In: Biology
Answer: As, locuses have been shown to be diploid, total no of allele wiil be twice total number of individuals.
Population 1: Black Hair (B1B1) = 18 ... 1
Brown Hair (B1B2) = 62 ... 2
Yellow Hair (B2B2) = 34 ... 3
from 1 and 2, frequency of B1 allele = [{(18*2) + 62}/ {2 * (18+62+34)}] = [{36 + 62}/ {2 * 114}] = [98 / 228] = 0.43 (approx)
from 2 and 3, in a similar way frequency of B2 allele = [{62 + (34 * 2)} / 228] = 0.57 (approx)
Thus, heterozygosity for population 1 is going to be = 1 - {(0.43)2 + (0.57)2} = 1 - 0.51 = 0.49.
For Population 2 :
B1B1 = 24 ... 1
B1B2 = 51 ... 2
B2B2 = 25 ... 3
Now, from 1 and 2, frequency of B1 allele is = (99/200) = 0.495
and from 2 and 3, frequency of B2 allele is = (101/200) = 0.505
So, the heterozygosity in population 2 is = 1 - {(0.495)2 - (0.505)2 } = 0.49995
The fixation index for population 1 is going to be = (1 / 228) = 0.0044 (approx.)
and for popultaion 2 = (1 / 200) = 0.005.