IB Biology HL Theme A: Unity and diversity questions

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.

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.

Answer: B. A universal genetic code and genes shared across the three domains support common ancestry and LUCA. A molecular clock estimates LUCA’s age, while fossils provide evidence that cells existed by the fossil date, not the exact beginning of life. Earlier lineages could have become extinct through competition with LUCA and its descendants. A: Treats the oldest known fossil as the exact origin of life; fossil evidence does not establish that life began then. C: Reverses the dating approaches: molecular clocks estimate LUCA’s age, while fossils provide evidence of ancient cells. D: Equates LUCA with the first life and treats its estimated age as excluding earlier forms. Remember: LUCA is the latest common ancestor of extant life, not necessarily the first life.

Answer: B. Chitin is characteristic of fungal cell walls. The absence of cross-walls and presence of many nuclei in continuous cytoplasm identify an aseptate fungal hypha.

A: associates a cell wall with plants but overlooks that plant walls contain cellulose, not chitin.
C: matches the multiple nuclei but overlooks that skeletal muscle cells have no cell wall.
D: identifies a fungus from the chitin wall but contradicts the observation that there are no cross-walls.
Remember: aseptate fungal hyphae contain multiple nuclei in continuous cytoplasm.

Answer: B. From 20 to 30 min, assembled phages inside bacteria fall from 20 to 0 while free infectious phages rise from 0 to 18. This indicates release; the phage DNA copy number remains unchanged during this interval.

A: selects the earlier increase in phage DNA, which indicates replication rather than release.
C: identifies the interval when assembled particles appear inside bacteria, indicating assembly before release.
D: assigns replication to an interval when the phage DNA copy number is unchanged.

Remember: distinguish replication, assembly and release by tracking what changes inside and outside the host cell.

Answer: A. The key routes an animal with no shell to the next distinction, where six legs identifies a water beetle. The eDNA barcode CTAG matches the freshwater shrimp reference barcode.

B: identifies the observed animal correctly but assigns the sample barcode to the pond snail, whose barcode is GACT.
C: matches the sample barcode correctly but selects the shell-present route for the observed animal.
D: swaps the key identification and the species identified from the eDNA barcode.
Remember: a dichotomous key uses observable features; an eDNA barcode is compared with reference barcodes.

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.

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.

Answer: B. EDGE prioritization considers both evolutionary distinctness and global endangerment. The tree shows that Species L has a longer branch not shared with close relatives than Species M: 5 units compared with 1 unit. Both are globally endangered, whereas Species K’s longer independent branch is labelled not globally endangered. Species L therefore fits both considerations.

A: selects the longest independent lineage but overlooks that Species K is not globally endangered.
C: uses the global-endangerment label but overlooks that Species M shares most of its recent lineage with Species N.
D: could result from misreading Species N’s status; it is not globally endangered and has only a short unshared branch.

Remember: EDGE combines evolutionary distinctness with global endangerment.