Question

In a dihybrid cross RRYY×rryy, F2 progeny of RRYY, RRYy, RrYY and RrYy occurs in the ratio of
A- 1:1:1:1
B- 9:3:3:1
C- 1:2:2:1
D- 1:2:2:4.
​

Answer 1

Answer:

The answer is B -9:3:3:1

Answer 2

The required ratio will be :

$\boxed{ \bold{9 : 3 : 3 : 1}}$

But why?

• First of all, a dihybrid cross is a genetic cross between two parents taking into account two different traits.

• So, the 1st dihybrid cross was made as follows; R/r for Round/Wrinked , Y/y for yellow/green

Parents:

$RRYY \: \: \: \: \: \: \: \: \times \: \: \: \: \: \: \: \: \: rryy$

$(dominant) \times (recessive)$

Now, the various gametes can be :

• RY
• Ry
• rY
• ry

Then, if we perform the PUNNET SQUARE cross (as shown in the diagram), we will get 4×4 = 16 squares.

Then , the ratio for various traits of progenies will be:

RRYY : RRYy : RrYY : rryy

$\implies 9 \: \: : \: 3 \: : \: \: 3 : \: 1$