Trophic Levels and Energy Transfer — The 10% Rule and Trophic Efficiency
Australia uses 54% of its total land mass for grazing livestock. Yet cattle convert only a tiny fraction of the energy in grass into beef. Understanding why — and calculating exactly how much energy is lost at each step — is the key to explaining why food chains are short, why apex predators are rare, and why land use matters for biodiversity.
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Q1. A steer consumes approximately 10,000 kJ of energy stored in grass each day. A student claims the steer will therefore gain 10,000 kJ of body mass energy per day. Is this claim correct? If not, where does the "missing" energy go?
Q2. Could a food chain realistically have 15 trophic levels? Why do natural food chains typically stop at 4 or 5 levels? Use your intuition about energy — not vocabulary you have memorised.
Key Terms
The feeding position of an organism in a food chain. T1 = producers; T2 = primary consumers; T3 = secondary consumers; T4 = tertiary consumers; T5 = quaternary consumers / apex predators.
The 10 percent rule of energy transfer between trophic levels
The percentage of energy transferred from one trophic level to the next. Typically ~10% in most ecosystems, though it ranges from 5% to 20%.
An approximation stating that only about 10% of the energy stored in biomass at one trophic level is converted into biomass at the next trophic level. The remaining ~90% is lost.
A graphical model showing the relationship between organisms at different trophic levels. Can represent numbers, biomass or energy.
The total mass of living material in a given area or volume, usually measured in g m⁻² or kg. Biomass pyramids show standing crop at each trophic level.
The flow of chemical energy stored in organic compounds from one organism to another when one is consumed. Energy flows one-way and is progressively lost as heat.
What Is Trophic Efficiency?
When a grasshopper eats grass, it does not convert 100% of the grass biomass into grasshopper biomass. Most of the energy is lost through three main pathways:
- Respiration: The organism breaks down glucose via cellular respiration to release ATP for movement, growth, temperature regulation and reproduction. A large fraction of the chemical energy in food is converted to heat and lost to the environment. This is the single largest energy loss — typically 60-90% of ingested energy.
- Egestion (faeces): Not all food is digestible. Cellulose in plant cell walls, chitin in insect exoskeletons and bone in vertebrates pass through the gut undigested and are egested. This energy is not absorbed by the consumer.
- Excretion (urine): Excess nitrogen from protein metabolism is converted to urea or uric acid and excreted. These waste products contain chemical energy that the consumer cannot use.
Only the energy that remains after these three losses is available for growth and reproduction — that is, for building new biomass that can be eaten by the next trophic level. This net energy is called production or assimilated energy.
100%
~60-90% lost as heat
~10-30% in faeces
~2-5% in urine
~5-20% available to next trophic level
Schematic only — actual percentages vary by organism, diet and environment.
This connects directly to your Module 1 knowledge: cellular respiration (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP + heat) is the mechanism by which energy is released from organic compounds. Every living cell respires continuously, which is why energy loss is unavoidable at every trophic level.
The 10% Rule — Worked Example
Trophic efficiency is calculated as:
In most ecosystems, trophic efficiency averages approximately 10%. This means that if producers (T1) capture 20,000 kJ of solar energy per square metre per year, only about 2,000 kJ will be stored in primary consumer biomass, about 200 kJ in secondary consumer biomass, and about 20 kJ in tertiary consumer biomass.
| Trophic Level | Example Organism (Australian Grassland) | Energy (kJ m⁻² yr⁻¹) | % Transferred |
|---|---|---|---|
| T1 — Producer | Native grasses (e.g. kangaroo grass, Themeda triandra) | 20,000 | — |
| T2 — Primary consumer | Red kangaroo (Osphranter rufus) | 2,000 | 10% |
| T3 — Secondary consumer | Dingo (Canis familiaris dingo) | 200 | 10% |
| T4 — Tertiary consumer | Wedge-tailed eagle (Aquila audax) | 20 | 10% |
Worked calculation: If a square metre of Australian grassland produces 20,000 kJ of grass energy per year, and trophic efficiency is 10% at each step:
- T2 = 20,000 × 0.10 = 2,000 kJ
- T3 = 2,000 × 0.10 = 200 kJ
- T4 = 200 × 0.10 = 20 kJ
After just three transfers, the available energy has fallen from 20,000 kJ to 20 kJ — a 1,000-fold reduction. This is why apex predators such as wedge-tailed eagles require enormous territories: each individual needs access to enough primary production to support the entire chain above it.
Energy Flows One-Way; Matter Is Cycled
This is one of the most important distinctions in ecology, and it is tested in almost every HSC exam:
Energy Flow
- Flows in one direction only: sun → producers → consumers → heat lost to space
- Cannot be recycled by ecosystems
- At each transfer, ~90% is lost (mostly as heat via respiration)
- Pyramids of energy are always upright
Matter Cycling
- Atoms (C, N, P, etc.) are recycled between biotic and abiotic components
- Decomposers and detritivores break down dead matter and release inorganic nutrients
- Producers re-absorb these nutrients and rebuild organic compounds
- No net loss of matter from the ecosystem (closed loop)
Students often confuse these two flows. A common error is to say that "energy is recycled by decomposers." Decomposers do recycle matter (nutrients), but they cannot recycle energy. The energy in dead organisms is also released as heat during decomposition — it does not return to the producer level.
Australian Anchor: Grazing and Land Use
Australia dedicates approximately 54% of its total land area to grazing livestock — predominantly cattle and sheep. This makes grazing the largest single land use in the country, far exceeding cropping, forestry and urban areas combined.
Why so much land? Because of trophic efficiency. Cattle are primary consumers (T2). To produce 1 kg of beef requires roughly 10 kg of feed (grain or pasture). At 10% efficiency, 90% of the energy in that feed is lost as heat, faeces and urine. A beef steer must consume vast quantities of grass to accumulate relatively little body mass.
This has profound implications for Australian biodiversity. Grazing modifies vegetation structure, reduces ground cover, increases soil compaction, and can lead to woody weed invasion. Understanding the 10% rule helps explain why shifting toward plant-based diets (eating producers directly) reduces land use pressure — you bypass the 90% energy loss at the T1→T2 transfer.
Common Errors to Avoid
[COMMON ERROR] "The 10% rule means exactly 10% is transferred every time."
Correction: 10% is an average approximation. Actual trophic efficiency ranges from approximately 5% (endothermic herbivores in cold climates with high metabolic costs) to 20% (ectothermic predators in warm aquatic environments with low metabolic costs). In HSC questions, you should use 10% unless a different value is specified, but you should also describe it as an approximation, not a fixed law.
[COMMON ERROR] "Energy is recycled by decomposers, so it returns to the producer level."
Correction: Decomposers break down dead organic matter and release nutrients (matter) as inorganic ions that producers can absorb. However, the energy in dead organisms is released as heat during decomposition and cellular respiration. Energy flows one-way and is not recycled. Only matter is cycled.
Copy Into Your Books
▼Respiration (heat), egestion (faeces), excretion (urine). Only production (growth + reproduction) becomes available biomass for the next trophic level.
Trophic efficiency = (energy at higher level / energy at lower level) × 100. Typically ~10% (range 5-20%).
Energy flows one-way (sun → producers → consumers → heat lost). Matter is cycled (decomposers return nutrients to soil/water; producers reabsorb them).
After 4-5 transfers, available energy is reduced by 10,000× or more. There is insufficient energy to support another viable population.
Trophic Efficiency Calculations
Australian Grazing and Land-Use Efficiency
- Native pasture productivity: 15,000 kJ m⁻² yr⁻¹
- A mature steer consumes pasture equivalent to 12,000 kJ m⁻² yr⁻¹ (the steer grazes an area, not a single square metre)
- Steer growth efficiency (energy converted to beef): approximately 3% of ingested energy (cattle are less efficient than the theoretical 10% because they are large endotherms with high metabolic costs)
- Average beef energy content: 6,000 kJ per kg
Test Your Understanding
1. A rabbit ingests 1,000 kJ of grass energy. Which of the following correctly identifies the fate of this energy?
2. In a grassland ecosystem, producers contain 40,000 kJ m⁻² yr⁻¹. Assuming 10% trophic efficiency at each level, what is the energy available to tertiary consumers?
3. Why are pyramids of energy always upright, regardless of the ecosystem?
4. An ecologist measures the energy content of organisms in a coral reef. She finds that primary consumers contain 5,000 kJ m⁻² yr⁻¹ and secondary consumers contain 1,000 kJ m⁻² yr⁻¹. What is the trophic efficiency between these two levels?
5. A student argues that because decomposers break down dead organisms and return nutrients to the soil, they must also return energy to the producer level. Evaluate this argument using your understanding of the difference between energy flow and matter cycling.
Short Answer Questions
6. The table below shows energy flow through a simplified Australian grassland food chain.
| Trophic Level | Organism | Energy (kJ m⁻² yr⁻¹) |
|---|---|---|
| T1 | Grasses | 25,000 |
| T2 | Grasshopper | 2,500 |
| T3 | Skink | ? |
| T4 | Brown falcon | 25 |
(a) Calculate the energy value for T3 (skink). Show your working. 1 MARK
(b) Calculate the overall trophic efficiency from T1 to T4. 1 MARK
(c) Explain why the trophic efficiency between T1 and T2 is approximately 10%, but the overall efficiency from T1 to T4 is much lower. 2 MARKS
7. Explain why natural food chains rarely exceed five trophic levels. In your answer, refer to the 10% rule, the three energy loss pathways, and explain why a sixth trophic level would be biologically unsustainable. Use a calculation to support your explanation. 5 MARKS
8. Using the Australian grazing case study from this lesson, evaluate whether reducing beef consumption would be an effective strategy for decreasing land-use pressure and protecting native biodiversity. In your answer, apply the 10% rule to compare the land required for a beef-based diet versus a plant-based diet, and discuss at least two ecological consequences of large-scale grazing on Australian ecosystems. 6 MARKS
Revisit Your Thinking
Return to your Think First responses at the start of this lesson.
- Q1 — steer energy: Did you identify that most energy is lost as heat via respiration, with smaller losses via faeces and urine? Did you know that only ~3-10% becomes new biomass in cattle?
- Q2 — 15 trophic levels: Did you predict that after 4-5 transfers the energy would be reduced by 10,000× or more, making a 15th level impossible? Can you now calculate: starting from 20,000 kJ at T1, what would T15 contain?
- Write the energy vs matter distinction from memory in one sentence each.
Comprehensive Answers
▼Activity 1 — Trophic Efficiency Calculations
(a) T2 = 50,000 × 0.10 = 5,000 kJ. T3 = 5,000 × 0.10 = 500 kJ. [1 mark for correct answer with working]
(b) T3 (secondary consumer) = 500 kJ. T2 = 500 × 10 = 5,000 kJ. T1 = 5,000 × 10 = 50,000 kJ. T1 for shark = 50,000 kJ. Total through T2 = 5,000 + 50,000 = 55,000 kJ (or simply 500 × 100 = 50,000 kJ through T2, and 50,000 × 10 = 500,000 kJ through T1). [1 mark for T2, 1 mark for T1]
(c) At 20% efficiency: T2 = 50,000 × 0.20 = 10,000 kJ. T3 = 10,000 × 0.20 = 2,000 kJ. T4 = 2,000 × 0.20 = 400 kJ. This is 20 times more than at 10% efficiency (20 kJ). Aquatic ectotherms (cold-blooded) have lower metabolic costs than terrestrial endotherms, so they convert a higher percentage of ingested energy into biomass. Warmer water temperatures also increase metabolic efficiency in fish. This means aquatic food chains can support more trophic levels (e.g. open ocean chains with 5-6 levels) compared to terrestrial chains (typically 3-4 levels). [2 marks for calculation, 1 mark for explanation]
Activity 2 — Australian Grazing
(a) Beef energy per kg = 6,000 kJ. Energy required in steer = 6,000 / 0.03 = 200,000 kJ of ingested energy per kg of beef. Pasture required = 200,000 / 15,000 = 13.3 m² per kg of beef. [1 mark for correct method, 1 mark for answer]
(b) 7,700,000 km² × 0.54 = 4,158,000 km² (approximately 4.2 million km²). [1 mark]
(c) Annual human need = 10,000 kJ/day × 365 = 3,650,000 kJ. Wheat area = 3,650,000 / 20,000 = 182.5 m² per year. Compared to 13.3 m² per kg of beef (a single meal for one person), eating plants directly requires approximately 50-100 times less land per unit of energy delivered. This is because bypassing the T1→T2 transfer avoids the ~90% energy loss. [2 marks for calculation, 1 mark for comparison, 1 mark for explanation]
(d) Large-scale grazing modifies vegetation structure (removing ground cover and palatable species), compacts soil (reducing water infiltration and increasing runoff), and can lead to woody weed invasion and erosion. It also reduces habitat for ground-dwelling fauna and fragments landscapes. Reducing beef demand decreases the economic incentive to convert native ecosystems to pasture, thereby protecting biodiversity. [2 marks for two valid consequences]
Multiple Choice
1. C — Most energy is lost via respiration, egestion and excretion. Only a small fraction (typically 3-20%) becomes biomass.
2. B — T2 = 4,000; T3 = 400; T4 = 40 kJ. Three transfers: 40,000 × (0.10)³ = 40.
3. D — Energy pyramids are always upright because energy is lost at each level. Options A and C confuse numbers/biomass with energy. Option B is wrong — energy is not recycled.
4. A — (1,000 / 5,000) × 100 = 20%. This is within the normal 5-20% range for aquatic ecosystems.
5. B — The argument confuses matter cycling with energy flow. Decomposers return nutrients (matter) but energy is lost as heat.
Short Answer Model Answers
Q6 (4 marks): (a) T3 = 2,500 × 0.10 = 250 kJ [1 mark]. (b) Overall efficiency = (25 / 25,000) × 100 = 0.1% [1 mark]. (c) The T1→T2 efficiency is ~10% (2,500/25,000 = 10%) [0.5 marks]. However, overall efficiency from T1 to T4 compounds the loss at each step: 10% × 10% × 10% = 0.1% [1 mark]. Each trophic level loses ~90% of the energy it receives, so after three transfers only 0.1% of the original energy remains [0.5 marks]. Total: 4 marks.
Q7 (5 marks): The 10% rule states that approximately 10% of energy is transferred between successive trophic levels [1 mark]. Three loss pathways: respiration (energy converted to heat during cellular respiration for ATP production), egestion (undigested material in faeces), and excretion (nitrogenous waste in urine) [1.5 marks — 0.5 each]. A sixth trophic level would contain only 20,000 × (0.10)⁵ = 0.2 kJ m⁻² yr⁻¹ [1 mark for calculation]. This is insufficient to support a viable population because organisms need enough energy for basal metabolism, movement, growth and reproduction [1 mark]. Even if the organism were extremely small and efficient, the energy available would be below the minimum viable threshold for any sustained population [0.5 marks]. Total: 5 marks.
Q8 (6 marks): The 10% rule means cattle (T2) convert only ~3-10% of grass energy into beef [1 mark]. Producing 1 kg of beef requires ~13 m² of pasture, while delivering the same energy from wheat (T1) requires only ~0.2 m² — a 50-100 fold difference [1 mark]. Therefore, a plant-based diet dramatically reduces land-use pressure [0.5 marks]. Ecological consequence 1: grazing removes ground cover and reduces habitat for ground-nesting birds, reptiles and small mammals, leading to population declines [1 mark]. Ecological consequence 2: hoof compaction reduces soil water infiltration, increases erosion and can cause dryland salinity (particularly in Western Australia where rising water tables bring salt to the surface) [1 mark]. Evaluated conclusion: reducing beef consumption is an effective strategy because it addresses the root cause — trophic inefficiency — while simultaneously reducing habitat destruction, soil degradation and greenhouse gas emissions (methane from enteric fermentation). However, it is not the only strategy needed; sustainable grazing management, protected areas and restoration are also required [1.5 marks]. Total: 6 marks.