Hardy-Weinberg principle?

Question

Any help is appreciated:

Cystic fibrosis is a fatal disease in humans caused by a deleterious recessive allele from a single locus. Individuals that suffer from cystic fibrosis die early in life, well before they can reproduce. In a fictional population of 10,000 humans, 1 is expected to have the disease. What is the allele frequency of the deleterious allele? How many people within the population would you expect to be a carrier of the disease without suffereing from it?

Answer 1

If 1/10000 have the disease, then this is the double recessive condition (also noting that cystic fibrosis is not s*x-linked).  So, let q be the frequency of the cystic fibrosis allele in the population:

q^2 = 1/10000 = 0.0001

q = 0.01

since the frequency of the normal allele is p, then:

p + q = 1

p = 1 - q

p = 1 - 0.01

p = 0.99

So, the frequency of the heterozygotes is:

2pq = 2  x  0.99  x  0.01 = 0.0198

So, the number of people in the population of 10,000 that are carriers is:

0.0198  x  10000 = 198 individuals are carriers

(carrying this idea further with this population, 1 has cystic fibrosis and 9,801 are homozygous normal).