Question

1 Determine the gamete genotype(s) for this trait that each parent contributes.

Father Genotype:_____
Mother Genotype:_____
2. Draw a Punnett square that has the same number of columns and the same number of rows as the number of alleles contributed for this trait by the gametes of each parent. See
p. 258 Figure 7 for an example.
3. Write the alphabetical letter for each allele from one parent just above each column, and write the alphabetical letter for each allele from the other parent just to the left of each row.
4. In the boxes within the table, write the genotype of the offspring resulting from each combination of male and female alleles.


Questions : 1. List the possible offspring phenotypes that could occur.
Questions : 2. Evaluate what is the phenotypic ratio of the possible offspring? What is the genotypic ratio of the possible offspring?

Answer 1

In the monohybrid cross of true-breeding pea plants with contrasting seed colors (yellow and green), the offspring phenotypes are all yellow seeds with a phenotypic ratio of 100% yellow. Since yellow is dominant, all offspring genotypes are Yy, giving a genotypic ratio of 100% heterozygous Yy.

Step-by-step explanation:

When demonstrating a monohybrid cross using true-breeding pea plants with yellow seeds (YY) and green seeds (yy), we find that the parents only contribute one type of allele each due to their homozygosity. The father's genotype for yellow seeds is YY and the mother's genotype for green seeds is yy. After creating a Punnett square, all offspring of this cross would be heterozygous (Yy) and show the yellow seed phenotype, due to the dominance of the Y allele over the y allele.

List of possible offspring phenotypes that could occur: Since all offspring are Yy, the only phenotype is yellow seeds. The phenotypic ratio in the offspring is 100% yellow-seeded plants. With respect to the genotypic ratio, it is 100% heterozygous (Yy) because that is the only genotype produced in this cross.