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)

188

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.

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