Question

Suppose that gene (b) is sex-linked, recessive, and lethal. A man and a woman who is heterozygous for this gene have many children. What would you predict the sex ratio of the children to be? What are the genotypes of the parents and children?

Answer 1

Answer:

Sex ratio of the children can be predicted by making observations using Mendelian crossing methods. The genotype of parents are  XX_(b ) and  XY and the children are XX,XY,XX_(b ),X_(b ) Y.

Explanation:

Given that the gene b is sex linked recessive and lethal. Being sex linked it will be associated with the X chromosome of affected or carrier individuals.  A man and woman who is heterozygous for this gene have children.  

Females have two X chromosomes while males have one X and one Y chromosome. Thus one affected X chromosome doesn’t make a female an affected individual. She will be a carrier. If there only one X chromosome of the male is affected he will be an affected individual.  

The genotype of the woman is   and the genotype of the man is XY.  

The cross can be represented as  

$XX \ (b ) \times XY$

XX   X_(b ) Y  XX_(b )   X_(b ) Y

Among the progenies the chance of the child being a female is ½ and being a male is also ½ among the two females one will be the carrier and the other is free of the lethal gene b.

Among the two males, one male will be affected by inheriting X_(b )chromosome from the mother and the other will not be affected.