In a marine estuary the recorded energy contents are: phytoplankton $8000{\text{kJm}}^{-2}$, zooplankton $600{\text{kJm}}^{-2}$, small fish $120{\text{kJm}}^{-2}$, large fish $24{\text{kJm}}^{-2}$. (a) Calculate the overall ecological efficiency from phytoplankton to large fish as a percentage. (b) State whether this overall efficiency follows the typical 10% rule.

Step-by-step solution

1. Compute efficiency of each trophic transfer:
2. Phytoplankton → zooplankton: $\frac{600}{8000}\times 100=7.5\%$.
3. Zooplankton → small fish: $\frac{120}{600}\times 100=20\%$.
4. Small fish → large fish: $\frac{24}{120}\times 100=20\%$.
5. Overall efficiency = product of the three percentages divided by 100 twice (or ratio of final to initial energy).
6. Overall = $\frac{24}{8000}\times 100=0.3\%$.
7. Alternatively, multiply step efficiencies as decimals: $0.075\times 0.20\times 0.20=0.003\to 0.3\%$.
8. Rounded to two significant figures, overall efficiency ≈ $0.30\%$ (or $0.375\%$ if using exact numbers).
9. Compare with the 10% rule: $0.30\%$ is much lower, indicating a less efficient energy transfer in this estuary.

Concept tested: Ecological pyramids and energy efficiency

Published 25 September 2026 · How this solution was prepared and checked