Question

One out of 10,000 babies born in North America is affected by cystic fibrosis, a recessive condition. Assuming that the North American human population is in Hardy-Weinberg equilibrium for this trait, what percentage of the population is heterozygous for this trait? (Remember the equation for a population in Hardy-Weinberg equilibrium: p2 + 2pq + q2 = 1.)A) 1% B) 0.01%C)2%

Answer 1

Answer:

C) 2%

Explanation:

According to Hardy-Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

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

Cystic fibrosis is a recessive condition hence here,

q² = 1/10000 = 0.0001

q = √0.0001 = 0.01

p = 1 - q

= 1 - 0.01

= 0.99

Percentage of the population heterozygous for trait = 2pq

= 2 * 0.01 * 0.99

= 0.0198

= 1.98 % = 2%