IB Biology HL Theme A: Unity and diversity questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceHard1 mark

For one star, a model places the Goldilocks zone—the orbital band where surface conditions could permit liquid water—at 1.20–2.00 × 10¹¹ m. A survey retains a site only if it is in this band, has liquid water and shows the expected solubility pattern for known hydrophilic and hydrophobic probes. Which site should be retained?

SiteOrbital distance / (10¹¹ m)Water observedHydrophilic probe in waterHydrophobic probe in water
A1.50LiquidDissolvedDissolved
B1.70IceNo liquid sampleNo liquid sample
C1.80LiquidDissolvedUndissolved
D2.20LiquidDissolvedUndissolved
Show answer and explanation

Answer: C. The modeled zone spans 1.20–2.00 × 10¹¹ m, so site C at 1.80 × 10¹¹ m is within it. Its sample is liquid, with the hydrophilic probe dissolved and the hydrophobic probe undissolved, matching water’s solvent properties. Many hydrophilic molecules dissolve in water, and most enzymes catalyse reactions in aqueous solution, so these observations support a possible medium for metabolism, but do not establish that life is present.
A: the location and liquid water fit, but dissolution of the known hydrophobic probe does not match the expected solvent pattern.
B: being in the zone does not make ice a liquid aqueous medium for enzyme catalysis.
D: the solvent observations fit, but 2.20 × 10¹¹ m is beyond the survey’s zone.
Remember: the Goldilocks zone makes surface liquid water possible; it does not guarantee water or life.

Question 2

Paper 2Hard16 marks

Two bacterial lineages, A and B, reproduce by binary fission. They can be distinguished by an inherited feature unrelated to gene marker M. At the start of an investigation, M was detected in all sampled A colonies and none of the B colonies. A mixed culture contained both lineages; a control culture contained B only. At each time point, 100 colonies of each lineage were sampled and tested for M.

01020300102030405060708090100Generations after start of cultureColonies carrying M (%)Percentage of sampled colonies carrying gene marker M○ Lineage A, mixed culture△ Lineage B, mixed culture□ Lineage B, B-only control
(a)

State two features of the biological species concept used to decide whether populations belong to the same species.

[2]
(b)

Describe the change in the percentage of lineage B colonies carrying M in the mixed culture over the 30 generations.

[2]
(c)

Suggest the most likely explanation for the appearance of M in lineage B, using all three series in Figure 1.

[3]
(d)

Two populations, P and Q, of another organism reproduce only asexually. Explain why the biological species concept cannot establish whether P and Q are the same or different species.

[3]
(e)

A student concludes that A and B must be one biological species because they share M after mixing, and that P and Q must be different species because they cannot interbreed. Assess these two conclusions, using Figure 1 and the biological species concept.

[6]

Question 3

Multiple choiceHard1 mark

A museum’s catalogue covers many species and must organize them for comparative studies while reflecting inferred ancestry. The table gives the branching sequence, from earliest split to latest, and current class assignments. A clade contains a common ancestor and all its descendants. Which proposal meets both aims?

InformationData
Branching sequenceS separates from P, Q and R; then R separates from P and Q; then P separates from Q
Current classesM: P and Q; N: R and S
Show answer and explanation

Answer: B. A clade contains a common ancestor and all its descendants. The branching sequence gives nested clades P+Q and P+Q+R (and the full set); R+S is not a clade because their common ancestor also has P and Q as descendants. Classification can still organize extensive diversity for comparison, while unranked clades avoid implying that fixed taxonomic ranks measure evolutionary divergence.
A: It assumes both traditional classes are clades; class N excludes P and Q even though they descend from the same ancestor as R and S.
C: It treats the limitation of fixed ranks as a reason to discard classification, although classification facilitates comparative study.
D: It assigns rank by group size; the branching pattern does not require nested clades to have class and order ranks.
Remember: clades express common ancestry, whereas taxonomic ranks are fixed categories.

Question 4

Multiple choiceHard1 mark

Researchers recorded responses to courtship displays between and within two species, then recorded the outcomes of resulting matings. Evolution is a change in heritable characteristics of a population. Which interpretation is supported?

PairingCourtship displays accepted (%)Matings producing viable offspring (%)Viable offspring that were fertile (%)
P × P849296
Q × Q819095
P × Q8880
Show answer and explanation

Answer: C. Courtship displays were accepted less often between species than within either species, indicating a behavioural barrier to mating. Although some interspecific matings produced viable offspring, none of the viable hybrids were fertile, so they could not pass alleles to later generations. The results indicate barriers to allele flow but do not show a change in heritable characteristics of a population over time.

A: Confuses viability with fertility; the hybrids were viable but could not reproduce.
B: Reverses the barrier stages: courtship affects mating before fertilization, while sterility occurs in hybrids.
D: Treats a difference between pairings as evidence of heritable change over time; the data do not establish that change.

Remember: Viable offspring can still be sterile and unable to transmit alleles to another generation.

Question 5

Paper 2Hard16 marks

Ecologists surveyed the same twelve 1 ha woodland plots in 2012, 2018 and 2024. Expert teams used the same method, visited at the same time of year and spent the same amount of time surveying each plot. The table shows the numbers of individuals recorded for six native bird species. A zero indicates that no individuals of that species were recorded.

Species201220182024
A405070
B302412
C20144
D1072
E850
F200

A citizen-science portal also collected observations in the surrounding landscape. A record is one submitted sighting, not a count of unique species. The portal did not use fixed observer-hours or fixed locations, and sightings were not independently checked.

YearRecords submittedContributorsHabitat types sampled
20121 200803
20245 1006009

Other estimates for the same landscape were:

Measure20122024
Human population42 00059 000
Native woodland remaining / ha1 200780
Woodland converted to agriculture or settlement since 2012 / ha0420
Plots occupied by an invasive alien plant / 1217
(a)

Calculate the percentage change in recorded bird species richness between 2012 and 2024.

[2]
(b)

Describe how species richness and evenness changed in the bird survey between 2012 and 2024.

[3]
(c)

State one benefit of citizen-science observations and explain why the increase in portal records alone does not establish an increase in biodiversity.

[4]
(d)

A conservation group concludes that biodiversity in this landscape is recovering and that any decline in the bird survey was caused solely by human population growth. Evaluate whether the data support these conclusions.

[7]

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