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How Nature Works in Harmony - Human-made ecosystems

Grade 8CBSE

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

🔑Concepts

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Human-made ecosystems, also known as artificial ecosystems, are environments created and managed by humans to meet specific needs, such as agriculture (croplands), aquariums, or botanical gardens.

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Unlike natural ecosystems, human-made ecosystems have low species diversity (SS) because humans selectively grow specific species and remove others (e.g., weeds).

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These ecosystems are often unstable and require constant external inputs such as water (H2OH_{2}O), fertilizers containing Nitrogen (NN), Phosphorus (PP), and Potassium (KK), and human labor to maintain balance.

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Energy Flow: Energy in these ecosystems still follows the Ten Percent Law, where only a fraction of energy is transferred from one trophic level to the next. If EE is the energy at the producer level, the next level receives 0.10×E0.10 \times E.

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Productivity: Human-made ecosystems like croplands often have higher productivity for a single species compared to natural forests, but they lack the self-sustaining nutrient cycling found in nature.

📐Formulae

En+1=En×10%=En10E_{n+1} = E_{n} \times 10\% = \frac{E_{n}}{10}

Net Primary Productivity (NPP)=Gross Primary Productivity (GPP)−Respiration (R)\text{Net Primary Productivity (NPP)} = \text{Gross Primary Productivity (GPP)} - \text{Respiration (R)}

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

💡Examples

Problem 1:

In a controlled wheat field (human-made ecosystem), the wheat plants capture 45,000 J45,000 \text{ J} of solar energy. Calculate the energy available to the primary consumers (insects) and the secondary consumers (birds) using Lindeman's Law.

Solution:

  1. Energy at Trophic Level 1 (Producers): E1=45,000 JE_1 = 45,000 \text{ J}.
  2. Energy at Trophic Level 2 (Insects): E2=45,000×0.1=4,500 JE_2 = 45,000 \times 0.1 = 4,500 \text{ J}.
  3. Energy at Trophic Level 3 (Birds): E3=4,500×0.1=450 JE_3 = 4,500 \times 0.1 = 450 \text{ J}.

Explanation:

According to the 10%10\% law, only 10%10\% of the energy stored in one trophic level is passed to the next. The rest is lost as heat during respiration and other metabolic processes.

Problem 2:

A farmer calculates the Gross Primary Productivity (GPP) of his maize field to be 12,500 kcal/m2/yr12,500 \text{ kcal/m}^2/\text{yr}. If the plants consume 3,250 kcal/m2/yr3,250 \text{ kcal/m}^2/\text{yr} for their own respiration (R), find the Net Primary Productivity (NPP) using vertical subtraction.

Solution:

Using the formula NPP=GPP−RNPP = GPP - R: 12500−32509250\begin{array}{r} 12500 \\ - 3250 \\ \hline 9250 \end{array} NPP = 9,250 kcal/m2/yr9,250 \text{ kcal/m}^2/\text{yr}.

Explanation:

NPP represents the actual biomass available for the consumers after the plants have used some energy for their own survival (respiration).