Ecology I: Fundamental Concepts, Population Ecology and Species Interactions

Section 1 of the GATE Ecology and Evolution paper, first half. Section 1 is the widest section of the paper, and it divides naturally into two subjects: how individuals and populations work — this chapter — and how communities and ecosystems work, which is the next. This chapter follows the syllabus order: the fundamental concepts (abiotic and biotic components, levels of organisation, niches, habitats, functional traits and ecophysiology); population dynamics and growth, density-dependent and density-independent; age-structured populations through life tables and Leslie matrices; metapopulations; and the interactions between species, with the Lotka–Volterra models of competition and predation and the direct and indirect effects that travel through trophic levels. Every numerical is worked in the text before the quiz asks for it.

1. Fundamental concepts: components, levels, niches, habitats, traits and ecophysiology

An ecosystem has abiotic components — light, temperature, water, soil, nutrients, pH, salinity — and biotic components: producers, consumers and decomposers, and the interactions among them. Ecology is organised in levels: the individual, the population (individuals of one species in one place), the species, the community (the populations of all species that co-occur), the ecosystem (the community with its abiotic environment and the fluxes of energy and matter between them) and the biome (a large region defined by its climate and its dominant vegetation form). Each level has properties the level below does not: a birth rate belongs to a population, not to an individual; diversity belongs to a community.

Three ideas of the niche
AuthorNiche asWhat it emphasises
Grinnell (1917)the habitat requirements of a specieswhere a species can live
Elton (1927)the functional role in the communitywhat a species does — what it eats and what eats it
Hutchinson (1957)an n-dimensional hypervolume of conditions and resourcesfundamental niche (without competitors) versus realised niche (with them)

The habitat is the place where an organism lives; the niche is its role and its requirements, so two species can share a habitat and differ in niche. Connell’s barnacles on the Scottish coast are the classic demonstration of the realised niche: Chthamalus can live across the whole intertidal zone, but is confined to the upper zone because Balanus outcompetes it lower down; removing Balanus extended Chthamalus downwards. Functional traits are measurable features of an organism that affect its performance or its effect on the ecosystem — specific leaf area, wood density, seed mass, body size, thermal tolerance — and they let ecologists compare communities by what their species do rather than by their names.

Ecophysiology asks how organisms cope with their abiotic environment. Homeostasis holds the internal state steady: regulators (mammals and birds for temperature, freshwater fish for osmotic balance) keep it constant at a cost, conformers let it follow the environment. Acclimation (acclimatisation in the field) is a reversible physiological adjustment within one individual’s lifetime, such as the rise in red-cell count at altitude; adaptation is a heritable change across generations produced by selection. The temperature sensitivity of a rate is summarised by Q₁₀ = (R₂/R₁)10/(T₂ − T₁): a respiration rate that rises from 2 to 5 units between 15 °C and 25 °C has Q₁₀ = 2.5. Physiological responses to a gradient are hump-shaped performance curves with a tolerance range, and Shelford’s law of tolerance says the distribution of a species is set by the factor it tolerates least well.

⚠️ Acclimation is not adaptation
A lowland visitor who makes more red cells after two weeks in Ladakh has acclimatised; the change reverses on returning and is not inherited. Populations that have lived at altitude for thousands of years and differ genetically in their oxygen physiology are adapted. A question that says "within the lifetime of an individual" and "reversible" is describing acclimation.

2. Population growth: exponential, logistic, density dependence and estimating N

With unlimited resources a population grows exponentially: dN/dt = rN, so N(t) = N₀ert, where r = b − d is the intrinsic rate of increase (per-capita births minus deaths per unit time). The doubling time is t_d = ln 2 / r = 0.6931/r; a population with r = 0.05 per day doubles every 13.86 days. In discrete generations the model is Nt+1 = λN_t, with λ, the finite rate of increase, related to r by λ = e^r and r = ln λ. λ = 1.2 per year means 20 % growth per year and r = ln 1.2 = 0.182 per year. The population grows if r > 0 (λ > 1), is stationary at r = 0 (λ = 1) and declines if r < 0 (λ < 1).

Resources run out, and the logistic model adds a brake: dN/dt = rN(1 − N/K), where K is the carrying capacity. Its solution is the S-shaped curve N(t) = K / [1 + ((K − N₀)/N₀)e−rt]. The per-capita growth rate (1/N)dN/dt = r(1 − N/K) falls linearly with N, from r at N → 0 to zero at N = K. The population growth rate dN/dt = rN − rN²/K is a downward parabola in N; setting its derivative r − 2rN/K to zero gives the maximum at N = K/2, where dN/dt = rK/4. This is the basis of maximum sustainable yield: with r = 0.4 per year and K = 1000, the fastest growth is 100 individuals per year at N = 500.

🧠 Work a logistic value in three moves
For N₀ = 100, K = 1000, r = 0.5 and t = 2: (K − N₀)/N₀ = 9; e−rt = e−1 = 0.3679; so N = 1000/(1 + 9 × 0.3679) = 1000/4.311 = 232. Checking: an exponential population would have reached 100e¹ = 272, and the logistic value must be smaller — the brake has already begun to act.

Density-dependent factors change the per-capita birth or death rate as density changes — competition for food or nest sites, disease transmission, predation that intensifies on a common prey — and can regulate a population towards an equilibrium. Density-independent factors act with the same per-capita force whatever the density — a frost, a flood, a drought — and cause fluctuations but cannot regulate. Delayed density dependence (a time lag between density and its effect, as in discrete-time models with large r) can turn a smooth approach to K into damped oscillations, cycles or chaos.

Population size is rarely counted directly. In mark–recapture, M animals are caught, marked and released; later a sample of n is caught, of which m are marked. If marked animals mix randomly, m/n = M/N, giving the Lincoln–Petersen estimate N̂ = Mn/m. Marking 120 fish, then catching 150 of which 30 are marked, gives N̂ = 120 × 150/30 = 600. The estimate assumes a closed population (no births, deaths or migration between samples), marks that are not lost and do not change behaviour, and equal catchability. For small samples Chapman’s form N̂ = (M + 1)(n + 1)/(m + 1) − 1 reduces the bias.

3. Age-structured populations: life tables and Leslie matrices

A cohort life table follows a group born at the same time. From the number alive at the start of each age class, nₓ, everything else follows: survivorship lₓ = nₓ/n₀; deaths dₓ = nₓ − nₓ₊₁; age-specific mortality qₓ = dₓ/nₓ; the average number alive during the interval Lₓ = (nₓ + nₓ₊₁)/2; the total individual-intervals still to be lived Tₓ = Σ Lₓ from x onwards; and life expectancy eₓ = Tₓ/nₓ. With the age-specific fecundity mₓ (female offspring per female in age class x) the table gives the net reproductive rate R₀ = Σ lₓmₓ, the average number of daughters a newborn female produces in her lifetime, and the generation time T = Σ x lₓmₓ / R₀. Then r ≈ ln R₀ / T.

A worked cohort life table (n₀ = 1000)
Age xnₓlₓqₓLₓTₓeₓmₓlₓmₓ
010001.000.5075012501.2500
15000.500.603505001.0021.0
22000.200.751251500.7530.6
3500.051.0025250.5020.1
400—00———

From the table: R₀ = 0 + 1.0 + 0.6 + 0.1 = 1.7, so each female replaces herself 1.7 times and the population grows. Σ x lₓmₓ = 1 × 1.0 + 2 × 0.6 + 3 × 0.1 = 2.5, so T = 2.5/1.7 = 1.47 years, and r ≈ ln 1.7/1.47 = 0.531/1.47 = 0.36 per year. Life expectancy at birth is e₀ = 1250/1000 = 1.25 years; at age 1 it is e₁ = 500/500 = 1.00 year. Notice that q rises with age here, which is a Type I pattern in the later classes.

The three survivorship curves (log lₓ against age)
TypeMortality patternExamples
I — convexlow early, concentrated in old agehumans in developed countries, large mammals
II — straight lineconstant probability of death at every agemany birds, some lizards, seed banks
III — concavevery high among the young, low for the few survivorsoysters, marine fish, most plants with many small seeds

The Leslie matrix projects an age-structured population: n(t + 1) = L n(t), where the top row holds the fecundities Fₓ and the sub-diagonal the survival probabilities Pₓ from class x to x + 1. With L = [[0, 2, 1.5], [0.5, 0, 0], [0, 0.4, 0]] and n(0) = (100, 50, 20): the new first class is 0 × 100 + 2 × 50 + 1.5 × 20 = 130; the second is 0.5 × 100 = 50; the third is 0.4 × 50 = 20, so n(1) = (130, 50, 20), total 200. One step more gives (2 × 50 + 1.5 × 20, 0.5 × 130, 0.4 × 50) = (130, 65, 20), total 215. Repeated projection converges to a stable age distribution (the dominant eigenvector) in which every class grows by the same factor each step, the dominant eigenvalue λ; λ > 1 means growth.

⚠️ Multiply rows into the column, not columns
Each entry of n(t + 1) is one row of L times the whole vector n(t). A common slip multiplies the fecundity row by the survival row, or puts the survival probabilities on the diagonal — which would mean individuals stay in their age class, something an age class by definition cannot do.

4. Metapopulation ecology

A metapopulation is a population of populations: local populations in discrete habitat patches, each of which can go extinct and be recolonised by dispersers from occupied patches. In Levins’ model the fraction of occupied patches p changes as dp/dt = c p(1 − p) − e p, colonisation of empty patches at rate c by occupied ones, minus extinction at rate e. At equilibrium p* = 1 − e/c; with e = 0.2 and c = 0.5 per year, 60 % of patches are occupied. The metapopulation persists only if c > e, and it can persist regionally even though every local population is doomed — persistence is a property of the network, not of any patch.

  • Source–sink dynamics: in a source habitat births exceed deaths (λ > 1) and surplus individuals emigrate; a sink (λ < 1) persists only through immigration. A sink can hold many individuals, so abundance is not a reliable guide to habitat quality.
  • Rescue effect: immigration into a declining local population lowers its extinction risk, so extinction rate falls as the fraction of occupied patches rises.
  • The Glanville fritillary butterfly on the Åland islands of Finland, studied by Hanski and colleagues, is the textbook metapopulation: hundreds of meadow patches with frequent local extinction and recolonisation.
  • Conservation corollary: destroying a fraction e/c of patches — even empty ones — can push the whole network to extinction, because empty patches are the targets of colonisation.

5. Interactions between species: types, Lotka–Volterra models and trophic effects

Interactions by their effect on each partner
InteractionEffect on A / BExample
Mutualism+ / +figs and fig wasps; mycorrhizal fungi and plant roots
Commensalism+ / 0epiphytic orchids on trees; cattle egrets following cattle
Competition− / −two barnacle species for rock space
Predation, herbivory, parasitism+ / −tiger and chital; tapeworm and host
Amensalism0 / −Penicillium inhibiting bacteria; a tree shading out herbs

"Symbiosis" in its broad, de Bary sense means any intimate living-together and so includes parasitism as well as mutualism; in many textbooks it is used for close mutualisms such as lichens. The Lotka–Volterra competition model gives each species logistic growth reduced by the other: dN₁/dt = r₁N₁(K₁ − N₁ − αN₂)/K₁ and dN₂/dt = r₂N₂(K₂ − N₂ − βN₁)/K₂, where α is the per-capita effect of species 2 on species 1 measured in species-1 equivalents. The zero-growth isoclines are straight lines: N₁ = K₁ − αN₂ and N₂ = K₂ − βN₁. Their relative positions give four outcomes: species 1 always wins, species 2 always wins, a stable coexistence equilibrium, or an unstable equilibrium where the winner depends on the starting numbers.

Solving the two isoclines together gives the interior equilibrium N₁* = (K₁ − αK₂)/(1 − αβ) and N₂* = (K₂ − βK₁)/(1 − αβ). Coexistence is stable when α < K₁/K₂ and β < K₂/K₁ — each species limits itself more than it limits its competitor. With K₁ = 500, K₂ = 400, α = 0.5 and β = 0.6: α < 1.25 and β < 0.8, so both conditions hold; 1 − αβ = 0.7; N₁* = (500 − 200)/0.7 = 428.6 and N₂* = (400 − 300)/0.7 = 142.9. Gause’s competitive exclusion principle — complete competitors cannot coexist — is the limiting case in which the niches do not differ.

The Lotka–Volterra predator–prey model is dN/dt = rN − aNP for prey and dP/dt = baNP − mP for predators, with attack rate a, conversion efficiency b and predator death rate m. Setting each to zero: prey stop changing when P* = r/a, predators when N* = m/(ba). With r = 0.6, a = 0.01, b = 0.1 and m = 0.2: P* = 60 and N* = 0.2/0.001 = 200. The model produces neutral cycles — predator peaks lag prey peaks by a quarter cycle and the amplitude is set by the starting point — which is structurally unstable; adding prey density dependence damps the cycles. The functional response (prey eaten per predator against prey density) is Holling’s Type I (linear), Type II (saturating, set by handling time) or Type III (sigmoid, which can stabilise prey at low density through prey switching or refuges).

🎯 Why the prey equilibrium is set by the predator’s parameters
N* = m/(ba) contains no prey parameter: it is the prey density at which predators just replace themselves. Enriching the prey’s environment (raising r) therefore raises the predator equilibrium P* = r/a, not the prey equilibrium — a result that surprises anyone expecting more food to mean more prey, and the seed of the "paradox of enrichment".

Effects also travel indirectly through food webs. In a trophic cascade a top predator suppresses herbivores and so releases plants: sea otters eat sea urchins, and where otters were hunted out, urchins grazed kelp forests into barrens. Keystone species have effects far out of proportion to their abundance — Paine’s removal of the starfish Pisaster from rocky shores let mussels monopolise space and cut the number of species. Apparent competition arises when two prey share a predator: more of one prey feeds more predators, which then eat more of the other, so the two prey harm each other without ever competing for a resource. Top-down (consumer) and bottom-up (resource) control both operate, and their relative strength differs between systems.

Key takeaways

  • The realised niche is the fundamental niche cut down by competitors and enemies; acclimation is reversible within a lifetime, adaptation is heritable across generations.
  • Exponential: N₀ert, doubling time ln 2/r, λ = e^r. Logistic: dN/dt = rN(1 − N/K), fastest growth rK/4 at N = K/2.
  • Life tables: lₓ = nₓ/n₀, qₓ = dₓ/nₓ, eₓ = Tₓ/nₓ, R₀ = Σlₓmₓ, T = Σxlₓmₓ/R₀; a Leslie matrix projects n(t + 1) = Ln(t) towards a stable age distribution growing at λ.
  • Levins: p* = 1 − e/c; Lincoln–Petersen: N̂ = Mn/m for a closed population with random mixing of marks.
  • Lotka–Volterra: stable coexistence needs α < K₁/K₂ and β < K₂/K₁; predator–prey equilibria are N* = m/(ba) and P* = r/a, with neutral cycles.

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. In Connell’s experiment, Chthamalus barnacles spread into the lower intertidal zone once Balanus was removed. The zone Chthamalus occupied before the removal is best described as its

    1. realised niche
    2. fundamental niche
    3. habitat niche in Grinnell’s sense only
    4. Eltonian trophic niche
    Show answer

    Answer: A — realised niche

    The zone occupied in the presence of the competitor is the realised niche; the larger zone it can occupy without Balanus is the fundamental niche. The experiment is the standard demonstration that competition shrinks a fundamental niche into a realised one.
  2. The respiration rate of an ectotherm is 2.0 units at 15 °C and 5.0 units at 25 °C. Its Q₁₀, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.5

    Q₁₀ = (R₂/R₁)10/(T₂ − T₁) = (5.0/2.0)10/10 = 2.5. The interval is exactly 10 °C, so the exponent is 1; for a 5 °C interval the ratio would have to be squared.
  3. A bacterial population grows exponentially with r = 0.05 per hour. Its doubling time in hours, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 13.86

    t_d = ln 2/r = 0.6931/0.05 = 13.86 hours. Answering 20 (1/r) or 10 (0.5/r) mistakes the mean waiting time or half of 1/r for the doubling time.
  4. A population with discrete annual breeding has a finite rate of increase λ = 1.2 per year. Its intrinsic rate of increase r per year, to three decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.182

    r = ln λ = ln 1.2 = 0.182 per year. Reading λ = 1.2 as r = 0.2 confuses the finite (discrete) rate with the instantaneous rate; they agree only when r is small.
  5. A population grows logistically with r = 0.5 per year and K = 1000, starting from N₀ = 100. Its size after 2 years, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 232

    N(t) = K/[1 + ((K − N₀)/N₀)e−rt] = 1000/(1 + 9e−1) = 1000/(1 + 3.311) = 1000/4.311 = 232. The exponential answer 100e = 272 ignores the brake; 1000/(1 + 9e−0.5) = 155 uses rt = 0.5 by forgetting t.
  6. A harvested fish stock follows logistic growth with r = 0.4 per year and K = 1000 tonnes. The maximum rate of growth of the stock, in tonnes per year, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 100

    dN/dt = rN(1 − N/K) peaks where its derivative r − 2rN/K = 0, at N = K/2 = 500, giving rK/4 = 0.4 × 1000/4 = 100 tonnes per year. rK = 400 would be the growth rate if the brake did not exist at N = K.
  7. In a pond, 120 fish are caught, marked and released. A week later 150 fish are caught, of which 30 carry marks. The Lincoln–Petersen estimate of the population is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 600

    N̂ = Mn/m = 120 × 150/30 = 600. The proportion marked in the second sample, 30/150 = 0.2, estimates M/N, so N = 120/0.2. Dividing the other way, 150 × 30/120 = 37.5, is the common inversion.
  8. In a cohort, 500 individuals are alive at the start of age class 1 and 200 at the start of age class 2. The age-specific mortality rate q₁, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.6

    q₁ = d₁/n₁ = (500 − 200)/500 = 300/500 = 0.6. Dividing the deaths by the original cohort of 1000 gives 0.3, which is d₁/n₀ — the fraction of the whole cohort dying at age 1, not the risk faced by those alive at age 1.
  9. A cohort life table has lₓ = 1, 0.5, 0.2 and 0.05 at ages 0, 1, 2 and 3, with mₓ = 0, 2, 3 and 2 respectively, and no survivors at age 4. The generation time T, in years to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.47

    R₀ = Σlₓmₓ = 0 + 1.0 + 0.6 + 0.1 = 1.7; Σxlₓmₓ = 1.0 + 1.2 + 0.3 = 2.5; T = 2.5/1.7 = 1.47 years. Stopping at R₀ = 1.7 answers a different question, and dividing 2.5 by the number of age classes gives 0.63.
  10. A cohort of 1000 has 500, 200, 50 and 0 survivors at the start of ages 1, 2, 3 and 4. Using Lₓ = (nₓ + nₓ₊₁)/2, the life expectancy at birth e₀, in years to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.25

    L₀ = 750, L₁ = 350, L₂ = 125, L₃ = 25, so T₀ = 1250 and e₀ = T₀/n₀ = 1250/1000 = 1.25 years. Summing nₓ instead of Lₓ gives 1750/1000 = 1.75, which counts every survivor as living the whole interval.
  11. A population has the Leslie matrix L = [[0, 2, 1.5], [0.5, 0, 0], [0, 0.4, 0]] and initial age vector n(0) = (100, 50, 20). The total population after two time steps is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 215

    n(1) = (2 × 50 + 1.5 × 20, 0.5 × 100, 0.4 × 50) = (130, 50, 20). n(2) = (2 × 50 + 1.5 × 20, 0.5 × 130, 0.4 × 50) = (130, 65, 20), total 215. Stopping after one step gives 200.
  12. An oyster releases millions of eggs; almost all larvae die, but the few that settle survive well for years. Its survivorship curve is

    1. Type III, concave
    2. Type I, convex
    3. Type II, a straight line on a log scale
    4. a J-shaped growth curve
    Show answer

    Answer: A — Type III, concave

    Very high mortality early and low mortality among survivors is the concave Type III curve. Type I describes low early mortality concentrated in old age, and Type II a constant risk at every age. A J-shaped curve is a growth curve, not a survivorship curve.
  13. In Levins’ metapopulation model, the colonisation rate is c = 0.5 per year and the local extinction rate is e = 0.2 per year. The equilibrium fraction of occupied patches, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.6

    Setting dp/dt = cp(1 − p) − ep = 0 for p > 0 gives p* = 1 − e/c = 1 − 0.2/0.5 = 0.6. The ratio e/c = 0.4 is the fraction left empty, a common confusion.
  14. Two species compete according to the Lotka–Volterra model with K₁ = 500, K₂ = 400, α = 0.5 (effect of species 2 on 1) and β = 0.6 (effect of 1 on 2). The equilibrium density of species 1, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 429

    N₁* = (K₁ − αK₂)/(1 − αβ) = (500 − 0.5 × 400)/(1 − 0.3) = 300/0.7 = 428.6, so 429. Coexistence is stable because α < K₁/K₂ = 1.25 and β < K₂/K₁ = 0.8. Forgetting the denominator gives 300.
  15. In the Lotka–Volterra predator–prey model dN/dt = rN − aNP, dP/dt = baNP − mP, with r = 0.6, a = 0.01, b = 0.1 and m = 0.2, the equilibrium predator density P* is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 60

    The prey equation is zero when r − aP = 0, so P* = r/a = 0.6/0.01 = 60. The prey equilibrium is N* = m/(ba) = 0.2/0.001 = 200; giving 200 answers for the wrong species.
  16. In the Lotka–Volterra competition model, which statements are correct?

    1. Stable coexistence requires each species to limit itself more than it limits the other
    2. The zero-growth isoclines are straight lines in the N₁–N₂ plane
    3. When α > K₁/K₂ and β > K₂/K₁, the outcome depends on the starting densities
    4. The species with the larger r always excludes the other
    Show answer

    Answer: A — Stable coexistence requires each species to limit itself more than it limits the other; B — The zero-growth isoclines are straight lines in the N₁–N₂ plane; C — When α > K₁/K₂ and β > K₂/K₁, the outcome depends on the starting densities

    Coexistence is stable when α < K₁/K₂ and β < K₂/K₁, meaning intraspecific competition exceeds interspecific. The isoclines N₁ = K₁ − αN₂ and N₂ = K₂ − βN₁ are straight. When both inequalities are reversed the interior equilibrium is a saddle and the winner depends on starting numbers. r sets how fast the outcome is reached, not which outcome occurs.
  17. Which of the following act as density-dependent factors on a population?

    1. Transmission of a contagious disease
    2. Competition for a limited number of nest holes
    3. A sudden hard frost that kills a fixed fraction of individuals
    4. A predator that switches to whichever prey is commonest
    Show answer

    Answer: A — Transmission of a contagious disease; B — Competition for a limited number of nest holes; D — A predator that switches to whichever prey is commonest

    Disease transmission rises with contact rate, competition for nest holes intensifies as they fill, and a switching predator takes a larger share of a prey as it becomes common — all change per-capita rates with density. A frost killing a fixed fraction regardless of density is density-independent.
  18. Two prey species never compete for food, yet an increase in one is followed by a decline in the other. Both are eaten by the same generalist predator, whose numbers track total prey. This is

    1. apparent competition
    2. exploitation competition
    3. a trophic cascade
    4. amensalism
    Show answer

    Answer: A — apparent competition

    A shared predator makes each prey harmful to the other through the predator’s numerical response — apparent competition, an indirect −/− effect with no shared resource. Exploitation competition needs a shared limiting resource; a trophic cascade runs down the food chain from predator to plant; amensalism is a direct 0/− effect.