VCE Biology 2022

Every question from the 2022 VCE Biology examination, the first exam on the study design introduced in 2022. 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 2022 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 2022 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.
• Study design: this paper was set on the study design first used in 2022, which is still current (the latest version, 1.2 from April 2026, changed only the legislation and codes it refers to). Every question is on content you can still be examined on.

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. The report noted that longer responses (three to six marks) made up a larger proportion of the paper, such as the 6-mark speciation answer in Question 8a, and that several questions list the points to cover.

Section A: Multiple choice

Q1

Myoglobin is a single polypeptide chain folded around one heme group. The structure shown is its Answer: C - tertiary structure: the overall 3D folding of one chain. Quaternary structure needs two or more chains. (61%; 19% chose D.)

Q2

A single-stranded molecule made of alternating exons and introns (50, 150, 70, 80, 120 and 40 bases) is most likely Answer: B - pre-mRNA: single-stranded and still containing introns. Mature mRNA has had its introns removed, and DNA is double-stranded. (86%)

Q3

Where in a cell would this molecule be found? Answer: A - the nucleus, where pre-mRNA is made and processed. (81%)

Q4

The length of the part that is translated. Answer: A - 240 bases: the exons only, 50 + 70 + 120. (77%)

Q5

A diagram of translation with molecules X, Y and Z and structure W. Answer: B - molecule X is a tRNA carrying an amino acid that will be added to the growing polypeptide. W is the ribosome, Y is mRNA read 5′ to 3′, and Z is the polypeptide, which has no codons. (88%)

Q6

An enzyme cuts a DNA sequence with a staggered cut. Answer: A - it is an endonuclease (a restriction enzyme): it cuts within the DNA, leaving sticky ends, not blunt ends. Ligase joins fragments. (79%)

Q7

An example of a GM trait that benefits people. Answer: C - increased resistance to toxins produced by fungi. (86%)

Q8

A graph of the percentage of US acres planted with five GM crops, 1996 to 2020. Answer: C - in 2013, Bt cotton and Bt corn were both at about 75%. (89%)

Q9

Where is the photosynthesis stage in the diagram (photosystems II and I with electron flow) located? Answer: B - the thylakoid membrane: it is the light-dependent stage. (75%)

Q10

What do V, W and Z represent? Answer: C - oxygen, ADP and NADPH. Water (U) is split to release H⁺ and oxygen (V); ADP (W) becomes ATP (X); NADP⁺ (Y) becomes NADPH (Z). (52%; 26% chose B.)

Q11

Light intensity against rate of photosynthesis, levelling off at a maximum X. What would increase X? Answer: B - a leaf with more chloroplasts. At the plateau light is no longer limiting; more chloroplasts means more chlorophyll and enzymes. A filter, a more distant light and a lower temperature would all lower the rate. (74%)

Q12

Qualitative data for choosing when to grow a new plant variety in Brisbane. Answer: C - a list of plant preferences from people living in Brisbane. The other options are numbers (quantitative). (64%)

Q13

A diagram of cellular respiration in an animal cell: which pathway is in Region 3 (the cytosol)? Answer: D - glycolysis. (79%)

Q14

Compounds R, T and V. Answer: C - pyruvate, carbon dioxide and water: glucose becomes pyruvate (R), the Krebs cycle releases carbon dioxide (T), and the electron transport chain (Region 1) produces water (V). (72%)

Q15

The carrier molecules taking electrons from the Krebs cycle to the electron transport chain are Answer: C - NADH and FADH₂. (70%)

Q16

The last electron acceptor in cellular respiration is Answer: B - oxygen, which accepts electrons at the end of the electron transport chain and forms water. (46%; 27% chose C.)

Q17

Which statement about enzymes is incorrect? Answer: C - enzymes do not increase the average energy of reactant molecules; they lower the activation energy needed. (54%; 23% chose D, a correct statement.)

Q18

Skin cells at different temperatures and oxygen levels: which tube has the most lactic acid after 24 hours? Answer: C - tube 3, at body temperature (37 °C) with low oxygen (5%), so the cells rely most on anaerobic fermentation. At 75 °C (tube 4) the enzymes denature. (57%; 20% chose B.)

Q19

A correct statement about the enzymes of photosynthesis and cellular respiration. Answer: B - a pH above the optimum for respiration may denature its enzymes. The pathways use different enzymes, enzyme concentration does affect rate, and low temperature slows enzymes without denaturing them. (78%)

Q20

Beneficial microorganisms inside a plant help it resist pathogens by Answer: A - making toxins that kill pathogenic microorganisms. Plants do not make antibodies or have a third line of defence. (60%)

Q21

Which labelled cell in the inflammatory response diagram is correctly named, with its role? Answer: D - cell 4 is a neutrophil, acting as a phagocyte. The published PDF leaves out this diagram for copyright reasons, so check the report's key rather than the image. (61%; 25% chose C.)

Q22

Immune cells tell self from non-self cells Answer: B - by the different antigens on non-self cells. (90%)

Q23

A measles infographic gives an average incubation period of 14 days (range 7 to 21 days). The minimum quarantine for an unvaccinated traveller returning from a country where measles is endemic is Answer: B - 21 days: the top of the range, so even the longest incubation would show up. (28%; 42% chose D and 29% chose C.)

Q24

A positive test for IgM antibodies tells a health official the traveller Answer: D - has been infected with the measles virus (IgM is made early in an infection). (82%)

Q25

The R0 for measles, from the infographic (one person can infect 12 to 18 susceptible people). Answer: D - 12 to 18. (90%)

Q26

If 200 fully vaccinated people over 12 months old are exposed, how many would be infected? Answer: C - 6: two doses of MMR are 97% effective, so about 3% of 200 could be infected. (65%; 22% chose A.)

Q27

Which cells respond first to the MMR vaccine? Answer: D - antigen-presenting cells, which take up the vaccine antigens before any T or B cells are activated. (68%)

Q28

A practical way to stop Hendra virus spreading from fruit bats to horses. Answer: B - cover the horses' feed and water containers, so bat body fluids cannot contaminate them. (71%)

Q29

A few koalas were introduced to French Island, and the island population now has low genetic diversity. Answer: B - the founder effect: a small group started a new population, carrying only part of the original gene pool. (64%; 29% chose A.)

Q30

The French Island koala population Answer: C - may be vulnerable to future changes in selection pressures, because low genetic diversity means fewer individuals may carry alleles that would help them survive. (90%)

Q31

Communicating all results, favourable or not, reflects which bioethical principle? Answer: C - integrity. (71%)

Q32

Which Koolasuchus cleelandi findings best explain why its fossils formed? Answer: B - findings 2 and 4: hard bony parts, and living in rivers carrying sediment that could bury it quickly. (85%)

Q33

How was the absolute age of the 125-million-year-old fossils most likely found? Answer: D - radioisotope dating of the rock around the fossils. Carbon-14 only works for about the last 50 000 years, and index fossils give a relative age. (52%; 27% chose B.)

Q34

Which barriers could give sympatric speciation? Answer: B - either barrier 2 (a new mating ritual) or barrier 3 (breeding several weeks later). Both isolate populations in the same area; a river or mountain (1 and 4) is geographic, so allopatric. (68%; 21% chose A.)

Q35

A trait selected for in breeding modern corn from teosinte. Answer: C - more seeds in the ear. (79%)

Q36

The arrangement of forelimb bones in a horse, a crocodile and a human is Answer: D - homologous: the same basic bones, modified for different uses. (74%)

Q37

A reasonable conclusion from these forelimbs. Answer: D - humans, crocodiles and horses share a common ancestor. (63%; 28% chose C.)

Q38

Two vertebrate phylogenies drawn with branches in different orders. Answer: C - Phylogeny II shows the same relationships as Phylogeny I. Rotating branches around a node does not change which groups share ancestors, and no group is more advanced. (71%)

Q39

A correct conclusion from Phylogeny I. Answer: A - frogs arose more recently than coelacanths, since the frog lineage branches off after the coelacanth split. Q is the ancestor of jawed vertebrates only, and R is the ancestor of ray-finned fishes and the lobe-finned groups, not of sharks. (65%; 21% chose C.)

Q40

Why can mitochondrial DNA reveal the migration of Aboriginal Australian populations? Answer: C - it is passed down the maternal line unchanged except by mutation, so lineages can be traced back. (71%)

Section B

Question 1 (7 marks)

Tryptophan (trp) is the least abundant amino acid in eukaryotic cells. A genetic code table is given.

a

In terms of the genetic code, what feature do trp and met share that the other 18 amino acids do not? (1 mark)

b

The DNA template strand reads TAC TTC GGC TCA GTA. Write the next triplet so that trp is the next amino acid, then the mRNA strand and the amino acid sequence. (3 marks)

c

E. coli makes its own tryptophan, controlled by the trp operon (genes trpE, trpD, trpC, trpB and trpA). Annotate a diagram to show how the genes are repressed when tryptophan levels are high. (3 marks)

Show worked solution

a. [1 mark]. Each is coded by only one codon (UGG for trp, AUG for met), so their code is not degenerate.

b. [3 marks]. The trp codon is UGG, so the template triplet is ACC.

DNA template	TAC	TTC	GGC	TCA	GTA	ACC
mRNA	AUG	AAG	CCG	AGU	CAU	UGG
Amino acid	met (start)	lys	pro	ser	his	trp

c. [3 marks]. The annotations should show:

• the promoter, where RNA polymerase binds, and the operator, between the promoter and trpE;
• when tryptophan is plentiful, tryptophan binds the repressor protein, changing its shape so that it becomes active;
• the active repressor binds the operator, blocking RNA polymerase, so the trp genes are not transcribed and the enzymes that make tryptophan are not produced.

From the report. Part b allowed consequential marks: a student who made a transcription error but then translated their own mRNA correctly could still earn the translation mark. Most students did well on b (58% full marks) and c (57%).

Question 2 (7 marks)

Reverse transcription PCR (RT-PCR) detects a virus by first converting its RNA to DNA. The ingredients are viral RNA, reverse transcriptase, primers, fluorescent probes, DNA polymerase and DNA building blocks.

a.i

Explain why RNA must be converted to DNA in RT-PCR. (1 mark)

a.ii

What is the purpose of DNA amplification in the PCR? (1 mark)

b

Explain why each stage of the PCR needs a different temperature. (3 marks)

c

The probes fluoresce when they attach to sequences found only in a particular virus. State two reasons RT-PCR cannot be used to identify all viruses. (2 marks)

Show worked solution

a.i. [1 mark]

PCR copies DNA: the DNA polymerase (Taq) needs a DNA template to replicate, so viral RNA must first be made into complementary DNA.

a.ii. [1 mark]

To make many identical copies of the viral DNA, so there is enough to detect and analyse.

b. [3 marks]

• Denaturation, about 94 °C: the high temperature breaks the hydrogen bonds between the bases, separating the DNA strands.
• Annealing, about 55 °C: cooling lets the primers bind to their complementary sequences on the single strands.
• Extension, about 72 °C: the optimal temperature for Taq polymerase, which adds nucleotides to the primers to build new strands.

c. [2 marks]. Any two: a complementary probe may not exist for that virus (for example, a new virus whose sequence is not yet known); not all viruses have an RNA genome; the virus's sequence may not be unique enough, so the probe could also bind other viruses' genetic material.

From the report. Answers to b were very clear. Part c was harder (13% full marks): the students who applied the idea of complementary sequences to this unfamiliar situation gave suitable reasons.

Question 3 (8 marks)

Diagrams compare carbon fixation in C3, C4 and CAM plants. Students found an unknown plant in the school grounds and compared it with the documented features of each type:

Feature	Unknown	C3	C4	CAM
ideal temperature	25 to 35 °C	15 to 25 °C	30 to 40 °C	over 40 °C
pathway to fix CO₂	C4 and Calvin	Calvin only	C4 and Calvin	C4 and Calvin
stomata open by day	yes	yes	yes	no
photorespiration	moderate to low	high	low	middle of day only
water loss by day	moderate	high	moderate	low
growth rate	moderate	moderate	fast	very slow

a. State whether the unknown plant is a C3, C4 or CAM plant, and justify your response. (5 marks)

b. Outline three additional steps that could extend the investigation and help identify the plant. (3 marks)

Show worked solution

a. [5 marks]. It is a C4 plant.

• It fixes carbon dioxide by the C4 pathway and the Calvin cycle, so it is not a C3 plant (which uses only the Calvin cycle).
• Its stomata are open during the day, like C4 plants and unlike CAM plants, which open them at night.
• Its ideal temperature (25 to 35 °C) overlaps the C4 range (30 to 40 °C) rather than the CAM range (over 40 °C).
• Its photorespiration is moderate to low, like a C4 plant and unlike a C3 plant (high).
• Its water loss is moderate, the same as a C4 plant (C3 is high, CAM is low).

b. [3 marks]. Any three steps that go beyond the data already collected:

• Look at a leaf cross-section under a microscope to check whether Rubisco and the Calvin cycle are in the bundle sheath cells, as in C4 plants.
• Compare its DNA (or its structure) with known C3, C4 and CAM plants.
• Test more samples of the plant to increase the sample size and reduce the effect of outliers.
• Change one controlled variable (such as temperature or carbon dioxide concentration) and measure the effect on the rate of photosynthesis, photorespiration or growth.

From the report. Part a was very well answered (70% full marks). In b, only 3% scored all three marks: many students suggested tests that had already been done and were in the table.

Question 4 (7 marks)

Chicken egg white protein (ovalbumin) is made by chicken cells during egg production.

a. Describe the role of organelles in exporting ovalbumin from a chicken's cells into an egg. (3 marks)

b. Demand for ovalbumin has led to intensive chicken farming, raising animal welfare concerns. Recombinant fungi can already make it, and using recombinant bacteria has been suggested. Discuss an ethical concern about intensive farming or recombinant protein production, propose a feasible solution, and state the ethical concept or approach addressed. (4 marks)

Show worked solution

a. [3 marks].

• Ribosomes on the rough endoplasmic reticulum make ovalbumin, and the rough ER transports it (in vesicles) to the Golgi apparatus.
• The Golgi apparatus modifies the protein and packages it into vesicles.
• The vesicles move to and fuse with the plasma membrane, releasing ovalbumin from the cell by exocytosis. (Mitochondria supply the ATP for this.)

b. [4 marks].

• Concern: intensive farming harms the welfare of the chickens.
• Cause and effect: chickens kept in cramped conditions cannot behave naturally and become stressed, which leads to more disease, injuries from pecking each other and early death.
• Ethical concept: non-maleficence (avoiding harm to the animals), or respect for their welfare.
• Solution: regulations requiring more space and areas where chickens can roam during the day, with a better diet; or making more ovalbumin by recombinant technology so fewer chickens are needed.

From the report. Answers needed all parts: the concern with its cause and effect, a named concept or approach, and a solution. Other accepted discussions included the concern that people may be uncomfortable eating recombinant protein, with transparent labelling (respect) as a solution.

Question 5 (8 marks)

Scientists used CRISPR-Cas9 with a designed single guide RNA (sgRNA) to edit a wheat gene linked to grain weight. After editing, the gene was not expressed, and the plants had heavier grains with more protein.

a

Explain how the scientists would have designed the sgRNA and outline its function in editing the gene. (3 marks)

b

The lack of expression could have come from inserting one nucleotide or deleting seven nucleotides. Explain how each edit could stop the gene being expressed. (3 marks)

c

Starch from wheat can be made into bioethanol by adding enzymes that break starch into monosaccharides and the facultative anaerobe Zymomonas mobilis. Explain how Z. mobilis is used in making bioethanol. (2 marks)

Show worked solution

a. [3 marks].

• Design: identify the target sequence in the grain-weight gene and make an sgRNA with a base sequence complementary to that target.
• The sgRNA binds to Cas9, forming a complex.
• The sgRNA then guides Cas9 to the target gene by complementary base pairing, where Cas9 cuts both DNA strands so the gene can be edited.

b. [3 marks]. Both edits cause a frameshift mutation, because neither 1 nor 7 is a multiple of 3. Every codon after the edit is read in a different reading frame, so the amino acid sequence changes, usually with an early stop codon. The result is a shortened, non-functional protein, so the gene is effectively not expressed.

c. [2 marks]. Under anaerobic conditions, Z. mobilis carries out anaerobic fermentation: it breaks down the glucose released from the starch into ethanol and carbon dioxide. The ethanol is then collected as bioethanol.

From the report. In a, students needed to know the difference between CRISPR's natural role in bacteria and its use for editing genes in other organisms. In b (17% full marks), students had to see that both edits cause a frameshift; some wrongly discussed block mutations.

Question 6 (7 marks)

Influenza in humans is caused by influenza A or B viruses.

a

Compare the roles of natural killer cells and cytotoxic T cells in the immune response to the influenza virus. (4 marks)

b

A diagram shows antigenic drift slightly changing the antigen structure of H3N2 influenza.

b.i

What change in the virus caused the change in antigen structure? (1 mark)

b.ii

Explain, in terms of antigenic drift, why flu vaccination is recommended every year rather than every few years. (2 marks)

Show worked solution

a. [4 marks].

• Similarity: both recognise body cells infected with the virus by their MHC class I markers and cause those cells to die (by apoptosis), which stops the virus replicating. Neither destroys the virus particles directly.
• Natural killer cells are part of the innate immune response: they are non-specific and act early, killing infected cells whose MHC I markers are abnormal.
• Cytotoxic T cells are part of the adaptive immune response: each is specific to one viral antigen presented on MHC I, and they act later, after activation and clonal expansion.

b.i. [1 mark]. A mutation in the viral gene coding for the antigen, changing its amino acid sequence.

b.ii. [2 marks]. Antigenic drift changes the surface antigens a little each season, so the antibodies and memory cells from earlier vaccination no longer recognise the new antigens. Each year's vaccine contains the current antigens, producing new memory cells specific to them for a fast, large response if the person is infected.

From the report. In a, many students wrongly said these cells destroy the pathogen itself rather than the infected cell. The comparison could use similarities, differences or both.

Question 7 (6 marks)

Some people have held chickenpox parties to infect children deliberately, despite warnings that chickenpox can be severe.

a

Name the type of immunity the infected children would develop. (1 mark)

b

One vaccine dose was 97% effective in the first year and 81% to 86% effective for years two to eight; most vaccinated children who caught chickenpox had a mild case. Why is the disease mild in most vaccinated people? (2 marks)

c

Chickenpox may have caused the fall in the Aboriginal Australian population after European arrival. A graph shows the estimated Aboriginal population falling sharply around 1789, rising slightly from 1790 to 1810, then falling again, while the colonist population grows. With reference to chickenpox, explain the large impact on Aboriginal Australians but not on colonists, and why the Aboriginal population increased from 1790 to 1810. (3 marks)

Show worked solution

a. [1 mark]

Natural active immunity.

b. [2 marks]

Vaccination produced memory cells. When the virus enters, these memory cells respond quickly, making large amounts of antibody rapidly, so the virus is neutralised before it can cause severe disease.

c. [3 marks]

• Most colonists had been exposed to chickenpox as children before arriving, so they had immunity (memory cells) and adults were protected.
• Chickenpox did not exist in Aboriginal communities before European arrival, so no one had immunity. Adults who caught it could become severely ill and die, causing the large fall.
• From 1790 to 1810, children who caught chickenpox had a mild illness and became immune. As these immune children grew up, the population increased.

From the report. The report named c as a question whose pointers each had to be addressed: why colonists were unaffected, why Aboriginal people were affected, and why numbers rose from 1790 to 1810. 36% scored all three marks.

Question 8 (9 marks)

The 13 species of Darwin's finches live only on the Galápagos Islands, about 1000 km west of South America. Their closest living relative is the dull-coloured grassquit on the mainland. The finches are similar in colour but differ in beak size and shape, habitat and diet.

a

Discuss how the different species arose from an ancestral mainland population. Name the type of speciation, identify the main selection pressure and explain how the islands enabled rapid speciation. (6 marks)

b

From a table of the species found on each island, identify the most widely distributed species. (1 mark)

c

Describe two types of evidence that would allow biologists to classify a species as extinct on a particular island. (2 marks)

Show worked solution

a. [6 marks]. Allopatric speciation.

• Finches from the ancestral mainland population reached the Galápagos, and groups settled on different islands. The ocean between the islands is a geographic barrier, so there was no gene flow between the island populations.
• Each population had variation in beak size and shape, and different mutations arose in each.
• The main selection pressure was food (diet): the type and size of seeds, insects and other food differed between islands and habitats.
• Finches whose beaks suited the local food survived and reproduced more, passing on their alleles, so allele frequencies changed differently on each island.
• The islands allowed rapid speciation because there were many isolated islands, each small founding population could change quickly by selection and genetic drift, and there were varied habitats and food sources with few competitors.
• Over many generations the gene pools diverged until the populations became separate species.

b. [1 mark]. Geospiza fuliginosa (found on 14 islands).

c. [2 marks]. Any two: no sightings of the species on the island despite thorough surveys over a long time; no physical evidence such as feathers, nests, eggs or DNA; its usual food source is no longer present.

From the report. Sympatric speciation was incorrect, though other points could still earn marks. Many students ended by explaining why the finches are different species, which was not asked: the question was how they arose. In c, many students gave reasons for extinction rather than evidence that it had happened.

Question 9 (6 marks)

A table gives the earliest and latest fossil dates and locations for nine hominin species: Australopithecus anamensis (4.2 to 3.8 million years ago, East Africa), A. afarensis (3.85 to 2.95 million, East Africa), A. africanus (3.3 to 2.1 million, South Africa), Homo habilis (2.4 to 1.4 million, East and South Africa), H. rudolfensis (1.9 to 1.8 million, East Africa), H. erectus (1.89 million to 110 000 years, Africa and Asia), H. floresiensis, H. neanderthalensis and H. sapiens.

a

Describe two trends in these fossils, from most ancient to most recent, in brain size and limb structure. (2 marks)

b.i

Apart from H. sapiens and H. neanderthalensis, identify two other occasions when two hominin species may have interbred, naming the species. (2 marks)

b.ii

Discuss the limitations of the evidence supporting your answer to b.i. (2 marks)

Show worked solution

a. [2 marks]. Brain size increased (a larger cranial capacity) from the australopithecines to Homo. Legs became longer relative to the arms (with more angled thigh bones), suited to efficient bipedal walking.

b.i. [2 marks]. Any two pairs that overlap in both time and place:

• A. anamensis and A. afarensis (East Africa, about 3.85 to 3.8 million years ago);
• A. africanus and H. habilis (South Africa, about 2.4 to 2.1 million years ago);
• H. habilis and H. rudolfensis (East Africa, about 1.9 to 1.8 million years ago);
• H. erectus and H. rudolfensis (East Africa, about 1.89 to 1.8 million years ago).

b.ii. [2 marks]. The table gives only broad regions, not exact sites, so the two species may never have met even if they lived in the same region at the same time. The dates come from a scarce and incomplete fossil record, and there is no DNA evidence for these species, so overlap in time and place does not show that they actually interbred.

From the report. In a, answers had to refer to brain size and limb structure; skull features such as the position of the foramen magnum were not accepted.

Question 10 (8 marks)

An article describes vaccines against cancer. Conventional vaccines, such as the HPV vaccine, contain a dead or inactivated pathogen or a protein from it, and take months to produce. An mRNA vaccine instead contains mRNA coding for an antigen found on a particular cancer cell; the person's cells make the antigen, which triggers an immune response. Changing the mRNA sequence could quickly give vaccines for other cancers, but cancer cells can evade immune cells and mutate, and one person's vaccine may not work for another.

a.i

According to the article, what do conventional vaccines contain, and how do they lead to a better immune response when a person meets the pathogen? (2 marks)

a.ii

Explain how mRNA vaccines differ from conventional vaccines. (2 marks)

b

Identify the advantages of developing mRNA vaccines for treating cancer, and outline an ethical concept to consider in developing them. (4 marks)

Show worked solution

a.i. [2 marks]

They contain a dead or inactivated version of the pathogen (or a protein from it). Its antigens trigger a primary response that produces memory cells, so a later encounter with the real pathogen brings a faster, larger antibody response.

a.ii. [2 marks]

mRNA vaccines contain mRNA, not the antigen. The person's own cells translate the mRNA to make the antigen, which is then released and triggers the immune response. Conventional vaccines deliver the antigen itself.

b. [4 marks]

• Advantages: mRNA vaccines can be produced quickly (conventional vaccines take months), and simply changing the nucleotide sequence could give vaccines for many different cancers. They produce antibodies specific to the cancer cells' antigen, not to healthy cells.
• Ethical concept, justice: a vaccine tailored to one person's cancer could be very expensive, so it should be made available fairly to everyone who needs it, not only those who can pay. (Also accepted: non-maleficence, ensuring the vaccine does not harm trial participants; respect, gaining informed consent; integrity, reporting all trial results honestly.)

From the report. In a.i, many students drew information from the article but did not apply it to the immune response. In b, students could recall the ethical concepts but struggled to apply them to developing vaccines.

Question 11 (7 marks)

A researcher tested Compound A on the Krebs cycle enzyme malate dehydrogenase (MDH), with malate as the substrate, at 37 °C for one minute:

Malate added (mmoles)	0	5	10	15	25
Rate without Compound A (mmol/min)	0.00	3.30	3.80	3.90	4.00
Rate with Compound A (mmol/min)	0.00	1.00	2.00	3.00	3.80

a

Plot a graph of the results, with labelled axes. (3 marks)

b

Compound A is a competitive inhibitor. Briefly explain how competitive inhibitors affect an enzyme's activity. (2 marks)

c

A repeat of the experiment gave curves of a similar shape but different results; it had been run at 20 °C. State the effect of this change and explain its cause. (2 marks)

Show worked solution

a. [3 marks]. Put the amount of malate on the x-axis with an even scale that includes 20, and the rate on the y-axis, with units on both axes. Plot both data sets and identify each with a key.

b. [2 marks]. A competitive inhibitor has a shape similar to the substrate (complementary to the active site), so it binds to the active site and competes with the substrate, stopping it binding there. Fewer enzyme-substrate complexes form, so the rate falls. (At high substrate amounts the substrate outcompetes the inhibitor, which is why the two curves nearly meet at 25 mmoles.)

c. [2 marks]. The rate of product formation would be lower at every amount of malate. At 20 °C, below the enzyme's optimum of about 37 °C, the enzyme and substrate molecules move more slowly, so there are fewer collisions between them and fewer enzyme-substrate complexes form each minute. (The enzyme is not denatured.)

From the report. In a, marks were lost for missing units, an uneven x-axis scale (some students left out 20) or no key.

Advice from the 2022 report

• Follow the pointers in the question. Longer responses (three to six marks) made up a larger proportion of this paper, and questions such as 4b and 7c listed the points to cover: address each one.
• Use the reading time to look through Section B and start planning answers.
• Answer the question asked. Question 8a asked how speciation happened, not why the finches are separate species, and Question 8c asked for evidence of extinction, not its causes.
• Use the space well. The number of lines hints at the length needed; blank space under a question or the extra page at the back can be used, but say so if you continue elsewhere.
• Write clearly in pen. Papers are scanned in colour and marked on screen, so pencil must be dark enough to scan.