1. Cystic fibrosis (CF), a life-threatening lung disease, is caused from the inheritance of a recessive allele from both parents for the CFTR gene. In order for this disease to manifest itself, one needs to be homozygous recessive for this autosomal gene. According to the World Health Organization (WHO), the frequency of CF in the human population is approximately 1 in 2000. (**keep 4 significant figures during calculations)

Question

1. Cystic fibrosis (CF), a life-threatening lung disease, is caused from the inheritance of a recessive allele from both parents for the CFTR gene. In order for this disease to manifest itself, one needs to be homozygous recessive for this autosomal gene. According to the World Health Organization (WHO), the frequency of CF in the human population is approximately 1 in 2000. (**keep 4 significant figures during calculations)

Using the Hardy-Weinberg equilibrium equation, what frequency of the population consists of heterozygous carriers of the CF allele? (1 pt)

What is the allelic frequency the recessive allele? (1 pt)

If a human population abides by Hardy-Weinberg equilibriums, out of a sample size of 2000, how many are likely to be carriers (round your answer to a whole number) SHOW ALL YOUR WORK using the H-W equation. (3 pts)

Answer 1

To find the frequency of the population that consists of heterozygous carriers of the CF allele, we can use the Hardy-Weinberg equilibrium equation:

p^2 + 2pq + q^2 = 1

where p is the frequency of the dominant allele and q is the frequency of the recessive allele.

Since CF is caused by inheriting a recessive allele from both parents, we know that q^2 represents the frequency of homozygous recessive individuals in the population, which is given as 1 in 2000 or 0.0005.

To find the frequency of the recessive allele (q), we take the square root of q^2:

q = sqrt(0.0005) = 0.0224 (rounded to four significant figures)

Now, to find the frequency of heterozygous carriers (2pq), we can substitute the known values into the equation:

2pq = 2 * 0.0224 * (1 - 0.0224) = 0.0438 (rounded to four significant figures)

Therefore, the frequency of the population that consists of heterozygous carriers of the CF allele is approximately 0.0438.

To find the allelic frequency of the recessive allele, we already found q to be 0.0224. Therefore, the allelic frequency of the recessive allele is also 0.0224.

To calculate the number of carriers in a sample size of 2000 individuals, we can multiply the sample size by the frequency of carriers:

Number of carriers = 2000 * 0.0438 = 87.6

As we need to round to a whole number, there will be approximately 88 individuals in the sample who are likely to be carriers of the CF allele.