HSC Exam Practice

Sample response. Trophic efficiency is the percentage of energy transferred from one trophic level to the next. It is calculated as: Trophic efficiency (%) = (energy at higher trophic level ÷ energy at lower trophic level) × 100.

Marking criteria. 1 mark — defines trophic efficiency as the proportion (or percentage) of energy transferred from one level to the next. 1 mark — states a correct formula linking the two energy levels (accept any algebraically equivalent form).

Sample response. The three energy loss pathways are: (1) respiration — chemical energy is converted to heat during cellular respiration to generate ATP; (2) egestion — undigested material (cellulose, chitin, bone) is lost in faeces; (3) excretion — nitrogenous waste (urea, uric acid) is eliminated in urine.

Marking criteria. 1 mark — names at least two of the three pathways correctly. 1 mark — names all three correctly (respiration / egestion / excretion). Accept synonyms such as ‘faeces’ for egestion, ‘heat loss’ for respiration.

Sample response. Energy flows one-way through ecosystems — from the sun to producers, then to consumers, and ultimately lost as heat to the environment. It cannot be recycled. Matter (atoms of C, N, P, etc.) cycles between biotic and abiotic components: decomposers break down dead organic material and release inorganic nutrients into the soil and water, which producers then reabsorb to build new organic compounds. Decomposers therefore recycle matter (nutrients) but do not recycle energy; the energy in dead organisms is released as heat during decomposition and dissipates permanently.

Marking criteria. 1 mark — states energy flows one-way and cannot be recycled (lost as heat). 1 mark — states matter (nutrients / atoms) is recycled between biotic and abiotic components. 1 mark — correctly identifies the role of decomposers as recycling matter/nutrients but not energy.

Sample response. Pyramids of energy are always upright because energy is lost (as heat via respiration) at every trophic level — the total energy passing through each successive level is always less than the level below, so the pyramid can never invert. Pyramids of numbers can be inverted when a single large individual (e.g. one oak tree) supports many small consumers (e.g. thousands of insects), because number does not directly reflect energy content.

Marking criteria. 1 mark — energy pyramids are always upright because energy is lost at every trophic level (mostly as heat) so successive levels always have less total energy. 1 mark — numbers pyramids can invert because number of organisms does not measure energy content (one large organism can host many small ones). Accept biomass inversion for partial credit.

Sample response. T2 = 80,000 × 0.10 = 8,000 kJ m⁻² yr⁻¹. T3 = 8,000 × 0.10 = 800 kJ m⁻² yr⁻¹. T4 = 800 × 0.10 = 80 kJ m⁻² yr⁻¹.

Marking criteria. 1 mark — correct method (three successive multiplications by 0.10 from T1). 1 mark — correct final answer of 80 kJ m⁻² yr⁻¹ with working shown.

Sample response (a). Energy content decreases steeply with each successive trophic level. T1 (algae) contains 18,000 kJ m⁻² yr⁻¹, T2 (coral) contains 2,700 kJ m⁻² yr⁻¹, T3 (parrotfish) 405 kJ m⁻² yr⁻¹, and T4 (reef shark) only 81 kJ m⁻² yr⁻¹. The decrease is approximately 15% between each pair of adjacent levels, representing a roughly 220-fold reduction from T1 to T4.

Marking criteria (a). 1 mark — describes the trend as a progressive, steep decrease in energy at each successive trophic level. 1 mark — supports the trend with at least one specific data value from the figure.

Sample response (b). Trophic efficiency (T1 to T2) = (2,700 ÷ 18,000) × 100 = 15%. This is higher than the 10% rule average. Ectothermic organisms (e.g. coral polyps, reef fish) do not expend metabolic energy to maintain a constant body temperature. As a result, a higher proportion of ingested energy can be directed towards growth and production rather than thermoregulation, reducing heat loss relative to endotherms and raising trophic efficiency above 10%.

Marking criteria (b). 1 mark — correct calculation: (2,700 ÷ 18,000) × 100 = 15%. 1 mark — identifies that 15% is above the 10% rule average and notes this is higher than typical. 1 mark — explains correctly that ectotherms do not expend energy on thermoregulation, so a greater fraction of ingested energy is available for production.

Sample response (c). Using the same 15% trophic efficiency, T5 would contain approximately 81 × 0.15 ≈ 12 kJ m⁻² yr⁻¹. Even using the more generous 15% efficiency, this energy is far too low to sustain a viable apex predator population. An individual large shark requires thousands of kJ per day; 12 kJ per square metre per year means the shark would need to forage over an impractically vast reef area. No self-sustaining T5 population could form because prey energy density is below the minimum viable threshold for reproduction and survival.

Marking criteria (c). 1 mark — correctly calculates T5 energy (accept range using 10–15% efficiency; any value between ~8 and 12 kJ m⁻² yr⁻¹). 1 mark — states that the energy is insufficient to support a viable T5 population. 1 mark — explains why in terms of metabolic energy demands or territory/prey-density requirements.

Sample response. Cattle are primary consumers (T2). The 10% rule states that only approximately 10% of energy in producers is incorporated into consumer biomass; in practice, cattle convert only about 3% of ingested grass energy into beef, because they are large, warm-blooded (endothermic) animals with very high metabolic costs for thermoregulation and movement. To produce 1 kg of beef (approximately 6,000 kJ), a steer must ingest 6,000 ÷ 0.03 = 200,000 kJ of grass. At a pasture productivity of 15,000 kJ m⁻² yr⁻¹, this requires 200,000 ÷ 15,000 ≈ 13 m² of pasture per kilogram of beef. By contrast, delivering the same 6,000 kJ directly from wheat (a T1 crop, ~20,000 kJ m⁻² yr⁻¹) requires only 6,000 ÷ 20,000 = 0.3 m² — a roughly 40-fold reduction in land use. The root cause is the 97% energy loss at the T1→T2 transfer in beef production.

Ecological consequence 1 — Vegetation structure modification and ground-cover loss. Livestock selectively consume palatable understorey plants, reducing plant species diversity and simplifying vegetation structure. This removes the microhabitat niches on which ground-nesting birds, reptiles and small marsupials depend, contributing to population declines. Australia has the world’s worst mammal extinction record, partly attributable to habitat loss driven by livestock grazing.

Ecological consequence 2 — Soil compaction and degradation. Cattle hooves compact soil, reducing pore space and water infiltration. This increases surface run-off, accelerates erosion, degrades water quality in streams, and reduces habitat for soil invertebrates and decomposer communities. Compacted soils also support fewer plant species, further simplifying ecosystem structure.

The ecological argument is strong: it correctly identifies trophic inefficiency as the mechanistic driver of the disproportionate land requirement for beef production, and the quantitative case for land savings is well supported. However, the argument has real limitations. Not all grazing land is suitable for cropping — much of Australia’s grazing land is arid or semi-arid and cannot produce grain. Dietary transition also carries social and economic costs for rural communities heavily reliant on the beef industry. Furthermore, removing cattle does not guarantee ecological recovery; active weed management and habitat restoration are also required.

In conclusion, reducing beef consumption is an ecologically sound and effective strategy for decreasing land-use pressure and could free up millions of square kilometres for biodiversity conservation. It addresses the root cause — the 10% rule energy loss — directly. However, it is a necessary but not sufficient solution: it must be paired with active restoration, sustainable land management and economic support for affected communities.

Marking criteria.

• 1 mark — Applies the 10% rule to explain beef production; notes cattle efficiency is approximately 3% (not 10%) because they are large endotherms with high metabolic costs.
• 1 mark — Quantifies the land area comparison between beef and a T1 crop (wheat or equivalent), showing a substantially larger land requirement for beef (accept any reasonable figures giving a ratio of ≥ 10-fold).
• 1 mark — First ecological consequence correctly identified and explained with a link to biodiversity impact (e.g. vegetation loss / habitat simplification / species extinction).
• 1 mark — Second ecological consequence correctly identified and explained (e.g. soil compaction, erosion, salinity, water quality degradation).
• 1 mark — Evaluates the strength of the argument (10% rule logic is sound; land savings are substantial and well supported by calculation).
• 1 mark — Identifies genuine limitations (arid land not suitable for cropping; social/economic costs; active restoration also required; not the only solution).
• 1 mark — Reaches a justified, nuanced conclusion that explicitly accepts or rejects the claim and applies ecological reasoning rather than a personal opinion.

Sample response. At each trophic level, approximately 90% of ingested energy is lost through three pathways: respiration (chemical energy converted to heat during cellular respiration to produce ATP — the largest loss, typically 60–90%); egestion (undigested material, such as cellulose and chitin, lost in faeces); and excretion (nitrogenous wastes such as urea lost in urine). As a result, only ~10% of energy at one level is available to build biomass at the next. Beginning with producers at 30,000 kJ m⁻² yr⁻¹:

• T2 = 30,000 × 0.10 = 3,000 kJ
• T3 = 3,000 × 0.10 = 300 kJ
• T4 = 300 × 0.10 = 30 kJ
• T5 = 30 × 0.10 = 3 kJ
• T6 = 3 × 0.10 = 0.3 kJ m⁻² yr⁻¹

A sixth trophic level would contain only 0.3 kJ m⁻² yr⁻¹. This is biologically unsustainable: any organism has a minimum metabolic requirement (for respiration, movement, growth and reproduction) that vastly exceeds 0.3 kJ per square metre per year. Even a very small predator would need to forage over an enormous area to obtain sufficient energy, meaning no viable self-sustaining population can form. This is why natural food chains typically stop at four to five trophic levels.

Marking criteria.

• 1 mark — Names all three energy loss pathways correctly (respiration, egestion, excretion).
• 1 mark — Correct calculation chain from T1 = 30,000 through to T6 = 0.3 kJ m⁻² yr⁻¹ (accept minor rounding).
• 1 mark — States that T6 energy (0.3 kJ m⁻² yr⁻¹) is insufficient to support a viable population.
• 1 mark — Explains ‘biologically unsustainable’ in terms of minimum metabolic requirements — available energy per unit area is below the threshold for an organism to survive, grow and reproduce.