Ecology & Ecosystems

Question 1 (12 marks)

(a) Apply 10% at each transfer (the "why": only ~10% of energy at one level is incorporated into the next): (4)

• Grass = $840{,}000$
• Grasshoppers = $840{,}000 \times 0.10 = 84{,}000$ kJ m⁻²yr⁻¹
• Frogs = $8{,}400$
• Snakes = $840$
• Hawks = $840{,}000 \times 0.1^4 = 84$ kJ m⁻²yr⁻¹

(1 mark per correct level after grass; award full 4 for reaching 84.)

(b) Energy at hawks = $84$ kJ m⁻²yr⁻¹, which is greater than the required $80$ kJ m⁻²yr⁻¹. (1) Therefore the ecosystem can support a breeding hawk population. (1) Margin is small ($84 - 80 = 4$ kJ), so it is marginal/vulnerable. (1)

(c) Any two: (2)

• Not all organisms/parts at a level are eaten (some die, decompose).
• Energy lost as heat during respiration.
• Energy lost in undigested material (faeces/egestion).
• Not all food is digestible/assimilated.

(d) Energy decreases by ~90% each level, so after ~4–5 levels too little energy remains to support another viable population (2); e.g. above the hawk level only ~8.4 kJ would remain, insufficient to sustain a further trophic level. (1)

Question 2 (10 marks)

(a) Pyramid of numbers: base very wide (2,000,000 phytoplankton), narrowing sharply upward (120,000 → 800 → 40). Upright. (3) (sketch tapering upward = 2; correct label upright = 1)

(b) Pyramid of biomass: base narrow (5 g phytoplankton), widening upward (15 → 40 → 60). Inverted. (3)

(c) Phytoplankton are individually tiny (low biomass each) but very numerous and reproduce/turn over rapidly. (1) So they are huge in number but small in total mass at any instant; consumers are fewer but individually larger, giving high biomass — hence numbers pyramid is upright while biomass pyramid is inverted. (1)

(d) The pyramid of energy is always upright (1), because energy is progressively lost (as heat/respiration/egestion) at each transfer, so each level always contains less energy than the one below — energy cannot be created. (1)

Question 3 (12 marks)

(a) $N_2$ has a strong, stable triple bond that plants lack the enzymes to break (1); plants can only absorb nitrogen as soluble ions (nitrate/ammonium), not as inert $N_2$ gas. (1)

(b) Process: nitrogen fixation (1); organism: nitrogen-fixing bacteria (e.g. Rhizobium / Azotobacter / cyanobacteria) (1); product: ammonia/ammonium ($NH_3/NH_4^+$) (later nitrified to nitrate). (1)

(c) Legumes have root nodules containing symbiotic Rhizobium bacteria that fix atmospheric $N_2$ into usable ammonium/nitrate (1); when the legume crop is ploughed in/decomposes, this fixed nitrogen enriches the soil (1), raising nitrate available for the following cereal crop. (1)

(d) Process: leaching / run-off carries dissolved nitrate and phosphate into the lake (eutrophication). (1) Two consequences (2 marks):

• Algal bloom blocks light → aquatic plants die.
• Decomposers respire and deplete dissolved oxygen → fish/animals die (hypoxia). (1 mark for naming eutrophication/run-off + up to 2 for consequences + 1 for O₂ depletion mechanism explicit = 4)

Question 4 (10 marks)

(a) New rock island = primary succession — starts on bare rock with no pre-existing soil or organisms (1.5). Burned woodland = secondary succession — soil, seeds and nutrients remain after disturbance (1.5).

(b) The burned woodland (secondary) reaches climax faster (1) because soil, nutrients, seed bank and root systems already exist, so colonisation and growth are much quicker. (1)

(c) Pioneer species: lichens (or mosses) (1). They can grow on bare rock, secrete acids that weather/break down rock (1), and on dying add organic matter, beginning soil formation that lets larger plants colonise. (1)

(d) Any two abiotic features of temperate deciduous forest: (2)

• Moderate rainfall (~750–1500 mm/yr).
• Four distinct seasons / moderate temperatures.
• Cold winters causing leaf fall; warm summers.
• Fertile, deep soils.

Question 5 (6 marks)

(a) Fundamental niche of Species A = the entire tidal zone (high + low shore) it can occupy in the absence of competition (1.5). Realised niche = the high shore only, the range it actually occupies when Species B is present. (1.5)

(b) Interaction: interspecific competition (1). It demonstrates the competitive exclusion principle — two species cannot occupy the exact same niche indefinitely (1); the superior competitor (B) excludes A from the low shore, restricting A's realised niche. (1)

[
  {"claim":"Energy at hawks = 840000 * 0.1**4 = 84 kJ", "code":"result = (840000*Rational(1,10)**4 == 84)"},
  {"claim":"Hawk energy 84 exceeds required 80 kJ so ecosystem supports hawks", "code":"result = (840000*Rational(1,10)**4 > 80)"},
  {"claim":"Grasshopper level at 6% efficiency = 50400 kJ", "code":"result = (840000*Rational(6,100) == 50400)"},
  {"claim":"Number pyramid producer:top ratio 2,000,000/40 = 50000 (upright, wide base)", "code":"result = (Rational(2000000,40) == 50000)"},
  {"claim":"Biomass increases upward (inverted): 5 < 15 < 40 < 60", "code":"result = (5 < 15 < 40 < 60)"}
]