In humans, freckles (F), widow’s peak (W) and attached earlobes (E) are all unlinked dominant traits. A man with freckles, widow's peak and attached earlobes marries a woman with freckles, straight hairline, and attached earlobes. Their first child has no-freckles, straight hairline, and detached earlobes. What is the probability that their second child has no-freckles, straight hairline, and attached earlobes? Select one:

The correct answer is - 3/32.

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.