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Ecology

1. Species, Communities, and Ecosystems

Species and Populations

Definition. A species is a group of organisms that can interbreed and produce fertile offspring under natural conditions (biological species concept).

Definition. A population is a group of organisms of the same species living in the same area at the same time.

Definition. A community is all the populations of different species living and interacting in a particular area.

Definition. An ecosystem is a community of living organisms (biotic factors) interacting with the non-living environment (abiotic factors: temperature, water, light, soil pH, mineral availability).

Autotrophs and Heterotrophs

  • Autotrophs (producers): organisms that synthesise organic compounds from inorganic sources. Photoautotrophs (plants, algae, cyanobacteria) use light energy; chemoautotrophs (some bacteria) use chemical energy.
  • Heterotrophs (consumers): organisms that obtain organic molecules by consuming other organisms. Includes herbivores, carnivores, omnivores, and decomposers.
  • Decomposers (saprotrophs): bacteria and fungi that break down dead organic matter, releasing inorganic nutrients back into the environment (mineralisation).

Trophic Levels and Food Chains

A food chain represents the linear transfer of energy from producers through consumers.

ProducerPrimaryConsumerSecondaryConsumerTertiaryConsumer\mathrm{Producer} \to \mathrm{Primary Consumer} \to \mathrm{Secondary Consumer} \to \mathrm{Tertiary Consumer}

A food web is a network of interconnected food chains, representing the complex feeding relationships in an ecosystem.

Ecological Niches

Definition. A niche is the role and position a species has in its environment: how it meets its needs for food and shelter, how it survives, and how it reproduces. It includes all biotic and abiotic interactions.

The competitive exclusion principle (Gause's principle) states that two species cannot coexist permanently in the same niche: one will outcompete the other. Species that coexist must occupy different niches (resource partitioning).

Fundamental niche: the full range of conditions a species can tolerate. Realised niche: the actual range occupied, limited by competition and other biotic factors.


2. Energy Flow

Laws of Thermodynamics Applied to Ecosystems

  1. First law: energy cannot be created or destroyed, only transformed. Solar energy is converted to chemical energy in photosynthesis.
  2. Second law: in every energy transfer, some energy is lost as heat (unavailable for further work). This limits the number of trophic levels.

Gross and Net Primary Productivity

Definition. Gross primary productivity (GPP) is the total amount of chemical energy fixed by photosynthesis per unit area per unit time.

Definition. Net primary productivity (NPP) is the energy remaining after subtracting the energy used in plant respiration:

NPP=GPPR\mathrm{NPP} = \mathrm{GPP} - R

where RR is the energy lost through respiration by the producers. NPP represents the energy available to herbivores (primary consumers).

Ecological Efficiency

Energy transfer between trophic levels is typically only 10%10\%--20%20\%. The remaining energy is lost as:

  • Heat from metabolic processes (respiration).
  • Undigested material in faeces.
  • Excretory products (urea).
  • Incomplete consumption (not all biomass at one level is eaten).

Pyramids of energy always have an upright shape because energy is lost at each transfer. Pyramids of numbers and biomass can be inverted (e.g., a single tree supporting many herbivorous insects).

Calculating Energy Transfer

Ecologicalefficiency=energyathighertrophiclevelenergyatlowertrophiclevel×100%\mathrm{Ecological efficiency} = \frac{\mathrm{energy at higher trophic level}}{\mathrm{energy at lower trophic level}} \times 100\%

3. Nutrient Cycles

The Carbon Cycle

Carbon circulates between the atmosphere, biosphere, hydrosphere, and lithosphere.

Key processes:

ProcessDescription
PhotosynthesisPlants, algae, and cyanobacteria convert CO2\mathrm{CO}_2 into organic compounds (carbohydrates).
Cellular respirationOrganisms release CO2\mathrm{CO}_2 by oxidising organic molecules.
CombustionBurning fossil fuels and biomass releases CO2\mathrm{CO}_2.
DecompositionDecomposers break down organic matter, releasing CO2\mathrm{CO}_2.
FossilisationOrganic matter is buried and slowly converted to fossil fuels (coal, oil, natural gas) over millions of years.
Ocean absorptionCO2\mathrm{CO}_2 dissolves in seawater; marine organisms incorporate carbon into shells (CaCO3\mathrm{CaCO}_3) and tissues.

Carbon reservoirs:

  • Atmosphere: 800\approx 800 gigatonnes of carbon (Gt C), primarily as CO2\mathrm{CO}_2.
  • Oceans: 38000\approx 38000 Gt C (dissolved CO2\mathrm{CO}_2, bicarbonate, carbonate, marine organisms).
  • Fossil fuels: 4000\approx 4000--1000010000 Gt C.
  • Terrestrial biosphere: 2000\approx 2000 Gt C (plant biomass, soil organic matter).

The Nitrogen Cycle

Nitrogen (N2\mathrm{N}_2) makes up 78%78\% of the atmosphere but is chemically inert and unavailable to most organisms. The nitrogen cycle converts N2\mathrm{N}_2 into biologically usable forms.

ProcessDescription
Nitrogen fixationConversion of N2\mathrm{N}_2 to NH3\mathrm{NH}_3 (ammonia). Performed by nitrogen-fixing bacteria (e.g., Rhizobium in root nodules of legumes; Azotobacter free-living in soil) and lightning.
NitrificationConversion of NH3\mathrm{NH}_3 / NH4+\mathrm{NH}_4^+ to NO2\mathrm{NO}_2^- (nitrite) by Nitrosomonas, then to NO3\mathrm{NO}_3^- (nitrate) by Nitrobacter.
AssimilationPlants absorb NO3\mathrm{NO}_3^- and NH4+\mathrm{NH}_4^+ through roots; animals obtain nitrogen by eating plants or other animals.
AmmonificationDecomposers convert organic nitrogen (proteins, nucleic acids) in dead matter and waste into NH3\mathrm{NH}_3.
DenitrificationConversion of NO3\mathrm{NO}_3^- back to N2\mathrm{N}_2 by denitrifying bacteria (e.g., Pseudomonas denitrificans), returning nitrogen to the atmosphere. Occurs in anaerobic conditions (waterlogged soils).

4. Climate Change

The Greenhouse Effect

Solar radiation reaches Earth's surface as short-wave radiation. Earth re-emits energy as long-wave infrared radiation. Greenhouse gases in the atmosphere absorb and re-emit this infrared radiation, warming the planet.

Key greenhouse gases:

GasSourcesRelative contribution
Carbon dioxide (CO2\mathrm{CO}_2)Fossil fuel combustion, deforestation, cement productionLargest
Methane (CH4\mathrm{CH}_4)Agriculture (rice paddies, livestock), landfills, natural gas leaksSecond largest
Nitrous oxide (N2O\mathrm{N}_2\mathrm{O})Agricultural fertilisers, industrial processesThird
Chlorofluorocarbons (CFCs)Refrigerants, aerosols (now largely banned by Montreal Protocol)Potent but declining

Evidence for Climate Change

  • Direct atmospheric measurements: CO2\mathrm{CO}_2 concentration at Mauna Loa Observatory has risen from 315  ppm\approx 315\;\mathrm{ppm} (1958) to over 420  ppm420\;\mathrm{ppm} (2024).
  • Ice core data: trapped air bubbles in Antarctic ice show CO2\mathrm{CO}_2 and temperature over the past 800000800000 years; current levels are unprecedented.
  • Global temperature records: average global temperature has risen by 1.1C\approx 1.1^\circ\mathrm{C} since pre-industrial times.
  • Sea-level rise: thermal expansion of seawater and melting of ice sheets contribute approximately 3.6  mm/year3.6\;\mathrm{mm/year}.
  • Ocean acidification: dissolved CO2\mathrm{CO}_2 forms carbonic acid, decreasing ocean pH (from 8.2\approx 8.2 to 8.1\approx 8.1 since pre-industrial times), threatening calcifying organisms (corals, shellfish).

Consequences

  • More frequent and severe extreme weather events (heatwaves, storms, droughts, flooding).
  • Range shifts and extinction of species unable to adapt or migrate.
  • Disruption of food webs and agricultural productivity.
  • Loss of polar ice habitats.
  • Spread of tropical diseases to higher latitudes.

5. Evolution

Evidence for Evolution

Evidence TypeDescription
Fossil recordFossils show gradual changes in organisms over geological time; transitional forms (e.g., Archaeopteryx between dinosaurs and birds).
Comparative anatomyHomologous structures: similar anatomy due to common ancestry (e.g., pentadactyl limb in vertebrates). Analogous structures: similar function but different origin (e.g., wings of birds vs insects).
Comparative embryologyEarly embryonic stages of vertebrates are remarkably similar (e.g., pharyngeal pouches, post-anal tail).
Molecular evidenceDNA and protein sequence comparisons; closely related species share more DNA sequences (e.g., humans and chimpanzees share 98.8%\approx 98.8\% of DNA).
BiogeographySpecies on islands resemble those on the nearest mainland (e.g., Darwin's finches on the Galapagos).

Natural Selection

Darwin's theory of natural selection (1859):

  1. Variation: individuals within a population show phenotypic variation, much of which is heritable (genetic).
  2. Competition: resources are limited; organisms compete for survival and reproduction.
  3. Differential survival and reproduction: individuals with advantageous traits (adaptations) are more likely to survive and reproduce, passing on their alleles to the next generation.
  4. Change in allele frequency: over generations, the frequency of advantageous alleles increases in the population. This is evolution.

Definition. Evolution is the change in allele frequency in a population over time.

Types of Selection

TypeEffect on distribution
Directional selectionFavours one extreme phenotype; the distribution shifts in that direction (e.g., antibiotic resistance in bacteria).
Stabilising selectionFavours the intermediate phenotype; reduces variation (e.g., human birth weight).
Disruptive selectionFavours both extremes over the intermediate; can lead to speciation (e.g., beak size in finches on islands with only large and small seeds).

Speciation

Definition. Speciation is the formation of new species through the evolution of reproductive isolation.

  • Allopatric speciation: populations are geographically separated; different selection pressures and genetic drift lead to reproductive isolation (most common mechanism).
  • Sympatric speciation: new species arise without geographic separation (e.g., polyploidy in plants, which creates instant reproductive isolation due to incompatible chromosome numbers).

Reproductive isolating mechanisms:

  • Pre-zygotic: temporal isolation (different breeding seasons), behavioural isolation (different courtship rituals), mechanical isolation (incompatible reproductive structures), ecological isolation (different habitats).
  • Post-zygotic: hybrid inviability (offspring do not survive), hybrid sterility (offspring are sterile, e.g., mules), hybrid breakdown (second-generation hybrids have reduced fitness).

Hardy-Weinberg Equilibrium

The Hardy-Weinberg principle describes a theoretical population in which allele frequencies remain constant from generation to generation (no evolution).

Conditions: no mutations, no migration, random mating, infinite population size (no genetic drift), no natural selection.

For a gene with two alleles (AA and aa) with frequencies pp and qq:

p+q=1p + q = 1 p2+2pq+q2=1p^2 + 2pq + q^2 = 1

where p2p^2 = frequency of AAAA, 2pq2pq = frequency of AaAa, q2q^2 = frequency of aaaa.

Deviations from Hardy-Weinberg proportions indicate that one or more evolutionary forces are acting on the population.


6. Classification

Taxonomy

The hierarchical system of classification (Linnaean system):

DomainKingdomPhylumClassOrderFamilyGenusSpecies\mathrm{Domain} \to \mathrm{Kingdom} \to \mathrm{Phylum} \to \mathrm{Class} \to \mathrm{Order} \to \mathrm{Family} \to \mathrm{Genus} \to \mathrm{Species}

Binomial nomenclature: each species is given a two-part Latin name: Genus species (e.g., Homo sapiens).

Three Domains

DomainCell typeCell wallGenetic materialExample
BacteriaProkaryoticPeptidoglycanCircular DNAE. coli
ArchaeaProkaryoticPseudopeptidoglycanCircular DNAHalobacterium
EukaryaEukaryoticVaries (cellulose, chitin, or absent)Linear chromosomes in nucleusAnimals, plants, fungi, protists

Dichotomous Keys

A dichotomous key is a tool for identifying organisms through a series of paired (couplet) statements, each leading to the next step or to an identification. At each step, the user chooses one of two alternatives based on observable characteristics.


Common Pitfalls

  • Confusing niche and habitat: habitat is the physical place; niche is the organism's role within that place.
  • Stating that energy is "recycled" in ecosystems: energy flows through and is ultimately lost as heat; only nutrients are recycled.
  • Assuming the greenhouse effect is entirely negative: the natural greenhouse effect keeps Earth habitable (33C\approx 33^\circ\mathrm{C} warmer than it would otherwise be); the concern is the enhanced greenhouse effect due to anthropogenic emissions.
  • Equating evolution with "progress": evolution is a change in allele frequency; it has no direction or goal.
  • Misapplying Hardy-Weinberg: the equilibrium only holds under ideal conditions never met in nature; it serves as a null model to detect evolutionary forces.

Practice Problems

Question 1: Energy Pyramid Calculation

A grassland ecosystem receives 10000  kJ/m2/year10000\;\mathrm{kJ/m^2/year} of solar energy. If GPP is 2000  kJ/m2/year2000\;\mathrm{kJ/m^2/year} and plant respiration accounts for 1200  kJ/m2/year1200\;\mathrm{kJ/m^2/year}, calculate the NPP. If ecological efficiency is 10%10\%, how much energy is available to secondary consumers?

Answer

NPP=GPPR=20001200=800  kJ/m2/year\mathrm{NPP} = \mathrm{GPP} - R = 2000 - 1200 = 800\;\mathrm{kJ/m^2/year}.

Energy available to primary consumers: 800  kJ/m2/year800\;\mathrm{kJ/m^2/year}.

Energy transferred to secondary consumers (10%10\% efficiency): 800×0.10=80  kJ/m2/year800 \times 0.10 = 80\;\mathrm{kJ/m^2/year}.

Question 2: Hardy-Weinberg Equilibrium

In a population of wildflowers, the allele for red flowers (RR) is dominant over white (rr). If 16%16\% of the population has white flowers, calculate the frequency of the RR allele, the frequency of heterozygous individuals, and the percentage of the population that is homozygous dominant.

Answer

White flowers are homozygous recessive (rrrr): q2=0.16q^2 = 0.16, so q=0.4q = 0.4.

p=1q=10.4=0.6p = 1 - q = 1 - 0.4 = 0.6.

Frequency of heterozygotes (RrRr): 2pq=2×0.6×0.4=0.482pq = 2 \times 0.6 \times 0.4 = 0.48 (48%48\%).

Frequency of homozygous dominant (RRRR): p2=0.62=0.36p^2 = 0.6^2 = 0.36 (36%36\%).

Check: 0.36+0.48+0.16=1.000.36 + 0.48 + 0.16 = 1.00.

Question 3: Natural Selection and Antibiotic Resistance

A population of 1000010000 bacteria is treated with an antibiotic that kills 99%99\% of them. The surviving 100100 bacteria carry a resistance gene. After 55 generations (binary fission), the population recovers to 32003200. Explain this in terms of natural selection.

Answer

The initial population contained genetic variation: a small fraction carried the resistance allele. The antibiotic acted as a strong selective pressure, killing susceptible bacteria (differential survival). The resistant bacteria survived and reproduced, passing the resistance allele to all offspring (heritability). After 55 generations of binary fission (25=322^5 = 32-fold increase from 100100), the population recovered to 32003200, all carrying the resistance gene. This demonstrates natural selection: the allele frequency of the resistance gene increased from 1%\approx 1\% to 100%\approx 100\% in five generations.

Question 4: Carbon Cycle Calculation

An area of tropical rainforest has an NPP of 2200  g  C/m2/year2200\;\mathrm{g\;C/m^2/year}. If deforestation removes 50%50\% of the trees, and assuming NPP is proportional to biomass, estimate the reduction in annual carbon uptake. Express your answer in tonnes of CO2\mathrm{CO}_2 per km2\mathrm{km}^2 per year.

Answer

Original NPP: 2200  g  C/m2/year2200\;\mathrm{g\;C/m^2/year}.

After 50%50\% deforestation: 2200×0.50=1100  g  C/m2/year2200 \times 0.50 = 1100\;\mathrm{g\;C/m^2/year}.

Reduction: 1100  g  C/m2/year1100\;\mathrm{g\;C/m^2/year}.

Convert to CO2\mathrm{CO}_2: CO2\mathrm{CO}_2 has molar mass 44  g/mol44\;\mathrm{g/mol}, C has molar mass 12  g/mol12\;\mathrm{g/mol}. Ratio: 44/12=3.6744/12 = 3.67.

Reduction in CO2\mathrm{CO}_2 uptake: 1100×3.67=4037  g  CO2/m2/year1100 \times 3.67 = 4037\;\mathrm{g\;CO_2/m^2/year}.

Convert to km2\mathrm{km}^2: 1  km2=106  m21\;\mathrm{km}^2 = 10^6\;\mathrm{m}^2.

Reduction: 4037×106  g/km2/year=4037  t  CO2/km2/year4037 \times 10^6\;\mathrm{g/km^2/year} = 4037\;\mathrm{t\;CO_2/km^2/year}.

Question 5: Classification Using a Dichotomous Key

An organism is unicellular, eukaryotic, photosynthetic, and has cellulose cell walls. Using the three-domain system and additional classification, determine its most likely kingdom and explain your reasoning.

Answer

The organism is eukaryotic (has a membrane-bound nucleus), placing it in Domain Eukarya.

Within Eukarya, the kingdoms are:

  • Animalia: multicellular, heterotrophic, no cell walls --- excluded.
  • Plantae: multicellular, autotrophic, cellulose cell walls --- excluded (the organism is unicellular).
  • Fungi: heterotrophic, chitin cell walls --- excluded (the organism is photosynthetic and has cellulose walls).
  • Protista: diverse kingdom containing unicellular eukaryotes. Photosynthetic protists with cellulose cell walls include algae (e.g., Chlorella, Chlamydomonas).

The organism most likely belongs to Kingdom Protista, specifically a photosynthetic alga. Note that some classification systems place algae within Plantae, but under the traditional IB convention, unicellular photosynthetic eukaryotes are classified as protists.


Worked Examples

Worked Example: Hardy-Weinberg Equilibrium with Selection

In a population of 50005000 peppered moths, the allele for dark colouration (DD) is dominant over the light allele (dd). The current genotype frequencies are: DD=0.36DD = 0.36, Dd=0.48Dd = 0.48, dd=0.16dd = 0.16. Industrial pollution darkens tree bark, giving dark moths a survival advantage. The relative fitness values are: wDD=1.0w_{DD} = 1.0, wDd=1.0w_{Dd} = 1.0, wdd=0.5w_{dd} = 0.5. Calculate the allele frequencies after one generation of selection.

Solution

Step 1: Initial allele frequencies. p=0.36+12(0.48)=0.36+0.24=0.60p = 0.36 + \frac{1}{2}(0.48) = 0.36 + 0.24 = 0.60 q=0.16+12(0.48)=0.16+0.24=0.40q = 0.16 + \frac{1}{2}(0.48) = 0.16 + 0.24 = 0.40

Step 2: Mean fitness of the population. wˉ=p2wDD+2pqwDd+q2wdd=(0.36)(1.0)+(0.48)(1.0)+(0.16)(0.5)=0.36+0.48+0.08=0.92\bar{w} = p^2 w_{DD} + 2pq \cdot w_{Dd} + q^2 w_{dd} = (0.36)(1.0) + (0.48)(1.0) + (0.16)(0.5) = 0.36 + 0.48 + 0.08 = 0.92

Step 3: New genotype frequencies after selection. f(DD)=p2wDDwˉ=0.360.920.391f(DD) = \frac{p^2 w_{DD}}{\bar{w}} = \frac{0.36}{0.92} \approx 0.391 f(Dd)=2pqwDdwˉ=0.480.920.522f(Dd) = \frac{2pq \cdot w_{Dd}}{\bar{w}} = \frac{0.48}{0.92} \approx 0.522 f(dd)=q2wddwˉ=0.080.920.087f(dd) = \frac{q^2 w_{dd}}{\bar{w}} = \frac{0.08}{0.92} \approx 0.087

Step 4: New allele frequencies. p=0.391+12(0.522)=0.391+0.261=0.652p' = 0.391 + \frac{1}{2}(0.522) = 0.391 + 0.261 = 0.652 q=0.087+12(0.522)=0.087+0.261=0.348q' = 0.087 + \frac{1}{2}(0.522) = 0.087 + 0.261 = 0.348

The frequency of the dark allele (DD) increased from 0.600.60 to 0.6520.652 in one generation due to directional selection favouring the dark phenotype. The light allele (dd) is declining but has not been eliminated because heterozygotes also carry and protect it.

Worked Example: Chi-Squared Test for Genetic Equilibrium

A population of 10001000 flowering plants is sampled for flower colour. The observed numbers are: 710710 red, 260260 pink, 3030 white. Under incomplete dominance, the expected Mendelian ratio from a cross of two heterozygotes (Rr×RrRr \times Rr) is 1:2:11:2:1. Use the chi-squared test at p=0.05p = 0.05 (critical value =5.99= 5.99 for 22 degrees of freedom) to determine whether the population is in Hardy-Weinberg equilibrium.

Solution

Expected values (total =1000= 1000; ratio 1:2:11:2:1): Red (RRRR): 250250, Pink (RrRr): 500500, White (rrrr): 250250

Chi-squared calculation: χ2=(710250)2250+(260500)2500+(30250)2250\chi^2 = \frac{(710 - 250)^2}{250} + \frac{(260 - 500)^2}{500} + \frac{(30 - 250)^2}{250} =4602250+(240)2500+(220)2250= \frac{460^2}{250} + \frac{(-240)^2}{500} + \frac{(-220)^2}{250} =211600250+57600500+48400250= \frac{211600}{250} + \frac{57600}{500} + \frac{48400}{250} =846.4+115.2+193.6=1155.2= 846.4 + 115.2 + 193.6 = 1155.2

Conclusion: χ2=1155.2\chi^2 = 1155.2 far exceeds the critical value of 5.995.99. We reject the null hypothesis. The population is not in Hardy-Weinberg equilibrium. The large excess of red flowers and deficiency of white flowers suggest strong selection against the white phenotype (e.g., herbivores preferentially consuming white flowers, or pollinator preference for red).

Worked Example: Calculating Ecological Footprint from NPP Data

A 10  km210\;\mathrm{km}^2 nature reserve has an average NPP of 1500  g  C/m2/year1500\;\mathrm{g\;C/m^2/year}. The reserve supports a population of 200200 deer, each consuming an average of 3.5  kg3.5\;\mathrm{kg} of plant biomass per day (dry weight, assume 50%50\% carbon by mass). Calculate the fraction of annual NPP consumed by the deer population.

Solution

Total annual NPP: 1500  g  C/m2/year×107  m2=1.5×1010  g  C/year=15000  t  C/year1500\;\mathrm{g\;C/m^2/year} \times 10^7\;\mathrm{m^2} = 1.5 \times 10^{10}\;\mathrm{g\;C/year} = 15000\;\mathrm{t\;C/year}

Total annual deer consumption (in carbon): 200×3.5  kg/day×365  days/year×0.50  (Cfraction)200 \times 3.5\;\mathrm{kg/day} \times 365\;\mathrm{days/year} \times 0.50\;\mathrm{(C fraction)} =200×3.5×365×0.5=127750  kg  C/year=127.75  t  C/year= 200 \times 3.5 \times 365 \times 0.5 = 127750\;\mathrm{kg\;C/year} = 127.75\;\mathrm{t\;C/year}

Fraction of NPP consumed: 127.75150000.0085\frac{127.75}{15000} \approx 0.0085 or 0.85%0.85\%

The deer population consumes less than 1%1\% of annual NPP, suggesting the ecosystem can support considerably more herbivory before NPP becomes limiting. However, this calculation does not account for other herbivores, decomposition losses, or the fact that only a portion of NPP is palatable.

Worked Example: Interpreting a Predator-Prey Graph

The following data show the population sizes of a predator (lynx) and its prey (hare) over a 1010-year period in a 500  km2500\;\mathrm{km}^2 boreal forest:

YearHare population (thousands)Lynx population
14025
26235
38555
46065
53045
61820
72212
84515
97840
109060

Describe the relationship between the two populations and explain the likely mechanism causing the oscillations.

Solution

The populations show cyclic oscillations with the predator population peaking slightly after the prey population. This lag is the signature feature of a Lotka-Volterra predator-prey dynamic.

Mechanism:

  1. When hare numbers are high (Year 1--3), food is abundant for lynx, which reproduce successfully. Lynx numbers increase.
  2. Increased lynx predation drives the hare population down (Year 3--6).
  3. With fewer hares, lynx face food scarcity; starvation and reduced reproduction cause lynx numbers to decline (Year 4--7).
  4. Reduced predation pressure allows the hare population to recover (Year 6--8), and the cycle repeats.

The period of oscillation is approximately 88--99 years, consistent with the classic Hudson's Bay Company fur trapping records for Canadian lynx and snowshoe hares. Additional factors such as hare density-dependent food limitation (they eat themselves out of food) also contribute to the hare cycles independently of predation.


Common Pitfalls (Expanded)

  • Confusing niche and habitat: habitat is the physical place where an organism lives; niche is the organism's role within that place (how it obtains resources, interacts with other species, and contributes to energy flow).
  • Stating that energy is "recycled" in ecosystems: energy flows through and is ultimately lost as heat at every trophic transfer; only nutrients (carbon, nitrogen, phosphorus) are recycled via biogeochemical cycles.
  • Assuming the greenhouse effect is entirely negative: the natural greenhouse effect keeps Earth approximately 33C33^\circ\mathrm{C} warmer than it would otherwise be, making it habitable. The concern is the enhanced greenhouse effect from anthropogenic emissions.
  • Equating evolution with "progress": evolution is a change in allele frequency with no direction or goal. A trait is "better" only in the context of the current environment.
  • Misapplying Hardy-Weinberg: the equilibrium only holds under ideal conditions (no mutation, no migration, random mating, infinite population, no selection). It serves as a null model to detect evolutionary forces, not as a description of real populations.
  • Confusing GPP and NPP: GPP is total photosynthetic output; NPP is what remains after plant respiration. NPP is the energy available to consumers.
  • Assuming all nitrogen is available to plants: atmospheric N2\mathrm{N}_2 is inert and unavailable. Plants can only use fixed forms: NH4+\mathrm{NH}_4^+, NO2\mathrm{NO}_2^-, and NO3\mathrm{NO}_3^-.

Exam-Style Problems

Problem 1: Data Analysis -- Carbon Cycle and Deforestation

A tropical rainforest has an average above-ground biomass of 200  t  C/ha200\;\mathrm{t\;C/ha} and an NPP of 1800  g  C/m2/year1800\;\mathrm{g\;C/m^2/year}. A logging company clears 500  ha500\;\mathrm{ha} of forest. Assume 60%60\% of the above-ground carbon is released as CO2\mathrm{CO}_2 immediately (through burning and decomposition), and the remaining 40%40\% is stored in timber products. (a) Calculate the immediate CO2\mathrm{CO}_2 release in tonnes. (b) Calculate the annual loss of CO2\mathrm{CO}_2 uptake capacity. (c) Discuss two long-term consequences for the local carbon cycle.

Problem 2: Extended Response -- Speciation Mechanisms

A river changes course over geological time, splitting a population of 5000050000 beetles into two isolated populations. The western population lives in a dry grassland; the eastern population inhabits a moist forest. Over 5000050000 generations, the populations diverge. Describe the likely mechanisms (natural selection, genetic drift, mutation) driving divergence in each habitat. Explain what additional evidence would be needed to confirm that the two populations are now separate species.

Problem 3: Quantitative -- Population Growth

A population of bacteria follows exponential growth with a per capita growth rate of r=0.8r = 0.8 per hour. Starting from an initial population of 500500 cells: (a) Calculate the population size after 66 hours. (b) Calculate the doubling time. (c) Explain why exponential growth cannot continue indefinitely and describe the shape of the resulting logistic growth curve.

Problem 4: Data Analysis -- Nitrogen Cycle and Eutrophication

A lake receives agricultural runoff containing 50  tonnes/year50\;\mathrm{tonnes/year} of nitrogen (as nitrate). The lake has a surface area of 10  km210\;\mathrm{km}^2 and an average depth of 5  m5\;\mathrm{m}. The natural nitrogen input from rainfall is 0.5  tonnes/year0.5\;\mathrm{tonnes/year}. (a) Calculate the nitrogen loading per unit area with and without agricultural runoff. (b) Explain the sequence of ecological events (algal bloom, oxygen depletion, fish kill) that would likely follow. (c) Describe two management strategies to reduce nitrogen loading.

Problem 5: Extended Response -- Evidence for Evolution

Using specific named examples, evaluate the relative strength of evidence from the fossil record, comparative anatomy, molecular biology, and biogeography in supporting the theory of evolution by natural selection. Discuss limitations of each evidence type and explain how multiple lines of evidence provide a more robust argument than any single type alone.

Problem 6: Quantitative -- Hardy-Weinberg and Genetic Drift

A small island population of 5050 lizards has allele frequencies for a colour gene: p=0.80p = 0.80 (allele GG, green) and q=0.20q = 0.20 (allele gg, grey). A storm kills 4040 lizards randomly, leaving 1010 survivors. If all survivors are green, what are the new allele frequencies? Explain why genetic drift has a stronger effect in small populations, and calculate the probability that the gg allele is completely lost in this bottleneck event (assuming the 4040 deaths were truly random with respect to genotype).

Problem 7: Extended Response -- Climate Change Impacts

Describe the mechanism by which increasing atmospheric CO2\mathrm{CO}_2 causes ocean acidification. Explain the impact of decreasing ocean pH on marine organisms that build calcium carbonate shells (e.g., corals, pteropods, bivalves), including the relevant chemical equilibrium. Discuss one cascading ecological consequence for the broader marine food web.

Problem 8: Data Analysis -- Ecological Efficiency and Biomass

A grassland ecosystem has the following biomass data at each trophic level:

Trophic levelBiomass (kJ/m2\mathrm{kJ/m^2})
Producers (grass)24000
Primary consumers (insects)2400
Secondary consumers (frogs)240
Tertiary consumers (snakes)24

(a) Calculate the ecological efficiency between each pair of adjacent trophic levels. (b) Explain why the biomass pyramid is upright in this ecosystem but can be inverted in aquatic systems. (c) If the producer biomass decreases by 50%50\% due to drought, predict the impact on the snake population and explain your reasoning.


If You Get These Wrong, Revise:


7. Ecosystem Dynamics (Extended)

Population Growth Models

Exponential growth:

When resources are unlimited, populations grow exponentially:

dNdt=rN\frac{dN}{dt} = rN

Solution: Nt=N0ertN_t = N_0 e^{rt}

where N0N_0 is the initial population size, rr is the intrinsic rate of increase, and tt is time.

Doubling time: td=ln2rt_d = \frac{\ln 2}{r}

Exponential growth cannot continue indefinitely because resources are finite.

Logistic growth:

The logistic equation (Verhulst, 1838) incorporates a carrying capacity (KK):

dNdt=rN(1NK)\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)

Solution: Nt=K1+(KN0N0)ertN_t = \frac{K}{1 + \left(\frac{K - N_0}{N_0}\right) e^{-rt}}

  • When NKN \ll K: population grows nearly exponentially (dNdtrN\frac{dN}{dt} \approx rN).
  • When N=KN = K: population growth stops (dNdt=0\frac{dN}{dt} = 0).
  • When N>KN > K: population declines.

The logistic curve is sigmoidal (S-shaped).

Survivorship Curves

Survivorship curves plot the number or proportion of individuals surviving to each age:

TypeShapeDescriptionExample
IConvex (drops steeply only at old age)Low mortality during early and middle life; most individuals survive to old age.Humans in developed countries; large mammals.
IIStraight (constant mortality rate)Constant mortality rate at all ages.Many bird species; some reptiles.
IIIConcave (high early mortality)High mortality in early life; those that survive to adulthood have lower mortality.Fish; invertebrates; many plants (seedling mortality).

Succession

Primary succession: colonisation of bare, lifeless substrate (e.g., volcanic rock, sand dunes, glacial moraine).

Stages (sand dune example):

  1. Pioneer community: lichens and algae colonise bare rock/sand. Lichens secrete acids that weather rock, beginning soil formation.
  2. Grass and herb stage: grasses and herbs colonise the thin soil. Their roots stabilise the substrate and add organic matter.
  3. Shrub stage: shrubs replace grasses as soil deepens.
  4. Tree stage: fast-growing trees (e.g., birch, pine) establish.
  5. Climax community: slow-growing, shade-tolerant trees (e.g., oak, beech) outcompete pioneers. The ecosystem reaches a relatively stable equilibrium.

Secondary succession: recolonisation of a disturbed habitat where soil remains (e.g., after fire, deforestation, abandoned farmland). Proceeds faster than primary succession because soil and some organisms persist.

Keystone Species

A keystone species has a disproportionately large effect on its ecosystem relative to its abundance. Its removal causes significant changes in community structure.

Examples:

  • Sea otters (keystone predator): prey on sea urchins. Without otters, sea urchin populations explode, overgrazing kelp forests, destroying the habitat for many species.
  • Beavers (ecosystem engineer): dam-building creates wetland habitats, fundamentally altering hydrology and creating niches for many species.
  • Fig wasps (mutualist): pollinate fig trees; the fig-wasp mutualism is essential for the survival of both species and many tropical frugivores that depend on figs.

Ecological Niches and Competitive Exclusion

Fundamental niche: the full range of environmental conditions a species can theoretically occupy.

Realised niche: the actual range occupied, limited by biotic interactions (competition, predation, parasitism) and abiotic factors.

Competitive exclusion principle (Gause's principle): two species cannot coexist indefinitely if they occupy exactly the same niche; one will outcompete the other.

Resource partitioning: coexisting species divide the available resources (e.g., different feeding times, different food sizes, different vertical stratification) to reduce competition.


8. Biogeochemical Cycles (Extended)

The Phosphorus Cycle

Unlike carbon and nitrogen, phosphorus has no significant gaseous phase. The cycle is sedimentary:

  1. Weathering: phosphate minerals in rocks are slowly released by weathering (PO43\mathrm{PO}_4^{3-}).
  2. Uptake by plants: plants absorb phosphate from soil through roots.
  3. Transfer through food chains: phosphorus passes from producers to consumers and decomposers.
  4. Return to soil: decomposition and excretion return phosphorus to the soil.
  5. Sedimentation: eroded phosphorus is transported to oceans and incorporated into marine sediments. Over geological time, these sediments are uplifted, completing the cycle (10610^6--10710^7 year timescale).

Human impact: phosphate mining for fertilisers has accelerated the release of phosphate from geological deposits. Agricultural runoff causes eutrophication of freshwater and marine ecosystems.

Eutrophication (Extended)

Stages of freshwater eutrophication:

  1. Nutrient enrichment: nitrate and phosphate from agricultural runoff enter a lake or river.
  2. Algal bloom: rapid growth of algae (cyanobacteria, green algae) at the water surface.
  3. Light attenuation: the dense algal layer blocks sunlight from reaching submerged plants.
  4. Plant death: submerged plants die due to insufficient light for photosynthesis.
  5. Decomposition: bacteria decompose dead algae and plants, consuming dissolved O2\mathrm{O}_2.
  6. Hypoxia/anoxia: dissolved O2\mathrm{O}_2 drops to levels that cannot support fish and invertebrates, causing a "dead zone."
  7. Biodiversity loss: fish kills and loss of aquatic species diversity.

9. Conservation Biology

Biodiversity Measurement

  • Species richness: the number of species in a given area.

  • Species evenness: how evenly individuals are distributed among species.

  • Shannon-Wiener diversity index: H=i=1spilnpiH' = -\sum_{i=1}^{s} p_i \ln p_i where pip_i is the proportion of individuals belonging to species ii and ss is the total number of species. HH' ranges from 00 (single species) to higher values (more diverse communities).

  • Simpson's diversity index: D=1i=1spi2D = 1 - \sum_{i=1}^{s} p_i^2 Measures the probability that two randomly selected individuals belong to different species.

Threats to Biodiversity

ThreatDescriptionExample
Habitat destructionConversion of natural habitats to agriculture, urban areas, and infrastructure.Deforestation of tropical rainforests; coral reef destruction.
Habitat fragmentationDivision of continuous habitat into patches, reducing connectivity and increasing edge effects.Road construction through forests; agricultural land division.
OverexploitationHarvesting species faster than they can recover.Overfishing; bushmeat trade; illegal wildlife trade.
PollutionContamination of air, water, and soil with harmful substances.Pesticide accumulation (DDT); oil spills; plastic pollution.
Climate changeGlobal warming, ocean acidification, changing precipitation patterns.Coral bleaching; range shifts; phenological mismatches.
Invasive speciesNon-native species that outcompete, prey on, or introduce diseases to native species.Cane toad in Australia; zebra mussel in North America; kudzu vine.

Conservation Strategies

  • In situ conservation: protecting species in their natural habitats (national parks, nature reserves, marine protected areas, wildlife corridors).
  • Ex situ conservation: protecting species outside their natural habitats (zoos, botanical gardens, seed banks, cryopreservation of gametes and embryos).
  • CITES: Convention on International Trade in Endangered Species regulates international trade.
  • Captive breeding programmes: for critically endangered species with very small wild populations (e.g., California condor, black rhino).
  • Sustainable management: balancing resource use with conservation (e.g., sustainable forestry, fisheries quotas, ecotourism).

Exam-Style Problems (Extended)

Problem 9: Quantitative -- Population Growth Models

A population of bacteria in a culture grows exponentially with r=0.6  h1r = 0.6\;\mathrm{h}^{-1} and N0=200N_0 = 200 cells. (a) Calculate the population size after 10 hours. (b) Calculate the doubling time. (c) If the carrying capacity of the culture is 10910^9 cells, at what time will the population reach 90%90\% of KK using the logistic equation? (d) Compare the logistic and exponential population sizes at t=20  ht = 20\;\mathrm{h}.

Problem 10: Data Analysis -- Shannon-Wiener Diversity Index

Two woodland communities are surveyed for butterfly species. Community A has 4 species with abundances: 80, 15, 3, 2. Community B has 5 species with abundances: 30, 28, 22, 12, 8. (a) Calculate the Shannon-Wiener diversity index (HH') for each community. (b) Calculate Simpson's diversity index (DD) for each community. (c) Which community is more diverse? Justify your answer with reference to both richness and evenness.

Problem 11: Extended Response -- Succession and Conservation

A volcanic island is formed by a submarine eruption. Over 500 years, the island is colonised by living organisms. (a) Describe the expected sequence of primary succession on the island, naming specific types of organisms at each stage and explaining how they modify the environment for the next stage. (b) Explain why the climax community on this island may differ from the climax community on a nearby mainland island. (c) If the island is later colonised by an invasive rat species, predict the impact on the native fauna and describe a conservation strategy to protect native species.

Problem 12: Extended Response -- Phosphorus Cycle and Agriculture

Explain why the phosphorus cycle is described as sedimentary rather than gaseous. Describe the role of phosphorus in biological systems (DNA, ATP, phospholipids, bone). Explain how the use of phosphate fertilisers in agriculture can lead to eutrophication in freshwater ecosystems. Discuss two strategies for reducing phosphate pollution from agricultural runoff.


Additional Worked Examples

Worked Example: Calculating Net Primary Productivity

A forest has a gross primary productivity (GPP) of 18000  kJ/m2/year18\,000\;\mathrm{kJ/m^2/year}. The plants in the forest respire at a rate of 12000  kJ/m2/year12\,000\;\mathrm{kJ/m^2/year}. (a) Calculate the net primary productivity (NPP). (b) If the ecological efficiency (transfer efficiency between trophic levels) is 10%10\%, calculate the net secondary productivity of herbivores. (c) If 60%60\% of NPP is used by decomposers, calculate the energy available to herbivores. (d) Explain why ecological efficiency is typically low.

Solution

(a) NPP=GPPRp=1800012000=6000  kJ/m2/year\mathrm{NPP} = \mathrm{GPP} - R_p = 18000 - 12000 = 6000\;\mathrm{kJ/m^2/year}

(b) Assuming herbivores consume 100%100\% of NPP (which is unrealistic; typically much less): NSP=NPP×ecological efficiency=6000×0.10=600  kJ/m2/year\mathrm{NSP} = \mathrm{NPP} \times \text{ecological efficiency} = 6000 \times 0.10 = 600\;\mathrm{kJ/m^2/year}

(c) If 60%60\% goes to decomposers: energy available to herbivores =6000×0.40=2400  kJ/m2/year= 6000 \times 0.40 = 2400\;\mathrm{kJ/m^2/year}. Secondary productivity of herbivores =2400×0.10=240  kJ/m2/year= 2400 \times 0.10 = 240\;\mathrm{kJ/m^2/year}.

(d) Ecological efficiency is typically low (55--20%20\%) because:

  • Not all biomass at one trophic level is consumed (some dies and goes to decomposers).
  • Not all consumed biomass is digested and absorbed (some is egested as faeces).
  • Of the absorbed energy, much is used for respiration (maintaining body temperature, movement, etc.) and is lost as heat.
  • Only the energy incorporated into new biomass (growth and reproduction) is available to the next trophic level. The "ten percent rule" is a useful approximation but actual efficiencies vary widely.

Worked Example: Population Growth with Immigration and Emigration

A population of birds on an island has a birth rate of 0.150.15 per individual per year and a death rate of 0.100.10 per individual per year. Immigration brings in 5050 birds per year and emigration removes 2020 birds per year. The current population is 20002000. (a) Calculate rr (per capita rate of increase) excluding migration. (b) Calculate the net change in population per year including migration. (c) If the carrying capacity is 50005000, calculate the population size after 5 years using the logistic equation (dN/dt=rN(1N/K)dN/dt = rN(1 - N/K)). Assume migration continues at the same rate.

Solution

(a) r=bd=0.150.10=0.05r = b - d = 0.15 - 0.10 = 0.05 per individual per year.

(b) Net change =rN+IE=0.05×2000+5020=100+30=130= rN + I - E = 0.05 \times 2000 + 50 - 20 = 100 + 30 = 130 birds per year.

(c) Using the logistic equation with migration: N0=2000N_0 = 2000, r=0.05r = 0.05, K=5000K = 5000, migration =+30= +30 per year.

Year 1: dN/dt=0.05×2000×(12000/5000)+30=100×0.6+30=60+30=90dN/dt = 0.05 \times 2000 \times (1 - 2000/5000) + 30 = 100 \times 0.6 + 30 = 60 + 30 = 90. N1=2090N_1 = 2090.

Year 2: dN/dt=0.05×2090×(12090/5000)+30=104.5×0.582+30=60.8+30=90.8dN/dt = 0.05 \times 2090 \times (1 - 2090/5000) + 30 = 104.5 \times 0.582 + 30 = 60.8 + 30 = 90.8. N2=2180.8N_2 = 2180.8.

Year 3: dN/dt=0.05×2180.8×(12180.8/5000)+30=109.04×0.564+30=61.5+30=91.5dN/dt = 0.05 \times 2180.8 \times (1 - 2180.8/5000) + 30 = 109.04 \times 0.564 + 30 = 61.5 + 30 = 91.5. N3=2272.3N_3 = 2272.3.

Year 4: dN/dt=0.05×2272.3×(12272.3/5000)+30=113.6×0.546+30=62.0+30=92.0dN/dt = 0.05 \times 2272.3 \times (1 - 2272.3/5000) + 30 = 113.6 \times 0.546 + 30 = 62.0 + 30 = 92.0. N4=2364.3N_4 = 2364.3.

Year 5: dN/dt=0.05×2364.3×(12364.3/5000)+30=118.2×0.527+30=62.3+30=92.3dN/dt = 0.05 \times 2364.3 \times (1 - 2364.3/5000) + 30 = 118.2 \times 0.527 + 30 = 62.3 + 30 = 92.3. N5=2456.6N_5 = 2456.6.

After 5 years: approximately 24572457 birds.

Note: with migration exceeding the net reproductive increase at small population sizes, the population grows more quickly than predicted by the logistic equation alone.

Worked Example: Carbon Cycle Calculations

A tropical rainforest covers 5×106  ha5 \times 10^6\;\mathrm{ha} and stores 200  t  C/ha200\;\mathrm{t\;C/ha} in biomass. It sequesters carbon at a rate of 6  t  C/ha/year6\;\mathrm{t\;C/ha/year}. (a) Calculate the total carbon stored in the forest. (b) Calculate the total CO2\mathrm{CO}_2 equivalent stored (molar mass: C=12\mathrm{C} = 12, CO2=44\mathrm{CO}_2 = 44). (c) If the forest is cleared for agriculture and 60%60\% of the biomass carbon is released as CO2\mathrm{CO}_2 within 5 years, calculate the annual CO2\mathrm{CO}_2 emission rate. (d) Explain how deforestation contributes to the enhanced greenhouse effect.

Solution

(a) Total carbon: 5×106×200=109  t  C=15 \times 10^6 \times 200 = 10^9\;\mathrm{t\;C} = 1 billion tonnes of carbon.

(b) CO2\mathrm{CO}_2 equivalent: 4412×109=3.67×109  t  CO2\frac{44}{12} \times 10^9 = 3.67 \times 10^9\;\mathrm{t\;CO}_2 =3.67= 3.67 billion tonnes of CO2\mathrm{CO}_2 equivalent.

(c) Carbon released: 0.60×109=6×108  t  C0.60 \times 10^9 = 6 \times 10^8\;\mathrm{t\;C}. CO2\mathrm{CO}_2 released: 6×108×4412=2.2×109  t  CO26 \times 10^8 \times \frac{44}{12} = 2.2 \times 10^9\;\mathrm{t\;CO}_2. Annual rate: 2.2×109/5=4.4×108  t  CO2/year=4402.2 \times 10^9 / 5 = 4.4 \times 10^8\;\mathrm{t\;CO}_2/year = 440 million tonnes/year.

(d) Deforestation contributes to the enhanced greenhouse effect by:

  • Releasing stored carbon as CO2\mathrm{CO}_2 when biomass is burned or decomposed.
  • Reducing the forest's capacity to absorb CO2\mathrm{CO}_2 through photosynthesis (loss of carbon sink).
  • CO2\mathrm{CO}_2 is a greenhouse gas that absorbs and re-emits infrared radiation, trapping heat in the atmosphere. Increased atmospheric CO2\mathrm{CO}_2 from deforestation amplifies the natural greenhouse effect, contributing to global warming and climate change.
  • Additionally, deforestation changes surface albedo (reflectivity), which can have regional climate effects.

Worked Example: Mark-Release-Recapture Population Estimation

A biologist captures, marks, and releases 8080 wood mice in a woodland. One week later, she captures 100100 mice, of which 1616 are marked. (a) Estimate the total population size using the Lincoln-Petersen index. (b) Calculate the 95%95\% confidence interval. (c) State three assumptions of this method. (d) Explain how each assumption could be violated and the effect on the population estimate.

Solution

(a) Lincoln-Petersen index: N=M×CR=80×10016=500N = \frac{M \times C}{R} = \frac{80 \times 100}{16} = 500 mice.

(b) 95%95\% confidence interval (approximate): SE=MC(MR)(CR)R3=80×100×64×84163\mathrm{SE} = \sqrt{\frac{MC(M-R)(C-R)}{R^3}} = \sqrt{\frac{80 \times 100 \times 64 \times 84}{16^3}} =430080004096=10500=102.5= \sqrt{\frac{43008000}{4096}} = \sqrt{10500} = 102.5

95%  CI=N±1.96×SE=500±1.96×102.5=500±20195\%\;\mathrm{CI} = N \pm 1.96 \times \mathrm{SE} = 500 \pm 1.96 \times 102.5 = 500 \pm 201 95%  CI=(299,701)95\%\;\mathrm{CI} = (299, 701)

(c) Assumptions:

  1. The marked individuals mix randomly with the unmarked population between sampling occasions.
  2. Marks are not lost and are always detectable.
  3. Births, deaths, immigration, and emigration are negligible between sampling occasions.
  4. All individuals have an equal probability of being captured in both samples.
  5. The population is closed (no migration).

(d) Violations and their effects:

  1. Non-random mixing: if marked individuals remain near the capture site, they may be overrepresented in the second sample, leading to an underestimate of NN.
  2. Mark loss: if marks are lost (e.g., tags fall off), fewer marked individuals are recaptured, leading to an overestimate of NN.
  3. Population change: if deaths occur between samples (especially of marked individuals), RR decreases, leading to an overestimate. If births occur, NN increases but the estimate is based on the original population size, leading to an underestimate.
  4. Unequal capture probability: trap-happy individuals (more likely to be recaptured) cause an underestimate; trap-shy individuals (avoid traps after first capture) cause an overestimate.
  5. Open population: immigration adds unmarked individuals, causing an overestimate of the original population.

Worked Example: Nitrogen Cycle and Fertiliser Application

A farmer applies 150  kg/ha150\;\mathrm{kg/ha} of ammonium nitrate fertiliser (NH4NO3\mathrm{NH}_4\mathrm{NO}_3, molar mass =80  g/mol= 80\;\mathrm{g/mol}) to a wheat field. (a) Calculate the mass of nitrogen applied per hectare. (b) If only 50%50\% of the nitrogen is taken up by the wheat crop, calculate the mass of nitrogen lost to the environment. (c) Explain the environmental consequences of nitrogen runoff into aquatic ecosystems. (d) Describe the process of denitrification and explain its importance in the nitrogen cycle.

Solution

(a) NH4NO3\mathrm{NH}_4\mathrm{NO}_3 contains 2 nitrogen atoms per molecule. Molar mass of N =14  g/mol= 14\;\mathrm{g/mol}. Mass fraction of N in NH4NO3\mathrm{NH}_4\mathrm{NO}_3: 2880=0.35=35%\frac{28}{80} = 0.35 = 35\%. Nitrogen applied: 150×0.35=52.5  kg  N/ha150 \times 0.35 = 52.5\;\mathrm{kg\;N/ha}.

(b) Nitrogen lost: 52.5×0.50=26.25  kg  N/ha52.5 \times 0.50 = 26.25\;\mathrm{kg\;N/ha}.

(c) Environmental consequences of nitrogen runoff:

  • Eutrophication: excess nitrogen stimulates rapid growth of algae (algal bloom), which blocks light and prevents photosynthesis by submerged plants. When algae die, decomposition by bacteria consumes dissolved oxygen, creating hypoxic or anoxic conditions that kill fish and other aquatic organisms.
  • Nitrate pollution of groundwater: nitrate (NO3\mathrm{NO}_3^-) is highly soluble and can leach into groundwater. High nitrate levels in drinking water cause methemoglobinaemia ("blue baby syndrome") in infants, in which nitrate is reduced to nitrite in the gut, converting haemoglobin to methaemoglobin, which cannot carry oxygen.
  • Nitrous oxide (N2O\mathrm{N}_2\mathrm{O}) emissions: a potent greenhouse gas (298×298\times the global warming potential of CO2\mathrm{CO}_2 over 100 years) produced by denitrifying bacteria in waterlogged soils.
  • Acidification and biodiversity loss: nitrogen deposition can acidify soils and freshwater, reducing plant and invertebrate diversity.

(d) Denitrification is the process by which denitrifying bacteria (e.g., Pseudomonas denitrificans, Paracoccus denitrificans) convert nitrate (NO3\mathrm{NO}_3^-) through nitrite (NO2\mathrm{NO}_2^-), nitric oxide (NO\mathrm{NO}), and nitrous oxide (N2O\mathrm{N}_2\mathrm{O}) to nitrogen gas (N2\mathrm{N}_2). It occurs under anaerobic conditions (e.g., waterlogged soils, sediments). Denitrification is important because it returns nitrogen to the atmosphere, completing the nitrogen cycle and preventing excessive accumulation of nitrate in soils and water. However, the intermediate N2O\mathrm{N}_2\mathrm{O} is a greenhouse gas and ozone-depleting substance, making the balance of complete vs incomplete denitrification environmentally significant.


Additional Common Pitfalls

  • Confusing gross primary productivity (GPP) and net primary productivity (NPP): GPP is the total energy fixed by photosynthesis; NPP = GPP - plant respiration. Only NPP is available to herbivores.
  • Assuming the "10% rule" applies universally: ecological efficiency varies from <1%<1\% (detritus food chains) to >30%>30\% (some aquatic systems with ectotherms). The "10% rule" is a rough guideline.
  • Confusing the greenhouse effect with global warming: the greenhouse effect is a natural process that keeps Earth habitable; the enhanced (anthropogenic) greenhouse effect is the additional warming caused by human activities increasing greenhouse gas concentrations.
  • Misidentifying the limiting nutrient in different ecosystems: phosphorus is typically limiting in freshwater; nitrogen in marine ecosystems; both can be limiting depending on the system.
  • Confusing species richness and species evenness: richness = number of species; evenness = relative abundance of each species. High diversity requires both high richness and high evenness.
  • Assuming all invasive species are harmful: some invasive species have neutral or even positive effects; the term "invasive" refers to spread, not necessarily impact.

Additional Exam-Style Problems with Full Solutions

Problem 13: Extended Response -- Biome Comparison

Compare and contrast tropical rainforest and tundra biomes with respect to: (a) climate (temperature, precipitation, growing season), (b) net primary productivity, (c) soil characteristics, (d) plant adaptations, (e) animal adaptations, and (f) vulnerability to climate change.

Answer 13
FeatureTropical RainforestTundra
Temperature2525--28C28^\circ\mathrm{C} year-round30-30 to 10C10^\circ\mathrm{C}; mean <10C<10^\circ\mathrm{C}
Precipitation20002000--10000  mm/year10000\;\mathrm{mm/year}150150--250  mm/year250\;\mathrm{mm/year}
Growing seasonYear-round5050--6060 days
NPPVery high (2200  g  C/m2/year\approx 2200\;\mathrm{g\;C/m^2/year})Very low (140  g  C/m2/year\approx 140\;\mathrm{g\;C/m^2/year})
SoilNutrient-poor (rapid decomposition, leaching); laterite soilsPermafrost; thin active layer; waterlogged in summer; low nutrient availability
Plant adaptationsButtress roots, drip tips, thin leaves, epiphytes, climbing plantsLow-growing (cushion plants), hairy leaves, dark pigments, clonal reproduction, shallow roots
Animal adaptationsArboreal locomotion, camouflage, specialised diets (coevolution with specific plants)Thick fur/feathers, fat deposits, migration, hibernation, seasonal colour change (e.g., arctic hare)
Climate vulnerabilityDrought stress, increased fire frequency, deforestation, loss of cloud forestsPermafrost thaw (releasing methane), shrubification (boreal shrubs replacing tundra vegetation), habitat loss for cold-adapted species

(f) Climate change vulnerability:

  • Tropical rainforest: increased frequency and severity of droughts can cause dieback and increase fire susceptibility; rising temperatures may exceed thermal tolerances of some species; shifts in rainfall patterns disrupt phenology (timing of flowering and fruiting).
  • Tundra: warming is occurring 22--3×3\times faster than the global average (Arctic amplification); permafrost thaw releases CO2\mathrm{CO}_2 and CH4\mathrm{CH}_4 (positive feedback for warming); treeline advance (boreal forest encroaching on tundra) reduces habitat for tundra species; loss of sea ice affects ice-dependent species (polar bears, seals).
Problem 14: Data Analysis -- Chi-Squared Test for Habitat Association

A researcher investigates whether a species of snail shows a preference for different habitat types. The observed and expected frequencies (based on the relative area of each habitat) are:

HabitatArea (ha)Expected proportionObserved snailsExpected snails
Woodland400.408572
Grassland350.355563
Marsh150.154027
Rocky100.102018
Total1001.00200180

(a) Calculate the chi-squared statistic. (b) Determine the degrees of freedom and the critical value at p=0.05p = 0.05. (c) State your conclusion. (d) Which habitat types are most preferred and avoided?

Answer 14

(a) The expected values should be based on 200200 total snails and the habitat proportions: Expected: Woodland =80= 80, Grassland =70= 70, Marsh =30= 30, Rocky =20= 20.

χ2=(8580)280+(5570)270+(4030)230+(2020)220\chi^2 = \frac{(85 - 80)^2}{80} + \frac{(55 - 70)^2}{70} + \frac{(40 - 30)^2}{30} + \frac{(20 - 20)^2}{20} =2580+22570+10030+0= \frac{25}{80} + \frac{225}{70} + \frac{100}{30} + 0 =0.313+3.214+3.333+0=6.860= 0.313 + 3.214 + 3.333 + 0 = 6.860

(b) df=41=3\mathrm{df} = 4 - 1 = 3. Critical value at p=0.05p = 0.05 for 3  df=7.8153\;\mathrm{df} = 7.815.

(c) χ2=6.860<7.815\chi^2 = 6.860 < 7.815. We fail to reject H0H_0 at p=0.05p = 0.05. The difference between observed and expected habitat use is not statistically significant at the 5%5\% level. However, the result is close to significant, suggesting a trend toward habitat preference that might be detected with a larger sample size.

(d) Despite failing to reach significance:

  • Most preferred: Marsh (observed 4040 vs expected 3030; deviation =+10= +10) and Woodland (observed 8585 vs expected 8080; deviation =+5= +5).
  • Most avoided: Grassland (observed 5555 vs expected 7070; deviation =15= -15).

The largest contributor to χ2\chi^2 is the grassland category (3.2143.214) and marsh category (3.3333.333).

Problem 15: Extended Response -- Human Impact on Ecosystems

Discuss the impact of human activities on three different ecosystems: (a) coral reefs (ocean acidification and warming), (b) tropical rainforests (deforestation), and (c) temperate grasslands (agricultural conversion). For each ecosystem: describe the impact, explain the ecological consequences, and evaluate one conservation strategy.

Answer 15

(a) Coral reefs -- ocean acidification and warming:

  • Impact: increased atmospheric CO2\mathrm{CO}_2 dissolves in seawater, lowering pH (acidification). Rising ocean temperatures cause coral bleaching (loss of symbiotic zooxanthellae).
  • Ecological consequences: acidification reduces the availability of carbonate ions needed for coral skeleton formation (calcification), weakening reef structure. Bleaching reduces coral growth and increases mortality. Coral reefs support approximately 25%25\% of all marine species; their decline leads to biodiversity loss, collapse of reef fisheries, loss of coastal protection (reefs attenuate wave energy), and economic losses for tourism-dependent communities.
  • Conservation strategy: marine protected areas (MPAs) restrict fishing and other extractive activities, allowing reef recovery. Evaluation: MPAs are effective where enforcement is strong and where they are large enough to protect entire reef ecosystems, but they do not address global threats (acidification, warming) and require ongoing political and financial commitment.

(b) Tropical rainforests -- deforestation:

  • Impact: logging, agriculture (soy, palm oil, cattle ranching), mining, and infrastructure development remove large areas of forest. Approximately 1010 million hectares of forest are lost annually.
  • Ecological consequences: habitat destruction and fragmentation cause species extinction (especially endemic species with small ranges); loss of carbon storage contributes to climate change; disruption of the water cycle (reduced evapotranspiration leads to regional drying); soil erosion and nutrient loss; loss of indigenous peoples' livelihoods and cultural knowledge.
  • Conservation strategy: sustainable forestry certification (e.g., FSC) ensures timber is harvested at rates that allow forest regeneration. Evaluation: certification can reduce deforestation rates and maintain some ecological functions, but enforcement is difficult, certified products may not command premium prices, and certification does not prevent conversion to agriculture (the main driver of deforestation).

(c) Temperate grasslands -- agricultural conversion:

  • Impact: conversion to cropland (wheat, maize, soy) and pasture has destroyed approximately 70%70\% of original temperate grasslands worldwide (e.g., North American prairie, Eurasian steppe).
  • Ecological consequences: loss of native grassland plant and animal species (e.g., prairie dogs, bison, ground-nesting birds); soil degradation from intensive tillage and monoculture; reduced soil carbon storage; loss of pollinator habitat; nitrogen and pesticide pollution from agricultural runoff.
  • Conservation strategy: restoration of native grassland through reseeding with native species and reintroduction of keystone herbivores (e.g., bison in North American prairie reserves). Evaluation: restoration can recover some ecosystem functions and biodiversity over decades, but fully restoring the deep, carbon-rich soils (mollisols) developed over millennia is not possible. Long-term monitoring and management (prescribed burning, grazing management) are required.

  • Evolution and natural selection: Review ./evolution-depth for adaptation, speciation, and Hardy-Weinberg equilibrium.
  • Plant biology and transport: Review ./plant-biology for transpiration, mineral uptake, and plant responses to environment.
  • Molecular biology and the carbon cycle: Review ./molecular-biology for photosynthesis and biochemical pathways.
  • Human physiology and environmental health: Review ./human-physiology for the effects of pollution on human health.
  • Genetics and population genetics: Review ./genetics for allele frequency calculations and genetic drift.

Supplementary: Biomes in Detail (HL Extension)

Tropical Rainforest

Climate: consistently warm (2525--28C28^\circ\mathrm{C}) and wet (20002000--10000  mm/year10000\;\mathrm{mm/year}), no dry season.

Structure: extremely high biodiversity (>50%>50\% of all terrestrial species). Distinct vertical stratification:

  • Emergent layer (4040--60  m60\;\mathrm{m}): tallest trees, exposed to high light and wind.
  • Canopy (2020--40  m40\;\mathrm{m}): continuous layer of foliage, intercepts most light (>95%>95\%).
  • Understorey (55--20  m20\;\mathrm{m}): shade-tolerant plants, limited light (<2%<2\%).
  • Forest floor: very little light (<1%<1\%), rapid decomposition, thin humus layer.

Nutrient cycling: most nutrients are stored in the biomass, not the soil. Decomposition is rapid due to warm, moist conditions. Nutrients are quickly taken up by roots (nutrient conservation). Soil is often nutrient-poor (oxisol/ultisol: heavily leached, high iron and aluminium content).

Adaptations: buttress roots (shallow soil, tall trees need support); drip tips on leaves (shed water quickly, prevent epiphyte growth); thin leaves with large surface area (maximise light capture in low light); epiphytes and lianas (compete for light in the canopy); cauliflory (flowers on trunk, for pollination by animals at lower levels).

Temperate Deciduous Forest

Climate: moderate temperatures (30-30 to 30C30^\circ\mathrm{C}), moderate precipitation (750750--1500  mm/year1500\;\mathrm{mm/year}), distinct seasons with a growing season of 44--66 months.

Structure: moderate biodiversity. Dominant trees: oak, maple, beech, birch. Clear vertical stratification (canopy, understorey, shrub layer, herb layer, forest floor).

Nutrient cycling: seasonal leaf fall creates a thick litter layer. Decomposition is slower than in tropical forests (cold winters slow microbial activity). Nutrients accumulate in the soil (alfisol/mollisol: fertile, rich in organic matter).

Adaptations: deciduous leaves (shed in autumn to reduce water loss and frost damage; costs of producing new leaves each spring offset by reduced maintenance); deep root systems (access groundwater during dry periods); vernal understorey plants (spring ephemerals that flower and set seed before the canopy closes); dormancy and cold hardiness (antifreeze proteins, supercooling).

Desert

Climate: extreme temperatures (>50C>50^\circ\mathrm{C} day, <0C<0^\circ\mathrm{C} night), very low precipitation (<250  mm/year<250\;\mathrm{mm/year}), high evapotranspiration.

Productivity: very low NPP (<90  g  C/m2/year<90\;\mathrm{g\;C/m^2/year}). Sparse vegetation, large bare areas.

Adaptations: succulence (water storage in fleshy tissues -- cacti store water in parenchyma); reduced leaf surface area (spines instead of leaves in cacti; reduces transpiration); CAM photosynthesis (stomata open at night, reducing water loss); deep taproots (accessing groundwater) or shallow widespread roots (capturing rainfall); ephemeral annuals (complete life cycle during brief rainy periods); light-coloured surfaces (reflecting heat); nocturnal behaviour in animals (avoiding daytime heat).

Tundra

Climate: very cold (30-30 to 10C10^\circ\mathrm{C}), low precipitation (150150--250  mm/year250\;\mathrm{mm/year}), permafrost (permanently frozen subsoil), short growing season (5050--6060 days), strong winds.

Structure: no trees (permafrost prevents deep root growth). Dominated by mosses, lichens, sedges, grasses, dwarf shrubs (willow, birch). Low biodiversity but high endemism.

Nutrient cycling: extremely slow decomposition (cold temperatures, waterlogged active layer). Deep layers of undecomposed organic matter (peat). Nutrient cycling is limited by low temperature and slow microbial activity.

Adaptations: low-growing cushion plants (avoid wind damage, benefit from insulating snow layer); hairy leaves (trap warm air); dark pigments (absorb more heat); shallow root systems (confined to active layer above permafrost); clonal reproduction (asexual reproduction is more reliable than sexual reproduction in harsh conditions); antifreeze compounds in cells.

Grassland/Savanna

Climate: tropical savanna: warm year-round, distinct wet and dry seasons (500500--1500  mm/year1500\;\mathrm{mm/year}). Temperate grassland: cold winters, hot summers (250250--750  mm/year750\;\mathrm{mm/year}).

Structure: dominated by grasses (C4 grasses in tropical savanna, C3 in temperate). Scattered trees in savanna (baobab, acacia). Deep, fertile soils (mollisol, chernozem) with high organic content from grass root systems.

Fire: a major ecological factor. Grasses recover quickly after fire (meristems at or below ground level). Fire prevents tree establishment, maintaining grassland. Fire also recycles nutrients (ash returns minerals to the soil).

Adaptations: deep root systems (22--3  m3\;\mathrm{m}); growth from basal meristems (grazing-resistant); silica deposits in leaves (deter herbivory); C4 photosynthesis (tropical grasses: efficient at high temperatures and low CO2\mathrm{CO}_2); tussock growth form (protects meristems from fire and grazing); drought deciduousness (trees shed leaves during dry season).

Aquatic Biomes (Brief)

Freshwater: lakes, rivers, wetlands. Low salt concentration. Productivity limited by light penetration (euphotic zone) and nutrient availability. Thermal stratification in deep lakes (epilimnion, thermocline, hypolimnion).

Marine: oceans cover 71%71\% of Earth's surface. Productivity concentrated in coastal upwelling zones and continental shelves (nutrient-rich). Open ocean is nutrient-poor but vast (low NPP per area but high total NPP). Coral reefs: highest marine biodiversity, built by calcium carbonate secretion by coral polyps (symbiosis with zooxanthellae).

Biome Map Summary

Equator : Tropical Rainforest
: Tropical Savanna (seasonal)
25 deg : Desert (subtropical high pressure)
35 deg : Mediterranean (hot, dry summers; mild, wet winters)
45 deg : Temperate Grassland
50 deg : Temperate Deciduous Forest
60 deg : Taiga (Boreal Forest)
70 deg : Tundra
Pole : Polar Ice (Antarctic/Arctic)

Conservation Biology Principles

Biodiversity hotspots: regions with high species richness and high levels of endemism that have lost at least 70%70\% of their original vegetation. There are 3636 recognised hotspots, including the Amazon, Madagascar, Sundaland, the Mediterranean Basin, and the Caribbean. These hotspots contain approximately 50%50\% of the world's plant species and 42%42\% of terrestrial vertebrate species on only 2.5%2.5\% of the Earth's land surface. Conservation resources are concentrated in hotspots for maximum impact.

Island biogeography and conservation design: the theory of island biogeography (MacArthur and Wilson, 1967) predicts that larger habitat patches and patches closer to other patches support more species. This has direct implications for reserve design: larger, connected reserves preserve more species than small, isolated ones. Wildlife corridors between habitat patches reduce the extinction risk of species with large territories.

Edge effects: the boundary between a habitat patch and the surrounding matrix (e.g., forest edge adjacent to agricultural land) has different environmental conditions (more light, wind, temperature fluctuations, higher predation). Edge effects reduce the effective habitat area for interior species. Small, narrow reserves have proportionally more edge than interior.

The SLOSS debate (Single Large Or Several Small): whether one large reserve or several small reserves of the same total area preserves more species. The answer depends on the species composition: a single large reserve protects interior species but risks losing everything to a single catastrophe; several small reserves spread risk but have more edge. Most conservation biologists now advocate for a network approach: a few large reserves connected by corridors, supplemented by smaller patches.

IUCN Red List categories: Extinct (EX), Extinct in the Wild (EW), Critically Endangered (CR), Endangered (EN), Vulnerable (VU), Near Threatened (NT), Least Concern (LC). Criteria include: population size reduction, geographic range, population size (small), population decline, probability of extinction.

CITES (Convention on International Trade in Endangered Species): regulates international trade in specimens of wild animals and plants. Appendix I (most endangered: trade banned), Appendix II (trade regulated), Appendix III (trade regulated at national level).

Keystone Species and Trophic Cascades

A keystone species is one whose impact on the community is disproportionately large relative to its abundance. Removing a keystone species causes dramatic changes in community structure.

Classic example -- starfish (Pisaster ochraceus): Paine (1966) experimentally removed starfish from intertidal rock pools. The starfish prey on mussels (Mytilus). Without starfish predation, mussels outcompeted other sessile species (barnacles, algae, sponges), reducing species diversity from 15 to 8 species. This was the first experimental demonstration of a keystone species.

Trophic cascade in Yellowstone: the reintroduction of wolves in 1995 triggered a trophic cascade: wolves reduced elk populations, which allowed willow and aspen to recover, which provided habitat for beavers, whose dams created wetlands benefiting fish, amphibians, and birds. This is an example of a top-down trophic cascade (predator \to herbivore \to plant \to ecosystem engineer).

Other keystone species examples: sea otters (prey on sea urchins, maintaining kelp forests); beavers (ecosystem engineers that create wetland habitat); elephants (modify savanna vegetation, creating grasslands for other species); figs (keystone resources in tropical forests, providing food for many species when other fruits are scarce).