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5.3.8Energy Transfers In and Between Organisms (A Level only)

Calculating Productivity and Efficiency — AQA A Level Biology 7402 Revision Notes & Topical Questions

5 min read 8 parts Lesson 20 of 30 1 quiz

📚 AQA A Level Biology 7402 Hub / Topical Revision
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5.3.8 Calculating Productivity and Efficiency — AQA A Level Biology 7402 notes: Calculate percentage efficiency of energy transfer between trophic levels and net production values from data.The percentage efficiency of energy transfer between two trophic levels compares the energ…

  • Exam-style question 1
  • Exam-style question 2
  • Percentage change
  • Write the correct equation: NPP = GPP − R; N = I − (F + R)
  • Identify the quantities and units given; convert so all use the same energy and area units
  • Substitute and calculate; keep working, since marks are given for method

Part 1 of 6Energy transfer efficiency

The percentage efficiency of energy transfer between two trophic levels compares the energy stored at one level with the energy that was available from the level below:

efficiency (%) = (energy available after the transfer / energy available before the transfer) × 100

For example, if herbivores store 1 500 kJ m-2 year-1 and the producers' NPP was 15 000 kJ m-2 year-1, efficiency = 1 500 / 15 000 × 100 = 10%.

State clearly which values you use; examination questions often give gross values at some levels and net at others. Use the question's definition, normally the net production of the lower level as the denominator and the net production (or biomass) of the upper level as the numerator.

Part 2 of 6Step-by-step approach to calculations

1
Write the correct equation: NPP = GPP − R; N = I − (F + R).
2
Identify the quantities and units given; convert so all use the same energy and area units.
3
Substitute and calculate; keep working, since marks are given for method.
4
If a percentage is asked, divide by the original amount and multiply by 100.
5
Give the answer to an appropriate number of significant figures, with units.

In a food chain, always check that the energy at a level is less than at the level below. If not, a value has been misread. Percentage change is calculated as (final − initial) / initial × 100.

Part 3 of 6Interpreting differences

Data often show different efficiencies at different transfers. Typical explanations:

Interpreting differences
Producer to herbivore tends to be lower, around 5–10%, because plants contain cellulose and lignin that are indigestible and some tissue is not eaten.
Herbivore to carnivore is often higher, around 10–20%, as animal tissue is more digestible and a larger proportion of the carcass is eaten.
Endotherms are less efficient than ectotherms because of heat loss.

The overall efficiency from sunlight to consumer is very low. At the global scale, only about 0.1–1% of sunlight energy striking a field ends up in a harvested animal product.

Part 4 of 6Improving efficiency in practice

Understanding where energy is lost shows how to increase yield.

Improving efficiency in practice
Food chains in agriculture can be shortened by humans eating crops directly.
Losses to pests can be reduced with pesticides or biological control, losses to competition with weeds are reduced by herbicides,
  • and animal respiration can be reduced by limiting movement and warming housing.
These methods increase the proportion of energy that ends up in human food,
  • but they have ethical and environmental implications, discussed in the next lesson.
StepExampleEfficiency
Light to producerSunlight to NPPUsually below 2%
Producer to primary consumerGrass to cattleAbout 10%
Primary to secondary consumerHerbivore to foxAbout 10–20%

Part 5 of 6Further practice and common traps

Try a longer chain.

Further practice and common traps
Sunlight falling on a lake is 4.0 × 106 kJ m-2 year-1.
Phytoplankton NPP is 28 000.
  • Zooplankton net production is 3 300, small fish 410 and large fish 38.
  • Efficiencies
    • sunlight to NPP = 28 000 / 4 000 000 × 100 = 0.70%;
      • phytoplankton to zooplankton = 3 300 / 28 000 = 11.8%;
      • zooplankton to small fish = 410 / 3 300 = 12.4%;
      • small fish to large fish = 38 / 410 = 9.3%.
Notice how the first value is far lower than the others,
  • which is usually explained by reflection, wavelengths not absorbed, and energy lost as heat in photosynthesis.

Traps to avoid

  • using a different time period for the numerator and denominator;
    • mixing kJ and J;
    • quoting the figure to many more decimal places than the data justify;
    • and confusing efficiency with the amount of energy lost.

Further practice and common traps
If asked for the energy lost, subtract.
If asked for a ratio, divide.
In an answer that asks you to explain a difference between efficiencies, state the numbers and then give two biological reasons such as digestibility, proportion eaten,
  • and respiratory heat loss.

Part 6 of 6Exam-style questions

  1. In a field, NPP of grass is 9 500 kJ m-2 year-1. Rabbits have net production of 760 kJ m-2 year-1 and foxes 95 kJ m-2 year-1. Calculate the efficiency of energy transfer from grass to rabbits and from rabbits to foxes. [3]3 marks

  2. Suggest why the efficiency from rabbits to foxes is higher than from grass to rabbits. [2]2 marks

Key terms

TermMeaning
Efficiency of energy transferPercentage of energy at one trophic level that is passed to the next.
Percentage change(final minus initial) divided by initial, times 100.
CellulosePlant cell wall polysaccharide that is hard for most animals to digest.
EndothermAnimal that maintains body temperature using metabolic heat.
EctothermAnimal that relies on external heat sources for body temperature.

Avoid these mistakes

1
Dividing by the wrong level. The denominator is the energy at the lower, source level.
2
Mixing GPP and NPP values within one calculation.
3
Quoting the answer without a percentage sign or units.
4
Rounding too early and carrying a rounding error to the final answer.

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