VCE Biology 2024

Every question from the 2024 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 2024 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 2024 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 questions. Several test the key science skills (models, validity, randomisation) as well as content, and three need graphs read accurately: bring a ruler.

Section A: Multiple choice

Q1

A molecule with two chains of amino acids and a quaternary structure (insulin). Which statement is correct? Answer: A - it is a protein, so its monomers are amino acids. (90%)

Q2

Structure and a possible role of an RNA molecule in a eukaryotic cell. Answer: D - RNA nucleotides contain ribose, and mRNA carries the instructions for protein synthesis to the ribosome. (84%)

Q3

In E. coli, attenuation prevents Answer: B - production of the trp enzymes, by stopping transcription before it is complete. When trp is plentiful, the leader mRNA forms a terminator loop and transcription stops before the structural genes. (56%)

Q4

The diagram that best represents a eukaryotic gene. Answer: D - promoter, then alternating exons and introns, then a terminator. Operators belong to prokaryotic operons. (61%)

Q5

A diagram of a bacteriophage injecting DNA into S. pyogenes. Which statement is correct? Answer: A - viral DNA is stored as spacers in the CRISPR array, a memory of the virus. (65%)

Q6

The function of CRISPR-Cas9 in S. pyogenes. Answer: B - to respond to specific viruses if they reinfect the cell, like a bacterial adaptive immune system. (84%)

Q7

GM canola carries cress genes, delivered by Agrobacterium plasmids, that make it herbicide resistant and produce modified omega-3 oils. Answer: A - the oil could be considered genetically modified, since it is made using proteins from the inserted cress genes. (69%)

Q8

Where in a mitochondrion (micrograph, four labelled locations) is the Krebs cycle? Answer: B - location 2, the matrix. (64%)

Q9

Relative ATP output at the four locations. Answer: A - high at location 1 (cristae, electron transport chain), low at location 2 (matrix, Krebs cycle) and low at location 3 (cytosol outside the mitochondrion, glycolysis), none at location 4 (the outer membrane). (52%)

Q10

Bat pancreas cells contain a lot of lactic acid. A valid conclusion? Answer: D - there was little or no oxygen, since lactic acid fermentation happens when oxygen is insufficient. (85%)

Q11

Which statement about the electron transport chain is correct? Answer: C - its outputs are water, ATP and NAD⁺, and NADH donates electrons and hydrogen ions. (77%)

Q12

Microalgae grown under white, red, blue and green light. A reasonable conclusion? Answer: D - under white light, cell density rose at every measurement. (86%)

Q13

A controlled variable for that experiment. Answer: C - the light intensity of all the lights. (84%)

Q14

Where is light energy used to split water? Answer: A - the grana (thylakoid membranes), in the light-dependent stage. (77%)

Q15

A graph shows carbon dioxide becoming less soluble in water as temperature rises. Which statement is correct? Answer: C - at 40 °C less carbon dioxide is dissolved, so Rubisco fixes carbon more slowly than at 20 °C. (68%)

Q16

Producing bioethanol from biomass. Answer: D - the biomass could be plants high in cellulose. Carbon dioxide is also produced, microorganisms are needed, and each step uses a different enzyme. (72%)

Q17

During a gout attack (urate crystals irritate a joint) Answer: C - the heat and swelling are caused by signalling molecules such as histamine acting on blood vessels. (58%; 23% chose B, whose reason is wrong.)

Q18

A study inviting people diagnosed with gout to complete an anonymous survey about gout attacks and alcohol. Answer: A - a correlational study using primary data; recording gender, age and ethnicity in a non-identifying way shows ethical standards being applied. (68%)

Q19

The complement system is a group of proteins that Answer: A - enhance inflammation by attracting phagocytes to the infection. They circulate inactive until triggered. (55%)

Q20

Roles of neutrophils only, and of both neutrophils and dendritic cells. Answer: B - only neutrophils secrete chemicals that kill pathogens; both are phagocytes. Neutrophils do not usually present antigen on MHC II. (59%)

Q21

A role of the lymphatic system. Answer: A - it transports lymphocytes to sites of infection. (67%)

Q22

Match type of immunity to strategy. Answer: C - passive: antibodies cross the placenta to a foetus. (70%)

Q23

A technique giving quantitative primary data on insecticides and mosquitoes. Answer: D - measuring the mass of dead mosquitoes after each spray. (88%)

Q24

How to check reproducibility. Answer: D - run the investigation on different days with different students (changed conditions). (80%)

Q25

In a cell-mediated immunity diagram, cell type 1 recognises antigen on an antigen-presenting cell and activates cell type 2. Cell type 1 is a Answer: C - helper T cell. (77%)

Q26

Which statement about the diagram is correct? Answer: B - cell type 2 (a cytotoxic T cell) acts on the infected cell to trigger its death. (65%)

Q27

Where would the antigen-presenting cell most likely meet cell type 1? Answer: D - in a lymph node. (69%)

Q28

Mice were bred over 20 generations: cage X with random mice returned to breed, cage Y with the longest-tibia mice returned. Independent and dependent variables? Answer: A - independent: the method of selecting mice to return; dependent: tibia length. (79%)

Q29

Why might cage X tibia length still change despite random selection? Answer: B - genetic drift: chance changes in allele frequency, which matter most in a small population. (27%; 48% chose A, but gene flow needs interbreeding between populations.)

Q30

Why did tibia length change in cage Y? Answer: D - selective breeding: the scientists chose which mice bred. (78%)

Q31

A diagram of fish-group diversity through geological time. Answer: C - the oldest group (jawless fish) evolved more than 485 million years ago. (84%)

Q32

How were spiny shark fossils (over 300 million years old) most likely dated? Answer: C - relative dating using superposition of rock layers. Carbon-14 only works to about 50 000 years. (55%)

Q33

Host ranges of influenza A, B, C and D. The likely result of antigenic drift? Answer: B - drift in influenza C would prompt public health measures, since people may have little immunity to the new variant. (68%)

Q34

Dolphin skeletons show the pelvis shrinking over time. In the most recent dolphin the pelvis is Answer: B - a vestigial structure. (85%)

Q35

Hylobates lar (a gibbon): fur, live young, family groups, no tail, swings by its arms. Answer: B - primate and hominoid. It is not bipedal, so not a hominin. (69%)

Q36

A marsupial phylogenetic tree. Answer: A - opossums and bandicoots split longer ago (about 77 million years) than bilbies and numbats (about 63 million years), so their DNA differs more. (89%)

Q37

Features Homo naledi is most likely to share with modern Homo. Answer: A - a well-developed thumb and long legs relative to arms. (68%)

Q38

Where to place H. naledi on a simplified hominin tree. Answer: A or D (both accepted) - position 1, since it has both Homo and Australopithecus features (A), or nowhere yet, because more fossil and DNA evidence is needed (D). (A 26%, D 36%)

Q39

Repeated measurements agree closely but differ a lot from reference values. Answer: C - precise but not necessarily accurate. (80%)

Q40

Antibiotic disks Q, R and S: only Q has no clear zone around it. The bacteria could be resistant to Answer: A - Q only. (66%; 31% chose D, confusing the clear zone with resistance.)

Section B

Question 1 (11 marks)

A diagram of a plant cell labels locations V, W, X, Y and Z.

a.i

Which location is the Golgi apparatus? (1 mark)

a.ii

Describe the role of the Golgi apparatus. (3 marks)

b

Describe the role of tRNA in making a protein. (2 marks)

c

Explain how an enzyme catalyses the breakdown of a protein. (2 marks)

d

Describe the role of vesicles in exporting protein from a cell. (1 mark)

e

Protein synthesis can be seen as a scientific model. Identify a limitation of this model and suggest a solution. (2 marks)

Show worked solution

a.i. [1 mark]

Location W.

a.ii. [3 marks]

The Golgi apparatus receives proteins (in vesicles) from the rough endoplasmic reticulum, modifies them (for example, by adding carbohydrate groups) and packages them into vesicles that carry them to the plasma membrane for secretion by exocytosis.

b. [2 marks]

Each tRNA carries a specific amino acid to the ribosome. Its anticodon pairs with the complementary codon on the mRNA, so amino acids are lined up in the order the mRNA specifies and joined into a polypeptide.

c. [2 marks]

The protein (substrate) binds to the enzyme's active site, which is complementary to it. The enzyme lowers the activation energy needed to break the peptide bonds between amino acids, so hydrolysis happens faster.

d. [1 mark]

Vesicles carry the protein from the Golgi to the plasma membrane and fuse with the membrane, releasing the protein outside the cell.

e. [2 marks]

Limitation: the model is simplified, showing protein synthesis as a single linear sequence and leaving out complexity such as gene regulation, alternative splicing and protein folding (or differences between prokaryotes and eukaryotes). Solution: use several models or more detailed versions, and refine the model as new research findings emerge.

From the report. In a.ii, answers had to be specific to the Golgi, not a general account of exocytosis. In c, "enzymes break down proteins" restated the stem. In d, many said vesicles fuse with the cell wall: they fuse with the plasma membrane. Part e was challenging: many gave a limitation of protein synthesis itself rather than of the model.

Question 2 (5 marks)

Intestinal epithelial cells (which absorb nutrients) and fat cells (which store fat) contain the same genes but have very different structures and functions. Describe an organism's proteome and use it to explain how organisms achieve such diverse cell types, referring to both cells.

Show worked solution

[5 marks]. The proteome is the complete set of proteins expressed by an organism (or a cell) at a given time.

• Every cell has the same genome, but gene regulation switches different genes on and off in different cell types, so each cell expresses only part of the genome and has its own set of proteins.
• Alternative splicing of pre-mRNA lets one gene make several different proteins, so the proteome is much larger than the number of genes.
• An intestinal epithelial cell expresses genes for membrane transport proteins and the structural proteins that build its microvilli, giving a large surface for absorbing nutrients.
• A fat cell expresses genes for enzymes that make and store lipids, and proteins that form a large lipid droplet, so it can store fat.
• The different proteins produced (and how they fold) give each cell type its own structure and function.

From the report. Many answers did not actually describe the proteome. Full marks needed specific reference to both cell types, and a range of ways different cells produce different proteins.

Question 3 (4 marks)

Yeast ferment a 45% glucose solution with no oxygen present. Over 500 hours ethanol rises and levels off at about 13%, while glucose falls and levels off at about 17%.

a. State two reasons the glucose concentration does not fall to zero. (2 marks)

b. A researcher suggests adding oxygen to make the yeast produce more ethanol and use more glucose. Describe the effect of adding oxygen on ethanol production, with a reason. (2 marks)

Show worked solution

a. [2 marks]. Any two: ethanol builds up to a level that is toxic to the yeast (or changes the pH), slowing or stopping fermentation; other factors become limiting, such as the number of yeast cells, coenzymes or nutrients; the yeast's ATP needs are already met; more time would be needed to use the remaining glucose.

b. [2 marks]. Ethanol production would decrease. With oxygen available, yeast carry out aerobic cellular respiration, which breaks glucose down fully to carbon dioxide and water, so less glucose goes through fermentation, the only pathway that makes ethanol. (The report also accepted that some fermentation can continue even with oxygen present.)

From the report. Part a was difficult (9% full marks): many wrote true statements about fermentation that did not explain why glucose remained. In b, "aerobic fermentation" is a contradiction.

Question 4 (11 marks)

A graph shows carbon dioxide fixation (mmol/hr) against light intensity (0 to 2000 lux) for a C4 and a C3 plant. Both rise roughly linearly and level off above about 1700 lux, the C4 plant at about 670 mmol/hr and the C3 plant at about 340 mmol/hr.

a

Explain why photosynthesis is considered a biochemical pathway. (3 marks)

b

Use the graph to support the claim that C4 plants growing next to C3 plants have an advantage. (3 marks)

c

Explain why CO₂ fixation in the C3 plant changes with light intensity and then plateaus above 1700 lux. (2 marks)

d

Explain how C4 plants are better adapted than C3 plants to reduce photorespiration. (3 marks)

Show worked solution

a. [3 marks]

Any three: it is a series of linked reactions (the light-dependent and light-independent stages); the outputs of one stage are inputs of the next (ATP and NADPH from the light-dependent stage drive the Calvin cycle); each step is catalysed by a specific enzyme (such as Rubisco) with coenzymes (NADP⁺/NADPH, ADP/ATP); the stages happen in different locations (thylakoid membranes, then stroma).

b. [3 marks]

At every light intensity the C4 plant fixes about twice as much CO₂: for example about 410 against 210 mmol/hr at 1000 lux, and about 670 against 340 mmol/hr at the plateau. Fixing carbon faster means the C4 plant carries out more photosynthesis and makes more glucose, so it can grow faster and outcompete the C3 plant for light and space.

c. [2 marks]

Rising section: more light speeds up the light-dependent stage, producing more ATP and NADPH for the light-independent stage, so more CO₂ can be fixed. Plateau: above about 1700 lux, light is no longer the limiting factor; another factor (such as the amount of Rubisco, which is saturated, or CO₂ concentration or temperature) limits the rate.

d. [3 marks]

In C4 plants, CO₂ is first fixed in mesophyll cells by an enzyme that does not bind oxygen, into a 4-carbon compound. This is carried into bundle sheath cells, where it releases CO₂, building up a high CO₂ concentration around Rubisco. So Rubisco in C4 plants rarely binds oxygen, whereas in C3 plants Rubisco sits in mesophyll cells exposed to air-level oxygen and binds O₂ more often, especially when stomata close in hot conditions.

From the report. In b, strong answers quoted data for both plants and linked it to more fixation; "C4 plants take up more CO₂" is not the same as fixing more. In c, many explained the plateau but not the initial rise, and some wrongly named light as the limiting factor at the plateau. In d, answers had to compare the two plant types; bundle sheath and mesophyll are different cells, not parts of one cell, and closing stomata by day describes CAM plants.

Question 5 (9 marks)

CRISPR-Cas9 was used to make a single nucleotide substitution in the barley gene Hv1, improving nitrogen uptake. Graphs compare control and transformed plants for chlorophyll content and nitrogen. A restriction enzyme BstN1 cuts non-transformed DNA once but does not cut transformed DNA; gel results are shown for a control and three plants, Hv1-A, Hv1-B and Hv1-C.

a.i

Read the chlorophyll and nitrogen values for control and transformed plants from the graphs. (1 mark)

a.ii

Explain how the change in either chlorophyll or nitrogen could affect the plant. (2 marks)

b

Explain why some barley seeds may not be transformed. (2 marks)

c

Using the gel, identify the transformed plant and justify your answer. (2 marks)

d

Describe how DNA fragments are separated in gel electrophoresis. (2 marks)

Show worked solution

a.i. [1 mark]

Chlorophyll: control 1.0 mg/g, transformed 1.5 mg/g. Nitrogen: control 0.2%, transformed 0.5%.

a.ii. [2 marks]

Chlorophyll: more chlorophyll lets the plant absorb more light energy, increasing the light-dependent stage and the rate of photosynthesis, so more glucose is made for growth and yield. (Or nitrogen: nitrogen is needed to make amino acids and nucleic acids, so more nitrogen allows more protein synthesis and growth.)

b. [2 marks]

Any two: the CRISPR-Cas9 components did not enter every cell; the sgRNA did not bind its target (for example, a mismatch), so Cas9 did not cut the gene; there was no PAM sequence next to the target; the cell repaired the cut without making the intended substitution; off-target editing happened instead.

c. [2 marks]

Hv1-C. It shows a single, larger band, so BstN1 did not cut its DNA: the substitution removed the recognition site. The control, Hv1-A and Hv1-B each show two smaller bands, because their DNA was cut once.

d. [2 marks]

DNA is negatively charged, so when a current is applied it moves through the gel towards the positive electrode. The gel acts as a mesh: smaller fragments move faster and further, so fragments separate by size.

From the report. In a.ii, many discussed the environmental benefit of less fertiliser rather than the effect on the plant. In b (13% full marks), the roles of sgRNA, Cas9 and the PAM site were poorly understood, and some wrongly discussed bacterial transformation and plasmids. In c, the band is larger because the DNA was not cut, not because a gene was inserted. In d, anode and cathode were often confused.

Question 6 (4 marks)

Trastuzumab and pertuzumab are monoclonal antibodies that both bind the HER2 receptor on breast cancer cells, at different sites.

a. Compare the variable regions of the two antibodies. (2 marks)

b. Identify two ways the antibodies could help treat HER2-positive breast cancer. (2 marks)

Show worked solution

a. [2 marks]. Both variable regions are complementary to the HER2 receptor, so both bind it. But they bind different sites (epitopes) on the receptor, so their variable regions have different shapes (amino acid sequences).

b. [2 marks]. Any two: they block the HER2 receptor, stopping growth signals, so the cancer cells stop dividing; they flag (opsonise) cancer cells so immune cells destroy them; they can deliver drugs or radioactive isotopes straight to cancer cells; they target only cancer cells, reducing side effects on healthy cells.

From the report. Answers had to refer to both antibodies in a. A common error was saying trastuzumab binds the membrane. In b, effects had to be on cancer cells, not "on cancer".

Question 7 (10 marks)

Measles is highly contagious and spreads by coughs and sneezes. Reported cases in Australia peaked at nearly 5000 in 1994 and then fell sharply to low levels by 2002.

a

Identify one action that may have caused the change from 1994, and explain how. (3 marks)

b

Measles is no longer endemic in Australia, but three cases were reported in Victoria in January 2024 and the places they had visited were published. Identify a likely source of infection and two benefits of listing the locations. (3 marks)

c

Describe two factors contributing to the emergence of new pathogens. (4 marks)

Show worked solution

a. [3 marks]

Vaccination (a national measles vaccination campaign). With more people immune, there are fewer susceptible hosts, so an infected person passes the virus to fewer people. This herd immunity breaks chains of transmission, so reported cases fall.

b. [3 marks]

Source: an infected traveller returning from (or visiting from) a country where measles is circulating. Benefits: people who were at those places know to watch for symptoms and get tested; they can isolate early to avoid spreading it; others can avoid those places (or check their vaccination status).

c. [4 marks]

Two factors, each described:

• Zoonosis / closer contact with animals (for example, clearing forests or the wildlife trade): pathogens in animals get more chances to jump to humans.
• Antigenic shift: two different virus strains infect the same host cell and swap genetic material, producing a new strain that people have no immunity to.

(Also accepted: climate change releasing or spreading pathogens; travel into previously unexplored areas.)

From the report. In a, mask wearing was not accepted as the cause of the drop. In b, "they visited a place with measles" was not a source. In c (7% full marks), antigenic drift and antibiotic resistance were not accepted, since they change existing pathogens rather than create new ones, and each description had to match its factor.

Question 8 (8 marks)

Researchers sampled 20 populations of Eucalyptus salubris in Western Australia and found two distinct groups living close together: one with thicker leaves, and one with thinner leaves found only in soils high in phosphorus. The groups differ genetically, and the researchers concluded they are separate species.

a

State the scientific methodology used to collect the samples. (1 mark)

b

Identify the evidence that the groups are separate species. (2 marks)

c.i

Identify the type of speciation that may be occurring. (1 mark)

c.ii

Explain what led to the genetic diversity between the two groups. (4 marks)

Show worked solution

a. [1 mark]

Fieldwork (sampling in the natural habitat).

b. [2 marks]

The groups have significantly different DNA sequences (molecular differences), and they have different structural features (tree appearance and leaf thickness) even where they grow close together.

c.i. [1 mark]

Sympatric speciation (the groups live in the same area).

c.ii. [4 marks]

• The original population already had variation in alleles for traits such as leaf thickness.
• Soils with high phosphorus imposed a different selection pressure. Trees with alleles for thinner leaves had a selective advantage there, survived and reproduced more, and passed those alleles on, so their frequency rose in that group.
• Gene flow between the groups was reduced, for example if they flowered at different times so insect pollinators rarely carried pollen between them, and seeds fall directly under the parent tree.
• Over many generations, different mutations accumulated in each group and the gene pools diverged.

From the report. Part a was answered poorly (39%). In b, restating the phosphorus information did not explain why they are separate species. In c.ii, avoid Lamarckian wording ("the trees adapted to phosphorus"), and answer the question asked: what created diversity between two existing groups, not a pre-learned speciation story.

Question 9 (9 marks)

A genome study of 121 Indigenous Australians from four remote communities (Tiwi Islands, Galiwin'ku, Titjikala and Yarrabah) was run under Indigenous governance, with communities invited and individuals consenting. DNA of 18 non-Indigenous Australians of European ancestry was also sequenced. In the 0.4% of the genome that varies, 46% of the sequence variation was found within a single community and only 2.8% was shared across all four. Indigenous Australians have the most DNA sequence variation found outside Africa. A tree dates the splits between the communities to roughly 26 000 to 35 000 years ago.

a

Give one reason the non-Indigenous DNA was sequenced. (1 mark)

b

Identify how the researchers applied the ethical principle of respect, with a reason. (2 marks)

c

Use the data to support two conclusions: each community has highly diverse DNA, and the variations differ between communities. (4 marks)

d

Explain how the results advance understanding of either human genomic diversity or the treatment of genetic disease. (2 marks)

Show worked solution

a. [1 mark]

As a comparison group, to show how much DNA sequence diversity there is within the Indigenous communities relative to another population.

b. [2 marks]

The research was carried out under Indigenous governance and custodianship of the data, with communities invited and individuals consenting. This respects the communities' right to make their own decisions and values their cultures and ways of knowing.

c. [4 marks]

• Highly diverse DNA within communities. Data: 46% of the sequence variation studied was found within individuals of the same community, and Indigenous Australians have the highest DNA variation outside Africa. Explanation: the populations are ancient, so there has been a very long time for mutations to accumulate.
• Different variations across communities. Data: only 2.8% of the variation was shared by all four communities. Explanation: the communities have been genetically isolated for at least 26 000 years with little gene flow, so each gained different mutations (and genetic drift acted separately in each).

d. [2 marks]. Genetic disease (for example): the study found many DNA variants not previously recorded, some specific to particular communities. Knowing these variants helps doctors identify genetic diseases that are more common in a community and design more relevant diagnosis and treatment for Indigenous patients, who were under-represented in earlier genome databases.

From the report. In a, a common wrong answer was that it tested for interbreeding. In c, the data had to relate to within-community versus between-community variation, not just quote numbers. Part d was difficult (15% full marks); answers about the treatment of genetic disease tended to score better.

Question 10 (9 marks)

A student tested whether mentioning herd immunity in vaccine advertising changes people's intention to vaccinate. 121 participants recruited through the school newsletter and social media were randomly split into three groups. All read a scenario about a new disease and a vaccine not yet fully tested for side effects; Group B also read about the individual benefit of vaccination, and Group C about the herd-immunity benefit. They rated their likelihood of vaccinating from 1 (not likely) to 5 (very likely), anonymously online. Average ratings were A 2.5, B 3.0 and C 3.5.

a

Analyse the results. (3 marks)

b.i

What strategy could the student recommend to health authorities? (1 mark)

b.ii

Propose another strategy to raise vaccination rates. (1 mark)

c.i

Outline why random placement of participants matters. (2 marks)

c.ii

Comment on the validity of the results, without referring to random placement. (2 marks)

Show worked solution

a. [3 marks]

Group A (scenario only) had the lowest average likelihood, 2.5 (between slightly and somewhat likely). Group B (individual benefit) averaged 3.0 (somewhat likely). Group C (herd-immunity benefit) had the highest average, 3.5 (between somewhat likely and likely). So any extra information increased willingness to vaccinate, and information about protecting others through herd immunity had the largest effect.

b.i. [1 mark]

Run advertising campaigns that explain the herd-immunity benefit of vaccination (protecting people who cannot be vaccinated).

b.ii. [1 mark]

A different kind of strategy, for example: free or subsidised vaccines, school vaccination programs, or "no jab, no play" requirements.

c.i. [2 marks]

Random placement spreads participants' existing beliefs and backgrounds (age, attitudes to vaccines) evenly across the groups. This reduces bias, so the information given is the only systematic difference between groups and any difference in ratings can be put down to it.

c.ii. [2 marks]

The results have limited validity. On the plus side, the groups differed only in the information given, Group A acted as a comparison, and anonymous answers reduce pressure to give a "good" answer. But participants came from a narrow group (a school newsletter and social media), so they may not represent the Australian population, and the ratings are subjective self-reports of intention, not actual vaccination.

From the report. Part a was well answered, but "3.5" is an average rating out of 5, not 3.5 people or 3.5%. In b.ii, another advertising or education strategy was not accepted, since it repeats b.i. Part c.i was challenging (5% full marks): answers had to apply terms such as bias and validity to this investigation, not just use them.

Advice from the 2024 report

• Use the marks and dot points. Each mark needs a separate point; in Question 4c many answered only half of a two-part question.
• Add biology, don't repeat the stem. Link data to knowledge, for example chlorophyll to the light-dependent stage (Question 5a.ii).
• Know the command terms such as compare, justify and outline, and answer the question asked rather than a pre-planned response (Question 8c.ii).
• Learn the key science skills. Models, reproducibility, validity, precision and randomisation were tested across both sections.
• Bring a ruler to read graphs accurately (Questions 4b, 5a and 10a).