Biology HL

1. 1.

   Two bird populations can interbreed and produce fertile offspring. A bacterium reproduces by cell division and can receive genes from other bacterial lineages. Apply the biological species concept to the birds, then explain two difficulties in applying the same criterion to the bacterium. Suggest an alternative type of evidence that can help classify bacterial lineages.

   [4 marks] · no calculator

   Worked explanation

   1. Apply the reproductive criterion where sexual reproduction provides relevant evidence.
   2. Binary fission and horizontal transfer do not provide the same mating boundary as the bird example.
   3. Different operational species criteria may be useful; lack of a universal criterion does not erase biological diversity.

   Common mistake: Horizontal gene transfer is not evidence that all bacteria form one freely interbreeding species.

   Marking points

   • Producing fertile offspring is consistent with the bird populations belonging to one biological species.
   • An interbreeding criterion cannot directly classify an asexually reproducing lineage by mating outcomes.
   • Horizontal gene transfer can move genes across lineage boundaries without sexual reproduction.
   • Classification of bacteria therefore needs other evidence such as genomic and phenotypic similarities, not a claim that bacteria cannot be classified.

   Examiner tip: Horizontal gene transfer is not evidence that all bacteria form one freely interbreeding species.

2. 2.

   Four hypothetical invertebrates have these observable features: A has six legs and wings; B has six legs and no wings; C has eight legs and no antennae; D has more than eight legs and antennae. Construct a dichotomous key identifying all four using only these supplied features.

   [4 marks] · no calculator

   Worked explanation

   1. Choose a feature dividing the supplied set into exclusive groups, for example six versus not six legs.
   2. Split A/B by wings and C/D by eight versus more than eight legs.
   3. Check each specimen reaches exactly one endpoint without relying on habitat or an unsupplied feature.

   Common mistake: A dichotomous key needs two exclusive choices at each decision, not a list of four organism descriptions.

   Marking points

   • First paired choice separates six legs (go to 2) from not six legs (go to 3).
   • Second paired choice separates wings present (A) from wings absent (B).
   • Third paired choice separates eight legs (C) from more than eight legs (D).
   • All alternatives are paired, observable, non-overlapping and lead unambiguously to one of A-D; equivalent valid keys earn credit.

   Examiner tip: A dichotomous key needs two exclusive choices at each decision, not a list of four organism descriptions.

3. 3.

   A hypothetical pond-water DNA sample contains a barcode matching a reference for frog species F. Explain how this can help identify biodiversity, why it does not prove a live frog was present at the instant of sampling, and give one control against contamination.

   [4 marks] · no calculator

   Worked explanation

   1. A sequence match supports identification only to the extent the reference and marker distinguish species reliably.
   2. Distinguish molecular detection from a count, location or viability measurement.
   3. Process a blank using the same sampling/extraction equipment and reagents to test for introduced DNA.

   Common mistake: Barcode read number is not automatically a count of living individuals and a positive match does not date the DNA's origin.

   Marking points

   • DNA sequences can be compared with validated reference barcodes to infer the source species.
   • Water sampling can detect traces without catching or observing the organism.
   • DNA may persist after departure/death or arrive from elsewhere, so a match is not direct evidence of a live frog at that instant.
   • A field or extraction blank processed alongside samples can detect contamination; matching negative controls strengthen confidence.

   Examiner tip: Barcode read number is not automatically a count of living individuals and a positive match does not date the DNA's origin.

4. 4.

   Hypothetical organisms A and B have 98% identity in one short DNA barcode. They have not been tested for interbreeding. Explain why the barcode can support an inference of relatedness but cannot by itself demonstrate that A and B are the same biological species. Identify one useful additional line of evidence.

   [3 marks] · no calculator

   Worked explanation

   1. A barcode is a limited sample of a genome, not a complete description of population boundaries.
   2. Apply the biological-species concept only where reproductive evidence is meaningful, rather than inventing a universal 98% cutoff.
   3. Choose independent evidence that can distinguish the proposed alternatives instead of simply resequencing the identical short fragment.

   Common mistake: There is no universal barcode percentage that makes two organisms the same biological species in every group.

   Marking points

   • Similarity in homologous DNA can support shared ancestry or relatedness.
   • A percentage from one short region is not a universal species threshold and does not establish reproductive compatibility or isolation.
   • Additional evidence could include multiple independent genetic regions and/or reproductive, ecological or morphological evidence appropriate to the organisms.

   Examiner tip: There is no universal barcode percentage that makes two organisms the same biological species in every group.

5. 5.

   Use this hypothetical dichotomous key: 1, wings present -> go to 2; wings absent -> group W. 2, one visible wing pair -> group X; two visible wing pairs -> go to 3. 3, hard protective forewings -> group Y; membranous forewings -> group Z. An intact specimen has two visible pairs of wings with hard forewings. Identify its group and explain one reason a damaged specimen could be misidentified.

   [3 marks] · no calculator

   Worked explanation

   1. Start at statement 1 and use only the supplied observable characters.
   2. The branch sequence ends at Y, regardless of an assumed familiar common name.
   3. A key depends on characters being observed correctly; missing parts do not imply the living organism originally lacked them.

   Common mistake: Do not skip the first couplet or infer taxonomic identity from a damaged character without checking it.

   Marking points

   • Follows wings present, then two wing pairs, then hard protective forewings.
   • Identifies group Y.
   • Damage could remove a wing or obscure the diagnostic character, leading to the wrong branch; inspect intact structures or use another character.

   Examiner tip: Do not skip the first couplet or infer taxonomic identity from a damaged character without checking it.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.