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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