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 ...

Answer 1

Let the allele of green seed color be G

Let the allele of white seed color be g

As the plant is heterozygous green its genotype is Gg.

When this plant is self crossed we get: Gg x Gg

	G	g
G	GG (green seeds)	Gg (green seeds)
g	Gg (green seeds)	gg (white seeds)

So, from the punnett square it is clear that the,

probability of getting green seeds= 3/4

probability of getting white seeds= 1/4.

A)

Probability of 4 green seeds and 2 white seeds

We multiply the probability of green seeds 4 times and white seeds 2 times.

=3/4 x 3/4 x 3/4 x 3/4 x 1/4 x 1/4

= 0.0197 = 0.020

B)

Probability of all green seeds

We multiply the probability of green sees 6 times

= 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4

= 0.1779 = 0.180

C)

Probability of at least one white seed

= 1 - Probability of all green seeds ( As in all cases other than in cases of all green seeds, there will be at least one white seed)

= 1 - 0.180

= 0.820