The disorder known as cystic fibrosis (CF) is inherited in a recessive autosomal manner. In populations of Northern European descent, the frequency of people with CF is about 1 in 2,500. Assuming Hardy-Weinberg equilibrium for this disorder, answer the following: A. Assuming Hardy-Weinberg equilibrium, what percentage of people in the population would be expected to be heterozygous carriers

The correct answer is -

A. 3.9% people in the population would be expected to be heterozygous carriers

Explanation:

The genotype frequency for the CF affected is 1/2,500,

= 1/2500

= 0.004. This is equal to q^2

here q = frequency of recessive alleles combination or CF people. (H-W equation => p^2+q^2+2pq=1)

The allele frequency is the square root of q^2 value, which equals 0.02. The frequency of the dominant allele or p would be

1 - 0.02 = 0.98 (p+q = 1)

The frequency for the CF combinantion =0.004

the dominant or unaffected people genotype

= (098)^2

= 0.96;

and for the heterozygote, it is

= 2pq

=2(0.98)(0.02)

= 0.039 or 3.9%

i'd say that the moon rock is about 2.2 billion years old.

yes? sounds right, but you didn’t ask a question.