Chapter 2: Ecosystem - Numerical Questions

Here are some numerical problems based on the principles of ecosystems, focusing on productivity and energy flow.

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Question 1: Net Primary Productivity (NPP)

The Gross Primary Productivity (GPP) of a forest ecosystem is 50,000 kcal/m²/year. The energy used by the producers for their respiration (R) is 15,000 kcal/m²/year. What is the Net Primary Productivity (NPP) of this ecosystem?

Solution:

• Gross Primary Productivity (GPP): 50,000 kcal/m²/year
• Respiration (R): 15,000 kcal/m²/year
• NPP = GPP - R
  • NPP = 50,000 - 15,000 = 35,000 kcal/m²/year

Therefore, the Net Primary Productivity of the ecosystem is 35,000 kcal/m²/year.

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Question 2: Energy Transfer (10% Law)

In a grassland ecosystem, the energy available at the producer level is 20,000 Joules. According to the 10% Law, how much energy will be available to the secondary consumers?

Solution:

• Energy at Producer Level (Trophic Level 1): 20,000 J
• Energy at Primary Consumer Level (Trophic Level 2):
  • 10% of 20,000 J = 0.10 * 20,000 = 2,000 J
• Energy at Secondary Consumer Level (Trophic Level 3):
  • 10% of 2,000 J = 0.10 * 2,000 = 200 J

Therefore, 200 Joules of energy will be available to the secondary consumers.

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Question 3: Photosynthetically Active Radiation (PAR)

If the total incident solar radiation on a plant is 1,000,000 Joules, and the plant is able to capture 5% of the Photosynthetically Active Radiation (PAR), which is 50% of the total incident radiation, how much energy is captured by the plant?

Solution:

• Total Incident Solar Radiation: 1,000,000 J
• Photosynthetically Active Radiation (PAR):
  • 50% of 1,000,000 J = 0.50 * 1,000,000 = 500,000 J
• Energy Captured by the Plant:
  • 5% of PAR = 0.05 * 500,000 J = 25,000 J

Therefore, the plant captures 25,000 Joules of energy.

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Question 4: Pyramid of Biomass

In a pond ecosystem, the biomass of phytoplankton is 10 kg. If the biomass of zooplankton is 50 kg, and the biomass of small fish is 100 kg, what is the shape of the pyramid of biomass?

Solution:

• Producer Level (Phytoplankton): 10 kg
• Primary Consumer Level (Zooplankton): 50 kg
• Secondary Consumer Level (Small Fish): 100 kg

Since the biomass increases at successive trophic levels (10 kg < 50 kg < 100 kg), the pyramid of biomass is inverted.

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Question 5: Trophic Level Energy Calculation

If the energy available to the tertiary consumers in a food chain is 15 kcal, how much energy was available at the producer level? (Assume the 10% Law of energy transfer).

Solution:

• This problem requires working backward from the tertiary consumers.
• Energy at Tertiary Consumer Level (Trophic Level 4): 15 kcal
• Energy at Secondary Consumer Level (Trophic Level 3):
  • Since only 10% of energy is transferred, the energy at the previous level is 10 times greater.
  • 15 kcal * 10 = 150 kcal
• Energy at Primary Consumer Level (Trophic Level 2):
  • 150 kcal * 10 = 1500 kcal
• Energy at Producer Level (Trophic Level 1):
  • 1500 kcal * 10 = 15,000 kcal

Therefore, 15,000 kcal of energy was available at the producer level.