VCE Biology 2025

Every question from the 2025 VCE Biology examination. Multiple-choice answers come with a short reason; Section B model answers sit behind a Show worked solution toggle.

How to use this page

• Questions are from the 2025 VCE Biology examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, data and diagrams.
• Answers are original ExamExplained model answers. Every multiple-choice answer matches the key in the 2025 Biology external assessment report (Word document), and every Section B answer was checked against the points the report accepted. Both files are listed on the VCAA Biology examinations page.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.
• Biology answers are marked on the points made, so the model answers below are written as one clear point per mark. Other correct points were also accepted; the report lists them.

Structure and timing

The paper is 120 marks in 150 minutes (plus 15 minutes reading time). That is 1.25 minutes per mark.

• Section A (40 marks): 40 multiple-choice questions. Aim for about 45 minutes.
• Section B (80 marks): 10 short-answer and extended-response questions, most built around an unfamiliar scenario, graph or study. Read the stem carefully: the report repeatedly penalised answers that ignored the context.

Section A: Multiple choice

Q1

Independent variable in an experiment on the effect of temperature on DNA structure. Answer: D - the temperature, which is what the experimenter changes. (87%)

Q2

Haemoglobin activity against temperature falls above 30 °C. What can be concluded? Answer: C - between 30 °C and 40 °C activity drops because the protein is denaturing, which disrupts its quaternary (and tertiary and secondary) structure. The primary structure is not changed by heat. (88%)

Q3

Two secretion pathways: Protein X goes from the rough ER straight to the plasma membrane; Protein Y passes through the Golgi. Answer: D - Protein X skips the Golgi, where proteins are modified, so it is not modified further after leaving the rough ER. Both are packaged in vesicles. (79%)

Q4

The DNA triplet GCA codes for an amino acid. The tRNA anticodon? Answer: B or C (both accepted). If GCA is on the template strand, the codon is CGU and the anticodon is GCA (B). If GCA is on the coding strand, the codon is GCA and the anticodon is CGU (C). The question did not say which strand. (B 38%, C 53%)

Q5

Which event ends transcription in a eukaryotic cell? Answer: D - RNA polymerase reaches a termination sequence on the DNA template. Intron removal is mRNA processing, and stop codons and polypeptide release belong to translation. (67%)

Q6

Match four DNA sections to prokaryotic gene, operon, eukaryotic structural gene and regulatory gene. Answer: C - S (promoter, no introns) is a prokaryotic gene; R (several genes under one promoter) is an operon; Q (exons and introns, spliced) is a eukaryotic structural gene; T (controls other genes) is a regulatory gene. (79%)

Q7

Purpose of single guide RNA in CRISPR-Cas9. Answer: C - it binds Cas9 and guides it to the matching target sequence, directing where the DNA is cut. Cas9 does the cutting. (90%)

Q8

A gel shows one large band for condition 1 (Cas9 only) and two smaller bands each for conditions 2 and 3. Where was CRISPR-Cas9 working? Answer: B - conditions 2 and 3, where the target DNA was cut into two pieces. (80%)

Q9

The hypothesis the researchers would propose. Answer: D - Cas9 can be guided to cut DNA when crRNA and tracrRNA are combined into a single guide RNA (condition 3 worked). Cas9 alone (condition 1) cut nothing. (81%)

Q10

Output alongside ethanol when yeast ferments waste starch, and the condition needed. Answer: B - carbon dioxide, in the absence of oxygen (anaerobic fermentation). (75%)

Q11

DNA profiles of a lamb, its mother and four possible fathers. Which is the father? Answer: C - every band the lamb did not get from its mother must come from its father. Father C is the only one with all three of those bands. (48%; 27% chose B.)

Q12

What makes DNA fragments move through a gel? Answer: B - DNA is negatively charged (from its phosphate groups), so it moves towards the positive electrode. (87%)

Q13

Which statement about PCR is correct? Answer: D - the cycle is repeated many times. The other options mix up the stages: Taq (DNA) polymerase, not RNA polymerase, works in extension; annealing is the cooler stage; primers attach during annealing. (72%)

Q14

Why does reducing leaf overlap speed up photosynthesis? Answer: D - more light reaches the grana (thylakoid membranes), where the light-dependent stage happens. (85%)

Q15

Which statement about cellular respiration is correct? Answer: B - theoretical and actual ATP yields differ. Yields are not fixed at 36 or 38, root cells do respire stored glucose, and C3 and C4 plants share the same respiration pathway. (54%)

Q16

Two steps in making vinegar. Answer: C - yeast ferments glucose to ethanol, then bacteria break the ethanol down to acetic acid. Yeast does not make glucose or lactic acid, and bacteria have no cristae. (66%)

Q17

Name two stages from a table of inputs and outputs. Answer: B - U (water in, oxygen out) is the light-dependent stage and V (glucose in, ATP out) is glycolysis. (63%)

Q18

Stage X converts pyruvate to lactic acid. Answer: A - this is anaerobic fermentation in animal cells. It happens in the cytosol and produces no ethanol or carbon dioxide. (70%)

Q19

A suitable way to investigate the rate of photosynthesis. Answer: B - a literature review of NADPH produced in stage S. Stage S (water in, NADPH out) is the light-dependent stage, and a literature review is a valid methodology using secondary data. The other options pair a method with a stage that is wrong for photosynthesis. (36%; 42% chose C, but glucose is an output of photosynthesis, not an input.)

Q20

Which factors give the same trend when plotted against the rate of photosynthesis? Answer: A - light intensity and carbon dioxide concentration both give a rise then a plateau. Temperature gives a rise to an optimum, then a sharp fall as enzymes denature. (67%)

Q21

Initial effect of a competitive inhibitor of ATP synthase. Answer: D - it occupies the enzyme's active site, so ADP cannot bind and builds up in the cell. (34%; 43% chose A, but the active site belongs to the enzyme, not to ATP.)

Q22

Essential oils and phenolic compounds in mint cells protect against pathogens. They act as Answer: A or B (both accepted) - they are chemical antimicrobial agents that inhibit pathogen growth (A), and as chemical barriers they are part of the plant's first line of defence (B). Plants have no second or third line. (A 26%, B 51%)

Q23

Which student offered an anecdote? Answer: D - student 4's personal recollection of her mother's remedy. (85%)

Q24

An example of a cellular pathogen, a non-cellular pathogen and an allergen. Answer: B - bacterium, virus, dog hair. (88%)

Q25

What happens in swollen lymph nodes during an infection? Answer: B - lymphocytes that recognise the antigen undergo clonal selection and expansion, which is why the nodes swell. (64%)

Q26

A feature of inflammation. Answer: D - vasodilation increases blood flow to the site, bringing more immune cells. (94%)

Q27

Match an innate immune molecule with its characteristic and function. Answer: C - complement proteins circulate in an inactive form and, once activated, can cause cell lysis. (58%; 26% chose A, but histamine is a signalling molecule, not an enzyme.)

Q28

Main role of plasma cells. Answer: A - to produce antibodies specific to an antigen. (82%)

Q29

Antibody levels start rising on day 13 of a graph of two exposures. When was the person first exposed? Answer: A - day 2. A primary response takes days to produce antibodies, so exposure came well before the rise. (27%; 69% chose B, the day antibodies first appear.)

Q30

Which statement about the antibody graph is correct? Answer: D - memory cells activated around day 33 by the second exposure give the fastest, largest antibody response. (63%)

Q31

Effect of high vaccination rates. Answer: A - they reduce spread between people, protecting those who are susceptible (herd immunity). (90%)

Q32

Which factor matches a new or re-emerging pathogen? Answer: C - people travelling and returning home can bring in a new pathogen. (84%)

Q33

Which change needs a new vaccine immediately? Answer: D - antigenic shift, which produces a new viral strain with very different antigens. (79%)

Q34

A phylogenetic tree of wolves, dingoes and five dog breeds. Answer: B - Great Danes and Boxers share a more recent common ancestor than Great Danes and dingoes, so their DNA differs less. (86%)

Q35

Which feature in the image identifies the lemur as a primate? Answer: B - an opposable thumb. Tails and fur are common to many mammals. (75%)

Q36

A fossil with both reptile and mammal features can be used to Answer: C - support the hypothesis that reptiles and mammals share a common ancestor (a transitional fossil). (79%)

Q37

Fossils in dated rock layers. Answer: B - fossil X is only in layer S, which lies between the 145 mya and 165 mya layers. (80%)

Q38

Evidence against the hypothesis that the pig-nosed turtle arrived in Australia recently. Answer: A - five-million-year-old fossils of it in Melbourne show it was in Australia long ago. (69%)

Q39

Why is the lettuce investigation not valid? Answer: D - temperature changes along with the amount of water, so it is an uncontrolled variable. (80%)

Q40

Most likely qualitative data. Answer: C - the shape of the leaves is descriptive; the others are measured quantities. (81%)

Section B

Question 1 (5 marks)

The insulin gene has three exons.

a. Draw the mRNA molecule produced by processing the insulin gene, labelling three modifications. (3 marks)

b. In making human insulin with recombinant bacteria, complete step 4 (DNA ligase ...) and step 8 (amino acid chains A and B ...). (2 marks)

Show worked solution

a. [3 marks]. Show a single strand with exons 1, 2 and 3 joined end to end, and label:

• introns removed and exons spliced together (the lines between the exons in the gene are introns);
• a methyl guanine cap at the 5′ end;
• a poly-A tail at the 3′ end.

Alternative splicing (for example, leaving out an exon) was also accepted as a labelled modification.

b. [2 marks]. Step 4: DNA ligase joins each insulin gene (A or B) into its plasmid by forming phosphodiester bonds, making recombinant plasmids. Step 8: amino acid chains A and B are joined together (by disulfide bonds) to form functional insulin.

From the report. The cap and tail had to be at opposite ends, though naming the 5′ and 3′ ends was not required. In b, a common misunderstanding was that both genes go into the same plasmid; genes A and B are inserted into separate plasmids.

Question 2 (11 marks)

Some bacteria make tryptophan (trp) using enzymes coded by the trp operon. A graph shows the activity of these enzymes; at 20 minutes the bacteria are moved from a medium with trp to one without it. A model simulation (line) and two experimental data sets (points) are shown. Activity is low until 20 minutes, rises, and is steady from about 60 minutes.

a

Explain how enzymes such as those coded by the trp operon catalyse reactions. (2 marks)

b

Using the graph, explain why activity is low from 0 to 20 minutes and steady after about 60 minutes, and how repression and attenuation work together to regulate the enzymes. (5 marks)

c

State two limitations of the model simulation that could explain why it does not match the experimental data. (2 marks)

d

Bacteria once had UGG as a stop codon and no codon for trp; UGG now codes for trp. Identify a consequence of this change and justify it. (2 marks)

Show worked solution

a. [2 marks]. Each enzyme has an active site with a shape complementary to its specific substrate(s), which bind to it. The enzyme lowers the activation energy needed to convert substrate into product, so the reaction goes faster.

b. [5 marks].

• 0 to 20 minutes (trp present): trp binds to the repressor protein and activates it. The active repressor binds the operator, blocking RNA polymerase, so the structural genes are rarely transcribed and enzyme activity stays low.
• After about 60 minutes (no trp): the repressor is inactive and cannot bind the operator, so the genes are transcribed and translated. Enzyme activity has risen and levels off because the enzymes being made balance those being lost, or other factors (such as substrate supply) now limit activity.
• Attenuation: when trp is plentiful, ribosomes translating the leader sequence move quickly past the trp codons, and the mRNA forms a terminator hairpin, stopping transcription early. When trp is scarce, the ribosome stalls at the trp codons, an anti-terminator loop forms instead, and transcription continues.
• Working together: repression is the main on/off switch, and attenuation is a second, finer control that stops transcription that escapes repression (the repressor detaches from time to time). Both prevent the cell wasting energy making trp enzymes when trp is available.

c. [2 marks]. Any two, for example: the model is simplified and cannot capture every real-world variable (such as changing temperature or pH); it assumes ideal conditions; it shows only a smoothed general trend and ignores real variation; it treats trp as simply present or absent rather than falling gradually; it cannot produce activity above 1.0, which the experimental data show.

d. [2 marks]. Genes that used to end at a UGG stop codon would now have trp added instead, so translation would continue and make longer polypeptides with a different primary structure. These proteins may fold differently and lose their function, which could reduce the bacteria's survival. (Other accepted consequences: the cell now needs trp and a tRNA carrying it; fewer nonsense mutations are possible.)

From the report. In a, "enzymes speed up reactions" alone scored nothing. In b, common errors were saying the repressor binds the promoter, that RNA polymerase "translates", mixing up the terminator and anti-terminator loops, and misreading the y-axis as trp concentration. In c, many gave limitations of the experimental data instead of the model. Part d was the hardest part of the question (6% full marks).

Question 3 (9 marks)

C3, C4 and CAM plants all use Rubisco.

a

For each plant type, mark where Rubisco is active (mesophyll or bundle sheath cells) and when its activity is greatest (day or night). (3 marks)

b

Explain why photorespiration increases with temperature in C3 plants. (2 marks)

c.i

A student testing this plans to control the sunlight and water each plant receives. Describe how to control each. (2 marks)

c.ii

The student proposes glucose in the plant as the dependent variable. Describe two reasons it is unsuitable. (2 marks)

Show worked solution

a. [3 marks].

Plant	Rubisco location	Rubisco most active
C3	mesophyll cells	daytime
C4	bundle sheath cells	daytime
CAM	mesophyll cells	daytime

CAM plants open their stomata and fix carbon dioxide at night, but Rubisco works in the day, using carbon dioxide released from stored malate.

b. [2 marks]

In hot conditions C3 plants close their stomata to reduce water loss, so carbon dioxide inside the leaf falls and oxygen builds up. C3 plants have no mechanism to concentrate carbon dioxide around Rubisco, so Rubisco binds oxygen instead of carbon dioxide more often, which is photorespiration.

c.i. [2 marks]

Light: place every plant the same distance from the same lamp (or measure and match light intensity) for the same number of hours, with other light sources excluded. Water: give each plant the same measured volume of water at the same time each day (or keep soil moisture the same).

c.ii. [2 marks]

Any two: glucose is not a product of photorespiration, so it does not measure it; glucose is used in cellular respiration and converted to starch or cellulose, so its level does not reflect how much was made; measuring glucose would mean damaging or destroying the plant and is hard to do accurately.

From the report. In a, a cross in both locations or both times for one plant scored nothing. In b, answers had to say that Rubisco (not "the plant") binds oxygen, and answers about denaturation were wrong. In c.i, restating "give the same amount of light" did not say how; use precise terms such as "volume" of water. In c.ii, many confused photorespiration with cellular respiration.

Question 4 (4 marks)

Rice has had its OsHXK1 gene edited with CRISPR-Cas9, raising leaf chlorophyll levels. Maize has had toxin genes from the bacterium Bacillus thuringiensis inserted to kill rootworm larvae.

a. State whether each crop is transgenic, genetically modified or both, with a justification. (2 marks)

b. Explain how editing the OsHXK1 gene could increase rice yield. (2 marks)

Show worked solution

a. [2 marks]. Rice: genetically modified (not transgenic), because its own gene was edited and no DNA from another species was added. Maize: both, because it contains genes from a different species (B. thuringiensis), which makes it transgenic, and every transgenic organism is also genetically modified.

b. [2 marks]. More chlorophyll means the leaves absorb more light energy, so the light-dependent stage and therefore the rate of photosynthesis increase. More glucose is made, which can be used for growth and stored in the grain, giving a higher yield.

From the report. In b, restating that chlorophyll increases, or saying growth increases without explaining how, did not score.

Question 5 (6 marks)

Neisseria meningitidis is a bacterium that causes meningococcal disease.

a. State two roles of neutrophils in the response to N. meningitidis. (2 marks)

b. Contrast the adaptive immune response to an extracellular pathogen such as N. meningitidis with the response to an intracellular pathogen such as influenza virus. (4 marks)

Show worked solution

a. [2 marks]. Any two: they migrate to the site of infection; they engulf and destroy the bacteria by phagocytosis; they release cytokines that attract other immune cells; they release toxic chemicals (such as defensins) that kill bacteria.

b. [4 marks]. Four contrasts, each stating both sides:

• Type of response: extracellular pathogens are cleared mainly by the humoral response, whereas intracellular pathogens need the cell-mediated response.
• Effector cells: B cells differentiate into plasma cells against the extracellular pathogen, whereas cytotoxic T cells act against the virus.
• How the pathogen is removed: antibodies bind the bacteria's antigens and neutralise, agglutinate or opsonise them, whereas cytotoxic T cells recognise infected body cells and trigger their death (apoptosis, using perforin and granzymes).
• Memory: memory B cells are the main memory cells after the humoral response, whereas memory T cells are formed in the cell-mediated response.

From the report. Many students did not know how the adaptive response differs between the two types of pathogen. Some confused "extracellular" with "not yet in the body" and compared the lines of defence. "Contrast" asks for differences, not similarities.

Question 6 (10 marks)

Pregnant women in Australia are offered a single dose of an RSV vaccine containing two RSV surface proteins, at 24 to 36 weeks of pregnancy. RSV spreads by respiratory droplets and can put infants in hospital.

a

Identify the type of immunity the babies receive, with a justification. (2 marks)

b

Suggest a reason for a booster in a later pregnancy. (2 marks)

c

Explain one economic benefit of vaccinating pregnant women. (2 marks)

d

Nirsevimab is a monoclonal antibody given to infants of unvaccinated mothers. Explain how it reduces the severity of RSV. (2 marks)

e

Other than vaccines and nirsevimab, identify one way to reduce the spread of RSV to these infants, with a justification. (2 marks)

Show worked solution

a. [2 marks]

Natural passive immunity. Passive, because the baby receives ready-made antibodies and does not make its own antibodies or memory cells. Natural, because the antibodies cross from the mother through the placenta (and later breast milk) without any treatment given to the baby.

b. [2 marks]

The mother's antibody levels (and memory cell numbers) decline over time. A booster raises her antibody levels again, so enough antibodies cross the placenta to protect the new baby from RSV in its first months.

c. [2 marks]

Fewer infants catch severe RSV, so there are fewer hospital admissions and less intensive care. Vaccination costs much less than hospital treatment, which saves money for the government and the health system (and reduces time off work for parents).

d. [2 marks]

The monoclonal antibodies have antigen-binding sites complementary to an RSV antigen, so they bind to the virus. This neutralises it (stops it entering cells) and marks it for phagocytosis or clumps viruses together, so fewer cells are infected and symptoms are milder.

e. [2 marks]

For example: carers and visitors wear masks around the infant, which blocks respiratory droplets carrying the virus. Or frequent handwashing and cleaning surfaces, which removes virus that could be transferred from hands and surfaces to the baby.

From the report. In a, "active" and "passive" (and "natural" and "artificial") contradict each other; pick one of each. In b, memory cells are not passed from mother to baby, and antigenic drift would need a new vaccine, not a booster. In c, claiming the baby would not need vaccinating later was biologically wrong. In d (26% full marks), antibodies bind antigens on the virus; they do not "kill" it. In e, "it reduces the spread" restates the question and is not a justification.

Question 7 (8 marks)

In 2007 a single colony of Asian honey bees reached Queensland. Its queen was already carrying stored sperm from matings in her native range and laid all the fertilised eggs. Within a decade there were 10 000 to 50 000 colonies.

a

Referring to genetic changes, state what the founder effect is and why the scientists are right to call this an example. (2 marks)

b

The new colonies gather food and organise better than the original one and have 481 DNA differences from it. Identify the process responsible and explain how it helped the colonies expand. (3 marks)

c

The Asian honey bee may be a selection pressure on native bees. Explain the potential genetic change in native bees. (3 marks)

Show worked solution

a. [2 marks]

The founder effect is when a small group starts a new population, carrying only a small, unrepresentative sample of the original gene pool, so the new population has reduced genetic diversity and different allele frequencies. It applies here because the whole Australian population came from one colony: the alleles of one queen and the males whose sperm she carried.

b. [3 marks]

Mutation. Mutations changed the DNA base sequence and created new alleles, adding genetic variation. Bees whose new alleles improved food gathering or colony organisation had a selective advantage: they survived and reproduced more and passed those alleles on, so over generations the favourable alleles became more common and the colonies spread.

c. [3 marks]

Native bees compete with the Asian honey bees for food and nest sites. Individuals with traits (alleles) that help them compete are more likely to survive and reproduce, passing those alleles on, so their frequency increases. Individuals without them die or breed less, so those alleles decrease or are lost. Overall, the native population may shrink and lose genetic diversity.

From the report. In a, many did not mention genetic change, or described a bottleneck instead. In b, natural selection or genetic drift were common wrong answers for the process: they change allele frequencies but do not create new DNA differences. In c, avoid Lamarckian wording such as "native bees adapt in order to survive"; the variation must already exist.

Question 8 (14 marks)

Researchers injected photosynthetic cyanobacteria into the hearts of tadpoles (which hatch in groups of thousands, have transparent skin and absorb oxygen through gills). In the dark with low oxygen, brain activity stopped after about 65 minutes; 15 to 20 minutes after the lights came on, it restarted. The work was approved by an ethics committee.

a

Propose two reasons tadpoles were chosen. (2 marks)

b.i

Other than enzymes, name two molecules the cyanobacteria need to produce oxygen. (2 marks)

b.ii

Describe the role of each in producing oxygen. (2 marks)

c

Explain why brain activity stopped in prolonged hypoxia. (2 marks)

d

Referring to a biochemical pathway, explain one reason for the delay before brain activity restarted. (2 marks)

e

Outline two events on the electron transport chain. (2 marks)

f

Student X says the experiment was ethical (beneficence); Student Y says it was not (non-maleficence). Say which you agree with and justify using the article. (2 marks)

Show worked solution

a. [2 marks]

Any two, each linked to the experiment: their transparent skin lets light reach the cyanobacteria inside them; they hatch in thousands, giving a large sample size; they take in dissolved oxygen through gills, so the oxygen in the water can be controlled to create hypoxia; they are small, cheap and quick to raise.

b.i. [2 marks]

Water and chlorophyll (NADP⁺, or ADP and Pi, were also accepted). Light is energy, not a molecule.

b.ii. [2 marks]

Water is split in the light-dependent stage, releasing oxygen as a by-product and supplying the electrons and hydrogen ions that load NADP⁺ to NADPH. Chlorophyll absorbs light energy, which drives the splitting of water.

c. [2 marks]

Neurons need a lot of ATP. Without oxygen they cannot carry out aerobic respiration (oxygen is the final electron acceptor in the electron transport chain), and anaerobic respiration makes only 2 ATP per glucose. Too little ATP is produced to keep the neurons active, so brain activity stops.

d. [2 marks]

After the light comes on, the cyanobacteria must first carry out the light-dependent stage of photosynthesis to make oxygen, and enough oxygen must build up in the tissues. Then the neurons need time to run aerobic respiration and make enough ATP before they can fire again.

e. [2 marks]

Any two: NADH and FADH₂ unload electrons and hydrogen ions; oxygen accepts the electrons and combines with hydrogen ions to form water; the energy released makes 26 or 28 ATP per glucose.

f. [2 marks]

For example, Student X: beneficence means the benefits of the research should outweigh the risks or harm. The research could lead to a way to raise oxygen levels in people in medical emergencies, saving lives, while the harm was limited because the tadpoles' brain activity was restarted. (Agreeing with Student Y is also fine if justified: non-maleficence means avoiding harm, and the tadpoles were injected in the heart and had brain activity deliberately stopped.)

From the report. In a, listing facts from the article without linking them to the experiment scored poorly. In b.i, carbon dioxide and glucose belong to the light-independent stage, not oxygen production. In c, answers that stopped at "no photosynthesis, so no oxygen" missed the link to respiration and ATP; low oxygen does not denature enzymes or kill the cells (activity restarted). In d, naming a pathway without linking it to oxygen or ATP for the brain scored poorly. In f, the justification had to match the student chosen.

Question 9 (5 marks)

The woodpecker finch lives on four Galápagos islands; San Cristobal is well separated from the other three by sea.

a. Using one example, describe how molecular homology can show relatedness between the populations. (2 marks)

b. The San Cristobal population may be a separate species. Referring to the map, explain the type of speciation. (3 marks)

Show worked solution

a. [2 marks]. Compare the DNA (or mtDNA) base sequence of a gene, or the amino acid sequence of a protein, between populations. The more similar the sequences, the more closely related the populations, because less time has passed for mutations to build up since they separated.

b. [3 marks]. Allopatric speciation. San Cristobal is separated from the other islands by a wide stretch of sea that the finches rarely cross, so the populations are geographically isolated with no gene flow. Each population gained different mutations, and different selection pressures on each island favoured individuals with different advantageous alleles, which survived and reproduced. Over time the gene pools diverged until the populations could no longer interbreed.

From the report. In a, the question was about populations, not species; answers had to be adapted. In b, "different selection pressures" alone was not enough; link it to individuals with an advantage surviving and reproducing. Sympatric speciation did not fit, since the populations are on different islands.

Question 10 (8 marks)

Fossilised footprints about 1.52 million years old were found in Kenya. Researchers concluded they are from hominins (Conclusion 1) and from two separate species (Conclusion 2). Separately, two studies dated interbreeding between Homo sapiens and Neanderthals: Study 1 (7 ancient genomes, about 45 000 years old) found 49 000 to 45 000 years ago; Study 2 (334 genomes: 275 present-day and 59 ancient) found 50 500 to 43 500 years ago. Study 2 also found present-day people carry a smaller share of Neanderthal DNA than ancient humans.

a

Describe evidence supporting each conclusion. (2 marks)

b

Give one reason for the different conclusions of the two studies, using the table. (2 marks)

c

Account for the lower percentage of Neanderthal DNA in present-day people. (2 marks)

d

Outline two benefits of using mtDNA to trace hominin migration. (2 marks)

Show worked solution

a. [2 marks]

Conclusion 1: the footprints show bipedal features, such as a big toe in line with the other toes (not opposable), a clear arch and a deep heel impression. Conclusion 2: the footprints differ in structure, not just size, such as a different big-toe angle, arch height or pattern of weight transfer from heel to toe, suggesting two different gaits.

b. [2 marks]

For example: sample size. Study 1 analysed only 7 genomes while Study 2 analysed 334, so Study 2's estimate is based on much more data (and a range of ages, from ancient to present-day), which could give a wider, better-supported date range.

c. [2 marks]

Neanderthals died out about 40 000 years ago, so no new Neanderthal DNA has entered the H. sapiens gene pool since. Over many generations of H. sapiens breeding with each other, Neanderthal segments were diluted, and many were selected against (and lost) because they were disadvantageous in H. sapiens.

d. [2 marks]

Any two: mtDNA is inherited only through the mother and does not recombine, so changes come only from mutation and lineages are easy to trace; it has a known, relatively high mutation rate, so differences can be used to estimate when groups split; there are many copies in each cell, so it survives better in old remains.

From the report. In a, "the footprints are different sizes" did not support two species, since a child and an adult of one species differ in size. In c, answers about interbreeding had to say which species and when. In d, describing how mtDNA traces migration, rather than why it is useful, scored nothing.

Advice from the 2025 report

• Answer in the context given. The report repeatedly penalised pre-learned answers that ignored the scenario (Questions 6b, 7c and 9a).
• Know the command terms. "Explain how" needs a mechanism; "contrast" needs differences; "justify" needs a reason beyond restating the question.
• Read graphs carefully. In Question 2b many mistook enzyme activity on the y-axis for trp concentration.
• Know inputs and outputs of each stage of photosynthesis and cellular respiration: they decided many marks in Section A and in Questions 3 and 8.
• Use the marks as a guide. Dot points are acceptable and help make one clear point per mark.