Question

In a dihybrid cross, 4000 offsprings are produced in the F${}_2$ generation. How many offspring will approximately be phenotypically different from the parental generation? Assume that the parents are pureline for the contrasting traits of two characters, that is, seed shape and seed color.

• A. 2500
• B. 1500
• C. 1000
• D. 750

Answer 1

The correct option is B 1500

Let us take the example of a dihybrid cross considering seed shape and seed colour.
Purelines refer to the true breeding individuals homozygous for the particular trait. Let us assume that the male is homozygous for round and yellow seeds while the female is homozygous for wrinkled and green seeds.


$\begin{array}{ccccc}\text{Gametes}& \text{RY}& \text{Ry}& \text{rY}& \text{ry}\\ \text{RY}& \text{RRYY}& \text{RRYy}& \text{RrYY}& \text{RrYy}\\ & \text{(Round, yellow)}& \text{(Round, yellow)}& \text{(Round, yellow)}& \text{(Round, yellow)}\\ \text{Ry}& \text{RRyy}& \text{RrYy}& \text{Rryy}& \\ & \text{(Round, yellow)}& \text{(Round, green)}& \text{(Round, yellow)}& \text{(Round, green)}\\ \text{rY}& \text{RrYY}& \text{RrYy}& \text{rrYY}& \text{rrYy}\\ & \text{(Round, yellow)}& \text{(Round, yellow)}& \text{(Wrinkled, yellow)}& \text{(Wrinkled, Yellow)}\\ \text{ry}& \text{RrYy}& \text{RrYy}& \text{rrYY}& \text{rrYy}\\ & \text{(Round, yellow)}& \text{(Round, green)}& \text{(Wrinkled, yellow)}& \text{(Wrinkled, green)}\end{array}$

Possible phenotypes of F${}_2$ offsprings:

$\begin{array}{cc}\text{Phenotypes}& \text{No. of offsprings}\\ \text{Round yellow}& \text{9}\\ \text{Round green}& \text{3}\\ \text{Wrinkled yellow}& \text{3}\\ \text{Wrinkled green}& \text{1}\end{array}$

Possible genotypes of F${}_2$ offsprings:

$\begin{array}{cc}\text{Genotypes}& \text{No.of offsprings}\\ \text{RRYY}& \text{1}\\ \text{RrYY}& \text{2}\\ \text{rrYY}& \text{1}\\ \text{RRYy}& \text{2}\\ \text{RrYy}& \text{4}\\ \text{rrYy}& \text{2}\\ \text{RRyy}& \text{1}\\ \text{Rryy}& \text{2}\\ \text{rryy}& \text{1}\end{array}$

Parental phenotypes: Round yellow and wrinkled green seeds
Non parental phenotypes: Round green and wrinkled yellow seeds
Parental phenotype : Non-parental phenotype = 10 : 6
$\Rightarrow$ Out of 16 plants, 6 plants will be different from the parental phenotypes.
$\Rightarrow$ Out of 4000 plants, $\left(6/16\right)\times 4000=1500$ plants will be different from parental phenotypes.
So, 1500 out of 4000 plants in F${}_2$ generation will be phenotypically different from parents.