Paleontology: Diversity and Extinction, Taphonomy, Functional Morphology, Morphometrics, Vertebrates and Paleoecology

Practice questions (18)

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 clay layer at a stratigraphic boundary contains an iridium anomaly, shocked quartz grains and spherules, and a buried crater about 180 km across has been dated to the same time. Which mass extinction is this evidence most directly associated with?

   1. The end-Ordovician extinction, attributed to Gondwanan glaciation and a fall in sea level
   2. The end-Permian extinction, attributed to Siberian Traps volcanism and ocean anoxia
   3. The end-Triassic extinction, attributed to volcanism of the Central Atlantic Magmatic Province
   4. The end-Cretaceous extinction, attributed mainly to a large impact at Chicxulub
   Show answer

   Answer: D — The end-Cretaceous extinction, attributed mainly to a large impact at Chicxulub

   Iridium is rare in crustal rock but abundant in meteorites, and shocked quartz, spherules and a crater of about 180 km are the signature of a large impact; this set of evidence belongs to the end-Cretaceous boundary at about 66 Ma. The end-Permian is the tempting wrong answer because it is the largest extinction, but its record points to flood-basalt volcanism and ocean chemistry, not shock features. The end-Triassic and end-Ordovician events are likewise tied to volcanism and glaciation.

2. In a measured section the last occurrences of 12 ammonite species are spread through the 4 m of rock below a boundary clay, although independent evidence shows that all 12 were eliminated in one brief event at the boundary. Which effect best explains the apparent gradual disappearance?

   1. The Signor–Lipps effect: with incomplete sampling the last preserved fossil of a taxon lies below its true last occurrence, so ranges seem to end early
   2. The Lazarus effect: taxa vanish from the record during the crisis interval and then reappear in younger rocks once the crisis has passed and conditions recover
   3. The pull of the Recent: younger rocks are better sampled and preserve more taxa, so ranges appear to extend upward towards the top of a section
   4. Time-averaging: shells of different ages are mixed within single beds, so older species are found reworked into the younger beds above them
   Show answer

   Answer: A — The Signor–Lipps effect: with incomplete sampling the last preserved fossil of a taxon lies below its true last occurrence, so ranges seem to end early

   Incomplete preservation means the last fossil found lies below the true last occurrence, so a single event is smeared into an apparent gradual decline: this is the Signor–Lipps effect. A Lazarus taxon needs a reappearance above the gap, which is not described here. The pull of the Recent concerns the better sampling of young rocks and exaggerates diversity near the present. Time-averaging would push apparent last occurrences upward (older shells in younger beds), the opposite of the smearing below the boundary.

3. A mass extinction removes 60% of the genera of a marine group. Species were lost at random, with the same probability in every genus, and most genera of the group contain several species. Which statement about species loss is correct?

   1. Species loss was about 60%, because species and genera are lost in equal proportions when extinction is random
   2. Species loss was below 60%, because a genus survives if any one of its species survives
   3. Species loss was well above 60%, because a genus disappears only when all of its species have disappeared
   4. Species loss cannot be related to genus loss without knowing the number of individuals per species
   Show answer

   Answer: C — Species loss was well above 60%, because a genus disappears only when all of its species have disappeared

   A genus is lost only if all k of its species are lost. With a random species loss fraction p the genus loss fraction is p^k, which for k > 1 is smaller than p. A genus loss of 60% therefore needs p^k = 0.6, so p = 0.6^(1/k) is well above 0.6 (for k = 3, p is about 0.84). The tempting answer, species loss below 60%, reverses this logic: surviving genera are the ones that retained at least one species, but extinct genera lost all of theirs, which makes the species percentage the larger.

4. A sandstone bed contains many cavities shaped like gastropod and bivalve shells but no shell material, while the associated articulate brachiopod shells (low-Mg calcite) are intact. Which explanation is most likely?

   1. Carbonisation drove off the volatile components of the molluscan shells and left only thin films of carbon on the bedding planes of the sandstone
   2. The aragonitic molluscan shells dissolved after burial and left moulds, while the low-Mg calcite brachiopod shells resisted dissolution
   3. Permineralisation filled the pores of the molluscan shells with silica, so that they weathered out later as solid casts
   4. Time-averaging mixed shells of several ages, and the molluscan shells were destroyed during the repeated reworking of the bed
   Show answer

   Answer: B — The aragonitic molluscan shells dissolved after burial and left moulds, while the low-Mg calcite brachiopod shells resisted dissolution

   Aragonite is the less stable CaCO3 polymorph and is dissolved or inverted early in diagenesis, whereas low-Mg calcite is stable. Mollusc shells (aragonite) therefore survive as external and internal moulds beside intact calcitic brachiopods. Carbonisation leaves carbon films, not shaped cavities. Permineralisation preserves the shell material itself, so the cavities would not be empty. Time-averaging blends ages but does not selectively delete one mineralogy.

5. A fossil shell has two unequal valves. A plane of bilateral symmetry cuts through the middle of each valve, and the larger valve has a small opening near its beak for a fleshy stalk. Which identification is best supported?

   1. A bivalve mollusc, because bivalves have unequal valves and each valve has its own plane of symmetry
   2. A gastropod, because the small opening near the beak is the siphonal canal of a snail with a bivalved shell
   3. An oyster-type bivalve, because one valve is larger than the other and cemented to the substrate by a stalk
   4. A brachiopod, because each valve is symmetrical on its own and the pedicle emerges through the larger valve
   Show answer

   Answer: D — A brachiopod, because each valve is symmetrical on its own and the pedicle emerges through the larger valve

   In brachiopods the plane of symmetry passes through both valves, so each valve is symmetrical, the valves differ in size, and the pedicle leaves through an opening in the pedicle valve. Bivalve molluscs have a symmetry plane between the two valves, so a valve alone is normally asymmetrical. Oysters are inequivalve, which makes them the tempting answer, but their valves are irregular and they have no pedicle opening. A gastropod has one shell, not two valves.

6. In a deep-sea core from the North Atlantic, the share of sinistrally coiled Neogloboquadrina pachyderma falls steadily from 90% at the base to 10% at the top, while the planktic : benthic foraminiferal ratio stays constant. Which interpretation is best?

   1. Surface water cooled upwards, because sinistral coiling in this species marks warm surface water rather than cold water
   2. Surface water warmed upwards, because this species coils mainly sinistrally in cold water and dextrally in warm water
   3. Water depth increased upwards, because sinistral individuals dominate in the shallow shelf waters inhabited by this species
   4. Salinity decreased upwards, because dextral coiling is restricted to brackish water in this species
   Show answer

   Answer: B — Surface water warmed upwards, because this species coils mainly sinistrally in cold water and dextrally in warm water

   In Neogloboquadrina pachyderma the dominant coiling direction tracks surface temperature: sinistral in cold water, dextral in warm. A fall from 90% to 10% sinistral therefore records warming upcore. A depth change would normally shift the planktic : benthic ratio, which stays constant here, and coiling in this species is not a salinity or depth indicator. The option claiming that sinistral coiling marks warm water reverses the temperature relationship.

7. Dinosaurs are divided into lizard-hipped and bird-hipped groups. Which statement about hip structure and the ancestry of birds is correct?

   1. Birds descend from theropod dinosaurs, which are lizard-hipped saurischians; the bird-hipped dinosaurs are the ornithischians
   2. Birds descend from the ornithischians, which is why that group is called bird-hipped and why it includes Archaeopteryx
   3. Lizard-hipped saurischians have the pubis swung back beside the ischium, as birds do, and are therefore the ancestors of birds
   4. Birds descend from pterosaurs, the flying members of the bird-hipped dinosaur group, which survived the end-Cretaceous event
   Show answer

   Answer: A — Birds descend from theropod dinosaurs, which are lizard-hipped saurischians; the bird-hipped dinosaurs are the ornithischians

   Birds evolved from small theropods, a lizard-hipped saurischian lineage; Archaeopteryx is a theropod-like bird of the Late Jurassic. The ornithischians are called bird-hipped because their pubis rotated backwards beside the ischium, a condition birds evolved independently, so the name misleads. Pterosaurs are not dinosaurs and became extinct at the end-Cretaceous.

8. A mudstone unit contains millions of individuals of only three species, all thin-shelled bivalves, gastropods and ostracods. Echinoderms, brachiopods, corals and cephalopods are absent from this unit although they occur in the beds above and below it. Which environment is most likely?

   1. A warm, clear, shallow reef flat with normal marine salinity and a rich, diverse fauna
   2. An open outer shelf with normal marine salinity, where all the stenohaline groups of the beds above and below would have flourished
   3. A brackish lagoon or estuary, where only euryhaline taxa tolerate the fluctuating salinity and can reach high abundance
   4. A cold, deep, well-oxygenated basin floor, where the stenohaline groups are excluded by low temperature alone
   Show answer

   Answer: C — A brackish lagoon or estuary, where only euryhaline taxa tolerate the fluctuating salinity and can reach high abundance

   Echinoderms, brachiopods, corals and cephalopods are stenohaline and need normal marine salinity. Their absence between beds that contain them, with a low-diversity, very abundant assemblage of tolerant bivalves, gastropods and ostracods, indicates fluctuating, brackish conditions such as a lagoon or estuary. A reef flat or an open shelf would carry the stenohaline groups, and the bed lies between marine beds, not a cold basin floor, so temperature alone does not explain it.

9. Which description fits a Konservat-Lagerstätte?

   1. A bed in which the hard parts of many generations are concentrated by currents, as in a bone bed or a shell bed
   2. A sandstone in which the shells have dissolved and survive only as external moulds and internal moulds
   3. A deposit in which fossils from several different habitats were mixed together by transport, so that the whole assemblage is allochthonous
   4. A deposit, such as the Burgess Shale or the Solnhofen Limestone, that preserves soft tissues or whole bodies in exceptional detail
   Show answer

   Answer: D — A deposit, such as the Burgess Shale or the Solnhofen Limestone, that preserves soft tissues or whole bodies in exceptional detail

   A Konservat-Lagerstätte is defined by exceptional preservation of soft tissue or whole bodies, as in the Burgess Shale, Chengjiang and Solnhofen. A concentration of hard parts, such as a bone bed, is a Konzentrat-Lagerstätte, which is the tempting confusion. Moulds and mixed transported assemblages describe other taphonomic outcomes and carry no claim of exceptional soft-tissue preservation.

10. Which of the following statements about mass extinctions are correct?

    1. Extinction is often selective: narrowly adapted, geographically restricted and reef-building taxa tend to suffer more than widespread generalists
    2. A Lazarus taxon is one that survives an extinction because its individuals are unusually long-lived
    3. Disaster taxa, such as the bivalve Claraia and the therapsid Lystrosaurus after the end-Permian event, briefly become abundant in the aftermath
    4. The percentage of genera lost in an extinction event is normally larger than the percentage of species lost in the same event
    Show answer

    Answer: A — Extinction is often selective: narrowly adapted, geographically restricted and reef-building taxa tend to suffer more than widespread generalists; C — Disaster taxa, such as the bivalve Claraia and the therapsid Lystrosaurus after the end-Permian event, briefly become abundant in the aftermath

    Extinction is selective, and disaster taxa bloom after the end-Permian crisis, so those two statements are correct. A Lazarus taxon is one that disappears from the fossil record and later reappears, not one that is long-lived. Because a genus is lost only when all its species are, the percentage of genera lost is normally smaller than the percentage of species lost, so the genus-versus-species statement is reversed.

11. Which of the following statements about invertebrate functional morphology are correct?

    1. Gastropod torsion is a 180° twist of the body mass during larval development and is distinct from the coiling of the shell
    2. Regular echinoids are bilaterally symmetrical, with the anus displaced from the apex to the posterior margin of the test
    3. A deep pallial sinus in a bivalve indicates long siphons and a deep-burrowing habit
    4. Ammonoid sutures are typically goniatitic in the Devonian to Permian, ceratitic in the Triassic and ammonitic in the Jurassic and Cretaceous
    Show answer

    Answer: A — Gastropod torsion is a 180° twist of the body mass during larval development and is distinct from the coiling of the shell; C — A deep pallial sinus in a bivalve indicates long siphons and a deep-burrowing habit; D — Ammonoid sutures are typically goniatitic in the Devonian to Permian, ceratitic in the Triassic and ammonitic in the Jurassic and Cretaceous

    Torsion is a larval twist unrelated to shell coiling; the pallial sinus marks the attachment of siphon retractors, so a deep sinus means long siphons and deep burial; and the goniatitic, ceratitic, ammonitic sequence is the usual pattern of increasing suture complexity. Regular echinoids are radially symmetrical with the anus at the apex; it is the irregular echinoids that have a displaced anus and secondary bilateral symmetry, so that statement is wrong.

12. Which of the following statements about vertebrate evolution are correct?

    1. Tiktaalik, a Late Devonian lobe-finned fish with a neck and wrist-like fin bones, is a transitional form between fishes and tetrapods
    2. Synapsids, the group containing the therapsids and mammals, have a single opening in the skull behind the eye on each side
    3. Pterosaurs and ichthyosaurs are dinosaurs that returned to the air and the sea respectively
    4. In the evolution of the mammalian ear, the articular and quadrate bones of the reptilian jaw joint became the malleus and incus
    Show answer

    Answer: A — Tiktaalik, a Late Devonian lobe-finned fish with a neck and wrist-like fin bones, is a transitional form between fishes and tetrapods; B — Synapsids, the group containing the therapsids and mammals, have a single opening in the skull behind the eye on each side; D — In the evolution of the mammalian ear, the articular and quadrate bones of the reptilian jaw joint became the malleus and incus

    Tiktaalik is the classic fish-tetrapod transitional form; synapsids are defined by one lower temporal opening; and the reptilian articular and quadrate became the malleus and incus as the mammalian jaw joint shifted to the dentary and squamosal. Pterosaurs (flying archosaurs) and ichthyosaurs (marine reptiles) are not dinosaurs, so that statement is wrong.

13. Which of the following statements about fossils and paleoenvironments are correct?

    1. A death assemblage is a snapshot of a single living community, because time-averaging is negligible in most shell beds
    2. A good index fossil is widespread, abundant, quickly evolving and not restricted to a single facies
    3. A woody dicot flora in which most leaves have toothed margins indicates a warmer climate than one with mostly entire margins
    4. Vertical dwelling burrows of the Skolithos type in clean sand typify a high-energy shoreface environment
    Show answer

    Answer: B — A good index fossil is widespread, abundant, quickly evolving and not restricted to a single facies; D — Vertical dwelling burrows of the Skolithos type in clean sand typify a high-energy shoreface environment

    An index fossil must be widespread, abundant, fast-evolving and independent of facies, and the Skolithos ichnofacies of vertical dwelling burrows marks high-energy sandy shorefaces. A death assemblage is usually time-averaged and may mix communities, so it is not a snapshot. Entire-margined leaves, not toothed ones, dominate warm climates, so the leaf statement is reversed.

14. A bed contains four species with 40, 30, 20 and 10 individuals. Using Shannon’s H = −Σ p_i ln p_i, with p_i the proportion of individuals of species i and the natural logarithm, the value of H is ____ (give the answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.28

    p = 0.40, 0.30, 0.20, 0.10. The terms −p ln p are 0.4 × 0.9163 = 0.3665, 0.3 × 1.2040 = 0.3612, 0.2 × 1.6094 = 0.3219 and 0.1 × 2.3026 = 0.2303. Their sum is 1.2799, which is 1.28. The maximum for four species would be ln 4 = 1.39, so the bed is fairly even.

15. In a stage lasting 4 Myr, 120 genera cross its bottom boundary. Of these, 30 also cross its top boundary and the other 90 become extinct within the stage. A further 40 genera originate and become extinct entirely inside the stage. Use q = −ln(N_bt / N_b) / Δt, where N_b is the number of taxa crossing the bottom boundary, N_bt the number crossing both boundaries and Δt the length in Myr. The per-capita extinction rate q, in per Myr, is ____ (give the answer to three decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.347

    N_b = 120 and N_bt = 30, so N_bt / N_b = 0.25 and −ln 0.25 = 1.3863. Dividing by Δt = 4 Myr gives q = 0.3466, which is 0.347 per Myr. The 40 genera confined to the stage cross neither boundary and are not counted, and using 90 (the number that went extinct) instead of the survivors is the usual slip.

16. A sample of N = 10 fossil individuals contains three species with 6, 3 and 1 individuals. Two individuals are drawn at random without replacement. Using E(S_n) = Σ [1 − C(N − N_i, n) / C(N, n)], where N_i is the count of species i, the expected number of species in the subsample of n = 2 is ____ (give the answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 1.6

    C(10, 2) = 45. For the species with 6 individuals, C(4, 2) = 6, so 1 − 6/45 = 0.8667. For 3 individuals, C(7, 2) = 21, so 1 − 21/45 = 0.5333. For 1 individual, C(9, 2) = 36, so 1 − 36/45 = 0.2. The sum is 1.6. Check: the chance that both drawn individuals share a species is (15 + 3 + 0)/45 = 0.4, so E(S_2) = 1 × 0.4 + 2 × 0.6 = 1.6.

17. In a brachiopod population two specimens are measured: x = 4 mm and y = 10 mm for the smaller, and x = 64 mm and y = 40 mm for the larger. Assume y = a x^b holds exactly. The predicted value of y, in mm, for a specimen with x = 100 mm is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 50

    The slope on a log–log plot is b = ln(40/10) / ln(64/4) = ln 4 / ln 16 = 0.5, which is below the isometric value of 1 for two lengths, so this is negative allometry. Then y(100) = 10 × (100/4)^0.5 = 10 × 5 = 50 mm. Using a straight-line (linear) extrapolation instead of the power law is the usual slip.

18. The lengths of n = 25 brachiopod shells have a mean of 30.0 mm and a standard deviation of 5.0 mm. For a 95% confidence interval of the mean use x̄ ± t s / √n with t = 2.064 (24 degrees of freedom). The full width of the interval, from lower to upper limit, in mm, is ____ (give the answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 4.13

    The standard error is s / √n = 5.0 / 5 = 1.0 mm. The half-width is t × SE = 2.064 mm, so the interval is 27.94 to 32.06 mm and its full width is 2 × 2.064 = 4.128, which is 4.13 mm. Reporting only the half-width (2.06) is the usual slip.