Answer to Question #108407 in Genetics for Erik

Question #108407
Assume that congenital dislocation of the lumbar joint is dependent upon the action of a dominant allele
with 25% penetrance (the frequency with which a dominant or homozygous recessive gene manifests itself
in the phenotype of an individual). Frequency of the disorder is 6:10,000. State the number of homozygous
individuals for the recessive allele.
Expert's answer

According to Hardy–Weinberg principle p2 + 2pq + q2 = 100% = 1,

where, p2 is the frequency of the AA genotype,

2pq - frequency of the genotype Aa,

q2 is the frequency of the aa genotype.

Then q2 = 1 - (p2 + 2pq)

However, the number of patients (6: 10,000) does not represent p2 + 2pq, but only 25% of carriers of allele A, and the true number of people with this allele is four times greater 24: 10,000.

Therefore p2 + 2pq = 24 : 10 000

Then q2 = 1 - (p2 + 2pq) = 1 - 24 : 10000 = 0,9976

State the number of homozygous individuals for the recessive allele = 9976 : 10000 or 1 : 10.

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