Based on the pedigree chart describing the inheritance of an autosomal-dominant disease, what is the probability that person A will give birth to an affected child if she mates with a wild-type male.
Based on the pedigree chart describing the inheritance of an autosomal-dominant disease, what is the probability that person A will give birth to an affected child if she mates with a wild-type male.
The correct option is B 50%
A trait that is autosomal-dominant will be expressed even if only one copy of the disease allele is present in an individual. For example, if “B” is the disease allele and “b” is the non-disease allele, a person will have the disease if its genotype is either BB or Bb. They will be healthy only if they have bb. Looking at person A’s parents, we know that the mother is “wild type,” meaning she doesn’t have the disease, and therefore must be bb. We know the father has at least one disease allele and that he passed the disease allele onto his daughter, person A. Person A is affected, so she must have at least one disease allele, but she also has an unaffected mother who we know would have passed her a non-disease allele. Person A must have a heterozygous genotype (Bb). If Person A with genotype Bb has children with a “wild type” male of genotype bb, then the probability of one of their children being affected can be determined by drawing a Punnett square. It would be seen that 50% of the children will have a disease allele and therefore be affected.
A trait that is autosomal-dominant will be expressed even if only one copy of the disease allele is present in an individual. For example, if “B” is the disease allele and “b” is the non-disease allele, a person will have the disease if its genotype is either BB or Bb. They will be healthy only if they have bb. Looking at person A’s parents, we know that the mother is “wild type,” meaning she doesn’t have the disease, and therefore must be bb. We know the father has at least one disease allele and that he passed the disease allele onto his daughter, person A. Person A is affected, so she must have at least one disease allele, but she also has an unaffected mother who we know would have passed her a non-disease allele. Person A must have a heterozygous genotype (Bb). If Person A with genotype Bb has children with a “wild type” male of genotype bb, then the probability of one of their children being affected can be determined by drawing a Punnett square. It would be seen that 50% of the children will have a disease allele and therefore be affected.
