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IB Biology A3.2: Classification and cladistics worked solutions

Worked solutions to every practice question in A3.2: Classification and cladistics of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

HL[1 mark]
The lion (Panthera leo) and the grey wolf (Canis lupus) both belong to the order Carnivora. Which statement must be true? (A) They belong to the same family. (B) They belong to the same class. (C) They belong to the same genus. (D) They belong to the same family and genus.
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A shared order forces every higher rank to match, so both are in the same class (B). Family and genus are lower ranks than order, so sharing the order does not force them to match: the lion is a felid and the wolf a canid.

Final answer

B

Question A2

HL[1 mark]
Two cladograms are drawn from the same sequence data. Tree 1 needs 9 changes and tree 2 needs 11 changes. Which statement is correct? (A) Tree 1 is preferred because it needs the fewest changes. (B) Tree 2 is preferred because it explains more changes. (C) The trees are equally probable because they use the same data. (D) Tree 2 is preferred because it needs more steps.
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Parsimony prefers the hypothesis that needs the fewest changes, so tree 1 is the more probable (A). The other options reverse or ignore the principle.

Final answer

A

Question A3

HL[1 mark]
Which molecule did Woese compare in 1977 to show that archaea have a line of descent separate from bacteria? (A) 16S ribosomal RNA (B) A messenger RNA for one enzyme (C) DNA polymerase (D) A transfer RNA for one amino acid
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Woese compared base sequences of 16S ribosomal RNA, which every cell has and which changes slowly enough to compare across all life (A). The other molecules are not universal markers used for the domains.

Final answer

A

Question A4

HL[1 mark]
Which group of tips on the cladogram forms a clade? [Graph: A cladogram with five tips. P and Q share a node; that node and R share a node; S and T share a node; the two groups meet at the root.] (A) Q and R (B) P, Q and R (C) R and S (D) P, Q, R and S
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A clade is a common ancestor and all its descendants. The ancestor of P, Q and R has only these three descendants (B). Q and R omit P, R and S come from separate branches, and P, Q, R and S omit T.

Final answer

B

Question A5

HL[1 mark]
Use the cladogram in A4. Which statement is correct? (A) S is more closely related to T than to R. (B) R is more closely related to S than to P. (C) P is more closely related to S than to Q. (D) Q is more closely related to T than to R.
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S and T share their own node, while S meets R only at the root, so S is closer to T (A). R shares a more recent node with P than with S (B false), P shares a more recent node with Q than with S (C false) and Q shares a more recent node with R than with T (D false).

Final answer

A

Question A6

HL[1 mark]
Which data give the best evidence of relationships among four species of bird? (A) The base sequence of the same gene in all four species (B) The base sequence of a different gene in each species (C) The colour of the feathers of each species (D) The number of eggs laid by each species
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Corresponding sequences of the same gene can be aligned, and the differences are inherited, so they show relationships (A). Different genes cannot be compared position by position, and feather colour and egg number can arise by convergence or vary with environment.

Final answer

A

Question A7

HL[1 mark]
Two species differ at 6 positions in a gene. Differences between the pair accumulate at 2 per million years. What is the estimated time since they diverged? (A) 0.3 million years (B) 3 million years (C) 8 million years (D) 12 million years
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Time = differences divided by rate = 6\div2=3 million years (B). The other values come from multiplying, adding or inverting the quantities.

Final answer

B

Question A8

HL[1 mark]
What is an advantage of classifying organisms by evolutionary relationships instead of by appearance? (A) Every member of a group shares a common ancestor, so groups reflect evolutionary history. (B) Organisms in a group always look the same. (C) The number of ranks is fixed. (D) No DNA or protein sequences are needed.
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Groups made of a common ancestor and all its descendants reflect real history (A). Members of a clade need not look alike, clades are unranked and the method uses sequence evidence.

Final answer

A

Question A9

HL[1 mark]
Two plant groups in different families both have fleshy, spiny stems. Sequence data show that they have separate ancestors. What best explains the similar stems? (A) Convergent evolution in similar environments (B) Divergent evolution from a recent common ancestor (C) A recent mutation shared by the two groups (D) Inheritance of acquired characteristics
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Similar structures in groups with separate ancestors arise by convergent evolution under similar selection pressures (A). A recent common ancestor or a shared mutation would put the groups on one branch, and acquired characteristics are not inherited.

Final answer

A

Tutorial practice

Question B1

HL[5 marks]
The cladogram shows the relationships of five mammals, from DNA sequences. [Graph: A cladogram with the cow branching first, then the hippopotamus, then the baleen whale, then the sperm whale and dolphin as the closest pair.] (a) State the organism most closely related to the sperm whale. [1] (b) Deduce, with a reason, whether the baleen whale or the hippopotamus is more closely related to the dolphin. [2] (c) A student claims that dolphins evolved from sperm whales. Evaluate the claim using the cladogram. [2]
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(a) Dolphin (1). (b) The baleen whale (1); it shares a more recent common ancestor with the dolphin than the hippopotamus does (1). (c) The claim is not supported (1): the sperm whale and the dolphin are both living tips that share a common ancestor, and neither is the ancestor of the other (1).

Final answer

(a) Dolphin. (b) Baleen whale: more recent shared node. (c) Not supported: both are living tips sharing an ancestor.

Question B2

HL[5 marks]
The first 12 bases of the same gene were aligned for four species: W \mathrm{ATGCCGTAACGT}, X \mathrm{ATGCCGTTACGT}, Y \mathrm{ATGACGTTACCT} and Z \mathrm{TTGACATTGACA}. Differences between W and X: 1; X and Y: 2; W and Z: 8; X and Z: 7. (a) Determine the number of differences between W and Y, and between Y and Z. [2] (b) State which two species are most closely related and why. [1] (c) Deduce the order in which the four species diverged, taking Z as the outgroup. [2]
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(a) W and Y differ at positions 4, 8 and 11: 3 differences (1). Y and Z differ at positions 1, 6, 9, 10 and 12: 5 differences (1). (b) W and X: they have the fewest differences (1), so they share the most recent common ancestor. (c) Z has the most differences from the others, so it diverged first (1); Y is next because it differs from X by 2 and from W by 3, and W and X separated most recently (1).

Final answer

(a) W and Y: 3; Y and Z: 5. (b) W and X: only one difference. (c) Z first, then Y, then W and X.

Question B3

HL[6 marks]
A gene was sequenced in three species P, Q and R. Fossils date the divergence of P and Q at 15 Myr ago, and they differ at 27 positions per 1000 bases. P and R differ at 63 positions per 1000 bases. (a) Calculate the rate of change for a pair, in differences per million years. [2] (b) Estimate the time since P and R diverged. [2] (c) Suggest two reasons why the estimate may be inaccurate. [2]
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(a) Rate = differences divided by time =27\div15 (1) =1.8 differences per Myr (1). (b) Time = 63\div1.8 (1) =35 Myr (1). (c) Any two: the mutation rate depends on the generation time (1); it depends on the population size (1); it depends on the intensity of selection on the gene (1); the rate calibrated on one pair or one gene may not hold for another (1).

Final answer

(a) 1.8 per Myr. (b) 35 Myr. (c) Generation time, population size or selection differ; genes change at different rates.

Question B4

HL[6 marks]
Four traits (1 present, 0 absent) were recorded in species P, Q, R and S and an outgroup O. Traits 1 to 4 in order: O \mathrm{0000}, S \mathrm{1000}, R \mathrm{1100}, Q \mathrm{1110}, P \mathrm{1111}. (a) Construct a cladogram for the five species. [3] (b) State the number of trait changes your cladogram needs. [1] (c) State the purpose of the outgroup. [1] (d) Explain why a tree that needs more changes is less probable. [1]
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(a) The outgroup O branches first (1). The traits nest: S then R then the pair Q and P, so S branches, then R, then Q and P are sister species (1); trait 1 is marked on the branch to S, R, Q and P, trait 2 to R, Q and P, trait 3 to Q and P and trait 4 on P alone (1). In nested form: (O,(S,(R,(Q,P)))). (b) 4 changes: each trait arises once (1). (c) The outgroup shows the ancestral state, so traits it lacks are derived (1). (d) By parsimony the hypothesis needing the fewest changes is the most probable, and any other tree would need a trait to arise more than once (1).

Final answer

(a) O first, then S, then R, then Q and P together. (b) 4. (c) It shows the ancestral state. (d) Parsimony: fewest changes is most probable.

Question B5

HL[4 marks]
The cladogram shows five species and an outgroup O. Trait X (square) and trait Y (circle) are present in the species marked. [Graph: A cladogram: the outgroup O branches first; A and B share a node; C shares a node with D and E; trait X present in C, D and E; trait Y present in A and E.] (a) State the branch on which trait X most probably arose. [1] (b) Explain why this is more probable than separate origins in C, D and E. [2] (c) Species A and E share trait Y. Suggest why this does not make A and E close relatives. [1]
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(a) The branch leading to the common ancestor of C, D and E (1). (b) One origin on that branch explains the trait in all three species with one change (1); separate origins would need three changes, so by parsimony the single origin is more probable (1). (c) A and E are on separate branches, so trait Y most probably arose independently by convergent evolution and is not inherited from a recent common ancestor (1).

Final answer

(a) The branch to the ancestor of C, D and E. (b) One change against three: parsimony. (c) Trait Y arose independently: convergence.

Question B6

HL[6 marks]
Many organisms are now classified by sequence evidence rather than structure. (a) Explain the importance of classifying organisms. [2] (b) Discuss the advantages of classifying by evolutionary relationships rather than by a hierarchy of fixed ranks. [4]
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(a) More than 2 million species are known, so a shared system lets scientists and the public communicate about the same organism (1); it organises organisms so that relatives can be found and features of a new species predicted from its group (1). (b) Any four: every member of a group descends from one common ancestor, so groups reflect evolutionary history (1); sequence evidence is objective and separates species that look identical (1); it can show that a structural classification was wrong because of convergence, as in the figwort family (1); shared ancestry predicts inherited traits (1); unranked clades avoid arbitrary boundaries between ranks (1); a balanced answer notes that sequences are hard to obtain in the field, where structure is more practical (1).

Final answer

(a) Shared names and a way to organise and predict. (b) Common ancestor; objective sequences; shows convergence; no arbitrary ranks; limit in the field.

Question B7

HL[5 marks]
Five species of aquatic plant were placed in family F (species 1, 2 and 3) and family G (species 4 and 5) because of leaf shape. The percentage difference in one gene is shown.\begin{array}{lccccc}\hline \text{}&\text{1}&\text{2}&\text{3}&\text{4}&\text{5}\\ \hline \text{1}&\text{}&\text{1.1}&\text{9.8}&\text{9.5}&\text{9.7}\\ \hline \text{2}&\text{}&\text{}&\text{10.1}&\text{9.9}&\text{10.0}\\ \hline \text{3}&\text{}&\text{}&\text{}&\text{1.4}&\text{1.6}\\ \hline \text{4}&\text{}&\text{}&\text{}&\text{}&\text{1.2}\\ \hline\end{array} (a) State the species most closely related to species 3. [1] (b) Explain how the data show that family F is not a clade. [2] (c) Suggest why species 3 has leaves like species 1 and 2. [1] (d) State how the classification should change. [1]
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(a) Species 4 (1.4 per cent) (1). (b) Species 3 differs from species 1 and 2 by 9.8 and 10.1 per cent (1) but from species 4 and 5 by only 1.4 and 1.6 per cent, so species 3 shares a more recent ancestor with 4 and 5 than with the rest of its family (1). (c) Convergent evolution: similar selection pressures produced similar leaves (1). (d) Species 3 should be reclassified with species 4 and 5 so that each group is a clade (1).

Final answer

(a) Species 4. (b) 3 is closer to 4 and 5 than to 1 and 2. (c) Convergent evolution. (d) Move species 3 into a group with 4 and 5.

Question B8

HL[5 marks]
The percentage difference in 16S ribosomal RNA sequences was compared for an archaeon, a bacterium and a eukaryote (illustrative values): archaeon and bacterium 31, archaeon and eukaryote 24, bacterium and eukaryote 33. (a) State which two organisms are the most similar. [1] (b) Explain how the data support placing archaea in a domain of their own. [2] (c) State the taxonomic level that sits above kingdom. [1] (d) Suggest why 16S ribosomal RNA is a suitable molecule to compare. [1]
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(a) The archaeon and the eukaryote (24 per cent) (1). (b) The archaeon differs from the bacterium by 31 per cent, almost as much as the eukaryote does (33 per cent) (1), and it is closer to the eukaryote (24 per cent) than to the bacterium, so archaea are as distinct from bacteria as eukaryotes are and are not part of the bacteria (1). (c) Domain (1). (d) Every cell has ribosomes, and the sequence changes slowly enough to align across all life (1).

Final answer

(a) Archaeon and eukaryote. (b) Archaea are as different from bacteria as eukaryotes are. (c) Domain. (d) In every cell and changes slowly.

Advanced practice

Question C1

HL[8 marks]
Two genes X and Y were compared in pairs of species whose divergence is dated by fossils. The values are the percentage difference between the pair.\begin{array}{lcccc}\hline \text{Time since divergence / Myr}&\text{10}&\text{20}&\text{30}&\text{40}\\ \hline \text{Gene X / \%}&\text{8.2}&\text{16.0}&\text{24.4}&\text{32.0}\\ \hline \text{Gene Y / \%}&\text{2.0}&\text{4.1}&\text{6.0}&\text{8.0}\\ \hline\end{array} (a) Calculate the rate of change of each gene, in per cent per Myr, from the 40 Myr pair. [2] (b) A new pair differs by 11.2% in gene X. Estimate the time since they diverged. [2] (c) Predict the difference between the new pair in gene Y. [1] (d) Gene Y changes more slowly than gene X. Discuss two reasons and the consequence for dating divergences with these genes. [3]
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(a) Gene X: 32.0\div40=0.80 per cent per Myr (1); gene Y: 8.0\div40=0.20 per cent per Myr (1). (b) Time =11.2\div0.80 (1) =14 Myr (1). (c) 14\times0.20=2.8 per cent (1). (d) Gene Y may code for a product whose structure is tightly constrained, so selection removes most changes (1); gene X may tolerate more neutral changes, which accumulate freely (1); because genes change at different rates each gene needs its own calibration, and the slower gene suits older divergences while the faster suits recent ones (1).

Final answer

(a) X 0.80, Y 0.20 per Myr. (b) 14 Myr. (c) 2.8 %. (d) Selection on an essential product; neutral changes; calibrate each gene.

Question C2

HL[9 marks]
Six traits (1 present, 0 absent) were recorded in five species and an outgroup O. Traits 1 to 6 in order: O \mathrm{000000}, A \mathrm{111001}, B \mathrm{111000}, C \mathrm{110010}, D \mathrm{100100}, E \mathrm{100110}. Two cladograms are proposed. [Graph: Two cladograms for outgroup O and species A to E. Tree 1 groups A with B, then C, then D with E. Tree 2 groups A with B and C with E, then D.] (a) State the minimum number of changes that trait 5 needs on tree 1 and suggest what this indicates. [2] (b) Calculate the total number of changes for the six traits on tree 1 and on tree 2. [3] (c) Deduce which tree is more probable and state the principle. [2] (d) Suggest one reason why trait 5 might arise independently in C and E, and one limit of the principle. [2]
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(a) Trait 5 is in C and E, which are not sister species on tree 1, so it needs 2 changes (1); this suggests it arose independently by convergent evolution (1). (b) Tree 1: traits 1, 2, 3, 4 and 6 need one change each and trait 5 needs two, so 5+2=7 (1). Tree 2: traits 1, 3, 5 and 6 need one change each and traits 2 and 4 need two each, so 4+4=8 (1). Working shown trait by trait (1). (c) Tree 1 (7 changes against 8) (1); parsimony: the hypothesis needing the fewest changes is the most probable (1). (d) C and E may face similar selection pressure, so a similar trait evolves separately (1); parsimony gives only the most probable tree, not a proof, and a trait with a high rate of change can mislead it (1).

Final answer

(a) 2; convergence. (b) Tree 1: 7; tree 2: 8. (c) Tree 1: parsimony. (d) Similar selection; the principle gives only the most probable tree.

Question C3

HL[8 marks]
Sixty herb species were placed in one family because all have two-lipped flowers. Researchers sequenced three chloroplast genes from every species and built a cladogram. The species fall into four groups, and each group is more closely related to species outside the family than to the other three groups. (a) Outline how sequence data are used to build a cladogram. [3] (b) Deduce what the cladogram shows about the original family. [2] (c) Suggest how two-lipped flowers could have appeared in all four groups. [2] (d) State one change that should be made to the classification. [1]
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(a) The same gene is sequenced in every species and the sequences are aligned (1); differences between species are compared, since each is inherited (1); species are grouped by shared changes into nested groups, choosing the tree that needs the fewest changes (1). (b) The family is not a clade: its members do not all descend from one common ancestor (1); the four groups have separate ancestors, each shared with species outside the family (1). (c) Similar selection, for example by the same type of pollinator, acted on each group (1), so the similar flowers arose independently by convergent evolution rather than by inheritance from one ancestor (1). (d) Split the family so that each of the four groups is placed with its closest relatives (1).

Final answer

(a) Same genes aligned; differences counted; species grouped by shared changes. (b) Not a clade. (c) Convergence. (d) Split into four clades.

Question C4

HL[8 marks]
The tree is drawn to scale. A molecular clock was calibrated with the pair A and C, which differ at 42 positions per 1000 bases. [Graph: A tree drawn to scale in millions of years before present. O diverges 50 Myr ago, D 32, C 21, and A and B 9 Myr ago.] (a) State how long ago D diverged from the ancestor of A, B and C. [1] (b) Calculate the rate of change for a pair, per Myr. [2] (c) Predict the number of differences between A and B and between A and O. [3] (d) The measured difference between A and O is 82 per 1000 bases. Suggest two reasons why the clock predicts more. [2]
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(a) 32 Myr (1). (b) A and C diverged 21 Myr ago, so rate =42\div21 (1) =2.0 per Myr (1). (c) A and B diverged 9 Myr ago: 9\times2.0=18 (1); A and O diverged 50 Myr ago: 50\times2.0=100 (1); the method is multiplication of the time at the shared node by the rate for the pair (1). (d) Any two: the lineage leading to O may have a longer generation time, so fewer mutations per Myr (1); more changes in that gene may have been removed by selection (1); the rate is not constant between lineages or over time (1).

Final answer

(a) 32 Myr. (b) 2.0 per Myr. (c) A and B: 18; A and O: 100. (d) Longer generation time or selection in a lineage; the rate is not constant.

Question C5

HL[6 marks]
A group of beetles was one family in 1930. A structural study in 1985 split it into three families. Sequence data in 2015 gave two families, one of which contains species from two of the 1985 families. Separately, Bacteria and Archaea are often called prokaryotes. (a) Using the dates, evaluate the claim that the boundaries between fixed ranks are arbitrary. [3] (b) Using the classical three-domain tree, deduce whether the prokaryotes form a clade. [3]
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(a) The family boundaries changed three times in 85 years as the evidence changed (1), which supports the claim that where the line is drawn is a human choice because variation is gradual (1); but the nested clades revealed by sequences stay the same whichever rank is named, so the groups themselves are not arbitrary (1). (b) The prokaryotes are Bacteria and Archaea (1). Their last common ancestor is the ancestor of all three domains, because Bacteria branch off first (1); Eukarya also descend from that ancestor but are not in the group, so it is not a clade (1).

Final answer

(a) Boundaries moved three times; evidence-based clades stable. (b) No: ancestor also gave Eukarya.

Connected questions

Question X1

Connected questions · HL[8 marks]
A3.2 + A4.1. Dolphins (mammals) and sharks (cartilaginous fish) both have a streamlined body and fins. Sequence data place dolphins closer to the hippopotamus than to sharks. (a) Distinguish between homologous and analogous structures. [2] (b) Explain why the shared body shape does not show that dolphins and sharks share a recent common ancestor, and what evidence would. [3] (c) Explain how natural selection could produce the same body shape in two unrelated groups. [3]
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(a) Homologous structures share an origin in a common ancestor even if their functions differ (1); analogous structures have similar functions but evolved separately (1). (b) The shape is an adaptation to moving fast in water, so it can arise by convergent evolution in separate lineages (1); similarity in structure is therefore not evidence of a recent common ancestor (1); sequences of the same gene, which are inherited, can be compared to find the true relationship (1). (c) Individuals vary in body shape (1); a streamlined shape lowers the cost of swimming, so those individuals survive and breed more (1); the allele frequencies for streamlining rise over generations in each group (1).

Final answer

(a) Homologous: same origin; analogous: same function, separate origin. (b) Convergence; sequences. (c) Same selection pressure in water.

Question X2

Connected questions · HL[7 marks]
A3.2 + D1.2. Part of the coding strand of one gene is shown in two species, with the amino acid for each codon. GCT and GCC code Ala; AAA and AAG Lys; CTG and CTC Leu; GAA Glu; GAT Asp; TGT Cys.\begin{array}{lccccc}\hline \text{}&\text{Codon 1}&\text{2}&\text{3}&\text{4}&\text{5}\\ \hline \text{Species 1}&\text{GCT}&\text{AAA}&\text{CTG}&\text{GAA}&\text{TGT}\\ \hline \text{Species 2}&\text{GCC}&\text{AAG}&\text{CTC}&\text{GAT}&\text{TGT}\\ \hline\end{array} (a) State the number of base differences and the number of amino acid differences. [2] (b) Explain, using the genetic code, why the base differences outnumber the amino acid differences. [2] (c) Explain why base sequences can show finer differences between close relatives than amino acid sequences. [2] (d) State why the same gene must be compared in both species. [1]
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(a) Base differences: codon 1 (T to C), codon 2 (A to G), codon 3 (G to C), codon 4 (A to T) = 4 (1); amino acid difference: only codon 4 (Glu to Asp) = 1 (1). (b) The genetic code is degenerate: several codons code for the same amino acid (1), so the changes in codons 1, 2 and 3 are silent and leave the protein unchanged (1). (c) Silent base changes accumulate without changing the protein (1), so they record differences between close relatives that the amino acid sequences do not show (1). (d) Corresponding positions in the same gene can be aligned and compared one by one (1).

Final answer

(a) 4 base differences; 1 amino acid difference. (b) Degenerate code. (c) Silent changes still show as base differences. (d) Matching positions can be aligned.

Question X3

Connected questions · HL[6 marks]
A3.2 + A2.1. Genes for 16S ribosomal RNA were found in all three domains and are used to trace the earliest divergences. (a) State the name given to the last common ancestor of all living organisms. [1] (b) Explain how a ribosomal RNA gene in all three domains supports a single common ancestor. [2] (c) State which domain shares a more recent common ancestor with Eukarya, using the classical tree. [1] (d) Explain why the sequence of 16S ribosomal RNA changes slowly. [2]
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(a) LUCA, the last universal common ancestor (1). (b) A gene with the same function and recognisably similar sequence in every domain is most simply explained by inheritance from one ancestor (1), rather than by separate origins in each domain (1). (c) Archaea (1). (d) The molecule is part of the ribosome, which every cell needs to make proteins (1), so most changes harm function and are removed by natural selection, leaving a slow rate of change (1).

Final answer

(a) LUCA. (b) A universal gene inherited, not invented several times. (c) Archaea. (d) Ribosome function constrains it; selection removes changes.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Classification and cladistics from the core ideas to full exam questions, with question types, worked examples and graded practice.

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