AQA A-level Biology: Energy Transfer and Productivity in Ecosystems (7402)

This guide teaches section 3.5.3, Energy and ecosystems, from the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015. The section is A-level only, so it is not part of the AS course. It sits in topic 5, which the specification assesses at A-level in the second written paper and the third. For the whole course use the AQA A-level Biology hub and the printable checklist. Companion pages: revision notes and practice questions. To find out where you stand first, try the free diagnostics. For wider exam technique see exam preparation.

What the section asks you to do

Specification idea	What you must be able to do
Biomass	Explain what biomass is and measure it as dry mass or mass of carbon per area
Calorimetry	Estimate the chemical energy store in dry biomass
GPP and NPP	Define both and use NPP = GPP - R
Consumers	Use N = I - (F + R)
Productivity	Give primary and secondary productivity as rates, for example kJ ha⁻¹ year⁻¹
Maths	Units (MS 0.1), percentage yields (MS 0.3), net productivity and efficiency of transfers (MS 2.4)
Farming	Appreciate how simplifying food webs and reducing respiratory losses raise efficiency

From carbon dioxide to biomass

In any ecosystem, plants synthesise organic compounds from carbon dioxide, taken from the atmosphere or, in water, from the water. Most of the sugars made are used by the plant as respiratory substrates. The remainder is used to make other biological molecules, such as cellulose, proteins and lipids. These molecules make up the biomass of the plant.

Biomass can be measured in two ways. One is the mass of carbon in the tissue. The other is the dry mass of tissue per given area. Dry mass is used because fresh tissue contains a variable amount of water, which holds no useful energy and changes with the weather.

Finding dry mass

1. Weigh a fresh sample from a measured area.
2. Dry it in an oven at a temperature that removes the water without burning the tissue.
3. Re-weigh at intervals until two readings are the same. This is constant mass, and it shows that all water has gone.
4. Divide the dry mass by the sampled area.

Worked example 1. A 250 g fresh sample is cut from a 0.50 m² plot. After drying to constant mass it weighs 62 g.

1. Water lost = 250 - 62 = 188 g, which is 188 ÷ 250 × 100 = 75.2% of the fresh mass.
2. Dry mass per unit area = 62 ÷ 0.50 = 124 g m⁻².
3. Convert to hectares. 1 ha = 10 000 m², so 124 × 10 000 = 1 240 000 g ha⁻¹ = 1 240 kg ha⁻¹.

Because the sample is destroyed, a small sample is taken from many places, and the result is scaled up. A single quadrat may not represent the field.

Calorimetry: the energy in biomass

The chemical energy store in dry biomass can be estimated by calorimetry. A weighed portion of dried tissue is burned, and the heat released warms a known mass of water. The energy transferred to the water is:

energy (J) = mass of water (g) × specific heat capacity of water × temperature rise (°C)

Use 4.18 J g⁻¹ °C⁻¹ for water. Dividing by the mass of tissue burned gives energy per gram of dry biomass.

Worked example 2. 0.50 g of dried plant tissue is burned beneath 50.0 g of water. The water warms from 20.0 °C to 54.0 °C.

1. Temperature rise = 54.0 - 20.0 = 34.0 °C.
2. Energy = 50.0 × 4.18 × 34.0 = 7 106 J.
3. Energy per gram = 7 106 ÷ 0.50 = 14 212 J g⁻¹ = 14.2 kJ g⁻¹ (3 significant figures).

The value is usually an underestimate. Some heat escapes to the air and to the apparatus, and combustion may be incomplete. Insulating the water, shielding the flame and stirring the water improve the result.

Linking the two methods. Suppose a crop makes 1 380 g m⁻² of new dry biomass in a year. At 14.2 kJ g⁻¹, that is 1 380 × 14.2 = 19 596 kJ m⁻² year⁻¹, and 19 596 × 10 000 = 1.96 × 10⁸ kJ ha⁻¹ year⁻¹ (3 significant figures). This is how a mass of tissue becomes an energy store per unit area.

Gross and net primary production

Gross primary production (GPP) is the chemical energy store in plant biomass, in a given area or volume. It is the total amount of energy that producers fix in organic compounds. Plants then use part of that energy in respiration, and lose it to the environment, mostly as heat. Net primary production (NPP) is the chemical energy store in plant biomass after respiratory losses to the environment have been taken into account.

NPP = GPP - R

R is the respiratory loss. The same relationship can be turned round: GPP = NPP + R, or R = GPP - NPP. NPP is the energy available for plant growth and reproduction. It is also available to other trophic levels, such as herbivores and decomposers. It is the starting point for the energy that passes along every food chain in the ecosystem.

Worked example 3. A field has GPP of 4.80 × 10⁸ kJ ha⁻¹ year⁻¹ and respiratory loss of 1.68 × 10⁸ kJ ha⁻¹ year⁻¹.

1. NPP = 4.80 × 10⁸ - 1.68 × 10⁸ = 3.12 × 10⁸ kJ ha⁻¹ year⁻¹.
2. NPP as a percentage of GPP = 3.12 ÷ 4.80 × 100 = 65.0%.
3. If the field receives 3.8 × 10¹⁰ kJ ha⁻¹ year⁻¹ of light energy, GPP is 4.80 × 10⁸ ÷ 3.8 × 10¹⁰ × 100 = 1.26% of the light received. This is a percentage yield (MS 0.3), and it shows that only a small fraction of incident light ends up in plant biomass.

Net production of consumers

Animals cannot photosynthesise. Their net production is what is left of their food energy after losses:

N = I - (F + R)

• I is the chemical energy store in ingested food.
• F is the chemical energy lost to the environment in faeces and urine.
• R is the respiratory loss to the environment.

N is the energy stored in new consumer biomass, which is available for growth and reproduction and for the next trophic level. Note that F and R are both losses; N is not the same as I - F.

Worked example 4. Cattle on a pasture ingest 2.40 × 10⁶ kJ ha⁻¹ year⁻¹. They lose 0.84 × 10⁶ kJ in faeces and urine and 1.15 × 10⁶ kJ in respiration.

1. F + R = 0.84 × 10⁶ + 1.15 × 10⁶ = 1.99 × 10⁶ kJ ha⁻¹ year⁻¹.
2. N = 2.40 × 10⁶ - 1.99 × 10⁶ = 4.1 × 10⁵ kJ ha⁻¹ year⁻¹.
3. N ÷ I × 100 = 4.1 × 10⁵ ÷ 2.40 × 10⁶ × 100 = 17.1%.

Rearranging. If I = 5.0 × 10⁶, F = 1.7 × 10⁶ and N = 6.5 × 10⁵ (all in kJ ha⁻¹ year⁻¹), then R = I - F - N = 5.0 × 10⁶ - 1.7 × 10⁶ - 0.65 × 10⁶ = 2.65 × 10⁶, so R = 2.65 × 10⁶ kJ ha⁻¹ year⁻¹.

Productivity and its units

Production is an amount stored. Productivity is the rate of production. Primary productivity is the rate of primary production, and secondary productivity is the rate of secondary production. Both are measured as biomass in a given area in a given time, for example kJ ha⁻¹ year⁻¹. For an aquatic habitat, the specification allows the store to be given for a volume, so kJ m⁻³ year⁻¹ makes sense.

Deriving a unit is a maths skill (MS 0.1). Energy per area per time gives kJ ha⁻¹ year⁻¹. Dry mass per area per time gives g m⁻² year⁻¹. Converting is a matter of scaling the area and the time.

Worked example 5. An estuary has NPP of 6 800 kJ m⁻² year⁻¹. In kJ ha⁻¹ year⁻¹ that is 6 800 × 10 000 = 6.8 × 10⁷. Going the other way, 6.8 × 10⁷ ÷ 10 000 = 6 800 kJ m⁻² year⁻¹. A rate per day would be divided by 365: 6 800 ÷ 365 = 18.6 kJ m⁻² day⁻¹.

Efficiency of energy transfer

The efficiency of a transfer compares what is stored in the next level with what was available:

efficiency (%) = energy stored in the next level ÷ energy available from the level below × 100

Both values must be in the same units and refer to the same area and time.

Worked example 6. In a grassland, NPP is 6.40 × 10⁶ kJ ha⁻¹ year⁻¹. Primary consumers have net production of 5.76 × 10⁵. Secondary consumers have 4.70 × 10⁴.

1. Producers to primary consumers = 5.76 × 10⁵ ÷ 6.40 × 10⁶ × 100 = 9.0%.
2. Primary to secondary consumers = 4.70 × 10⁴ ÷ 5.76 × 10⁵ × 100 = 8.16%.
3. Overall, producers to secondary consumers = 4.70 × 10⁴ ÷ 6.40 × 10⁶ × 100 = 0.734%.

The overall figure is lower than either step, because losses compound.

Farming and the efficiency of energy transfer

Farming aims to put as much of the energy captured by plants as possible into food that humans eat. The specification names two approaches.

Simplifying food webs to reduce energy losses to non-human food chains. Herbicides kill weeds that would compete with the crop for light, water and ions. Pesticides kill insects that would eat the crop. Energy that would have been stored in weeds and pests stays in the crop, so more of the NPP ends up in a human food chain.

Reducing respiratory losses within a human food chain. Livestock kept in pens or barns move less and, if the building is warm, use less energy to keep their body temperature constant. Less respiration means a bigger N, so more energy goes into growth. Feeding animals a high-energy diet and slaughtering them when growth slows also help.

Worked example 7. Per animal over a rearing period, compare two systems (energy in MJ).

System	I	F	R
Indoors, restricted movement	74.0	14.8	49.2
Free-range	88.0	17.6	63.4

Indoors: N = 74.0 - (14.8 + 49.2) = 10.0 MJ, and 10.0 ÷ 74.0 × 100 = 13.5%. Free-range: N = 88.0 - (17.6 + 63.4) = 7.0 MJ, and 7.0 ÷ 88.0 × 100 = 7.95%.

The indoor system converts a larger share of its food into biomass because R is a smaller part of I. These practices have costs: lower animal welfare, loss of habitat and wildlife where weeds and insects are removed, and possible pollution. A good evaluation states both the gain in efficiency and the cost.

Common errors

• Writing NPP = GPP + R. Respiration is a loss, so it is subtracted.
• Writing N = I - F, or adding R to the other losses twice.
• Mixing units, such as kJ m⁻² with kJ ha⁻¹, in one efficiency calculation.
• Leaving off the time unit. A rate needs “year⁻¹” or similar.
• Using fresh mass as biomass. Biomass is measured as dry mass.
• Saying energy is “used up”. It is transferred to the environment, mostly as heat.
• Saying herbicides “kill the food chain”. Say that they remove competing plants, so more NPP is stored in the crop.
• Giving one-sided answers on farming. State the efficiency gain and a cost.

Official syllabus

AQA, AS and A-level Biology specification (7401 and 7402), first teaching 2015, section 3.5.3 Energy and ecosystems (A-level only), with mathematical skills MS 0.1, MS 0.3 and MS 2.4.