OxfordAQA International AS and A-level Biology: Energy transfer, productivity and farming (9610)

This guide covers sections 3.3.4.1 (Energy transfer) and 3.3.4.2 (Energy and human food production) of the OxfordAQA International AS and A-level Biology specification, Version 5.1, for International A-level exams from May/June 2018 onwards. Both sections sit in Unit 3, Populations and genes, which the specification lists as International A-level only. Use the course hub and the printable checklist to track the whole course. Companion pages: revision notes and practice questions. Not sure where you stand? Take a free diagnostic first.

What the specification asks you to do

Section	What you must be able to do
3.3.4.1	Describe energy transfer through trophic levels in food chains and food webs
3.3.4.1	Explain why transfer from producers to consumers, and between consumers, is inefficient
3.3.4.1	Make quantitative calculations about energy transfer between trophic levels
3.3.4.1	Relate pyramids of number, biomass and energy to food chains and webs
3.3.4.2	Define GPP and NPP and use NPP = GPP - R
3.3.4.2	Use N = I - (F + U + R) for the net production of consumers
3.3.4.2	Explain how pesticides, biological agents, integrated systems and reduced respiratory losses raise efficiency
3.3.4.2	Argue scientifically about productivity and evaluate economic, environmental and ethical data

Trophic levels and food chains

A trophic level is a feeding position in a food chain. Producers occupy the first level. They capture light energy and store it as chemical energy in organic molecules. Primary consumers eat producers, secondary consumers eat primary consumers, and so on. A food web joins many chains together. One species can sit at more than one level, for example an omnivore that eats both plants and herbivores.

Only the chemical energy stored in an organism’s tissues can pass to the next level when it is eaten.

Why transfer is inefficient

Only a small fraction of the energy at one level reaches the next. At each step, energy is lost in four main ways:

1. Not all of an organism is eaten. Roots, bones, shells, fur and whole individuals that die uneaten stay in the food web as dead material.
2. Not all of what is eaten is absorbed. Undigested material leaves in the faeces, and it still holds chemical energy.
3. Some absorbed material is excreted, for example urea in urine, which still holds chemical energy.
4. Much absorbed energy is used in respiration. It drives movement, active transport and keeping warm, and it is finally lost as heat. A warm-blooded animal loses a large share this way.

Producers lose further energy because most light reaching a plant is not used in photosynthesis. Energy lost as heat cannot be recaptured, so energy flows one way through a food chain.

Because a fraction is lost at every step, food chains are short: too little energy remains to support many levels.

Quantitative energy transfer

Efficiency of transfer between trophic levels:

efficiency (%) = (energy transferred to the next level ÷ energy available at the level below) × 100

Always divide by the lower level, which is the one the energy came from. Energy values are best given per unit area per year, for example kJ m⁻² year⁻¹, so that different ecosystems and levels can be compared.

Worked example 1. On a kelp forest, the energy stored per year is 12 000 kJ m⁻² in kelp (producer), 1 500 kJ m⁻² in sea urchins (primary consumer) and 180 kJ m⁻² in sea otters (secondary consumer).

1. Kelp to urchins: 1 500 ÷ 12 000 × 100 = 12.5%.
2. Urchins to otters: 180 ÷ 1 500 × 100 = 12.0%.
3. Overall, kelp to otters: 180 ÷ 12 000 × 100 = 1.5%. This is not 12.5% + 12.0%. Efficiencies multiply, they do not add.
4. Energy not passed from kelp to urchins: 12 000 - 1 500 = 10 500 kJ m⁻² year⁻¹.

Worked example 2. A crop stores 5 000 kJ in its edible parts. If it is fed to cattle, which store 8% of it, and people eat the beef, which holds 10% of the cattle’s energy as food for people, how much reaches people?

1. Energy in cattle: 5 000 × 0.08 = 400 kJ.
2. Energy reaching people: 400 × 0.10 = 40 kJ.

An extra level leaves far less energy for people, so shorter chains give more food per hectare.

Pyramids of number, biomass and energy

A pyramid shows the amount at each trophic level, with producers as the lowest bar. Bars are drawn to scale, each of equal height.

Pyramid	Measures	Typical shape	Limitation
Number	Individual organisms per area	Narrows upward, but can be inverted	Ignores body size
Biomass	Dry mass of living material per area at one moment	Usually narrows upward, can be inverted	A snapshot, not a rate
Energy	Energy transferred per area per year	Always narrows upward	Needs detailed data

Pyramid of number. One oak tree supports thousands of caterpillars, which are eaten by a few blue tits, which in turn feed one sparrowhawk. The first level is a single tree, so the pyramid is widest in the middle. A chain of parasites on one host is also inverted. Counting treats a tree and a caterpillar as equal, so size is ignored.

Pyramid of biomass. Biomass is mass of living material, measured as dry mass because the water content of fresh tissue varies. A biomass pyramid for a woodland is a clear pyramid. In some aquatic ecosystems it is inverted.

Worked example 3. In a stretch of open sea the standing biomass is 6 g m⁻² of phytoplankton and 21 g m⁻² of zooplankton, but the annual production is 840 g m⁻² of phytoplankton and 168 g m⁻² of zooplankton.

1. Standing biomass is inverted: 21 is greater than 6, so the zooplankton bar is wider.
2. The phytoplankton replace themselves rapidly. 6 ÷ 840 × 365 = 2.6 days, so a small stock is replaced very fast.
3. Zooplankton production (168) is 20% of phytoplankton production (840), so the pyramid of production is not inverted.
4. A biomass pyramid shows what is present at one time, not how fast it is replaced.

Pyramid of energy. This shows energy transferred in a given area over a given time, so it is a rate. It is never inverted over a full year, because the energy in each level comes from the level below and some is lost at every step. It is the most reliable of the three for comparing levels.

In a food web some species feed at several levels, so each organism must be assigned to a level before a pyramid can be drawn.

Gross and net primary production

Primary production is the energy fixed by producers.

• Gross primary production (GPP) is the chemical energy store in plant biomass in a given area or volume, in a given time.
• Net primary production (NPP) is the chemical energy store in plant biomass after respiratory loss (R) has been taken into account.

NPP = GPP - R

NPP is the energy available to primary consumers and decomposers. Units are usually kJ m⁻² year⁻¹ for land or kJ m⁻³ year⁻¹ for water.

Worked example 4. A grassland has a GPP of 9 800 kJ m⁻² year⁻¹ and plant respiration of 4 410 kJ m⁻² year⁻¹.

1. NPP = 9 800 - 4 410 = 5 390 kJ m⁻² year⁻¹.
2. Respiration as a share of GPP: 4 410 ÷ 9 800 × 100 = 45%.
3. So 55% of the energy fixed is stored as new biomass.

Rearranging. If NPP and GPP are given, R = GPP - NPP. If NPP and R are given, GPP = NPP + R.

Net production of consumers

A consumer cannot fix energy. Its net production is the chemical energy it stores as new body tissue and offspring.

N = I - (F + U + R)

where I is the chemical energy of the food ingested, F is the energy in faeces, U is the energy in urine and R is the energy lost in respiration. Everything not stored is lost in F, U or R.

Worked example 5. A pig ingests food containing 36 000 kJ per day. Its faeces contain 5 400 kJ and its urine 1 800 kJ. It loses 21 600 kJ in respiration.

1. F + U + R = 5 400 + 1 800 + 21 600 = 28 800 kJ.
2. N = 36 000 - 28 800 = 7 200 kJ per day.
3. Net production as a percentage of intake: 7 200 ÷ 36 000 × 100 = 20%.
4. Respiration took 21 600 ÷ 36 000 × 100 = 60% of the intake.

Respiration is the largest loss, so the quickest way to increase N is to reduce R.

Farming and efficiency of energy transfer

Farmers aim to maximise the energy that ends up in the crop or animal that people eat. The specification gives two approaches.

Reducing energy losses and simplifying food webs

• Chemical pesticides. Insecticides kill insects that eat crop tissue, herbicides kill weeds that compete with the crop for light, water and mineral ions, and fungicides kill fungal pathogens. More of the energy fixed stays in the crop, and the non-human food web is simpler.
• Biological agents. Use living organisms to reduce pests. One example is the parasitic wasp Encarsia formosa, which parasitises whitefly in glasshouses. Another is the bacterium Bacillus thuringiensis, whose toxin kills caterpillars. Pests are reduced without a chemical residue.
• Integrated systems. These combine methods, for example crop rotation, a pest-resistant variety, a biological agent and a targeted chemical only when pests pass a threshold. They cut losses while using fewer chemicals.

Reducing respiratory losses

In a food chain involving humans, the animal’s respiration is the main loss. Farmers reduce it by limiting movement, for example by housing animals indoors, and by keeping the animals warm so that less energy is used to maintain body temperature.

Worked example 6. A fish farm raises salmon with intake 25 000 kJ, faeces 3 250 kJ, urine 1 500 kJ and net production 5 250 kJ per fish over the growing period.

1. R = 25 000 - (3 250 + 1 500 + 5 250) = 15 000 kJ.
2. Calmer water cuts R by 15%: 15 000 × 0.85 = 12 750 kJ.
3. New N = 25 000 - (3 250 + 1 500 + 12 750) = 7 500 kJ.
4. Net production rises from 21.0% to 30.0% of intake, a 42.9% increase in N.

Arguing and evaluating

You are expected to apply ecological principles to explain how a practice affects productivity, and to judge data using three lenses.

• Economic: yield gain against cost, and price received.
• Environmental: effects on non-target species, pollution, resistance, introduced species.
• Ethical: animal welfare, and the balance between cheap food and the freedom to behave naturally.

Worked example 7. Cabbage is sold at £180 per tonne. Three treatments were compared, with no control giving 14 t per hectare.

Treatment	Yield (t ha⁻¹)	Cost (£ ha⁻¹)
Chemical spray	31	420
Biological agent	27	610
Integrated system	29	540

1. Extra revenue: chemical (31 - 14) × 180 = £3 060, biological £2 340, integrated £2 700.
2. Return per £ spent: chemical 3 060 ÷ 420 = 7.3, biological 3.8, integrated 5.0.
3. Chemical spray looks best economically.
4. Environmental effects are not in the table. Chemicals may harm non-target insects and encourage resistance, which may cut future yields.
5. A balanced answer says the chemical spray gives the best return, but the integrated system may be the better long-term choice, as it protects the wider ecosystem and still gives a high yield.

Common errors

• Dividing by the wrong level when finding efficiency. Divide by the energy that was available.
• Adding efficiencies for a multi-step chain. Multiply them.
• Saying “energy is used up”. Energy is transferred, and most ends as heat.
• Forgetting that urine carries energy in N = I - (F + U + R).
• Evaluating with one lens only. Name economic, environmental and ethical points.

Official syllabus

OxfordAQA International AS and A-level Biology, Version 5.1, for International A-level exams May/June 2018 onwards, sections 3.3.4.1 and 3.3.4.2.