The allele frequency for the spider syndrome allele in a flock of sheep is 0.41. The...

The allele frequency for the spider syndrome allele in a flock of sheep is 0.41. The normal allele N is dominant to the n allele, and homozygous nn lambs do not survive to reproduce. In the flock, all NN and Nn animals are retained, and all nn animals are removed from the herd. Determine the allele and genotypic frequencies in the flock following a single generation of selection.

Expert Solution

Answer:

Frequency of recessive allele n for spider syndrome is 0.35. This frequency can be designated q .
Accordingi to hardy weinberg equilbirum p+ q = 1 , where p is the frequency of the normal dominant N allele.
- p + 0.35 = 1
p = 1 - 0.35 = 0.65
p² + 2pq + q² = 1
p² is the frequency of NN genotype individuals
2pq is the frequency of heterzygotes Nn
q² is the frequency of nn genotype individuals

the frequency of genotype are:

NN = 0.65 x 0.65 = 0.4225
Nn = 2 x 0.65 x 0.35 = 0.455
nn = 0.35 x 0.35 = 0.1225
lets assume the population consists of 1000 before selection
expected number of NN genotype = 1000 x 0.4225 = 422.5
expected number of Nn genotype = 0.455 x 1000 = 455
expected number of nn genotype = 0.1225 x 1000 = 122.5
followimg selection all 122.5 nn genotype individuals would die.
Now the population will consist of 422.5 NN genotype and 455 Nn genotype sheeps.
allele frequency of N allele = number of NN sheeps x 2 + number of Nn sheeps / total number of sheeps left x 2

Note:we are multiplying number of homozygous NN sheeps by 2 because each sheep contains 2 N alleles.

Heterozygous sheeps contains only one N allele. The sheep population is diploid and each sheeps contains 2 alleles of a particulat gene

allele frequency of N = 422.5 x 2 + 455 / ( 422.5 + 455) x 2 = 845 + 455 / 1755 = 0.74
allele frequency of n = number of nn sheeps x 2 + number of Nn sheeps / total number of sheeps after selection x 2 = 0 + 455 / ( 422.5 + 455 ) x 2 = 455 / 1755 = 0.26
NN genotype frequency = 0.74 x 0.74 = 0.55
Nn genotype frequency = 2 x 0.74 x 0.26 = 0.38
nn genotype frequency = 0.26 x 0.26 = 0.07
The allelic frequency of N and n after selection is 0.74 and 0.26

The genotypic frequency of NN , Nn and nn after selection is 0.55, 0.38 and 0.07 respectively.

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gladiator answered 11 months ago

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