A healthy 31-year-old woman comes to the office as she and her husband desire a second child. The husband is infertile and the patient's son, who was conceived via donor insemination, was recently diagnosed with glycogen storage disease type II (Pompe disease). This rare autosomal recessive disease is known to affect 1 in 40,000 of the general population. Genetic testing confirms that the patient is a carrier for the disease. A different sperm donor is selected with no personal or family history of Pompe disease; however, his carrier status is unknown. What is the probability of the patient having an affected child with the new sperm donor?

Question

A healthy 31-year-old woman comes to the office as she and her husband desire a second child.  The husband is infertile and the patient's son, who was conceived via donor insemination, was recently diagnosed with glycogen storage disease type II (Pompe disease).  This rare autosomal recessive disease is known to affect 1 in 40,000 of the general population.  Genetic testing confirms that the patient is a carrier for the disease.  A different sperm donor is selected with no personal or family history of Pompe disease; however, his carrier status is unknown.  What is the probability of the patient having an affected child with the new sperm donor?

Quizplus · Accepted Answer

11ec160b_3754_107b_9249_b31f2e090343_MD0001_00 To be affected by an autosomal recessive disorder, an offspring must inherit 2 copies of the mutant allele (a), a copy from the mother and another copy from the father (in this case, the sperm donor). In this example, genetic analysis confirms that the mother is a carrier (Aa), which gives her a 50% chance of passing 1 mutant allele (a) to her child. The sperm donor does not have Pompe disease, but his carrier status is unknown (ie, he has either an AA or Aa genotype). Given that he has no family history of the disease, his risk of being a carrier is the same as that of the general population. Hardy-Weinberg analysis can be used to estimate the frequency of alleles and genotypes in the general population: The total gene pool is given by (p + q) = 1. By convention, p = normal allele (A) frequency and q = mutant allele (a) frequency in the population of interest. Disease frequency is equivalent to the proportion of homozygous recessive individuals. These affected individuals have 2 copies of the mutant allele (aa); therefore, the frequency of affected individuals = q × q = q². Carrier frequency is equal to the proportion of heterozygous individuals. Carriers have only 1 mutant allele that may be inherited in 2 different ways (ie, Aa or aA); consequently, the probability of being a carrier = 2pq. For rare autosomal recessive disorders, p ≈ 1; therefore, the probability approximates to 2 times the frequency of the mutant allele, or 2q. In this case, 1 in 40,000 individuals is affected by the condition in the general population, so q² = 1/40,000 and q = 1/√40,000 = 1/200. Thus, the probability of the sperm donor being a carrier = 2q = 2 ×(1/200) = 1/100. Given that a carrier sperm donor would have a 50% chance of passing on the mutant allele (a), the probability of the child being affected is: P (mother gives recessive allele) × P (donor is a carrier) × P (donor gives recessive allele) = (1/2 × 1/100 ×1/2) = 1/400 (Choices E and F) 1/40,000 represents the disease frequency (q²), not the mutant allele frequency (q). Educational objective: The Hardy-Weinberg equation can be used to estimate carrier frequency, disease frequency, and frequency of different genotypes if at least 1 of these values is known. __________ References: A century of Hardy-Weinberg equilibrium. (http://www.ncbi.nlm.nih.gov/pubmed/18498203) Prevalence estimation for monogenic autosomal recessive diseases using population-based genetic data. (http://www.ncbi.nlm.nih.gov/pubmed/25893794)

A Healthy 31-Year-Old Woman Comes to the Office as She