A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste the chemical phenylthiocarbamide (autosomal dominant; common allele), but their mothers could not. Give the genotypes of the couple. If the genes assort independently and the couple has four children, what is the probability of all of them being brachydactylous

The correct answer is -

A) BbTt x bbTt

B) .5^4 or 1/16 or 0.0625

Explanation:

As it is given that Brachydactylus and PTC tasters both traits are autosomal dominant conditions which mean only one allele would be enough.

For Branchydactylus and For tasting :  man and women will be heterogeneous.

Hence,

The Genotype of man = BbTt

The Genotype of wife = bbTt

b. Answer = 0.0625

For Branchydactylus:

Bb X bb

Possible genotypes:

B b

bBb( Brachydactylus)bb(normal)

bBb (Brachydactylus)bb (normal)

The probability of a single child being Branchydactylus = 2/4 = 0.5

So,  Probability of all 4 child being Branchydactylus = .5 x .5 x .5 x .5 = 0.0625