VCE Biology 2021

Every question from the 2021 VCE Biology examination, the last exam on the old 2016 to 2021 study design. Multiple-choice answers come with a short reason; Section B model answers sit behind a Show worked solution toggle. Questions on topics that have since left the course are flagged, so you know what to skip or adapt.

How to use this page

• Questions are from the 2021 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 2021 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.

Which questions still apply

This paper was set on the VCE Biology study design for 2016 to 2021 (Units 3 and 4 from 2017). The study design used since 2022 is still the current one: the version on the VCAA site (version 1.2, April 2026) changed only the legislation and codes it refers to, not the content. Its Unit 3 and 4 topics differ in places, so parts of this paper test content that is no longer examined.

• Outside the current Units 3 and 4: Section A Questions 9 to 11 (apoptosis and caspases, now Unit 1), 13 and 14 (cell signalling), 20 (polyploidy, now Unit 2), 21, 23, 24 and 27 (the timeline of life, mass extinctions, fossilisation and the BMP4 gene) and 38 (the antiviral drug Relenza); Section B Questions 1a (membrane transport, now Unit 1), 5 (cell signalling), 7d (DNA hybridisation) and 8a (cultural evolution).
• Partly outside: Section A Questions 1, 18, 22 and 25, and Section B Questions 1b and 7b. The notes on each question say what still applies.
• Everything else (gene expression, photosynthesis and respiration, immunity, DNA tools, speciation, relatedness, hominins and the science skills) is still examinable and good practice.

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, from one-mark definitions to a 5-mark immune-response summary (Question 4a). Several are built on unfamiliar data or articles, such as the insulin history in Question 10 and the antibiotic experiment in Question 11.

Section A: Multiple choice

Q1

Diagrams show amino acids joined by bond X and nucleotides joined by bond Y. Forming bonds X and Y Answer: B - involves condensation polymerisation: each bond releases a water molecule, so water is a product, not a reactant, and making the bonds uses energy rather than releasing it. (84%) Study design: monomers and polymers are still in Unit 3, but the current design no longer names condensation polymerisation.

Q2

To make the mRNA that the FISH probe detects, a cell must Answer: C - create a nucleotide chain with a ribose-phosphate backbone. RNA contains uracil, not thymine (D), the poly-A tail is added to the RNA, not to the gene (B), and transcription uses RNA polymerase, not DNA polymerase (A). (44%; 27% chose B.)

Q3

One aspect of producing the new molecule (a protein) from this mRNA. Answer: A - rRNA and its associated proteins (the ribosome) create a binding site for the mRNA. tRNA delivers amino acids, not nucleotides; codons are on the mRNA and anticodons on the tRNA; and the small subunit starts translation, not transcription. (36%)

Q4

The probe sequence and the number of amino acids coded by the 18-base target mRNA. Answer: D - 3′ CGGAGCGGAATGCCTGCA 5′, coding 6 amino acids. The probe is DNA, so it pairs A in the mRNA with T, not U; 18 bases is 6 codons. (37%; 53% chose C, which puts U in a DNA probe.)

Q5

Correct information about each stage of photosynthesis. Answer: D - light provides the energy for the light-dependent stage, and ATP (with NADPH) provides it for the light-independent stage. The Calvin cycle uses carbon dioxide, NADPH (not NADH) is the carrier, and the stroma and grana are the other way round. (69%)

Q6

The trend in the rate of photosynthesis as carbon dioxide concentration rises. Answer: C - the rate rises quickly, then levels off when another factor (such as light) becomes limiting. (86%)

Q7

Which observation supports the hypothesis that Rubisco uses the same active site for oxygen (Reaction 1) and carbon dioxide (Reaction 2)? Answer: B - the rate of Reaction 2 decreases when oxygen concentration increases: oxygen competes with carbon dioxide for the same site. (47%; 20% chose A.)

Q8

Predicted against actual chlorophyll a concentration. Answer: C - at 0.8 mg/L the predicted and actual values were the same (the point sits at 0.8, 0.8). (92%)

Q9

The C. elegans cell in Diagram 2 (Ced proteins activated after a death signal) is undergoing Answer: A - apoptosis. (96%) Study design: apoptosis is now Unit 1 content, not examined in Units 3 and 4.

Q10

The Ced-3 protein is most likely a Answer: D - caspase, an enzyme that cleaves proteins to dismantle the cell. (70%) Study design: outside the current course; caspases are not named in it.

Q11

The purpose of most programmed cell deaths in C. elegans embryos. Answer: D - to ensure the developing organs form correctly, removing cells that are not needed. (50%) Study design: outside the current Units 3 and 4 (Unit 1).

Q12

Which organelle transports growth hormone (GH) to the plasma membrane of pituitary cells? Answer: A - vesicles carry it from the Golgi apparatus to the membrane. (71%; 18% chose C.)

Q13

Where are the GH receptors on liver and fat cells? Answer: C - on the plasma membrane. GH is a peptide (hydrophilic) hormone that cannot cross the membrane. (69%) Study design: receptor position and signal transduction are no longer in the course.

Q14

Liver cells secreting IGF-1 in response to GH is called Answer: B - a cellular response, the final step after reception and transduction. (57%) Study design: outside the current course (the stimulus-response model of cell signalling).

Q15

Neutrophils Answer: B - migrate to the site of infection and engulf pathogens (phagocytosis). Mast cells release histamine. (69%)

Q16

A chemical barrier that stops pathogens entering the body. Answer: C - lysozymes, enzymes in tears and saliva that break down bacterial cell walls. Histamine, interferons and complement act after entry. (79%)

Q17

A role of the lymphatic system in an immune response. Answer: C - transporting antigen-presenting cells, including dendritic cells, to the lymph nodes. (70%)

Q18

Multiple sclerosis (MS) is tracked by NfL protein fragments released when nerves are damaged. Patients with progressive MS would have Answer: A - more T helper lymphocytes, since MS is an autoimmune disease in which T cells attack the myelin around nerves. More damage releases more NfL, not less (B). (59%) Study design: partly outside; the current design mentions autoimmune disease only as a target for monoclonal antibody treatment, and MS is no longer a named example.

Q19

The main advantage of conjugated monoclonal antibodies (with an anti-cancer drug attached). Answer: B - they reduce the drug's effects on non-cancerous cells, because the antibody delivers it only to cells with the target antigen. (63%)

Q20

Plant species 2 has four copies of each chromosome where species 1 has two. Answer: B - polyploidy: whole extra sets of chromosomes. Aneuploidy and trisomy involve single extra chromosomes; triploidy would be three sets. (77%) Study design: outside the current Units 3 and 4; chromosome number and abnormalities are Unit 2.

Q21

The order in which organisms first appear in the fossil record. Answer: D - green algae, fish, mammals, flowering land plants. The report notes mammals appeared about 210 million years ago and flowering plants about 130 million years ago. (44%; 48% chose B.) Study design: outside the current course, which no longer lists the timeline of major life forms.

Q22

From the graph, the minimum species loss for a major mass extinction. Answer: C - 74%: the smallest major event plotted is about 75%, while other events reach 50% at most. (88%) Study design: the graph reading is still good practice, but mass extinctions are no longer in the course.

Q23

Mass extinction events Answer: A - allow new species to evolve to occupy the niches left empty. Not every surviving species recovers (B). (68%; 21% chose B.) Study design: outside the current course.

Q24

Why are sea star fossils rare? Answer: D - their soft bodies decompose quickly, before they can be fossilised. (72%) Study design: outside the current course (the conditions and steps of fossilisation are no longer listed).

Q25

Two photographs of arthropods in amber (each with a 100 μm scale bar) and a drawing with a 50 μm bar. Answer: C - the fly in Image A is longer than the 100 μm scale bar. Amber preserves the body itself, so these are not trace fossils, the drawn arthropod in Image C is about twice the 50 μm bar in height (more than 90 μm), and resin engulfs an organism directly rather than after burial in sediment. (64%) Study design: using a scale bar is still a key skill; types of fossils are no longer listed.

Q26

Denisovan fossils are known only from Siberia and Tibet, but Denisovan DNA occurs in present-day populations on several continents, including Australia. Answer: D - the Denisovans, or their descendants, were more widespread than the fossil evidence suggests. (88%)

Q27

The BMP4 gene in African cichlid fish is responsible for Answer: B - the amount of bone laid down in the jaws of embryos. (77%) Study design: outside the current course (BMP4 and master genes are not in it).

Q28

The method and enzyme used to copy DNA from a crime scene. Answer: A - the polymerase chain reaction, using Taq polymerase. (72%)

Q29

Classes switched between antimicrobial gel and soap and water; no difference in illness was found. The purpose was most likely to Answer: D - decide whether buying gel sanitiser for students was worthwhile. (79%)

Q30

A requirement of Step 1 (isolating mitochondrial DNA from echidna quills) or Step 2 (amplifying it). Answer: D - PCR usually needs two primers, one for each strand. The thermocycler cycles through several temperatures rather than holding 37 °C, and mtDNA comes from mitochondria, not nuclei. (46%; 31% chose C.)

Q31

For the experiment to be valid, the student would need to Answer: A - make sure the amplified DNA is from the correct locus of the mitochondrial genome, so it measures what it claims to. Another observer getting the same results (B) is reproducibility. (43%; 28% chose B.)

Q32

What could affect a conclusion from the gel? Answer: C - uncertainty in the results for sample D: its two lanes (4 and 8) show bands at different positions, while the other samples' lanes agree. (51%; 35% chose A.)

Q33

Twenty-base mtDNA sequences from four echidnas; one is from New Guinea. Answer: B - sample B differs from the others at four positions, while A, C and D differ from each other at only one position each. (90%)

Q34

Modelled cervical cancer incidence with vaccination and screening. Answer: B - vaccination is the largest factor in the long-term fall: it alone takes incidence from about 20 to under 3 per 100 000, and screening adds only a little more. (75%)

Q35

Frog pregnancy tests: what could cause a false positive? Answer: D - a female frog laying eggs naturally around the time of the injection. (88%)

Q36

In a doxycycline trial, the group receiving standard care only was the Answer: A - control group. (93%)

Q37

An advantage of testing sewage for viral particles. Answer: B - infection in a community can be identified quickly, even before people are tested. (86%)

Q38

The MERS virus makes no neuraminidase, so the drug Relenza would Answer: C - be unlikely to help MERS patients recover, since Relenza works by blocking neuraminidase. (72%) Study design: outside the current course (rational drug design and Relenza have been removed).

Q39

An appropriate treatment for the control seedlings. Answer: A - spraying with the buffered detergent solution (no hormone) at the same intervals, so any effect can be put down to the hormones rather than the detergent. (15%; 49% chose B, distilled water.)

Q40

A conclusion from the BA and TIBA bar graph. Answer: D - high TIBA (50 mg/L) gives shorter stems than the control but many more lateral shoots (about 80 against about 10). Low BA shortened stems (A is wrong), high BA stems were under a quarter of the control length (B), and BA did not affect lateral shoots, so TIBA could not reduce that effect (C). (47%; 35% chose A.)

Section B

Question 1 (5 marks)

A diagram shows part of a plasma membrane. Oxygen is given as an example of a small hydrophobic substance that crosses the phospholipid bilayer.

a. Name one small hydrophilic and one small hydrophobic substance that can move through the bilayer, with an example of a situation when each moves. (2 marks)

b. Using one example, explain how very large substances or particles move into or out of a cell without channel or carrier proteins. Choose a direction. (3 marks)

Study design: part a is now Unit 1 content and is not examined in Units 3 and 4. In part b, exocytosis (out of a cell) is still part of the protein secretory pathway in Unit 3; endocytosis is no longer in the course.

Show worked solution

a. [2 marks].

Substance	Type	Situation
Water	hydrophilic	Water diffuses into a leaf cell, where it is used in photosynthesis.
Carbon dioxide	hydrophobic	Carbon dioxide diffuses out of a cell carrying out aerobic respiration.

Also accepted: ethanol (hydrophilic) diffusing out of a fermenting yeast cell; carbon dioxide diffusing into a photosynthesising cell; a steroid hormone (hydrophobic) diffusing into a target cell to bind a receptor inside.

b. [3 marks]. Out of a cell, by exocytosis. For example, a plasma cell secreting antibodies: the antibodies are packaged by the Golgi apparatus into vesicles, which move to the plasma membrane and fuse with it, releasing the antibodies outside the cell. This bulk transport uses energy (ATP).

(Into a cell, by endocytosis or phagocytosis: for example, a macrophage engulfing a bacterium. The plasma membrane extends around the particle and pinches off to form a vesicle inside the cell.)

From the report. Part a needed a specific situation, not a general statement: "water moves by osmosis" with no example was a common wrong answer. Only 17% scored both marks.

Question 2 (9 marks)

A yeast cell (Saccharomyces cerevisiae) contains about 6000 different proteins.

a

Give one type of protein that may change in concentration when conditions change, and explain why the change is necessary. (3 marks)

b

State the immediate end product of transcription, RNA processing, translation and the breakdown of protein. (2 marks)

c

A table gives four yeast genes: HSP82 (a proton pump), ADR1 (a transcriptional factor), GCN4 (controls RNA polymerase) and AQY1 (a water channel). Identify the regulatory genes and justify your answer. (2 marks)

d

The PFY1 gene is shown as a promoter P, then sections A, B (an intron) and C. Name sections A and C and outline their function. (2 marks)

Show worked solution

a. [3 marks]. An enzyme. If the concentration of a substrate changes (for example, a new sugar becomes available in the yeast's surroundings), the cell makes more of the enzyme that breaks it down, so it can use the new food source. When the substrate runs out it makes less, which saves ATP and amino acids for other proteins.

(Also accepted: structural proteins to repair organelles damaged by a temperature change; more or less repressor protein when substrate availability changes; more or fewer transcription factors when the need for protein synthesis changes.)

b. [2 marks]

Transcription: pre-mRNA. RNA processing: mature mRNA. Translation: a polypeptide. Breakdown of protein: amino acids.

c. [2 marks]

ADR1* and *GCN4. Their products are regulatory proteins that control whether other genes are expressed (switching transcription on or off), whereas HSP82 and AQY1 are structural genes whose products do a job in the cell.

d. [2 marks]

A and C are exons. They are the coding sections: they are kept in the mature mRNA after the intron is removed and are translated into the amino acid sequence of the protein.

From the report. In c, many identified the genes but just repeated the table as the justification, which was not accepted. Part d was poorly answered: 66% scored zero.

Question 3 (6 marks)

Scientists measured the metabolic activity of mammalian cells, kept at 37 °C and pH 7.4 in a glucose solution, by measuring glucose uptake over 30 minutes.

a. Explain why glucose uptake could be used to measure the cells' metabolic activity. (2 marks)

b. The experiment was repeated in low-oxygen conditions. Would glucose uptake be higher, lower or the same? Justify your response. (4 marks)

Show worked solution

a. [2 marks]. Glucose is the main input to cellular respiration, which breaks it down to make ATP. ATP supplies the energy for the cell's metabolic activities, so the more active the cells are, the faster they use glucose and take up more.

b. [4 marks]. Higher.

• With little oxygen, the electron transport chain (which needs oxygen as the final electron acceptor) and so the Krebs cycle slow down, and less aerobic respiration takes place.
• The cells rely more on glycolysis followed by anaerobic fermentation (to lactic acid in mammals).
• Anaerobic respiration yields only 2 ATP per glucose, far fewer than aerobic respiration.
• To make enough ATP, the cells must break down more glucose, so they take up glucose faster.

The report accepted any of the three predictions with a matching justification: lower (anaerobic respiration produces lactic acid, which can damage the cells and slow their metabolism) or the same (glycolysis does not need oxygen, so glucose is still broken down, just with less ATP made).

From the report. Answers had to name the correct type or stage of cellular respiration. Only 10% scored all four marks.

Question 4 (10 marks)

A live, attenuated varicella zoster (chickenpox) vaccine is given to children at 18 months under the National Immunisation Program.

a

Summarise the immune response in the child that results in long-term protection from chickenpox. (5 marks)

b

Unvaccinated people over 14 are advised to have two doses at least four weeks apart. What is the benefit of two doses instead of one? Justify your response. (2 marks)

c

Outline similarities and/or differences between active and passive ways of acquiring immunity. (3 marks)

Show worked solution

a. [5 marks].

1. Antigen-presenting cells (macrophages or dendritic cells) engulf the attenuated virus and present its antigens on their surface (on MHC class II markers).
2. They travel through the lymphatic system to a lymph node, where they present the antigen to a helper T cell with a complementary receptor.
3. The activated helper T cell releases cytokines that activate the B cell whose receptors match the same antigen (clonal selection).
4. The B cell divides (clonal expansion) and differentiates into plasma cells, which secrete antibodies specific to the varicella zoster antigen.
5. Some become long-lived memory B cells (and memory T cells). If the child meets the real virus later, these memory cells produce a fast, large secondary response that stops chickenpox developing.

b. [2 marks]. The first dose triggers a primary response that makes only a small number of memory cells. The second dose triggers a secondary response that makes many more memory cells and more antibodies, so any later infection is met by a faster, larger antibody response.

c. [3 marks].

• Similarity: both provide antibodies that protect the person against a specific pathogen.
• Active immunity: the person's own immune system makes antibodies and memory cells after meeting the antigen (through infection or vaccination). It takes time to develop but is long-lasting.
• Passive immunity: the person receives ready-made antibodies (across the placenta, in breast milk or by injection). It works immediately but is short-term, because no memory cells are made.

From the report. Part a was very well answered (30% scored all five marks). In c, some students confused active and passive with natural and artificial immunity.

Question 5 (6 marks)

a

Outline what is meant by cell signalling. (2 marks)

b

Two diagrams show short-distance signalling: in Mode X, a signalling cell releases molecules that reach a nearby target cell; in Mode Y, an action potential in a signalling cell releases molecules onto a target cell. Name an example of the signalling molecule and target cell for each mode. (2 marks)

c

When a hive is disturbed, guard honey bees release a banana-smelling substance from glands in the abdomen while fanning their wings. Name the type of signalling molecule and give a reason for the fanning. (2 marks)

Study design: the whole of Question 5 is outside the current course. Cell signalling (signalling molecules, reception and transduction) is no longer in Units 3 and 4.

Show worked solution

a. [2 marks]. Cell signalling is the transfer of information to (or within) a cell by a signalling molecule. The target cell detects the molecule with a specific receptor (reception), and this leads to a response in that cell.

b. [2 marks].

Mode	Signalling molecule	Target cell
X	a cytokine (such as an interleukin from a helper T cell)	a B cell
Y	a neurotransmitter (such as acetylcholine)	a neuron or a muscle cell

Other accepted pairs for X included histamine and a nearby phagocyte, and interferon and a cell near a virus-infected cell.

c. [2 marks]. A pheromone (an alarm pheromone). Fanning spreads the pheromone through the air, so other bees detect it more quickly and come to defend the hive.

From the report. In b, a common error was naming a dendritic cell as the neurotransmitter's target (confusing it with the dendrites of a neuron). Part c was well answered (67% full marks).

Question 6 (6 marks)

Octopus populations live on the east and west coasts of Australia and on the north coast of New Zealand. Molecular and morphological data show all share a distant common ancestor, and the east and west Australian populations are now distinct species.

a. Describe the process that may have led to the two distinct species from the common ancestor. (4 marks)

b. No significant genetic differences were found between the east Australian and New Zealand octopus. Young octopi can be carried long distances by currents, and adults have been seen on floating wood. Give two reasons for the lack of genetic diversity between these populations. (2 marks)

Show worked solution

a. [4 marks] (allopatric speciation).

• The ancestral population was split into east and west populations by a geographic barrier (the Australian continent and the distance around it), so there was no gene flow between them.
• Different mutations arose in each population, adding different new alleles.
• The two coasts had different selection pressures (such as water temperature, food and predators), so natural selection changed allele frequencies differently in each population.
• Over many generations the changes accumulated until the gene pools were so different that the two populations became separate species.

b. [2 marks].

• Gene flow still occurs: young octopi carried by currents, and adults rafting on wood, move alleles between Australia and New Zealand, so the gene pools stay similar.
• The populations share a relatively recent common ancestor, and they may face similar selection pressures, so there has been little time or reason for differences to build up.

From the report. In a, some students explained why the populations count as separate species instead of describing the process; stating that they cannot produce fertile offspring earned no mark. In b, some answers wrongly discussed variation within a population.

Question 7 (10 marks)

Lemurs are primates found in the wild only on Madagascar.

a

Identify two structural features common to most primates that distinguish them from other mammals. (2 marks)

b

The biologists used mitochondrial DNA (mtDNA) as a molecular clock. Give two advantages of using mtDNA this way. (2 marks)

c

Describe the phylogenetic relationships between Lemur catta, Eulemur coronatus and Eulemur flavifrons shown on the tree (time axis 23.0 million years ago to the present). (2 marks)

d

Outline the main steps of DNA hybridisation and explain how it supports the relationships in the tree. (4 marks)

Study design: a and c are still in the course. In b, the current design no longer names the molecular clock (it uses mtDNA to trace human migration), though comparing DNA sequences is still examined. Part d is outside the course: DNA hybridisation has been removed.

Show worked solution

a. [2 marks]

Any two: opposable thumbs (or big toes); grasping (prehensile) hands; forward-facing eyes; nails instead of claws; a large brain relative to body size.

b. [2 marks]

Any two: mtDNA mutates at a relatively fast and predictable rate, so differences build up at a steady rate that can be converted to time; it is inherited from the mother only, so it does not recombine and every difference comes from mutation; there are many copies per cell, so it is easier to extract, even from small or old samples.

c. [2 marks]

All three species share a common ancestor about 20 million years ago. E. coronatus and E. flavifrons share a more recent common ancestor, about 8 to 9 million years ago, so they are more closely related to each other than either is to L. catta.

d. [4 marks]

• Heat DNA from each of two lemur species (to about 94 °C) to separate it into single strands.
• Mix the single strands from the two species and let them cool, so they re-form double strands as hybrid DNA wherever the bases are complementary.
• Reheat the hybrid DNA and measure the temperature at which it separates.
• The more complementary base pairs, the more hydrogen bonds, so the higher the separation temperature, the more closely related the species. Hybrid DNA of E. coronatus and E. flavifrons would separate at a higher temperature than hybrid DNA of either with L. catta, matching the tree.

From the report. In a, features that are not structural, or that belong only to hominins or to all mammals, were common errors; "stereoscopic vision" and "five digits" were not accepted. In c, some students misread the scale. In d, a common wrong answer described PCR.

Question 8 (6 marks)

Cave drawings are argued to show cognitive changes unique to Homo sapiens that allowed cultural evolution.

a

Describe how ancient cave drawings give evidence of cultural evolution in H. sapiens. (2 marks)

b

Drawings on stalagmites in the Ardales Cave in Spain are covered by a transparent mineral layer. Uranium-thorium dating shows some of this layer is 65 000 years old. What can be concluded about the age of the drawings? Justify your response. (2 marks)

c

A table gives H. sapiens arriving in Europe 45 000 years ago and H. neanderthalensis living there from 130 000 to 30 000 years ago. How does the Ardales discovery challenge the view that cave drawing first arose in H. sapiens? (2 marks)

Study design: part a is outside the current course, which tracks hominin trends in brain size and limb structure rather than cultural evolution. Parts b and c (dating and changing interpretations of the fossil evidence) still apply.

Show worked solution

a. [2 marks]

Any two: the drawings show symbolic representation and complex thought; they are a way of passing information between people and across generations by learning rather than genes; drawings become more complex over time, showing changes in cultural practices.

b. [2 marks]

The drawings are older than 65 000 years. The mineral layer formed on top of the drawings, so the drawings must already have been there when it began to form.

c. [2 marks]

H. sapiens only arrived in Europe about 45 000 years ago, but the drawings are more than 65 000 years old, so H. sapiens could not have made them. H. neanderthalensis was living in Europe at that time, so Neanderthals probably made the drawings, meaning cave art did not first arise in H. sapiens.

(Also accepted: H. sapiens may have been in Europe earlier than the fossils found so far show, or the dating may be unreliable.)

From the report. In b, some students saw that the layer was on top of the drawings but then said the drawings must be younger than it.

Question 9 (6 marks)

A table describes three genetically modified (GM) crops grown in Australia: GM cotton (several bacterial genes inserted), GM canola (genes from two bacterial species inserted, giving herbicide tolerance) and GM safflower (some of its own genes silenced).

a

Select one crop and justify whether or not it could be described as transgenic. (1 mark)

b

GM canola seeds spilt along roadsides grow into plants resistant to glyphosate, the herbicide normally used there. Suggest one practical solution. (1 mark)

c

A new GM canola with extra omega-3 fatty acids is being developed as a sustainable alternative to fish. Discuss one social and one biological implication of using it. (4 marks)

Show worked solution

a. [1 mark]

GM cotton is transgenic, because it contains genes from another species (bacteria). (Or: GM canola is transgenic for the same reason; GM safflower is not transgenic, because no genes from another species were added.)

b. [1 mark]

Spray the plants with a different herbicide that the GM canola is not resistant to (or remove them by mowing, hand-pulling or controlled burning).

c. [4 marks]

• Social: GM canola oil could be cheaper and easier to get than fish, so more people get enough omega-3. Better nutrition improves health and reduces demand on the health system. (Or: some consumers may refuse GM food, so demand falls and farmers earn less.)
• Biological: pollen from the GM canola could cross-pollinate non-GM canola crops, spreading the modified genes into them and changing their genomes. (Or: less fishing for omega-3 could let fish populations recover; or a GM crop with little genetic variation could be wiped out if a new selection pressure such as a disease appears.)

From the report. The accepted social implications were effects on people (farm income, access to nutrition, demand on the health system); the biological ones were effects on genomes, genetic variation or fish populations. Only 14% scored all four marks.

Question 10 (7 marks)

An article describes making human insulin in Escherichia coli in the 1970s. The researchers knew insulin's amino acid sequence and structure, and knew E. coli plasmids contain an EcoRI recognition site (GAATTC). Safety rules managed the risks, and the insulin was tested in cell cultures, animals and then clinical trials. By 2020, insulin genes could be made in hours by machine, then cloned and expressed in about a week.

a

What is meant by recombinant DNA in this article? (1 mark)

b

How would the researchers have used the amino acid sequence to engineer a human insulin gene, and how might it differ from the natural gene in human DNA? (2 marks)

c

Outline the steps needed to clone and express the human insulin gene in bacteria in 2020. (3 marks)

d

What ethical response would be expected if clinical trials showed recombinant insulin was much more effective than insulin from cows or pigs? (1 mark)

Show worked solution

a. [1 mark]

DNA made by joining DNA from two different sources: here, the human insulin gene inserted into bacterial (plasmid) DNA.

b. [2 marks]

The researchers would work backwards from the amino acid sequence, using the genetic code to choose a codon for each amino acid, and chemically make that DNA sequence. The engineered gene would differ from the natural one because it would have no introns (bacteria cannot remove introns), and because the code is degenerate, the codons chosen might not be the ones in the natural gene.

c. [3 marks]

• Synthesise the insulin gene from its known sequence (or isolate it from human DNA).
• Cut the plasmid (and the ends of the gene) with the same restriction enzyme (such as EcoRI) to give complementary sticky ends, and join the gene into the plasmid with DNA ligase, making a recombinant plasmid.
• Transform E. coli with the recombinant plasmid (for example, by heat shock), then grow the bacteria in culture, where they express the gene and make insulin to be collected and purified.

d. [1 mark]. Stop the trial and offer the more effective recombinant insulin to the participants who were getting animal insulin.

(Also accepted: stop producing insulin from animals to protect their welfare; release the results to the public.)

From the report. Question 10 was poorly answered. In a, only 23% scored the mark: the idea of recombination was not well understood in this context. In b, 80% scored zero.

Question 11 (9 marks)

Two students tested antibiotic resistance in E. coli. On Day 0, one plate had 0 μg/mL ampicillin (control) and another 1 μg/mL (experimental). Each day, bacteria from each plate were moved to a new plate with more ampicillin (2, then 4, then 6 μg/mL). Plates were incubated at 37 °C for 24 hours, then refrigerated and photographed at the end. Figure 1 shows their prediction: the experimental group keeps a few colonies to Day 3, while the control dwindles to none. Figure 2 shows the actual growth.

a

Identify any two controlled variables. (2 marks)

b

Write a suitable hypothesis for this experiment. (2 marks)

c.i

Analyse the results in Figure 2. (3 marks)

c.ii

Explain whether the results in Figure 2 support the predictions in Figure 1. (2 marks)

Show worked solution

a. [2 marks]

Any two: the volume of culture spread (1 mL); the type of nutrient agar; the size of the plates; the incubation temperature (37 °C); the incubation time (24 hours); the batch of ampicillin.

b. [2 marks]

If E. coli is first exposed to a low concentration of ampicillin (1 μg/mL), then more E. coli will grow at increasingly higher ampicillin concentrations than E. coli that was not exposed first.

c.i. [3 marks]

Any three:

• On Day 0, more colonies grew on 0 μg/mL (control) than on 1 μg/mL.
• On Day 1, 2 μg/mL did not reduce growth in either group: both plates were covered.
• On Days 2 and 3, 4 and 6 μg/mL greatly reduced growth in both groups, so the higher concentrations killed many bacteria.
• By Day 3, the experimental plate had more colonies than the control, so more ampicillin-resistant bacteria were present after the early 1 μg/mL exposure.
• All plates grew more than the students predicted.

c.ii. [2 marks]. Only partly. In support, colonies decreased over time in both groups, and on Day 3 the experimental group had more colonies than the control. But there was much more growth than predicted on every plate, and the control still grew at 6 μg/mL where no growth was predicted, so resistant bacteria survived whether or not they had been exposed first.

(The report accepted "supports" or "does not support" with two matching points, or one point for each side.)

From the report. In b, many students did not identify the independent variable (prior exposure to a low concentration) and wrote in general terms about ampicillin's effect on E. coli.

Advice from the 2021 report

• Use the reading time to look through Section B, pick out what each question needs and start planning. The 5-mark summary in Question 4a rewarded clear, logical steps.
• Answer the question asked. Question 6a asked how the species formed, not why they count as separate species; Question 1a asked for a situation, not a general statement.
• Add biology, don't repeat the stem. Copying the table was not a justification in Question 2c.
• Read scales and data carefully. Question 7c needed the time axis, and Questions 39 and 40 needed the control and the graph read accurately.
• Write in the spaces provided. Papers are scanned and marked online; if you must continue elsewhere, make it easy to find.