IB Biology HL Theme A: Unity and diversity questions

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?

Site	Orbital distance / (10¹¹ m)	Water observed	Hydrophilic probe in water	Hydrophobic probe in water
A	1.50	Liquid	Dissolved	Dissolved
B	1.70	Ice	No liquid sample	No liquid sample
C	1.80	Liquid	Dissolved	Undissolved
D	2.20	Liquid	Dissolved	Undissolved

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

Multiple choiceEasy1 mark

During molecular visualization of chromatin, a student selects a DNA–protein unit for closer study. Which arrangement in the model represents a nucleosome?

Show answer and explanation

Answer: C. A nucleosome has DNA wrapped around a core of eight histone proteins, with an additional histone attached to linker DNA.

A: uses seven histone proteins in the core instead of eight. B: places the additional histone on the core rather than on linker DNA. D: omits the additional histone attached to linker DNA.

Remember: distinguish the eight-protein core from the additional linker-associated histone.

Question 3

Paper 2Easy4 marks

Figure 1 shows the number of possible DNA base sequences for strands of different lengths.

(a)

Use Figure 1 to state the number of possible DNA base sequences for a strand of four nucleotide positions.

[1]

(b)

Use the pattern in Figure 1 to explain how a short DNA molecule can store a large amount of information.

[1]

(c)

Explain how complementary base pairing, based on hydrogen bonding, allows genetic information to be replicated and expressed.

[2]

Question 4

Paper 2Easy6 marks

A research group summarized its estimates for the ages of the first living cells and the last universal common ancestor (LUCA) in Figure 1. The estimates are ranges in Ga, where 1 Ga is one billion years before the present. The estimate for the first living cells was based on fossilized evidence in ancient seafloor hydrothermal-vent precipitates; the estimate for LUCA was based on conserved sequences from genomic analysis.

(a)

Using Figure 1, state the estimated age range for each of the first living cells and LUCA.

[2]

(b)

State the interval in which the two estimates overlap.

[1]

(c)

State what the fossilized evidence in ancient seafloor hydrothermal-vent precipitates indicates.

[1]

(d)

State what conserved sequences from genomic analysis indicate about LUCA.

[1]

(e)

Using Figure 1, state what the age estimates show about the time scale over which life has been evolving.

[1]

Question 5

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.

(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 6

Paper 2Easy5 marks

A botanical team reviewed a classification in which flowering plant species A and B were grouped together because their flowers have a similar shape. The team compared the same 12-base region of DNA in the three species.

Species pair	Number of base differences
A and B	12
A and C	2
B and C	10

(a)

Identify the pair with the most similar DNA sequences.

[1]

(b)

Does the DNA evidence support classifying A and B as the closest relatives among these three species? Justify your answer using one comparison from the table.

[2]

(c)

Suggest why classification based on flower shape could have been misleading.

[2]

Question 7

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?

Information	Data
Branching sequence	S separates from P, Q and R; then R separates from P and Q; then P separates from Q
Current classes	M: 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 8

Paper 2Easy6 marks

Researchers record the percentage of beetles in a population with an inherited pale-colour characteristic.

Generation	Beetles with pale colour / %
1	32
5	47

One adult beetle develops greater leg-muscle mass through repeated exercise. This change is not passed to its offspring.

(a)

Calculate the increase in the percentage of beetles with pale colour from generation 1 to generation 5, in percentage points.

[2]

(b)

State what is meant by evolution.

[2]

(c)

Explain why the change in leg-muscle mass in the adult beetle does not count as evolution.

[2]

Question 9

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?

Pairing	Courtship displays accepted (%)	Matings producing viable offspring (%)	Viable offspring that were fertile (%)
P × P	84	92	96
Q × Q	81	90	95
P × Q	8	88	0

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 10

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.

Species	2012	2018	2024
A	40	50	70
B	30	24	12
C	20	14	4
D	10	7	2
E	8	5	0
F	2	0	0

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.

Year	Records submitted	Contributors	Habitat types sampled
2012	1 200	80	3
2024	5 100	600	9

Other estimates for the same landscape were:

Measure	2012	2024
Human population	42 000	59 000
Native woodland remaining / ha	1 200	780
Woodland converted to agriculture or settlement since 2012 / ha	0	420
Plots occupied by an invasive alien plant / 12	1	7

(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]

IB Biology HL Theme A: Unity and diversity questions

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