Chapter 3: Biodiversity and Conservation - Numerical Questions

Here are some numerical problems based on the principles of biodiversity and conservation, focusing on the species-area relationship.

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Question 1: Species-Area Relationship

The species-area relationship is given by the equation log S = log C + Z log A. If the value of the regression coefficient (Z) is 0.2, the Y-intercept (C) is 0.5, and the area (A) is 100 square kilometers, what is the species richness (S)?

Solution:

• Given Equation: log S = log C + Z log A
• Given Values:
  • Z = 0.2
  • C = 0.5
  • A = 100
• Substitute the values into the equation:
  • log S = log(0.5) + 0.2 * log(100)
  • log S = -0.301 + 0.2 * 2
  • log S = -0.301 + 0.4
  • log S = 0.099
• To find S, we take the antilog of both sides:
  • S = 10^0.099
  • S ≈ 1.256

Therefore, the species richness (S) is approximately 1.26.

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Question 2: Calculating the Regression Coefficient (Z)

In a study of a forest ecosystem, it was found that the species richness (S) is 10 when the area (A) is 1000 square kilometers. If the Y-intercept (C) is 2, what is the value of the regression coefficient (Z)?

Solution:

• Given Equation: log S = log C + Z log A
• Given Values:
  • S = 10
  • A = 1000
  • C = 2
• Substitute the values into the equation:
  • log(10) = log(2) + Z * log(1000)
  • 1 = 0.301 + Z * 3
• Solve for Z:
  • 1 - 0.301 = 3Z
  • 0.699 = 3Z
  • Z = 0.699 / 3
  • Z ≈ 0.233

Therefore, the value of the regression coefficient (Z) is approximately 0.233.

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Question 3: Impact of Area on Species Richness

For a particular group of organisms, the species-area relationship is described by the equation S = 5A^0.3. If the area is increased from 100 km² to 200 km², what is the percentage increase in species richness?

Solution:

• Given Equation: S = 5A^0.3
• Case 1: Area (A1) = 100 km²
  • S1 = 5 * (100)^0.3
  • S1 = 5 * (10²)^0.3
  • S1 = 5 * 10^0.6
  • S1 ≈ 5 * 3.981 = 19.905
• Case 2: Area (A2) = 200 km²
  • S2 = 5 * (200)^0.3
  • S2 = 5 * (2 * 100)^0.3
  • S2 = 5 * 2^0.3 * 100^0.3
  • S2 ≈ 5 * 1.231 * 3.981 = 24.50
• Percentage Increase:
  • Increase = S2 - S1 = 24.50 - 19.905 = 4.595
  • Percentage Increase = (Increase / S1) * 100
  • Percentage Increase = (4.595 / 19.905) * 100 ≈ 23.08%

Therefore, the percentage increase in species richness is approximately 23.08%.