Hardy-Weinberg Equilibrium Problems p is the frequency of the dominant allele in

Question

Hardy-Weinberg Equilibrium Problems

p is the frequency of the dominant allele in the population

q is the frequency of the recessive allele in the population

p2 is the frequency of the homozygous dominant genotype

q2 is the frequency of the homozygous recessive genotype

2pq is the frequency of the heterozygous genotype

p + q = 1

p2 + 2pq + q2 = 1 Find q2 q p p2 2pq

A rare disorder of cystic fibrosis (characterized by abnormal transport ofchloride and sodium across an epithelium, leading to thick, viscous secretions) occurs every 1 in 3,300 births. To express this trait one must be homozygous recessive (ff).

The cystic fibrosis allele is represented by f. The normal allele is F. Suppose both parents have alleles Ff. The possible combinations of alleles in the children are FF, Ff, Ff and ff. The alleles ff will cause the disease. So, although the parents do not have cystic fibrosis, they can produce children with the disease. The parents are called 'carriers' of the disease

What is the frequency of the population that carries the trait? (Ff)

Explanation / Answer

Cystic fibrosis is an autosomal recessive disorder. This means that two copies of a mutated allele (f) must be present for the occurrence of disease i.e ff will cause the disease.

Given that, number of affected children are every 1 in 3,300 births.
Therefore, frequency of homozygous recessive genotype in the population (q2)= 1/3300 or 0.0003
So, frequency of the recessive allele in the population q = square root of 0.0003= 0.017 ~ 0.02

Frequency of the dominant allele in the population,p = 1- q = 1 - 0.02 = 0.98
Frequency of the heterozygous genotype (carriers), 2pq = 2 X 0.01 X 0.98 = 0.0392 ~ 0.04

The result will come 0.0334 if we take q =0.017 without rounding off.

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