Answer:
The ans will be [tex]\frac{3}{8}[/tex]
Explanation:
As phenylketonuria is a recessive autosomal disease, and both the parents are carrier, so they both carry one defective gene and one good gene.
If they produced 4 children, the probability of diseased and non diseased child will be as follows.
diseased: [tex]\frac{1}{4}[/tex]
non diseased: [tex]\frac{3}{4}[/tex]
And the probability of male and female child will be [tex]\frac{1}{2}[/tex].
So, the probability of phenotypically normal first girl child will be [tex]\frac{3}{4}[/tex] × [tex]\frac{1}{2}[/tex] ⇒ [tex]\frac{3}{8}[/tex]
