Answer:
B- 0.4
Step-by-step explanation:
Using the Hardy-Weinberg equation-
p^{2}+2pq+q^{2}= 1
First population, where p = 0.2, q = 1 - 0.2, therefore, q = 0.8
Inserting the values into the equation for the first population-
(0.2 x 0.2) + 2pq + (0.8 x 0.8)= 1
0.04 + 0.64 + 2pq = 1
0.68 + 2pq = 1
2pq = 1 - 0.68
pq = 0.32/2
pq = 0.16
repeating equation for second population, where p = 0.6, q = 1 - 0.6, therefore, q = 0.4
Inserting values....
(0.6 x 0.6) + 2pq + (0.4 x 0.4) = 1
0.36 + 0.16 + 2pq = 1
0.52 + 2pq = 1
2pq = 1 - 0.52
pq = 0.48/2
pq = 0.24
Adding the two results to determine the heterozygosity of the locus of the combined population, would give-
0.16 + 0.24 = 0.4