VCE Biology 2023

Every question from the 2023 VCE Biology examination, the second exam on the 2022 to 2026 study design. 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 2023 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 2023 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): 11 questions, with more extended responses (three to six marks) than earlier papers, such as the 6-mark comparison in Question 1b.

Section A: Multiple choice

Q1

Match a nucleic acid with its role. Answer: B - tRNA carries amino acids to the ribosome. (86%)

Q2

Structures of two nucleotides, one with uracil and one with thymine. Answer: D - the thymine nucleotide, like every nucleotide, has a pentose sugar (deoxyribose). Uracil is found in RNA, not DNA. (71%)

Q3

A diagram of gene expression: a gene, then molecules E, F and G across steps 1 to 3. Answer: D - the circle on molecule F is the modified guanine (methyl) cap added during mRNA processing. (39%)

Q4

How many codons code for a 100-amino-acid polypeptide? Answer: C - 100: one codon per amino acid. 300 is the number of nucleotides. (33%; 39% chose D.)

Q5

Roles of organelles R, S, T and U in exocytosis. Answer: B - S, the rough endoplasmic reticulum, folds the protein and transports it on towards the Golgi. (74%)

Q6

What would students expect when comparing proteins in different human cells? Answer: B - proteins in the same cell can differ in structure and function. (75%)

Q7

Systematic errors Answer: D - shift readings away from the true value by a consistent amount each time. They affect accuracy, not precision. (71%)

Q8

Most likely route for a rust fungus to enter wheat. Answer: A - through the stomata, openings in the leaf surface. (72%)

Q9

Wheat carrying rust-resistance genes from rye and wild grasses is Answer: B - transgenic: it contains genes from another species. (82%)

Q10

Electrophoresis is used to Answer: C - separate DNA fragments. (78%)

Q11

A cancer drug binds to protease as shown. The protease is Answer: D - inhibited. (66%)

Q12

Publishing all results, favourable or not, aligns most with Answer: C - integrity (honest reporting). (84%)

Q13

Where in a chloroplast micrograph is the Calvin cycle? Answer: B - location B, the stroma. (69%)

Q14

A correct statement about the Calvin cycle. Answer: A - it is light-independent (it does not need light directly, though it relies on ATP and NADPH made in the light). (85%)

Q15

Molecules involved in a light-independent reaction. Answer: D - carbon dioxide, ATP and NADPH. NADH belongs to cellular respiration. (74%)

Q16

Where does a light-dependent reaction occur? Answer: C - the thylakoid membrane. (82%)

Q17

Enzymes of glucose metabolism are Answer: C - proteins. (68%)

Q18

White blood cells at four temperatures, with and without oxygen: when is lactate formation highest (of the options)? Answer: B - 20 °C without oxygen. (77%)

Q19

Why is more glucose used without oxygen at 35 °C? Answer: D - fermentation makes far less ATP per glucose, so the cell must break down more glucose to meet its needs. (61%)

Q20

Why are glucose use and lactate formation the same with or without oxygen at 50 °C? Answer: A - most glycolysis enzymes have denatured, so both rates are near zero. (91%)

Q21

Cells that release histamine in an allergic response. Answer: A - mast cells. (91%)

Q22

How do antigen-presenting cells present allergens to T cells? Answer: D - on major histocompatibility complex (MHC) proteins. (53%)

Q23

Tea-tree oil reduces skin inflammation. It inhibits the Answer: B - innate immune response (inflammation is part of the second line of defence). (70%)

Q24

Over nine years, prescriptions fell more in the allergen immunotherapy (AIT) group than in the control group, and the gap was maintained. Answer: A - AIT has long-term, sustained effects. (49%; 38% chose C, but the stronger conclusion is that the AIT group's advantage continues and grows across all nine years.)

Q25

A reasonable conclusion about the control group. Answer: C - it had a similar number of participants, as a fair comparison group should. (77%)

Q26

A correct statement about making monoclonal antibodies. Answer: A - the antibodies from one hybrid cell (hybridoma) line are identical. (60%)

Q27

Why humanise a mouse monoclonal antibody? Answer: B - to reduce the chance of the patient's immune system reacting against the antibody. (75%)

Q28

Why can a bispecific antibody bind two targets? Answer: D - it has two different-shaped variable regions. (81%)

Q29

An event that may occur in the lymphatic system. Answer: A - a dendritic cell presents antigen to helper T cells (in a lymph node). (71%)

Q30

Selective breeding of Atlantic salmon for growth and disease resistance. Answer: C - alleles for the desired phenotypes increase in frequency over generations. (90%)

Q31

A vaccine containing inactivated diphtheria toxin gives Answer: A - artificial active immunity. (78%)

Q32

The most likely reason for diphtheria outbreaks. Answer: B - loss of herd immunity when children are not vaccinated. (83%)

Q33

Why was the 2015 to 2016 Zika outbreak an epidemic, not a pandemic? Answer: A - cases were confined to specific regions. (89%)

Q34

The least effective way to control Zika. Answer: B - face masks: Zika spreads by mosquito bites, not droplets. (71%)

Q35

How to improve accuracy. Answer: C - calibrate the equipment; poor calibration is a systematic error. Repeats improve precision. (55%; 38% chose B.)

Q36

Least useful evidence for a phylogenetic tree of mammals. Answer: D - presence of hair, since all mammals have it. (73%)

Q37

A finch phylogenetic tree. The pair with the most recent common ancestor? Answer: D - small and medium ground finches, which join at the most recent branch point. (89%)

Q38

Four nested circles showing hominins and other groups, from largest to smallest. Answer: A - mammals, primates, hominoids, hominins. (73%)

Q39

Brain mass against body mass (log scales) with an average-mammal line. Which species is Homo sapiens? Answer: B - species 2, which lies furthest above the line (a far larger brain than expected for its body mass). (77%)

Q40

Carrying angles of the femur in H. sapiens, australopithecines and chimpanzees. Answer: D - the large carrying angle of australopithecines supports bipedalism, by bringing the knees under the centre of gravity. (72%)

Section B

Question 1 (8 marks)

The structure of tryptophan (trp) is shown. In E. coli the trp operon is shown with the trpR gene, promoter (P), operator (O), leader (trpL) containing the attenuator, and the structural genes trpE to trpA.

a. To which group of biological monomers does tryptophan belong? Justify using the diagram. (2 marks)

b. Compare repression and attenuation as ways of regulating the structural genes. Refer to the diagram. (6 marks)

Show worked solution

a. [2 marks]. Amino acids. The structure has an amine group (NH₂) and a carboxyl group (COOH) on the same carbon, plus a variable R group (tryptophan's ring side chain).

b. [6 marks].

• Both regulate the structural genes trpE to trpA when trp levels are high, so the cell does not waste energy and materials making trp it already has.
• Repression: the trpR gene codes for a repressor protein. When trp is plentiful, trp binds the repressor and changes its shape so it can bind the operator (O). This blocks RNA polymerase at the promoter (P), so transcription does not start.
• Attenuation: acts on transcription that has already started (for example, when the repressor briefly detaches). RNA polymerase transcribes the leader (trpL), which contains trp codons, while a ribosome translates it. With plenty of trp, the ribosome moves quickly past these codons and the leader mRNA folds into a terminator hairpin at the attenuator, so transcription stops early, before the structural genes.
• Differences: repression acts at the operator and prevents transcription from starting, while attenuation acts at the leader/attenuator and stops transcription part-way; repression uses a repressor protein, while attenuation depends on ribosome movement and mRNA folding; repression is the main on/off switch, and attenuation is a finer second control.

From the report. Students could compare the two directly or describe each in turn. Only high-trp conditions needed discussion.

Question 2 (7 marks)

Plants of an unknown species were grown under different temperatures (20 or 40 °C), light colours, CO₂ levels and humidities, and starch formation was measured. At 40 °C starch formation was higher than at 20 °C in every matching condition, often nearly double (for example 2.8 against 1.5 mg/day in white light, 5% CO₂, 70% humidity). Green light gave 0.3 to 0.4 mg/day and blue light 1.4 to 2.7 mg/day.

a. Is the plant most likely C3, C4 or CAM? Justify. (3 marks)

b. Why does green or blue light, rather than white light, change starch formation? Justify. (4 marks)

Show worked solution

a. [3 marks]. C4. Starch formation is highest at 40 °C, higher than at 20 °C in every matching condition and nearly double in several, and stays high even at low CO₂ (2.4 mg/day at 1% CO₂). C4 plants concentrate CO₂ around Rubisco in bundle sheath cells, so photorespiration stays low at high temperatures, and they photosynthesise well in hot conditions. A C3 plant would be expected to lose efficiency at 40 °C through photorespiration. (CAM was also accepted if linked to high temperature and the higher starch formation at low humidity.)

b. [4 marks].

• White light contains all visible wavelengths, including red, green and blue.
• Chlorophyll absorbs mainly red and blue light and reflects most green light.
• Under green light very little light is absorbed, so the light-dependent stage produces little ATP and NADPH, little glucose is made and starch formation is very low (0.3 mg/day).
• Blue light is absorbed, so photosynthesis continues and starch formation is only slightly lower than in white light (1.4 against 1.8 mg/day), since the red wavelengths are missing.

From the report. Choosing C3 scored nothing in a. Part b needed four separate points.

Question 3 (10 marks)

People infected with Streptococcus pyogenes make specific antibodies and usually recover. A vaccine study deliberately infected participants' throats. In a follow-up, group A received a vaccine and group B a salt-solution placebo; 10 days later memory cells were measured and throats were swabbed with the bacteria; after 14 more days antibodies and white blood cells were measured. Of four participants, participant 1 had a sore throat, low memory cells at stage 2 and medium antibody at stage 3; participants 2 to 4 had medium or high memory cells and high antibody.

a

Describe how the specific antibodies act during infection and lead to recovery. (3 marks)

b

Suggest a bioethical principle the first study may have violated, and explain. (2 marks)

c

Which participant most likely received the placebo? Explain. (3 marks)

d

Some Mycobacterium tuberculosis strains resist many drugs. Outline two ongoing challenges for treatment. (2 marks)

Show worked solution

a. [3 marks]

The antibodies bind to specific antigens on the bacteria. This neutralises them, clumps them together (agglutination) and marks them for phagocytosis (opsonisation), and can activate complement proteins. Phagocytes then engulf and destroy the bacteria, so their numbers fall and the person recovers.

b. [2 marks]

Non-maleficence (avoid or minimise harm). The researchers deliberately infected healthy people, knowingly causing a sore throat and a risk of more serious illness.

c. [3 marks]

Participant 1. They had a low memory cell count at stage 2, before being swabbed, showing no earlier exposure to the antigen, as expected after a salt-solution placebo. After the swab they developed a sore throat and had only medium antibody and memory cell levels: a slower, smaller primary response. The other participants had higher memory cells before the swab and higher antibody afterwards, consistent with vaccination.

d. [2 marks]

Any two: new drugs must be developed, which takes a lot of time and money; treatment is longer, more expensive and less successful; resistance can spread further if drugs are overprescribed or courses are not finished; resistant strains make outbreaks harder to control.

From the report. Terms such as agglutination did not need to be named in a, as long as the process was described. Part b was well answered (80% full marks).

Question 4 (6 marks)

Coenzymes play a role in glucose metabolism. Red blood cells at 37 °C consumed 1.6 nmoles of glucose and produced 3.2 nmoles of lactate (per 10⁷ cells per hour), whether oxygen was present or absent.

a. Describe the roles of coenzymes in forming ATP. (3 marks)

b. Explain why the results are the same with or without oxygen. (3 marks)

Show worked solution

a. [3 marks]. During glycolysis and the Krebs cycle, the coenzymes NAD⁺ and FAD are loaded with electrons and hydrogen ions, becoming NADH and FADH₂. They carry these to the electron transport chain, where they are unloaded. The energy released drives the joining of ADP and Pi to form ATP, and the unloaded coenzymes return to be used again.

b. [3 marks]. Mature red blood cells have no mitochondria, so they cannot carry out the Krebs cycle or electron transport chain, even when oxygen is present. They rely on glycolysis followed by lactic acid fermentation in the cytosol, so glucose use and lactate production do not change with oxygen. Each glucose gives two pyruvate, each converted to lactate, which is why lactate (3.2) is exactly twice the glucose used (1.6).

From the report. Part b was difficult (8% full marks).

Question 5 (6 marks)

Citrus canker bacteria activate the citrus gene CsWRKY22; its protein binds to the gene CsLOB1, whose transcription produces proteins that cause the disease. Infected trees could be turned into biofuel: steam explosion breaks cells open (36% at 20 °C up to 94% at 250 °C), then the mixture is cooled to 35 °C and yeast is added.

a

Using the flow chart, explain how CRISPR-Cas9 could produce canker-resistant plants. (4 marks)

b.i

Why use 250 °C? (1 mark)

b.ii

Why lower the temperature to 35 °C and add yeast? (1 mark)

Show worked solution

a. [4 marks].

• Design a single guide RNA (sgRNA) complementary to a target sequence in the CsWRKY22 gene (or in CsLOB1), next to a PAM sequence.
• The sgRNA combines with Cas9 and guides it to the target.
• Cas9 cuts both DNA strands at the target. The cell's error-prone repair adds or deletes nucleotides, knocking out the gene.
• Without a working CsWRKY22 protein, CsLOB1 is not activated (or, if CsLOB1 is knocked out, it produces no functional protein), so the proteins that cause canker are not made and the plants resist the disease.

b.i. [1 mark]. At 250 °C the most cells (94%) are broken open, releasing the most glucose (and other carbohydrates) for fermentation.

b.ii. [1 mark]. Yeast ferments the released glucose to ethanol (the biofuel), and 35 °C is close to the optimum temperature for the enzymes of anaerobic fermentation, which would be denatured at 250 °C.

From the report. In b.ii, "the optimum temperature for the yeast" was not enough: link it to the yeast producing ethanol or to its fermentation enzymes.

Question 6 (11 marks)

Human insulin has two polypeptide chains, A (21 amino acids) and B (30 amino acids), joined by two disulfide bridges, and can be made with recombinant plasmids. A graph shows the activity of four insulins over 24 hours: lispro peaks at about 7 units within about 2 hours and is finished by about 5 hours; regular insulin peaks at about 6 units near 5 hours and lasts about 12 hours.

a

Outline how human insulin is made using recombinant plasmids. (5 marks)

b

Which insulin is rapid acting compared with regular insulin? Use data. (3 marks)

c.i

What level of protein structure does insulin have? (1 mark)

c.ii

What should scientists consider when changing insulin's primary structure? Justify. (2 marks)

Show worked solution

a. [5 marks].

1. Obtain the human genes for chain A and chain B separately.
2. Cut plasmids and the genes with the same restriction enzyme, giving matching sticky ends.
3. Insert the chain A gene into some plasmids and the chain B gene into other plasmids; DNA ligase joins them, forming recombinant plasmids.
4. Put the plasmids into bacteria by heat shock (transformation), and select transformed bacteria, for example by growing them on antibiotic plates when the plasmid carries an antibiotic resistance gene.
5. Grow the bacteria in large volumes; they express the genes, making chain A or chain B. Extract and purify the chains and join A and B with disulfide bridges to make functional insulin.

b. [3 marks]

Lispro. It starts acting almost immediately and peaks at about 2 hours (about 7 units), compared with about 5 hours (about 6 units) for regular insulin, and its effect is over after about 5 hours, compared with about 12 hours.

c.i. [1 mark]

Quaternary (two polypeptide chains).

c.ii. [2 marks]

Changes must not alter the tertiary (and quaternary) shape of the regions insulin needs, especially the part that binds the insulin receptor and the cysteines that form the disulfide bridges. Otherwise the insulin may fold differently and no longer function.

From the report. In a, genes A and B go into different plasmids. In b, nearly everyone named lispro; the third mark went to those who read the graph correctly. Part c.ii was difficult (12% full marks).

Question 7 (8 marks)

The helmeted honeyeater survives in only three small wild populations east of Melbourne and is critically endangered. Since 1989 Zoos Victoria has run a captive-breeding program, releasing over 250 birds and swapping eggs between wild and captive populations each year.

a

Explain how genetic diversity may change over time through genetic drift. (3 marks)

b

How is genetic diversity in the wild populations being maintained? (2 marks)

c

Define a consequence-based approach to bioethics and relate it to Zoos Victoria's work. (3 marks)

Show worked solution

a. [3 marks]

Genetic drift is a random change in allele frequencies from one generation to the next, caused by chance (which individuals happen to survive and breed). Its effect is strongest in small, isolated populations like these. By chance, some alleles can be lost completely and others become fixed, so genetic diversity decreases over time, leaving the birds less able to adapt to change.

b. [2 marks]

Swapping eggs creates gene flow: birds hatched from swapped eggs carry alleles from other populations into each wild population. This adds alleles that may have been lost by drift, maintaining diversity.

c. [3 marks]

A consequence-based approach judges an action by its outcomes: the right action gives the greatest overall benefit and the least harm. For Zoos Victoria, the benefits are more birds in the wild, greater genetic diversity and a lower risk of extinction. The harms to minimise include stress or health risks to birds from captivity and handling, and not releasing more birds than the habitat's food and nesting sites can support. Because the benefits outweigh the harms, the program is justified under this approach.

From the report. In c, answers needed to consider both positive outcomes and negative effects.

Question 8 (4 marks)

Neanderthals lived in Europe and Asia. Recent research found about 0.5% Neanderthal DNA in people of African ancestry, compared with 1.7% in Europeans and 1.8% in Asians.

a. Suggest two ways people of African ancestry could have some Neanderthal DNA. (2 marks)

b. Suggest why Europeans and Asians have more Neanderthal DNA than people of African origin. (2 marks)

Show worked solution

a. [2 marks]. Any two: some H. sapiens left Africa, interbred with Neanderthals in Europe or Asia, and their descendants later migrated back into Africa; ancestors of modern Africans may have interbred with Neanderthals (or their relatives) before H. sapiens left Africa; some shared DNA may be inherited from the common ancestor of both species.

b. [2 marks]. Neanderthals lived in Europe and Asia, not Africa. H. sapiens who moved into Europe and Asia lived alongside Neanderthals for a long time, with many opportunities to interbreed, so more Neanderthal DNA entered their gene pools. Populations that stayed in Africa had little or no contact with Neanderthals.

From the report. Only about a quarter of students scored full marks in a, and a third in b.

Question 9 (5 marks)

Two palm species on Lord Howe Island have different flowering patterns: Howea forsteriana males and females flower mainly in weeks 1 to 6, while H. belmoreana flowers mainly in weeks 7 to 13. Graphs show abundance of each species against soil pH and altitude.

a

Identify two factors from the table that may have contributed to speciation. (2 marks)

b.i

Complete the ideal pH and altitude range for each species. (1 mark)

b.ii

The common ancestor colonised acidic volcanic soils about 6 million years ago; H. forsteriana diverged between 2 million and 500 000 years ago. Explain how the pH and altitude differences could have led to speciation. (2 marks)

Show worked solution

a. [2 marks]

The two species flower at different times (H. forsteriana peaks in weeks 2 to 5, H. belmoreana in weeks 8 to 11), and the peak flowering of males of one species does not overlap with that of females of the other. So there is little cross-pollination between them (temporal reproductive isolation).

b.i. [1 mark]

H. forsteriana: pH 8, altitude 31 to 60 m. H. belmoreana: pH 6, altitude above 121 m.

b.ii. [2 marks]

Some descendants of the ancestor, which was adapted to acidic soils at high altitude, spread onto basic soils at low altitude. There, different selection pressures favoured different alleles by natural selection, and the change in soil shifted their flowering time. With flowering no longer overlapping, gene flow stopped between the two groups (reproductive isolation), and their gene pools diverged until H. forsteriana became a separate species, even on the same small island.

From the report. Part a was difficult (16% full marks). In b.ii, many discussed H. belmoreana instead of how H. forsteriana diverged from the common ancestor.

Question 10 (8 marks)

Volunteers collected water at over 1500 river sites in Victoria to detect platypus environmental DNA (eDNA). Six samples per site were tested by PCR with a platypus-specific DNA probe, and at least two positives were needed to record the platypus as present. A map compares predicted platypus occurrence with the eDNA results.

a

Give two possible sources of platypus eDNA. (2 marks)

b

Explain two factors to consider when sampling water to reduce error. (2 marks)

c

State why PCR was used. (2 marks)

d

Using the map, describe one difference between predicted occurrence and the eDNA detections, and justify it. (2 marks)

Show worked solution

a. [2 marks]

Any two: skin cells, fur, faeces, urine, saliva or mucus shed into the water.

b. [2 marks]

Avoid contamination: wear gloves and use sterile containers, so DNA from people, other animals or other sites does not enter the sample and cause false results. Use a consistent method: collect the same volume and number of samples in the same way at each site, and label each with location, date and time, so results are comparable and correctly recorded.

c. [2 marks]

eDNA is present in tiny amounts. PCR amplifies the platypus DNA, making millions of copies of the target sequence, so there is enough to detect with the probe.

d. [2 marks]

For example: platypuses were detected at some sites in areas predicted to be absent or unknown. Platypuses are shy and mostly active at night, so earlier surveys by sighting or trapping may have missed them. (Or: no platypus DNA at some sites where they were predicted to be widespread, perhaps because populations have declined through drought or habitat damage.)

From the report. In a, giving a location instead of a source was not accepted.

Question 11 (7 marks)

In Lenski's long-term experiment, 12 genetically identical E. coli populations were kept in flasks of 10 mL growth medium containing glucose and citrate (which E. coli normally cannot use). Every 24 hours, 0.1 mL of each culture was transferred to 9.9 mL of fresh medium. After 27 years (about 65 000 generations) results were reported.

a

State two controlled factors. (2 marks)

b

At about 30 000 generations a mutation let some bacteria use citrate. What effect would you expect on later generations, and why? (2 marks)

c

Describe an experiment to investigate the effect of glucose concentration on the frequency of the citrate-using phenotype. (3 marks)

Show worked solution

a. [2 marks]

Any two: the volume of growth medium (10 mL); the volume transferred each day (0.1 mL); the time between transfers (24 hours); the incubation temperature; the starting strain of E. coli; the amounts of glucose and citrate in the medium.

b. [2 marks]

The citrate-using mutants have an extra food and energy source that the others cannot use, so they face less competition and reproduce more. Through natural selection the frequency of the mutant allele increases in later generations, and citrate users come to dominate that population.

c. [3 marks]

• Independent variable: glucose concentration, for example 0%, 0.05%, 0.1% and 0.2%, with the same citrate concentration in every flask.
• Controlled variables: same volume of medium, same starting sample of post-30 000-generation bacteria, same temperature and same 24-hour transfer routine. Include a control with no glucose (citrate only).
• Dependent variable and repeats: after a set number of generations, measure the percentage of citrate-using bacteria (for example, by plating samples on citrate-only agar and counting colonies). Use several flasks for each concentration and run the experiment over many generations for reliable results.

From the report. Because the independent and dependent variables were given in the stem, they could earn at most one mark in c; the other marks came from controlled variables, a control group and repetition.

Advice from the 2023 report

• Know the task words such as compare, justify and outline, and respond to them.
• Plan extended answers. Longer responses (3 to 6 marks) made up much of Section B; set them out as clear separate points.
• Refer to the data. Several questions (2, 3c, 6b) required answers to use the table or graph given.
• Use the space well. Blank space under a question can be used, and if you continue at the back, say so.
• Write in pen. Papers are scanned and marked on screen.