In pea plants, the gene for the color of the seed has two alleles. in the following punnett square shown below, the dominant allele (y) represents yellow and the recessive allele (y) represents green. punnett square showing a grid that is blank with 4 empty boxes, two columns and two rows. the female parent plant across the horizontal side, top, of the grid indicates a yellow phenotype and a genotype of yy. the male parent plant along the vertical side of the grid indicates a yellow phenotype and a genotype of yy. based on the punnett square, what percentage of offspring would be expected to have green seeds?

Parents : Yy (female)  x  YY (male)

Gametes: Y  and y (from female) ; Y and Y (from male)

F1 generation: YY (yellow); Yy (yellow) ; YY (yellow); Yy (yellow)

So we will observe yellow seeds only in the F1 generation. Since there is no homozygous recessivie"yy" genotype in F1 generation, 0% of the offspring would be expected to have green seeds.

None of the offspring is going to have green seeds.

The male parent = YY

And they are yellow

The female parent = Yy

and they are also yellow.

We do this to get the phenotype of the other generation. We match Yy with YY

YY * Yy

        Y                y

Y     YY              Yy

Y     YY              Yy

This gives a generation with 50 percent having the fathers YY and the other 50% having Yy

Read more on  the punnet square here:

link

answer; i believe that the correct answer is ; males are hemizygous for the x chromosome;