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This lesson covers energy flow through ecosystems, including food chains, food webs, trophic levels, productivity (GPP and NPP), and energy transfer efficiency calculations, as required by the Edexcel A-Level Biology specification (9BI0), Topic 10 -- Ecosystems.
Energy enters most ecosystems as sunlight and flows through the ecosystem via feeding relationships. At each stage, energy is transferred but also lost, primarily as heat from respiration.
Key principles:
flowchart TB
Sun["Sunlight\n(1,000,000 kJ)"] -->|"~1-3% captured\nby photosynthesis"| T1["Producers (T1)\nGPP = 20,000 kJ"]
T1 -->|"Respiration\n12,000 kJ"| Heat1["Heat"]
T1 -->|"NPP = 8,000 kJ\navailable to next level"| T2["Primary Consumers (T2)\nIngested: 1,600 kJ"]
T1 -->|"Not eaten /\nDead matter"| Dec["Decomposers"]
T2 -->|"Faeces & urine\n400 kJ"| Dec
T2 -->|"Respiration\n1,040 kJ"| Heat2["Heat"]
T2 -->|"Net production\n160 kJ"| T3["Secondary Consumers (T3)"]
T3 -->|"Respiration"| Heat3["Heat"]
Dec -->|"Respiration"| Heat4["Heat"]
A trophic level is the position of an organism in a food chain or food web.
| Trophic Level | Organisms | Energy Source |
|---|---|---|
| Primary producers (T1) | Photosynthetic organisms (plants, algae, cyanobacteria) | Sunlight (photosynthesis) |
| Primary consumers (T2) | Herbivores | Producers |
| Secondary consumers (T3) | Carnivores that eat herbivores | Primary consumers |
| Tertiary consumers (T4) | Top predators; carnivores that eat other carnivores | Secondary consumers |
| Decomposers | Bacteria and fungi | Dead organic matter at all trophic levels |
Note that decomposers do not fit neatly into a single trophic level -- they feed on dead organic matter from all levels. Some organisms are omnivores and feed at multiple trophic levels (e.g. humans eat both plants and animals).
A food chain shows a single linear pathway of energy transfer:
Producer→Primary consumer→Secondary consumer→Tertiary consumer
Example: Grass --> Rabbit --> Fox --> Eagle
A food web shows the interconnected food chains in an ecosystem. Food webs are more realistic because:
Common Misconception: Students often think arrows in food chains show "what eats what." In fact, the arrows show the direction of energy transfer -- from the organism being eaten to the organism doing the eating. Grass --> Rabbit means energy flows FROM grass TO rabbit.
GPP is the total rate of energy fixation (photosynthesis) by producers in an ecosystem, measured per unit area per unit time.
GPP=Total energy fixed by photosynthesis
Typical units: kJ m^{-2} year^{-1}
However, producers use some of this energy for their own respiration (R). The energy remaining after respiration is available for growth, reproduction, and transfer to the next trophic level.
NPP is the rate of energy storage in plant biomass after respiratory losses have been subtracted.
NPP=GPP−R
Where R = energy lost through respiration by producers
NPP represents the energy available to primary consumers and decomposers.
| Ecosystem | GPP (kJ m^{-2} year^{-1}) | NPP (kJ m^{-2} year^{-1}) | NPP/GPP Ratio |
|---|---|---|---|
| Tropical rainforest | ~90,000 | ~37,000 | ~41% |
| Temperate deciduous forest | ~50,000 | ~25,000 | ~50% |
| Temperate grassland | ~20,000 | ~8,000 | ~40% |
| Open ocean | ~5,000 | ~2,500 | ~50% |
| Desert | ~1,500 | ~600 | ~40% |
Exam Tip: You must be able to use the formula NPP = GPP - R. Remember that GPP is the total energy fixed; NPP is what remains for growth and the next trophic level. NPP is always less than GPP. The NPP/GPP ratio varies because different ecosystems have different respiratory demands (e.g. tropical forests have high respiration due to high temperatures).
For consumers, the energy budget is:
N=I−(F+R)
Where:
The efficiency of energy transfer between trophic levels is typically 5--20% (often quoted as approximately 10% as a rule of thumb).
Efficiency=Energy available at trophic level nEnergy available at trophic level n+1×100
| Transfer | Typical Efficiency | Explanation |
|---|---|---|
| Sunlight to producers | 1--3% | Most sunlight misses leaves, is reflected, or is wrong wavelength |
| Producers to primary consumers | 5--15% | Much plant biomass is not eaten; cellulose is hard to digest; high respiration |
| Primary to secondary consumers | 10--20% | Animal tissue is more digestible than plant tissue |
| Secondary to tertiary consumers | 10--20% | Similar efficiency to above |
| Energy Loss | Explanation |
|---|---|
| Respiration | The largest loss at every trophic level; energy is released as heat during metabolic reactions |
| Not all biomass is eaten | Some organisms die without being consumed; some parts (bones, shells, bark) are not eaten |
| Faeces and urine | Not all ingested food is assimilated; some is egested in faeces or excreted |
| Not all sunlight is captured | Much sunlight misses leaves, is reflected, or is the wrong wavelength for photosynthesis |
Endotherms (birds, mammals) use a large proportion of their ingested energy for thermoregulation (maintaining body temperature). This means they lose more energy through respiration, so the efficiency of energy transfer to the next trophic level is lower (~5--10%).
Ectotherms (reptiles, fish, invertebrates) do not use metabolic energy for thermoregulation, so a higher proportion of ingested energy is available for growth. Efficiency of transfer is higher (~15--20%).
This is why farming ectotherms (e.g. fish farming) is more energy-efficient than farming endotherms (e.g. cattle).
Question: In a grassland ecosystem, the GPP of the grass is 20,000 kJ m^{-2} year^{-1}. The grass uses 12,000 kJ m^{-2} year^{-1} for respiration. Calculate the NPP.
Answer:
NPP=GPP−R=20,000−12,000=8,000 kJ m−2 year−1
Question: The NPP of producers in an ecosystem is 8,000 kJ m^{-2} year^{-1}. The net production of the primary consumers is 800 kJ m^{-2} year^{-1}. Calculate the efficiency of energy transfer from producers to primary consumers.
Answer:
Efficiency=8,000800×100=10%
Question: A population of rabbits ingests 5,000 kJ m^{-2} year^{-1}. They lose 1,200 kJ in faeces and urine and 3,200 kJ through respiration. Calculate the net production available for the next trophic level.
Answer:
N=I−(F+R)=5,000−(1,200+3,200)=5,000−4,400=600 kJ m−2 year−1
Question: In a marine ecosystem, phytoplankton have an NPP of 10,000 kJ m^{-2} year^{-1}. The energy transfer efficiency from phytoplankton to zooplankton is 12%, and from zooplankton to small fish is 15%. Calculate the energy available to small fish.
Answer:
Step 1: Energy available to zooplankton (T2): Energy at T2=10,000×10012=1,200 kJ m−2 year−1
Step 2: Energy available to small fish (T3): Energy at T3=1,200×10015=180 kJ m−2 year−1
Only 180 kJ m^{-2} year^{-1} out of the original 10,000 kJ reaches the small fish -- just 1.8% of the original NPP. This illustrates why food chains are short: very little energy reaches the higher trophic levels.
A pyramid of energy (also called a pyramid of productivity) shows the rate of energy flow at each trophic level. Each bar represents the energy available at that level per unit area per unit time.
Pyramids of energy are always upright because energy is lost at each trophic level.
| Pyramid Type | Always Upright? | What It Shows | Limitation |
|---|---|---|---|
| Pyramid of numbers | No (can be inverted, e.g. one tree supporting many insects) | Number of individuals at each trophic level | Does not account for organism size |
| Pyramid of biomass | Usually (can be inverted in aquatic ecosystems due to rapid phytoplankton turnover) | Mass of organisms at each trophic level at a point in time | Snapshot only -- misses rapid turnover |
| Pyramid of energy | Always yes | Rate of energy flow at each trophic level | Difficult and time-consuming to measure |
Exam Tip: Pyramids of energy are the most reliable representation of trophic structure because they are always upright and account for the rate of production, not just the standing crop. A biomass pyramid in the ocean can appear inverted because phytoplankton reproduce so rapidly that their standing biomass at any moment is less than the zooplankton feeding on them, even though phytoplankton produce more biomass per year.
Understanding energy transfer efficiency has practical implications:
| Food Source | Land Required per 1,000 kcal | Water Required per 1,000 kcal |
|---|---|---|
| Grain (wheat) | ~1.1 m^2 | ~500 litres |
| Chicken | ~3.2 m^2 | ~1,500 litres |
| Pork | ~4.5 m^2 | ~2,300 litres |
| Beef | ~8.9 m^2 | ~5,000 litres |
This data shows why reducing meat consumption -- particularly red meat -- is one of the most effective ways individuals can reduce their environmental footprint. This links directly to the conservation and sustainability topics.
The Edexcel 9BI0 specification places energy transfer in Topic 5: On the Wild Side — Photosynthesis, Energy and Ecosystems, on Paper 2 (Energy, Exercise and Coordination). This lesson is the quantitative core of Topic 5: lesson 1 supplies the trophic players (producers, primary, secondary, tertiary consumers, decomposers) and the ecosystem framing; lessons 3–4 (carbon and nitrogen cycles) trace the matter that accompanies energy flow but, unlike energy, cycles rather than flows; the photosynthesis lessons of Topic 5 supply the energy entry point — light energy fixed as chemical bond energy in glucose at rate GPP; the respiration lessons of Topic 5 supply the dominant loss term R at every trophic level, returning chemical energy to heat. Statements concern: light as the ultimate energy source for most ecosystems; the definitions and calculation of gross primary productivity (GPP) and net primary productivity (NPP); the relationship NPP=GPP−R; energy transfer efficiency between trophic levels (typically 5–25%, with the rule-of-thumb ~10%); reasons for energy losses (respiration, heat, faeces, urine, uneaten material); pyramids of numbers, biomass and energy and why energy pyramids never invert; and the implications for food-chain length, agriculture and biofuels (refer to the official Pearson Edexcel 9BI0 specification document for exact wording). Topic 1 (lipids and carbohydrates) supplies the storage forms in which energy moves between trophic levels.
Question (8 marks):
A temperate UK meadow is monitored for one year. Solar radiation incident on the meadow surface totals 3,500,000kJm−2yr−1. Producers fix energy at a gross primary productivity of GPP=10,000kJm−2yr−1. Producer respiration totals R=4,000kJm−2yr−1.
(a) Define GPP and NPP and calculate NPP for this meadow. (2)
(b) Calculate the percentage of incident solar radiation captured as GPP, and the percentage of GPP retained as NPP. Comment on each value. (3)
(c) Assuming a primary-consumer trophic transfer efficiency of 10%, predict the energy available to primary consumers as new biomass. State two assumptions of the 10% rule and explain why measured trophic transfer efficiencies range from 5% to 25%. (3)
Solution with mark scheme:
(a) M1 (AO1.1) — GPP is the total chemical energy fixed by producers via photosynthesis per unit area per unit time; NPP is the energy retained as new producer biomass after subtracting producer respiration: NPP=GPP−R. A1 (AO2.2) — NPP=10,000−4,000=6,000kJm−2yr−1.
(b) M1 (AO2.2) — fraction of solar radiation fixed as GPP =10,000/3,500,000=2.86×10−3≈0.29%. M1 (AO2.2) — fraction of GPP retained as NPP =6,000/10,000=60%. A1 (AO3.1a) — both values are typical: less than 1% of incident solar energy is fixed by terrestrial producers because most light is reflected, transmitted, of unsuitable wavelength, or hits non-photosynthetic surfaces; the 60% NPP retention is typical for a temperate meadow where roughly 40% of fixed energy is lost to producer respiration to maintain metabolism.
(c) M1 (AO2.2) — energy to primary consumers as new biomass ≈0.10×6,000=600kJm−2yr−1. M1 (AO1.2) — assumptions: (i) all NPP is potentially available to primary consumers (in reality much falls as litter to decomposers without being grazed); (ii) the 10% transfer is uniform across primary-consumer species (in reality it varies with consumer thermoregulation, digestive efficiency and metabolic rate). A1 (AO3.1a) — measured efficiencies range 5–25% because endotherms (mammals, birds) lose more energy to maintaining body temperature than ectotherms (fish, reptiles, invertebrates), so endothermic transfer efficiencies cluster nearer 5–10% while ectothermic efficiencies (e.g. fish in aquaculture) can reach 20–25%; digestion of plant cellulose is also less efficient than digestion of animal tissue, so transfer to primary consumers is typically less efficient than to higher carnivores.
Total: 8 marks.
Question (6 marks): Explain the difference between gross primary productivity and net primary productivity, and use the concept of cumulative energy losses to explain why food chains rarely exceed four or five trophic levels.
Mark scheme decomposition by AO:
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