Question

Cystic fibrosis (CF) is a recessive autosomal disorder. In certain populations of Northern European descent, the number of people born with this disorder is about 1 in 2500. Assuming Hardy-Weinberg equilibrium for this trait: What are the frequencies for the normal and CF alleles? What are the genotype frequencies of homozygous normal, heterozygous, and homozygous affected individuals? Assuming random mating, what is the probability that two phenotypically unaffected heterozygous carriers will choose each other as mates?

Answer 1

The probability that two phenotypically unaffected heterozygous carriers will choose each other as mates is 0.0015 (approximately 0.15%)

What is the probability that two phenotypically unaffected heterozygous carriers will choose each other as mates?

Let p represent the frequency of the normal allele and q represent the frequency of the CF allele.

Since CF is recessive, the frequency of individuals with CF (q²) needs to be 1 in 2500:

q² = 1 / 2500

q ≈ 0.02 (approximately 2%)

Therefore, the frequency of the normal allele is:

p = 1 - q ≈ 0.98 (approximately 98%)

Hardy-Weinberg equilibrium assumes random mating and no other evolutionary forces affecting allele frequencies.

Based on the allele frequencies, the genotype frequencies can be calculated using the binomial theorem:

Homozygous normal (p²): p² ≈ (0.98)² ≈ 0.9604 (approximately 96.04%)

Heterozygous (2pq): 2pq ≈ 2 * 0.98 * 0.02 ≈ 0.0392 (approximately 3.92%)

Homozygous affected (q²): q² ≈ (0.02)² ≈ 0.0004 (approximately 0.04%)

Assuming random mating, the probability of any two individuals choosing each other is equally likely.

Therefore, the probability that two phenotypically unaffected heterozygous carriers will choose each other as mates is simply the square of the heterozygous frequency:

p(heterozygous x heterozygous) = (0.0392)² ≈ 0.0015 (approximately 0.15%)