Energy transfer and nutrient recyclingEdexcel A-Level Biology B: Revision notes
Section 1
Gross and net primary productivity
Gross primary productivity (GPP) is the rate at which plants fix light energy in photosynthesis. Plants use part of this in respiration (R), lost as heat. The rest is net primary productivity (NPP), the energy stored in new plant biomass and available to consumers and decomposers.
NPP = GPP − R
Example: GPP 20 000 and R 8 000 kJ m⁻² year⁻¹ gives NPP = 12 000 kJ m⁻² year⁻¹, which is 60% of GPP.
Section 2
Efficiency of energy transfer
Efficiency (%) = energy transferred to the next level ÷ energy available at the previous level × 100
Example: producers 31 200 and primary consumers 3 380 kJ m⁻² year⁻¹: efficiency = 3 380 ÷ 31 200 × 100 = 10.8%.
Transfers are typically about 10% (often 5 to 20%), which is why food chains are short and why there are few top predators.
Divide the energy in the higher level by the lower level, not the other way round, and always multiply by 100.
Section 3
Why energy is lost at each level
- Not all of the organism is eaten (roots, bones, shells).
- Not all of what is eaten is digested and absorbed: some is lost in faeces.
- Energy is lost in urine (nitrogenous waste).
- Most is lost in respiration, as heat, for movement, keeping warm and active transport.
Energy is not destroyed; it is transferred to the environment. Endotherms lose more as heat. Farmers raise efficiency by restricting movement, warming the animals and giving digestible feed.
Section 4
Microorganisms and nutrient recycling
Dead organisms and waste contain nutrients locked in organic compounds. Saprobionts (bacteria and fungi) release enzymes onto the dead material (extracellular digestion), absorb the soluble products and respire, releasing CO₂ and mineral ions.
In the nitrogen cycle, saprobionts convert nitrogen compounds into ammonium ions (ammonification). Nitrifying bacteria convert ammonium to nitrite and then nitrate, which plants absorb. Without these microorganisms nutrients would stay locked in dead matter and growth would stop.
Section 5
Conditions for decay
Decay is fastest when it is warm, moist, well aerated and near neutral pH. In waterlogged, acidic or cold soils (a bog) decay is slow, because oxygen is limited and enzymes work slowly, so peat accumulates and nutrients stay locked up. Plants there have low nutrient availability and low productivity.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Energy transfer and nutrient recycling
- In a grassland, the gross primary productivity (GPP) of the plants is 20 000 kJ m⁻² year⁻¹. The plants lose 8 000 kJ m⁻² year⁻¹ as heat in respiration.Calculate the NPP of the grassland as a percentage of its GPP.2 marks
- The energy stored in the biomass at three trophic levels of a lake food chain was measured over one year. Producers: 31 200 kJ m⁻² year⁻¹. Primary consumers: 3 380 kJ m⁻² year⁻¹. Secondary consumers: 405 kJ m⁻² year⁻¹.Calculate the efficiency of energy transfer from the primary consumers to the secondary consumers. Give your answer to one decimal place.2 marks
- A farmer rears beef cattle in two ways. Group X graze freely outdoors on pasture. Group Y are kept in heated barns with little room to move and are fed on concentrated feed. After 12 months, the cattle in group Y have gained more body mass per unit of food energy than those in group X.Suggest three reasons why the cattle in group Y convert more of their food energy into body mass than those in group X.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).