IB Biology HL Theme A: Unity and diversity questions

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. 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.