Ecology II: Community Ecology, Ecosystem Structure and Function, Biomes and the Biogeography of India
1. Community assembly, structure and succession
A local community is assembled from a regional species pool through a series of filters. Dispersal limits which species arrive at all; environmental (habitat) filtering removes those that cannot tolerate local conditions, so co-occurring species tend to share the traits the environment demands; and biotic interactions — competition, predation, facilitation — sort the survivors, with limiting similarity predicting that species too alike in resource use cannot coexist. Priority effects mean the order of arrival can change the final community. Community structure is described by the number of species, their relative abundances, the dominant species, the physical layering (stratification of a forest into canopy, understorey, shrub and ground layers) and the trophic organisation.
Succession is the directional change in community composition over time at one site. Primary succession starts on ground with no soil and no biological legacy — bare rock, a lava flow, a retreating glacier’s moraine, a new sand dune — with pioneers such as lichens and mosses that begin soil formation. Secondary succession follows a disturbance that leaves soil and a seed bank — an abandoned field, a burnt or felled forest — and is much faster. Successions on dry substrates are xeroseres and in water hydroseres, which fill a pond through submerged, floating, reed-swamp, sedge-meadow and woodland stages. Clements viewed succession as a predictable progression to a single climatic climax, the community as a superorganism; Gleason’s individualistic view holds that each species responds independently to the environment, so communities are loose, shifting assemblages.
| Model | Early species’ effect on later ones | Example |
|---|---|---|
| Facilitation | make the site more suitable (soil, nitrogen, shade) | nitrogen-fixing alder on Glacier Bay moraines |
| Tolerance | neither help nor hinder; later species tolerate lower resources | shade-tolerant trees growing up beneath pioneers |
| Inhibition | hold the site until they die or are damaged | algal mats on intertidal boulders |
2. Diversity: alpha, beta, gamma, richness, evenness and the indices
Whittaker distinguished alpha (α) diversity, the diversity within one site or habitat; gamma (γ) diversity, the diversity of a whole region or landscape; and beta (β) diversity, the turnover in species composition between sites. In his multiplicative form β = γ/ᾱ: if a landscape holds 30 species and its sites average 12, β = 2.5 — the landscape is equivalent to 2.5 completely distinct communities. Species richness S is simply the number of species; evenness is how equally individuals are spread among them; relative abundance pᵢ = nᵢ/N is each species’ share; and a rank–abundance plot (log abundance against rank) shows both at once, steep for a dominated community and shallow for an even one.
| Index | Formula | Meaning and range |
|---|---|---|
| Shannon H′ | H′ = −Σ pᵢ ln pᵢ | uncertainty of the species of a random individual; 0 for one species, maximum ln S when all are equal |
| Pielou evenness J′ | J′ = H′/ln S | 0 to 1; 1 when all species are equally abundant |
| Simpson D (dominance) | D = Σ pᵢ² (or Σ nᵢ(nᵢ − 1)/[N(N − 1)]) | probability that two individuals drawn at random are the same species; rises with dominance |
| Simpson diversity | 1 − D, or 1/D | 1 − D is the probability they differ (0 to 1); 1/D is the effective number of equally common species |
Worked example: four species with 40, 30, 20 and 10 individuals, so p = 0.4, 0.3, 0.2 and 0.1. Shannon: −[0.4 ln 0.4 + 0.3 ln 0.3 + 0.2 ln 0.2 + 0.1 ln 0.1] = 0.3665 + 0.3612 + 0.3219 + 0.2303 = 1.280. The maximum for four species is ln 4 = 1.386, so J′ = 1.280/1.386 = 0.92. Simpson: D = 0.16 + 0.09 + 0.04 + 0.01 = 0.30, so 1 − D = 0.70 and 1/D = 3.33 — the community is as diverse as 3.33 equally common species, fewer than its actual four because it is uneven.
3. Species–area relationships, diversity gradients and island biogeography
The number of species rises with area as a power law, S = cA^z, which is a straight line on log–log axes: log S = log c + z log A, slope z. For isolated islands z is typically about 0.2 to 0.35; for nested samples within a continuous mainland it is lower, about 0.1 to 0.2, because mainland samples are topped up by immigrants from the surrounding area. Finding z from two islands: 20 species on 10 km² and 50 on 1000 km² give z = log(50/20)/log(1000/10) = log 2.5/log 100 = 0.398/2 = 0.20. A useful rule of thumb follows: with z ≈ 0.3, a tenfold increase in area roughly doubles the species number, because 10^0.3 = 2.0.
Diversity gradients: species richness is highest in the tropics and declines towards the poles for most groups — the latitudinal diversity gradient. Candidate explanations include the greater area of the tropics, higher energy and productivity, a longer time for diversification in climatically stable regions, higher speciation or lower extinction rates, tropical niche conservatism, and stronger biotic interactions; no single one is agreed. Along altitudinal gradients richness often peaks at mid-elevations rather than declining monotonically, partly because of geometric constraints (the mid-domain effect) and partly because of climate. The Western Ghats and the eastern Himalaya show richness peaks well below their summits.
The equilibrium theory of island biogeography (MacArthur and Wilson, 1963 and 1967) treats island richness as a balance between immigration, which falls as the island fills (fewer arrivals are new species), and extinction, which rises as more species crowd into limited space. The curves cross at the equilibrium richness S*. Near islands have higher immigration curves and large islands lower extinction curves, so large, near islands hold the most species; the equilibrium is dynamic, with continual turnover. In a linear version, I = I₀(1 − S/P) and E = E₀S/P for a mainland pool of P species, and setting I = E gives S* = PI₀/(I₀ + E₀): with P = 100, I₀ = 10 and E₀ = 15 species per year, S* = 1000/25 = 40. Simberloff and Wilson’s defaunation of small mangrove islands in Florida, which recovered to similar richness with different species, tested the theory directly.
4. Productivity, food webs and energy flow
Gross primary productivity (GPP) is the rate at which producers fix energy by photosynthesis; net primary productivity (NPP) = GPP − R_a, where R_a is autotrophic respiration, is what is available to consumers and decomposers. Net ecosystem productivity subtracts heterotrophic respiration as well, and is positive for an ecosystem accumulating carbon. Secondary productivity is the rate of biomass production by consumers. Per unit area, tropical rainforests, swamps and marshes, and coral reefs and estuaries are among the most productive ecosystems, and deserts and the open ocean among the least; the open ocean nonetheless contributes a large share of global NPP because of its area.
| Efficiency | Definition | Typical values |
|---|---|---|
| Consumption | energy ingested by level n / production of level n − 1 | low in forests (much wood is uneaten), high in plankton |
| Assimilation | assimilated / ingested | about 20–50 % for herbivores, about 80 % for carnivores |
| Production | new biomass / assimilated | a few per cent in endotherms, much higher in ectotherms |
| Trophic transfer (Lindeman) | production of level n / production of level n − 1 | about 10 %, ranging roughly 5–20 % |
Worked example: if producers have an NPP of 20 000 kJ m⁻² y⁻¹ and herbivore production is 1500 kJ m⁻² y⁻¹, the trophic transfer efficiency is 1500/20 000 = 7.5 %. At a constant 10 %, the top of a chain producer → herbivore → carnivore → top carnivore receives 0.1³ = 0.1 % of NPP, which is why food chains rarely exceed four or five links. The pyramid of energy is always upright; the pyramid of numbers can be inverted (a tree carrying thousands of insects) and the pyramid of biomass can be inverted in the open ocean, where a small standing stock of fast-turning-over phytoplankton supports a larger standing stock of zooplankton. Energy flows one way through an ecosystem and is lost as heat at each step; matter cycles.
Food chains are the grazing chain (living plants → herbivores → carnivores) and the detritus chain (dead organic matter → detritivores and decomposers → their predators); in most terrestrial ecosystems most NPP passes through the detritus chain. A food web links the chains. Connectance (the fraction of possible links realised), omnivory and the number of trophic levels describe its structure.
5. Biogeochemical cycles and ecosystem properties
The carbon cycle moves carbon between the atmosphere (as CO₂), living and dead biomass, soils, the ocean (the largest active reservoir, mostly as dissolved inorganic carbon) and rocks and fossil fuels. Photosynthesis draws CO₂ down and respiration and decomposition return it; the ocean exchanges CO₂ with the air and moves it to depth by the solubility and biological pumps. Burning fossil fuels and clearing forests add carbon to the atmosphere faster than the land and ocean sinks take it up, which is why atmospheric CO₂ rises.
| Step | Transformation | Agents |
|---|---|---|
| Fixation | N₂ → NH₃ (by nitrogenase, which oxygen inactivates) | Rhizobium in legume nodules, Frankia, free-living Azotobacter, cyanobacteria such as Anabaena and Nostoc; lightning; the Haber–Bosch process |
| Ammonification | organic N → NH₄⁺ | decomposer bacteria and fungi |
| Nitrification | NH₄⁺ → NO₂⁻ → NO₃⁻ (aerobic) | Nitrosomonas (first step), Nitrobacter (second step) |
| Denitrification | NO₃⁻ → N₂O → N₂ (anaerobic) | Pseudomonas and other facultative anaerobes in waterlogged soils |
| Anammox | NH₄⁺ + NO₂⁻ → N₂ (anaerobic) | planctomycete bacteria, important in oceans |
Carbon and nitrogen have large atmospheric pools and are gaseous cycles; phosphorus has no significant gaseous phase and follows a slow sedimentary cycle from rock weathering to soils, water and ocean sediments, which is why phosphorus often limits freshwater productivity and why phosphate run-off drives eutrophication. Among ecosystem properties, stability has several components: resistance is the ability to stay unchanged under a disturbance, resilience the speed (or ability) of return afterwards, and variability the size of fluctuations over time. A grassland can be low in resistance to drought but highly resilient; a forest can resist a dry year but take centuries to recover from clearing. Holling also used resilience for the size of disturbance a system can absorb before shifting to a different state, such as a clear lake turning turbid.
6. Biomes, biogeographic realms and the biogeography of India
Biomes follow climate: Whittaker’s diagram places them by mean annual temperature and precipitation — tropical rainforest (hot, wet all year), tropical seasonal forest and savanna (hot, with a dry season), desert (dry at any temperature), temperate grassland, temperate deciduous forest, temperate rainforest, boreal forest or taiga (cold, conifer-dominated) and tundra (too cold for trees, with permafrost). Mountains compress the same sequence into altitudinal belts. Wallace divided the world’s land fauna into six zoogeographic realms: Palaearctic, Nearctic, Neotropical, Ethiopian (Afrotropical), Oriental (Indomalayan) and Australian. Wallace’s line, between Bali and Lombok and between Borneo and Sulawesi, marks the sharp change from Oriental to Australian faunas that follows the edge of the Sunda shelf.
India lies mainly in the Oriental realm with Palaearctic elements in the Himalaya and Trans-Himalaya and Ethiopian (African) and Malayan affinities in its fauna. It has about 2.4 % of the world’s land area and holds an estimated 7–8 % of recorded species. Rodgers and Panwar’s classification divides it into ten biogeographic zones: the Trans-Himalaya, the Himalaya, the Desert, the Semi-arid zone, the Western Ghats, the Deccan Peninsula, the Gangetic Plain, the North-East, the Coasts and the Islands. Champion and Seth classified India’s forests into 16 type groups, from tropical wet evergreen and tropical moist and dry deciduous forests, through tropical thorn, littoral and swamp forests (including mangroves such as the Sundarbans), to subtropical, montane temperate, subalpine and alpine types.
- Western Ghats: wet evergreen forests with high endemism, especially of amphibians, freshwater fish and flowering plants; the lion-tailed macaque, the Nilgiri tahr and the purple frog (Nasikabatrachus sahyadrensis) are endemics.
- Himalaya and North-East: steep altitudinal zonation from subtropical to alpine; the red panda, the hoolock gibbon (India’s only ape) and many rhododendrons and orchids.
- Desert and Semi-arid: the Thar with the great Indian bustard, blackbuck and chinkara; the Gir forest holds the only wild Asiatic lions.
- Islands: the Andaman and Nicobar Islands have strong Malayan affinities and many endemics, such as the Nicobar megapode and the Narcondam hornbill.
Key takeaways
- Communities are assembled by dispersal, environmental filtering and interactions; primary succession starts without soil, secondary succession with it, by facilitation, tolerance or inhibition.
- β = γ/ᾱ; H′ = −Σp ln p with J′ = H′/ln S; Simpson D = Σp² is dominance, 1 − D and 1/D are diversity.
- S = cA^z with z ≈ 0.2–0.35 for islands; island richness is an equilibrium of falling immigration and rising extinction, highest on large, near islands.
- NPP = GPP − R_a; trophic transfer efficiency is about 10 %; energy flows one way and the energy pyramid is never inverted.
- Nitrosomonas takes ammonium to nitrite, Nitrobacter nitrite to nitrate, denitrifiers nitrate to N₂; India has ten biogeographic zones and lies mainly in the Oriental realm.
Practice questions (16)
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.
Succession on an abandoned agricultural field, where soil and a seed bank remain, is
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Answer: A — secondary succession
Succession that starts with soil and a biological legacy after a disturbance is secondary, and it is faster than primary succession, which begins on bare substrate such as rock or lava. A hydrosere starts in open water.On glacial moraines, nitrogen-fixing alders enrich the soil and make it possible for spruce to establish later. This supports which of Connell and Slatyer’s models?
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Answer: A — Facilitation
Early species that make the site more suitable for later ones are facilitators. Under tolerance the early species have no effect on the establishment of later ones, and under inhibition they hold the site against them.A community has four species with 40, 30, 20 and 10 individuals. Its Shannon index H′ (natural logarithms), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.28
H′ = −Σp ln p = 0.4(0.916) + 0.3(1.204) + 0.2(1.609) + 0.1(2.303) = 0.367 + 0.361 + 0.322 + 0.230 = 1.28. Using log₁₀ gives 0.56, and forgetting the minus sign gives −1.28.For the same community of four species with 40, 30, 20 and 10 individuals, Pielou’s evenness J′ = H′/ln S, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.92
H′ = 1.280 and ln 4 = 1.386, so J′ = 1.280/1.386 = 0.92. Dividing by S = 4 instead of ln S gives 0.32, and dividing by log₁₀ 4 mixes bases.A community has relative abundances 0.4, 0.3, 0.2 and 0.1. Simpson’s diversity index 1 − D, where D = Σpᵢ², to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.70
D = 0.16 + 0.09 + 0.04 + 0.01 = 0.30, the probability that two random individuals are the same species, so 1 − D = 0.70. Reporting 0.30 gives the dominance form D, and 1/D = 3.33 is the reciprocal form.A landscape contains 30 species in all, and its individual sites contain on average 12 species. Whittaker’s multiplicative beta diversity is ____.
Numerical answer — type the value.
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Answer: 2.5
β = γ/ᾱ = 30/12 = 2.5, the number of completely distinct communities the landscape is equivalent to. The additive form γ − ᾱ = 18 counts species not found in an average site and is a different measure.An island of 10 km² holds 20 species of reptiles and an island of 1000 km² in the same archipelago holds 50. Assuming S = cA^z, the exponent z, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.20
z = log(S₂/S₁)/log(A₂/A₁) = log 2.5/log 100 = 0.398/2 = 0.20. Taking the ratio of areas instead of its logarithm, or using (50 − 20)/(1000 − 10), treats a power law as a straight line on arithmetic axes.In a linear island-biogeography model with a mainland pool of P = 100 species, immigration is I = 10(1 − S/P) and extinction is E = 15S/P species per year. The equilibrium number of species on the island is ____.
Numerical answer — type the value.
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Answer: 40
Setting I = E: 10(1 − S/100) = 15S/100, so 10 = 25S/100 and S* = PI₀/(I₀ + E₀) = 1000/25 = 40. Taking 100 × 10/15 = 67 ignores the fall in immigration as the island fills.According to the equilibrium theory of island biogeography, which statements are correct?
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Answer: A — Large islands have lower extinction rates than small ones at the same richness; B — Islands near the mainland have higher immigration rates than distant ones; C — At equilibrium the species composition keeps changing through turnover
Area lowers the extinction curve and nearness raises the immigration curve. The equilibrium is dynamic: immigration and extinction continue at equal rates, so species turn over while the number stays roughly constant. The last statement confuses a dynamic with a static equilibrium.Producers in a grassland have a net primary productivity of 20 000 kJ m⁻² y⁻¹, and herbivore production is 1500 kJ m⁻² y⁻¹. The trophic transfer efficiency, in per cent to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 7.5
Efficiency = production of herbivores/production of plants = 1500/20 000 = 0.075 = 7.5 %. This lies within the usual range of about 5–20 % around Lindeman’s 10 %; 13.3 is the inverted ratio expressed as a number.In the open ocean the standing biomass of zooplankton can exceed that of phytoplankton. The best explanation is that
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Answer: A — phytoplankton turn over very rapidly, so a small standing stock sustains a high production
A pyramid of biomass records standing stock, not production. Phytoplankton divide within days and are grazed almost as fast, so their small stock supports a larger, slower-turning-over zooplankton stock. The pyramid of energy (production) remains upright; no efficiency can exceed 100 %.Which pairings of a nitrogen-cycle step with an organism that carries it out are correct?
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Answer: A — Oxidation of ammonium to nitrite — Nitrosomonas; B — Oxidation of nitrite to nitrate — Nitrobacter; C — Fixation of N₂ in root nodules of legumes — Rhizobium
Nitrification is a two-step aerobic oxidation: Nitrosomonas takes NH₄⁺ to NO₂⁻, Nitrobacter takes NO₂⁻ to NO₃⁻. Rhizobium fixes N₂ in legume nodules. Denitrification is anaerobic, carried out by bacteria such as Pseudomonas in waterlogged soils — not by the nitrifier Nitrosomonas in aerated soil.After a severe drought, grassland A lost 60 % of its biomass and recovered fully in one season; forest B lost 10 % but took 15 years to recover. Which description is correct?
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Answer: A — A has low resistance and high resilience; B has high resistance and low resilience
Resistance is how little a system changes under disturbance (B changed less); resilience is how quickly it returns (A returned faster). Stability has several components and the two systems rank oppositely on them, which is why "more stable" without a component is not a complete answer.Which of the following are among the ten biogeographic zones into which India is classified?
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Answer: A — Trans-Himalaya; B — Western Ghats; C — Deccan Peninsula
Rodgers and Panwar’s ten zones are the Trans-Himalaya, Himalaya, Desert, Semi-arid, Western Ghats, Deccan Peninsula, Gangetic Plain, North-East, Coasts and Islands. The Sunda Shelf is the continental shelf of South-East Asia whose edge Wallace’s line follows; it is not an Indian zone.India lies mainly within which of Wallace’s zoogeographic realms?
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Answer: A — Oriental (Indomalayan)
Most of India is Oriental; Palaearctic elements occur in the Himalaya and Trans-Himalaya, and the fauna shows some African and Malayan affinities. The Australian realm begins east of Wallace’s line in Indonesia.A reserve network is to be designed for a species group whose species–area exponent is z = 0.3. If a reserve’s area is increased tenfold, the expected number of species it holds changes by a factor of about
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Answer: A — 2
S₂/S₁ = (A₂/A₁)^z = 10^0.3 = 2.0 — a tenfold increase in area roughly doubles richness. Multiplying 10 × 0.3 = 3 confuses the exponent with a coefficient, and 10 would be the answer only for z = 1.