Mrs. and mr. smith both have widow’s peaks (dominant). their first child also has a widow’s peak, but their second child doesn’t. mr. smith accuses mrs. smith of being unfaithful to him. is he necessarily justified? why or why not? work the genetics problem predicting the frequencies of the versions of this trait among their prospective children.

Mr. Smith is not justified, Mrs. Smith can not be blamed necessarily for being unfaithful

Explanation:

Widow's peak is a dominant trait which means that it will be expressed both in homozygous and heterozygous condition. If "A" is the dominant allele and "a" is the recessive allele, the trait can be represented as AA or Aa.

Both Mr. Smith and Mrs. smith expressed the trait so they can either have AA or Aa genotype. If even one of them had AA genotype all the offspring would have the trait. But if both of them are heterozygous for it:

     A     a

A  AA  Aa

a  Aa   aa

There is 75 % probability of the child to have the trait (AA or Aa) but there is also 25% probability that the child does not express the trait (aa). Their second child belongs to this category and hence Mrs. Smith can not be blamed for being unfaithful.

answer///cannot be totally predicted;

chris is feeling fitt not fat