Question

assume that the frequency of an allele that causes an autosomal recessive disease is 0.03. what is the probability of a person having the disease when his or her parents are unrelated (i.e., mating is random)?

Answer 1

The probability of a person having an autosomal recessive disease when his or her parents are unrelated is very low, approximately 0.1%.

Step-by-step explanation:

The probability of a person having an autosomal recessive disease when his or her parents are unrelated can be calculated using the Hardy-Weinberg equation. Given that the frequency of the disease-causing allele is 0.03, we can assume the frequency of the recessive allele (q) is 0.03 and the frequency of the dominant allele (p) is 0.97 (since p + q = 1).

The probability of a person being a carrier of the disease-causing allele (2pq) would be 2 x 0.03 x (1 - 0.03) = 0.0558, which is approximately 0.056. This means that the probability of a person being homozygous recessive and having the disease (q²) would be 0.03 x 0.03 = 0.0009, which is approximately 0.001 (0.1%).

Therefore, the probability of a person having the disease when his or her parents are unrelated is very low, approximately 0.1%.

Answer 2

The probability of a person having an autosomal recessive disease when the frequency of the recessive allele is 0.03 and parents are unrelated is 0.09%.

Step-by-step explanation:

The probability of a person having an autosomal recessive disease when his or her parents are unrelated and mating is random can be determined using the Hardy-Weinberg equation. Given the frequency of the recessive allele (q) is 0.03, we must first calculate the probability that an individual is homozygous recessive, which is q2 (0.032). This equates to a probability of 0.0009 or 0.09%. Therefore, the probability of a person having the autosomal recessive disease in such a scenario is 0.09%.