Cambridge AS & A Level Biology 9700/22 is the May/June 2026 Paper 2 (AS Level Structured Questions), variant 2. It lasts 1 hour 15 minutes, is worth 60 marks and tests the AS Level content. It had six compulsory questions, and the biggest, on collagen in the gas exchange system, carried 16 marks. Several questions were set in unfamiliar contexts, from bacteriophage therapy to salt-stressed oil-seed plants. 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.

9700/22 at a glance

Question	Topic	Marks	Main command word
1	Prokaryotes and viruses: E. coli, T-phage structure and size, phage therapy	11	Name, suggest
2	Collagen in the gas exchange system: cartilage, telomeres, mitosis, the interstitium, collagen structure	16	Explain, describe
3	Cell signalling: methyl jasmonate and salt stress in Brassica napus	4	State, explain
4	Immunity: the hybridoma method, mutation, HIV and the monoclonal antibody VRC01	14	Explain
5	Transport: a double circulation diagram, water in blood and phloem	9	Complete, explain
6	Enzymes: pH and temperature effects on lactate dehydrogenase	6	State, explain
	Total	60

Question-by-question walkthrough

Question 1: E. coli and bacteriophages (11 marks)

A transmission electron micrograph showed a T-phage attached to the outside of an E. coli cell.

• Double-membrane organelle missing from E. coli (1 mark): mitochondrion, chloroplast or nucleus. Misspellings of nucleus that could be another word, such as “nuculus”, were rejected.
• Naming structures A and B (2 marks): A is the capsid (protein coat was accepted). B is the cell wall, which could be called the murein or peptidoglycan wall. A cellulose or chitin cell wall was rejected, and so was describing the capsid as having an envelope or membrane.
• Actual length of the T-phage (2 marks): measure line X–Y and the T-phage on the micrograph. The actual X–Y diameter is 0.9 µm, which is 900 nm, so actual length = (T-phage image length ÷ X–Y image length) × 900. The scheme accepted 152 to 186 nm as a whole number. A correct value not rounded to the nearest whole nm scored 1 mark, and so did a slightly out-of-range whole number caused by a small measuring error.
• Why the T-phage attaches only to E. coli (2 marks): one mark for specificity, or complementary shapes. The second was for the detail of the interaction: the tail fibres or receptors on the phage binding to a named site on E. coli, such as receptors in the cell surface membrane, the outer membrane or membrane glycoproteins. Comparing it to enzyme-substrate or antibody-antigen binding was rejected.
• Which disease phage therapy is least likely to treat (1 mark): malaria, because it is caused by a protoctist (a eukaryote), not a bacterium. Cholera and TB are bacterial. The mark was rejected if “virus” appeared anywhere in the answer.
• Benefits of phage therapy (3 marks): any three from a list of eleven, including: it does not harm human cells; fewer side effects or allergic reactions; it kills only the pathogen causing the disease, so beneficial bacteria such as the gut microbiome are left alone; phages multiply inside the bacteria being treated, so a single dose or shorter course may be enough; faster recovery; and it can treat antibiotic-resistant bacteria and reduces reliance on antibiotics. “Specific to bacteria” on its own was ignored.

Question 2: collagen in the gas exchange system (16 marks)

The context was the extracellular matrix (ECM) around cells in the gas exchange system, made by fibroblasts and chondrocytes.

• The tissue containing chondrocytes (1 mark): cartilage.
• Features and role of cartilage (3 marks): any three from: C-shaped (incomplete) rings in the trachea, or irregular plates in the bronchi; another structural detail, such as the surrounding perichondrium or chondrocytes sitting in spaces called lacunae; flexible; supports the airway; and keeps the airways open so they do not collapse during breathing. If you named the wrong tissue in the first part, you could still score 1 mark here for a correct description of the tissue you named.
• Why telomere length matters in dividing fibroblasts (2 marks): the ends of chromosomes shorten each time DNA replicates, telomeres stop genes being lost, and keeping them allows the cells to carry on dividing by mitosis. “Telomeres are lost” on its own was ignored, and “genetic material” was too vague for the genes point.
• Spindle structures absent from plant cells (1 mark): centrioles (centrosomes accepted). The mark was rejected if microtubules were also given.
• A chromosome at telophase (2 marks): any two from: a single DNA molecule (the sister chromatids are now separate chromosomes); DNA wound around histone proteins; one centromere; condensed at first, then beginning to uncoil; detaching from the spindle. Saying the chromosome is simply “decondensed” was rejected unless you made clear it happens towards the end of telophase.
• Cells of the capillary wall (1 mark): endothelial cells. “Squamous epithelial cells” was rejected.
• Effect of more fibroblast activity (2 marks): the interstitium becomes thicker, which increases the diffusion distance for oxygen, so less oxygen diffuses into the blood per unit time. “More collagen” and “longer” were ignored; you had to describe the change in thickness.
• The bond between amino acid residues (1 mark): peptide bond. “Polypeptide bond” was rejected.
• Why glycine and alanine matter in collagen (3 marks): they are small amino acids, with R-groups of H (glycine) and CH₃ (alanine). This lets the three polypeptides pack tightly together and wind into the triple helix, held by hydrogen bonds between the polypeptides, with room for the amine groups to form those bonds. Answers about an alpha helix, or about covalent or disulfide cross-links here, were rejected.

Question 3: methyl jasmonate and salt stress (4 marks)

Seedlings of the oil-seed crop Brassica napus were grown in soil with four salt concentrations, and half were sprayed on their shoots with methyl jasmonate (MeJA). Table 3.1 gave the mean root and shoot dry mass for each treatment.

• Conclusions (up to 3 marks): higher salt concentration reduces dry mass; MeJA gives more growth when salt is present; and MeJA reduces growth when there is no salt (root dry mass 0.19 g with MeJA against 0.53 g without). One mark was available for a data comparison, for example root dry mass at 110 mmol dm⁻³ salt of 0.14 g without MeJA and 0.21 g with it.
• Explanation (the rest): MeJA is transported to the roots in the phloem, binds to receptors on target cells in the roots and triggers a protective response. A suggested response, such as more water uptake by osmosis or pumping sodium ions out of cells, also scored.
• The rule that decides full marks: the scheme required an attempt at both the conclusions and the explanation to reach 4. A long analysis of the table with no explanation of how a leaf spray reaches the roots tops out at 3.

Question 4: monoclonal antibodies and HIV (14 marks)

• The immune cell that makes antibody (1 mark): the plasma cell (B-lymphocyte was given benefit of the doubt).
• Order of the hybridoma method (2 marks): T → S → Q → R → P: inject the mammal with the antigen, wait for an immune response, remove spleen cells and fuse them with myeloma cells, culture on a special medium so only hybridomas survive, then screen for the antibody wanted. One mark was available if only step P was out of place.
• Why myeloma cells without purine nucleotides cannot divide (2 marks): they cannot make DNA, so DNA replication (S phase) cannot happen. The second mark was for a consequence: no sister chromatids form, the cell cannot pass the checkpoint into mitosis, or complementary base pairing cannot take place.
• How a mutation in gene HPRT gives a non-functioning enzyme (3 marks): the base sequence changes; a substitution changes an amino acid, or a deletion or insertion causes a frameshift, possibly creating an early stop codon; the primary structure changes; so the active site changes shape and is no longer complementary to its substrate. “Does not function” earned nothing, and so did describing insertions or deletions of amino acids instead of bases.
• Full name of HIV (1 mark): human immunodeficiency virus. It had to be spelled correctly.
• Why the antibody infusion must be repeated (2 marks): the antibodies are broken down over time; it is passive immunity, so no immune response is triggered and no memory cells are made; and a high-risk person may be exposed to HIV again and again.
• Why a mixture of monoclonal antibodies could work better than VRC01 alone (3 marks): each monoclonal antibody is specific to one shape; different HIV strains may have slightly different gp120, because HIV mutates; a mixture can bind to more sites or more variants, so it protects against more strains. Answers about treating someone who is already infected were capped at 2.

Question 5: transport in mammals and plants (9 marks)

• Letters on the double circulation diagram (4 marks): the chamber receiving blood from the vena cava was H, the chamber pumping into the pulmonary artery was G, an elastic artery with little smooth muscle was A, C or K, and the vein carrying oxygenated blood was B. If you gave more than one letter for the artery, all had to be correct.
• Why high specific heat capacity suits blood (2 marks): blood resists changes in temperature, which helps keep body temperature stable and gives cells a steady environment. A second mark could come from what a stable temperature protects: enzyme activity, protein structure, solubility of solutes or the temperature of tissue fluid. “Optimum temperature” on its own was ignored.
• The role of water in mass flow in phloem (3 marks): water is the solvent for sucrose and other assimilates; it enters the sieve tubes at the source by osmosis because sucrose lowers the water potential; this raises the hydrostatic pressure there, creating a pressure gradient to the sink; water leaves at the sink, lowering the pressure; and cohesion between water molecules keeps the sap moving as a continuous stream. “Lower water potential gradient” was rejected; the point is the water potential inside the sieve tube.

Question 6: lactate dehydrogenase (6 marks)

LDH from the bacterium Lactobacillus delbrueckii is used in a blood test for the liver enzyme ALT, which is carried out at room temperature.

• Optimum pH and temperature (2 marks): pH 7.5 and 55 °C (51–57 °C accepted).
• How pH affects the enzyme (3 marks): at pH 7.5 the active site works at its maximum rate. Either side of the optimum, hydrogen and ionic bonds are disrupted and the charges on R-groups change, so the active site changes shape and fewer enzyme-substrate complexes form; at the extremes some molecules are partly denatured. Saying peptide or disulfide bonds break was rejected.
• Why the test is not run at 55 °C (1 mark): any one reason, such as room temperature being easier, ALT possibly denaturing at 55 °C, the reaction going too fast to measure accurately, or the bacterial LDH possibly being a different form of the enzyme.

Where the marks are on 9700/22

• Explain questions carried 28 of the 60 marks, including “suggest and explain” and “state and explain”: 1(d)(i), 2(b), 2(e)(ii), 2(f)(ii), 3, 4(c)(i), 4(c)(ii), 4(d)(ii), 4(d)(iii), 5(b)(ii) and 6(b). Each is marked as a list of separate points.
• Recall carried 17 marks: the “name”, “state” and “complete” parts, from organelles and tissues to the hybridoma sequence and the circulation letters. These should be close to full marks with secure knowledge.
• Describe questions carried 7 marks (cartilage, the telophase chromosome and specific heat capacity), and suggest questions 6 (phage binding, phage therapy and the test temperature).
• The only calculation was the 2-mark T-phage length, which depended on careful measuring and rounding to a whole number.
• Question 2 alone was worth 16 marks. It mixed gas exchange, the cell cycle and protein structure, so a gap in any one of them shows up here.

How to self-mark 9700/22 with the mark scheme

1. Learn the abbreviations first. A semicolon separates marking points and a slash shows alternatives. A means accept, R reject and I ignore; AW is alternative wording, ora means “or reverse argument”, ecf is error carried forward, AVP is an alternative valid point and bod means benefit of the doubt. An underlined word must appear in your answer.
2. Mark point by point. For every “any three from” list, tick the separate points your answer contains and stop at the maximum.
3. Read the R and I notes. “Polypeptide bond”, “squamous epithelial cell” for the capillary wall and “lower water potential gradient” in the phloem answer all sounded close but were rejected.
4. Check the special rules. Question 3 needs both conclusions and an explanation for 4 marks, Question 4(d)(iii) is capped at 2 if you wrote about an infected person, and Question 4(d)(i) needs the correct spelling.
5. For the calculation, check your measurements. Your answer is right if it is a whole number from 152 to 186 nm; a correct unrounded value scores 1.
6. Apply the list rule. For “name one” questions, only the first answer counts, and two answers that contradict each other score nothing.

Sit 9700/22 as a timed mock

The paper allows 75 minutes for 60 marks, about one and a quarter minutes per mark. A workable plan:

Block	Marks	Time
Question 1	11	13 minutes
Question 2	16	19 minutes
Question 3	4	6 minutes
Question 4	14	17 minutes
Question 5	9	10 minutes
Question 6	6	7 minutes
Check, especially the calculation and spellings	–	3 minutes

Question 3 has extra time for its marks because you need to read Table 3.1 before you write. Plan it as two short paragraphs: conclusions with data first, then how MeJA reaches the roots.

Other May/June 2026 9700 papers

• 9700 Paper 12 (Multiple Choice)
• 9700 Paper 32 (Advanced Practical Skills)
• 9700 Paper 42 (A Level Structured Questions)
• 9700 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9700/22 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

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Frequently asked questions

What topics were on 9700 Paper 22 May/June 2026? +

Six questions: bacteriophages and E. coli; collagen in the gas exchange system, with cartilage, telomeres, mitosis and protein structure; methyl jasmonate and salt stress in a crop; monoclonal antibodies, mutation and HIV; the double circulation and water as a transport medium; and the effect of pH and temperature on lactate dehydrogenase.

How many marks is 9700/22 and how long is it? +

It is out of 60 marks and lasts 1 hour 15 minutes. All six questions are compulsory, and Question 2, on collagen in the gas exchange system, is the biggest at 16 marks.

What was the answer to the T-phage calculation on 9700/22? +

The mark scheme accepted any whole number from 152 to 186 nm, depending on your ruler measurements. Divide the T-phage's image length by the image length of line X–Y, then multiply by the actual diameter, 0.9 µm, which is 900 nm. A correct value not rounded to a whole number scored 1 of the 2 marks.

How was the 4-mark MeJA question on 9700/22 marked? +

Conclusions from Table 3.1 could score up to 3 marks, such as salt reducing dry mass, MeJA giving more growth when salt was present, and MeJA reducing growth with no salt. The explanation, that MeJA travels to the roots in the phloem and binds to receptors on target cells, scored the rest. You had to attempt both parts to reach 4.

Is 9700/22 an AS Level or A Level paper? +

Paper 2 tests the AS Level content. AS candidates sit it with Papers 1 and 3, and full A Level candidates also sit Papers 4 and 5.

Where can I download 9700/22 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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