A healthy man married a healthy woman whose father had no sweat glands (a recessive trait linked to a gene)

A healthy man married a healthy woman whose father had no sweat glands (a recessive trait linked to a gene) and the mother and her ancestors were healthy. How many sons of these parents may not have sweat glands?

Let’s designate the chromosome that carries the gene that causes the absence of sweat glands in humans as Xp. Then the chromosome that carries the gene that causes the normal development of the sweat glands will be designated XP.

The woman’s father had no sweat glands, let’s write it as Xp Y. He produced two types of sperm – Xp and Y. To his daughter, he could only pass on the X chromosome containing the pathological gene – Xp.

A woman is healthy, therefore, having inherited a chromosome with a normal gene from a healthy mother, she is a heterozygote – XPXp. Such a woman produces two types of germ cells – XP and Xp.

A healthy man – XP Y. His body produces two types of sperm – XP and Y.

The possible offspring of this married couple will be represented by the following options:

heterozygote girls with sweat glands (XPXp) – 25%;

boys with sweat glands (XP Y) – 25%;

homozygote girls with sweat glands (XPXP) – 25%;

boys without sweat glands (Xp Y) – 25%.

Answer: 25% of sons will not have sweat glands.