Question

Question: The seeds in bush bean pods are each the product of an independent fertilization event. Green seed color is dominant to white seed color in bush beans. If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny plant will consist of 2 green and 4 white seeds?

Express your answer to three decimal places (example: 0.123)

Answer 1

Given that green seed color (G) is dominant to white seed color (g) in bush beans.

A heterozygous plant with green seeds self-fertilizes:

So, the cross will be: Gg X Gg

Result: 1 GG: 2 Gg: 1gg

Note that: Probability of having green seeds = ¾ and Probability of having white seeds = ¼

Probability of having 2 green out of 6 = 2/6 * ¾ = 6/24 = 1/4

Chances of having 4 white seeds out of 6 = 2/3 * 1/4 = 2/12 = 1/6

Chances of having 2 green and 4 white out of 6 = 1/4 * 1/6 = 1/24 = 0.042