Question

Try this one: A homozygous yellow seeded plant is crossed with a green seeded plant. 1. Establish the genotypes: 2. Set up the square X 3. Determine the phenotypic ratios: out of 4 are yellow out of 4 are green Yellow To determine the parent genotypes, consider what heterozygous and homozygous means. You might make a chart. Green Genotype Phenotype YY Yy yy​

Answer 1

In a monohybrid cross between homozygous yellow-seeded (YY) and green-seeded (yy) pea plants, all F₁ offspring are heterozygous (Yy) with a yellow phenotype. A self-cross of the F₁ generation would yield F₂ genotypic ratios of 1 YY:2 Yy:1 yy and a phenotypic ratio of 3 yellow:1 green.

Step-by-step explanation:

To demonstrate a monohybrid cross, consider the example of true-breeding pea plants with yellow versus green seeds. The dominant seed color is yellow, so the parental genotypes are YY (yellow seeds) and yy (green seeds). We use a Punnett square to predict the outcomes of the cross.

The Punnett square setup would have YY along one side and yy along the other, representing the allocation of alleles into gametes. Upon fertilization, all offspring in the F₁ generation would be Yy, exhibiting the yellow phenotype due to the dominant yellow allele.

Thus, the phenotypic ratio in the F₁ generation is 4 yellow:0 green, since all are Yy with yellow seeds. If these F₁ heterozygotes were self-crossed to produce an F₂ generation, the Punnett square would show genotypic ratios of 1:2:1 for YY:Yy:yy, with a phenotypic ratio of approximately 3 yellow:1 green.