2024 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

What 'monophyletic' means

A clade is an ancestor plus every descendant. Nobody from that branch is left out.
Palaeoptera is drawn that way: one ancestor, all of its descendants inside the group.
Any option that leaves a descendant out, or that groups by a look rather than a branch, is not monophyletic.
Answer A.

Question 2

Insect tree with time on the axis

The vertical axis is time. A lower node is an older split. A split nearer the present is a later divergence.
A can be checked: the node joining Hemiptera and Zygentoma sits near 415 million years. If A misreads that node, drop it.
An extant clade is one whose line reaches today. C: Megaloptera is equally related to Neuroptera and Raphidioptera if they meet at the same node.
D is the one that matches the axis: Diptera splits off near 300 million years ago, Lepidoptera near 200. Lepidoptera diverged later. Answer D.

Question 3

Homoplasy: same look, different origin

Homoplasy means a similarity that was not inherited from the shared ancestor.
Pangolin scales and snake scales evolved separately. Their last common ancestor did not hand that trait to both.
That pair is the example of homoplasy. Answer A.

Question 4

What makes two butterflies one species?

The biological species test is reproduction: they can mate and the offspring can also reproduce.
Sex chromosomes make males and females genetically a bit different. They are still one species. B fails.
Sharing a habitat, or sharing a food, is true of many species at once. C and D fail.
Answer A.

Question 5

A flatworm that hijacks a snail

The broodsac wiggles to look like a caterpillar so a bird eats it. That is the transmission trick. If A says something else, drop it.
A frog is not a host in this life cycle, so the worm is not adapted to live out its life in a frog.
A successful parasite takes nutrients without killing the host. A dead host is a dead food supply.
A human who accidentally swallows it is not a host either. The option that respects the real host list is D.

Question 6

Archosaur tree: dinosaurs, birds, crocodiles

Birds, Archaeopteryx, ornithischians and sauropodomorphs share the dinosaur ancestor. Living birds are dinosaurs. An option that throws birds out of Dinosauria is wrong.
Crocodiles and dinosaurs are both archosaurs. They share an older ancestor. Dinosaurs did not evolve from crocodiles. That is the statement to keep. Answer B.
Flight evolved twice on this tree: once in pterosaurs, once in paravians. One origin is wrong.
A line that stops at 66 million years is an extinction. Crocodiles and living birds reach the present, so they are still with us.

Question 7

Did a beetle really gain a bacterial gene?

A shared metabolic pathway can evolve twice. The pathway alone is not proof of gene transfer. A fails.
Bacteria living in the gut are not automatically donors of genes into the beetle genome. B and C fail.
Compare sequences. A beetle gene should look like other insect genes. If this one looks like a bacterial gene instead, horizontal transfer is the serious explanation.
Answer D.

Question 8

How C4 and CAM avoid photorespiration

Both C4 and CAM first store CO2 in a four-carbon compound so Rubisco meets less oxygen. That is what the diagram shows. A.
An option that claims this does not reduce photorespiration fights the diagram. B fails.
In C4 the Calvin cycle is in a different cell (spatial sheath). In CAM it is in the same cell at a different time of day. 'Spatially separated in both' is wrong. C fails.
CAM releases the stored CO2 and runs the Calvin cycle in the light. Night fixation does not replace daytime photosynthesis. D fails.

Question 9

Which moss parts are haploid?

Gametophyte tissues are haploid (n). Sporophyte tissues are diploid (2n).
Spore and antheridium belong to the gametophyte generation.
Capsule and seta (the stalk) belong to the sporophyte.
The option that pairs them that way is B.

Question 10

Why a plant bothers to warm itself

The heat is small, about 35°C. Match each job to that size.
(I) Warmth helps scent molecules leave the flower. True.
(II) 35°C does not sterilise bacteria. False.
(III) That warmth can melt a thin cover of snow. True.
(IV) Ordinary plant tissue is not waiting for this heat in order to work. False.
I and III. Answer C.

Question 11

A dwarf plant and gibberellin

Look at figure B: adding GA makes the dwarf grow. So the plant can still detect GA.
A broken receptor would ignore added GA. The receptor is intact. C fails.
The experiment does not show a general metabolic collapse. A and B are not supported.
Only added GA works, so the dwarf cannot make its own. The biosynthesis gene is the faulty one. Answer D.

Question 12

What speeds up transpiration?

Transpiration is evaporation through stomata.
Wind, bright light and high temperature all speed that evaporation.
Humid air slows it, because the air is already holding water.
The option that lists the three promoters, and not humidity, is A.

Question 13

A root growing downward toward water

Tropism is a growth response with a direction.
Growth toward water is positive hydrotropism.
That is the advantage: the root is more likely to find water. Answer C.

Question 14

How a microtubule drug stops a tumour cell dividing

Microtubules are the spindle.
Without a spindle, chromosomes cannot line up at metaphase and sister chromatids cannot be pulled apart at anaphase.
The cell does not finish division. That is B. Do not pick a vague 'it poisons metabolism' option.

Question 15

Other reasons the same drug hurts a tumour

Cancer cells crawl on microtubules. Blocking that limits invasion.
New blood vessels also need cell movement. Blocking it limits the tumour's blood supply.
The option that includes movement and vessel growth, not only the spindle, is D.

Question 16

Steroid hormones versus insulin

Insulin and glucagon are peptides. They bind a surface receptor and do not enter the cell.
Steroid hormones cross the membrane, bind inside, and change which genes are transcribed. Making new protein takes longer than switching an enzyme on.
The option that states both of those is C.

Question 17

A histone and long-term memory

Histones decide which stretches of DNA are open for transcription. They do not change the DNA letters.
H2BE is linked to keeping synaptic genes available, so the connection can stay strengthened.
Answer A.

Question 18

The wrong explanation of a histone knockout

The question asks for the incorrect statement.
Knocking out H2BE changes how tightly DNA is packed. It does not delete the genes for the synapses.
Lower synaptic activity comes from those genes being harder to read, not from the DNA being cut out.
That wrong explanation is C.

Question 19

Getting a drug out of a clathrin coat

Clathrin coats the vesicle that brings the drug in. The coat then has to come off.
Until it comes off, the vesicle cannot fuse and release the drug into the cell.
Speeding up uncoating releases more drug. Answer A.

Question 20

Blocking entry of a coronavirus

Many viruses, including SARS-CoV-2, enter through the cell's own endocytosis machinery.
If that route is blocked, the virus does not reach the cytoplasm to copy itself.
The route is the host's, so it stays available even when the viral surface protein changes. Answer C.

Question 21

The serious result of horizontal gene transfer

HGT adds genes. It does not reduce bacterial diversity.
The genes can move between species that are not close relatives.
The consequence that matters in the clinic is antibiotic resistance jumping to a new species. Answer C.

Question 22

What a bacteriophage can and cannot do

A phage sticks to a matching site on the bacterial surface.
When new phages are built they sometimes package a piece of bacterial DNA and carry it to the next cell. That is transduction.
A phage cannot copy itself with its own machinery. It uses the bacterium's. Any option that says the phage replicates alone is wrong.
The accurate statement is B.

Question 23

The incorrect sentence about quorum sensing

Bacteria make the signal all the time. Group behaviour starts only when the signal passes a threshold, which means the crowd is dense enough.
'As soon as any autoinducer is detected, they switch' is too early. That sentence is the incorrect one.
Answer A.

Question 24

Ways to quench quorum sensing

Quorum quenching means stopping the crowd behaviour, not killing every cell.
You can remove the proteins that carry out the group behaviour, block or destroy the receptor, or add a mimic that jams the receptor.
Blocking the receptor that detects the external signal is also quenching.
The option that lists real blocks of the signal, rather than 'add more signal', is B.

Question 25

Tolerant bacteria versus susceptible ones

Tolerant cells need a longer exposure to die to the same extent.
A normal antibiotic course often fails to clear tolerant and persistent cells.
Persistent populations are mixed: some cells die as easily as susceptible cells, others survive like tolerant cells.
The option that includes that mix, rather than 'every persistent cell is invincible', is D.

Question 26

Seal pups: begging versus practising foraging

Under 45 days, pups beg for 40–60% of the time. Older pups beg no more than 14%. Light and heavy pups both do this. (I) is in the data.
(IV) Heavier young pups beg less and forage more than lighter young pups. The preamble says practice builds skill, so their higher foraging later is a fair inference.
(II) Foraging time does rise with age, but heavier young pups already forage more than lighter ones. A statement that body condition does not matter at that age is false.
(III) Food obtained by the young pups is not tied to body condition in the data. Do not infer it.
I and IV. Answer B.

Question 27

A salmon smell experiment: which controls are real?

(I) is the wrong reason. Swapping which arm gets the odour each day stops the fish choosing a favourite arm. It is not about something else.
(II) Leftover chemical would contaminate the next trial, especially in the odour group. The chamber water has to be changed. True.
(III) The test is the odour. Sex and size must stay the same. True.
II and III. Answer C.

Question 28

Spines on nightshade: which claim is not supported?

The question is which statement cannot be inferred. B is that one.
A is supported: the three species differ in spine mass.
C is supported: pull the spines off and caterpillars finish the leaves faster, because they can move from leaf to leaf.
D is supported: the two spiny species change a lot when spines are removed; the species with few spines changes little.
B generalises a result that only holds for two of the three species. That is the one to reject.

Question 29

Mouth-brooding: what the delays show

(I) Non-caring females spawn again after 14 days. Caring females, once the brood is removed, show the same gap. The extra 10 days of care matches the extra wait. They are not mating while the young are in the mouth. True.
(II) Caring females invest less in the second clutch (lower weight). Non-caring females invest more the second time. Care costs energy. True.
(III) A heavier brood could be bigger eggs, more eggs, or both. Size and number are not given. False.
(IV) Nobody measured feeding. False.
I and II. Answer C.

Question 30

IL-11 and how long mice live

(I) Compare the gap between treated and control curves in females versus males after week 120. The female gap is wider. True.
(II) There is no sharp survival drop between weeks 40 and 60. False.
(III) On the pooled graph, 50% survival is about 120 weeks for wild type and about 150 for the deletion. That rise is about 25%. True.
(IV) The mutant curve itself does not crash between weeks 40 and 60. False.
I and III. Answer A.

Question 31

Which follow-up would copy the IL-11 result?

The result to copy is: less IL-11 signalling, longer survival.
A receptor blocker reduces signalling. That is A.
Options that add IL-11, or boost the receptor, do the opposite. Reject them.

Question 32

Melatonin and bone

Melatonin inhibits osteoclasts (bone removal) and promotes osteoblasts (bone building). It is made in the dark.
A long day means less melatonin and less building. A is the wrong direction if it predicts stronger bone.
More osteoblast activity takes calcium out of the blood and into bone. Blood calcium would fall, not rise. B fails.
Hours of sleep are not the same variable as darkness and melatonin. C is not supported.
A melatonin supplement would favour building and slow osteoporosis. Answer D.

Question 33

Spirometry loops: restrictive versus obstructive

The grey loop is normal.
A loop with the same shape but a smaller volume is restrictive: the lung holds less. That is the red line in one of the panels.
A loop that still reaches a large volume but breathes out slowly is obstructive. The problem is flow, not the total volume.
The option that matches those two red loops to the right diseases is A. The others swap them, or call obstruction a volume problem.

Question 34

Diarrhoea and bicarbonate

Bicarbonate is lost, so the blood becomes more acidic.
The lungs compensate by faster breathing, which blows off CO2.
The kidneys compensate by keeping bicarbonate and excreting more H+.
The option that includes both compensations is D.

Question 35

An athlete breathing too fast

Rapid breathing washes out CO2.
Less CO2 means less carbonic acid.
The blood becomes more alkaline. Answer B.
Do not pick 'the blood becomes more acidic'. That is the diarrhoea question, not this one.

Question 36

The second heart sound

The second heart sound is the semilunar valves closing. Point X is that moment: the start of isovolumetric relaxation.
Those valves shut so blood does not fall back from the aorta and the pulmonary artery.
The atrioventricular valves are still shut while the atria are beginning to fill.
Answer C.

Question 37

Why atrial pressure creeps up during ejection

During ventricular ejection the ventricle is emptying, but the atria are relaxed.
Veins keep delivering blood into the closed atria (venae cavae, pulmonary veins).
Atrial pressure therefore rises. Answer A.

Question 38

ABC model: what happens when C is removed

Normal flower, from outside in: sepals (A), petals (A+B), stamens (B+C), carpels (C).
Without C the flower makes sepals, and also petals in a region that should have been further in. A activity has spread into the zone C used to occupy.
So C normally stops A from being expressed there.
A, B and C as worded cannot explain both the normal petals or stamens and this mutant. Answer D.

Question 39

Two unlinked genes, incomplete dominance, how many F2 looks?

The parents are homozygous mutants at two different unlinked genes. Every F1 fly is heterozygous at both, so the F1 all look the same.
Incomplete dominance means three phenotypes per gene: +/+, +/m, m/m.
Two genes assort independently: 3 eye phenotypes × 3 wing phenotypes = 9 combinations.
Answer B.

Question 40

Is this virus DNA or RNA, single or double stranded?

A small mismatch in the percentages is normal measurement error. Do not call it single-stranded because of a 1% gap.
A ≈ T and C ≈ G, and purines ≈ pyrimidines ≈ 50%. That is base pairing: double stranded.
T is about 34%. RNA uses U, not T. A trace of U is contamination.
Double-stranded DNA. Answer C.

Question 41

Why one ethnic group shows up in a marathon final

New variants appear at random. Sport does not write them into the DNA. A and B fail.
One Olympic final is not generations of survival selection. It does not change allele frequencies in the population. D fails.
The environment those ancestors lived in selected the variants over a very long time. The Games are one sample of people who already carry them. Answer C.

Question 42

Ordering three arginine enzymes

None of the mutants grows on glutamate alone, and all three fail to make arginine. Glutamate is the starting point and every enzyme is needed after it.
Arg-1 still fails when citrulline or ornithine is given. It must catalyse the last step, after those compounds. Only arginine itself would rescue it.
Arg-2 fails on glutamate and on ornithine, but is rescued once citrulline is provided. It makes citrulline.
Arg-3 is rescued by ornithine, so it makes ornithine.
Order: glutamate → ornithine (Arg-3) → citrulline (Arg-2) → arginine (Arg-1). Answer B.

Question 43

A rare disease pedigree, and a probability

Test X-linked recessive first. An affected father would pass his only X to every daughter and the condition would show in the males the key mentions. The pedigree does not do that. Not A.
Autosomal recessive and rare: you would need carriers marrying in at every generation. That is unlikely. Not B.
Autosomal dominant fits: the affected people are heterozygotes and the trait passes down the affected line.
III-2 is heterozygous, so the next child has a 1/2 chance of inheriting the allele. Chance that child is a boy is 1/2. Both: 1/4. Answer D, not the option that stops at 1/2.

Question 44

Three Paramecium species alone and together

Alone, all three grow well. The medium is not the problem.
Together, P. aurelia stays highest but grows less than it did alone. P. caudatum crashes. P. bursaria hangs on at a lower level.
That is competition: one species suffers most, one dominates, one coexists at a reduced number.
Answer B.

Question 45

Opossum shrimp in Flathead Lake

Read the arrows. Piscivores eat fish. Planktivores (kokanee, lake trout) eat zooplankton. Herbivores eat the algae.
Opossum shrimp eat the same zooplankton as the kokanee, so zooplankton fall.
The option that follows that competition, rather than 'the shrimp feed the salmon', is D.
If an option says kokanee increase because food is suddenly abundant, check it against the shared zooplankton. The shrimp are taking that food.

Question 46

Diversity is not the same as evenness

Richness: how many species. Habitat B has 7, habitat A has 5. B is richer.
Evenness: whether the individuals are spread fairly across species. A is more even. B has a few common species and several rare ones.
A more even spread is the more balanced community, even with fewer species.
The option that separates richness from evenness is C.

Question 47

An introduced mud snail

The two snails meet through a shared food, the diatoms. Batillaria grazes diatoms faster.
Both suffer parasites and density-independent death. Cerithidea suffers more of both.
The invader can therefore push the native snail down and change how many diatoms are left.
Answer D.

Question 48

Energy from beetles to toads

Use the toad's own budget, not a memorised 10% unless the numbers give 10%.
Ingested 10,000. Heat loss 7,000. Waste 2,000. Left for growth: 1,000.
1,000 / 10,000 = 10%. Answer A.

Question 49

What trophic level is the cane toad?

Plants are level 1. Beetles that eat the plants are level 2 (primary consumers).
A toad that eats only those beetles is one step further: level 3, a secondary consumer.
Answer C. Do not count the toad as level 2 just because it is the predator in the story.

Question 50

Choosing a virus to control cane toads

It has to infect and kill the toad, or the population will not fall.
It also has to spare humans and the other animals you did not mean to hit.
A virus that is lethal but sloppy, or safe but harmless to the toad, fails.
Answer B.