Animal Diversity: Systematics, Classification and Phylogenetic Relationships

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. A zoologist erects a new taxon at the rank of family, based on the type genus Testudo. Under the International Code of Zoological Nomenclature, which name has the correct form for a family?

   1. Testudinoidea
   2. Testudininae
   3. Testudinidae
   4. Testudinaceae
   Show answer

   Answer: C — Testudinidae

   Zoological family names end in -idae, so the family is Testudinidae. Testudinoidea carries the superfamily ending -oidea and Testudininae the subfamily ending -inae; both are tempting because they are valid zoological endings, but for other ranks. The ending -aceae marks a family in botanical nomenclature, not in zoology.

2. The only name-bearing specimen of a moth species, designated in the original description, was destroyed in a museum fire, and no other specimen of the original type series survives. A specialist designates a newly collected specimen as the name-bearer. What is this specimen called?

   1. A lectotype
   2. A neotype
   3. A paratype
   4. A syntype
   Show answer

   Answer: B — A neotype

   A neotype is designated when the name-bearing type has been lost or destroyed. A lectotype is the tempting answer, but it is chosen from surviving syntypes of the original series, and none survive here. Paratypes and syntypes are specimens of the original type series, so a newly collected specimen cannot be either.

3. A bilaterally symmetrical, triploblastic animal has a complete gut, a body cavity formed from the blastocoel with mesoderm lining only the body wall, only longitudinal muscles in its body wall, and a cuticle that it moults as it grows. To which phylum does it belong?

   1. Annelida
   2. Platyhelminthes
   3. Arthropoda
   4. Nematoda
   Show answer

   Answer: D — Nematoda

   A blastocoel-derived pseudocoelom, longitudinal muscles only and a moulted cuticle together identify Nematoda. Arthropoda is the tempting choice because arthropods also moult (both are ecdysozoans), but arthropods are coelomates with a haemocoel and jointed appendages. Annelids have a true coelom with circular and longitudinal muscles, and flatworms are acoelomate with a blind gut.

4. In the classical embryological scheme, which of the following are characteristic of deuterostomes as opposed to protostomes?

   1. The blastopore becomes the anus and the mouth forms secondarily
   2. Early cleavage is radial and indeterminate
   3. The coelom forms by splitting of solid mesodermal bands (schizocoely)
   4. A trochophore larva occurs in the life cycle
   Show answer

   Answer: A — The blastopore becomes the anus and the mouth forms secondarily; B — Early cleavage is radial and indeterminate

   Deuterostomes (echinoderms, hemichordates, chordates) classically show the blastopore becoming the anus and radial, indeterminate cleavage. Schizocoely and the trochophore larva are protostome features, typical of annelids and molluscs.

5. Blastomeres are separated at the four-cell stage in embryos of two marine invertebrates. In species X each isolated blastomere forms a complete though small larva; in species Y each forms only part of a larva, with specific structures missing. Which interpretation is most consistent with these results?

   1. X has radial, indeterminate cleavage as in a sea urchin, and Y has spiral, determinate cleavage as in a marine snail
   2. X has spiral, determinate cleavage as in a marine snail, and Y has radial, indeterminate cleavage as in a sea urchin
   3. Both have spiral cleavage, and the difference arises because only X forms its coelom by enterocoely
   4. Both are deuterostomes, and the difference arises because only Y develops a pseudocoelom
   Show answer

   Answer: A — X has radial, indeterminate cleavage as in a sea urchin, and Y has spiral, determinate cleavage as in a marine snail

   Indeterminate cleavage means early blastomeres keep their full developmental potential, so each can form a whole larva, as in sea urchins (deuterostomes). The determinate spiral cleavage of protostomes such as snails fixes each blastomere's fate early, so an isolated cell gives only its allotted part. The reversed pairing is the tempting distractor; the mode of coelom formation and a pseudocoelom have nothing to do with the potency of early blastomeres.

6. Which structure is correctly matched with the phylum in which it is characteristically found?

   1. Flame cells (protonephridia) — Annelida
   2. Colloblasts on the tentacles — Cnidaria
   3. Radula in the buccal cavity — Arthropoda
   4. Water vascular system — Echinodermata
   Show answer

   Answer: D — Water vascular system — Echinodermata

   The water vascular system with tube feet is unique to echinoderms. Flame cells are the protonephridia of flatworms (annelids have metanephridia), colloblasts belong to ctenophores (cnidarians have cnidocytes), and the radula is the feeding organ of molluscs.

7. Which of the following groups is monophyletic?

   1. Reptiles as traditionally defined: turtles, lizards, snakes and crocodilians, but not birds
   2. Fishes as traditionally defined: jawless, cartilaginous and bony fishes, but not tetrapods
   3. Warm-blooded vertebrates: birds and mammals grouped together on the basis of endothermy
   4. Archosauria: crocodilians, birds and all other descendants of their latest common ancestor
   Show answer

   Answer: D — Archosauria: crocodilians, birds and all other descendants of their latest common ancestor

   Archosauria contains a common ancestor and all its descendants, so it is a clade. Traditional reptiles and fishes are paraphyletic because they leave out some descendants (birds; tetrapods). Grouping birds with mammals by endothermy is polyphyletic: endothermy evolved independently in the two lineages, so the group rests on a convergent character.

8. A study of relationships among five species of the cat family uses a species of the dog family as outgroup. One character scored is “presence of mammary glands”, which the dog and all five cats show. For resolving relationships among the five cats, this character is best described as —

   1. A synapomorphy that unites all five cats as a clade separate from the dog
   2. A symplesiomorphy that is uninformative about relationships among the five cats
   3. An autapomorphy of the cat family that separately diagnoses each of the five species
   4. A homoplasy, because mammary glands arose independently in the cats and in the dog
   Show answer

   Answer: B — A symplesiomorphy that is uninformative about relationships among the five cats

   The state is present in the outgroup and in every ingroup species, so outgroup comparison makes it ancestral (plesiomorphic) for the cats, and a shared ancestral state groups nothing. It is a synapomorphy only at the level of Mammalia as a whole, which is the tempting misreading. It is not unique to one species, so not an autapomorphy, and mammary glands in cats and dogs are homologous, not convergent.

9. Which of the following statements about reading phylogenetic trees are correct?

   1. Rotating the two branches that arise from a node does not change the relationships shown
   2. In a phylogram, branch lengths are proportional to the amount of character change
   3. An unrooted tree shows which of its taxa is ancestral to the others
   4. Two tips are sister groups only if they are printed next to each other
   Show answer

   Answer: A — Rotating the two branches that arise from a node does not change the relationships shown; B — In a phylogram, branch lengths are proportional to the amount of character change

   Relationships are carried by the nodes, so swapping branches about a node leaves the tree unchanged, and phylogram branch lengths do scale with change. An unrooted tree shows no direction of descent, and living tips are not ancestors of one another anyway. Adjacency of tips on the page means nothing because branches rotate freely; sister groups are defined by an immediate common ancestor.

10. Which marker is most suitable for resolving relationships among animal phyla that diverged more than 500 million years ago?

    1. The mitochondrial COI barcode region, because it varies between closely related species
    2. Microsatellite loci, because their high mutation rate gives many alleles per locus
    3. The nuclear 18S rRNA gene, because its slow rate of change avoids saturation
    4. The mitochondrial control region, because it is the fastest-evolving part of the mitochondrial genome
    Show answer

    Answer: C — The nuclear 18S rRNA gene, because its slow rate of change avoids saturation

    Deep divergences need slowly evolving sequences; otherwise repeated substitutions at the same sites erase the signal (saturation). 18S rRNA was the marker that first defined the modern animal tree. COI is tempting because it is the standard animal barcode, but its fast rate suits species identification, not phyla; microsatellites and the control region evolve faster still and suit populations.

11. A distance matrix is computed for four species whose lineages have evolved at very different rates, one about three times faster than the others. Which method is most likely to recover a wrong topology from this matrix, and why?

    1. UPGMA, because it assumes that all lineages evolve at the same rate and builds an ultrametric tree
    2. Neighbour-joining, because it assumes a clock and forces every tip to be equidistant from the root
    3. Maximum likelihood, because it cannot accommodate different branch lengths in different lineages
    4. The bootstrap, because resampling sites with replacement creates rate differences among lineages
    Show answer

    Answer: A — UPGMA, because it assumes that all lineages evolve at the same rate and builds an ultrametric tree

    UPGMA clusters by averaged distances under a strict-clock assumption, so a fast lineage looks distant from its true sister and is placed wrongly. Neighbour-joining is the tempting option, but it was designed precisely to correct for unequal rates and returns an unrooted tree. Maximum likelihood estimates a separate length for every branch, and the bootstrap measures support; it is not a tree-building method.

12. Parsimony groups two distantly related, fast-evolving species together with strong bootstrap support. Maximum likelihood with a well-fitting substitution model, after adding slower-evolving relatives of each species, places them in separate clades. What is the most likely explanation?

    1. The two species are true sisters, and maximum likelihood undercounts the substitutions they share
    2. Long-branch attraction: parallel changes on two long branches were read by parsimony as synapomorphies
    3. Strong bootstrap support guarantees a correct clade, so the parsimony result must be preferred
    4. Parsimony corrects better than likelihood for repeated substitutions at a site, so it is right
    Show answer

    Answer: B — Long-branch attraction: parallel changes on two long branches were read by parsimony as synapomorphies

    On long branches many sites change by chance to the same state; parsimony, which has no model of multiple hits, counts these parallel changes as shared derived states and joins the long branches. A model-based likelihood analysis and extra taxa that break the long branches remove the artefact. The bootstrap measures consistency within the data, not correctness, so it can strongly support an artefact, and parsimony has no correction for multiple hits at all.

13. Which of the following conclusions are supported by molecular phylogenies of animals?

    1. Arthropods and nematodes both belong to the moulting clade Ecdysozoa
    2. Annelids and arthropods together form a clade, Articulata, united by segmentation
    3. Tunicates, rather than cephalochordates, are the sister group of vertebrates
    4. Flatworms are the sister group of all other bilaterians, as their acoelomate plan suggests
    Show answer

    Answer: A — Arthropods and nematodes both belong to the moulting clade Ecdysozoa; C — Tunicates, rather than cephalochordates, are the sister group of vertebrates

    Ecdysozoa (arthropods, nematodes and other moulting animals) and the tunicate–vertebrate pairing are well supported by molecular data. Articulata is rejected: annelids are lophotrochozoans and arthropods ecdysozoans, so segmentation arose more than once or was lost. Flatworms nest within Lophotrochozoa, so their acoelomate condition is not evidence of a basal position.

14. How many distinct rooted, fully bifurcating trees are possible for 6 taxa?

    Numerical answer — type the value.

    Show answer

    Answer: 945

    Rooted trees for n taxa = (2n − 3)!! = 1 × 3 × 5 × … × (2n − 3). For n = 6 this is 9!! = 1 × 3 × 5 × 7 × 9 = 945. Using the unrooted formula (2n − 5)!! gives 105, the usual slip; equivalently, the rooted count for 6 taxa equals the unrooted count for 7.

15. Four taxa W, X, Y and Z have eight aligned sites; each string gives the bases of W, X, Y and Z in that order: site 1 AAAA, site 2 GGAA, site 3 CCTT, site 4 TCTC, site 5 AAAG, site 6 TTGG, site 7 CTTC, site 8 AGCT. Using Fitch parsimony, the length (total number of steps) of the most parsimonious unrooted tree is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 11

    Steps per site on W+X | Y+Z: 0, 1, 1, 2, 1, 1, 2, 3 = 11. On W+Y | X+Z: 0, 2, 2, 1, 1, 2, 2, 3 = 13. On W+Z | X+Y: 0, 2, 2, 2, 1, 2, 1, 3 = 13. Sites 2, 3 and 6 favour W+X, site 4 favours W+Y and site 7 favours W+Z; site 5 (singleton) costs 1 and site 8 (four states) costs 3 on every tree. The shortest tree has length 11. Leaving out the uninformative sites gives 7, which is not the tree length.

16. Two aligned mitochondrial sequences of 400 sites differ at 60 sites. Using the Jukes–Cantor correction d = −(3/4) ln(1 − 4p/3), where p is the proportion of differing sites, the estimated number of substitutions per site is ____ (to three decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.167

    p = 60/400 = 0.15. 4p/3 = 0.2, so 1 − 4p/3 = 0.8 and ln 0.8 = −0.2231. d = −0.75 × (−0.2231) = 0.1674, i.e. 0.167 substitutions per site. The corrected value exceeds p (0.150) because some sites have changed more than once; reporting p itself is the usual slip.

17. For four taxa W, X, Y and Z, which aligned site (bases listed for W, X, Y and Z in that order) is parsimony-informative?

    1. AAAA
    2. AAAG
    3. AGAG
    4. AGCT
    Show answer

    Answer: C — AGAG

    A site is informative only if at least two states each occur in at least two taxa. AGAG (W and Y share A, X and Z share G) costs 1 step on the tree pairing W with Y and 2 on the other two, so it discriminates. A constant site costs 0 everywhere, a singleton (AAAG) 1 everywhere and a site with four different bases 3 everywhere, so none of them prefers any tree.