Trophic levels and biomass transfer (HT)
Higher-tier students must be able to identify trophic levels in a food chain, draw or interpret pyramids of biomass, and calculate the efficiency of biomass transfer between levels.
Trophic levels
Each step in a food chain is a trophic level:
- Level 1: Producers — plants and algae.
- Level 2: Primary consumers — herbivores.
- Level 3: Secondary consumers — carnivores eating herbivores.
- Level 4: Tertiary consumers — top predators.
Decomposers (bacteria, fungi) are not numbered — they break down dead material at every level.
Pyramids of biomass
A pyramid of biomass is a horizontal-bar diagram, with the producer at the bottom and each higher trophic level on top, scaled to its biomass (total mass of organisms, usually in g or kg per unit area).
Pyramids of biomass narrow towards the top because biomass decreases at each trophic level. This is because:
- Some material is eaten but not absorbed (egested in faeces).
- Some absorbed material is lost in urine (waste from amino acid breakdown).
- Most absorbed energy is used in respiration to release energy for movement, growth and heat. This produces CO₂ and water.
- Only the remaining biomass is incorporated into new tissue — and so available to the next trophic level.
This is why food chains rarely have more than 4 or 5 levels — there isn't enough energy left to support another consumer.
Efficiency of biomass transfer (the key calculation)
Efficiency of transfer = (biomass of higher trophic level / biomass of lower trophic level) × 100
Or in some questions:
Efficiency = (energy at higher level / energy at lower level) × 100
The typical efficiency between trophic levels is roughly 10 % — although it varies. Plants → herbivores might be 5–15 %; herbivores → carnivores up to 20 %.
Worked example
A field of grass has biomass 8000 kg. The cows feeding on it have a total biomass of 800 kg. The biomass of the bacteria living in the cows' guts is not relevant. Calculate the efficiency of transfer from grass to cows.
Efficiency = (800 / 8000) × 100 = 10 %.
Worked example with energy
A producer fixes 2,000,000 kJ of energy per year. The herbivores in the same area gain 200,000 kJ per year. The carnivores gain 30,000 kJ.
- Producer → herbivore = (200,000 / 2,000,000) × 100 = 10 %
- Herbivore → carnivore = (30,000 / 200,000) × 100 = 15 %
Why is efficiency low?
Because most of the energy taken in by an organism is used in respiration (and lost as heat) rather than stored as biomass. Organisms also lose energy in faeces and urine.
Why does this matter for food security?
Eating lower in the food chain (more plants, less meat) means more humans can be fed from the same land area. A field of wheat feeds far more people than the same area used for cattle to provide meat.
⚠Common mistakes
- "Energy is destroyed." Energy is not destroyed — it's transferred to the surroundings as heat (still in agreement with conservation of energy).
- Drawing pyramids upside down. Producer biomass is bigger than that of consumers; the pyramid has its widest base at the bottom.
- Forgetting the units. Biomass is mass; energy is in joules. The two go hand-in-hand but aren't identical.
- Using "10 %" as a rule — it's an approximation; real values vary.
Links
Builds on B7.2 (food chains), B7.3 (carbon cycle / respiration). Connects to B7.9 (food security and sustainable agriculture).
AI-generated · claude-opus-4-7 · v3-deep-biology