# On a small island, there is a close relationship between the various organisms of the five trophic levels.

**On a small island, there is a close relationship between the various organisms of the five trophic levels. The only source of energy is sunlight with a photosynthesis efficiency of 1.5%. Determine the total annual energy of the Sun entering this ecosystem. It is known that the predators of the second order of this community can feed 45 parasites with a mass of 0.2 g each, and for 1 kg of their mass there is 5 * 10 ^ 4 kJ of energy.**

1) Let’s make a diagram of the food chain of five trophic levels:

producers (organic producers) – “consumers of the 1st

order -> predators of the 1st order (consumers of the 2nd order) – »predators of the 2nd order (consumers of the 3rd order) -> parasites (consumers of the 4th order).

2) We calculate the mass of all parasites: 0.2 g x 45 = 9 g.

3) Let’s calculate how much energy is contained in the bodies of all parasites:

1000 g (1 kg) contains 5 • 104 kJ 9 g – x kJ x = 450 kJ.

3) Taking into account Lindemann’s rule (10%), let’s calculate how much energy must be stored at the first trophic level to feed the entire ecosystem:

450 kJ • 10 • 10 • 10 • 10 = 4,500,000 kJ = 4.5 • 10 ^ 6 kJ.

The obtained result is only 1.5% (according to the condition of the problem, the efficiency of photosynthesis = 1.5%).

4) We find the total annual energy of the Sun entering this ecosystem:

4.5 • 10 ^ 6 kJ corresponds to 1.5% x – 100% x = 3 • 10 ^ 8 kJ.

Answer. 3 • 10 ^ 8 kJ.