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Ecology - Energy Flow

Grade 11IBBiology

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, which is captured by chlorophyll in producers (autotrophs) through the process of photosynthesis: 6CO2+6H2O→lightC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{light} C_6H_{12}O_6 + 6O_2.

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Light energy is converted into chemical energy in carbon compounds (such as glucose, lipids, and proteins). This energy flows through the food chain by means of feeding.

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Energy released from carbon compounds by respiration is used in living organisms and converted to heat. The general equation for aerobic respiration is C6H12O6+6O2→6CO2+6H2O+ATPC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + ATP.

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Living organisms cannot convert heat energy into any other form of energy (e.g., they cannot turn heat back into chemical energy).

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Energy is lost from ecosystems in the form of heat, which is eventually radiated into the atmosphere and then into space.

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Energy losses between trophic levels restrict the length of food chains and the biomass of higher trophic levels. Approximately 90%90\% of energy is lost at each level.

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Pyramids of energy represent the amount of energy converted to new biomass at each trophic level and are expressed in units of kJ⋅m−2⋅yr−1kJ \cdot m^{-2} \cdot yr^{-1}.

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Nutrients (such as CC, NN, and PP) are recycled within an ecosystem, whereas energy is not recycled; it flows through and is eventually lost.

📐Formulae

Net Primary Production (NPP)=Gross Primary Production (GPP)−Respiration (R)Net\ Primary\ Production\ (NPP) = Gross\ Primary\ Production\ (GPP) - Respiration\ (R)

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

Unit of Energy Flow=kJ⋅m−2⋅yr−1\text{Unit of Energy Flow} = kJ \cdot m^{-2} \cdot yr^{-1}

💡Examples

Problem 1:

In a specific grassland ecosystem, the primary producers generate 20,000 kJ⋅m−2⋅yr−120,000\ kJ \cdot m^{-2} \cdot yr^{-1}. If the primary consumers receive 2,000 kJ⋅m−2⋅yr−12,000\ kJ \cdot m^{-2} \cdot yr^{-1}, calculate the percentage efficiency of energy transfer between these two trophic levels.

Solution:

Efficiency=2,00020,000×100=10%Efficiency = \frac{2,000}{20,000} \times 100 = 10\%

Explanation:

To find the efficiency, divide the energy at the higher trophic level by the energy at the lower trophic level and multiply by 100100 to get the percentage.

Problem 2:

Explain why the total biomass of a tertiary consumer (e.g., an Aquila chrysaetosAquila\ chrysaetos or Golden Eagle) is significantly lower than the biomass of the producers in its ecosystem.

Solution:

Energy is lost as heat during cellular respiration, and some biomass is not consumed (bones, fur) or is lost as waste (feces/urea). Since only ≈10%\approx 10\% of energy is transferred, there is insufficient energy to support a large biomass at the 4th4^{th} trophic level.

Explanation:

Energy flow is inefficient. As energy is lost at every step, the total available energy at the top of the pyramid is a small fraction of the original 100%100\% captured by producers, limiting the number of individuals and total biomass that can be sustained.

Energy Flow Grade 11 Notes & Examples | IB Biology