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: $Pr(G)=\frac{3}{4}$

the probability for the white plant (pp) will be: $Pr (W)= \frac{1}{4}$

what is the probability that 6 seeds in a single pod of the progeny plant will consist of 4 green and 2 white seeds?

$Pr(4G2W)$ = $(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})$

The probability includes the use of permutation:

$Permutation(P)=\frac{6!}{(4!*2!)} * (\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})$

= $15*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})$

= 0.2966

≅ 0.3

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 = $\frac{1}{4}$

The probability that 6 seeds will consist of all white seeds will be = $(\frac{1}{4})^6$

= 2.44 × 10⁻⁴

= 0.000244

≅ 0.0002

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 :

$Pr(5 seeds)=(\frac{1}{4})^5*(\frac{3}{4})^1 *6$

= 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.

$Pr(6seeds)=(\frac{1}{4})^6$

= 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

3 0

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saw5 [17]

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