Question

A child is born to a couple, one of whom is heterozygous for an autosomal dominant disease. The other parent is homozygous normal. What would be the child's chances of having the disease? (Use a Punnett square to figure this out).

1) 0
2) 25
3) 50
4) 75

Answer 1

A child born to one parent who is heterozygous for an autosomal dominant disease and one parent who is homozygous normal has a 50% chance of inheriting the disease, as determined by a Punnett square.

Step-by-step explanation:

Understanding Autosomal Dominant Inheritance

When considering the inheritance of an autosomal dominant disease, where one parent is heterozygous and the other is homozygous normal, we use a Punnett square to determine the probabilities of the offspring having the disease. The heterozygous parent (let's denote the alleles as Nn, where 'N' is the disease allele and 'n' is the normal allele) has a 50% chance of passing on the disease allele, while the homozygous normal parent (nn) can only pass on the normal allele. Therefore, the Punnett square would have two rows, one for each of the heterozygous parent's possible alleles (N and n), and two columns, which only contain the normal allele n from the homozygous parent.

As such, the squares would show genotypes Nn, nn, Nn, and nn, leading to two out of four cases where the disease allele is passed on. Thus, there is a 50% chance that any child from this union will inherit the autosomal dominant disease.