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Ecology and Conservation - Food Chains and Food Webs

Grade 6IB

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

🔑Concepts

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A food chain is a linear sequence of organisms through which nutrients and energy pass as one organism eats another. The direction of the arrow →\rightarrow represents the flow of energy from the prey to the predator.

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Producers (Autotrophs) form the base of every food chain. They convert solar energy into chemical energy via 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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Consumers (Heterotrophs) are organisms that eat other organisms. They are classified as Primary Consumers (Herbivores), Secondary Consumers (Carnivores/Omnivores), and Tertiary Consumers (Top Predators).

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Trophic Levels describe the position of an organism in a food chain. T1T_1 represents producers, T2T_2 primary consumers, T3T_3 secondary consumers, and so on.

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Food Webs are complex networks of interconnected food chains within an ecosystem, showing that most organisms eat more than one type of food and are eaten by more than one type of predator.

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The 10% Rule states that only about 10%10\% of the energy stored in organic matter at one trophic level is transferred to the next level. The remaining 90%90\% is lost as heat during metabolic processes like respiration: C6H12O6+6O2→6CO2+6H2O+ATP (Energy)+HeatC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{ATP (Energy)} + \text{Heat}.

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Decomposers (e.g., fungi and bacteria) break down dead organic matter, returning essential nutrients like Nitrogen (NN) and Phosphorus (PP) back to the soil.

📐Formulae

Energy at Next Trophic Level=Energy at Current Level×0.10\text{Energy at Next Trophic Level} = \text{Energy at Current Level} \times 0.10

Percentage Energy Transfer Efficiency=(Energy at Trophic Level n+1Energy at Trophic Level n)×100%\text{Percentage Energy Transfer Efficiency} = \left( \frac{\text{Energy at Trophic Level } n+1}{\text{Energy at Trophic Level } n} \right) \times 100\%

6CO2+6H2O→lightC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{light}} C_6H_{12}O_6 + 6O_2

💡Examples

Problem 1:

In a meadow ecosystem, the producers (grass) generate 20,000 kJ20,000 \text{ kJ} of energy. Calculate the amount of energy available to the secondary consumers (frogs) in the following food chain: Grass→Grasshopper→Frog→Snake\text{Grass} \rightarrow \text{Grasshopper} \rightarrow \text{Frog} \rightarrow \text{Snake}.

Solution:

200 kJ200 \text{ kJ}

Explanation:

Using the 10%10\% rule: The Primary Consumer (Grasshopper) receives 20,000×0.10=2,000 kJ20,000 \times 0.10 = 2,000 \text{ kJ}. The Secondary Consumer (Frog) receives 2,000×0.10=200 kJ2,000 \times 0.10 = 200 \text{ kJ}.

Problem 2:

Identify the trophic levels in the following food chain: Phytoplankton→Zooplankton→Small Fish→Shark\text{Phytoplankton} \rightarrow \text{Zooplankton} \rightarrow \text{Small Fish} \rightarrow \text{Shark}.

Solution:

T1:Phytoplankton,T2:Zooplankton,T3:Small Fish,T4:SharkT_1: \text{Phytoplankton}, T_2: \text{Zooplankton}, T_3: \text{Small Fish}, T_4: \text{Shark}

Explanation:

Phytoplankton are the producers (T1T_1). Zooplankton are herbivores/primary consumers (T2T_2). Small fish are secondary consumers (T3T_3). The shark is the tertiary consumer or apex predator (T4T_4).

Problem 3:

If 5,000 units5,000 \text{ units} of energy are available at the producer level, and only 5 units5 \text{ units} reach the quaternary consumer, what is the total percentage of the original energy that reached the top of this 5-level chain?

Solution:

0.1%0.1\%

Explanation:

Calculation: 55000×100=0.1%\frac{5}{5000} \times 100 = 0.1\%. This demonstrates why food chains rarely exceed 4 or 5 levels, as there is insufficient energy to support higher levels.

Food Chains and Food Webs - Revision Notes & Key Formulas | IB Grade 6 Science