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

Answer 1

Answer:

The hardy weinberg equation is

Explanation:

p square + 2 PQ + q square

Answer 2

The correct answer is :

Explanation:

• Hardy-Weinberg Equilibrium is :

     $p^{2} {+} 2pq + q^{2} = 1$ and $p+q=1$

                          , where p = Dominant allele's frequency.

                                       q = Recessive allele's frequency.

                                       $p^{2}$ = Homozygous dominant's frequency.

                                       $q^{2}$ = Homozygous recessive's frequency.

                                       $2pq$ = Heterozygous dominant's frequency.

• Given, $q^{2} = \frac{1}{10000} = 0.0001$,
• Square rooting both sides, $q = 0.01$.
• Therefore, $p = 1 - 0.01 = 0.99$.
• The frequency of hererozygotes is: $2pq = 2\times0.99\times0.01 =0.0198$.
• $0.0198$ is the required answer.