Question

The answer is e. I don't understand how to solve for it.

Cystic fibrosis is a rare recessive genetic condition. Suppose that the allele that causes Cystic fibrosis has a frequency of 0.001 in a certain population. If the population can be assumed to be at Hardy Weinberg equilibrium approximately how many heterozygous carriers of the disease will be born for every individual born with the disease?

1. 10
2. 100
3. 999
4. 1000
5. 1998

Answer 1

Question

According to the Hardy -weinberg equation

p2 + 2pq+ q2 =1

Where, p , is the frequency of dominant allele in the population.

q is the frequency of recessive allele in the population.

2pq denote the frequency of heterozygous carriers individuals.

Now, in this question, frequency of allele causing cystic fibrosis is 0.001.

Please note that cystic fibrosis is caused by recessive allele.

Therefore, in this question, the

Frequency of recessive allele(q) is given = 0.001(given)

So, we can calculate the frequency of dominant allele is calculated using the formuale

p+ q=1

Therefore, p =1-q

= 1-.001= .999

So, to calculate, heterozygous individuals frequency can be calculated by formulae

=2pq

= 2* .001*0.999= .01998

Therefore, number of individuals will be .01998*100= 1.998

Therefore, number of individual per thousand people would be:

1.998*1000= 1998

Hence, option e is correct.