14 free IB Biology questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.
Calculate the Simpson reciprocal index, , for meadow A, where is the total number of organisms and the number of each species.
Answer: 4.78
The Simpson reciprocal index rises with both species richness and evenness; its maximum equals the number of species when all are equally abundant.
The cladogram shows the evolutionary relationships between five species, V to Z. Which group of species forms a clade?
Answer: A
V, W and X: Correct. A clade is a common ancestor and all of its descendants; the node joining X to the V–W branch has exactly V, W and X as descendants (A3.2.4, A3.2.7).
W and X: This leaves out V, which shares a more recent common ancestor with W than X does.
X and Y: The ancestor of X and Y also gave rise to V and W, so leaving them out does not make a clade.
V, W and Y: This leaves out X, which descends from the same common ancestor as V, W and Y.
To test a clade, pick a node and include every terminal branch beyond it; leaving out any descendant makes the group not a clade.
The ringed seal (Pusa hispida) and the black-throated loon (Gavia arctica) both spend time in air and in cold water. Which property in the table best explains why a warm-blooded animal loses body heat much faster in water than in air at the same temperature?
| Property | Air | Water |
|---|---|---|
| Thermal conductivity / W m⁻¹ K⁻¹ | 0.026 | 0.60 |
| Specific heat capacity / J g⁻¹ K⁻¹ | 1.0 | 4.2 |
| Density / kg m⁻³ | 1.2 | 1000 |
| Viscosity / mPa s | 0.018 | 1.0 |
Answer: C
Specific heat capacity, because water warms up rapidly when it is in contact with warm skin: Misreads the property: a high specific heat capacity means water warms up slowly, not rapidly.
Density, because the buoyancy of water reduces the energy needed to support the body: Buoyancy is a real consequence of density, but it affects support, not heat loss.
Thermal conductivity, because water carries heat away from the skin much faster than air: Correct. Water’s thermal conductivity is about 23 times that of air (0.60 ÷ 0.026), so heat is conducted away from the body far faster; this is why the seal relies on a thick blubber layer (A1.1.6).
Viscosity, because water resists movement and so increases the work of swimming: True of viscosity, but it explains the energy cost of swimming, not the rate of heat loss.
Match each physical property of water to its consequence: buoyancy (density), drag (viscosity), heat loss (thermal conductivity), stable temperature (specific heat capacity).
The graph shows oxygen dissociation curves for adult and foetal haemoglobin. In the placenta, the partial pressure of oxygen is about 4 kPa. Which statement explains the importance of the difference between the curves?
Answer: A
Foetal haemoglobin has a higher affinity, so it takes up oxygen released by adult haemoglobin: Correct. The foetal curve lies to the left (80% vs 60% saturated at 4 kPa): at the placental pO₂, adult haemoglobin releases oxygen that the higher-affinity foetal haemoglobin picks up (B3.1.11, B3.1.13).
Foetal haemoglobin has a lower affinity, so it releases oxygen into the mother’s blood at the placenta: Reversed: the foetal curve is to the left, so it has a higher affinity and gains oxygen from the mother.
Adult haemoglobin is more saturated at 4 kPa, so it keeps its oxygen while foetal haemoglobin gives oxygen up: Misread values: at 4 kPa adult haemoglobin is 60% saturated and foetal haemoglobin 80%.
Both curves are S-shaped, so oxygen moves equally in both directions across the placenta: Both are sigmoid because of cooperative binding, but their positions differ, so there is net transfer to the foetus.
A curve further left means higher affinity for oxygen: foetal haemoglobin loads oxygen at partial pressures where adult haemoglobin unloads it.
The trace shows a student breathing normally, then breathing in as deeply as possible and breathing out as far as possible, before returning to normal breathing. What is the student’s vital capacity?
Answer: C
0.5 dm³: This is the tidal volume, the volume of a normal breath (3.0 − 2.5 dm³).
2.0 dm³: This is only the inspiratory reserve volume (5.0 − 3.0 dm³), the extra volume above a normal breath in.
4.0 dm³: Correct. Vital capacity = maximum volume after the deepest breath in (5.0 dm³) − minimum after the fullest breath out (1.0 dm³) = 4.0 dm³ (B3.1.6).
5.0 dm³: This is the peak reading, not the difference between the deepest breath in and the fullest breath out.
Vital capacity = tidal volume + inspiratory reserve + expiratory reserve = the range from maximum inhalation to maximum exhalation.
Cubes of agar with side lengths of 1 cm and 4 cm were placed in a dye that diffuses into agar. Which row gives the surface area-to-volume ratio of the 4 cm cube and the cube whose centre would become coloured first?
Answer: C
6 : 1; the 1 cm cube: 6 : 1 is the ratio for the 1 cm cube; the 4 cm cube has 96 cm² ÷ 64 cm³ = 1.5 : 1.
1.5 : 1; the 4 cm cube: The ratio is right, but the larger cube has the smaller SA:V and the longer diffusion distance, so it colours last.
1.5 : 1; the 1 cm cube: Correct. 4 cm cube: surface area = 6 × 16 = 96 cm², volume = 64 cm³, so SA:V = 1.5 : 1. The 1 cm cube (6 : 1) has more surface per unit volume and a shorter distance to the centre, so it colours first (B2.3.6).
0.67 : 1; the 1 cm cube: This is volume ÷ surface area (64 ÷ 96), the inverse of the ratio asked for.
As a cube gets larger, volume rises faster (side³) than surface area (6 × side²), so SA:V falls; this limits cell size.
Which conclusion is best supported by these results?
| Treatment | Mean elongation / mm (± SD, n = 10) |
|---|---|
| 1. Weak buffer, pH 6.0, no auxin | 0.5 ± 0.1 |
| 2. Weak buffer, pH 6.0, with auxin | 2.4 ± 0.3 |
| 3. Strong buffer, pH 6.0, no auxin | 0.5 ± 0.1 |
| 4. Strong buffer, pH 6.0, with auxin | 0.7 ± 0.2 |
| 5. Strong buffer, pH 4.5, no auxin | 2.6 ± 0.3 |
Answer: B
Auxin promotes elongation by raising the pH of the cell wall, which loosens cross-links between cellulose molecules: Reversed: holding the pH at 6.0 with a strong buffer (treatment 4) blocks the effect of auxin, while a low pH of 4.5 alone (treatment 5) promotes elongation.
Auxin promotes elongation mainly by acidifying the cell wall, which loosens cross-links between cellulose molecules: Correct. Treatments 2 and 4 differ only in buffer strength: preventing acidification with a strong buffer largely removes auxin's effect (2.4 vs 0.7 mm), and the strong buffer alone does not slow growth (treatment 3 = treatment 1). An acidic medium without auxin (5) causes as much elongation as auxin (2). This supports auxin stimulating H⁺ secretion into the cell wall, lowering its pH and loosening cross-links so turgor can stretch the cells.
Low pH destroys auxin, so auxin is not needed for coleoptile elongation in the plant: Treatment 5 contained no auxin at all; the data show acid can replace auxin, not that acid destroys it.
Auxin promotes growth only by increasing the rate of cell division in the segments: The measured response is elongation within 3 hours and is mimicked by acid alone, which fits cell expansion rather than division.
Acid-growth mechanism: auxin → H⁺ pumped into the cell wall → lower pH → cross-links between cellulose molecules loosen → cells elongate.
Which explains the fall in CO₂ concentration between May and September of each year?
Answer: A
Photosynthesis by vegetation in the northern hemisphere exceeds respiration during the growing season: Correct. In the northern summer, plants in the large land area of the northern hemisphere fix more CO₂ by photosynthesis than is released by respiration, so CO₂ falls; in winter respiration and decomposition exceed photosynthesis and CO₂ rises.
Respiration by soil decomposers increases in the summer months: Increased respiration would release CO₂ and raise the concentration, not lower it.
More CO₂ dissolves in the oceans as they warm during the summer: Gases are less soluble in warmer water, so warming oceans would release CO₂ rather than absorb more.
Combustion of fossil fuels stops during the summer months: Combustion continues all year and explains the long-term rise, not the seasonal fall.
Keeling Curve: the annual fluctuation reflects seasonal photosynthesis vs respiration in the northern hemisphere; the long-term rise reflects combustion of fossil fuels and biomass.
Algae photosynthesizing in bright light were suddenly deprived of carbon dioxide while the light remained on. Which row shows the immediate changes in the concentrations of glycerate 3-phosphate (GP) and RuBP in the chloroplasts?
Answer: C
GP: increases; RuBP: decreases: This is the pattern when the light is switched off: ATP and reduced NADP run out, so GP is not reduced and accumulates while RuBP is not regenerated.
GP: decreases; RuBP: decreases: RuBP is still regenerated using ATP from the light-dependent reactions but is no longer consumed by carbon fixation, so it rises.
GP: decreases; RuBP: increases: Correct. Without CO₂, Rubisco cannot convert RuBP to GP, so GP is no longer made while it is still being reduced to triose phosphate (GP falls). Light-dependent reactions still supply ATP, so triose phosphate continues to regenerate RuBP, which accumulates.
GP: increases; RuBP: increases: GP is made only by carboxylation of RuBP, which requires CO₂, so GP cannot increase.
No CO₂: GP falls, RuBP rises. No light: GP rises, RuBP falls. Reason from which step of the Calvin cycle has stopped.
The graph shows the initial rate of a reaction with no inhibitor and with inhibitor P or inhibitor Q at a fixed concentration. Which conclusion about the inhibitors is best supported by the data?
Answer: A
P competes with the substrate for the active site; Q binds to an allosteric site and changes the shape of the active site: Correct. The effect of P becomes smaller as substrate concentration rises (73 vs 89 at 16 mmol dm⁻³), as substrate out-competes a competitive inhibitor. Q lowers the rate by about half at every concentration, so adding substrate cannot overcome it: non-competitive (allosteric) inhibition.
P binds to an allosteric site and changes the active site; Q competes with the substrate for the active site: Reversed: it is P whose effect is overcome by more substrate, which is the signature of competitive inhibition.
P and Q both compete with the substrate for the active site, but Q binds more strongly: Q's maximum rate stays at about half the uninhibited rate even at high substrate, which is not the pattern of a competitive inhibitor.
P binds irreversibly to the active site; Q binds reversibly to the active site: An irreversible (mechanism-based) inhibitor would not have its effect reduced by adding substrate, as P does.
Competitive inhibition: the effect is reduced at high substrate concentration. Non-competitive inhibition: the maximum rate is lowered whatever the substrate concentration.
A female fruit fly from a cross between a true-breeding grey, normal-winged fly and a true-breeding black, vestigial-winged fly was test crossed with a black, vestigial-winged male (ggnn). Which conclusion is supported by the results?
| Offspring phenotype | Number of flies |
|---|---|
| Grey body, normal wings | 415 |
| Black body, vestigial wings | 405 |
| Grey body, vestigial wings | 92 |
| Black body, normal wings | 88 |
Answer: A
The genes are linked; grey-vestigial and black-normal flies are recombinants formed by crossing over in the female: Correct. Unlinked genes would give a 1 : 1 : 1 : 1 ratio. Instead the parental combinations (G with N, g with n) make up 820 of 1000 offspring, so the genes are on the same chromosome. The 180 grey-vestigial and black-normal flies carry new combinations produced by crossing over between the loci in the female.
The genes are linked; grey-normal and black-vestigial flies are recombinants formed by crossing over: Confuses parental and recombinant classes: grey-normal and black-vestigial match the original P-generation combinations and are the most frequent.
The genes are unlinked, because all four phenotypes appear among the offspring: Linked genes also give all four phenotypes when crossing over occurs; the unequal proportions, not the presence of four classes, are the evidence.
The genes are linked and no crossing over occurred between them: Without crossing over only the two parental phenotypes would appear; the 180 recombinant flies show crossing over did occur.
In a dihybrid test cross, a large excess of parental-type offspring over the 1 : 1 : 1 : 1 expected indicates linkage; the small classes are recombinants.
What pattern of inheritance does the pedigree show for this disorder?
Answer: D
Autosomal dominant: An affected individual with a dominant allele must have at least one affected parent; II-1 and III-1 have unaffected parents.
X-linked recessive: III-1 (an affected son of unaffected parents) fits this, but II-1, an affected daughter of an unaffected father, rules it out.
X-linked dominant: A dominant allele cannot appear in children of two unaffected parents.
Autosomal recessive: Correct. II-1 is affected although neither parent (I-1, I-2) is, so the allele is recessive and both parents are carriers. II-1 is a female whose father is unaffected; for an X-linked recessive allele an affected daughter must have an affected father, so the gene is autosomal.
Affected child of unaffected parents → recessive; an affected daughter with an unaffected father → not X-linked, so autosomal.
A plant cell has a solute potential of −750 kPa and a pressure potential of +250 kPa. It is placed in a sucrose solution with a water potential of −600 kPa. Which describes the net movement of water?
Answer: B
Into the cell, because the solute potential of the cell (−750 kPa) is lower than the water potential of the solution: Uses ψs alone and ignores the pressure potential; the direction depends on the whole water potential of the cell.
Out of the cell, because the water potential of the cell (−500 kPa) is higher than that of the solution: Correct. kPa. Water moves from higher to lower water potential, so it moves from the cell (−500 kPa) into the solution (−600 kPa), and the pressure potential of the cell falls.
Into the cell, because the water potential of the cell (−1000 kPa) is lower than that of the solution: Sign error: the pressure potential was subtracted instead of added (−750 − 250).
Into the cell, because water moves from the solution (−600 kPa) to the higher water potential of the cell (−500 kPa): Correct values but reversed rule: water moves from higher to lower water potential, not towards the higher value.
Calculate the cell's (keep the signs), then water moves from higher (less negative) to lower (more negative) ψ.
The genotype frequencies do not fit the Hardy–Weinberg equation. Which is the most likely explanation?
| Genotype | AA | Aa | aa |
|---|---|---|---|
| Number of plants | 290 | 220 | 90 |
Answer: C
The population is very large, so chance effects are small: A large population is one of the Hardy–Weinberg conditions; it would help keep the population in equilibrium.
Heterozygous plants survive better than either homozygote: Heterozygote advantage would give more heterozygotes than expected, but there are fewer (220 vs about 267).
The plants often self-pollinate, so mating is not random: Correct. With p = 0.67 and q = 0.33, about 267 AA, 267 Aa and 67 aa are expected. There are fewer heterozygotes and more homozygotes than expected, the pattern produced by self-pollination (inbreeding), which breaks the random-mating condition.
A high rate of mutation is converting allele A into allele a: Mutation changes allele frequencies very slowly and would not by itself produce a deficit of heterozygotes.
Compare observed with p², 2pq and q². A deficit of heterozygotes suggests non-random mating (e.g. self-fertilization); an excess suggests selection favouring heterozygotes.
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