Question

The enzyme glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive gene on the X chromosome in humans. A phenotypically normal woman (whose father had G6PD) is married to a normal man. (a) What fraction of their sons would be expected to have G6PD? (b) If the husband had G6PD, would it make a difference in your answer in part (a)? Why or why nor?

Answer 1

a)1/2

b. No

Step-by-step explanation:

A).Assuming the allele for the G6PD deficiency is represented with (g). Therefore this will be carried on one of the XX- chromosomes of the mother.Since her father has the defective allele on his X-chromosomes of XY-chromosomes , she must have obtained the defective allele from this X allele inherited from her father. Definitely she is a carrier.( the genetic status for this disease was not stated for her mother).

Therefore during mating based on Monohybrid cross,

two Sons will be produced. One will have this defective allele( XY(g)) while the other will be normal.(XY,normal)

this will be fraction of 1/2.

the same proportion of inheritance should be observed in the the daughters since the mother of the wife is assumed to be normal.

B. No. it will not make any differences because it is only the Y-chromosomes that determines male sex .Fathers do not give X -chromosomes to male children, and since it is the X-chromosomeswhich is normal in the father that bears this defective gene,it will make no difference.

Answer 2

1/2 , no difference to the fraction of the sons

Step-by-step explanation:

Since the disorder is inherited recessive gene on the X. The woman will be a carrier since her father had G6PD. Therefore her genotype is X Xh

The normal man genotype will be XY

The genotype of their offspring will be

• Daughters XX, XXh
• Sons XY, XhY

Therefore 1 out of the 2 sons(1/2) will have the disorder

If the father had the disorder, it will not change the fraction of the sons because the disorder is on the X chromosome. However one of the daughters will have it XhXh