More practice questions

  1. Q4. What is the significance of the slope of regression in a species–area relationship? (NCERT Exercise 4, page 227.)

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    The slope Z of the line log S = log C + Z log A tells how fast species richness rises with area. Within a region it is remarkably constant, 0.1 to 0.2, whatever the taxonomic group or region (plants in Britain, birds in California, molluscs in New York state). Among very large areas such as entire continents it is much steeper, 0.6 to 1.2 (1.15 for frugivorous birds and mammals of tropical forests on different continents), so at that scale species richness rises much faster with area (NCERT page 220).

  2. Q5. What are the major causes of species losses in a geographical region? (NCERT Exercise 5, page 227.)

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    The evil quartet: (i) habitat loss and fragmentation, the most important cause, as when tropical rain forests, once over 14 per cent of the land surface, shrank to no more than 6 per cent; (ii) over-exploitation, which caused the extinction of Steller’s sea cow and the passenger pigeon; (iii) alien species invasions, such as the Nile perch in Lake Victoria, which led to the extinction of more than 200 cichlid species; (iv) co-extinctions, when species obligately associated with an extinct species, such as its parasites or its coevolved pollinator, also die out (NCERT pages 222–223).

  3. Q6. How is biodiversity important for ecosystem functioning? (NCERT Exercise 6, page 227.)

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    Ecologists have no definitive answer; for decades they believed that communities with more species tend to be more stable, a stable community being one that does not vary too much in productivity from year to year, is resistant or resilient to occasional disturbances and resists invasions by alien species. Tilman’s plots with more species showed less year-to-year variation in total biomass and higher productivity. Paul Ehrlich’s rivet popper hypothesis compares species to the rivets of an airplane: losing a few may not matter at once, but losing many, or key species (the rivets on the wings), weakens the whole ecosystem. Loss of biodiversity may reduce plant production, lower resistance to perturbations such as drought, and increase variability in processes such as productivity, water use and pest and disease cycles (NCERT pages 220–222).

  4. Q7. The species diversity of plants (22 per cent) is much less than that of animals (72 per cent). What could be the explanations to how animals achieved greater diversification? (NCERT Exercise 9, page 227.)

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    NCERT leaves this open; it asks the same question about insects, more than 70 per cent of all animals (page 218). Possible explanations (our reasoning): animals move and so can use far more ways of life, in more niches; insects are small, breed fast in short generations and are adapted to many particular foods, so one plant species can support several insect species; and animals evolve with their food plants, predators and parasites, as in co-evolution, which multiplies specialised forms.

  5. Q8. For a group with Z = 0.2, by how many times does species richness rise when the area studied grows tenfold? And by what fraction does it fall if the area shrinks to a tenth?

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    From S = CAZ: richness × 100.2 ≈ 1.58 for a tenfold area; for a tenth of the area, richness × 10−0.2 ≈ 0.63, a fall of about 37 per cent (our arithmetic from NCERT’s equation, page 220).

  6. Q9. Tropical rain forests once covered more than 14 per cent of the earth’s land surface and now cover no more than 6 per cent. At least what fraction of their former area has been lost?

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    (14 − 6) ÷ 14 = 8 ÷ 14 ≈ 0.57: at least 57 per cent, more than half (our arithmetic from NCERT page 222).

  7. Q10. How does the present sixth extinction differ from the five earlier mass extinctions?

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    In its rate and its cause: present extinction rates are estimated to be 100 to 1000 times faster than in pre-human times, and human activities are responsible (NCERT page 222).

Equation check

For log10 S = 2 + 0.15 log10 A, the slope is Z = 0.15 and the vertical intercept is log10 C = 2, so C = 100. Thus S = 100 A0.15, with A measured in the area units used to fit the equation; the constant C changes if those units change (our arithmetic).