To answer these exercises we need to do the proper punnet squares, in the first case we have a supersmart pure bread that is to say dominant homozygous, and a supersmart hybrid that is to say heterozygous. So the alleles and squares are as follows.
Alleles
W: supersmart
w: normal inteligence
Genotypes
Male: WW
Female: Ww
Parental cross:
WW x Ww
Punnet square
[tex]\begin{bmatrix}{\square} & {W} & {W} \\ {W} & {WW} & {WW} \\ {w} & {Ww} & {Ww}\end{bmatrix}[/tex]
Genotypic and phenotypic frequencies
Supersmart homozygous WW: 2:4= 2/4= 50%
Supersmart heterozygous Ww: 2:4= 2/4= 50%
Therefore phenotypically all the offspring would be super smart (100%) and 0% would be normal smarts
Now we can solve the second square
Here we have a male of normal smarts so is ww, and the female is homozygous recessive so is also ww
Genotypes
Male: ww
Female: ww
Parental cross:
ww x ww
Punnet square
[tex]\begin{bmatrix}{\square} & {w} & {w} \\ {w} & {ww} & {ww} \\ {w} & {ww} & {ww}\end{bmatrix}[/tex]
Genotypic and phenotypic frequencies
Normal smarts ww: 4:4= 4/4= 100%
Therefore all the offspring would be normal smarts (100%) as both parents are recessive, and 0% super smarts.
