosecurity statements.
Question 30 (11 marks)
PAI-1 protein is encoded by the SERPINE1 gene in humans. Anopheles mosquitoes have been genetically modified to express PAI-1, which blocks malarial Plasmodium entering the mosquito gut and reduces malaria transmission.
(a) Describe a process that could be used to produce mosquitoes which express PAI-1. (4 marks)
(b) "Genetic technologies are beneficial for society." Evaluate this statement. (7 marks)
Show worked solution
(a) [4 marks]. Use recombinant DNA technology to make a transgenic mosquito:
- Use a restriction enzyme to cut the SERPINE1 gene out of human DNA, leaving sticky ends.
- Cut a bacterial plasmid (vector) with the same restriction enzyme so the ends are complementary, then join the SERPINE1 gene into the plasmid using DNA ligase.
- Transform bacteria with the recombinant plasmid and let them reproduce to amplify many copies of the gene.
- Insert the gene into the mosquito's germline by micro-injecting it into mosquito egg cells (or using a viral vector), so mosquitoes that develop from those eggs carry SERPINE1, express PAI-1, and pass the trait to their offspring.
(b) [7 marks]. Genetic technologies bring real benefits but also genuine risks, so the statement is largely, but not unconditionally, supported.
Benefits. In recombinant DNA technology the human insulin gene has been transferred into bacteria, which then produce large amounts of human insulin cheaply, so people with diabetes have reliable, affordable treatment and longer, better-quality lives. Gene technologies such as CRISPR and the transgenic mosquitoes above can reduce vector-borne diseases like malaria and dengue, lowering disease incidence and easing pressure on health systems. GM crops (for example Bt cotton) raise yields and cut insecticide use, increasing food supply.
Costs and concerns. These technologies also carry unintended health effects, environmental risks (for example a released gene spreading through wild populations), and ethical concerns about patenting living things and equitable access, since poorer farmers and patients may be priced out.
Judgement. On balance the benefits to health, food supply and disease control are substantial and well evidenced, so genetic technologies are mostly beneficial for society, provided the environmental and ethical risks are regulated and monitored.
Marker's note. In (a) recognise this needs a transgenic organism (germline modification), name the steps (restriction enzymes, ligase, plasmid vector, micro-injection), and use precise terminology rather than "place into mosquito". In (b) name specific technologies with real examples, weigh benefits against risks at a societal level (health, environmental, ethical, economic), and keep one consistent judgement throughout.
Question 31 (6 marks)
A pedigree shows the inheritance of CAMT, a rare inherited disorder.
(a) What type of inheritance is shown in the pedigree? Justify your answer. (3 marks)
(b) A CAMT mutation produces the amino acid sequence Glutamine - Tyrosine - Isoleucine - Aspartic acid. The unaffected individual's template strand is GTC ATA CAG CTG. Using the codon chart, explain the type of mutation which causes CAMT. (3 marks)
Show worked solution
(a) [3 marks]. Autosomal recessive. Unaffected parents in the pedigree have an affected child, so the allele must be recessive and was carried, unexpressed, by both parents. Because affected individuals are both male and female and the trait is not passed in a sex-linked pattern (an affected daughter is not explained by a single X from the father), the gene is autosomal. So each unaffected carrier parent is heterozygous (Bb) and an affected child is homozygous recessive (bb), having inherited one recessive allele from each parent.
(b) [3 marks]. The template strand GTC ATA CAG CTG transcribes to the mRNA codons CAG UAU GUC GAC, which the chart reads as Glutamine - Tyrosine - Valine - Aspartic acid in the unaffected individual. In the CAMT sequence the third amino acid is Isoleucine, coded by AUC. So the GUC codon has become AUC: a single base (G to A) has changed. This is a point mutation (a substitution), and because it swaps one amino acid for another it is a missense mutation.
Marker's note. In (a) use the precise term "autosomal recessive" and justify with named individuals from the pedigree (unaffected parents with an affected child), drawing a Punnett square if it supports the case. In (b) use the codon chart to translate the unaffected sequence, identify the single changed base, and name it a point/substitution mutation.
Question 32 (7 marks)
A population lives across three regions A, B and C; A and B are joined by a road bridge and C is isolated. Community B developed an environmental disease. An epidemiological study determined the risk of developing the disease by age at exposure (graph shows risk over years after exposure for ages 10, 20 and 30 at exposure).
Design an epidemiological study that could be used to produce the results shown in the graph. Justify the features of your design.
Show worked solution
[7 marks]. A long-term cohort (prospective) study tracking groups by age at exposure.
- Sample and groups. Recruit large numbers of people from regions A, B and C, sorted into cohorts by age at first exposure (about 10, 20 and 30 years old) to match the three lines on the graph. Large samples and equal numbers of males and females improve validity by reducing the effect of chance and confounding by sex.
- Comparison and control. Communities A and B are connected by the bridge and mix freely, so both can be exposed; isolated community C serves as an unexposed control to show the background rate of the disease. Comparing exposed and control groups lets you attribute risk to the exposure.
- Duration. Follow the cohorts for about 55 years, as shown on the x-axis, recording new cases over time so the rising risk for each age group can be plotted.
- Controlling confounders. Collect data on diet, lifestyle, occupation and other exposures, so any difference in disease risk can be linked to age at exposure rather than to another variable. Correlating these data identifies the true risk factor.
- Justification overall. Large samples, a control region, a long follow-up and recording confounders together make the results reliable and valid, and reproduce the age-related risk curves shown.
Marker's note. Use the stimulus (the three regions, the bridge, the age groups and the timescale) to justify each design feature, and show how the features support reliability and validity, rather than describing epidemiology in general.
Question 33 (9 marks)
A diagram shows cell-division processes in two related individuals, with germline and somatic cells labelled (cells R, S in Individual 1; cells U, V, W, X and mutations A, B in Individual 2).
(a) Compare the cell division processes carried out by cells R and S in Individual 1. (3 marks)
(b) Explain the relationship between Individuals 1 and 2. (2 marks)
(c) A and B are two separate mutations. Analyse how mutations A and B affect the genetic information present in cells U, V, W and X. (4 marks)
Show worked solution
- (a) [3 marks]
- Cell R (a somatic cell) undergoes mitosis, producing two daughter cells that are genetically identical to the parent and to each other, with the full (diploid) chromosome number. Cell S (a germline cell) undergoes meiosis, producing gametes with half the chromosome number (haploid) and, through crossing over and independent assortment, genetic variation. Similarity: both are preceded by DNA replication, in which the genetic content is doubled before division.
- (b) [2 marks]
- Individual 1 is a parent of Individual 2: a gamete produced by Individual 1's germline cells fused with another gamete at fertilisation to form the zygote that developed into Individual 2, who therefore inherited a set of chromosomes from Individual 1.
- (c) [4 marks]
- Mutation A is a germline mutation and mutation B is a somatic mutation.
- Cell U (the early cell line before either mutation) carries neither mutation A nor B.
- Because A occurs in a germline cell, it can be passed to offspring: cell V could carry mutation A but not B.
- Cell W, formed after both mutations, could carry both A and B.
- Cell X, in the germline line, could carry mutation A (heritable) but not the somatic mutation B.
A germline mutation (A) can be inherited by future generations, while a somatic mutation (B) is confined to the body cells in which it arose and is not passed on.
Marker's note. In (a) compare the division processes themselves (chromosome number, identical versus varied daughters), not just the outcomes. In (b) state the parent-offspring link via a gamete. In (c) use the key to track which of U, V, W and X carry the heritable germline mutation A versus the non-heritable somatic mutation B.
Question 34 (6 marks)
Two graphs show changes in allele frequencies in two separate populations of the same species (population 20 and population 2000); each line is an introduced allele.
(a) Explain why the fluctuations of the allele frequencies are more pronounced in the small population, compared to the larger population. (2 marks)
(b) Evaluate the effects of gene flow on the gene pools of the two populations. (4 marks)
Show worked solution
(a) [2 marks]. A smaller population has a smaller gene pool and lower genetic diversity, so each individual makes up a larger share of the allele frequencies. Random events (genetic drift) therefore change the frequencies by a large proportion, producing the pronounced fluctuations, whereas in the large population the same random events are averaged out across many more individuals.
(b) [4 marks]. Gene flow (here, the introduced alleles) adds genetic variation to a gene pool.
- Small population (20): the introduced allele is a large fraction of a small, low-diversity gene pool, so its frequency swings sharply. If the new allele is favourable it can spread quickly and improve the population's ability to adapt; if unfavourable, selection removes it rapidly, as seen in the steep changes in the small-population graph.
- Large population (2000): the same introduced allele is a tiny fraction of a large, already diverse gene pool, so it changes allele frequencies only slightly and gradually, as the flatter lines show.
- Judgement: gene flow generally has a positive effect by increasing genetic diversity, but its impact is far greater in the small population, while in the large population the effect is small but still adds variation. So the effect of gene flow depends on population size and environmental conditions.
Marker's note. In (a) link small population size to a small gene pool and to a larger effect of drift. In (b) recognise that the same gene flow was introduced into both populations, that gene flow changes the gene pool (not the reverse), and that population size governs how strongly it affects allele frequencies; reach a clear evaluation.
General marker feedback
Stronger responses across the paper: read the question carefully and addressed every component; planned extended responses so the information flowed logically; integrated relevant scientific terms; engaged with the stimulus and referred to it directly rather than rewriting it; showed all working with correct units in calculations; and reviewed the response so it stayed succinct and on the question.
Use this paper well
- Sit the paper under exam conditions (180 minutes, 100 marks).
- Mark yourself against the official NESA marking notes.
- Compare against the Biology hub to find the syllabus dot points this paper tested.
