Question

May you please be able to help me on this

Answer 1

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

`\begin{bmatrix}{\square} & {W} & {W} \\ {W} & {WW} & {WW} \\ {w} & {Ww} & {Ww}\end{bmatrix}`

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

`\begin{bmatrix}{\square} & {w} & {w} \\ {w} & {ww} & {ww} \\ {w} & {ww} & {ww}\end{bmatrix}`

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.