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Organisms and their Environment - Food chains and food webs

Grade 12IGCSEBiology

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

🔑Concepts

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Energy flows through ecosystems starting from the Sun. Producers (autotrophs) convert light energy into chemical energy via photosynthesis: 6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2.

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Trophic levels define the position of an organism in a food chain. These include Producers (T1T_1), Primary Consumers (T2T_2), Secondary Consumers (T3T_3), and Tertiary Consumers (T4T_4).

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Energy transfer between trophic levels is inefficient; typically only about 10%10\% of the energy is passed on. The rest is lost as heat during respiration (C6H12O6+6O2→6CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}), through excretion, or as undigested material.

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Food webs represent interconnected food chains, showing that most consumers are generalists. This complexity increases ecosystem stability against the loss of a single species.

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Pyramids of Biomass represent the total dry mass of organisms at each level, usually measured in g/m2g/m^2 or kg/m2kg/m^2. Unlike pyramids of numbers, these are almost always upright in terrestrial ecosystems.

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Bioaccumulation refers to the buildup of toxins (e.g., heavy metals like Hg2+Hg^{2+} or pesticides like DDTDDT) within an organism, while Biomagnification describes the increasing concentration of these toxins at higher trophic levels.

📐Formulae

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

NPP=GPP−RNPP = GPP - R

Percentage of energy lost=100%−Efficiency of Energy Transfer\text{Percentage of energy lost} = 100\% - \text{Efficiency of Energy Transfer}

💡Examples

Problem 1:

In a grassland ecosystem, the producers fix 20,000 kJ/m2/year20,000 \text{ kJ/m}^2\text{/year} of energy. The primary consumers receive 2,400 kJ/m2/year2,400 \text{ kJ/m}^2\text{/year}. Calculate the efficiency of energy transfer from the producers to the primary consumers.

Solution:

Efficiency=2,40020,000×100=12%\text{Efficiency} = \frac{2,400}{20,000} \times 100 = 12\%

Explanation:

To find the efficiency, we divide the energy stored in the higher trophic level by the energy available in the previous level and multiply by 100 to get a percentage.

Problem 2:

A forest ecosystem has a Gross Primary Productivity (GPPGPP) of 3.5 kg/m2/year3.5 \text{ kg/m}^2\text{/year}. If the autotrophs consume 1.2 kg/m2/year1.2 \text{ kg/m}^2\text{/year} through cellular respiration (RR), what is the Net Primary Productivity (NPPNPP)?

Solution:

NPP=3.5−1.2=2.3 kg/m2/yearNPP = 3.5 - 1.2 = 2.3 \text{ kg/m}^2\text{/year}

Explanation:

NPPNPP represents the actual biomass available to consumers after the producers have met their own metabolic needs via respiration.

Problem 3:

Explain why food chains rarely exceed four or five trophic levels.

Solution:

Due to the 10%10\% rule of energy transfer, the energy available at each successive level decreases exponentially (En≈E1×0.1n−1E_n \approx E_1 \times 0.1^{n-1}).

Explanation:

By the time energy reaches a quaternary consumer, the total energy remaining is usually insufficient to support a viable population of a higher trophic level.

Food chains and food webs - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology