Question

*Huntingdon's disease is a fatal disorder characterized by progressive deterioration of the nervous system. The symptoms of this disease usually begin to develop in middle age. It is caused by a dominant allele (H). A man heterozygous for the Huntington's allele marries a woman who has the homozygous recessive genotype. They plan to have children. What is the probability that they will have a child who develops Huntington's disease later in life? Use a Punnett Square to help answer the question.

Answer 1

Using a Punnett square, we can determine that a child of a heterozygous man (Hh) for Huntington's disease and a homozygous recessive woman (hh) has a 50% chance of inheriting the disease.

Step-by-step explanation:

Probability of Inheriting Huntington's Disease

Huntington's disease is caused by a dominant allele (H) and is characterized by progressive deterioration of the nervous system. Since Huntington's is an autosomal dominant disorder, a person needs only one copy of the defective allele to express the disease. Given that a heterozygous man (Hh) mates with a homozygous recessive woman (hh), we can predict the inheritance pattern of their offspring using a Punnett square:

• Man (Hh): Heterozygous genotype, carries one dominant Huntington allele (H) and one normal allele (h).
• Woman (hh): Homozygous recessive genotype, carries two normal alleles (hh).

A Punnett square allows us to visualize the genotypic possibilities for their children:

• Hh (Heterozygous: Carries the Huntington's disease allele) - 50% chance
• hh (Homozygous recessive: Does not carry the disease allele) - 50% chance

Therefore, there is a 50% chance that a child of this couple will develop Huntington's disease later in life, since they have a 50% chance of receiving the dominant Huntington allele from their father.