The number of individuals of different species in two communities P and Q is given below.

​Based on the gi​ven data, select the correct stat​ement?​

Correct option is B

Step-by-step Reasoning:

To assess species diversity, we consider:

1. Species richness (number of different species) – both P and Q have 10 species → equal richness

2. Evenness – how individuals are distributed across species

   • If a community has a more even distribution, it is more diverse.

Visual Pattern from the Table:

• In Community P, some species dominate:

  • A = 59, C = 44, J = 30 → these 3 species alone = 133 out of total 196

• In Community Q, distribution is more even:

  • Highest = C (23), lowest = G (1), most others between 12–23

Let’s compute Simpson’s Diversity Index (D) for both, which is commonly used:

$$D = 1 - \sum \left( \frac{n_i(n_i - 1)}{N(N - 1)} \right)$$

$$\text{For Community P:} \\\text{Total } N_P = 196 \\\sum n_i(n_i - 1)_P = 59(58) + 12(11) + 44(43) + 20(19) + 11(10) + 10(9) + 2(1) + 5(4) + 3(2) + 30(29) \\= 3422 + 132 + 1892 + 380 + 110 + 90 + 2 + 20 + 6 + 870 = \mathbf{6924}$$

$$D_P = 1 - \frac{6924}{196(195)} = 1 - \frac{6924}{38220} \approx 1 - 0.1812 = \mathbf{0.8188}$$

$$\text{For Community Q:} \\\text{Total } N_Q = 148 \\\sum n_i(n_i - 1)_Q = 21(20) + 20(19) + 23(22) + 12(11) + 19(18) + 14(13) + 1(0) + 13(12) + 13(12) + 12(11) \\= 420 + 380 + 506 + 132 + 342 + 182 + 0 + 156 + 156 + 132 = \mathbf{2406}$$

$$D_Q = 1 - \frac{2406}{148(147)} = 1 - \frac{2406}{21756} \approx 1 - 0.1106 = \mathbf{0.8894}$$

 Conclusion:

• Community Q has a higher Simpson’s diversity index than P

• This is due to more even distribution among species

 Correct Answer: 2. Community Q has higher species diversity than P

 Information Booster:

• Simpson's Diversity Index gives higher value for evenly distributed species.

• Both communities have the same species richness (10 species), but evenness differs.

• Diversity is a function of both richness and evenness.

​