IB Biology A3.1: Diversity of organisms worked solutions

B is supported because the first word assigns genus. A repeated second word across genera does not identify one species. Names alone establish neither habitat nor chromosome count.

Write Pisum sativum: initial capital for the genus, lowercase epithet and italics for both words [1]. The genus is Pisum [1].

C tests successful reproduction in the next generation. Similar features, mating behaviour or offspring survival alone do not establish fertility.

There are 7\times2=14 chromosome objects [1]. Replication supplies two chromatids per object, so there are 14\times2=28 chromatids [1]. Joined sisters are counted as one chromosome.

A adds two structural matching criteria to length. Chromosomes need not be next to one another in the source micrograph; colour density alone is not a band-pattern comparison.

Add all chromosome objects: 44+1+1=46 [1]. The X and Y provide an XY complement, predicting male under the supplied model [1]. This does not test every gene variant.

D includes all three DNA-containing compartments and does not restrict the definition to genes that are expressed. A population allele collection is a gene pool.

750\,\mathrm{Mbp}/1000=0.750\,\mathrm{Gbp} [1]. DNA amount includes non-coding regions and does not count genes, so gene number cannot be deduced from this value [1].

Opposite leaves send you from 1b to decision 2; the toothed edge then selects 2b, ending at D [1]. The decision sequence is an identification route, not an ancestry statement.

A species may be absent from the reference set, so its read cannot be assigned reliably [1]. Add a reference from an independently identified specimen, or analyse an additional informative barcode region [1]. Do not invent a species identification from the best poor match.

(a) The shared genus is Larus [1]. (b) The complete binomials differ, so the labels assign distinct species [1]. A biological test asks whether representative members can interbreed and produce fertile offspring; sharing a genus is insufficient [1].

(a) Different shell bands support a morphological distinction [1], but offspring survival demonstrates viability, not fertility, so a biological-species conclusion is premature [1]. (b) Rear hybrid offspring to reproductive maturity and test whether they produce offspring [1]. Use comparable within-population crosses as controls, with similar age and conditions, so a failure is not merely caused by the procedure [1].

(a) Match a with e, b with d and c with f [1 for the complete pairing]. (b) Compare length, centromere position and staining-band pattern; award [2] for two precise matching criteria beyond simply calling them the same shape. (c) There are six chromosome objects [1]. Joined sister chromatids do not double this count.

The total is 2+3+2+2=9 [1]. Group II contains three rather than the usual two chromosomes, identifying the extra chromosome [1]. The sex group is XY, predicting male under the stated model [1]. No parental origin follows from this record.

(a) R has two unreplicated sets, so its haploid value is 1.8/2=0.9\,\mathrm{Gbp} [1]. S converts to 1200/1000=1.2\,\mathrm{Gbp} [1]. (b) A larger DNA amount need not mean more genes because non-coding sequence contributes [1]. Complexity needs a stated independent measure; these two records alone cannot rank it [1].

(a) 120/8=15, so the time is reduced by a factor of 15 [1]. (b) Sequencing determines DNA base order [1]. (c) Identify variants in the individual patient [1], then use validated associations with drug response or risk to inform that patient care [1]. Sequence alone does not establish a variant function or prescribe treatment.

(a) Gametes contain 16/2=8 and 18/2=9, so the hybrid has 8+9=17 chromosomes [1]. (b) Some parental chromosomes may lack matching homologues [1]; irregular meiotic segregation can then produce unbalanced gametes, lowering fertility [1]. (c) The count alone does not establish the actual pairing behaviour, so sterility is not certain [1].

One complete key is: 1a alternate leaves: go to 2; 1b opposite leaves: go to 3 [1]. 2a smooth: W; 2b toothed: X [1]. 3a smooth: Y; 3b toothed: Z [1]. Test each endpoint using every specimen, preferably with another observer using the key unaided, and revise any ambiguous decision [1]. Each alternative has a destination and the paired statements do not overlap.

(a) Between-population success is 6/10\times100=60\%; controls give 9/10\times100=90\% [1]. Their difference is 90-60=30 percentage points [1]. (b) Distinct wing patterns support a morphological split [1], but overlapping dimensions and viable broods do not establish a complete reproductive barrier or descendant fertility [1]. (c) Rear descendants to maturity and test reproductive success [1], with comparable within-population controls, repeated crosses and matched conditions [1]. The small cross sample supports a provisional judgement rather than a firm species boundary.

(a) F contains two unreplicated sets: 6.4/2=3.2\,\mathrm{Gbp} [1]. I converts to 800/1000=0.8\,\mathrm{Gbp} [1]. (b) 3.2/0.8=4 [1]. (c) F has more DNA than I yet a lower score, providing a counterexample to an invariably increasing size-complexity claim [1]. Three chosen species and one scoring system cannot establish a universal relationship [1]. (d) Sequence corresponding regions in the taxa [1], then compare aligned differences across suitable regions to test relationship hypotheses [1]. Genome size itself is not an alignment.

(a) Ordinary diploid totals are 2\times12=24 and 2\times11=22 [2, one each]. (b) Conserved blocks occur in the predicted joined order [1]. Internal end-like repeats or the inactive centromere provide an independent feature expected after joining two former chromosomes [1]. (c) Loss of a chromosome could also lower the number [1]. Mapping the retained ancestral sequence blocks in the descendant tests fusion against simple loss [1]. This is a testable reconstruction, not proof of the cause of a species split.

(a) The strains reproduce without sexual crosses, so an interbreeding/fertility test is inapplicable [1]. Horizontal transfer moves a gene between groups outside parent-to-offspring inheritance, further weakening a simple reproductive boundary [1]. (b) Equal barcode matches leave the two references unresolved [1]. A transferred resistance gene does not establish whole-genome identity or one species, because the rest of the genome may have a different history [1]. (c) A positive blank is evidence of possible contamination [1]; collect independent samples with clean equipment and fresh blanks, and use additional informative regions or verified specimens [1]. (d) DNA amount or read number is affected by shedding, persistence and amplification, so it cannot directly count living individuals [1].

(a) The normal haploid number is 24/2=12 [1]. Joining two chromosomes into one reduces the count by one without requiring loss of their sequence blocks [1]. (b) The fused chromosome must associate with the corresponding regions on two unfused partners [1], so segregation can create unbalanced chromosome combinations and reduce the production of functional gametes [1]. (c) Test actual fertility or offspring chromosome balance rather than declaring inevitable sterility [1]. Evidence that gene exchange is persistently restricted is needed before attributing a species boundary to the rearrangement [1].

(a) 4/500\times100=0.8\% of the aligned positions differ [1]. (b) Equal size states DNA amount, not base order [1]. A 500-base fragment is not a representative whole-genome estimate by itself [1]. (c) Sample multiple individuals from each population [1] and measure alternative-base frequencies at the aligned sites to determine whether the variants are common within either population [1]. (d) Compare additional corresponding regions with reliable references or an outgroup to test evolutionary relationship hypotheses [1]; no species boundary follows from the four differences alone.

(a) The rare plant may occur locally [1], but DNA transported from the upstream population could also explain the result [1]. (b) Take replicated samples along the watercourse and look for local specimens [1]; identify them using verified diagnostic traits and the tested key, providing independent morphological evidence [1]. (c) Use paired non-overlapping statements about the same feature and test the route on both known species [1]. (d) A detection does not estimate abundance because DNA concentration depends on shedding, transport, decay and laboratory amplification, not just organism number [1].