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4.1.10 Efficiency of biomass transfers

Efficiency measures biomass transfer

Definition

Biomass transfer efficiency

Biomass transfer efficiency is the percentage of biomass available at one trophic level that becomes new biomass at the next trophic level.

  1. Only part of the biomass available at one trophic level is converted into new biomass at the next level.
  2. Transfer efficiency describes how large that transferred part is compared with the starting amount.
  3. For a biomass transfer, use efficiency=biomass transferred to the next levelbiomass available at the previous level×100%\text{efficiency}=\dfrac{\text{biomass transferred to the next level}}{\text{biomass available at the previous level}}\times100\%efficiency=biomass available at the previous levelbiomass transferred to the next level​×100%.
  4. Place the amount transferred in the numerator and the starting amount in the denominator.
  5. Use quantities measured on a comparable basis, including the same area and time period where those are specified.
  6. The masses must have the same units before division, so convert units first if necessary.
  7. Because identical units cancel in the ratio, an efficiency is given as a percentage, not in grams or kilograms.
Example
  • Of 800 kg800\,\mathrm{kg}800kg of producer biomass available over a stated period, 120 kg120\,\mathrm{kg}120kg becomes new primary-consumer biomass.
  • The transferred fraction is 120800=0.15\dfrac{120}{800}=0.15800120​=0.15.
  • The percentage transfer efficiency is 0.15×100%=15%0.15\times100\%=15\%0.15×100%=15%.
  • The amount not transferred into that consumer biomass is 800−120=680 kg800-120=680\,\mathrm{kg}800−120=680kg.
  • The percentage not transferred is 100%−15%=85%100\%-15\%=85\%100%−15%=85%.

Use the supplied values

  1. The often-used figure of about 10%10\%10% is an illustration, not a fixed value that applies to every ecosystem or feeding step.
  2. Differences in food digestibility, the proportion eaten and the amount used in respiration can produce different transfer efficiencies.
  3. An efficiency for one step cannot automatically be used for another step unless the question states that they are equal.
  4. Use the relevant data to calculate each transfer rather than replacing the data with an assumed percentage.
Common Mistake

Do not invert the fraction: dividing the larger starting biomass by the smaller transferred biomass gives a result above 100%100\%100% and does not calculate the transfer efficiency.

Convert units before dividing

  1. A mass stated in kilograms can be converted to grams using 1 kg=1000 g1\,\mathrm{kg}=1000\,\mathrm{g}1kg=1000g.
  2. Writing the conversion before the substitution makes it clear which quantities are being compared.
  3. Keep unrounded values during intermediate steps, then round the final result to the precision requested.
Example
  • A transfer produces 450 g450\,\mathrm{g}450g of new consumer biomass from 3.0 kg3.0\,\mathrm{kg}3.0kg of available food biomass.
  • Convert the starting mass: 3.0 kg=3000 g3.0\,\mathrm{kg}=3000\,\mathrm{g}3.0kg=3000g.
  • Substitute comparable masses: efficiency=4503000×100%=15%\text{efficiency}=\dfrac{450}{3000}\times100\%=15\%efficiency=3000450​×100%=15%.
  • The same result follows from converting 450 g450\,\mathrm{g}450g to 0.450 kg0.450\,\mathrm{kg}0.450kg and calculating 0.4503.0×100%\dfrac{0.450}{3.0}\times100\%3.00.450​×100%.

Calculate an unknown biomass

  1. To calculate the biomass transferred, multiply the starting biomass by the efficiency expressed as a decimal fraction.
  2. Use biomass transferred=starting biomass×percentage efficiency100\text{biomass transferred}=\text{starting biomass}\times\dfrac{\text{percentage efficiency}}{100}biomass transferred=starting biomass×100percentage efficiency​.
  3. To calculate the starting biomass required, rearrange to starting biomass=biomass transferredpercentage efficiency/100\text{starting biomass}=\dfrac{\text{biomass transferred}}{\text{percentage efficiency}/100}starting biomass=percentage efficiency/100biomass transferred​.
  4. A percentage must be divided by 100100100 before it is used as a multiplier, so 12%12\%12% becomes 0.120.120.12.
Example
  • With a starting biomass of 250 kg250\,\mathrm{kg}250kg and an efficiency of 12%12\%12%, the transferred biomass is 250×0.12=30 kg250\times0.12=30\,\mathrm{kg}250×0.12=30kg.
  • If 30 kg30\,\mathrm{kg}30kg is the transferred biomass and the efficiency is 12%12\%12%, the starting biomass is 300.12=250 kg\dfrac{30}{0.12}=250\,\mathrm{kg}0.1230​=250kg.
  • The reverse calculation gives a larger starting mass because only part of it was transferred.

Repeated transfers reduce available biomass

  1. Each additional feeding step retains only a fraction of the biomass available at the previous level.
  2. For successive transfers, multiply the fractional efficiencies rather than adding the percentage efficiencies.
  3. This means that the overall efficiency across several steps is smaller than the efficiency of an individual step.
Example
  • Suppose 2000 kg2000\,\mathrm{kg}2000kg of producer biomass supplies a first transfer with efficiency 20%20\%20% and a second transfer with efficiency 15%15\%15%.
  • The new primary-consumer biomass is 2000×0.20=400 kg2000\times0.20=400\,\mathrm{kg}2000×0.20=400kg.
  • The new secondary-consumer biomass is 400×0.15=60 kg400\times0.15=60\,\mathrm{kg}400×0.15=60kg.
  • The overall transfer from producers to secondary consumers is 602000×100%=3%\dfrac{60}{2000}\times100\%=3\%200060​×100%=3%.
  • Equivalently, 0.20×0.15=0.030.20\times0.15=0.030.20×0.15=0.03, which is 3%3\%3%, not 35%35\%35%.

Energy and biomass data

  1. An energy-transfer efficiency uses the same ratio method, but both quantities must be energy values, such as values measured in kJ\mathrm{kJ}kJ.
  2. Do not divide a biomass value in g\mathrm{g}g by an energy value in kJ\mathrm{kJ}kJ and label the result a percentage transfer efficiency.
  3. The retained energy-pyramid diagram shows the declining energy supply through feeding levels and supports the explanation for limited food-chain length.
  4. A snapshot of biomass and a measured transfer over time describe different things, so interpret the quantities exactly as the data labels specify.
  5. When efficiencies are positive but below 100%100\%100%, energy and biomass available through that feeding route become progressively smaller, rather than suddenly disappearing at a fixed level.

Transfer losses limit food-chain length

  1. Biomass contains chemical energy that consumers need for maintenance, movement, growth and reproduction.
  2. At each feeding step, material fails to become new consumer biomass through the routes explained previously, leaving less biomass and associated energy for another level.
  3. Eventually the supply reaching a further level may be too small to maintain a viable population of consumers.
  4. This helps explain why food chains usually contain only a few trophic levels, rather than extending indefinitely.
  5. There is no single compulsory maximum number of levels: the amount produced at the base and the efficiencies of the transfers both affect what can be supported.
  6. Large top predators often occur at low densities and range over large areas to obtain enough prey.
  7. The percentages in this illustration show one possible sequence of transfers and should not replace measured values in a calculation.

A trophic level pyramid showing producers at the base (100% energy) followed by primary, secondary, tertiary, and quaternary consumers. The diagram illustrates the 10% rule of energy transfer, showing a decrease in biomass and energy at each higher level, with energy lost as heat.

Exam technique
  • To explain a limit on food-chain length, link repeated inefficient transfers to too little biomass and energy reaching a further level to support a population.
  • For a calculation, identify the two levels being compared, show the substituted values and state the final percentage.
Self review
  • Which biomass belongs in the denominator of a transfer-efficiency calculation?
  • Why must the two masses use the same units?
  • How is a percentage converted into a multiplier?
  • Why are successive transfer efficiencies multiplied rather than added?
  • How do inefficient transfers limit the length of a food chain?
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Biomass transfer efficiency is the percentage of biomass available at one trophic level that becomes new biomass at the next trophic level. The starting biomass belongs in the denominator, while the biomass transferred belongs in the numerator.

efficiency=biomass transferred to the next levelbiomass available at the previous level×100% \text{efficiency}=\frac{\text{biomass transferred to the next level}}{\text{biomass available at the previous level}}\times100\% efficiency=biomass available at the previous levelbiomass transferred to the next level​×100%

Both biomass values must cover a comparable area and time period and must use the same units. The units cancel during division, so efficiency is reported as a percentage rather than a mass.

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Biomass transfer efficiency is the percentage of available biomass that becomes [     ].

4.1.10 Efficiency of biomass transfers Revision Guide

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Revision notes for OCR GCSE Biology 4.1.10 Efficiency of biomass transfers: explanations and worked examples.

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