Cystic fibrosis is an autosomal recessive disorder. In one population, the frequ
ID: 43624 • Letter: C
Question
Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004. Assuming that this population is under Hardy-Weinberg equilibrium, calculate all allele frequencies and genotype frequencies. Enter your numerical answers in the boxes below. Express each answer to four decimal places. Frequency of the A1 allele = Frequency of the A2 allele = Frequency of homozygous dominant individuals = Frequency of heterozygous individuals = Frequency of homozygous recessive individuals =Explanation / Answer
The Hardy weinberg formula is as follows:
p+q=1
&
p2+2pq+q2= 1
where p = dominant allele frequency
q= recessive allele frequency
2pq= frequency of heterozygous individuals
p2= frequency of homozygous dominant individuals
q2= frequency of homozygous recessive individuals
A2A2= 0.0004 (given)
i.e. q2= 0.0004
thus, q= 0.02
q= frequency of recessive allele A2= 0.02
frequency of p (allele A1) = 1-q= 0.98
frequency of homozygous dominant = A1A1 = p2= 0.9604
frequency of heterozygous individuals=2pq= 2X 0.02X 0.98= 0.0392
frequency of homozygous recessive =q2= 0.0004 (given)
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