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BU-105 / L06 · lecture

Ecological Efficiency and the 10% Rule

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BEFORE YOU START

You should understand the concept of trophic levels and recall that energy flows unidirectionally through an ecosystem, dissipating as heat at each step.

Why this matters

Ecological efficiency quantifies the energetic cost of biological organisation.

Understanding how little energy is transferred between trophic levels explains why food chains are short, why top predators are rare, and why human agriculture must be highly efficient to support large populations. This lecture moves beyond the qualitative idea of energy loss to define the specific metrics used to calculate these losses, distinguishing between production efficiency and trophic-level transfer efficiency.

The Energetic Budget of an Organism

Imagine a grasshopper eating a leaf. The grasshopper does not convert every calorie of that leaf into new muscle or egg mass. A significant portion of the energy is used immediately for maintenance: keeping its heart beating, its muscles moving, and its body temperature regulated (if it were an endotherm). This maintenance cost is the metabolic rate. The energy that remains after this metabolic cost is paid is what can be used for growth and reproduction. This is the biological production. Ecological efficiency is fundamentally a ratio: it asks how much of the energy available at one level is actually captured and converted into biomass at the next level. It is not a measure of how well an organism "works," but a measure of how much energy is "wasted" in the thermodynamic sense of being dissipated as heat or excreted as waste. This distinction is crucial because it separates the internal physiology of an individual from the external ecological flow of energy through a community.

Defining Ecological Efficiency

Ecological efficiency (\( \varepsilon \)) is defined as the ratio of energy assimilated or produced at one trophic level to the energy available at the preceding level. There are two primary ways to calculate this, depending on whether we are looking at the efficiency of an individual organism's metabolism or the efficiency of the entire trophic level transfer. The first is the production efficiency (\( P_e \)), which is the ratio of net production (\( P \)) to assimilation (\( A \)). This tells us how effectively an organism converts the food it eats into new biomass. The second is the trophic-level transfer efficiency (\( T_e \)), which is the ratio of production at level \( n \) to production at level \( n-1 \). This is the metric most relevant to food chains. Production efficiency depends mainly on whether the consumer is an endotherm (low \( P_e \), because so much energy goes into keeping warm) or an ectotherm (higher \( P_e \)); diet matters more for assimilation efficiency, which is lower for herbivores eating fibrous plant tissue than for carnivores eating easily digested animal tissue.

\[ P_e = \frac{P_n}{A} \times 100\% \]

Where \( P_n \) is net production (growth plus reproduction) and \( A \) is assimilation (ingestion minus egestion). Note that assimilation is not the same as ingestion; it is the energy actually absorbed into the body. The difference between ingestion and assimilation is egestion, the energy lost in faeces. This distinction is vital for accurate calculation, as using ingestion in the denominator would underestimate the organism's metabolic efficiency.

The 10% Rule and Trophic Transfers

The "10% rule" is a heuristic, not a law. It states that, on average, only about 10% of the energy from one trophic level is transferred to the next. This means that if producers (level 1) fix 10,000 kJ of solar energy, primary consumers (level 2) will only gain about 1,000 kJ, and secondary consumers (level 3) will gain only about 100 kJ. This decline is described by the formula from the previous lecture: \( E_n \approx E_0\,\varepsilon^{\,n-1} \). The value of \( \varepsilon \) varies widely in nature. In aquatic systems, where organisms are often small and have high surface-area-to-volume ratios, transfer efficiencies can be as high as 20-30%. In terrestrial systems, where large endothermic predators are common, efficiencies can be as low as 1-5%. The 10% figure is a convenient average for teaching, but it must be treated as a variable parameter, not a constant. The key insight is that because \( \varepsilon \ll 1 \), the number of trophic levels is inherently limited. You cannot have a food chain with ten levels because the energy available at the top would be vanishingly small, insufficient to support a viable population of predators.

\[ E_{\text{top}} = E_{\text{base}} \times (0.1)^{\text{levels}-1} \]

This equation illustrates the rapid decline in available energy. For a five-level chain, the top level receives only \( 0.1^4 = 0.0001 \) or 0.01% of the base energy. This mathematical constraint is why ecosystems are structured as pyramids of energy, where the base is always the widest.

Factors Influencing Efficiency

Several biological and environmental factors determine the actual value of \( \varepsilon \) in a specific ecosystem. The first is the metabolic strategy of the consumer. Endotherms (birds and mammals) have much higher metabolic rates than ectotherms (reptiles, fish, insects). An endotherm must burn a large fraction of its assimilated energy just to maintain body temperature, leaving less for growth. Therefore, the production efficiency of an endotherm is typically 1-3%, whereas an ectotherm may have a production efficiency of 10-40%. The second factor is the quality of the food. Plant material is often rich in cellulose, which is difficult to digest. Herbivores have lower assimilation efficiencies than carnivores because a larger fraction of plant energy is egested. Carnivores eat animal tissue, which is protein-rich and easily digestible, leading to higher assimilation efficiencies. The third factor is the age and life stage of the organism. Juveniles have higher growth rates and thus higher production efficiencies than adults, who allocate more energy to maintenance and reproduction. Finally, environmental conditions such as temperature and oxygen availability can affect metabolic rates, thereby influencing efficiency. In cold environments, ectotherms have lower metabolic rates, which can paradoxically increase their production efficiency because they burn less energy for maintenance.

\[ A = I - E_g \]

Where \( I \) is ingestion and \( E_g \) is egestion. The assimilation efficiency (\( A_e \)) is \( A/I \). The trophic-level transfer efficiency is the product of three steps: the fraction of the lower level\'s production that is eaten (consumption efficiency \( C_e \)), the fraction of that which is assimilated (\( A_e \)), and the fraction of that which becomes new biomass (\( P_e \)): \( T_e = C_e \times A_e \times P_e \). A low value in any one step gives a low overall efficiency. For example, a herbivore population that eats 40% of plant production, assimilates 50% of what it eats and converts 20% of that into growth has \( T_e = 0.4 \times 0.5 \times 0.2 = 4\% \).

The concept

◆ CONCEPT · T-022 · full treatment on its own page

Trophic levels and energy flow

Energy dissipates as it passes up a food chain.

\[ E_n \approx E_0\,\varepsilon^{\,n-1}, \qquad \varepsilon \ll 1\ (\text{commonly} \sim 10\%\text{ per transfer}) \]
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Tier 3 ยท watch out

A common misconception is that the 10% rule applies to biomass rather than energy. Biomass pyramids can be inverted in aquatic systems (e.g., phytoplankton biomass is lower than zooplankton biomass because phytoplankton turn over rapidly), but energy pyramids are never inverted. Energy is always lost as heat, so the energy pyramid must always be upright. Confusing biomass with energy leads to incorrect conclusions about the sustainability of top predators. C

Worked example

Problem. A grassland has net primary production of 5000 kJ m-2 yr-1. Herbivores eat 40% of it (consumption efficiency), assimilate 60% of what they eat and convert 20% of what they assimilate into new biomass. Find the herbivores' ingestion, assimilation and net production, their trophic transfer efficiency, and the production available to carnivores if the next transfer is 10%.

  1. Ingestion: \(I = 0.40 \times 5000 = 2000\ \mathrm{kJ\,m^{-2}\,yr^{-1}}\); the other 3000 goes to decomposers as uneaten plant matter.
  2. Assimilation: \(A = 0.60 \times 2000 = 1200\ \mathrm{kJ\,m^{-2}\,yr^{-1}}\); egestion is \(2000 - 1200 = 800\).
  3. Net production: \(P = 0.20 \times 1200 = 240\ \mathrm{kJ\,m^{-2}\,yr^{-1}}\); respiration is \(1200 - 240 = 960\).
  4. Trophic transfer efficiency: \(T_e = 240/5000 = 0.048 = 4.8\%\), which equals \(0.40 \times 0.60 \times 0.20\).
  5. Carnivore production: \(0.10 \times 240 = 24\ \mathrm{kJ\,m^{-2}\,yr^{-1}}\).
\[ P_{\text{herb}} = 240,\quad T_e = 4.8\%,\quad P_{\text{carn}} = 24\ \ \mathrm{kJ\,m^{-2}\,yr^{-1}} \]

Units check. Every flow is in kJ m-2 yr-1 and the efficiencies are dimensionless, so products of the two keep the flow units.

CHECK YOURSELF

Three quick tests

  1. If the production efficiency of an organism is 10% and its assimilation is 500 kJ, what is its net production?
    Answer50 kJ. Calculation: \( 0.10 \times 500 = 50 \).
  2. Why are energy pyramids always upright, while biomass pyramids can be inverted?
    AnswerEnergy is lost as heat at each trophic level, so it must decrease. Biomass can increase if the lower level has a high turnover rate.
  3. Calculate the energy available at the fourth trophic level if the base is 10,000 kJ and efficiency is 10%.
    Answer10 kJ. Calculation: \( 10,000 \times (0.1)^3 = 10 \).
UNIT

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NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Concepts cited: T-022 · Threads: systems, energy