Answer:
1/4 or 25%
Step-by-step 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%.