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 ninety per cent is used by the organisms during life processesThe key reactions that all living organisms complete..
This is an example of a food chain:
phytoplankton 鈫 zooplankton 鈫 herring 鈫 sea lion
To complete this calculation, we divide the amount from the higher trophic level by the amount from the lower trophic level and multiply by one hundred. That is, we divide the smaller number by the bigger one (and multiply by one hundred).
\(\text{Percentage efficiency transfer = }\frac{\text{biomass in higher trophic level}}{\text{biomass in lower trophic level}}\times 100\)
\(\text{Percentage efficiency transfer = }\frac{\text{1.3 kg}}{\text{14.6 kg}}\times 100 = 8.9\%\)
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 |
Question
Calculate the efficiency of the transfer of energy between each trophic level.
| Trophic level | Organism | Total biomass (kg) | Percentage biomass transferred |
| 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 |
| Percentage biomass transferred | No transfer |
| Trophic level | 2 |
|---|---|
| Organism | Zooplankton |
| Total biomass (kg) | 1.3 |
| Percentage biomass transferred | 8.9 |
| Trophic level | 3 |
|---|---|
| Organism | Herring fish |
| Total biomass (kg) | 0.15 |
| Percentage biomass transferred | 11.5 |
| Trophic level | 4 |
|---|---|
| Organism | Sea lions |
| Total biomass (kg) | 0.017 |
| Percentage biomass transferred | 11.3 |
Calculating efficiency
Question
A bullock has eaten 10 kg of biomass in the form of grass, and excreted 6.3 kg in the form of faeces, urine and gas. The increase in its body tissues is 0.4 kg. So how much biomass has been used up in respiration?
\(\text{The biomass used in respiration = }\text{10 - 6.3 - 0.4 = 3.3 kg}\)
Only 0.4 kg of the original biomass available to the bullock is available to the next stage, which might be humans. The efficiency of this transfer is:
\(\text{Efficiency = }\frac{\text{0.4}}{\text{10}}\times 100 = 4\%\)