Answer:
The correct answer is - 3/32.
Step-by-step explanation:
We know that freckles F is dominant over no freckles f, similarly, Widow's peak W is dominant over straight hairline w and attached earlobes are dominant over free earlobes.
So, the combination of alleles for each phenotype would be:
No freckles = ff , Freckles = FF, Ff
Free earlobes = ee , Attached earlobes = EE, Ee
Straight hairline = ww , Widow's peak = WW, Ww
In the question, it is given that the first child has no-freckles, straight hairline, and detached earlobes that would be - ffwwee genotype and as it is given that all alleles are in recessive which means both parents must have recessive alleles.
So, the correct genotype of the man will be FfWwEe and the woman will be FfwwEe.
(man = freckles, widow's peak, and attached earlobes (F_W_E_) and woman = freckles, straight hairline, and attached earlobes (F_w_E_))
then cross between: FfWwEe × FfwwEe
The probability of the freckles and no freckles would be-
Ff × Ff = 3/4 freckles, 1/4 no freckles
The probability of the widows and straight hairline would be-
Ww × ww = 1/2 widows peak, 1/2 straight hairline
The probability of the free earlobes and attached earlobes would be-
Ee × Ee = 3/4 attached, 1/4 free earlobes
Combined possibility of all three phenotypes: no freckles, straight, attached would be = 1/4 × 1/2 × 3/4 = 3/32
Thus, the correct answer is - 3/32.