What would be the heterozygosity of these two populations ? Field rodent Populat

Question

What would be the heterozygosity of these two populations ?
Field rodent Population 1 Black hair (genotype B1B1)=18 Brown hair (genotype B1B2)=62 Yellow hair (genotype B2B2)=34
Field rodent population 2 Black hair 24 Brown hair 51 Yellow hair 25
What would be the fixation index between the 2 populations ?
On a single graph, depict the trends in heterozygosity in each subpopulation, and the trend in fixation index, after each round in migration (reciprocal migration beginning at a level of m=0.05) how would these curves change if the rate of migration were m=0.01? And m=0.10 ?
What would be the heterozygosity of these two populations ?
Field rodent Population 1 Black hair (genotype B1B1)=18 Brown hair (genotype B1B2)=62 Yellow hair (genotype B2B2)=34
Field rodent population 2 Black hair 24 Brown hair 51 Yellow hair 25
What would be the fixation index between the 2 populations ?
On a single graph, depict the trends in heterozygosity in each subpopulation, and the trend in fixation index, after each round in migration (reciprocal migration beginning at a level of m=0.05) how would these curves change if the rate of migration were m=0.01? And m=0.10 ?

Field rodent Population 1 Black hair (genotype B1B1)=18 Brown hair (genotype B1B2)=62 Yellow hair (genotype B2B2)=34
Field rodent population 2 Black hair 24 Brown hair 51 Yellow hair 25
What would be the fixation index between the 2 populations ?
On a single graph, depict the trends in heterozygosity in each subpopulation, and the trend in fixation index, after each round in migration (reciprocal migration beginning at a level of m=0.05) how would these curves change if the rate of migration were m=0.01? And m=0.10 ?

Explanation / Answer

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.

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