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IB Biology · full-length mock · 2 hours

IB Biology Paper 1 mock. The whole syllabus, at the real length.

A complete Paper 1 (Higher level) built from the 2025 guide: 75 marks across 44 questions, in 2 hours. It follows the real paper's structure and rubric below, and every question is original and tagged to the syllabus. Sit it on screen against the clock and mark yourself, print it for a class, or build a fresh one with nothing repeated.

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2 hours · 75 marks · 36% of the final grade.

  1. Paper 1A40 marks 40 multiple-choice questions. Answer all. One best answer per question; no marks lost for wrong answers.
  2. Paper 1B35 marks 4 data-based questions. Longer than SL, 8 to 10 marks each, and drawing on additional higher level content.

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The 40 multiple-choice questions are spread across all 40 subtopics in proportion to each theme, and about 30% of them come from additional higher level material, the AHL share of teaching hours. The four data-based questions sit in the 8 to 10 mark range and are chosen to sum as close to 35 as the bank allows.

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Biology · Paper 1
Higher level · mock examination · Paper 1A multiple choice and Paper 1B data-based
120 minutes75 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers Whole syllabus: all 40 subtopics
  • Do not open this examination paper until instructed to do so.
  • Answer all questions.
  • A calculator is required for this paper.
  • Answers must be written within the answer boxes provided.
Name:
Paper 1A · Multiple choice

Answer all questions. Choose the one best answer for each question.

1.

The table shows the boiling points of water and of three other compounds whose molecules are of similar size. Which is the best explanation for the high boiling point of water?

[1]
CompoundFormulaRelative molecular massBoiling point / °C
waterH₂O18100
ammoniaNH₃17−33
methaneCH₄16−162
hydrogen sulfideH₂S34−60
  1. Boiling point rises with relative molecular mass, and water has the heaviest molecules of the four
  2. Oxygen shares electrons equally with hydrogen, so water molecules pack more tightly than the others
  3. Hydrogen bonds between polar water molecules must be broken before the molecules can escape as gas
  4. More energy is needed to break the covalent bonds within water molecules than within the others
2.

How is the presence of large amounts of liquid water on the early Earth commonly explained?

[1]
  1. Water formed from the burning of methane
  2. The oceans have always existed unchanged
  3. Water was delivered by asteroids and comets
  4. Water was made by early photosynthetic cells
3.

During transpiration, water is pulled up through xylem vessels as a continuous column under tension. Which property of water prevents the column from breaking?

[1]
whole plant water lost from the leaves a continuous column of water, under tension from root to leaf one xylem vessel
  1. The high specific heat capacity of water, which prevents it from vaporizing
  2. Adhesion of water molecules to the lignin of the vessel walls
  3. The low viscosity of water, which allows it to flow through narrow vessels
  4. Cohesion between water molecules resulting from hydrogen bonding
4.

Which statements about DNA are correct?

  • I. It is (usually) a double helix.
  • II. It contains the base uracil.
  • III. It has a sugar–phosphate backbone.
[1]
  1. I and II only
  2. I and III only
  3. II and III only
  4. I, II and III
5.

A student drew a diagram of a cell from an electron micrograph and added the annotation shown for one structure. Which annotation follows the convention of including a function?

[1]
  1. Rough endoplasmic reticulum: synthesizes proteins for secretion
  2. Cell wall: rigid outermost layer surrounding the plant cell
  3. Golgi apparatus: flattened membrane sacs near the nucleus
  4. Mitochondrion: approximately 1 μm wide, oval in section
6.

Cheese is made using cultures of the bacterium Lactococcus lactis. A bacteriophage that infects L. lactis has a lytic cycle lasting 40 minutes from attachment to lysis, and each infected cell releases 50 new phages. A single phage enters a culture containing a very large number of the bacteria. Assuming that every phage released infects a new cell straight away, how many phages are present 160 minutes after the first infection?

[1]
  1. 2 500
  2. 6.25 × 10⁶
  3. 200
  4. 1.25 × 10⁵
7.

The diagram compares the forelimb bones of four mammals. What do these limbs provide evidence for?

[1]
human armbat wingwhale flipperhorse foreleghumerusradius and ulnacarpalsmetacarpal and phalangesForelimb skeletons of four mammals, not drawn to the same scale
  1. Structures that evolved independently in each of the four groups
  2. Analogous structures produced by the same selection pressures
  3. Common ancestry, because the same bones occur in the same arrangement
  4. Convergent evolution, because the limbs carry out different functions
8.

The tree shows the evolutionary relationships of five lizard species, P to T, drawn to a time scale. All five species are classified as Critically Endangered, and a conservation charity can fund a programme for only one of them. According to the rationale of the EDGE of Existence programme, which species should be chosen?

[1]
Species P Species Q Species R Species S Species T 100 80 60 40 20 0 Time / millions of years ago (0 = present day)
  1. Any of the five, because all five are Critically Endangered, so their EDGE priority must be equal
  2. Species T, because it has not shared a common ancestor with any of the other four species for 100 million years
  3. Species Q, because it belongs to the largest group of related species, so saving it protects genes shared by many species
  4. Species R, because it belongs to the most recently formed pair of species, so it is the most highly evolved of the five
9.

Which is an indirectly anthropogenic cause of the loss of mixed dipterocarp forest in Southeast Asia?

[1]
  1. Burning of felled trees and undergrowth by local farmers to clear the ground for crops, farms and settlements
  2. Felling of the largest dipterocarp trees by logging companies for export overseas as tropical hardwood timber
  3. Fires that spread further through the forest during the more severe droughts linked to global climate change
  4. Clearing of forest land by plantation companies and planting it in rows with oil palm for palm oil production
10.

Starch is digested to glucose in the gut. What type of reaction is this, and what is consumed?

[1]
  1. Oxidation, consuming oxygen
  2. Condensation, consuming water
  3. Hydrolysis, consuming water
  4. Condensation, releasing water
11.

Which of the following is NOT considered an organelle?

[1]
  1. A mitochondrion
  2. A ribosome
  3. A lysosome
  4. The cytoplasm
12.

The figure shows two cells, X and Y, drawn as rectangular blocks. Both cells have a volume of 64 µm³. What is the surface area-to-volume ratio of cell Y?

[1]
cell X cell Y 4 µm 4 µm 4 µm 8 µm 1 µm 8 µm both drawn to the same scale
  1. 1.25 : 1
  2. 1.5 : 1
  3. 2.0 : 1
  4. 2.5 : 1
13.

Stem cells in the deepest layer of the skin divide throughout life. Millions of new specialized skin cells are produced every day, yet the number of stem cells in the skin stays about the same. Which explanation accounts for this?

[1]
  1. Each stem cell divides only once and is then replaced by a new stem cell carried in the blood
  2. Specialized skin cells divide to form new stem cells as quickly as the stem cells are used up
  3. On average, one daughter cell of each division stays a stem cell and the other differentiates
  4. Both daughter cells of every division differentiate, but the stem cells divide very rapidly
14.

How is the concentration gradient of oxygen across the gill surface of a bony fish maintained?

[1]
  1. The fish holds water in its mouth until the oxygen concentrations in water and blood become equal, then releases it
  2. Oxygen is stored in the gill tissue during periods of rest and released into the blood during activity
  3. Cells of the gill surface actively transport oxygen from the water into the blood using ATP
  4. Water is continuously pumped over the gills while blood flows continuously through the gill capillaries
15.

An inflatable cuff round a volunteer's upper arm was inflated just enough to compress the veins of the arm but not its arteries. The blood pressure in skin capillaries of the hand on that arm was measured before and during inflation.
What do the data predict about tissue fluid in the hand while the cuff is inflated?

[1]
CuffBlood pressure at arteriole end of capillaries / kPaBlood pressure at venule end of capillaries / kPa
deflated4.41.6
inflated5.24.6
  1. Tissue fluid builds up, as the higher venule-end pressure lets less of it drain back
  2. Tissue fluid returns faster, as higher venule-end pressure draws it back into the blood
  3. Tissue fluid stops forming, as compressing the veins halts blood flow in the capillaries
  4. Tissue fluid is unchanged, as the pressure at the arteriole end changes only slightly
16.

During contraction of a sarcomere, which of the following shorten?

  • I. The sarcomere (Z-disc to Z-disc)
  • II. The individual actin and myosin filaments
  • III. The light band (region of actin only)
[1]
one sarcomere of a relaxed muscle fibre, seen from the side Z-disc Z-disc thin filament thick filament sarcomere light band dark band light band
  1. I and II only
  2. I and III only
  3. II and III only
  4. I, II and III
17.

The forelimb of a whale contains the same bones as a human arm, but it forms a flipper. Which set of changes adapts the forelimb as a flipper for swimming?

[1]
  1. Loss of all the arm bones, leaving a flexible fin of skin supported by thin rays
  2. Short arm bones, long fingers enclosed in one paddle, and an elbow that barely bends
  3. Extra joints at the elbow and wrist so that the limb can rotate freely in all directions
  4. Long, mobile arm bones with separate webbed fingers that can spread to grip prey
18.

Relaxed muscle fibres were stretched and the force resisting the stretch was measured. The fibres were then treated with an enzyme that cuts titin molecules but leaves actin and myosin intact. When stretched again, the treated fibres offered far less resistance. What would be expected in a living muscle with its titin damaged in this way, when its antagonistic muscle contracted strongly?

[1]
  1. Its sarcomeres could be overstretched until actin and myosin no longer overlap
  2. Its myosin filaments would be stretched so that they became longer than normal
  3. It would contract to resist the pull, as titin normally stops it from contracting
  4. It could not relax, because titin normally detaches the myosin heads from actin
19.

Caterpillars of two moth species, V and W, were reared either on leaves of a plant variety that produces a toxic alkaloid or on leaves of a variety that produces none. The activity of a detoxifying enzyme in the gut of each species was also measured.

Which conclusion is best supported by the data?

[1]
SpeciesSurvival on alkaloid-free leaves / %Survival on leaves with alkaloid / %Gut detoxifying enzyme activity / arbitrary units
V888542
W90123
  1. The alkaloid gives the plant equal protection against both moth species
  2. Species W survives on leaves with alkaloid by avoiding eating the leaves
  3. The alkaloid is a physical defence that stops both species biting the leaves
  4. Species V has a metabolic adaptation for feeding on the defended plant
20.

Cristae greatly increase the surface area of the inner mitochondrial membrane. Which process benefits directly from this increase?

[1]
matrix outer membrane inner membrane, folded into cristae space between the two membranes a mitochondrion in section
  1. Oxidative phosphorylation, because more electron transport chains and ATP synthase complexes can be embedded
  2. The link reaction, because pyruvate is decarboxylated on the inner membrane before entering the matrix
  3. Glycolysis, because more glucose can be bound to the membrane surface and broken down to pyruvate
  4. The Krebs cycle, because its enzymes are attached to the cristae and more can be held in a larger membrane
21.

In pyruvate dehydrogenase deficiency, the enzyme complex that catalyses the link reaction does not work, and pyruvate and lactate build up in the tissues. Patients are often given a diet that is high in fat and very low in carbohydrate. Why does this diet help?

[1]
  1. Fats can be respired anaerobically, so ATP can be made without any pyruvate being used
  2. Fatty acids are broken down to acetyl groups, which coenzyme A carries into the Krebs cycle
  3. Fatty acids are converted into glucose, which can then be respired to pyruvate in glycolysis
  4. With less carbohydrate, pyruvate enters the Krebs cycle directly instead of through the link reaction
22.

During photosynthesis, carbon dioxide is converted to glucose. What is the source of the hydrogen needed for this conversion?

[1]
  1. Atmospheric hydrogen gas absorbed by the leaves
  2. The breakdown of chlorophyll molecules
  3. Glucose stored in the chloroplast
  4. The splitting of water by photolysis
23.

Epinephrine (adrenaline) causes liver cells to break down glycogen, the smooth muscle around arterioles supplying the gut to contract, and the smooth muscle around arterioles in skeletal muscles to relax. What explains how one signalling chemical can cause these different responses?

[1]
  1. Each tissue responds in its own way to the concentration of epinephrine that reaches it in the blood
  2. Epinephrine is converted into a different signalling chemical in each tissue before it binds to a cell
  3. The cells have different receptors for epinephrine, linked to different signal transduction pathways
  4. Epinephrine enters each cell and binds to different genes in the nucleus of each type of cell
24.

Which sequence of events occurs when an action potential arrives at a cholinergic synapse?

[1]
a synapse, in section presynaptic neuron postsynaptic neuron cleft vesicles receptors in the membrane
  1. Acetylcholine is pumped into the presynaptic cell → receptors close → K⁺ enters
  2. Na⁺ is released into the cleft → binds receptors → depolarizes the membrane
  3. Acetylcholine is released by the postsynaptic cell → diffuses backwards → blocks the axon
  4. Ca²⁺ channels open → vesicles release acetylcholine → postsynaptic ligand-gated channels open
25.

Local anaesthetics such as lidocaine block voltage-gated sodium channels in sensory neurons. Why does this stop pain?

[1]
  1. Potassium ions flood into the neuron and block depolarization
  2. Action potentials cannot be initiated or propagated without Na⁺ influx
  3. The resting potential of the sensory neuron is abolished
  4. The anaesthetic destroys the pain receptors in the sensory endings
26.

Where are synapses found?

[1]
  1. Between two neurons and between a neuron and an effector cell such as a muscle fibre
  2. Between adjacent muscle fibres, so that they contract together as one unit
  3. Only between sensory receptors and the sensory neurons that carry impulses from them to the CNS
  4. Only between the dendrites of one neuron and the axon terminal of another neuron in the brain
27.

The diagram shows a transverse section through part of a nerve, drawn from a light micrograph. Which row identifies the structures labelled X, Y and Z?

[1]
RowXYZ
A.myelin sheathprotective sheath of the nerveunmyelinated nerve fibre
B.protective sheath of the nervemyelin sheathunmyelinated nerve fibre
C.protective sheath of the nervemyelin sheathcell body of a motor neuron
D.protective sheath of the nervecell body of a neuronmyelinated nerve fibre
X Y Z
  1. Row B
  2. Row D
  3. Row A
  4. Row C
28.

Which organisms in the tundra food web are secondary consumers?

[1]
lichensgrasses and sedgeslemmingarctic harecaribousnowy owlarctic foxwolf
  1. Lichens, grasses and sedges
  2. Lemming, arctic hare and caribou
  3. Snowy owl, arctic fox and wolf
  4. Snowy owl, lemming and lichens
29.

Protein R is made in human cells grown in culture. Three groups of cells were given different treatments at time 0, and the amount of protein R in each group was measured over the next four hours. Which conclusion is supported by the results in the graph?

[1]
01234020406080100120time after treatment / hamount of protein R / % of time 0 no treatmentprotein synthesis inhibitorprotein synthesis inhibitor + proteasome inhibitor
  1. Protein R is broken down fastest in the cells whose proteasomes are inhibited
  2. Protein R is broken down mainly by enzymes other than those in proteasomes
  3. Protein R is broken down by proteasomes and continually replaced by synthesis
  4. The proteasome inhibitor makes the cells synthesize protein R more quickly
30.

In which stage are sister chromatids being separated?

[1]
CEAFBDSix stages of meiosis in a cell with 2n = 4, not in order. Maternal chromosomes green (solid), paternal pink (dashed).
  1. B
  2. A
  3. E
  4. C
31.

How does the rate at which an mRNA molecule is degraded affect gene expression?

[1]
  1. A short-lived mRNA is translated more quickly, so more of the protein is made
  2. A short-lived mRNA is translated fewer times, so less of the protein is made
  3. A long-lived mRNA is translated only once, because ribosomes cannot reuse it
  4. The lifetime of an mRNA does not affect how much protein is made from it
32.

Water moves from regions of higher water potential to regions of lower water potential. Which statement explains this movement in terms of energy?

[1]
  1. Solute molecules chemically attract water molecules across the membrane
  2. Water tends to move so that its potential energy decreases, as such changes occur spontaneously
  3. Water molecules gain potential energy as they move, making the movement irreversible
  4. Water at lower water potential has higher kinetic energy, drawing molecules towards it
33.

The figure shows the chromosomes from one dividing cell of a Tasmanian devil (Sarcophilus harrisii), arranged in numbered pairs. Sex is determined in Tasmanian devils in the same way as in humans. What can be concluded about the animal that the cell came from, and why?

[1]
1234567Chromosomes from one dividing cell, arranged in numbered pairs
  1. Male, since the two chromosomes that make up pair 7 are very different in size from each other
  2. Female, since every one of the fourteen chromosomes in the cell can be placed in one of seven pairs
  3. Female, since pair 7 includes a chromosome that is about as long as the chromosomes of pair 6
  4. Male, since the cell holds fourteen chromosomes, which is the diploid number for this species
34.

The pedigree chart shows a rare condition caused by a dominant allele. Which observation shows that the allele is on an autosome rather than on the X chromosome?

[1]
IIIIIII-1I-2II-1II-2II-3II-4II-5III-1III-2III-3III-4III-5unaffectedaffectedsquares male, circles female
  1. Both males and females are affected by the condition
  2. The affected parents II-4 and II-5 have an affected son, III-4
  3. The condition appears in every generation of the family
  4. The affected male I-1 has an affected son, II-4
35.

Athletes are advised not to begin vigorous physical activity soon after eating a large meal. Which is the best explanation, based on how blood supply to organs changes during activity?

[1]
  1. Cardiac output falls, so every organ receives less blood than at rest
  2. Blood flow to the brain falls, so the muscles are poorly coordinated
  3. Blood flow to the gut falls, so food is digested and absorbed slowly
  4. Digestive enzymes are denatured as the core body temperature rises
36.

In Endler's experiments, guppies were kept in ponds with or without predatory fish. In the ponds without predators, male guppies became more brightly coloured over several generations. Which conclusion is supported by this observation?

[1]
the pools used in Endler's guppy experiment pool with predators with predatory fish pool without predators no predators guppies (small, green) were moved between the pools and followed for several generations
  1. Sexual selection by females favoured brightly coloured males
  2. Mutation rates increased in the absence of predatory fish
  3. Predation by other fish favoured duller, camouflaged males
  4. Bright colours were induced directly by the gravel in the pond
37.

In evolutionary biology, what does the "fitness" of an individual measure?

[1]
  1. Its physical strength and stamina
  2. Its contribution of offspring
  3. Its total lifespan
  4. Its overall body size
38.

The diagram shows the concentration of DDT in a lake food chain. By approximately what factor is DDT more concentrated in the osprey than in the phytoplankton?

[1]
fish-eating bird (osprey)25.0ppm DDTtertiary consumerlarge fish (pike)2.1ppm DDTsecondary consumersmall fish (minnow)0.5ppm DDTprimary/secondary consumerzooplankton0.04ppm DDTprimary consumerphytoplankton0.005ppm DDTproducerlake water0.00002ppm DDTMean DDT concentration in the tissues of organisms from one lake, parts per million (ppm)
  1. 5
  2. 50
  3. 500
  4. 5000
39.

Which changes are expected as primary succession proceeds on newly exposed rock? I. Species diversity increases II. Primary production increases III. Food webs become more complex

[1]
  1. I and II only
  2. I and III only
  3. II and III only
  4. I, II and III
40.

After heavy nutrient input, a shallow lake flips from clear water with rooted plants to turbid, algae-dominated water, and stays turbid even when nutrient input falls. What does this persistence illustrate?

[1]
the same shallow lake in two states, drawn in cross-section sunlight sunlight rooted plants lake bed algae lake bed clear state turbid state
  1. An alternative stable state with hysteresis
  2. A simple linear response to nutrients
  3. Negative feedback restoring the old state
  4. Immediate and complete recovery
Paper 1B · Data-based questions

Answer all questions. Answers must be written within the answer boxes provided.

41.

Most cells of a strain of the bacterium Escherichia coli are killed by a virus that infects bacteria, but a very few cells in a large culture are resistant to it, and their descendants are resistant too. Two hypotheses were proposed for how resistant cells arise. Hypothesis 1: when the cells meet the virus, contact with it causes a resistance mutation in a small proportion of them. Hypothesis 2: resistance mutations arise at random while the cells are growing and dividing, before they meet the virus, and each one is passed on to all the descendants of the cell in which it arose. A classic experiment from the 1940s tested these hypotheses. In set 1, ten small cultures were each started from about 100 cells and grown separately in identical conditions until each held about 2 × 10⁸ cells. Each whole culture was then spread on its own plate of agar containing the virus. In set 2, one large culture was grown in the same conditions to the same density, and ten samples, each the same volume as one small culture, were taken from it and spread on ten more plates containing the virus. Only resistant cells can grow into colonies on these plates. The table shows the number of colonies on each plate.

PlateSet 1: colonies from one separate small cultureSet 2: colonies from one sample of the large culture
109
235
3011
417
5584
608
726
8010
9117
1016
(a)Calculate the mean number of colonies per plate for set 1.[1]
(b)Compare the results of set 1 with the results of set 2.[2]
(c)Explain why the researchers included set 2 in this investigation.[2]
(d)Explain which hypothesis the results of set 1 support.[3]
(e)Outline how results like these support the current understanding of mutation.[2]
42.

Researchers measured glucose uptake by cultured fat cells (adipocytes) exposed to increasing insulin concentrations for 30 minutes. A second set of cells was first treated with an antibody that binds to the insulin receptor's extracellular domain, blocking the hormone-binding site. Uptake is shown relative to untreated cells with no insulin (= 1.0).

Insulin concentration / nmol dm⁻³Relative glucose uptake (no antibody)Relative glucose uptake (+ receptor antibody)
01.01.0
0.11.61.0
13.11.1
104.81.2
1005.01.3
(a)Describe the effect of insulin concentration on glucose uptake in cells without the antibody.[2]
(b)Calculate the percentage increase in glucose uptake caused by 10 nmol dm⁻³ insulin in cells without the antibody, compared with no insulin.[1]
(c)Explain why glucose uptake plateaus at high insulin concentrations.[2]
(d)Explain the results for the cells treated with the receptor antibody.[2]
(e)State the general term for a chemical that binds specifically to a receptor and triggers its response.[1]
43.

Beetroot cells contain a red pigment inside the vacuole. A student hypothesises that increasing the concentration of ethanol (an organic solvent) in the surrounding water increases the permeability of beetroot cell membranes. More pigment released gives a higher absorbance reading.

(a)Design an investigation to test this hypothesis. Include the independent, dependent and controlled variables and a suitable control.[6]
(b)Explain, in terms of membrane structure, why a high concentration of ethanol is expected to increase pigment release.[2]
44.

Each year a surveillance laboratory sequences the gene for haemagglutinin, a surface protein of influenza A, in viruses collected from patients. The number of amino acid differences between the circulating virus and the strain used in that year's vaccine was recorded, together with the effectiveness of the vaccine measured in the same season.

SeasonAmino acid differences from the vaccine strainVaccine effectiveness / %
1261
2452
3928
41219
5357
(a)Describe the relationship between the number of amino acid differences and vaccine effectiveness.[2]
(b)Estimate the vaccine effectiveness expected in a season with 6 amino acid differences.[1]
(c)Suggest how the haemagglutinin of the circulating virus comes to differ from that of the vaccine strain.[2]
(d)Identify two variables, other than the vaccine strain, that could affect the measured effectiveness of a vaccine.[2]
1.
2.
(e)Suggest how surveillance data of this kind are used when choosing the strain for next season's vaccine.[2]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. A1.1.2 [1]
  • C: unequal sharing of electrons makes water molecules polar, so hydrogen bonds form between neighbouring molecules and much energy is needed to separate them; boiling separates molecules and does not break covalent bonds (A); hydrogen sulfide is heavier but boils at −60 °C (B); the sharing is unequal (D);
2. A1.1.7 [1]
  • C: bombardment by icy bodies plus outgassing supplied water; Earth's orbit (Goldilocks zone) and gravity let it persist as liquid;
3. A1.1.3 [1]
  • D: hydrogen bonds between water molecules hold the column together under tension; A adhesion helps water cling to the walls but does not hold water molecules to each other; C heat capacity affects temperature change, not the ability to withstand tension; D low viscosity eases flow but does nothing to stop the column separating;
4. A1.2.3, A1.2.6, A1.2.7 [1]
  • B: uracil occurs in RNA; DNA contains thymine;
5. A2.2.11 [1]
  • A: an annotation must state the structure's function. Only this option pairs the identification with what the structure does (protein synthesis for secretion); the others describe appearance or position only;
6. A2.3.3 [1]
  • B: 160 minutes allows four lytic cycles, and each cycle multiplies the number of phages by 50, so 50⁴ = 6 250 000 = 6.25 × 10⁶ (which is why one phage can destroy a starter culture within hours); A adds 50 phages per cycle (50 × 4), as if only one cell were infected each time; B is the number after two cycles; C is the number after three cycles;
7. A4.1.4 [1]
  • C: the bones correspond one to one in the same arrangement despite serving different functions, which is best explained by inheritance from a common ancestor with a pentadactyl limb;
8. A4.2.8 [1]
  • B: T has evolved separately from the other four species for 100 million years, so it is the most evolutionarily distinct; its loss would remove the most unique evolutionary history, and when endangerment is equal the more distinct species has the higher EDGE priority; B fails because all living species have been evolving for the same length of time, and a recent split means R has a close relative (S); C fails because genes that Q shares with its relatives would survive in them; D fails because EDGE combines endangerment with evolutionary distinctiveness, which differs between these species;
9. A4.2.4 [1]
  • C: the droughts are made more severe by climate change, which is caused by human greenhouse-gas emissions far from the forest, so the fire damage is only indirectly anthropogenic; B, C and D are direct human actions that remove the forest where they take place;
10. B1.1.3 [1]
  • C: glycosidic bonds are broken by hydrolysis, with one water molecule used per bond;
11. B2.2.1 [1]
  • D: organelles are discrete subcellular structures with specific functions; the cytoplasm (and the cell wall) are not organelles;
12. B2.3.7 [1]
  • D: surface area of Y = 2 × (8 × 8) + 4 × (8 × 1) = 128 + 32 = 160 µm², and 160 ÷ 64 = 2.5 : 1, compared with 96 ÷ 64 = 1.5 : 1 for cube X, so flattening raises the ratio without changing the volume; A counts only one face of each pair (80 µm²); B is the ratio of cube X; C counts only the two large faces (128 µm²);
13. B2.3.2 [1]
  • C: stem cells can divide repeatedly while keeping a stem cell population (self-renewal) and can also differentiate, so if on average one daughter remains a stem cell and the other differentiates, the number of stem cells stays constant while specialized cells are produced; A fails because specialized skin cells do not normally become stem cells again; B fails because stem cells divide many times, not once; C fails because if both daughters always differentiated, the stem cells would soon be used up however fast they divided;
14. B3.1.3 [1]
  • D: ventilation with water and a continuous flow of blood keep bringing oxygen-rich water to the surface and carrying oxygenated blood away, so the gradient is never lost; A would abolish the gradient rather than maintain it; B fails because oxygen crosses by diffusion, not active transport; D fails because gills do not store oxygen;
15. B3.2.11 [1]
  • A: fluid is still forced out at the arteriole end (pressure 5.2 kPa, higher than before), but the pressure at the venule end rises from 1.6 to 4.6 kPa, so less tissue fluid drains back into the capillaries and it accumulates in the tissues; (blood still flows, since pressure still falls from arteriole end to venule end);
16. B3.3.2 [1]
  • B: filaments slide past each other but do not themselves shorten; the sarcomere and the light band narrow as overlap increases;
17. B3.3.10 [1]
  • B: a short, flattened humerus, radius and ulna and long digits (with extra finger bones) inside a single stiff paddle, with little movement at the elbow, form a flipper used for steering and stability;
18. B3.3.3 [1]
  • A: titin resists stretching and so prevents overstretching of the sarcomere; without it, the pull of the antagonist could draw the thin filaments out beyond the ends of the thick filaments, so no cross-bridges could form;
19. B4.2.9 [1]
  • D: V survives almost as well with the alkaloid (85 %) as without it (88 %) and has high detoxifying enzyme activity, so it can detoxify the plant's secondary compound; the alkaloid is a chemical, not a physical, defence (A), it kills most W but few V (B), and W survives poorly on these leaves (C);
20. C1.2.15 [1]
  • A: glycolysis is in the cytosol; link reaction and Krebs cycle are in the matrix; only the membrane-bound stage gains from more membrane;
21. C1.2.11 [1]
  • B: fatty acids are broken down to 2C acetyl groups, which are transferred by coenzyme A to the Krebs cycle without passing through pyruvate or the link reaction; pyruvate cannot enter the Krebs cycle directly, fats cannot be respired anaerobically, and glucose made from fat would still give pyruvate that cannot be used;
22. C1.3.2 [1]
  • D: water is split using light energy (photolysis), providing hydrogen for the reduction of CO₂ to glucose and releasing oxygen as a waste product;
23. C2.1.7 [1]
  • C: the response depends on the receptor and the transduction pathway that its binding sets off, not on the ligand alone; different types of epinephrine receptor start different sequences of reactions, so the same ligand triggers glycogen breakdown in one cell, contraction in another and relaxation in a third; one concentration cannot explain opposite responses (B), epinephrine is not converted (C), and as a hydrophilic amine it does not enter cells (D);
24. C2.2.7 [1]
  • D: Ca²⁺ influx triggers exocytosis of ACh; ACh diffuses across the cleft, opens ligand-gated (Na⁺) channels, and is then broken down by acetylcholinesterase;
25. C2.2.8 [1]
  • B: depolarization depends on voltage-gated Na⁺ channels; blocking them prevents impulse initiation/conduction in the treated region, so pain signals never reach the brain;
26. C2.2.5 [1]
  • A: synapses are junctions between neurons and between neurons and effectors (muscle, gland);
27. C3.1.8 [1]
  • A: X is the protective sheath round the whole nerve, Y is the myelin sheath round a myelinated fibre and Z is a fibre with no myelin; a nerve is a bundle of nerve fibres, so it contains no cell bodies;
28. C4.2.10 [1]
  • C: secondary consumers eat primary consumers; the owl, fox and wolf all feed on the herbivores (lemming, hare, caribou), which in turn eat the producers;
29. D1.2.19 [1]
  • C: when protein synthesis is blocked the amount of R falls quickly, so R is being broken down all the time; adding the proteasome inhibitor almost prevents the fall, so the breakdown is carried out by proteasomes; because the amount stays constant in untreated cells, the breakdown must be balanced by continual synthesis of new R; A and C contradict the proteasome-inhibitor results, and in that group synthesis is blocked, so the inhibitor cannot be speeding up synthesis (D);
30. D2.1.9 [1]
  • A: in stage B (anaphase II) single chromatids are moving to opposite poles; in stage A (anaphase I) whole chromosomes, still made of two chromatids, are separated;
31. D2.2.3 [1]
  • B: degradation limits the number of rounds of translation; stability is a control point;
32. D2.3.9 [1]
  • B: water potential is a measure of the potential energy of water, and matter moves spontaneously from higher to lower potential energy;
33. D3.2.11 [1]
  • A: as in humans, a female has two X chromosomes of the same size and a male has one X and a much smaller Y, so a pair of unequal chromosomes shows the animal is male; A misses that pair 7 is unmatched, B reads one X-sized chromosome as a sign of a female although a male carries one X too, and D gives a reason that is equally true of a female's body cells;
34. D3.2.13 [1]
  • D: a father passes his Y chromosome, not his X, to his sons, so an allele on the X chromosome can never pass from father to son; A and B are equally true of an X-linked dominant allele, and III-4 could have received an X-linked allele from his affected mother II-5 (D);
35. D3.3.11 [1]
  • C: during vigorous activity arterioles supplying the gut constrict while those supplying skeletal muscles dilate, so the gut receives less blood and digestion and absorption slow; cardiac output rises rather than falls, blood flow to the brain stays about the same in all states, and body temperature does not rise enough to denature enzymes;
36. D4.1.8 [1]
  • A: with no predators, the only selection pressure acting on colour is female mate choice, so the increase in brightness shows that bright males had greater reproductive success; predation was absent in these ponds, colour is inherited rather than induced, and mutation rate is not affected by predators;
37. D4.1.5 [1]
  • B: fitness = reproductive success; a weak but prolific individual is fitter (in this sense) than a strong childless one;
38. D4.2.9 [1]
  • D: 25 ÷ 0.005 = 5000;
39. D4.2.13 [1]
  • D: species diversity, primary production and food-web complexity all increase during primary succession (along with plant size and nutrient cycling);
40. D4.2.3 [1]
  • A: turbidity shades out rooted plants that stabilized the clear state; the new state is self-maintaining, making restoration hard;
41. D1.3.5
  • (a) [1]
    • 7.6;
  • (b) [2]
    • the mean numbers of colonies per plate are similar (7.6 and 7.3);
    • set 1 varies far more than set 2: from 0 to 58 colonies, compared with 4 to 11;
    • four of the ten plates in set 1 have no colonies and one has 58, whereas every plate in set 2 has a similar number (at least 4), OWTTE;

    At least one point must be a difference. Accept a mean for set 2 of 7.3 or 7.

  • (c) [2]
    • all the samples in set 2 come from one culture with a single history of growth, so they differ only by chance in which cells were sampled, OWTTE;
    • so set 2 shows how much the counts vary from sampling and plating alone / acts as a control, OWTTE;
    • so variation in set 1 that is much larger than in set 2 must come from differences in what happened as each culture grew, not from sampling, OWTTE;
    • set 2 also shows that the plates and method give consistent counts for cultures of this density, OWTTE;
  • (d) [3]
    • hypothesis 2;
    • under hypothesis 1, every cell spread on the plates would have the same small chance of becoming resistant on meeting the virus, so the separate cultures would give similar numbers of colonies, like the samples in set 2, OWTTE;
    • under hypothesis 2, a mutation can arise at any time as a culture grows, and is then copied into all the descendants of the cell in which it arose;
    • a mutation early in the growth of a culture gives a large number of resistant cells (e.g. the plate with 58 colonies), whereas one that arises late gives only a few, OWTTE;
    • in some small cultures no resistance mutation arose at all, giving plates with no colonies (4 of 10), OWTTE;
    • so the numbers vary greatly between separate cultures, as in set 1, which only hypothesis 2 predicts, OWTTE;

    Naming hypothesis 2 without reasoning scores no marks: [3] needs reasoning about the variation between plates, not about the mean.

  • (e) [2]
    • mutations occur at random / are not caused or directed by the conditions in which they would be useful, OWTTE;
    • the virus does not cause resistance: it kills the cells without it, leaving those that already carried a resistance mutation, OWTTE;
    • no natural mechanism is known for making a deliberate change to a particular base in order to change a trait (such as resistance);
    • (NOS) the two hypotheses made different testable predictions about the variation between cultures, so the experiment could reject hypothesis 1, OWTTE;
    • mutation is the original source of the variation, and conditions such as the virus then act on that variation, OWTTE;

    Content pivot. Nature of science: a classic experiment that distinguished two hypotheses by their predictions.

42. C2.1.11
  • (a) [2]
    • uptake increases with insulin concentration;
    • (but) levels off/plateaus at high concentrations (between 10 and 100 nmol dm⁻³), OWTTE;
  • (b) [1]
    • 380 %, from (4.8 − 1.0) ÷ 1.0 × 100;

    Do not accept 480 % or a fold change: the question asks for a percentage increase.

  • (c) [2]
    • (essentially) all receptors are occupied/saturated at high hormone concentration;
    • (so) further insulin cannot increase the response / the number of receptors (or transporters) becomes limiting, OWTTE;
  • (d) [2 max]
    • the antibody blocks the hormone-binding site (extracellular domain);
    • insulin cannot bind, so the (tyrosine kinase) receptor is not activated / no signalling cascade;
    • (so) GLUT4 is not moved to the membrane and uptake stays near 1.0, OWTTE;
  • (e) [1]
    • ligand / (an) agonist;
43. B2.1.2
  • (a) [6]
    • independent variable: concentration of ethanol surrounding the beetroot (e.g. 0, 10, 20, 30, 40 %);
    • dependent variable: pigment released, measured as absorbance of the surrounding solution using a colorimeter;
    • method: cut equal-sized beetroot discs/cubes, wash off surface pigment, place equal numbers in equal volumes of each ethanol concentration for a fixed time, then measure absorbance;
    • control: discs in water with 0 % ethanol (to show pigment release without solvent);
    • controlled variables: size/surface area and number of discs, volume of solution, temperature, time, beetroot source/variety;
    • repeat each concentration ≥3 times and calculate a mean, OWTTE;

    Award for IV, DV+measurement, method, control, controlled variables, replication.

  • (b) [2]
    • ethanol dissolves in / disrupts the (hydrophobic) phospholipid bilayer / dissolves membrane lipids;
    • the membrane becomes more permeable / develops gaps, so pigment leaks out of the vacuole and cell, OWTTE;
44. A2.3.6
  • (a) [2]
    • as the number of amino acid differences increases, vaccine effectiveness decreases (a negative relationship);
    • the fall is large, from 61 % at 2 differences to 19 % at 12 differences;
  • (b) [1]
    • about 42 %;

    Accept any value in the range 35-48 %.

  • (c) [2 max]
    • the influenza genome is RNA and is copied without proofreading, so mutations occur at a high rate;
    • variants whose haemagglutinin is not recognized by antibodies in the population survive and reproduce (natural selection);
    • these variants spread, so the circulating virus drifts away from the strain chosen for the vaccine months earlier, OWTTE;
  • (d) [2]
    • the age or health of the people vaccinated (immune response is weaker in the elderly);
    • the proportion of the population vaccinated / how early in the season they were vaccinated;
    • previous exposure of the population to related strains;
    • how effectiveness is defined and measured (laboratory-confirmed infection, hospital admission);

    Mark the first two only.

  • (e) [2 max]
    • sequences from viruses circulating now are compared to identify which variants are increasing in frequency;
    • the strain chosen is the one predicted to be dominant next season, so that the number of amino acid differences is as small as possible;
    • the decision must be made months in advance to allow time for manufacture, so it is a prediction that can turn out wrong, OWTTE;

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