Answer:
Frequency of B allele is 0.6681
Explanation:
If p represents the frequency of dominant allele and q represents the frequency of recessive allele, according to Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p² = frequency of homozygous dominant genotype
q² = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
Given that number of recessive chestnut horse = 28
Total horses = 226 + 28 = 254
frequency of b² genotype = 28/254 = 0.1102
frequency of recessive b allele = √0.1102 = 0.3319
So, frequency of B allele =
1 - 0.3319 = 0.6681
Hence frequency of B allele is 0.6681
The frequency of the B allele is : 0.6681
p = frequency of dominant allele
q = frequency of recessive allele
therefore p + q = 1
Hardy -weinberg equilibrium : p² + 2pq + q² = 1
where :
p² = frequency of h0m0zygous dominant genotype
q² = frequency of h0m0zygous recessive genotype
2pq = frequency of heterozygous genotype
Given that :
Recessive chestnut horse ( q ) = 28
Dominant horse ( p ) = 226
Therefore Total number of horses = P + q
= 226 + 28 = 254
Next step : determine the frequency of recessive b allele
Frequency of b² genotype = Recessive horse / Total number of horses
= 28 / 254
= 0.1102
Hence the frequency of recessive b allele = [tex]\sqrt{0.1102}[/tex] = 0.3319
Given that the frequency of recessive b allele = 0.3319
Frequency of B allele = 1 - 0.3319
= 0.6681
Hence we can conclude that the frequency of B allele is 0.6681
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