Question

This pedigree shows the inheritance pattern of a disease caused by a paternally imprinted allele in a family. Genetic testing of individuals I-1 and II-5 indicates they are carriers of the disease allele. No one else in the family agreed to be tested. If first cousins III-4 and III-5 have a child together (IV-1) what is the chance the child will have the disease? (assume complete penetrance)

Answer 1

The chance that the child (IV-1) will have the disease is 0% because they have received two normal alleles.

Step-by-step explanation:

To determine the chance the child (IV-1) will have the disease, we need to consider the inheritance pattern of the disease and the genotypes of the parents.

Since individuals I-1 and II-5 are carriers of the disease allele, they are heterozygous (Aa) for the disease gene.

Cousins III-4 and III-5 are both unaffected, so they must be homozygous for the normal allele (AA).

If cousins III-4 and III-5 have a child (IV-1), the parents will each contribute one allele to the child.

Since both parents are homozygous for the normal allele, the child will receive one normal allele from each parent, making the child genotype AA.

Therefore, the chance that the child (IV-1) will have the disease is 0% because they have received two normal alleles.

Answer 2

If both first cousins III-4 and III-5 are carriers of the disease allele, there is a 25% chance that their child (IV-1) will have the disease.

Step-by-step explanation:

To determine the chance that the child (IV-1) of first cousins III-4 and III-5 will have the disease, we need to understand the inheritance pattern shown in the pedigree. Based on the information provided, the disease is caused by a paternally imprinted allele, which means it is only expressed when inherited from the father. Therefore, if both III-4 and III-5 are carriers of the disease allele, there is a 25% chance that their child (IV-1) will inherit the disease allele from III-4's father. This would result in the child having the disease.