Question

n babies have the disease, what are the expected frequencies of the dominant (A1) and recessive (A2) alleles according to the Hardy-Weinberg equation? Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. If 9 in 10,000 newborn babies have the disease, what are the expected frequencies of the dominant (A1) and recessive (A2) alleles according to the Hardy-Weinberg equation? f(A1) = 0.9800, f(A2) = 0.0200 f(A1) = 0.9604, f(A2) = 0.0392 f(A1) = 0.9997, f(A2) = 0.0003 f(A1) = 0.9700, f(A2) = 0.0300

Answer 1

Answer:

f(A1) = 0.9700, f(A2) = 0.0300

Explanation:

According to Hardy-Weinberg equilibrium equation:

p + q = 1

p² + 2pq + q² = 1

where,

p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of dominant homozygous genotype

2pq = frequency of heterozygous genotype

q² = frequency of recessive homozygous genotype

q² = 9/10000 = 0.0009

A2 = q

= √0.0009 = 0.03

A1 = p

= 1-q

= 1-0.03 = 0.97