A tall pea plant with yellow seeds is heterozygous for height and seed color ([tex]TtYy[/tex]). This plant is crossed with a pea plant heterozygous for height but homozygous recessive for seed color ([tex]Ttyy[/tex]). If 145 plants are produced, how many are expected to be tall and have yellow seeds?
Answer (3)
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The question is asking to predict the phenotype of the offspring from a cross between two pea plants, where one is heterozygous for height and seed color with genotype TtYy and the other is heterozygous for height but homozygous recessive for seed color with genotype Ttyy.
To answer this, we need to set up a Punnett square to predict the genotypes and phenotypes of the offspring.
In a Punnett square, we list all possible gametes that each parent can produce, and then we cross them to find the genotypic ratio of the offspring. For the first parent (TtYy), the gametes are TY, Ty, tY, and ty. For the second parent (Ttyy), the gametes are TY, Ty, tY, and ty.
When we cross these, we will get various combinations, with 9/16 of the offspring expected to exhibit the dominant phenotypes for both traits (tall and yellow seeds), considering the independent assortment and dominance of the alleles.
Since there are 145 offspring, we would expect approximately 9/16 of them to be tall with yellow seeds. Doing the calculation, 145 * (9/16) = approximately 81.56, which can be rounded down to 81 offspring expected to be tall and have yellow seeds.
By performing a dihybrid cross between the given genotypes, we calculate that approximately 54 out of 145 plants will be tall and have yellow seeds. This is based on the genetic ratios derived from the Punnett square analysis for height and seed color. The expected ratio of tall yellow plants is 3 out of 8 in this specific cross.
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