2025 paper: how to approach each question

Open the official paper beside this page. Try the question for about three minutes, then read the card. The green letter is the published answer. The steps are the route, not a copy of the question.

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Question 1

Zebrafish wnt8a: mother's copy and the embryo's own copy

Read the four percentages only: wild type 100, zygotic loss 100, maternal loss 100, both lost 40.
(I) and (II) are true because losing only one copy still gives 100% normal embryos. Neither copy is essential on its own.
(III) is true because losing both drops normal embryos to 40%. The pattern is disrupted.
(IV) is not in the table. Nothing here moves BMP or shows which protein is upstream. Cross out every option that includes (IV).
I, II and III remain. That is B.

Question 2

Webbed digits and a caspase inhibitor

Write the three rows: control 10% webbed (caspase 100), low dose 35% (caspase 60), high dose 85% (caspase 15).
(I) and (II) follow the trend: less caspase, more webbing. Planned cell death separates the digits.
(III) says webbing happens only when death is completely stopped. The low dose has not stopped it (caspase is still 60) and webbing is already 35%. False.
(IV) is true because the untreated control is already 10% webbed.
Keep I, II and IV. That is D.

Question 3

Where cells die in a developing limb

This is a TUNEL stain: it marks nuclei of cells that are dying. Compare place and stage, not the species name.
Death between the digits rises from stage 26 to 30, which fits digit separation. Death is not the same in every region, so it is stage-specific and region-specific.
Digit-tip death falls (about 8% to 3%). At stage 30 the tip is lower than the joint regions, and the joints are still rising.
All four statements match that pattern. Answer A.

Question 4

Old muscle stem cells put into a young mouse

Find the regeneration index: young stem cells about 95, aged stem cells about 35, including when the aged cells sit in a young host.
(I) and (III) are the numbers. Old cells still do badly in a young body.
(II) blames the young environment. The young environment supports regeneration, so the fault travelled with the old cells. False.
(IV) says the young host fully repairs the old cells. 35 is not a rescue. False.
I and III only. Answer A.

Question 5

Young serum and aged stem cells

Compare each serum with each cell type. Young serum lifts Delta-positive aged cells from about 15% to 35%.
In the same serum, young cells activate more than aged cells. That is (II) and (IV).
35% is only a partial lift, so the aged cells can still change. They are not locked. That is (III).
All four are supported. Answer D.

Question 6

Whose DNA does a donor egg pass on?

Separate the nucleus from the mitochondria before you read the options.
(I) Mitochondria copy their own DNA. They do not need the nucleus to divide. True.
(II) A few mitochondria can stick to a nucleus that has been moved into a donor egg. True.
(III) Blood groups are nuclear genes. The donor egg supplies mitochondrial DNA, not a new blood group. False.
(IV) Sperm mitochondria sit in the neck. The head that fuses brings the nucleus. False.
I and II only. Answer A.

Question 7

A drug and the mitotic index

Mitotic index means the percentage of cells currently in mitosis.
A sharp rise after the drug means cells enter mitosis and cannot finish. The spindle is blocked. (I) is true.
(II) compares 12 hours with the control: 0.5% versus 2% is a fall of about three quarters, not a small change. False if the option says otherwise.
A later fall can mean the stuck cells have died and broken up. (III). A flat control means entry and exit are balanced. (IV).
I, III and IV. Answer C.

Question 8

Ribosomes under stress

Translation starts at the 5′ end and walks toward the 3′ end.
Stress lowers ribosome density at the 5′ end: new ribosomes are not starting. (I).
Ribosomes already on the mRNA can finish the journey, so the 3′ end need not fall as much. (II).
(III) Ribosome profiling measures translation, not how much mRNA is in the cell.
(IV) The graph is the change under stress, not a map of normal density along the mRNA.
I and II. Answer A.

Question 9

How insulin leaves the cell

Fusion events: glucose takes them from about 3 to 28. (I) is true.
A calcium chelator cuts fusion, so calcium is required. (II).
(III) says SNARE proteins do not matter, or the reverse of the data. Blocking SNAREs reduces fusion, so they are required. False.
(IV) Type 2 diabetes is poor insulin release when glucose is high. Losing calcium or SNAREs can produce that. True on these results.
I, II and IV. Answer B.

Question 10

A lysosomal enzyme and pH

A normal lysosomal enzyme is most active in acid. Read the wild-type peak: about 100% at pH 4.5. (I).
At that same pH the mutant is about 30% of wild type. (II).
(III) The mutant does not peak at pH 7. Its peak is at pH 6. False.
(IV) From pH 4.5 to pH 7 the mutant falls less steeply than the wild type. True.
I, II and IV. Answer D.

Question 11

Which iris muscle opens the pupil?

Two muscles, opposite jobs. Do not memorise a nerve. Match muscle to pupil size.
Dilator (radial) muscle contracts → pupil wider → more light in.
Sphincter (circular) muscle contracts → pupil narrower → less light in.
The option that states both directions is B.

Question 12

A stiff artery and the two pressures

Systolic pressure is while the heart squeezes. Diastolic is while it rests.
A stiff wall is hard to stretch, so systolic pressure rises. That part of A is fair, but check the option that is actually correct about diastole.
When the heart relaxes, the stiffened vessel stays wide, so diastolic pressure falls. That is B.
Mean arterial pressure ≈ diastolic + ⅓(systolic − diastolic). An equal rise and fall does not leave the mean unchanged, because diastole is weighted more. C fails.
Pulse pressure = systolic − diastolic, so it gets bigger, not smaller. D fails.

Question 13

Four plots of the circulation

Label the plots before the options: (a) diameter of one vessel, (b) total cross-sectional area, (c) pressure, (d) speed.
A is false: pressure is lowest in the veins, but speed there is not the lowest.
B is false: arteries and veins can have similar total area while arterial blood is faster. Area is not the only cause.
C is false: from capillaries to veins, pressure falls while single-vessel diameter rises. Wider does not mean lower pressure along this whole path.
D is the pair that move as mirror images: diameter is smallest at the capillaries, total area is greatest there. Answer D.

Question 14

Walls, speed, and where capillaries are missing

The stem says 'and your knowledge', so DSE facts are allowed if they fit the paragraph.
(I) Active muscle needs a dense capillary bed. True.
(II) Vein walls are thinner and stretchier, which suits low-pressure return. True.
(III) Slow flow in thin capillaries gives time for exchange. True.
(IV) says arterial muscle relaxing raises pressure. Relaxing widens the vessel and pressure falls. False.
(V) Lens and cornea have no capillaries; they have to stay clear. True.
I, II, III and V. That is B.

Question 15

Two transcription factors and cell migration

Set the control bar as 1. Then read bar 2 (TF-1 only), bar 3 (TF-2 only), bar 4 (both).
TF-1 alone lowers migration. TF-2 alone raises it. They are not both drivers, so B is false. They are not a simple tug-of-war either: together the bar stays as high as TF-2 alone, so A is false.
C matches bar 2: TF-1 suppresses migration when TF-2 is absent.
D says the two pathways are independent. Independent effects would still show TF-1 pulling the double-expressed bar down. It does not. TF-2's effect covers TF-1. Answer C.

Question 16

Which transfusion clashes?

Write the rule once: the recipient's plasma antibodies attack antigens on the donor red cells.
A person with only anti-A can still receive B cells. A person with no ABO antibodies (group AB) can receive any ABO cells.
Anti-Rh(D) alone does not attack A antigen.
The clash is the recipient who has both anti-B and anti-Rh(D), given red cells that carry Rh(D). Answer D.

Question 17

Safest red cells when the group is unknown

You need cells with no A, no B, and no Rh(D) antigen.
That is group O, Rh negative.
Answer C.

Question 18

Voles, risk, and food left behind

GUD is food remaining. A high GUD means the animal ate less and gave up sooner. ΔGUD is the gap between two patches.
The risky mixed landscape has the highest mean GUD, so it has the lowest feeding success, not the highest. A fails.
That same landscape has the biggest ΔGUD: the voles pile onto the safe patch. Effort is uneven. B fails.
The safe uniform landscape has the smallest ΔGUD: both patches are used evenly. That is C.
The two uniform landscapes do not give the same mean GUD. The safe one is eaten down further. D fails.

Question 19

Cricket song and what females have heard before

Do not pick the option that says females are born with a fixed taste, and do not pick 'she copies the last song'.
Naive females do prefer more long chirps. After hearing very good songs, the reference song becomes less attractive. After hearing poor songs, the reference song becomes more attractive.
Experience shifts the judgement in a relative way. The song features still matter, so the choice is not random.
Answer C.

Question 20

Larger lizard toepads on invaded islands

The useful comparison is wild versus lab-reared, invaded versus uninvaded.
The toepad difference is still there when lizards from invaded islands are raised in the same lab. It travelled with the animals. That points to a genetic difference. A.
If it were only plasticity from the island environment, lab rearing would erase it. B fails.
The same direction in the wild and in the lab is not random noise, and it is not a trait that appears only outdoors. C and D fail.

Question 21

Who approaches the female after a fight?

Count the sequence types. Type 1 (aggressor then goes to the same female) is about 53%. Type 3 (a different female) is about 9%.
A 53 versus 9 split is not chance, and it is not the aggressor bouncing into females at random. B and C fail.
Female choice alone does not explain why the aggressor specifically follows the female the losing male just approached. A fails.
The pattern fits a redirect: the losing male briefly contacts a female and the attacker switches onto her. Answer D.

Question 22

Bigger groups of guppies

Two curves: time to find food, and chance of feeding. Both should be read against group size.
Discovery gets faster and feeding probability keeps rising up to the largest group tested. Benefits are still ahead of scramble competition in this range. A.
A slightly flatter slope at the end is not a fall. B fails.
Feeding probability goes up, not down, so C and D fail.

Question 23

XXX offspring from a mutant male worm

The male is the mutant parent. He has one X (XO). The abnormal daughters are XXX.
A sperm that delivers two copies of the single X, plus the egg's X, makes XXX. Two copies of one chromosome are sister chromatids. They separate at meiosis II.
Meiosis I separates different chromosomes, and this male has only one X to begin with. The female parent is wild type. Mitosis in a precursor is not the division named by the result.
Answer C, meiosis II in the male.

Question 24

Nutrients, gene A, gene B, and cancer

Draw the chain before the options. Nutrients inhibit A. A, at full strength, inhibits B. B promotes division and cancer.
The homozygous x mutant gets cancer on poor food and on rich food, so A is not doing its braking job. A hyperactive A (option A) would brake harder and cause less cancer, not more.
Removing B as well as A would remove the promoter of cancer. Option B says proliferation would not be inhibited, which fights the chain.
Plenty of nutrients already switches A down, so B is free to drive cancer. Inhibiting B in that situation blocks the driver. That is C.
Lots of B product promoting cancer would be a gain of function, not a recessive loss. D fails.

Question 25

Mutant mitochondrial DNA in one family

mtDNA comes from the mother. The father is 0% and stays out of it.
The children are 89%, 65% and 15% from a mother at 85%. The percentage can rise or fall. It is not diluted in every child, so A is false.
The drop to 15% in II-3 is the random split of mitochondria when eggs are made. That is B.
II-1 is a son. His mitochondria will not pass to his children. C is false.
II-2 is a daughter at 65% and currently unaffected. Her eggs can still land above or below 65% by chance. 'Unlikely' is too strong. D is false.

Question 26

Mouse coat colours: a complementation test

Wild-type dark-brown babies mean the two mutations are in different genes. They filled in for each other. Mutant babies mean the two mutations failed to complement: same gene.
M1 fails to complement M2, M5 and M6. M3 complements M1 and also complements M4. M4 complements M1. That is three loci. (I) is true.
M5's white wins over the other alleles of that same gene (M5 × M1 is white, M5 × M6 is white). M2 is in that gene, so M2 × M5 should be white. (II) is true.
M6 and M3 are different genes, so the babies should be wild type, not silvery gray. (III) is false.
M7 that complements every one of M1 to M6 is a fourth gene. (IV) is true.
I, II and IV. Answer D.

Question 27

A family blood-group puzzle: which gift cannot work?

Mary can give to everyone and can receive from nobody in the family. She is group O (ii). Michael is A and Bonnie is B, and they produced an O child, so both are heterozygous: Michael IAi, Bonnie IBi.
Patrick can receive from Bonnie (B), Mary (O) and Terence, but not from Michael (A). The only group that fits is B.
Timothy cannot give to Bonnie or to Patrick, so his cells carry A antigen. The pair of statements that is impossible fits Timothy as AB, not as plain A.
(I) Michael (A) into Terence fails: a B or O recipient has anti-A.
(III) Timothy (AB) into Michael (A) fails: Michael's plasma has anti-B.
(II) Patrick into Timothy works if Timothy is AB. (IV) either parent into an AB patient works. Those two can happen.
The pair that cannot happen is I and III. Answer C.

Question 28

Vestigial wings that follow the X chromosome

Sons show the mother's phenotype (all vestigial). Daughters show the father's phenotype (all normal). That split by sex is X linkage.
The father has one X, carrying the dominant normal allele. The mother has two X chromosomes, both recessive vestigial.
A and C are autosomal wordings. They would give both phenotypes in both sexes.
B calls the female hemizygous. Females have two X chromosomes. The hemizygous parent is the male. Answer D.

Question 29

A virus test: sensitivity and specificity

Sensitivity: of the people who really have it, what fraction test positive. Specificity: of the people who do not, what fraction test negative.
DFA sensitivity 65% is higher than culture 51.8%, so DFA finds more true cases. (IV).
Using either positive result raises sensitivity to 81.3%. (I).
Each test alone has specificity 99.6%. The combined 'either positive' rule is 99.2%, a small drop, not a collapse. Read (III) against the exact wording; the key treats the 'much worse' version as false.
Both single tests are highly specific. (II).
The surviving set is I, II and IV. Answer D.

Question 30

What M cells in the gut actually do

M cells sample the gut contents and hand Salmonella to immune cells. They are a doorway, not a killer.
(I) true: they take pathogens in from the lumen.
(II) false: macrophages do the killing, not the M cell.
(III) false: antibodies come from B cells.
(IV) true: after the hand-off, T cells and B cells can be switched on.
I and IV. Answer B.

Question 31

Antibiotic discs and the millimetre cut-offs

Use the standard printed next to each drug. Do not compare millimetres across different drugs as if they shared one cut-off.
Gentamicin ≥ 15 mm is susceptible. (I).
Erythromycin at 12 mm and penicillin G at 15 mm fall in the resistant zone for those drugs. (II).
(III) A 31 mm penicillin zone is susceptible, so a statement calling it resistant is false.
(IV) Treat with a drug this organism is still susceptible to, rather than training a resistant strain. True.
I, II and IV. Answer B.

Question 32

Why Gram-positive stays purple

Both types take up crystal violet at the start. (I) is false if it says only one type takes it up.
Gram-positive keeps the purple dye because of the thick wall. Gram-negative loses it in the alcohol step because the peptidoglycan is thin, then safranin stains those cells pink. (II) and (III).
(IV) The stain sorts bacteria into two big groups. It does not name the species. False if it claims it does. It is still useful for a first antibiotic choice.
II and III. Answer B.

Question 33

Hand sanitiser graphs

Find the control bar (PBS) before you praise a sanitiser.
(I) Sanitiser C kills all the bacteria within 15 seconds.
(II) Every sanitiser has cleared the microbes by 1800 seconds (30 minutes).
(III) C, D and E, the ones with the extra agent, all finish the job in 15 seconds.
(IV) A kills more than B.
All four. Answer D.

Question 34

A keystone score

The top score can be a sum across many jobs. That species need not be first in every single job. (I) is false.
(II) A keystone has a large effect across several functions for its numbers. True.
(III) Species that join many functions also tend to connect those functions to each other. True.
II and III. Answer C.

Question 35

Kiwi traits that match a night-time forest floor

(I) Hair-like feathers break up the outline on the forest floor. True.
(II) Foraging by day would show them to predators. The useful habit is the night. False.
(III) A large egg produces a chick that can feed and run. True.
(IV) Nostrils at the tip of the beak, and a strong sense of smell, fit hunting by scent in the dark. True.
I, III and IV. Answer C.

Question 36

Which kiwi plans actually raise wild numbers?

(I) A zoo population does not, by itself, add birds in the forest. False.
(II) Lift eggs, hatch them safe from predators, release the juveniles. That does raise wild survival. True.
(III) Cloning is not how you rebuild a species that can still breed. False.
(IV) Stoats, cats, ferrets and dogs take the chicks. Trapping and fenced sanctuaries are why numbers have risen. True.
II and IV. Answer C.

Question 37

Three shapes of survivorship

Type I: most survive until old age, then the curve drops. Humans, not plants. (I) is false.
(II) Birds: after adulthood the chance of dying stays fairly steady for a long time. True.
(III) The late steep drop means death is concentrated at old age. True.
II and III. Answer C.

Question 38

Why those curves have those shapes

(I) Parental care keeps early death low. True for the human-shaped curve.
(II) Most seeds land in a bad spot, so young plants die in large numbers. True for the plant-shaped curve.
(III) If most young die, the species stays in the game by making a lot of young. True.
All three. Answer D.

Question 39

The whale tree

A lineage drawn through to the present is still alive. Check A against the time axis before you keep it.
B is the equal-relatedness point: Odontocetes and Mysticetes sit on one split, so each is equally related to Dorudon.
Echolocation is marked on the toothed-whale branch only. It is not a trait of the whole group. C fails if it spreads the trait too wide.
Read Indohyus's habitat off its own node. D fails if it moves that label.
Answer B.

Question 40

Hearts drawn on the animal tree

Fish: one circuit. Do not give them the mammal plan.
Mixing of oxygenated and deoxygenated blood is a feature of the turtle and lizard hearts on this figure, not of the crocodile. That is B.
Check where the shunt is drawn before you accept C. Birds and humans share the four-chamber plan; if an option denies that, it fails.
Answer B.

Question 41

The tree of life

Eukaryotes and archaea share a more recent ancestor than either does with bacteria, on a standard three-domain tree. A can be true.
The base of the tree is one universal ancestor. B can be true.
The earliest eukaryotic branch on the figure is the diplomonads. C can be true.
Branch length on this drawing is not a count of how many individuals exist. D is the false statement, and the question asks for it.

Question 42

What actually blocks speciation?

One ancestor can split into more than two species. A is possible.
Interbreeding mixes the populations back together and works against speciation. B describes a real obstacle.
A changed environment is a source of selection. C is fair.
The false statement is the one that says natural selection plays no part in speciation. Answer D.

Question 43

Darwin's finches and their foods

The tree does not say which lineage appeared first. Any option that reads birth-order off the drawing is false. That is A.
Same genus does not force the same diet. B is a true caution.
Count the insect-eating tips versus the fruit-eating tips. C follows the labels.
Diet is not the only pressure, so insect-eaters need not be stuck as one species. D is fair.
The false statement is A.

Question 44

Fruit hormones: which sentence is false?

The question asks for the false sentence. Trace each arrow.
No pollination → auxin stays off → GA stays off. A is true.
Auxin and cytokinin are both drawn on cell division. B is true.
No pollination → ABA is not switched off → ABA keeps suppressing fruit growth. C is true.
Ethylene is drawn as controlling GA, and GA is drawn on cell expansion. 'Ethylene does not affect cell expansion' is false. Answer D.

Question 45

Nutrient concentration and growth: which sentence is false?

The y-axis is growth. The experiment measures dry mass, so water is not part of the nutrient figure. A says water is included. That is the false sentence.
In the deficiency zone, more mineral goes with more growth. B is true.
In the adequate zone the curve flattens, so something else (for example photosynthesis) is limiting. C is true.
In the toxic zone more mineral goes with less growth. D is true.
Answer A.

Question 46

Sunflower florets: which sentence is false?

The question asks for the false sentence.
The large outer florets are the ray florets; their job is to attract insects. The inner disc florets carry stigma, style, anther and ovary.
Each disc floret has its own ovary. One head therefore makes many fruits, not a single fruit.
'A single fruit will be formed' is the false sentence. Answer D.

Question 47

Moss: which sentence is false?

Spores are made by meiosis, so they are not genetically identical. A statement that they differ is true.
The stalk is the sporophyte (2n). The leafy stem is the gametophyte (n). 'Same genetic makeup' is false. Answer B.
The sporophyte grows on the gametophyte and takes food from it. True.
The stalk lifts the capsule so spores can disperse. True.

Question 48

Chlorophagy: which sentence is false?

The diagram puts damaged chloroplasts into the vacuole to be broken down and reused. They are not spat out by exocytosis. A is false.
ATG proteins are required, so an ATG mutant cannot run the process. B is true.
The hydrolases that finish the job are in the vacuole. C matches the diagram.
Dark, and also strong light damage, are drawn as triggers. D matches the figure.
Answer A.

Question 49

Water potential up a 20 m tree

Water potential = pressure potential + osmotic potential + gravity potential. Water moves toward the lower (more negative) total.
(I) Positive pressure inside a living cell is turgor. True for the vacuoles in the table.
(II) Pure water would have osmotic potential 0. Cell and soil water contain solutes, so osmotic potential stays negative. True.
(III) Compare the gravity column with the other two. On the values in the table it is small enough that the question accepts neglecting it.
All three. Answer D.

Question 50

Two Atriplex species: which sentence is false?

The species whose photosynthesis levels off is no longer limited by light. That is triangularis, the C3 plant.
Rosea keeps the C4 features from the introduction: it copes with hotter, brighter places and uses bundle sheath cells to concentrate CO2.
The false sentence is the one that swaps the labels and calls rosea the C3 plant. Answer B.