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Biodiversity and Conservation — Class 12 CBSE Biology revision notes

A 4-page visual revision sheet for Biodiversity and Conservation, written for Class 12 CBSE Biology. Formulas, diagrams and the traps that cost marks, laid out to be read the night before rather than the term before. Free, and it opens inside CapCompass.

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Biodiversity and Conservation in one look

counting species is the easy part, keeping them is not

Three levels of biodiversity

  • genetic diversity: variation within one species, e.g. rice and mango varieties across India
  • species diversity: how many species live in a region, from a patch to the whole planet
  • ecological diversity: variety of ecosystems in a region, e.g. deserts, wetlands, rainforests, coral reefs
  • a plan aimed at only one level can let the other two quietly erode

How many species are there

  • roughly 1.7 to 1.8 million species have been formally described so far
  • a working estimate for the true total, described plus undescribed, is about 7 million species
  • over 70% of described animal species are insects, by far the most species-rich group
  • described fungal species alone outnumber the combined total of fishes, amphibians, reptiles and mammals

Patterns across the globe: the latitudinal gradient

  • species diversity is generally highest near the equator and falls steadily toward the poles
  • tropics stayed relatively undisturbed over a long evolutionary time, letting speciation build up diversity
  • temperate regions, by contrast, were repeatedly reshaped by glaciation, resetting that process
  • tropical climates are more stable and less seasonal, favouring niche specialisation among species
  • tropics receive more solar energy, supporting higher productivity and so more species

Reading the species-area relationship

Q. For a hypothetical island, the species-area relationship is log S = log C + Z log A, with C = 5 and Z = 0.25. Estimate the number of species S for an area A of 10,000 units.

  1. log A = log 10,000 = 4 (10,000 is 10 to the power 4)
  2. Z log A is 0.25 times 4, which is 1
  3. log C is log of 5, about 0.70
  4. log S = log C + Z log A = about 0.70 + 1 = 1.70
  5. S = 10 raised to the power 1.70
  6. since log 5 is about 0.70, this equals 10 x 5

Answer: about 50 species for this area, given these assumed C and Z values

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