Biology  by Bradford
IB Biology · full-length mock · 1 hour 30 minutes

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

A complete Paper 1 (Standard level) built from the 2025 guide: 55 marks across 34 questions, in 1 hour 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.

What the real paper requires

1 hour 30 minutes · 55 marks · 36% of the final grade.

  1. Paper 1A30 marks 30 multiple-choice questions. Answer all. One best answer per question; no marks lost for wrong answers.
  2. Paper 1B25 marks 4 data-based questions. Each built on a data set, an experiment or a micrograph, ending in a question that leaves the data and tests taught content.

On the front of the paper

How this mock is built

The 30 multiple-choice questions are spread across all 40 subtopics in proportion to each theme, with subtopics taking turns so nothing is over-drawn. The four data-based questions are one per theme where the bank allows, each in the 5 to 7 mark range the real paper uses, chosen to sum as close to 25 as possible.

The code on the paper rebuilds this exact mock and its markscheme. Build a fresh mock draws a new one from the same rules, so a second sitting is never the first one memorised.

Other mocks

The mock

Take it on screen → Build a fresh mock Paper code BbB-GAD______wAAZS2L
Biology · Paper 1
Standard level · mock examination · Paper 1A multiple choice and Paper 1B data-based
90 minutes55 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.

Water rises up the narrow spaces within the cellulose cell walls of a leaf. Which property of water is chiefly responsible?

[1]
  1. Adhesion of water to the cellulose surfaces
  2. The high density of water
  3. The low viscosity of water
  4. Evaporation of water from the leaf surface
2.

In almost all organisms, from bacteria to humans, the codon GUG codes for the amino acid valine. What is the most reasonable explanation for the genetic code being shared across all life?

[1]
  1. Each group of organisms evolved the same code independently by convergent evolution
  2. The code is determined by the laws of chemistry, so no other code is possible
  3. The code was inherited from a common ancestor of all living things
  4. Organisms frequently exchange their genetic codes with each other
3.

How many chromosomes are present in the karyogram?

[1]
12345678910111213141516171819202122XYKaryogram of one individual (chromosomes arranged by size and centromere position)
  1. 45
  2. 46
  3. 47
  4. 48
4.

The wing of a bat and the flipper of a whale contain the same pattern of bones (the pentadactyl limb) despite being used differently. What are these structures an example of?

[1]
  1. Analogous structures resulting from convergent evolution
  2. Structures with no evolutionary significance
  3. Homologous structures resulting from divergent evolution
  4. Structures acquired during each animal's lifetime
5.

What is the purpose of a seed bank?

[1]
  1. To sell rare seed varieties commercially
  2. To store seeds of many plant species long-term
  3. To breed entirely new species of crop plant
  4. To supply farmers with seed for each growing season
6.

Which property of glucose makes it suitable as the main respiratory substrate transported in blood?

[1]
  1. It is insoluble, so it does not affect the osmotic balance of blood
  2. It is a large polymer that stores a great deal of energy per molecule
  3. It is soluble in water because of its many hydroxyl groups
  4. It cannot cross membranes, so it stays in the blood until needed
7.

How are amino acids joined to form a polypeptide?

[1]
  1. By hydrogen bonds formed between adjacent R-groups
  2. By ester bonds formed in condensation reactions
  3. By glycosidic bonds formed in hydrolysis reactions
  4. By peptide bonds formed in condensation reactions
8.

Which row correctly predicts how each substance crosses a plasma membrane?

[1]
RowOxygenGlucoseSodium ions
A.simple diffusionchannel/carrier proteinchannel protein / pump
B.channel proteinsimple diffusionsimple diffusion
C.pumppumpsimple diffusion
D.exocytosisendocytosiscarrier
  1. Row A
  2. Row B
  3. Row C
  4. Row D
9.

A gene from a eukaryote was inserted into a bacterium. The gene was transcribed, but the protein produced was faulty. Which statement is the most likely explanation, based on the way the two types of cell are organized?

[1]
  1. The bacterium has no ribosomes of its own, so the mRNA was translated by ribosomes that had entered with the gene
  2. The bacterial cell wall prevented the mRNA from leaving the cell to reach ribosomes in the surrounding medium
  3. The bacterium lacks a nucleus, so the DNA could not be transcribed into mRNA in the correct sequence
  4. The bacterium has no nuclear compartment, so ribosomes translated the mRNA before it had been modified
10.

As a cell grows, which quantity increases fastest?

[1]
  1. Volume
  2. Surface area
  3. Surface area-to-volume ratio
  4. Membrane thickness
11.

Which structures does carbon dioxide pass through, in order, when leaving a leaf during the day?

[1]
  1. Stoma → mesophyll cell → air space
  2. Mesophyll cell → stoma → air space
  3. Mesophyll cell → air space → stoma
  4. Air space → stoma → mesophyll cell
12.

On a hot, windy day, transpiration from a well-watered plant increases sharply. Why?

[1]
  1. Heat speeds evaporation and wind removes humid air
  2. Hot air contains no water vapour at all
  3. Wind physically pushes water up the xylem
  4. Stomata always open wider in windy conditions
13.

Some unicellular organisms are facultative mixotrophs. What does "facultative" indicate?

[1]
  1. They are restricted to heterotrophic nutrition
  2. They digest food externally like saprotrophs
  3. They can switch between autotrophic and heterotrophic nutrition depending on conditions
  4. They must use autotrophic and heterotrophic nutrition simultaneously at all times
14.

A tenfold dilution series of a substrate solution is prepared. Tube 1 holds the undiluted 1.0 mol dm⁻³ stock. Each following tube is made by adding 1.0 cm³ from the previous tube to 9.0 cm³ of distilled water. What is the substrate concentration in tube 5?

[1]
  1. 1 × 10⁻³ mol dm⁻³
  2. 1 × 10⁻⁴ mol dm⁻³
  3. 1 × 10⁻⁵ mol dm⁻³
  4. 2 × 10⁻¹ mol dm⁻³
15.

What are the final products of aerobic cell respiration in humans?

[1]
  1. Carbon dioxide and water
  2. Lactate and carbon dioxide
  3. Glucose and oxygen
  4. Ethanol and carbon dioxide
16.

The absorption spectrum of chlorophyll and the action spectrum of photosynthesis have peaks at similar wavelengths. What does this similarity indicate?

[1]
  1. The wavelengths absorbed by chlorophyll are the wavelengths that drive photosynthesis
  2. Chlorophyll is the only pigment involved in photosynthesis, so no other pigment contributes
  3. Green light is used more efficiently than red or blue light for photosynthesis in the chloroplast
  4. The rate of photosynthesis is independent of the wavelength of light that strikes the leaf
17.

Acetylcholinesterase in the synaptic cleft rapidly breaks down acetylcholine. Why is this essential?

[1]
  1. To prevent acetylcholine from ever reaching the receptors
  2. To allow the synapse to transmit impulses in both directions
  3. So that each impulse produces only a brief postsynaptic response
  4. To supply the ATP needed for vesicle exocytosis
18.

Blood CO₂ rises during exercise. How does this lead to faster, deeper breathing?

[1]
  1. The cerebellum detects the rise in CO₂ and inhibits the ventilation muscles
  2. The rise in CO₂ stretches the alveoli, triggering deeper breaths
  3. Chemoreceptors detect the rise and the medulla stimulates the ventilation muscles
  4. CO₂ stimulates stretch receptors in the leg muscles, which signal the lungs
19.

Oak trees, the insects that feed on them, insect-eating birds, fungi and soil bacteria live together in a woodland. Which term describes all of these interacting populations?

[1]
  1. A species
  2. A population
  3. An ecosystem
  4. A community
20.

Which row correctly compares photoautotrophs and chemoautotrophs?

[1]
RowEnergy source of photoautotrophEnergy source of chemoautotrophExample of a chemoautotroph
A.lightoxidation of inorganic substancesiron-oxidizing bacteria
B.lightconsuming other organismsfungi
C.oxidation of inorganic substanceslightalgae
D.heatlightnitrifying bacteria
  1. Row A
  2. Row B
  3. Row C
  4. Row D
21.

How does complementary base pairing ensure that DNA replication produces identical copies?

[1]
  1. The bases on the new strand are identical to those on the template
  2. Free nucleotides are only added where they pair with the template
  3. Hydrogen bonds transfer the base sequence directly to the new strand
  4. Any base can pair with any other base on the template strand
22.

During translation, what determines which amino acid is added next to the growing polypeptide?

[1]
  1. Pairing of the mRNA codon with a tRNA anticodon
  2. The ribosome selects amino acids at random
  3. The amino acid sequence of the last protein
  4. The order of the ribosomes on the mRNA
23.

At one position in the human genome, 60 % of people have the base C and 40 % have the base T. The position lies within the coding sequence of a gene, yet everyone produces an identical polypeptide from the gene. Which statement is correct?

[1]
  1. The variant is not a single-nucleotide polymorphism, because it has no effect on the polypeptide produced
  2. The variant is a single-nucleotide polymorphism that arose by base substitution, and the two codons specify the same amino acid
  3. The variant is a single-nucleotide polymorphism that arose by insertion of a single base, which the ribosome skips over during translation
  4. The variant must lie in an intron of the gene, because any change within a codon always changes the amino acid
24.

Why must the nucleus divide before the cytoplasm during eukaryotic cell division?

[1]
  1. So that the cytoplasm has time to increase in volume before division
  2. So that each daughter cell receives a complete copy of the genome
  3. Because the nucleus is too large to be split by a cleavage furrow
  4. Because the mitochondria must replicate before the cytoplasm divides
25.

A pollen tube grows from structure II to an ovule in structure III and delivers a male nucleus. Which process is completed when the nuclei fuse?

[1]
IIIIIIIVpetalsepalfilamentstylereceptacleHalf of an insect-pollinated flower
  1. Pollination
  2. Fertilization
  3. Germination
  4. Seed dispersal
26.

Arctic foxes grown at warm temperatures develop darker fur than genetically identical foxes grown at cold temperatures, even though both have the same genotype. What term describes this ability of a single genotype to produce different phenotypes in different environments?

[1]
  1. Codominance
  2. Incomplete dominance
  3. Polygenic inheritance
  4. Phenotypic plasticity
27.

A person with type 1 diabetes produces no insulin. Which step in the diagram fails after this person eats a meal?

[1]
Blood glucose at set point≈ 5 mmol dm⁻³Blood glucose rises(e.g. after a meal)β cells of pancreassecrete insulinLiver and muscle take upglucose; glycogen madeglucose falls backBlood glucose falls(e.g. during exercise)α cells of pancreassecrete glucagonLiver breaks glycogendown; glucose releasedglucose rises backabovebelow
  1. Liver and muscle cells taking up glucose and making glycogen
  2. α cells of the pancreas secreting glucagon
  3. The liver breaking down glycogen and releasing glucose
  4. Blood glucose falling during exercise
28.

A population of bacteria is repeatedly exposed to an antibiotic. Over time, resistant strains dominate. Which explanation is correct?

[1]
  1. The antibiotic caused the resistance mutations
  2. All bacteria gradually became slightly resistant
  3. Rare pre-existing resistant variants survived
  4. Bacteria learned tolerance and taught others
29.

Why does the concentration of DDT rise at each step in the food chain?

[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. DDT is stored in fat and not excreted, so a consumer retains the DDT from all the prey it eats
  2. DDT is a nutrient that consumers absorb from the water at each level
  3. Organisms at higher trophic levels have a larger surface area in contact with the water
  4. DDT is synthesized by the tissues of consumers from simpler compounds
30.

Which consequences for human health and agriculture are expected as insect vectors and pests spread poleward?
I. Diseases such as malaria/dengue reaching new regions
II. Crop pests surviving milder winters in higher numbers
III. Fewer pest generations per year everywhere

[1]
  1. I and II only
  2. II and III only
  3. I and III only
  4. I, II and III
Paper 1B · Data-based questions

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

31.
TreatmentFall in blood glucose after 1 h / mmol dm⁻³
Saline (control)0.2
Normal insulin4.6
Mutant insulin (one amino-acid change at binding site)0.9
(a)Analyse the effects of the three treatments on blood glucose.[2]
(b)Using your knowledge of protein structure (Theme A), explain why the mutant insulin is much less effective.[3]
(c)State the purpose of the saline control.[1]
32.

A student investigated the effect of glucose concentration on the rate of anaerobic respiration in a yeast suspension by counting the number of carbon dioxide bubbles released per minute through a delivery tube. The results are shown in the table.

Glucose concentration / % w/vMean bubbles per minute
00
29
417
624
825
1025
(a)Describe the relationship between glucose concentration and the rate of carbon dioxide production shown in the table.[2]
(b)Explain the shape of the relationship you described in (a).[2]
(c)Identify one variable that should be controlled in this investigation.[1]
(d)State the gas being collected and name the pathway by which it is produced in this investigation.[2]
33.

A student used a simple calorimeter to compare the energy content of foods. A weighed sample of each food was burned beneath a boiling tube containing 20 cm³ of water, and the temperature rise of the water was recorded. Three samples of each food were tested and the mean energy released per gram was calculated.

FoodMain componentMean energy released / kJ g⁻¹
Peanutlipid19.7
Breadcarbohydrate9.8
Dried applecarbohydrate10.4
Butterlipid24.1
(a)Calculate how many times more energy per gram is released by butter than by bread.[1]
(b)State the conclusion about lipids and carbohydrates supported by these data.[1]
(c)The energy values measured with this apparatus are lower than the true values. Suggest two reasons why.[2]
(d)Outline one reason why birds preparing to migrate store fuel as lipid rather than as carbohydrate.[2]
34.

A student calibrated the eyepiece graticule of a light microscope. The diagram shows the field of view with the eyepiece graticule, which carries 100 arbitrary divisions, lying above a stage micrometer ruled in millimetres. No magnification is stated on the diagram.

020406080100 0.00.51.0 Eyepiece graticule Stage micrometer arbitrary divisions scale / mm
(a)Determine the distance, in µm, represented by one division of the eyepiece graticule.[2]
(b)A cell measured with the same objective spans 26 eyepiece graticule divisions. Calculate its length in µm.[1]
(c)The student could read the graticule only to the nearest half division. State the uncertainty of the length in (b), in µm.[1]
(d)Explain why the eyepiece graticule alone cannot give a measurement in µm.[2]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
Rebuild or edit this exact paper (and its markscheme): biologybybradford.com/exam-maker?code=BbB-GAD______wAAZS2L

Show the markscheme

Markscheme BbB-GAD______wAAZS2L

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. A1.1.4 [1]
  • A: capillary action depends on adhesion to polar surfaces (with cohesion following behind);
2. A1.2.10 [1]
  • C: the (near-)universal genetic code is most simply explained by inheritance from a universal common ancestor; a later change to the code would alter almost every protein and be lethal, so it has been conserved;
3. A3.1.7 [1]
  • C: 22 pairs plus X and Y would give 46; the extra copy of chromosome 21 makes 47;
4. A4.1.4-A4.1.5 [1]
  • C: same underlying structure, different function = homology, evidence of divergence from a common ancestor; analogous = similar function, different structure (convergence);
5. A4.2.7 [1]
  • B: seeds are stored dried and cold so they remain viable for decades; this ex situ germplasm storage insures against loss of species and crop varieties (genetic diversity), allowing future reintroduction/breeding;
6. B1.1.4 [1]
  • C: glucose is small and polar (hydroxyl groups), so it dissolves and is transported easily; it is also chemically stable and readily oxidised;
7. B1.2.2 [1]
  • D: peptide bonds form between the amine group of one amino acid and the carboxyl group of the next, by condensation, during translation on ribosomes;
8. B2.1.4 [1]
  • A: small non-polar O₂ crosses freely; polar glucose needs a transporter; charged Na⁺ needs channels or pumps;
9. B2.2.2 [1]
  • D: in a eukaryote the nuclear envelope separates transcription from translation, so mRNA can be modified before it meets ribosomes; in a prokaryote ribosomes can bind the mRNA immediately, so the unmodified message is translated; A fails because bacteria have their own ribosomes; B fails because translation occurs inside the cell, not in the medium; C fails because the stem states that transcription did occur;
10. B2.3.6 [1]
  • A: volume scales with the cube of length, surface area with the square, so volume outpaces area and the SA:V ratio falls;
11. B3.1.7 [1]
  • C: CO₂ produced/present in mesophyll cells diffuses into the (spongy) air spaces and exits through open stomata; (during daylight net flow of CO₂ is inward, but any CO₂ leaving takes this path);
12. B3.2.7 [1]
  • A: temperature raises the evaporation rate and the air's capacity for vapour; wind removes the humid boundary layer at the leaf surface, so vapour diffuses out faster;
13. B4.2.5 [1]
  • C: facultative mixotrophs use whichever mode conditions favour (photosynthesis in light, feeding when prey is abundant); obligate mixotrophs must use both;
14. C1.1.8 [1]
  • B: tube 1 is the undiluted stock, so only four tenfold dilutions have been carried out: 1.0 × 10⁻⁴ mol dm⁻³; A and C are the off-by-one errors of counting three or five dilution steps;

Design move: near-miss discrimination — the two adjacent powers of ten are the distractors, so the tube must be counted, not the transfers.

15. C1.2.5 [1]
  • A: glucose + oxygen → carbon dioxide + water (with ATP produced);
16. C1.3.6 [1]
  • A: correlation between absorption and rate shows chlorophyll captures the light energy used; accessory pigments explain the differences;
17. C2.2.7 [1]
  • C: clearing the transmitter ends each signal; without it the postsynaptic membrane stays depolarized/blocked (as with nerve agents and some insecticides);
18. C3.1.15 [1]
  • C: CO₂/pH-sensitive chemoreceptors (in the medulla and arteries) feed the medulla's respiratory centre, which increases signals to the diaphragm and intercostal muscles, driving faster, deeper ventilation;
19. C4.1.10 [1]
  • D: a community is all the populations of different species living and interacting in an area; including the abiotic environment as well would make it an ecosystem;
20. C4.2.7 [1]
  • A: photoautotrophs (plants, algae, cyanobacteria) use light energy, whereas chemoautotrophs (e.g. iron-oxidizing or nitrifying bacteria) use energy released by exothermic oxidation of inorganic chemicals;
21. D1.1.2 [1]
  • B: pairing rules (A with T, G with C) mean the template dictates the new sequence exactly (new strand is complementary, so the copy of the partner strand is identical);
22. D1.2.7 [1]
  • A: each tRNA has an anticodon and carries the corresponding amino acid; codon–anticodon pairing at the ribosome orders the amino acids;
23. D1.3.2 [1]
  • B: a single-nucleotide polymorphism (SNP) is a position where a base substitution has produced two variants common in the population, and because the genetic code is degenerate the two codons can code for the same amino acid; A fails because a SNP is defined by the base difference, not by its effect; C fails because SNPs are substitutions and ribosomes cannot skip bases; D fails because the stem places the variant in the coding sequence and most codon changes at the third position are silent;
24. D2.1.1 [1]
  • B: nuclear division (mitosis/meiosis) distributes the replicated chromosomes; cytokinesis then splits the cytoplasm;
25. D3.1.8 [1]
  • B: pollination is only the transfer of pollen to the stigma; fertilization is the fusion of male and female gametes inside the ovule, producing the embryo;
26. D3.2.6 [1]
  • D — phenotypic plasticity is the capacity to develop different traits, from the same genotype, suited to the environment experienced;
27. D3.3.3, D3.3.4 [1]
  • A: without insulin the target cells are not stimulated to take up glucose, so blood glucose stays high after a meal; the glucagon loop is unaffected;
28. D4.1.4 [1]
  • C: mutation is random and prior; the antibiotic changes the selection pressure, so pre-existing resistant variants are favoured;
29. D4.2.9 [1]
  • A: DDT is persistent and fat-soluble; each consumer eats many prey over its life and keeps their DDT, so the toxin becomes more concentrated with each transfer;
30. D4.3.7 [1]
  • A: warming extends vector/pest ranges and (usually) increases generations per year; III is backwards;
31. D3.3.3
  • (a) [2]
    • normal insulin lowers blood glucose strongly (4.6), whereas the mutant insulin lowers it only slightly (0.9);
    • saline causes almost no change (0.2), so the injection itself is not responsible, OWTTE;
  • (b) [3]
    • insulin is a protein whose three-dimensional shape (including its binding site) is determined by its amino-acid sequence;
    • the single amino-acid change alters the shape of the receptor-binding region;
    • (so) the mutant insulin binds its receptor poorly, triggering little signalling, so cells take up/store less glucose, OWTTE;
  • (c) [1]
    • to show that the injection procedure/fluid alone does not lower blood glucose, so any fall is due to the insulin, OWTTE;
32. C1.2.6
  • (a) [2]
    • rate of CO2 production increases with glucose concentration up to about 6–8 %;
    • above about 8 % glucose concentration, the rate levels off (no further increase), OWTTE;
  • (b) [2 max]
    • at low glucose concentrations, glucose (substrate) availability limits the rate of respiration, so more glucose increases the rate;
    • at higher concentrations another factor becomes limiting, e.g. the amount/activity of respiratory enzymes (or oxygen for the aerobic component), OWTTE;
  • (c) [1]
    • temperature (accept: yeast concentration/mass, volume of suspension, time allowed);
  • (d) [2]
    • carbon dioxide;
    • produced by anaerobic respiration / alcoholic fermentation in yeast;
33. B1.1.11
  • (a) [1]
    • 24.1 ÷ 9.8 ≈ 2.5 (times);

    Accept 2.4–2.5.

  • (b) [1]
    • lipids release (roughly) twice as much energy per gram as carbohydrates;
  • (c) [2]
    • heat is lost to the surroundings/air/tube (not all transferred to the water);
    • incomplete combustion of the sample / some sample remains unburned / heat absorbed by the glass, OWTTE;

    Mark the first two answers only.

  • (d) [2]
    • lipid stores (about) twice the energy per gram / is stored without (associated) water;
    • (so) the bird carries less mass for the same energy, which matters for flight, OWTTE;

    Content pivot.

34. A2.2.2
  • (a) [2 max]
    • reads from the diagram that 40 eyepiece divisions align with 0.5 mm of the stage micrometer;
    • converts to 500 µm and divides by 40;
    • 12.5 (µm);

    Award full marks for a correct answer with no working. Accept 12.5 µm per division. ECF from a misread alignment.

  • (b) [1]
    • 325 (µm);

    ECF from part (a). Units not required.

  • (c) [1]
    • ± 6.25 (µm) / ± 6.3 (µm);

    ECF from part (a). Accept ± half of the value found in (a).

  • (d) [2 max]
    • the graticule divisions are arbitrary / have no fixed size of their own;
    • the size a division represents depends on the magnification (of the objective in use), so it changes between objectives;
    • so it must be compared with a scale of known length, the stage micrometer, before any reading means anything, OWTTE;

    Do not accept 'the graticule is not accurate'. Accept 'it is not to scale' only with a reason.

Practice by subtopic →  ·  Open the exam maker →