Question

A family is affected by a rare genetic disease with a Mendelian pattern of inheritance. The father is affected, and the mother is unaffected by this disease. They have three children: an affected son and daughter and an unaffected son. If they have a fourth child, what is the probability that this child will be affected by the disease?

100% 50% 33% 25% 0%

Answer 1

Answer: The inheritance pattern of single gene diseases are referred to as Mendelian pattern of inheritance.

In the given case since the male and female both are equally affected hence it must not be Y linked disease. It is not autosomal dominant disease otherwise all the children would have been affected.

It is not X-linked dominant trait otherwise all the daughters would be affected

It must be a autosomal recessive trait or must be X-linked recessive trait where though the mother is not affected she is the carrier.

1)If it is an autosomal recessive trait then

Let the allele for dominant(normal) trait be ‘H’

Allele for recessive trait be ‘h’

Then the genotype of the father would be’ hh’

Genotype of the mother would be ‘Hh’

The cross would be

Gamates	H	h
h	Hh Not affected	hh affected
h	Hh Not affected	hh affected

By this cross it can be clearly seen that the probability of the fourth child affected by this disease is 0%

2)If it is an X-linked recessive trait then

Let the allele for dominant(normal) trait be XH

Allele for recessive trait be Xh

Then the genotype of the father would be’ XhY

Genotype of the mother would be XHXh

The cross would be

Gamates	XH	Xh
Xh	XHXh Not affected daughter	XhXh Affected daughter
Y	XHY Not affected son	XhY Affected son

By this cross it can be clearly seen that the probability of the fourth child affected by this disease is 0%