07-24-2006, 07:14 AM
Patients with sickle cell disease are homozygous for HbS. Because in the case described only one potential parent has the HbS allele, the only way their child could inherit two HbS genes is if one of the father's normal Hb genes underwent a new mutation in a sperm cell, an extremely unlikely event. Therefore, their child does not have a reasonable chance of inheriting the disease (choice A). However, there is a 1/4 chance that their child would be a carrier. Such heterozygous carriers may have mild symptoms and are said to have the sickle cell trait.
The probability that two heterozygous carriers of the trait will have a child with the disease is 1/4 (choice B); that is, 25% of their children will have the disease, while 50% will have the trait and another 25% will be normal.
Assuming there is no knowledge of family histories or hemoglobin (Hb) patterns, choice C 1/144 represents the probability that two carriers will meet. That is, the carrier rate among African Americans is approximately 1 in 12, and 1/12 × 1/12 equals 1/144.
Choice (D) 1/576 corresponds to the likelihood of a child having the disease if two African Americans having no knowledge of family histories or Hb patterns had children. As indicated in choice C the random chance of two African-American heterozygotes meeting is 1/144. Since only 25% of their children will have the disease (see choice B above) those having the disease will equal 1/144 × 1/4, or 1/576.
Choice E is not a sensible representation of the probability of a child inheriting sickle cell anemia.