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Interaction and interdependence - Transfer of energy and matter

Grade 12IBBiology

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

🔑Concepts

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Energy enters most ecosystems as sunlight, which is captured by autotrophs (producers) through photosynthesis and converted into chemical energy in carbon compounds such as C6H12O6C_{6}H_{12}O_{6}.

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Energy flows through food chains via feeding, moving from producers to primary consumers, then to secondary and tertiary consumers.

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Energy is lost from ecosystems as heat resulting from cellular respiration. This heat cannot be converted back into chemical energy, which is why energy flow is linear and not cyclical.

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The 10% Rule: Generally, only about 10%10\% of the energy available at one trophic level is passed to the next. The remaining 90%90\% is lost through heat, excretion, and unconsumed parts.

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Nutrients (matter) such as Carbon (CC), Nitrogen (NN), and Phosphorus (PP) are finite and must be recycled within an ecosystem through the action of saprotrophs and detritivores.

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Pyramids of energy represent the rate of energy flow. These pyramids are always upright because energy is lost at each successive trophic level. Units are typically kJ m−2 yr−1kJ \, m^{-2} \, yr^{-1}.

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Net Primary Productivity (NPPNPP) represents the actual amount of energy available to consumers after the producer has met its own metabolic needs via respiration (RR).

📐Formulae

NPP=GPP−RNPP = GPP - R

Trophic Efficiency=Energy at higher levelEnergy at lower level×100\text{Trophic Efficiency} = \frac{\text{Energy at higher level}}{\text{Energy at lower level}} \times 100

Energy Loss=Energy In−Energy Out\text{Energy Loss} = \text{Energy In} - \text{Energy Out}

💡Examples

Problem 1:

In a forest ecosystem, the Gross Primary Productivity (GPPGPP) is measured at 35,000 kJ m−2 yr−135,000 \, kJ \, m^{-2} \, yr^{-1}. If the respiration rate (RR) of the producers is 18,500 kJ m−2 yr−118,500 \, kJ \, m^{-2} \, yr^{-1}, calculate the Net Primary Productivity (NPPNPP).

Solution:

NPP=GPP−RNPP = GPP - R NPP=35,000−18,500=16,500 kJ m−2 yr−1NPP = 35,000 - 18,500 = 16,500 \, kJ \, m^{-2} \, yr^{-1}

Explanation:

The NPP is calculated by subtracting the energy used by the plants for their own biological processes (respiration) from the total energy they captured (GPP).

Problem 2:

Calculate the efficiency of energy transfer between a primary consumer level with 4,000 kJ m−2 yr−14,000 \, kJ \, m^{-2} \, yr^{-1} and a secondary consumer level with 320 kJ m−2 yr−1320 \, kJ \, m^{-2} \, yr^{-1}.

Solution:

Efficiency=3204,000×100\text{Efficiency} = \frac{320}{4,000} \times 100 Efficiency=0.08×100=8%\text{Efficiency} = 0.08 \times 100 = 8\%

Explanation:

The efficiency is the percentage of energy that successfully transfers from one trophic level to the next. In this case, 8%8\% of the energy from the primary consumers was transferred to the secondary consumers.

Problem 3:

If a food chain starts with 1,000,000 J1,000,000 \, J of sunlight energy, but producers only capture 1%1\% of that energy, and there is a 10%10\% transfer efficiency to primary consumers, how much energy reaches the primary consumer?

Solution:

Energy Captured by Producers=1,000,000×0.01=10,000 J\text{Energy Captured by Producers} = 1,000,000 \times 0.01 = 10,000 \, J Energy to Primary Consumers=10,000×0.10=1,000 J\text{Energy to Primary Consumers} = 10,000 \times 0.10 = 1,000 \, J

Explanation:

First, we calculate the energy stored by producers (1%1\% of solar energy). Then, we apply the 10%10\% transfer rule to find the energy available to the next level.