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Here is the complete 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:
A) 4 green and 2 white seeds?
B) all white seeds?
C) at least 5 white seeds?
Answer:
A) 0.3
B) 0.0002
C) 0.001
Explanation:
Let the allele for Green plant be = P
Let the allele for white plant be = p
Let the heterozygous plant be Pp, if self fertilization occurs we have:
P p
P PP Pp
p Pp pp
From the punnet square above, we have the following cross;
PP, Pp, Pp, pp
Since P is dominant to p; we have a phenotypic ratio of 3:1 between the dominant phenotype ( PP, Pp, Pp ) and recessive phenotype ( pp )
NOW, the probability for the green plant(i.e PP or Pp) will be:
the probability for the white plant (pp) will be:
A)
what is the probability that 6 seeds in a single pod of the progeny plant will consist of 4 green and 2 white seeds?
=
The probability includes the use of permutation:
=
= 0.2966
≅ 0.3
B)
What is the probability that 6 seeds in a single pod of the progeny plant will consist of all white seeds?
Since probability for white seed =
The probability that 6 seeds will consist of all white seeds will be =
= 2.44 × 10⁻⁴
= 0.000244
≅ 0.0002
C)
What is the probability that 6 seeds in a single pod of the progeny plant will consist of at least 5 white seeds?
For at least 5 white seeds, the probability will consist of 5 white seeds and 6 white seeds.
For 5 white seeds; we have:
5 White and 1 Green Since there are 6 permutations.
The probability For 5 white seeds will therefore be :
= 7.32 × 10⁻⁴
= 0.000732
For 6 white seeds; we have:
Only one permutation for 6 whites since there will no be any presence of any green seed.
= 2.44 × 10⁻⁴
= 0.000244
∴
The total probability for at least 5 white seeds = 0.000732 + 0.000244
= 9.76 × 10⁻⁴
= 0.000976
≅ 0.001