The abilify to taste phenylthiocarbamide (prc) is determined by a dominant gene (T). Individuals who can taste pTC are called tasters. A man who is a taster and has a non-taster mother marries a woman who is a taster. She has five siblings, three ofwhom are non-tasters. what are the chances that the children of this marriage will be nonasters?

1/4 or 25%

Explanation:

A man who is a taster and has a non taster mother is heterozygous and will have the genotype Tt.

A woman who is a taster will have the genotype Tt or TT. However, the woman has 5 siblings out of which 3 are non-tasters. It means that her parent had 6 children out of which 3 are tasters and 3 are non-tasters (1:1).

A 1:1 result is usually from a cross involving one homozygous non-taster and one heterozygous taster. Hence;

Tt x tt = Tt, Tt, tt and tt

It thus means that the taster woman is heterozygous with genotype Tt.

A marriage between Tt and Tt:

Tt x Tt = TT, 2Tt and tt

Recall that the trait is is a dominant one, hence TT and Tt are tasters while tt is a non-taster.

The chances of the children being a non-taster therefore is 1/4 or 25%.

The chances of producing children that will be non-tasters is 1/4.

Explanation:

The ability to taste is determined by T which can either be TT or Tt. Non tasters can only be inherited in the recessive condition (tt).

A man who is a taster with a non taster mother is heterozygous for the trait (Tt). He marries a woman who is a taster (she is a TT or Tt since she has siblings that are non tasters). To have a higher percentage of non-tasters in her (the woman) family, it means, one of the parent is heterozygous for the trait and the other is homozygous recessive for the trait. Thus, the taster woman will be an heterozygote.

Now the taster man that is heterozygous for the trait marries this heterozygote woman, the chances of producing children that will be nontasters is 1/4.