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Ecology - Energy flow

Grade 12IBBiology

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

🔑Concepts

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Most ecosystems rely on a supply of energy from sunlight. Photoautotrophs (producers) convert light energy into chemical energy in carbon compounds such as C6H12O6C_6H_{12}O_6 through photosynthesis.

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Energy flows through food chains by means of feeding. When consumers ingest producers or other consumers, the chemical energy stored in carbon compounds is transferred.

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Energy is lost from ecosystems in several ways: through cellular respiration as heat, as excretion (e.g., urea), or as unconsumed material (e.g., bones or hair).

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Living organisms cannot convert heat energy back into other forms of energy (such as light or chemical energy). Consequently, energy is not recycled and must be continuously supplied by the sun.

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The inefficiency of energy transfer (typically only 10%10\% is passed to the next level) limits the length of food chains and the biomass of higher trophic levels.

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Pyramids of energy represent the flow of energy over time. The units used are energy per unit area per unit time: kJ m−2 yr−1kJ \, m^{-2} \, yr^{-1}.

📐Formulae

NPP=GPP−RNPP = GPP - R

Trophic Efficiency=Energy at trophic level n+1Energy at trophic level n×100\text{Trophic Efficiency} = \frac{\text{Energy at trophic level } n+1}{\text{Energy at trophic level } n} \times 100

C=P+R+U+FC = P + R + U + F

💡Examples

Problem 1:

In a forest ecosystem, the primary producers generate 30,000 kJ m−2 yr−130,000 \, kJ \, m^{-2} \, yr^{-1} of energy. Calculate the energy available to the tertiary consumers, assuming an average energy transfer efficiency of 10%10\% at each trophic level.

Solution:

  1. Primary Consumers (Level 2): 30,000×0.10=3,000 kJ m−2 yr−130,000 \times 0.10 = 3,000 \, kJ \, m^{-2} \, yr^{-1} \ 2. Secondary Consumers (Level 3): 3,000×0.10=300 kJ m−2 yr−13,000 \times 0.10 = 300 \, kJ \, m^{-2} \, yr^{-1} \ 3. Tertiary Consumers (Level 4): 300×0.10=30 kJ m−2 yr−1300 \times 0.10 = 30 \, kJ \, m^{-2} \, yr^{-1}.

Explanation:

Energy transfer is calculated by multiplying the energy of the previous trophic level by the efficiency rate (expressed as a decimal, 0.100.10 for 10%10\%). Note that the energy decreases by a power of 10 at each step.

Problem 2:

If a plant has a Gross Primary Productivity (GPPGPP) of 120 mg O2 L−1 h−1120 \, mg \, O_2 \, L^{-1} \, h^{-1} and its Respiration (RR) rate is 45 mg O2 L−1 h−145 \, mg \, O_2 \, L^{-1} \, h^{-1}, find the Net Primary Productivity (NPPNPP).

Solution:

NPP=120−45=75 mg O2 L−1 h−1NPP = 120 - 45 = 75 \, mg \, O_2 \, L^{-1} \, h^{-1}

Explanation:

The NPPNPP represents the actual energy (or biomass) available to consumers after the producer has met its own metabolic needs through respiration.

Energy flow - Revision Notes & Key Diagrams | IB Grade 12 Biology