Chapter 2, Problem 31CONQ

A true-breeding plant with round and green seeds was crossed to a true-breeding plant with wrinkled and yellow seeds. The ${\text{F}}_{\text{1}}$ plants were allowed to self-fertilize. What is the probability of obtaining the following plants in the ${\text{F}}_{\text{2}}$ generation: two that have round, yellow seeds; one with round, green seeds; and two with wrinkled, green seeds?

Summary Introduction

To review:

The probability of obtaining the following plants in the F₂ generation: two that have round and yellow seeds, one with round and green seeds, and two with wrinkled and green seeds.

Introduction:

True breeding plants are those plants which produce offsprings of the same type with same traits. Plant with genotype YY is an example of true breeding plant. Here Y represents the seed color that is yellow and it is easily visible that both the alleles are dominant forms and hence are true breeding.

Explanation of Solution

Genotypes to be used; RR: round, rr: wrinkled, Rr: round, YY: yellow, yy: green and Yy: yellow. Round is dominant over wrinkled seed whereas yellow seeds are dominant over green seeds. F₁ plants are allowed to self-fertilize:

Gametes	R	r
Y	RY	rY
y	Ry	ry

Results of selfing: RY, rY, Ry and ry. Further, crossing for F₂ generation:

Gametes	RY	rY	Ry	ry
RY	RRYY	RrYY	RRYy	RrYy
rY	RrYY	rrYY	RrYy	rrYy
Ry	RRYy	RrYy	RRyy	Rryy
ry	RrYy	rrYy	Rryy	rryy

Probability of round, yellow seeds:

$\begin{array}{c} p=\frac{\text{Number }\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{of round yellow seeds}}{\text{Total number of seeds produced}}\\ =\frac{9}{16}\end{array}$

Probability of round, green seeds:

$\begin{array}{c}q=\frac{\text{Number }\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{of round green seeds}}{\text{Total number of seeds produced}}\\ =\frac{3}{16}\end{array}$

Probability of wrinkled, green seeds:

$\begin{array}{c} r=\frac{\text{Number }\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{of wrinkled green seeds}}{\text{Total number of seeds produced}}\\ =\frac{1}{16}\end{array}$

Calculation with the help of multinomial equation:

$\begin{array}{c}n={\text{5, number of total plants in F}}_{\text{2}}\text{ according to given condition}\\ a=\text{2, power of }p\\ b=1,\text{power of }q\\ c=2,\text{power of }r\end{array}$

Upon substituting the values in the equation:

$\begin{array}{c}P=\frac{n!}{a!b!c!}{\left(p\right)}^a{\left(q\right)}^b{\left(r\right)}^c\\ =\frac{5!}{2!1!2}{\left(\frac{9}{16}\right)}^2{\left(\frac{3}{16}\right)}^1{\left(\frac{1}{16}\right)}^2\end{array}$

$\begin{array}{c}=\left(30\right)\left(\frac{81}{256}\right)\left(\frac{3}{16}\right)\left(\frac{1}{256}\right)\\ =\left(30\right)\frac{243}{1048576}\\ =0.007\end{array}$

Conclusion

$\text{Therefore, it can be concluded that the probability for given condition is 0}\text{.007}\text{.}$