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Ecology and Conservation - Energy Pyramids

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An Energy Pyramid is a graphical representation designed to show the biomass or bio-productivity at each trophic level in a given ecosystem.

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Energy enters most ecosystems as sunlight and is converted into chemical energy by producers (autotrophs) through the process of photosynthesis: 6CO2+6H2O+light energy→C6H12O6+6O26CO_2 + 6H_2O + \text{light energy} \rightarrow C_6H_{12}O_6 + 6O_2.

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The 10% Rule: On average, only about 10%10\% of the energy stored as biomass in one trophic level is passed on to the next level.

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Energy is measured in Joules (JJ) or Kilojoules (kJkJ). As energy moves up the pyramid, the total amount of available energy decreases.

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Energy loss occurs primarily through metabolic heat (QQ), movement, growth, and waste products that are not consumed by the next level.

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Trophic levels are organized from the bottom up: Producers (Level 1), Primary Consumers (Level 2), Secondary Consumers (Level 3), and Tertiary Consumers (Level 4).

📐Formulae

Eavailable=Elower_level×0.10E_{available} = E_{lower\_level} \times 0.10

Energy Efficiency=(Energy at current levelEnergy at previous level)×100%\text{Energy Efficiency} = \left( \frac{\text{Energy at current level}}{\text{Energy at previous level}} \right) \times 100\%

Elost=Etotal−EtransferredE_{lost} = E_{total} - E_{transferred}

💡Examples

Problem 1:

In a local grassland ecosystem, the total energy produced by the plants (producers) is calculated to be 20,000 kJ20,000\text{ kJ}. According to the 10%10\% rule, how much energy is available for the secondary consumers (e.g., snakes that eat mice)?

Solution:

Step 1: Calculate energy for Primary Consumers: 20,000 kJ×0.10=2,000 kJ20,000\text{ kJ} \times 0.10 = 2,000\text{ kJ}. Step 2: Calculate energy for Secondary Consumers: 2,000 kJ×0.10=200 kJ2,000\text{ kJ} \times 0.10 = 200\text{ kJ}.

Explanation:

At each step, we multiply the available energy by 0.100.10 (or divide by 1010). The secondary consumers receive only 200 kJ200\text{ kJ} from the original 20,000 kJ20,000\text{ kJ} available at the base.

Problem 2:

If a Tertiary Consumer level contains only 5 J5\text{ J} of energy, how much energy was likely present at the Producer level, assuming a perfect 10%10\% transfer rate at each of the three steps?

Solution:

Eproducer=5 J×10×10×10=5,000 JE_{producer} = 5\text{ J} \times 10 \times 10 \times 10 = 5,000\text{ J}

Explanation:

To find the energy at a lower level, we perform the inverse operation. Since there are three levels of separation (Producer →\rightarrow Primary →\rightarrow Secondary →\rightarrow Tertiary), we multiply by 1010 three times.