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Organisms and their Environment - Energy flow and food webs

Grade 11A LevelBiology

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

🔑Concepts

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The Sun is the principal source of energy input to biological systems, where light energy is captured by producers during photosynthesis to create chemical energy in the form of C6H12O6C_6H_{12}O_6.

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Energy flow is linear and non-cyclical; it enters ecosystems as light and is eventually lost to the environment as heat energy.

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A food chain shows the transfer of energy from one organism to the next, beginning with a producer (Trophic Level 1).

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Food webs are networks of interconnected food chains, representing the complex feeding relationships within an ecosystem.

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Energy is lost at each trophic level through metabolic processes such as aerobic respiration (C6H12O6+6O2→6CO2+6H2O+energyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{energy}), excretion, and as heat.

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Only approximately 10%10\% of the energy from one trophic level is passed to the next, which limits the number of trophic levels in a food chain to usually no more than five.

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Pyramids of numbers represent the count of individuals at each level, while Pyramids of Biomass represent the dry mass of living material (g/m2g/m^2). Pyramids of Energy represent the total energy available (kJ/m2/yearkJ/m^2/year) and are always upright.

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Decomposers (fungi and bacteria) play a vital role by breaking down dead organic matter and returning nutrients to the soil, though the energy they extract is also eventually lost as heat.

📐Formulae

Efficiency of Energy Transfer=Energy available after transferEnergy available before transfer×100%\text{Efficiency of Energy Transfer} = \frac{\text{Energy available after transfer}}{\text{Energy available before transfer}} \times 100\%

6CO2+6H2O→light, chlorophyllC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{light, chlorophyll}} C_6H_{12}O_6 + 6O_2

C6H12O6+6O2→6CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}

💡Examples

Problem 1:

In a simple food chain: Grass →\rightarrow Grasshopper →\rightarrow Frog →\rightarrow Snake. If the Grass produces 50,000 kJ50,000\text{ kJ} of energy, calculate the energy available to the Snake, assuming a 10%10\% efficiency of transfer at each step.

Solution:

50 kJ50\text{ kJ}

Explanation:

Energy at Trophic Level 1 (Grass) = 50,000 kJ50,000\text{ kJ}. At Trophic Level 2 (Grasshopper), energy = 50,000×0.10=5,000 kJ50,000 \times 0.10 = 5,000\text{ kJ}. At Trophic Level 3 (Frog), energy = 5,000×0.10=500 kJ5,000 \times 0.10 = 500\text{ kJ}. At Trophic Level 4 (Snake), energy = 500×0.10=50 kJ500 \times 0.10 = 50\text{ kJ}.

Problem 2:

A field of wheat contains 8,000 kg8,000\text{ kg} of biomass. The locusts consuming the wheat have a total biomass of 800 kg800\text{ kg}. Calculate the efficiency of biomass transfer from the wheat to the locusts.

Solution:

10%10\%

Explanation:

Using the efficiency formula: 800 kg8,000 kg×100=10%\frac{800\text{ kg}}{8,000\text{ kg}} \times 100 = 10\%. This follows the general ecological rule that roughly 90%90\% of biomass/energy is lost between levels.