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Biology: Life and Systems - Ecology and Interactions in Ecosystems

Grade 8IB

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

🔑Concepts

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Ecosystem Dynamics: An ecosystem consists of biotic (living) factors like plants and animals, and abiotic (non-living) factors such as H2OH_2O, temperature, CO2CO_2, and sunlight.

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Trophic Levels: Organisms are categorized into levels: Producers (T1T_1), Primary Consumers (T2T_2), Secondary Consumers (T3T_3), and Tertiary Consumers (T4T_4).

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Energy Flow and the 10% Rule: Energy enters ecosystems via photosynthesis (6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2). Only about 10%10\% of the energy from one trophic level is passed to the next; the rest is lost as heat or used for metabolic processes.

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Ecological Pyramids: Graphical representations of the relationship between different organisms in an ecosystem, including Pyramids of Numbers, Biomass (measured in g/m2g/m^2 or kg/m2kg/m^2), and Energy (kJ/m2/yearkJ/m^2/year).

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Symbiotic Relationships: Close interactions between species including Mutualism (+/++/+), Commensalism (+/0+/0), and Parasitism (+/−+/-).

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Population Density: A measure of the number of individuals of a species per unit area or volume, expressed as NA\frac{N}{A}.

📐Formulae

Energy Transfer Efficiency=(Energy at higher trophic levelEnergy at lower trophic level)×100%\text{Energy Transfer Efficiency} = \left( \frac{\text{Energy at higher trophic level}}{\text{Energy at lower trophic level}} \right) \times 100\%

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

Population Density=NA\text{Population Density} = \frac{N}{A}

6CO2+6H2O+light energy→C6H12O6+6O26CO_2 + 6H_2O + \text{light energy} \rightarrow C_6H_{12}O_6 + 6O_2

💡Examples

Problem 1:

In a local grassland ecosystem, the primary producers (grass) generate 20,000 kJ/m2/year20,000\text{ kJ/m}^2/\text{year}. According to the 10%10\% rule, calculate the energy available to the tertiary consumers (e.g., hawks).

Solution:

20 kJ/m2/year20\text{ kJ/m}^2/\text{year}

Explanation:

Using the 10%10\% rule: Primary Consumers receive 10%10\% of 20,000=2,000 kJ20,000 = 2,000\text{ kJ}; Secondary Consumers receive 10%10\% of 2,000=200 kJ2,000 = 200\text{ kJ}; Tertiary Consumers receive 10%10\% of 200=20 kJ200 = 20\text{ kJ}.

Problem 2:

A biologist counts 450450 individual milkweed plants in a rectangular field measuring 50 m50\text{ m} by 30 m30\text{ m}. Calculate the population density.

Solution:

0.3 plants/m20.3\text{ plants/m}^2

Explanation:

First, calculate the total area: A=50 m×30 m=1500 m2A = 50\text{ m} \times 30\text{ m} = 1500\text{ m}^2. Then use the formula Density=NA=4501500=0.3 plants/m2\text{Density} = \frac{N}{A} = \frac{450}{1500} = 0.3\text{ plants/m}^2.

Problem 3:

Identify the type of interaction: A tapeworm lives in the intestines of a mammal, absorbing nutrients and causing the host to become weak.

Solution:

Parasitism (+/−+/-)

Explanation:

In this interaction, the tapeworm (parasite) benefits by obtaining nutrients, while the mammal (host) is harmed by the loss of nutrients and potential illness.

Ecology and Interactions in Ecosystems Revision - Grade 8 Science IB