Final answer:
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.