Animal Ecology: Ecosystems, Populations, Diversity, Zoogeography and Conservation
2. Biogeochemical cycles
Nutrients cycle between living and non-living pools. Gaseous cycles (carbon, nitrogen, oxygen) have a large atmospheric reservoir and recover quickly; sedimentary cycles (phosphorus; sulfur is mostly sedimentary) have their reservoir in rocks and are slow. Animals move nutrients by feeding, excretion (ammonia, urea, uric acid) and migration: seabird guano and spawning salmon carry marine nitrogen and phosphorus onto land.
| Cycle | Step | Agents and notes |
|---|---|---|
| Carbon | CO₂ fixed by photosynthesis, returned by respiration, decomposition and combustion | Oceans dissolve CO₂ and store carbonate in sediments and shells |
| Nitrogen | Fixation N₂ → NH₃ | Nitrogenase (O₂-sensitive) in Rhizobium, Azotobacter, cyanobacteria; also lightning and the industrial Haber–Bosch process |
| Nitrogen | Ammonification, then nitrification NH₄⁺ → NO₂⁻ → NO₃⁻ | Decomposers release NH₄⁺; Nitrosomonas oxidises it to nitrite, Nitrobacter nitrite to nitrate |
| Nitrogen | Denitrification NO₃⁻ → N₂ | Anaerobic soils and sediments (Pseudomonas, Paracoccus) |
| Phosphorus | Weathering of rock → PO₄³⁻ → organisms → sediments | No significant gaseous phase; often limits freshwater production, so P runoff drives eutrophication |
| Sulfur | SO₄²⁻ ⇄ H₂S | Desulfovibrio reduces sulfate anaerobically; Thiobacillus and Beggiatoa oxidise sulfide |
3. The niche, competition and animal distribution
The niche has three classic meanings: Grinnell’s habitat niche (where a species lives), Elton’s functional niche (its role in the food web) and Hutchinson’s n-dimensional hypervolume of all conditions and resources in which it can persist. Hutchinson separated the fundamental niche (possible without enemies or competitors) from the narrower realized niche (actually occupied). On the Scottish shore Connell found that the barnacle Chthamalus survives lower down when Balanus is removed, so competition sets its lower limit, while desiccation sets the upper limit of Balanus.
Gause’s competitive exclusion principle: two species cannot coexist indefinitely on the same limiting resource; in his cultures Paramecium aurelia drove P. caudatum out. Coexistence therefore needs niche differentiation: resource partitioning (MacArthur’s warblers feeding in different zones of the same spruce trees) and character displacement (beak sizes of Galápagos ground finches diverge where two species meet). This is how the niche contributes to diversity: the finer resources are divided, the more species can be packed into a community.
- Limits to distribution: Liebig’s law of the minimum (the scarcest factor limits) and Shelford’s law of tolerance (both too little and too much limit); add dispersal barriers, competitors, predators and history.
- Dispersion within a range is clumped, random or uniform; for quadrat counts, variance ÷ mean is > 1, = 1 or < 1 respectively.
- Ecogeographical rules in endotherms: Bergmann (larger body in colder climates), Allen (shorter ears, tails and limbs in colder climates), Gloger (darker pigmentation in warm, humid regions).
4. Population dynamics
A population changes by births, deaths, immigration and emigration. With unlimited resources it grows exponentially: dN/dt = rN, so Nₜ = N₀e^(rt) and doubling time = ln 2 ÷ r (r = 0.07 per year doubles in about 10 years). With a carrying capacity K it grows logistically: dN/dt = rN(1 − N/K), an S-shaped curve whose growth rate peaks at N = K/2, at rK/4. Density-dependent factors (food, disease, predation) bite harder as N rises and regulate; density-independent factors (frost, floods) do not. Lotka–Volterra predator–prey equations give coupled cycles with the predator lagging the prey, as in the snowshoe hare and lynx fur records.
A life table follows a cohort: nₓ alive at age x, lₓ = nₓ/n₀ (survivorship), dₓ = nₓ − nₓ₊₁, qₓ = dₓ/nₓ (age-specific mortality) and mₓ (daughters per female). The net reproductive rate R₀ = Σ lₓmₓ; R₀ > 1 means growth. Survivorship curves: type I (low mortality until old age: humans, large mammals), type II (constant mortality: many birds), type III (heavy early mortality: oysters, most fishes). These track r-selected species (many small young, little care, unstable habitats) and K-selected ones (few large young, much care, populations near K).
5. Species diversity
Diversity has two parts: richness (number of species, S) and evenness (how equally individuals are shared). Whittaker’s scales: α (within one habitat), β (turnover between habitats) and γ (a whole region). With pᵢ the proportion of individuals in species i, the Shannon index H′ = −Σ pᵢ ln pᵢ (natural log unless stated) is 0 for one species and maximal, ln S, when all are equally common; evenness J = H′/ln S. Simpson’s index D = Σ pᵢ² is the chance that two individuals drawn at random (with replacement) belong to the same species, so it measures dominance; diversity is reported as 1 − D or 1/D. Always check which form, and which log base, a question uses.
Patterns: species number rises with area as S = cA^z (z is larger for isolated islands); MacArthur and Wilson’s island biogeography balances immigration (falling as the island fills, lower for distant islands) against extinction (rising with S, higher on small islands), so large near islands hold most species. Diversity generally rises towards the tropics. A keystone species has an effect far beyond its abundance: removing the starfish Pisaster from rocky shores let mussels monopolise space and diversity fell (Paine).
6. Zoogeography
Distributions are explained by dispersal across barriers and by vicariance, the splitting of a once-continuous range by geology: lungfishes in South America, Africa and Australia are relics of Gondwana. Wallace (1876) divided the land into six zoogeographical realms. Wallace’s line, between Bali and Lombok and between Borneo and Sulawesi, follows a deep-water channel that marks the eastern limit of many Oriental mammals; Weber’s line, farther east, is where Oriental and Australian faunas balance.
| Realm | Area | Characteristic animals |
|---|---|---|
| Palaearctic | Europe, North Africa, Asia north of the Himalaya | Few endemic families; shares much with the Nearctic (together Holarctic) |
| Nearctic | North America to the Mexican highlands | Pronghorn, mountain beaver |
| Neotropical | South and Central America | Sloths, armadillos, anteaters, New World monkeys, rheas |
| Ethiopian (Afrotropical) | Africa south of the Sahara, Madagascar | Giraffe, hippopotamus, aardvark, lemurs of Madagascar |
| Oriental | India, South-East Asia to Wallace’s line | Gibbons, tree shrews, colugos |
| Australian | Australia, New Guinea, nearby islands | Monotremes, most marsupials, emu, cassowary; few native placentals besides bats and rodents |
India lies mainly in the Oriental realm, with Palaearctic elements in the Trans-Himalaya and Himalaya and some African affinities in the dry west. Within India, Rodgers and Panwar (1988) recognised ten biogeographic zones: Trans-Himalaya, Himalaya, Desert, Semi-arid, Western Ghats, Deccan Peninsula, Gangetic Plain, North-East, Coasts and Islands.
7. Conservation biology and India’s biodiversity hotspots
The main threats are habitat loss and fragmentation, invasive species, pollution, overexploitation and climate change. Small populations add genetic risks: drift and inbreeding depression depend on the effective population size, which for unequal sexes is Nₑ = 4NₘN_f/(Nₘ + N_f); 10 breeding males and 40 females give Nₑ = 1,600/50 = 32, not 50. Demographic and genetic problems feed each other in an extinction vortex. The IUCN Red List grades risk as Extinct, Extinct in the Wild, Critically Endangered, Endangered, Vulnerable (the last three are “threatened”), Near Threatened, Least Concern and Data Deficient. Conservation is in situ (national parks, wildlife sanctuaries, biosphere reserves with core, buffer and transition zones, corridors) or ex situ (zoos, captive breeding, gene and gamete banks). In India the Wildlife (Protection) Act 1972, Project Tiger (1973) and the Biological Diversity Act 2002 frame this work.
A biodiversity hotspot (Norman Myers’ idea, as defined by Conservation International) is a region with at least 1,500 endemic vascular plant species that has lost at least 70% of its original habitat. The global count has grown with revisions, from 25 (2000) to 34 (the 2004–05 revision) to 36 in the current scheme. India includes parts of four: the Himalaya, Indo-Burma (the North-East), the Western Ghats and Sri Lanka, and Sundaland (through the Nicobar Islands). The Western Ghats are especially rich in endemic amphibians.
8. Ecotoxicology
Ecotoxicology studies how pollutants affect organisms, populations and ecosystems. Toxicity is measured by dose–response tests: LD₅₀/LC₅₀ (dose or concentration killing half the test animals in a fixed time; lower means more toxic), EC₅₀ for a sub-lethal effect, and NOEC (highest concentration with no observed effect) for chronic exposure. Three terms must be kept apart: bioconcentration is uptake from water alone (BCF = C_organism ÷ C_water); bioaccumulation is build-up in an organism from all sources over its life; biomagnification is the rise in concentration from prey to predator (BMF = C_predator ÷ C_prey). Magnification needs a chemical that is persistent, lipophilic (high octanol–water partition coefficient) and poorly excreted: DDT and its metabolite DDE, PCBs and methylmercury qualify; water-soluble, readily metabolised chemicals do not.
- DDE thinned the eggshells of fish- and bird-eating raptors such as the peregrine falcon, so eggs broke during incubation and populations crashed.
- Methylmercury biomagnified in fish caused Minamata disease, a neurological poisoning of people eating fish from Minamata Bay, Japan.
- Diclofenac, a veterinary anti-inflammatory drug left in livestock carcasses, causes kidney failure and visceral gout in Gyps vultures; it drove the collapse of Indian vulture populations and its veterinary use was banned in India in 2006, with meloxicam as a vulture-safe substitute.
- Biomarkers reveal exposure early: inhibition of acetylcholinesterase by organophosphate and carbamate insecticides, and imposex (female snails growing male organs) caused by tributyltin antifouling paint, an endocrine disruptor.
Key takeaways
- Energy flows one way and only about 10% passes each trophic link, so the energy pyramid is always upright and chains are short; nutrients cycle, with phosphorus lacking a gaseous phase and nitrogen moved by fixers, nitrifiers (Nitrosomonas, Nitrobacter) and denitrifiers.
- The realized niche is narrower than the fundamental niche; competitive exclusion forces niche differentiation, which is how niches add to diversity.
- Nₜ = N₀e^(rt); logistic growth dN/dt = rN(1 − N/K) peaks at K/2; Lincoln–Petersen N = Mn/m; R₀ = Σ lₓmₓ.
- Shannon H′ = −Σ pᵢ ln pᵢ rises with richness and evenness; Simpson D = Σ pᵢ² measures dominance, so diversity is 1 − D or 1/D.
- Wallace’s six realms; India is mainly Oriental and has ten biogeographic zones. Of the 36 hotspots in the current scheme India shares four: Himalaya, Indo-Burma, Western Ghats and Sri Lanka, Sundaland (Nicobars).
- Persistent lipophilic chemicals biomagnify (BMF = C_predator ÷ C_prey), hitting top predators: DDE and eggshells, methylmercury and Minamata, diclofenac and vultures.
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.
The net primary productivity of a grassland is 25,000 kJ m⁻² yr⁻¹. Assuming 10% transfer efficiency between successive trophic levels, how much energy (kJ m⁻² yr⁻¹) is available to tertiary consumers?
Show answer
Answer: C — 25
NPP is the producers’ output. Primary consumers (herbivores) get 10% = 2,500; secondary consumers 250; tertiary consumers 25 kJ m⁻² yr⁻¹. Three transfers separate producers from tertiary consumers. The tempting 250 counts only two transfers, as if tertiary consumers were the third trophic level rather than the fourth; 2.5 adds a fourth transfer.In the open ocean the standing biomass of zooplankton often exceeds that of the phytoplankton on which they feed. Which explanation is correct?
Show answer
Answer: B — Phytoplankton have a small standing crop but divide so fast that their yearly production exceeds consumer needs
A biomass pyramid shows standing crop at one moment, not production over time. Phytoplankton are eaten almost as fast as they grow, so their standing crop is small, but their high turnover gives an annual production larger than that of the zooplankton. The pyramid of energy (production per year) therefore stays upright; an inverted energy pyramid, the tempting choice, would violate the second law of thermodynamics. Photosynthetic phytoplankton cannot live below the photic zone.Which of the following statements about biogeochemical cycles are correct?
Show answer
Answer: A — Phosphorus has no significant atmospheric reservoir, so it enters food chains mainly through the weathering of rocks; B — Denitrification, which returns N₂ to the atmosphere, proceeds mainly under anaerobic conditions
Phosphorus cycles through rocks, soil, water and organisms with no important gaseous form, and denitrifying bacteria use nitrate as an electron acceptor when oxygen is absent, releasing N₂. Nitrification is two-step: Nitrosomonas oxidises NH₄⁺ to NO₂⁻ and Nitrobacter oxidises NO₂⁻ to NO₃⁻. Nitrogenase is irreversibly damaged by O₂, which is why heterocysts, leghaemoglobin and high respiration rates protect it.A population of 200 animals grows exponentially with an intrinsic rate of increase r = 0.2 per week (continuous growth, Nₜ = N₀e^(rt)). Its size after 15 weeks, to the nearest whole number, is ____.
Numerical answer — type the value.
Show answer
Answer: 4017
rt = 0.2 × 15 = 3. Nₜ = 200 × e³ = 200 × 20.0855 = 4017.1, which rounds to 4017. Using discrete doubling or (1 + r)ᵗ = 1.2¹⁵ = 15.41 instead of e^(rt) would give about 3081, the usual slip.A population follows logistic growth, dN/dt = rN(1 − N/K), with r = 0.4 per year and carrying capacity K = 5,000. Its growth rate, in individuals per year, when N = 1,000 is ____.
Numerical answer — type the value.
Show answer
Answer: 320
dN/dt = 0.4 × 1000 × (1 − 1000/5000) = 400 × 0.8 = 320 individuals per year. Forgetting the braking term gives 400 (pure exponential growth). For comparison, the maximum rate occurs at N = K/2 = 2,500 and equals rK/4 = 500 per year.To estimate the frog population of a pond, 120 frogs are caught, marked and released. A week later 150 frogs are caught, of which 18 carry marks. Assuming a closed population and equal catchability, the simple Lincoln–Petersen estimate (N = Mn/m) of the population is ____.
Numerical answer — type the value.
Show answer
Answer: 1000
The marked fraction in the second sample estimates the marked fraction of the whole population: 18/150 = 120/N, so N = 120 × 150 ÷ 18 = 18,000 ÷ 18 = 1000. If marked frogs became trap-shy, fewer would be recaptured and N would be overestimated.A cohort of 500 newborn females of an insect is followed. The numbers alive at the start of age classes 0, 1, 2, 3 and 4 are 500, 300, 150, 50 and 0. The mean numbers of female offspring per female (mₓ) in age classes 0, 1, 2 and 3 are 0, 1.0, 2.0 and 1.5. The net reproductive rate R₀ = Σ lₓmₓ, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.35
lₓ = nₓ/n₀: l₀ = 1.0, l₁ = 300/500 = 0.6, l₂ = 150/500 = 0.3, l₃ = 50/500 = 0.1. Products lₓmₓ: 0, 0.6 × 1.0 = 0.6, 0.3 × 2.0 = 0.6, 0.1 × 1.5 = 0.15. R₀ = 0 + 0.6 + 0.6 + 0.15 = 1.35, so each female replaces herself 1.35 times and the population grows. Summing mₓ alone (4.5) ignores that most females die before breeding.Plotted as log lₓ against age, a species’ survivorship falls very steeply in early life and then flattens, so the few that survive the earliest stage live long. Which organism most typically shows such a type III curve?
Show answer
Answer: D — An oyster, which releases millions of planktonic larvae and gives no parental care
Type III survivorship goes with very many small offspring and no care, so almost all die as larvae: oysters and most marine fishes are the standard examples. Elephants and well-cared-for human populations show type I (most survive to old age); the songbird losing a constant fraction each year is the type II pattern, the most tempting distractor because its curve also falls from the start, but as a straight line on a log scale.On a rocky shore the barnacle Chthamalus occupies the upper intertidal zone and Balanus the lower. When Balanus is removed from the lower zone, Chthamalus settles and survives there; when Chthamalus is removed from the upper zone, Balanus still fails there because it dries out. Which conclusion follows?
Show answer
Answer: A — Competition from *Balanus* restricts *Chthamalus* to a realized niche narrower than its fundamental niche
Chthamalus can live in the lower zone (its fundamental niche includes it) but does so only when Balanus is absent, so interspecific competition shrinks its realized niche, as Gause’s exclusion principle predicts. The tempting reverse claim fails because removing Chthamalus does not let Balanus survive higher up: physical stress, not competition, sets the upper limit of Balanus. Desiccation is worse higher on the shore, not lower, and larvae settling lower clearly survive once Balanus is gone.A sample of 100 beetles contains four species with 40, 30, 20 and 10 individuals. Using the Shannon index H′ = −Σ pᵢ ln pᵢ (natural logarithm), H′ to two decimal places is ____.
Numerical answer — type the value.
Show answer
Answer: 1.28
pᵢ = 0.4, 0.3, 0.2, 0.1. Terms −pᵢ ln pᵢ: 0.4 × 0.9163 = 0.3665; 0.3 × 1.2040 = 0.3612; 0.2 × 1.6094 = 0.3219; 0.1 × 2.3026 = 0.2303. Sum H′ = 1.2799 ≈ 1.28. The maximum for four species is ln 4 = 1.386, so evenness J = 0.92. Using log₁₀ instead of ln would give 0.56.Two communities each contain four species and 100 individuals: X has 25, 25, 25 and 25; Y has 70, 10, 10 and 10. Using Simpson’s diversity 1 − D, with D = Σ pᵢ², which statement is correct?
Show answer
Answer: C — X scores 0.75 and Y 0.48, so X is more diverse because it is more even
X: D = 4 × 0.25² = 0.25, so 1 − D = 0.75. Y: D = 0.7² + 3 × 0.1² = 0.49 + 0.03 = 0.52, so 1 − D = 0.48. Richness is equal, so the difference is evenness, and X is more diverse. The tempting 0.25 versus 0.52 reports D itself, which measures dominance and rises as diversity falls; the 0.52 for Y in the remaining wrong option is also D, not 1 − D.Wallace’s line, a classic zoogeographical boundary, runs between
Show answer
Answer: B — Bali and Lombok, and Borneo and Sulawesi, marking the eastern limit of many Oriental mammals
Wallace noticed the abrupt faunal change across the deep strait between Bali and Lombok, continuing between Borneo and Sulawesi: placental mammals of Asian type (such as many primates, cats and ungulates) largely stop there. New Guinea is wrong because it shares marsupials with Australia across a shallow shelf; both lie east of the line in the Australian realm. Madagascar’s separation is real but is not Wallace’s line.Which of the following pairings of a zoogeographical realm (Wallace’s scheme) with animals characteristic of it are correct?
Show answer
Answer: A — Neotropical realm: sloths, armadillos and anteaters; B — Australian realm: monotremes and most living marsupials
Sloths, armadillos and anteaters (Xenarthra) are the signature Neotropical mammals, and the platypus, echidnas and most marsupials belong to the Australian realm. Gibbons and tree shrews are Oriental (South and South-East Asia), and lemurs (Madagascar) and the aardvark (sub-Saharan Africa) are Ethiopian, not Nearctic.Across the range of a widespread mammal species, populations in colder regions have shorter ears, tails and limbs than those in warmer regions. This pattern is described by
Show answer
Answer: D — Allen’s rule, linking shorter extremities to colder climates
Allen’s rule concerns appendages: shorter extremities reduce surface area and heat loss in the cold, while long ears and limbs help shed heat in hot climates. Bergmann’s rule is the tempting distractor because it also concerns heat balance in cold climates, but it is about overall body size, not the length of appendages.Under Conservation International’s current scheme of 36 global biodiversity hotspots, which of the following statements are correct?
Show answer
Answer: A — A region qualifies if it holds at least 1,500 endemic vascular plant species and has lost at least 70% of its original habitat; B — India includes parts of four hotspots: the Himalaya, Indo-Burma, the Western Ghats and Sri Lanka, and Sundaland
Hotspots are defined by two thresholds together, plant endemism (at least 1,500 endemic vascular plants) and threat (at least 70% of original habitat lost), so richness alone is not enough. India shares four hotspots in this scheme. The Nicobar Islands lie in Sundaland; it is the Andaman Islands that fall in Indo-Burma.A captive-breeding programme for a pheasant has 8 breeding males and 24 breeding females. Using Nₑ = 4NₘN_f/(Nₘ + N_f), its effective population size is
Show answer
Answer: A — 24
Nₑ = 4 × 8 × 24 ÷ (8 + 24) = 768 ÷ 32 = 24. The census count of 32 breeders is the tempting answer, but with a skewed sex ratio each male contributes half the genes of the next generation from a small pool, so the population loses genetic diversity as fast as an ideal population of only 24.A persistent organochlorine is dissolved in a lake at 0.002 µg L⁻¹. Phytoplankton take it up from water with a bioconcentration factor of 2,000 L kg⁻¹, and the biomagnification factor is 6 at each of the next three predator–prey steps (phytoplankton → zooplankton → small fish → large fish). The concentration in the large fish, in µg kg⁻¹, is ____.
Numerical answer — type the value.
Show answer
Answer: 864
Phytoplankton: 2,000 L kg⁻¹ × 0.002 µg L⁻¹ = 4 µg kg⁻¹. Three magnification steps: 4 × 6 × 6 × 6 = 4 × 216 = 864 µg kg⁻¹. Applying the factor 6 four times (5,184) or only twice (144) are the usual slips; count the arrows after the producer.The collapse of Gyps vulture populations in India from the 1990s onwards was traced mainly to which cause?
Show answer
Answer: C — Diclofenac residues in livestock carcasses, which caused kidney failure and visceral gout
Vultures feeding on carcasses of cattle treated with the anti-inflammatory drug diclofenac developed acute kidney failure, seen as visceral gout (urate deposits) at post-mortem; India banned its veterinary use in 2006. DDT eggshell thinning is the tempting distractor because it is the classic raptor case, but it affected fish- and bird-eating raptors such as peregrines, and the vulture deaths were of adults with kidney failure, not breeding failure.