- 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.
If a widow's peak is dominant, people with both the heterozygous and homozygous genotypes (allele combinations) would result in the widow's peak phenotype (physical attribute). Homozygous genotypes contain two of the same alleles (homo= same). Heterozygous genotypes contain a dominant along with a recessive allele (hetero= different).
If both parents are heterozygous for the trait of widow's peaks, then Mr. Smith is not necessarily unjustified.
Let's use the allele (W) to represent widow's peak and (w) to represent someone without a widow's peak. Then, we can set up the following Punnett square to represent to heterozygous parents for the trait:
W WW Ww
w Ww ww
The genotype frequencies wold be:
25% homozygous dominant (WW)
50% heterozygous (Ww)
25% homozygous recessive (ww)
The phenotype frequencies would be:
75% show the widow's peak trait (Ww or WW- recall that if a dominant allele is present, then it will show).
25% without widow's peaks (it IS possible that Mrs. Smith has remained faithful).