Calculating efficiency of biomass transfers
The efficiency of biomassThe dry mass of an organism. transfer is a measure of the proportion of biomass transferred from a lower trophic levelThe position of an organism in a food chain, food web or pyramid. to a higher one. Usually around ten per cent of biomass is transferred between trophic levels in a healthy ecosystemThe living organisms in a particular area, together with the non-living components of the environment. and the remaining 90 per cent is used by the organisms during life processesThe key reactions that all living organisms complete..
Calculating efficiency
This is an example of a food chain:
phytoplankton → zooplankton → herring → sea lion
What is the efficiency of this transfer?
To complete this calculation, we divide the amount from the higher trophic level (the zooplankton) by the amount from the lower trophic level (the phytoplankton) and multiply by one hundred. That is, we divide the smaller number by the bigger one (and multiply by one hundred).
\(\text{percentage efficiency transfer} = \frac{\text{biomass in higher trophic}}{\text{biomass in lower trophic level}} \times 100\)
\(\text{percentage efficiency transfer} = \frac{1.3~\text{kg}}{14.6~\text{kg}} \times 100 = 8.9\%\)
Question
The amount of biomass contained within each trophic level is shown in the table below.
| Trophic level | Organism | Total biomass (kg) |
| 1 | Phytoplankton | 14.6 |
| 2 | Zooplankton | 1.3 |
| 3 | Herring fish | 0.15 |
| 4 | Sea lions | 0.017 |
| Trophic level | 1 |
|---|---|
| Organism | Phytoplankton |
| Total biomass (kg) | 14.6 |
| Trophic level | 2 |
|---|---|
| Organism | Zooplankton |
| Total biomass (kg) | 1.3 |
| Trophic level | 3 |
|---|---|
| Organism | Herring fish |
| Total biomass (kg) | 0.15 |
| Trophic level | 4 |
|---|---|
| Organism | Sea lions |
| Total biomass (kg) | 0.017 |
Calculate the efficiency of the transfer of energy between each trophic level.
| Trophic level | Organism | Total biomass (kg) | Transfer efficiency (%) |
| 1 | Phytoplankton | 14.6 | No transfer |
| 2 | Zooplankton | 1.3 | 8.9 |
| 3 | Herring fish | 0.15 | 11.5 |
| 4 | Sea lions | 0.017 | 11.3 |
| Trophic level | 1 |
|---|---|
| Organism | Phytoplankton |
| Total biomass (kg) | 14.6 |
| Transfer efficiency (%) | No transfer |
| Trophic level | 2 |
|---|---|
| Organism | Zooplankton |
| Total biomass (kg) | 1.3 |
| Transfer efficiency (%) | 8.9 |
| Trophic level | 3 |
|---|---|
| Organism | Herring fish |
| Total biomass (kg) | 0.15 |
| Transfer efficiency (%) | 11.5 |
| Trophic level | 4 |
|---|---|
| Organism | Sea lions |
| Total biomass (kg) | 0.017 |
| Transfer efficiency (%) | 11.3 |
Question
A bullock has eaten 10 kg of biomass in the form of grass. It excreted 2.3 kg in the form of urine and gas and egested 4 kg in the form of faeces. The increase in mass of its body tissues is 0.4 kg.
a. How much biomass has been used up in respiration?
b. The bullock is eaten by humans. What is the efficiency of this transfer?
Input - The bullock ate 10 kg of grass.
Stored - The biomass stored in the body tissues was 0.4 kg
Lost - The total biomass lost by egestion and excretion was 2.3 + 4 = 6.3 kg. We need to calculate the biomass lost by respiration.
\(\text{Total input} = \text{total biomass stored} + \text{total biomass lost}\)
\(\text{biomass used in respiration} = 10 - 6.3 - 0.4 = 3.3~\text{kg}\)
Only 0.4 kg of the original biomass available to the bullock is available to humans for food. The efficiency of this transfer is:
\(\text{efficiency} = \frac{0.4}{10} \times 100 = 4%\)
