IB Biology SL Paper 1 Practice Paper (45 minutes, 30 marks, multiple choice) with mark scheme

2026-10-29

Practice paper | IB DP | Biology SL | Paper 1

IB Biology SL Paper 1 is a pure multiple-choice paper: 30 questions, 45 minutes, one mark each, 30 marks in total, no calculator. The questions here are spread evenly over the four syllabus themes and every option is explained in the answer PDF.

IB Biology SL Paper 1 practice paper: 30 multiple-choice questions, 45 minutes, 30 marks, no calculator, with a mark scheme explaining every option.

Download　Question paper PDF (for students)　|　Answer PDF (every option explained)

Paper details　Level: IB DP (SL, years 1–2) | Subject: Biology SL | Time: 45 minutes | Marks: 30 (no calculator)

Paper structure and marks

Section	Questions	No. of questions	Marks	Content
Section A	Questions 1–8	8	8	Water and molecules, cells, enzymes, membrane transport
Section B	Questions 9–16	8	8	Human physiology and plant transport (circulation, gas exchange, kidney, nerves, muscle, digestion)
Section C	Questions 17–22	6	6	Ecology: energy flow, population estimates, carbon cycle, greenhouse effect, interactions
Section D	Questions 23–30	8	8	Genetics, molecular biology, evolution and classification
Total	1–30	30	30	All themes above

Questions (30 multiple choice)

Question 1 | Water and life (1 mark)

Which property of water is mainly responsible for keeping body temperature stable during vigorous exercise?

A　High specific heat capacity (about 4.2 J g⁻¹ K⁻¹)

B　High latent heat of vaporisation

C　High surface tension

D　It is a good solvent

Question 2 | Cells and microscopy (1 mark)

A cell measures 45 mm across on a micrograph. Its actual diameter is 30 μm. What is the magnification?

A　150×

B　1 500×

C　15 000×

D　0.67×

Question 3 | Cell size and exchange (1 mark)

Two cubes have sides of 2 cm and 4 cm. What are their surface-area-to-volume ratios (SA:V)?

A　3:1 and 1.5:1

B　1.5:1 and 3:1

C　6:1 and 3:1

D　3:1 and 6:1

Question 4 | Water potential and osmosis (1 mark)

A plant cell has ψp = 0 and ψs = −450 kPa and is placed in a solution of water potential −200 kPa. What is the net movement of water and the final state of the cell?

A　Water leaves the cell and the cell plasmolyses

B　Water enters the cell and it becomes turgid

C　No net movement — the water potentials are equal

D　Water enters and the cell finally bursts

Question 5 | Protein structure (1 mark)

A protein loses its function after heating. The main reason is that:

A　The molecular mass halves

B　Peptide bonds are hydrolysed into amino acids

C　Hydrogen and ionic bonds break and the 3-D shape changes, while peptide bonds stay intact

D　The amino acid sequence is rearranged

Question 6 | Enzyme kinetics (1 mark)

An enzyme-catalysed reaction runs at 4.0 μmol min⁻¹ at 20 °C and the enzyme has Q₁₀ = 2. What is the rate at 30 °C, before any denaturation?

A　2.0 μmol min⁻¹

B　4.0 μmol min⁻¹

C　8.0 μmol min⁻¹

D　16.0 μmol min⁻¹

Question 7 | Membrane transport (1 mark)

Mineral ions inside a root-hair cell are already more concentrated than in the soil solution, yet uptake continues. This process is:

A　Active transport, using ATP and carrier proteins

B　Endocytosis

C　Osmosis

D　Facilitated diffusion

Question 8 | Resolution (1 mark)

An electron microscope rather than a light microscope is used to see organelles inside a cell mainly because the electron microscope:

A　Can view living cells

B　Shows the true colours of the cell

C　Has higher magnification but lower resolution

D　Has higher resolution, so points closer together can be distinguished

Question 9 | Circulatory system (1 mark)

An athlete at rest has a stroke volume of 70 cm³ and a heart rate of 72 beats min⁻¹. What is the cardiac output?

A　0.5 dm³ min⁻¹

B　5.04 dm³ min⁻¹

C　50.4 dm³ min⁻¹

D　142 cm³ min⁻¹

Question 10 | Gas exchange (1 mark)

Which of the following is NOT a feature that adapts alveoli for gas exchange?

A　A thin wall — a single layer of epithelial cells, about 0.5 μm

B　A rich capillary network that maintains the concentration gradient

C　A thick layer of mucus on the inner surface to keep it moist

D　A large total surface area, about 70 m²

Question 11 | Kidney and excretion (1 mark)

In which part of the nephron is glucose completely reabsorbed?

A　Bowman’s capsule

B　Proximal convoluted tubule

C　Loop of Henle

D　Collecting duct

Question 12 | Ventilation (1 mark)

A person has a tidal volume of 500 cm³ and a breathing rate of 15 per minute. The pulmonary ventilation rate is:

A　33.3 cm³ min⁻¹

B　515 cm³ min⁻¹

C　7 500 cm³ min⁻¹

D　75 000 cm³ min⁻¹

Question 13 | Transport in plants (1 mark)

The main force that pulls water up the xylem from root to leaf is:

A　Active transport by companion cells

B　Root pressure

C　Transpiration pull (cohesion–tension)

D　Capillary action

Question 14 | Muscles and movement (1 mark)

What is the relationship between the biceps and the triceps at the elbow joint?

A　Sphincters controlling a tube

B　Cardiac muscle working involuntarily

C　Synergists that contract together for the same movement

D　Antagonists: one contracts while the other relaxes, giving movement both ways

Question 15 | Digestion (1 mark)

What does bile do in fat digestion?

A　Neutralises stomach acid and digests protein

B　Absorbs amino acids into the villi

C　Hydrolyses triglycerides into fatty acids and glycerol

D　Emulsifies fat into small droplets, increasing the surface area for lipase

Question 16 | Nerve conduction (1 mark)

A myelinated neurone conducts impulses faster than an unmyelinated one because:

A　The action potential jumps from one node of Ranvier to the next

B　A smaller axon diameter means lower resistance

C　Myelin conducts electricity directly along the axon

D　More synapses allow faster relay

Question 17 | Energy flow (1 mark)

Producers in an ecosystem fix 250 000 kJ m⁻² per year and each trophic transfer is 10% efficient. How much energy reaches the secondary consumers (third trophic level)?

A　25 000 kJ m⁻² yr⁻¹

B　2 500 kJ m⁻² yr⁻¹

C　250 kJ m⁻² yr⁻¹

D　250 000 kJ m⁻² yr⁻¹

Question 18 | Population estimates (1 mark)

Researchers capture and mark 40 snails, then release them. A week later they catch 50 snails, 8 of which are marked. The estimated population size is:

A　200

B　250

C　320

D　90

Question 19 | Carbon cycle (1 mark)

Which process removes carbon dioxide from the atmosphere and fixes it as organic matter?

A　Respiration

B　Photosynthesis

C　Burning fossil fuels

D　Decomposition

Question 20 | Greenhouse effect (1 mark)

The enhanced greenhouse effect works because:

A　The Sun’s total output increases

B　A thinner ozone layer lets more ultraviolet light reach the ground

C　Greenhouse gases absorb outgoing infrared radiation and re-emit it, reducing heat loss

D　The proportion of oxygen in the atmosphere rises

Question 21 | Population growth (1 mark)

In logistic (S-shaped) growth, the growth rate is greatest when the population is about:

A　Close to zero

B　2K

C　Close to the carrying capacity K

D　K/2

Question 22 | Species interactions (1 mark)

Which of these is an example of mutualism?

A　A tapeworm and a pig

B　A lion and an antelope

C　Head lice and humans

D　Rhizobium bacteria and leguminous plants

Question 23 | Meiosis (1 mark)

Homologous chromosomes pair and exchange segments (crossing over) during:

A　Prophase I

B　Metaphase I

C　Anaphase II

D　Telophase II

Question 24 | Mendelian genetics (1 mark)

In the cross AaBb × aabb (the two genes are on different chromosomes) there are 400 offspring. The expected numbers of the four phenotypes are:

A　100 of each

B　300 : 100 : 100 : 0

C　225 : 75 : 75 : 25

D　200 : 100 : 100 : 0

Question 25 | Population genetics (1 mark)

A recessive disease affects 4% of a population (q² = 0.04) which is in Hardy–Weinberg equilibrium. The frequency of carriers (heterozygotes) is:

A　0.04

B　0.16

C　0.32

D　0.64

Question 26 | DNA replication (1 mark)

The Meselson–Stahl ¹⁵N experiment showed that DNA replication is:

A　Conservative

B　Semi-conservative

C　Dispersive

D　Random

Question 27 | Protein synthesis (1 mark)

A polypeptide has 150 amino acids. What is the minimum number of mRNA bases that encode it, including the stop codon and ignoring any other untranslated region?

A　50

B　150

C　450

D　453

Question 28 | Mutation (1 mark)

The molecular cause of sickle-cell anaemia is:

A　A single base substitution changes amino acid 6 of the β-globin chain from glutamate to valine

B　Deletion of the whole haemoglobin gene

C　An abnormal chromosome number (aneuploidy)

D　Methylation of haemoglobin that abolishes its function

Question 29 | Natural selection and resistance (1 mark)

The rise of antibiotic-resistant bacteria in hospitals is explained in Darwinian terms because:

A　Resistance depends only on dose and has nothing to do with inheritance

B　Bacteria actively make resistance genes on exposure to the antibiotic

C　Resistant variants already exist in the population; the antibiotic kills sensitive cells, so resistant ones survive and reproduce

D　The antibiotic directly mutates bacteria into resistant forms

Question 30 | Classification and naming (1 mark)

Which statement about the scientific name Panthera leo is correct?

A　Both are family names

B　Scientific names are written in English and need no italics

C　Panthera is the species name and leo is the genus

D　Panthera is the genus, leo the specific epithet; both are italicised

Answer sheet

Q	Q	Q
1	11	21
2	12	22
3	13	23
4	14	24
5	15	25
6	16	26
7	17	27
8	18	28
9	19	29
10	20	30

Answers and mark scheme

Question 1 (1 mark)

A1　Correct answer：A

A ✓　A small temperature change for a given heat gain or loss buffers body temperature — correct.

B　High latent heat of vaporisation explains heat loss by sweating; it is a cooling mechanism, not the buffer itself.

C　Surface tension is unrelated to thermoregulation; it matters for water in narrow tubes.

D　Being a solvent matters for transport and metabolism, not for thermal buffering.

Question 2 (1 mark)

A1　Correct answer：B

A　Treating 45 mm as 45 μm and dividing by 30 — the units were never converted.

B ✓　45 mm = 45 000 μm, and 45 000 ÷ 30 = 1 500 — correct.

C　A factor of 10 too big, from using 1 mm = 100 μm.

D　The formula inverted (actual ÷ image), giving a value below 1.

Question 3 (1 mark)

A1　Correct answer：A

A ✓　2 cm: surface 6 × 4 = 24 cm², volume 8 cm³, ratio 3:1; 4 cm: 96 ÷ 64 = 1.5:1 — correct.

B　The two cubes are swapped; doubling the side halves SA:V, it does not double it.

C　6:1 belongs to a 1 cm cube, not a 2 cm one.

D　The 4 cm cube has SA:V 1.5:1, not 6:1.

Question 4 (1 mark)

A1　Correct answer：B

A　Efflux needs a lower outside water potential; −200 kPa is higher than −450 kPa, so the direction is reversed.

B ✓　Cell water potential −450 kPa is below the solution’s −200 kPa, so water enters until ψp balances it — correct.

C　The two potentials differ by 250 kPa; they are not equal.

D　A plant cell has a wall: it gains turgor pressure instead of bursting.

Question 5 (1 mark)

A1　Correct answer：C

A　The mass is unchanged; only the shape changes.

B　Hydrolysis needs a protease or prolonged strong acid; heating does not cut peptide bonds.

C ✓　Primary structure is unchanged; the weaker bonds holding secondary and tertiary structure break first, distorting the active site — correct.

D　Sequence is set by the gene; heat does not reorder it, that would be a mutation.

Question 6 (1 mark)

A1　Correct answer：C

A　Dividing by 2 assumes warming destroys half the activity, the opposite of Q₁₀.

B　An unchanged rate over 10 °C would mean Q₁₀ = 1.

C ✓　Q₁₀ = 2 means the rate doubles per 10 °C: 4.0 × 2 = 8.0 — correct.

D　Multiplying by 2 twice treats it as a 20 °C rise.

Question 7 (1 mark)

A1　Correct answer：A

A ✓　Accumulating against the gradient requires energy; root-hair cells supply it with many mitochondria — correct.

B　Endocytosis takes in macromolecules or particles, not single ions.

C　Osmosis moves water only, not ions.

D　Facilitated diffusion runs down the gradient and cannot accumulate ions against it.

Question 8 (1 mark)

A1　Correct answer：D

A　Electron microscopy needs a vacuum and fixed, dehydrated specimens, so living cells cannot be viewed.

B　Images are greyscale; any colour is added afterwards.

C　It is the other way round: electron microscopes resolve far better.

D ✓　The electron beam has a far shorter wavelength (resolution about 0.1 nm), which reveals organelle detail — correct.

Question 9 (1 mark)

A1　Correct answer：B

A　Reads 5 040 cm³ as 500 cm³ before converting.

B ✓　70 × 72 = 5 040 cm³ min⁻¹ = 5.04 dm³ min⁻¹ — correct.

C　Decimal slip: 1 dm³ is 1 000 cm³, not 100 cm³.

D　Adds the two figures (70 + 72) instead of multiplying.

Question 10 (1 mark)

A1　Correct answer：C

A　A thin wall shortens the diffusion distance, an adaptation.

B　Rich blood flow removes oxygen and delivers carbon dioxide, holding the gradient — an adaptation.

C ✓　Thick mucus would lengthen the diffusion path and block exchange; the alveolar surface has only a very thin film of fluid — correct.

D　A large area is exactly what raises the diffusion rate, so it is an adaptation.

Question 11 (1 mark)

A1　Correct answer：B

A　Bowman’s capsule filters; glucose is filtered out there, not taken back.

B ✓　The proximal tubule has microvilli and many mitochondria and reabsorbs glucose completely by sodium–glucose co-transport — correct.

C　The loop of Henle builds the medullary osmotic gradient and reclaims water and some salt.

D　The collecting duct adjusts water reabsorption under ADH; healthy urine contains no glucose.

Question 12 (1 mark)

A1　Correct answer：C

A　Divides 500 by 15 — the formula inverted.

B　Adds the two figures (500 + 15).

C ✓　Tidal volume × rate = 500 × 15 = 7 500 cm³ min⁻¹ — correct.

D　A factor of 10 too big from a unit slip.

Question 13 (1 mark)

A1　Correct answer：C

A　Companion cells energise the phloem, not the xylem.

B　Root pressure lifts water only a few tens of centimetres, not to the top of a tall tree.

C ✓　Transpiration in the leaves creates negative pressure; cohesion holds the water column together so it is pulled up — correct.

D　Capillary action is a minor contributor in wide vessels.

Question 14 (1 mark)

A1　Correct answer：D

A　Sphincters are rings of muscle around tubes such as the gut and bladder neck.

B　Both are skeletal muscles under voluntary control.

C　They pull in opposite directions, so they are not synergists.

D ✓　Muscle can only pull, so flexion and extension need a pair of antagonists working in turn — correct.

Question 15 (1 mark)

A1　Correct answer：D

A　Neutralising chyme is a side effect of bile; proteins are digested by proteases.

B　Absorption happens across the villus epithelium by diffusion and active transport, not by bile.

C　Hydrolysis is done by pancreatic lipase; bile contains no hydrolytic enzyme.

D ✓　Bile salts are emulsifiers: they break large fat globules into small ones, giving lipase far more surface — correct.

Question 16 (1 mark)

A1　Correct answer：A

A ✓　Myelin insulates the axon so the action potential is regenerated only at the nodes of Ranvier — saltatory conduction — correct.

B　A larger diameter lowers resistance; myelinated axons are usually the thicker ones.

C　Myelin is lipid and insulates; it prevents charge leakage rather than conducting.

D　Each synapse adds a delay of about 0.5 ms, so more synapses means slower.

Question 17 (1 mark)

A1　Correct answer：B

A　That is the primary consumer level — only one 10% step.

B ✓　250 000 × 10% = 25 000 (primary consumers), then × 10% = 2 500 kJ m⁻² yr⁻¹ — correct.

C　One 10% step too many — that is the fourth trophic level.

D　No transfer loss applied at all (assuming 100% efficiency).

Question 18 (1 mark)

A1　Correct answer：B

A　Uses 40 × 50 ÷ 10, mis-copying the recapture number.

B ✓　N = (marked first time × second sample) ÷ marked in second sample = 40 × 50 ÷ 8 = 250 — correct.

C　An arithmetic slip in the ratio, not a valid method.

D　Adds the two catches (40 + 50) and ignores the recapture proportion.

Question 19 (1 mark)

A1　Correct answer：B

A　Respiration releases carbon dioxide — the reverse direction.

B ✓　Photosynthesis reduces carbon dioxide to carbohydrate using light — the main carbon-fixation route.

C　Combustion returns stored carbon to the atmosphere as carbon dioxide.

D　Decomposers respire and also release carbon dioxide.

Question 20 (1 mark)

A1　Correct answer：C

A　Solar output is stable on a decadal scale and is not the driver.

B　Ozone depletion is a separate issue affecting UV, skin and eyes, not the greenhouse effect.

C ✓　Carbon dioxide and methane absorb infrared and re-emit part of it downwards, so net heat loss falls — correct.

D　Oxygen is not a greenhouse gas and its share is not rising.

Question 21 (1 mark)

A1　Correct answer：D

A　With very few individuals the absolute growth rate is small, not maximal.

B　2K is beyond carrying capacity and cannot persist.

C　Near K resources run short and the growth rate approaches zero.

D ✓　The inflection point of the S-curve is at K/2, where growth is fastest; it then slows under limiting factors — correct.

Question 22 (1 mark)

A1　Correct answer：D

A　The tapeworm absorbs nutrients in the host gut — parasitism.

B　The lion kills and eats the antelope — predation.

C　Lice feed on blood at the host’s expense — parasitism.

D ✓　The bacteria fix nitrogen for the plant and receive sugar and shelter — both benefit, correct.

Question 23 (1 mark)

A1　Correct answer：A

A ✓　In prophase I the homologues form bivalents and non-sister chromatids exchange segments, giving recombination — correct.

B　In metaphase I the bivalents line up at the equator; crossing over is already done.

C　Anaphase II separates sister chromatids.

D　By telophase II four haploid nuclei have formed.

Question 24 (1 mark)

A1　Correct answer：A

A ✓　AaBb makes four gamete types in equal numbers and aabb contributes only ab, so the four phenotypes appear 1:1:1:1, i.e. 100 each — correct.

B　That is a mis-applied test-cross ratio and does not even total 400.

C　225:75:75:25 is the 9:3:3:1 outcome of AaBb × AaBb, not a test cross.

D　That assumes linkage, but the question states the genes are on different chromosomes.

Question 25 (1 mark)

A1　Correct answer：C

A　That is q² itself — the disease frequency, not the carrier frequency.

B　That is p² (0.8²), the homozygous dominant frequency.

C ✓　q = 0.2 and p = 0.8, so 2pq = 2 × 0.8 × 0.2 = 0.32 — correct.

D　That is p² + 2pq, everyone showing the dominant phenotype.

Question 26 (1 mark)

A1　Correct answer：B

A　Conservative replication keeps one intact parental duplex, which the first generation did not show.

B ✓　Generation one was all hybrid density (one old strand, one new) and generation two was half hybrid, half light — semi-conservative, correct.

C　Dispersive replication shifts density gradually and gives no clean two-band pattern.

D　“Random” is not a recognised mode of replication.

Question 27 (1 mark)

A1　Correct answer：D

A　Divides by 3 instead of multiplying — 150 ÷ 3.

B　Assumes one base per amino acid, ignoring the triplet code.

C　150 × 3 = 450 covers only the amino-acid codons and omits the stop codon.

D ✓　150 codons × 3 = 450 bases, plus 3 bases of stop codon = 453 — correct.

Question 28 (1 mark)

A1　Correct answer：A

A ✓　A GAG → GTG substitution is a missense mutation; the protein polymerises at low oxygen — correct.

B　Whole-gene deletion abolishes the protein altogether, a different severe anaemia, not sickle cell.

C　Sickle cell is a single-gene point mutation; the chromosome number is normal.

D　Methylation is epigenetic regulation, not the cause here.

Question 29 (1 mark)

A1　Correct answer：C

A　Resistance is inherited via chromosomal genes or plasmids; dose only sets the strength of selection.

B　Resistance genes arise by mutation or horizontal transfer beforehand, not on demand after exposure.

C ✓　Variation comes first and selection second: the antibiotic is the selective pressure, so resistance alleles rise in frequency — correct.

D　Antibiotics do not direct mutation; mutations are random.

Question 30 (1 mark)

A1　Correct answer：D

A　The family (Felidae) is a rank above the genus and is not either of these words.

B　Scientific names are Latin (or latinised) and must be italicised under the codes.

C　The order is reversed: genus first, specific epithet second.

D ✓　Binomial nomenclature: genus capitalised, epithet lower case, both italicised (underlined in handwriting) — correct.

Grade guide (out of 30, this paper only)

Grade	Marks	Description
7	26–30	Secure across all four themes; numerical items and conceptual judgements both accurate.
6	22–25	Mostly correct; loses marks on one or two physiological mechanisms or genetic ratios.
5	18–21	Basic ideas clear, but energy flow, the nephron or molecular detail is muddled.
4	14–17	Copes with basic cell and genetics items; more errors in physiology and ecology.
3	10–13	Isolated facts only; cannot link concepts to explain phenomena.
1–2	0–9	Re-read all four themes and rebuild from definitions and diagram questions.

Suggested timing (45 minutes)

Stage	Minutes
Questions 1–10 (water, molecules, cells, enzymes, membranes)	15
Questions 11–20 (human physiology, gas exchange, energy flow)	15
Questions 21–30 (ecology, genetics, molecular biology, evolution)	13
Review: transfer answers to the answer sheet and check for gaps	2

Notices

•　Time allowed: 45 minutes. 30 multiple-choice questions, one mark each, 30 marks in total. No calculator.

•　Each question has exactly one correct answer. Multiple answers or a blank score zero; there is no negative marking.

•　Record all answers on the answer sheet. Notes written on the question paper are not marked.

•　For the numerical items (magnification, surface-area-to-volume ratio, cardiac output, energy transfer) show your working in the margin so it can be checked.

•　The mark scheme explains every option, including why each distractor is wrong.

Same series | DSE practice papers | DSE Biology Paper 1

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All questions are original and are not IB past-paper questions. The grade bands above are this paper’s own guide: IB Biology grades are awarded on Paper 1, Paper 2 and the Internal Assessment combined.

IB Biology SL Paper 1 Practice Paper (45 minutes, 30 marks, multiple choice) with mark scheme