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In a Mendelian dihybrid cross, the proba...

In a Mendelian dihybrid cross, the probability of getting seeds with genotype Rryy, RrYy, rrYy and RrYY in `F_(2)` generation is respectively

A

`(2)/(16):(4)/(16):(1)/(8):(1)/(8)`

B

`(2)/(16):(2)/(16):(2)/(16):(2)/(16)`

C

`(4)/(16):(4)/(16):(2)/(16):(2)/(16)`

D

`(1)/(8):(1)/(4):(2)/(8):(1)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
A
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