Biology  by Bradford
IB Biology · full-length mock · 2 hours 30 minutes

IB Biology Paper 2 mock. The whole syllabus, at the real length.

A complete Paper 2 (Higher level) built from the 2025 guide: 80 marks across 14 questions, in 2 hours 30 minutes. 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 30 minutes · 80 marks · 44% of the final grade.

  1. Section A48 marks One long data question, then short-answer questions. Answer all. Longer than SL and drawing on additional higher level content.
  2. Section B32 marks Three extended-response questions. Answer two. Each is three parts from different areas summing to 15, plus 1 mark for construction.

On the front of the paper

How this mock is built

Section A opens with a long data question, then fills to 48 marks evenly across the syllabus with about 30% additional higher level material. Section B offers three themes and marking on screen counts the best two you answer, plus their construction marks, so the total matches the cover.

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Biology · Paper 2
Higher level · mock examination
150 minutes80 marks (including construction 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.
  • Section A: answer all questions.
  • Section B: answer two questions.
  • A calculator is required for this paper.
  • Answers must be written within the answer boxes provided.
Name:
Section A

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

1.

Western mosquitofish (Gambusia affinis) were collected from a stream in which large predatory fish were present and were used to found two new populations in isolated ponds containing no predators. After 15 generations, researchers caught 30 adult males from the source stream and from each pond and recorded the number of dark pigment spots on the body and the body length of each male. To test whether any differences were heritable, offspring of each population were reared from birth in identical aquaria in the laboratory and the number of spots was counted when they became adults. Mean values are shown with the standard error (SE).

PopulationMean number of spots per wild-caught male ± SEMean body length of wild-caught males / mm ± SEMean number of spots per laboratory-reared male offspring ± SE
Source stream (predators present)3.2 ± 0.424.8 ± 0.63.4 ± 0.5
Pond 1 (no predators)9.1 ± 0.725.9 ± 0.78.6 ± 0.8
Pond 2 (no predators)8.4 ± 0.625.4 ± 0.68.0 ± 0.7
(a)State the independent variable in this investigation.[1]
(b)Calculate the percentage increase in the mean number of spots per wild-caught male between the source stream and Pond 1.[2]
(c)Deduce, with a reason, whether the difference in mean body length between wild-caught males from the source stream and from Pond 1 is statistically significant.[2]
(d)Explain why the researchers reared offspring of each population in identical conditions in the laboratory.[2]
(e)Deduce, using the data, whether the change in the number of spots in the pond populations is an example of evolution. Give a reason for your answer.[2]
(f)Suggest how the absence of predators led to the increase in the number of spots in the pond populations.[3]
(g)The theory of evolution by natural selection is supported by a very large body of evidence, yet it is still referred to as a theory. Explain why.[3]
2.

Compare and contrast allopatric and sympatric speciation.

[4]
3.

Cabbage, kale, broccoli, cauliflower, kohlrabi and Brussels sprouts are all varieties of one species, Brassica oleracea, produced by farmers over the last few thousand years from a wild coastal plant.

(a)Outline how these varieties provide evidence for evolution.[3]
(b)Suggest why the varieties are all still classified as one species.[1]
4.

Explain how the primary structure of a protein determines its three-dimensional shape.

[3]
5.

Membrane proteins have hydrophobic amino acids on the surfaces that contact the bilayer core, while soluble cytoplasmic proteins have hydrophilic surfaces. Explain this difference.

[2]
6.

Blood flows through veins at low pressure. Explain how blood is nevertheless returned to the heart.

[3]
7.

The herbicide DCMU blocks the transfer of electrons out of photosystem II. Predict its effect on oxygen production and on the Calvin cycle, giving reasons.

[3]
8.

Chemiosmosis in the thylakoids of chloroplasts produces ATP.

(a)Outline how the chain of electron carriers in the thylakoid membrane contributes to the production of ATP.[2]
(b)Distinguish between cyclic and non-cyclic photophosphorylation.[2]
9.

Libraries of knockout organisms, in which a different single gene has been made inoperative in each strain, are available for model species such as the mouse (Mus musculus) and the yeast Saccharomyces cerevisiae. Suggest advantages to researchers of having access to such a library.

[3]
10.

A deletion removes 300 consecutive bases from the middle of the coding sequence of a gene for an enzyme. Explain why the enzyme produced is nevertheless likely to cease functioning, even though the number of bases removed is a multiple of three.

[3]
11.

Non-disjunction of one chromosome pair can occur either in anaphase I or in anaphase II of meiosis. Predict, with reasons, the chromosome numbers of the four cells produced in each case, for a cell with a normal haploid number n.

[4]
Section B

Answer two questions. Answers must be written within the answer boxes provided. One additional mark is available for the construction of your answers for each question.

12.

Molecular techniques have transformed how biology is studied and applied.

(a)Outline how PCR and gel electrophoresis are used together to study DNA samples.[4]
(b)Explain how eukaryotic cells regulate which of their genes are expressed.[7]
(c)Outline how comparisons of base sequences are used to classify organisms.[4]
13.

The community found in a place depends on its abiotic conditions, and it changes through time both by succession and in response to a rapidly warming climate.

(a)Outline the changes that occur during primary succession, using a named example.[5]
(b)Outline how abiotic factors determine which biome develops in an area.[3]
(c)Discuss the effects of climate change on the timing of biological events and on the life cycles of species in an ecosystem.[7]
14.

Tissues are supplied reliably only if blood is kept flowing under pressure, held constant in composition and sealed in when vessels are damaged.

(a)Outline how a cut in the skin is sealed by blood clotting.[3]
(b)Distinguish between the structure of arteries and the structure of veins in relation to their functions.[4]
(c)Explain how blood glucose concentration is regulated and how this regulation fails in type 1 and type 2 diabetes.[8]

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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Markscheme BbB-GQH______wAAZbiB

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. A4.1.1, A4.1.6
  • (a) [1]
    • presence or absence of predators (in the habitat / population);

    Do not accept 'the population' or 'the pond' without reference to predators.

  • (b) [2]
    • (9.1 − 3.2) / 3.2 × 100;
    • 184 %;

    Accept 184–185 %. Award [2] for the correct answer with no working. ECF for a correct method with an arithmetic slip.

  • (c) [2]
    • not (shown to be) significant;
    • because the ranges of mean ± SE overlap / 24.2–25.4 mm and 25.2–26.6 mm overlap, OWTTE;

    The second mark requires reference to overlap of the standard-error ranges. Do not award for 'the difference is small' alone.

  • (d) [2 max]
    • to find out whether the difference in number of spots is heritable / genetic, rather than caused by the environment (e.g. diet, water, light) of each habitat;
    • identical conditions control the environmental variables, so any remaining difference must be inherited, OWTTE;
    • evolution is change in heritable characteristics, so only a heritable difference would count as evolution / acquired changes are not evolution;
  • (e) [2 max]
    • yes, it is evolution;
    • laboratory-reared offspring of pond fish still have many more spots than offspring of stream fish (8.0–8.6 compared with 3.4) and the SE ranges do not overlap, so the characteristic is heritable;
    • the change occurred in the characteristics of a population over generations (not within an individual), OWTTE;
  • (f) [3 max]
    • in the source stream, males with many spots are more conspicuous / more easily seen and eaten by predators (before they reproduce);
    • so in the stream, alleles for many spots are removed / selected against, keeping the mean number low;
    • in the ponds there are no predators, so spotted males survive as well as plain males;
    • spotted males are preferred by females / have greater mating success, so they leave more offspring and alleles for many spots increase in frequency over the generations, OWTTE;
  • (g) [3 max]
    • in science a theory is a well-supported explanation that predicts and explains a broad range of observations (not a guess), OWTTE;
    • it is impossible to formally prove that any scientific theory is true / to prove it corresponds exactly with reality;
    • (because) future evidence could in principle contradict it / scientific knowledge is provisional;
    • the theory is accepted as a pragmatic truth because it is consistent with all the evidence so far and works in practice;
    • it is very unlikely ever to be falsified, given the breadth of the supporting evidence (e.g. sequence data, fossils, observed selection);

    Content pivot (NOS). Do not accept 'because it has not been proved' as an unqualified statement.

2. A4.1.8 [4]
  • both begin with one species / one population and end with two reproductively isolated species that can no longer interbreed to give fertile offspring;
  • both require a barrier to gene flow and divergence under different selection (and/or drift);
  • allopatric speciation involves a geographical barrier separating populations, whereas sympatric speciation occurs without geographical separation (populations remain in the same area);
  • in sympatric speciation isolation arises by other means (e.g. different host/timing, or polyploidy giving instant isolation), whereas in allopatric it arises from physical separation, OWTTE;

At least one similarity and one difference required. Award converse.

3. A4.1.3
  • (a) [3 max]
    • farmers chose plants with desired traits (e.g. large leaves, swollen stems, enlarged flower buds) to breed from / selective breeding / artificial selection;
    • the chosen traits are heritable, so they were passed on and became more pronounced over many generations, OWTTE;
    • the heritable characteristics of each population changed / each variety diverged from the wild plant (and from the other varieties);
    • large differences arose in only a few thousand years, showing how rapidly evolutionary change can occur;
    • if selection by humans can produce this much change, selection by the environment (natural selection) over much longer times could produce even greater change, OWTTE;
  • (b) [1 max]
    • they can (still) interbreed / cross-pollinate to produce fertile offspring;
    • they are not reproductively isolated from each other / gene flow between them is still possible, OWTTE;
4. B1.2.7 [3 max]
  • the primary structure is the sequence of amino acids (coded by the gene);
  • the sequence determines where each type of R-group sits along the chain;
  • R-group interactions (hydrogen bonds / ionic bonds / disulfide bridges / hydrophobic interactions) form between particular residues;
  • (so) the chain folds into a specific/stable three-dimensional (tertiary) conformation, OWTTE;
5. B1.2.10 [2 max]
  • proteins fold so that surfaces match their (chemical) environment;
  • in membranes, non-polar/hydrophobic R-groups face the (hydrophobic) hydrocarbon tails, anchoring the protein in the bilayer;
  • in the cytoplasm, polar/charged R-groups face the (aqueous) surroundings, keeping the protein soluble (with hydrophobic R-groups buried inside), OWTTE;
6. B3.2.5 [3 max]
  • contraction of (skeletal) muscles around the veins squeezes them, pushing blood along;
  • (pocket) valves prevent backflow, so blood moves only toward the heart;
  • the wide lumen of veins offers low resistance to flow;
  • breathing movements / pressure changes in the thorax also assist return, OWTTE;
7. C1.3.11-C1.3.13 [3]
  • oxygen production stops, because electrons are not being drained from photosystem II, so photolysis of water is not needed / halts;
  • no proton gradient is built and no ATP or NADPH is made in the light-dependent reactions;
  • so the Calvin cycle stops, as it depends on ATP and NADPH to reduce glycerate 3-phosphate and regenerate RuBP, OWTTE;
8. C1.3.12
  • (a) [2 max]
    • excited electrons emitted by a photosystem are passed along the chain of electron carriers;
    • energy released as the electrons pass along the chain is used to pump protons / H⁺ from the stroma into the thylakoid lumen;
    • (this creates) a proton gradient / a higher concentration of protons in the lumen than in the stroma;
    • protons flow back to the stroma through ATP synthase, which phosphorylates ADP to ATP, OWTTE;

    Do not accept 'electrons are pumped across the membrane'.

  • (b) [2 max]
    • in cyclic photophosphorylation the electrons come from photosystem I, whereas in non-cyclic photophosphorylation they come from photosystem II;
    • in cyclic photophosphorylation the electrons return to photosystem I after passing along the carriers, whereas in non-cyclic photophosphorylation they are passed on (to photosystem I and then to NADP) and are replaced by electrons from photolysis of water;
    • cyclic photophosphorylation produces only ATP, whereas non-cyclic photophosphorylation produces ATP and reduced NADP / NADPH (and oxygen);

    Credit only explicitly paired statements. Award converse statements.

9. D1.3.8 [3 max]
  • a strain lacking a gene of interest can be obtained without having to produce it, saving time and cost;
  • the function of the gene can be investigated by comparing the phenotype of the knockout with that of the normal (wild-type) organism;
  • the same strains are used by many laboratories, so results can be compared and repeated / reliability is improved;
  • the effects of knocking out every gene in the genome can be surveyed systematically / genes of unknown function can be screened;
  • knockouts of different genes can be combined by crossing to study interactions between genes, OWTTE;

Mark the first three only.

10. D1.3.3 [3 max]
  • because 300 is a multiple of three, the reading frame is not shifted / the codons after the deletion are read as before;
  • (but) 100 amino acids are missing from the middle of the polypeptide / the polypeptide is much shorter;
  • the missing amino acids may include those that form the active site / that make the bonds (R-group interactions) needed for correct folding;
  • (so) the polypeptide folds into a different three-dimensional (tertiary) shape;
  • (so) the active site is lost / no longer complementary to the substrate, and the enzyme cannot catalyse its reaction, OWTTE;

Do not accept a frameshift as the reason. Accept "a major deletion" or "a large deletion" as the category of change.

11. D2.1.10 [4 max]
  • anaphase I non-disjunction: both homologous chromosomes (of the pair) move to the same pole;
  • (so) all four gametes are abnormal: two have an extra chromosome (n+1) and two are missing one (n−1);
  • anaphase II non-disjunction: meiosis I is normal, but sister chromatids fail to separate in one of the two cells;
  • (so) two gametes are normal (n), one is n+1 and one is n−1;
  • (in either case) fertilization of an n+1 gamete produces a trisomic zygote (e.g. trisomy 21 / Down syndrome), OWTTE;

Award any 4; the contrast between all four abnormal (anaphase I) and only two abnormal (anaphase II) is required for full marks.

12.
  • (a) D1.1 [4 max]
    • PCR amplifies a chosen sequence: cycles of denaturation (~95 °C), primer annealing, and extension;
    • heat-stable (Taq) polymerase copies the target, doubling it each cycle (millions of copies from traces);
    • primers determine which sequence is amplified;
    • gel electrophoresis separates DNA fragments by size, DNA (negatively charged) moves toward the positive electrode;
    • smaller fragments travel further, giving a band pattern;
    • applications credited: DNA profiling (forensics/paternity) / diagnosis / sequencing preparation, OWTTE;
  • (b) D2.2 [7 max]
    • most control is at transcription;
    • transcription factors bind specific (promoter) sequences;
    • promoting or blocking the binding/activity of RNA polymerase;
    • enhancers (possibly distant, looped into contact) increase transcription when bound;
    • DNA (cytosine) methylation of promoters silences genes;
    • histone modifications (acetylation/methylation) loosen or compact chromatin, changing accessibility;
    • these epigenetic patterns differ between cell types and are inherited through cell division;
    • the environment/signals can alter transcription-factor activity and epigenetic marks;
    • mRNA stability/degradation (and translation rate) tune how much protein is made, OWTTE;
  • (c) A3.2 [4 max]
    • the same gene/sequence is aligned across the species being classified;
    • (numbers of) base differences between each pair are counted;
    • fewer differences indicate a more recent common ancestor / closer relationship;
    • a cladogram is constructed (most parsimonious arrangement), grouping species into clades;
    • differences accumulate at a roughly constant rate (molecular clock), so divergence times can be estimated / groups shown not to be clades are reclassified, OWTTE;

Plus 1 mark for the construction of the answer: clear enough to be understood without re-reading, succinct, with little or no repetition or irrelevant material.

13.
  • (a) D4.2.13, D4.2.12 [5 max]
    • primary succession begins on newly formed substrate with no soil or previous community, e.g. bare rock after a volcanic eruption / moraine left by a retreating glacier / sand dunes;
    • pioneer species (lichens / mosses / bacteria) colonize, tolerating harsh conditions, weathering the rock and adding organic matter when they die to form soil;
    • as the soil deepens and holds more water and nutrients, larger plants replace the pioneers: grasses and herbs, then shrubs, then trees / increase in plant size;
    • primary production increases as larger plants capture more light;
    • species diversity increases (more habitats and niches) and food webs become more complex;
    • nutrient cycling increases as biomass and decomposer activity increase;
    • each community changes the abiotic environment (shade, soil, humidity) and the biotic environment, making conditions suitable for the next stage, until a climax community forms, OWTTE;

    Accept any suitable terrestrial example.

  • (b) B4.1.6, B4.1.7 [3 max]
    • for any combination of temperature and rainfall (pattern) one type of natural ecosystem is likely to develop / biome distribution can be shown on a graph with these two variables as axes;
    • a biome is a group of ecosystems with similar communities because of similar abiotic conditions (and convergent evolution);
    • e.g. hot desert: very low rainfall and high temperatures / tundra: very low temperatures and low precipitation;
    • e.g. tropical forest: high temperature and rainfall all year / temperate forest: moderate temperature and rainfall with seasons / taiga: long cold winters / grassland: rainfall too low to support forest;
    • similar biomes on different continents contain unrelated species with similar adaptations because of convergent evolution, OWTTE;

    Award the example points for any two biomes with both temperature and rainfall correctly characterized.

  • (c) D4.3.10, D4.3.11 [7 max]
    • phenology is the study of the timing of biological events, e.g. flowering / budburst / bird migration / nesting / emergence of insects;
    • timing is cued by variables such as temperature and photoperiod (day length);
    • warming advances events cued by temperature, e.g. earlier budburst / flowering / earlier peak of caterpillar biomass;
    • events cued by photoperiod do not shift, because day length is unaffected by warming;
    • (so) synchrony between interacting populations can be disrupted / a mismatch arises when one species uses temperature and the other uses photoperiod;
    • e.g. Arctic mouse-ear chickweed grows earlier in warm springs but reindeer migration (cued by photoperiod) arrives at the same time, so calves miss the peak of food and survival falls;
    • e.g. great tits: caterpillar biomass peaks earlier in warmer springs but egg-laying has advanced less, so chicks hatch after peak food and breeding success falls;
    • warmer conditions allow some insects to complete more life cycles per year, e.g. spruce bark beetle two generations instead of one, so populations grow faster and outbreaks kill more trees;
    • some species benefit (longer growing season / range expansion) whereas others decline, so effects are not uniformly negative;
    • natural selection may allow populations to adapt, e.g. earlier breeding in some great tit populations, but change may be too fast for long-lived species, OWTTE;

    Accept common or scientific names. A balanced discussion is expected: at least one point on species that benefit or adapt is needed for full marks.

Plus 1 mark for the construction of the answer: clear enough to be understood without re-reading, succinct, with little or no repetition or irrelevant material.

14.
  • (a) C3.2.3 [3 max]
    • damage to a blood vessel exposes platelets to the damaged tissue / collagen, so platelets stick to the site and to each other;
    • platelets release clotting factors;
    • the clotting factors trigger a cascade of reactions, each step activating the next / amplifying the response;
    • the cascade produces thrombin, which rapidly converts soluble fibrinogen into insoluble fibrin;
    • fibrin forms a mesh of fibres that traps erythrocytes / red blood cells to form a clot, sealing the cut and preventing blood loss and entry of pathogens, OWTTE;

    Accept "prothrombin is converted to thrombin" as part of the cascade point. No further details of the cascade are required.

  • (b) B3.2.3, B3.2.5 [4 max]
    • arteries have thick walls, whereas veins have thin walls (relative to the diameter of the vessel);
    • arteries have a narrow lumen (relative to the wall), whereas veins have a wide lumen;
    • arteries have thick layers of smooth muscle and elastic tissue to withstand high blood pressure, whereas veins have little muscle and elastic tissue because pressure is low;
    • elastic tissue in arteries stretches during systole and recoils, maintaining / evening out the pressure between heartbeats, whereas veins do not maintain pressure;
    • veins have valves to prevent backflow, whereas arteries have no valves because pressure keeps blood moving forward;
    • the flexible wall of a vein allows it to be compressed by skeletal muscle action, squeezing blood towards the heart, whereas the wall of an artery resists compression / stays circular in section, OWTTE;

    Award marks only for paired, comparative statements. Accept converse wording.

  • (c) D3.3.3, D3.3.4 [8 max]
    • blood glucose concentration is monitored by endocrine cells in the pancreas / islets of Langerhans;
    • when glucose rises (after a meal) β-cells secrete insulin into the blood;
    • insulin is transported in the blood to target cells, e.g. liver / muscle / adipose cells;
    • insulin stimulates uptake of glucose by cells (more glucose transporters in their membranes) and conversion of glucose to glycogen in liver and muscle / to fat;
    • (so) blood glucose falls back towards the set point;
    • when glucose falls (between meals / during exercise) α-cells secrete glucagon;
    • glucagon stimulates the liver to break down glycogen to glucose and release it into the blood, raising blood glucose;
    • (this is) negative feedback: each hormone reverses the change that caused its secretion / the two hormones have antagonistic effects;
    • type 1 diabetes: β-cells are destroyed by the person’s own immune system (autoimmune), so little or no insulin is produced, usually beginning in childhood;
    • type 2 diabetes: target cells become resistant to insulin / receptors fail to respond, and insulin secretion may later decline; risk factors include obesity, diets rich in sugar and fat, lack of exercise, genetic factors and age;
    • type 1 is treated with insulin injections (and monitoring of blood glucose), whereas type 2 is treated / prevented by diet, exercise and weight loss (and drugs); in both, persistently high blood glucose causes glucose in urine / thirst / damage to blood vessels, OWTTE;

    Accept "glycogen breakdown" for glycogenolysis. Do not accept "glucagon converts glucose to glycogen".

Plus 1 mark for the construction of the answer: clear enough to be understood without re-reading, succinct, with little or no repetition or irrelevant material.

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