Cambridge AS & A Level Biology 9700/42 is the May/June 2026 Paper 4 (A Level Structured Questions), variant 2. It lasts 2 hours, is worth 100 marks and is sat by every candidate taking the full A Level. It had ten compulsory questions across the A Level topics. Question 3, on tree-frog speciation, carried 13 marks, and the biggest single part was a 7-mark description of ADH in Question 8. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all five 9700 papers from this series, see A Level Biology 9700 May/June 2026 past papers: what came up and how it was marked.

9700/42 at a glance

Question	Topic	Marks	Main command word
1	Motor neurone structure, the action potential	8	Identify, outline
2	Gibberellin in barley seeds, DELLA proteins, dwarf peas	8	Identify, suggest
3	Tree-frog speciation, genetic techniques, variation	13	Outline, explain, suggest
4	Genetic engineering of crops and food supply	10	Explain, calculate
5	Sex-linked feather colour in pigeons, test cross	9	Complete, state, describe
6	Glucose uptake, ATP use in neurones, caffeine	10	Calculate, suggest
7	Classifying the southern damselfly, conservation	12	Complete, describe, suggest
8	Proximal convoluted tubule, ADH and osmoregulation	11	Complete, describe
9	Hexokinase deficiency, Krebs cycle, anaerobic respiration	11	Suggest, outline, describe
10	Chloroplast structure, chromatography, accessory pigments	8	Outline, calculate
	Total	100

Question-by-question walkthrough

Question 1: the motor neurone (8 marks)

• Letters on Figure 1.1 (4 marks): F, a dendrite on the cell body, forms a synapse with an intermediate neurone; A, an axon terminal, is part of the neuromuscular junction; B, the myelin sheath, provides insulation; B and C (the myelin sheath and a node of Ranvier) together allow saltatory conduction.
• Events of an action potential (4 marks, any four): voltage-gated Na⁺ channels open and Na⁺ moves in; this depolarises the membrane; the threshold is reached; voltage-gated K⁺ channels open and K⁺ moves out; the membrane repolarises; it then hyperpolarises. “Frequency” was rejected for the threshold point.

Question 2: gibberellin in barley seeds (8 marks)

• Labelling the seed (2 marks): A, where gibberellin is released, is the embryo; B, the target cells that make amylase, is the aleurone layer.
• Proteins C, D and E (3 marks): C is the gibberellin receptor (GID1 was accepted); D is the DELLA protein (repressor); E is a transcription factor (PIF, phytochrome interacting factor, was also accepted).
• Region of DNA and gene product (2 marks): RNA polymerase binds to the promoter, and the gene product is amylase (mRNA for amylase was accepted).
• Peas that can’t make active gibberellin (1 mark): they are shorter, or dwarf.

Question 3: tree-frog speciation (13 marks)

Northern (N) and southern (S) populations of the green-eyed tree-frog were separated by a dry area for about 5 million years and met again in a contact zone 6500 years ago.

• Genetic techniques (3 marks, any three): PCR to amplify the DNA; gel electrophoresis to separate and compare DNA fragments; microarrays to compare genomes; databases or bioinformatics to compare base sequences.
• How separation leads to speciation (4 marks, any four): this is allopatric speciation; with no contact there is reproductive isolation and no gene flow; different mutations occur in each population; there are different selection pressures; morphological, behavioural or ecological differences build up; so the populations can no longer interbreed to produce fertile offspring.
• What the mating calls and failed eggs show (3 marks): N and S are now different species, or speciation has happened (1); the different mating calls show behavioural isolation (1); the fertilised eggs from an S female and N male are not viable, so no offspring are produced (1).
• Variation in body length (3 marks): genes and environment both play a part (1); body length is controlled by many genes (polygenes) (1); a named environmental factor such as diet or food availability, competition for food, temperature or disease (1).

Question 4: GM crops (10 marks)

• What genetic engineering of crops means (3 marks): deliberately changing the DNA or genome (1) to modify or improve a named characteristic of the crop (1) by adding, deleting or replacing a gene, or by gene editing such as CRISPR (1).
• Percentage change, 2000 to 2012 (2 marks): from Figure 4.1, (170 − 42.5) ÷ 42.5 × 100 = 300%. Any answer from 295 to 305 scored both marks; a wrong answer with this working scored 1.
• How GM crops improve food supply, with examples (5 marks, any five): higher yield or productivity; better (named) quality; herbicide tolerance, so less competition from weeds and more light and water for the crop; insect or pest resistance, so less crop damage and less food wasted; an economic advantage; and tolerance of a named factor such as disease, frost, high temperature, wind or poor soil.

Question 5: feather colour in pigeons (9 marks)

Female pigeons are ZW and males are ZZ. The colour gene sits on the Z chromosome only and has three alleles: ash-red, dominant to blue, dominant to brown.

• Genetic diagram (4 marks): for an ash-red male and a blue female to have a brown chick, the parents must be Z^AZ^Br and Z^BlW (1). The gametes are Z^A and Z^Br from the male and Z^Bl and W from the female (1). The offspring genotypes are Z^AZ^Bl, Z^AW, Z^BlZ^Br and Z^BrW (1), giving an ash-red male, an ash-red female, a blue male and a brown female (1, colour and sex matched). If the male parent’s genotype was wrong, offspring genotypes correctly worked from your gametes could still earn 1 mark (ecf).
• Type of inheritance (2 marks): sex linkage (1) and multiple alleles, or a dominance hierarchy (1).
• Finding an ash-red male’s genotype (3 marks): cross him with a brown female (1). If all the offspring are ash-red, he is Z^AZ^A; if any are blue, he is Z^AZ^Bl; if any are brown, he is Z^AZ^Br (any two of these, 1 each).

Question 6: ATP in brain neurones (10 marks)

• How glucose enters neurones (1 mark): facilitated diffusion.
• Calculation (2 marks): 4.7 × 10⁹ ÷ 9.8 × 10⁶ = 479.59…, so 480 muscle cells. An unrounded answer scored 1, and if the standard form went wrong, evidence of dividing 4.7 by 9.8 scored 1.
• Why neurones use so much ATP (4 marks, any four): many impulses and many synapses; the sodium-potassium pump (active transport of Na⁺ and K⁺; “channel” was rejected); setting up ion gradients; maintaining the resting potential; making or recycling neurotransmitter such as acetylcholine; making and moving vesicles, or exocytosis and endocytosis; making acetylcholinesterase. “Exocytosis of vesicles” was rejected; it is the neurotransmitter that is released by exocytosis.
• Caffeine and sleep (3 marks, any three): caffeine binds to the adenosine receptors, blocking adenosine (“active site” was rejected); more ATP is made while awake or less during sleep; postsynaptic activity is not inhibited; so sleep is not promoted and the person stays awake.

Question 7: the southern damselfly (12 marks)

• Classification table (2 marks): the missing rank names class and family (1), and the genus Coenagrion with the species mercuriale (1).
• Animalia against Plantae (4 marks, any four comparisons): no chloroplasts or chlorophyll against chloroplasts; heterotrophic against autotrophic (photosynthesis); no cell walls against cell walls; small, temporary vacuoles against a large, permanent vacuole; glycogen as the store against starch; able to move against no locomotion; a nervous system against none. Each mark needs both sides of the comparison.
• Why it is endangered in the UK (4 marks, any four): habitat loss, including less open vegetation or slow-flowing water (“deforestation” was ignored); named water pollution; less food or prey, or more competition for food; more predation or a new predator; named climate change such as flooding, drought or warmer water; a new disease.
• Why mark-release-recapture may be inaccurate (2 marks, any two): flying adults are hard to catch and mark; adults live only a few weeks, so they may die before recapture, or new adults emerge in between; migration; a small population.

Question 8: the kidney and ADH (11 marks)

• Table 8.1 (4 marks): in Figure 8.1, P is the folded membrane of the tubule cell on the blood side, S the microvilli facing the lumen, R the junction between the two cells and Q the capillary wall. Na⁺/K⁺ pumps are at P; cotransport proteins are at S; glucose passes through P and S; and R prevents leakage of fluid.
• The role of ADH (7 marks, any seven): more ADH is released (“secreted” was accepted, “produced” was ignored); from the posterior pituitary; ADH binds to receptors; on the cell surface membrane of collecting duct cells; cAMP is produced as a second messenger; an enzyme cascade amplifies the signal; vesicles containing aquaporins move to and fuse with the luminal membrane; so there are more aquaporins in it; the collecting duct becomes more permeable to water; more water moves out of the collecting duct into the cells, tissue fluid and blood; by osmosis, down the water potential gradient; and a smaller volume of more concentrated urine is produced.

Question 9: respiration (11 marks)

• Hexokinase deficiency (4 marks): up to three effects on red blood cells (less glucose phosphorylated; less glucose diffusing in because the concentration gradient is less steep; less glycolysis or pyruvate; less ATP; loss of red blood cells, anaemia or less oxygen transport) plus one symptom: tiredness, shortness of breath, fainting or dizziness.
• The Krebs cycle (4 marks, any four): the acetyl group (from acetyl CoA) combines with oxaloacetate to form citrate; decarboxylation releases carbon dioxide; dehydrogenation (oxidation) of intermediates; reduced NAD and reduced FAD are produced; ATP is made by substrate-linked phosphorylation; oxaloacetate is regenerated.
• Anaerobic respiration in muscle (3 marks, any three): glycolysis still makes pyruvate and a net 2 ATP by substrate-linked phosphorylation; pyruvate is reduced by reduced NAD (“pyruvate is dehydrogenated” was rejected); lactate is formed; NAD is regenerated so glycolysis can continue. Naming lactate dehydrogenase was an accepted extra point.

Question 10: the chloroplast (8 marks)

• Structure and function (4 marks, any four): the stroma is the site of the light-independent stage (Calvin cycle); the thylakoid membranes are the site of the light-dependent stage, containing the photosystems, pigments, electron transport chain and ATP synthase; grana hold many photosystems and give a large surface area for absorbing light; the thylakoid space builds up a high concentration of protons for the proton gradient; the chloroplast DNA codes for proteins; ribosomes translate mRNA into proteins.
• Rf of pigment B (1 mark): 0.42.
• Identifying the pigments (1 mark): A, with Rf 0.96, is carotene; B is chlorophyll b.
• Role of pigments A and B (2 marks, any two): they are accessory pigments (“secondary pigments” was ignored); they absorb wavelengths that chlorophyll a in the reaction centre does not; so they extend the range of wavelengths absorbed; and they pass the energy to the reaction centre.

Where the marks are on 9700/42

• Energy and respiration: 21 marks (Questions 6 and 9).
• Control and coordination: 16 marks (Questions 1 and 2).
• Selection and evolution: 13 marks (Question 3).
• Classification, biodiversity and conservation: 12 marks (Question 7).
• Homeostasis: 11 marks (Question 8).
• Genetic technology: 10 marks (Question 4).
• Inheritance: 9 marks (Question 5).
• Photosynthesis: 8 marks (Question 10).

Ten parts worth 4 marks or more were marked from “any n from” lists: 1(b), 3(a)(ii), 4(c), 6(c), 7(b), 7(c)(i), 8(b), 9(a), 9(b) and 10(a). Together they carried 44 marks, so writing one precise point per mark matters more here than anywhere else. The three calculations, 4(b), 6(b) and 10(b)(i), were worth 5 marks.

How to self-mark 9700/42 with the mark scheme

1. Learn the abbreviations first. A semicolon separates marking points, a slash shows alternatives, A means accept, R means reject, I means ignore, AW means alternative wording, ora means or reverse argument, ecf means error carried forward and AVP means alternative valid point. An underlined word must appear in your answer.
2. Number the points in your long answers. For 8(b), tick each of the twelve points you actually made and stop at seven.
3. Check the reject list. “Active site” in 6(d), “channel” for the sodium-potassium pump, and “pyruvate is dehydrogenated” in 9(c) lose marks, and “produced” for ADH release is ignored.
4. Mark comparisons as pairs. In 7(b), a feature of animals with no matching plant feature does not score.
5. Mark the genetic diagram in four parts, and give yourself the ecf mark only if your offspring genotypes really follow from your own gametes.
6. Check rounding. 480 must be a whole number, and 300% has a range of 295 to 305.

Sit 9700/42 as a timed mock

The paper allows 120 minutes for 100 marks, a little over one minute per mark. A workable plan:

Block	Marks	Time
Question 1	8	9 minutes
Question 2	8	9 minutes
Question 3	13	15 minutes
Question 4	10	12 minutes
Question 5	9	11 minutes
Question 6	10	12 minutes
Question 7	12	14 minutes
Question 8	11	13 minutes
Question 9	11	13 minutes
Question 10	8	9 minutes
Check calculations and key terms	–	3 minutes

Before writing 8(b), jot the ADH sequence in the margin, from the pituitary to the urine, so each step becomes one sentence.

Other May/June 2026 9700 papers

• 9700 Paper 12 (Multiple Choice)
• 9700 Paper 22 (AS Level Structured Questions)
• 9700 Paper 32 (Advanced Practical Skills)
• 9700 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9700/42 and its mark scheme on the Learning Portal →
• All 9700 May/June 2026 papers: overview and marking patterns
• Try a free A Level Biology quiz

Need help with a 9700 topic? Book a free trial lesson with an A Level Biology tutor →

Frequently asked questions

How many marks is 9700/42 and how long is it? +

It is out of 100 marks and lasts 2 hours. All ten questions are compulsory. Question 3, on speciation in tree-frogs, is the biggest at 13 marks, and the single biggest part is the 7-mark ADH answer in Question 8(b).

What were the calculation answers on 9700/42 May/June 2026? +

The GM crop area rose by about 300% from 2000 to 2012 ((170 − 42.5) ÷ 42.5 × 100, with 295 to 305 accepted). One brain neurone uses as much ATP as 480 muscle cells (4.7 × 10⁹ ÷ 9.8 × 10⁶, rounded to a whole number). The Rf value of pigment B was 0.42.

How was the pigeon genetic diagram on 9700/42 marked? +

Four marks: the parental genotypes (an ash-red male carrying the brown allele and a blue female), the gametes, the four offspring genotypes, and the phenotypes matched to colour and sex. The cross produces ash-red males and females, a blue male and a brown female. Part (b) named the inheritance as sex-linked with multiple alleles.

How many points did the ADH answer on 9700/42 need? +

Seven, from a list of twelve. They run from more ADH released from the posterior pituitary, through binding to receptors on collecting duct cells, cAMP and vesicles of aquaporins fusing with the luminal membrane, to water leaving the collecting duct by osmosis and a smaller volume of concentrated urine.

What did 9700/42 ask about caffeine? +

Caffeine has a similar shape to adenosine, so it binds to adenosine receptors and blocks adenosine. Postsynaptic activity is not inhibited, so sleep is not promoted and the person stays awake. Saying caffeine binds to an active site was rejected.

Where can I download 9700/42 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Biology 9700 past papers page.

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