Stratigraphy: Principles, Correlation, Quantitative and Isotope Methods, Sequence Stratigraphy and Boundary Problems in India

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 flat-lying conglomerate rests, across an erosion surface, on steeply dipping, folded sandstone and shale beds. This contact is a(n)

   1. disconformity, because an erosion surface separates the two rock units
   2. angular unconformity, because the older beds are tilted or folded relative to the younger beds
   3. nonconformity, because the older rocks have been deformed before the younger ones were laid down
   4. paraconformity, because a long time gap separates the two units even though they differ in dip
   Show answer

   Answer: B — angular unconformity, because the older beds are tilted or folded relative to the younger beds

   When younger beds rest on tilted or folded older beds across an erosion surface, the contact is an angular unconformity. A disconformity needs parallel beds on both sides. A nonconformity needs igneous or metamorphic rock below, and here the older rocks are bedded sedimentary rocks. A paraconformity has parallel beds and no visible erosion, so it cannot involve a difference in dip.

2. A borehole cuts a conformable succession (no erosion surface). From the base upward it passes from fluvial sandstone to estuarine muddy sand to shallow-marine shale. What does Walther’s law imply?

   1. The shoreline migrated seaward during deposition, because the environment at the top lay seaward of the one at the base, so the whole succession records a regression
   2. The three environments never coexisted: each replaced the previous one over the whole area at the same moment, so the contacts are time lines
   3. The succession must hide two unconformities, because the law forbids a vertical change of facies without a break in deposition
   4. The environments lay side by side, fluvial landward, estuarine between and marine seaward, and the shoreline migrated landward during deposition, a transgression
   Show answer

   Answer: D — The environments lay side by side, fluvial landward, estuarine between and marine seaward, and the shoreline migrated landward during deposition, a transgression

   Walther’s law says that facies which follow one another in a conformable vertical section lay side by side when deposited. Fluvial, estuarine and marine environments are arranged from land to sea, so a base-to-top change from fluvial to marine means the marine environment advanced over the land: a landward shoreline shift, or transgression. The tempting answer reverses this, because the top environment lies seaward of the base environment only after the shoreline has moved landward. The law permits conformable vertical changes and does not demand unconformities.

3. A lithologically uniform sandstone formation is traced for 150 km from the sea towards the land. Volcanic ash beds above and below it show that the formation is 1.5 Myr younger at the landward end. What is the correct inference?

   1. The formation is diachronous, probably a transgressive sandstone, so its base and top are not time lines
   2. The ash-bed ages must be in error, because a lithologically uniform formation is always deposited at a single time
   3. The formation is a single stage throughout, because formations are chronostratigraphic units of fixed age
   4. The landward end must be a different formation, because a single formation cannot extend for 150 km
   Show answer

   Answer: A — The formation is diachronous, probably a transgressive sandstone, so its base and top are not time lines

   Formations are lithostratigraphic units defined by rock type, not by age, and a transgressing shoreline deposits the same sandstone progressively younger landward. The ash beds are isochronous markers and are the more reliable evidence of age. Chronostratigraphic units, not formations, have fixed ages, and a formation may extend for hundreds of kilometres.

4. Which fossil plant assemblage characterises the Permian Lower Gondwana strata (Talchir to Raniganj) of peninsular India?

   1. The Dicroidium flora, which marks the Triassic of the Southern Hemisphere
   2. The Ptilophyllum flora, which marks the Jurassic and Early Cretaceous of the Rajmahal region
   3. The Glossopteris flora, with Gangamopteris, the seed-fern assemblage of the Gondwana lands
   4. The Archaeopteris flora, which marks the Devonian forests of the Northern Hemisphere
   Show answer

   Answer: C — The Glossopteris flora, with Gangamopteris, the seed-fern assemblage of the Gondwana lands

   The Glossopteris flora, with Gangamopteris, is the characteristic Permian plant assemblage of Gondwana, including the Talchir to Raniganj beds of India. The Dicroidium flora is Triassic, the Ptilophyllum flora is Jurassic to Early Cretaceous, and Archaeopteris is a Devonian plant of a different time and region.

5. In a graphic-correlation plot with section X on the horizontal axis and section Y on the vertical axis, the fossil datums fall on a straight line of slope 2.0, then on a vertical segment, then on a line of slope 2.0 again. Which interpretation is correct?

   1. X accumulated twice as fast as Y, and the vertical segment marks a hiatus or a condensed interval in section Y
   2. Y accumulated twice the thickness of X in equal time, and the vertical segment is time present in Y but missing or condensed in X
   3. Y is twice as old as X, and the vertical segment marks an unconformity that is present in both sections at the same time
   4. The slope of 2.0 shows that the two sections correlate perfectly, and the vertical segment is only a measurement error in section Y, not a geological signal
   Show answer

   Answer: B — Y accumulated twice the thickness of X in equal time, and the vertical segment is time present in Y but missing or condensed in X

   The slope of the line of correlation is the thickness in Y per unit thickness in X over the same time, so a slope of 2.0 means Y accumulated twice as much rock. On a vertical segment Y gains thickness while X does not change: the time is present in Y but missing (hiatus) or very condensed in X. The option that makes X the faster section reverses both the rates and the section that has the gap; the others confuse slope with age or with perfect correlation.

6. A deep-sea benthic foraminiferal δ¹⁸O record shows high values at about 20 ka and low values at about 125 ka. Which reading is correct?

   1. The 20 ka interval was warm and ice-free, because high δ¹⁸O always means warm water
   2. Both intervals were glacial, because benthic δ¹⁸O is controlled only by local salinity
   3. The 20 ka interval was glacial, with ice sheets storing light oxygen-16 on land, and about 125 ka was a warm interglacial
   4. The 125 ka interval was glacial, because growing ice sheets remove oxygen-18 from the ocean and so lower the δ¹⁸O of its calcite
   Show answer

   Answer: C — The 20 ka interval was glacial, with ice sheets storing light oxygen-16 on land, and about 125 ka was a warm interglacial

   Evaporation-precipitation fractionation stores the lighter ¹⁶O in ice sheets, so glacial oceans, and the calcite formed in them, are enriched in ¹⁸O and show high δ¹⁸O; colder water adds to the effect. High values at about 20 ka match the last glacial maximum (MIS 2) and low values at about 125 ka match the last interglacial (MIS 5e). The option claiming that growing ice removes oxygen-18 reverses the fractionation: ice removes the light isotope, not the heavy one.

7. Two samples from one Miocene bed are analysed for ⁸⁷Sr/⁸⁶Sr to obtain an age. Sample P is recrystallised micrite with a high Mn/Sr ratio. Sample R is an unrecrystallised calcite brachiopod shell with low Mn and Fe. P gives a ratio well above R. Which procedure is correct?

   1. Use sample R, because diagenetic fluids with a different Sr composition have probably reset the ratio of the altered micrite
   2. Use sample P, because greater alteration shows that the carbonate has fully equilibrated with Miocene seawater at the time of burial
   3. Average the two ratios, because both samples record seawater at slightly different times within the bed
   4. Use sample P, because micrite holds more strontium than shell calcite and its ratio is therefore more precise
   Show answer

   Answer: A — Use sample R, because diagenetic fluids with a different Sr composition have probably reset the ratio of the altered micrite

   A high Mn/Sr ratio and recrystallised texture signal diagenetic alteration, in which fluids carrying a different, usually more radiogenic, Sr composition replace the original ratio. Well-preserved low-Mn, low-Fe calcite of a brachiopod keeps the seawater value and is the sample to use. Equilibrating with later fluids is not equilibrating with Miocene seawater, averaging mixes a good value with a bad one, and the amount of Sr does not make an altered ratio more precise.

8. Which combination best identifies the maximum flooding surface in a well section?

   1. A sharp-based, coarsening-upward shoreface sandstone with a low gamma-ray response and abundant cross-bedding, marking a forced regression
   2. A coarse lag deposit above an erosion surface that cuts into the beds beneath it, marking incision by rivers
   3. A thick blocky sandstone with a low gamma-ray response and an erosional base, marking a lowstand fan
   4. A thin condensed, glauconitic, organic-rich mudstone with a gamma-ray peak, between retrograding beds below and prograding beds above
   Show answer

   Answer: D — A thin condensed, glauconitic, organic-rich mudstone with a gamma-ray peak, between retrograding beds below and prograding beds above

   The maximum flooding surface lies at the turnaround from retrogradation to progradation, where sediment supply to the shelf is least. It is marked by a condensed section: slow deposition, glauconite, organic-rich mud, a gamma-ray peak and abundant microfossils. A sharp-based coarsening-upward sandstone belongs to the falling-stage tract, a coarse lag on an erosion surface marks the sequence boundary or a ravinement, and a blocky low-gamma sandstone is a lowstand deposit.

9. Why is the Permian–Triassic boundary in the Gondwana basins of peninsular India hard to tie to the global marine definition?

   1. The Gondwana rocks are strongly metamorphosed, which destroys every fossil and every isotopic signal that might have been used for correlation
   2. The succession is continental and lacks the marine zone fossils that define the global boundary, so other evidence is needed
   3. The Raniganj and Panchet formations have identical lithology and fossils, so the two cannot be told apart in any section
   4. India lay near the Equator in the Permian, where no sediments accumulated, so the boundary interval is absent from the basins
   Show answer

   Answer: B — The succession is continental and lacks the marine zone fossils that define the global boundary, so other evidence is needed

   The global boundary is defined in marine sections by conodont and isotope evidence. The Raniganj–Panchet transition is continental, so the marine marker is missing and correlation relies on plants, vertebrates such as Lystrosaurus and ages, and on marine sections such as Spiti and the Guryul Ravine in Kashmir. The Gondwana rocks are not strongly metamorphosed, the two formations differ in lithology and fossils, and Gondwana India was in high southern latitudes.

10. Which of the following statements about stratigraphic principles and units are correct?

    1. A GSSP defines a chronostratigraphic boundary at a specified point in a specified section
    2. For most of the Precambrian the boundaries are defined by agreed numerical ages rather than by GSSPs
    3. A formation is the basic chronostratigraphic unit, defined by the time interval in which its rocks formed
    4. Uniformitarianism requires that the rates of geological processes have always equalled their present rates
    Show answer

    Answer: A — A GSSP defines a chronostratigraphic boundary at a specified point in a specified section; B — For most of the Precambrian the boundaries are defined by agreed numerical ages rather than by GSSPs

    A GSSP fixes a boundary at a point in a section, and Precambrian boundaries are mostly agreed numerical ages (GSSA). A formation is the basic lithostratigraphic unit, defined by rock type, not by time. Uniformitarianism says that the same physical laws and processes operate through time, but rates may differ, so the uniform-rates statement overstates it.

11. Which of the following statements about the stratigraphy of India are correct?

    1. In the Damodar valley the Talchir Formation, which begins with a glacial tillite, lies below the coal-bearing Barakar Formation
    2. In the Gondwana succession the Panchet Formation lies below the Raniganj Formation
    3. The groups of the Vindhyan Supergroup, in ascending order, are Semri, Kaimur, Rewa and Bhander
    4. The Siwalik Group is a Miocene to Pleistocene continental foreland succession, dated using mammals and magnetostratigraphy
    Show answer

    Answer: A — In the Damodar valley the Talchir Formation, which begins with a glacial tillite, lies below the coal-bearing Barakar Formation; C — The groups of the Vindhyan Supergroup, in ascending order, are Semri, Kaimur, Rewa and Bhander; D — The Siwalik Group is a Miocene to Pleistocene continental foreland succession, dated using mammals and magnetostratigraphy

    Talchir (tillite at the base) underlies the coal-bearing Barakar; the Vindhyan groups ascend Semri, Kaimur, Rewa, Bhander; and the Siwaliks are a continental foreland molasse dated by mammals and magnetic polarity. Panchet (Early Triassic) lies above Raniganj (Late Permian), so the statement that reverses them is wrong.

12. Which of the following statements about isotope stratigraphy are correct?

    1. Seawater ⁸⁷Sr/⁸⁶Sr dating is most precise where the curve is steep and monotonic, as for much of the Cenozoic
    2. A negative δ¹³C excursion can mark the release of isotopically light carbon, as at the Paleocene–Eocene Thermal Maximum
    3. Even-numbered marine isotope stages are interglacial intervals with low benthic δ¹⁸O
    4. Because the ocean mixes in about a thousand years while the residence time of Sr is several Myr, the Sr ratio is the same worldwide at any one time
    Show answer

    Answer: A — Seawater ⁸⁷Sr/⁸⁶Sr dating is most precise where the curve is steep and monotonic, as for much of the Cenozoic; B — A negative δ¹³C excursion can mark the release of isotopically light carbon, as at the Paleocene–Eocene Thermal Maximum; D — Because the ocean mixes in about a thousand years while the residence time of Sr is several Myr, the Sr ratio is the same worldwide at any one time

    Sr dating needs a steep monotonic curve; negative δ¹³C excursions mark light-carbon release such as the PETM; and the Sr ratio is globally uniform because the residence time far exceeds the mixing time. Even-numbered marine isotope stages are glacial with high δ¹⁸O, and the odd-numbered ones are interglacial, so that statement is reversed.

13. Which of the following statements about sequence stratigraphy are correct?

    1. When accommodation is created faster than sediment is supplied (A/S above 1), the shoreline progrades seaward
    2. Sequence boundaries are drawn at the same position in every sequence-stratigraphic model
    3. Where eustatic sea level falls at 25 m/Myr but the shelf subsides at 40 m/Myr, relative sea level still rises and accommodation is still created
    4. The maximum flooding surface separates the retrogradational transgressive systems tract from the progradational highstand systems tract
    Show answer

    Answer: C — Where eustatic sea level falls at 25 m/Myr but the shelf subsides at 40 m/Myr, relative sea level still rises and accommodation is still created; D — The maximum flooding surface separates the retrogradational transgressive systems tract from the progradational highstand systems tract

    Relative sea level change is eustasy plus subsidence, so a 25 m/Myr eustatic fall against 40 m/Myr of subsidence is a net rise of 15 m/Myr, and the maximum flooding surface is the turnaround between the retrogradational TST and the progradational HST. A/S above 1 produces retrogradation, not progradation, and models place the sequence boundary differently.

14. A section contains two dated ash beds: the lower, at 20 m, is 6.00 Ma and the upper, at 56 m, is 4.50 Ma. A turbidite 6 m thick, deposited geologically instantaneously, lies between 25 m and 31 m. Apart from the turbidite, deposition was at a constant rate, and all thicknesses are decompacted. A fossil horizon lies at 39 m. The age of this horizon is ____ Ma (give the answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 5.35

    Between the ash beds the thickness is 36 m, of which 6 m is the instantaneous turbidite, leaving 30 m deposited in 6.00 − 4.50 = 1.50 Myr, a rate of 20 m/Myr. The fossil is 39 − 20 = 19 m above the lower ash, of which the 6 m turbidite lies below it, so 13 m of time-bearing rock separates it from the ash. That is 13 / 20 = 0.65 Myr, and the age is 6.00 − 0.65 = 5.35 Ma. Ignoring the turbidite gives 5.21 Ma.

15. A trilobite species is known from H = 9 horizons spread at random through its true range, and the observed range is R = 30 m of section. The probability that the true end lies beyond the observed end by more than a fraction a of R is (1 + a)^−(H − 1). The one-sided 95% confidence interval for the position of the true last occurrence extends beyond the highest known occurrence by ____ m (give the answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 13.6

    For 95% confidence the exceedance probability is 0.05, so (1 + a)^−8 = 0.05 and 1 + a = 0.05^(−1/8) = 1.4542, giving a = 0.4542. The interval therefore extends a × R = 0.4542 × 30 = 13.63, which is 13.6 m beyond the highest known occurrence. Using H instead of H − 1 as the exponent gives a smaller, wrong value.

16. Assume that seawater ⁸⁷Sr/⁸⁶Sr rises linearly from 0.708200 at 20 Ma to 0.708800 at 10 Ma. A well-preserved brachiopod calcite measures 0.708520 in an analytical run in which the reference standard, whose accepted value is 0.710250, measured 0.710270. After correcting the sample for the same offset, its age is ____ Ma.

    Numerical answer — type the value.

    Show answer

    Answer: 15

    The standard read 0.710270 against an accepted 0.710250, an offset of +0.000020, so the corrected sample ratio is 0.708520 − 0.000020 = 0.708500. On the linear segment the ratio rises 0.000600 over 10 Myr, so 0.708500 is (0.708500 − 0.708200) / 0.000600 = 0.5 of the way from 20 Ma towards 10 Ma, an age of 15 Ma. Without the correction the age would be 20 − 5.33 = 14.7 Ma.

17. A benthic foraminiferal δ¹⁸O record rises from 3.4‰ in an interglacial to 4.6‰ in a glacial. Assume that sea level fell by 70 m, that each 10 m of sea-level fall raises seawater, and hence calcite, δ¹⁸O by 0.10‰, and that the rest of the change is due to temperature, with calcite δ¹⁸O changing by 0.25‰ per °C. The cooling of the deep water is ____ °C.

    Numerical answer — type the value.

    Show answer

    Answer: 2

    The total change is 4.6 − 3.4 = 1.2‰. The ice-volume part is 7 × 0.10 = 0.70‰, leaving 1.2 − 0.70 = 0.50‰ for temperature. At 0.25‰ per °C this is 0.50 / 0.25 = 2.0 °C of cooling. Attributing all 1.2‰ to temperature would give 4.8 °C, the usual overestimate.

18. A third-order sequence, 150 m thick, accumulated in 2.4 Myr and consists only of lowstand, transgressive and highstand systems tracts. The lowstand tract lasted 0.4 Myr and accumulated at 15 m/Myr; the transgressive tract lasted 0.6 Myr and accumulated at 30 m/Myr; the highstand tract occupies the rest of the time. The average accumulation rate of the highstand tract is ____ m/Myr.

    Numerical answer — type the value.

    Show answer

    Answer: 90

    The lowstand tract is 0.4 × 15 = 6 m thick and the transgressive tract 0.6 × 30 = 18 m. The highstand tract therefore holds 150 − 6 − 18 = 126 m. Its duration is 2.4 − 0.4 − 0.6 = 1.4 Myr, so its average rate is 126 / 1.4 = 90 m/Myr, the highest of the three, as expected for a progradational highstand wedge.