Question

Suppose you perform a cross between two true-breeding stocks of garden peas, with respect to two characteristics of their flowers. The first parent had red, axial flowers and the second had white, terminal flowers; all F1 individuals were like the first parent. If you obtained 1000 F2 offspring by allowing the F1s to self-hybridize, about how many of them would expect to have red, terminal flowers? (Assume that the genes assort independently)

Answer 1

You can expect roughly 188 red, terminal flowered offspring in the F2 generation from the given dihybrid cross, based on a 9:3:3:1 phenotypic ratio typical for independently assorting genes.

Step-by-step explanation:

The question involves a dihybrid cross, given that there are two traits being considered (flower color and flower position), with red and axial being dominant over white and terminal respectively. The F1 generation displaying all red, axial flowers suggests that red and axial traits are dominant. The F2 generation, resulting from the F1's self-hybridization, will display a 9:3:3:1 phenotypic ratio, typical for dihybrid crosses of independently assorting genes.

For red, terminal flowers, the expected phenotype ratio is 3/16 of the F2 generation (since red is dominant and terminal is recessive). Thus, the calculation would be 3/16 of 1000 F2 offspring, equating to 187.5, which can be rounded to approximately 188 red, terminal offspring.

Answer 2

188

Step-by-step explanation:

Let's assume that the allele for red flower is "R" and the allele for white flower is "r". Similarly, allele "T" is responsible for axial flower and allele "t" gives terminal flower.

Genotype of first parent with red, axial flower = RRTT

Genotype of second parent with white terminal flower = rrtt

F1 progeny = RrTt

A cross between RrTt and RrTt obtains F2 progeny in 9:3:3:1.

Since the ratio of plants with red terminal flower is = 9/16.

Out of 1000 F2 plants, the number of plants with red, terminal flowers = 9/16 x 1000 = 188.