Biology • Year 11 • Module 4 • Lesson 4
Trophic Levels and Energy Transfer — The 10% Rule
Apply the 10% rule and energy-flow concepts to data tables, calculation scenarios, and the Australian grazing case study.
1. Energy flow through an Australian woodland food chain
The table below shows energy values at three trophic levels in an Australian woodland ecosystem. Some values are missing. Use the 10% rule to answer the questions. 8 marks
| Trophic level | Organism | Energy (kJ m−2 yr−1) | Trophic efficiency (%) |
|---|---|---|---|
| T1 — Producer | Eucalyptus woodland understorey | 40,000 | — |
| T2 — Primary consumer | Eastern grey kangaroo | ? | 10% |
| T3 — Secondary consumer | Dingo | ? | 10% |
| T4 — Tertiary consumer | Wedge-tailed eagle | ? | 10% |
1.1 Calculate the energy values for T2, T3 and T4. Show all working. 3 marks
1.2 Calculate the overall trophic efficiency from T1 to T4 as a percentage. 2 marks
1.3 Explain why only 40 kJ m−2 yr−1 is available at T4, even though T1 started with 40,000 kJ. Reference the three energy loss pathways in your answer. 3 marks
2. Interpret a graph — energy at each trophic level
The bar chart below shows the energy content (kJ m−2 yr−1) at each trophic level in a freshwater lake ecosystem. 6 marks
Figure 2: Energy at each trophic level in a hypothetical freshwater lake (illustrative data).
2.1 Describe the trend shown in the graph. 2 marks
2.2 Calculate the trophic efficiency from T1 to T2. Does this match the 10% rule? Explain why ectothermic (cold-blooded) freshwater organisms can sometimes achieve efficiencies slightly above 10%. 3 marks
2.3 A fifth trophic level (T5) is proposed. Using the data pattern, estimate how much energy (kJ m−2 yr−1) would be available at T5, and explain whether this could support a viable predator population. 1 mark
3. Australian grazing case study — beef vs plants
A Queensland cattle station has the following energy data. Use these values to answer the questions below. 8 marks
- Native pasture productivity: 15,000 kJ m−2 yr−1
- Steer growth efficiency (energy converted to beef): 3% of ingested energy
- Average beef energy content: 6,000 kJ per kg
- Wheat productivity: 20,000 kJ m−2 yr−1
- Average human daily energy requirement: 9,000 kJ per day
3.1 Calculate how many square metres of pasture are needed to produce 1 kg of beef. Show all working. 3 marks
3.2 A human eating wheat directly (T1) needs 9,000 kJ per day. Calculate how many square metres of wheat are needed to feed one person for one year (365 days). 3 marks
3.3 Compare your answers to 3.1 and 3.2. Explain why a plant-based diet requires significantly less land than a beef-based diet, using the 10% rule. 2 marks
4. Identify errors in a student's energy flow diagram
A student drew the following description of energy flow: "Producers absorb all the sun's energy. Primary consumers gain 100% of the producer's energy. Decomposers return all this energy back to the producers, so energy is recycled in ecosystems just like matter." 4 marks
4.1 Identify two biological errors in the student's description and write a correction for each. 4 marks (1 per error identified, 1 per correction)
Error 1:
Correction 1:
Error 2:
Correction 2:
Q1.1 — Energy calculations
T2 = 40,000 × 0.10 = 4,000 kJ m−2 yr−1 [1 mark].
T3 = 4,000 × 0.10 = 400 kJ m−2 yr−1 [1 mark].
T4 = 400 × 0.10 = 40 kJ m−2 yr−1 [1 mark].
Q1.2 — Overall trophic efficiency T1 to T4
Overall efficiency = (T4 ÷ T1) × 100 = (40 ÷ 40,000) × 100 = 0.1%. [1 mark for correct formula, 1 mark for correct answer].
Q1.3 — Why only 40 kJ remains at T4
At each trophic level, approximately 90% of the energy in food is lost through three pathways: (1) respiration — chemical energy is converted to heat during cellular respiration to produce ATP [1 mark]; (2) egestion — undigested material is lost in faeces; (3) excretion — nitrogenous waste (urea/uric acid) is eliminated in urine [1 mark]. Only the remaining ~10% is available as production (new biomass) for the next level. After three such transfers, only 0.1% of the original 40,000 kJ (= 40 kJ) remains [1 mark].
Q2.1 — Trend in the graph
Energy content decreases sharply with each successive trophic level [1 mark]. The decrease is not linear but exponential — T2 is approximately 15% of T1, T3 is about 14% of T2, and T4 is about 10% of T3. The bars decrease steeply, with T1 dominating and T4 nearly invisible on the scale used [1 mark].
Q2.2 — Trophic efficiency T1 to T2 and ectotherm explanation
Trophic efficiency = (4,500 ÷ 30,000) × 100 = 15% [1 mark]. This is slightly above the 10% rule average. Ectothermic (cold-blooded) organisms such as fish and invertebrates do not use metabolic energy to maintain a constant body temperature, so a higher proportion of ingested energy can be allocated to growth (production) rather than heat generation. This reduces metabolic heat loss compared to endotherms (warm-blooded animals), resulting in higher trophic efficiency [2 marks].
Q2.3 — Energy at T5
T5 ≈ 63 × 0.10 = ~6.3 kJ m−2 yr−1. This is far too little energy to support a viable predator population — a single adult predator would need a vast foraging area to meet its metabolic demands, making a self-sustaining T5 population biologically unsustainable in this ecosystem [1 mark].
Q3.1 — Pasture required per kg of beef
Energy needed in the steer to produce 1 kg of beef = 6,000 kJ ÷ 0.03 = 200,000 kJ of ingested energy [1 mark].
Area of pasture = 200,000 kJ ÷ 15,000 kJ m−2 yr−1 = 13.3 m2 of pasture per kg of beef [2 marks for method and answer].
Q3.2 — Wheat area for one person per year
Annual human energy need = 9,000 kJ day−1 × 365 days = 3,285,000 kJ yr−1 [1 mark].
Wheat area = 3,285,000 ÷ 20,000 = 164.25 m2 yr−1 to feed one person directly from wheat [2 marks for method and answer].
Q3.3 — Comparison and 10% rule explanation
Producing 1 kg of beef requires ~13.3 m2 of pasture. A single person eating beef-derived energy at this rate would need far more land than the 164 m2 needed to supply their full year of energy from wheat. A plant-based diet requires dramatically less land because it bypasses the T1→T2 energy transfer, avoiding the ~90% (or ~97% for cattle) energy loss at that step [2 marks].
Q4.1 — Errors in student's description
Error 1: "Primary consumers gain 100% of the producer's energy." This is wrong — only approximately 10% (range 5–20%) of energy in producers is available to primary consumers. The remaining ~90% is lost as heat (respiration), in faeces (egestion) and urine (excretion).
Correction 1: Primary consumers gain only approximately 10% of the energy stored in producers; the rest is lost to the environment, primarily as heat via respiration.
Error 2: "Decomposers return all this energy back to the producers, so energy is recycled." This confuses energy with matter. Decomposers return nutrients (matter), not energy.
Correction 2: Decomposers break down dead organic matter and release inorganic nutrients (matter) that producers can re-absorb. The energy in dead organisms is released as heat during decomposition and cannot be returned to the producer level — energy flows one-way and is not recycled.