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Q.

The number of F2 plants similar to F1 phenotype and genotypes respectively in a total of 128 plants in F2 progeny of a mendelian dihybrid cross are


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a

  48, 32

b

72, 32

c

48, 24

d

72, 48 

answer is B.

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Detailed Solution

Concept- A dihybrid cross occurs when two different genes that differ in two observed traits are crossed. Lets assume there are two genes, A and B. Now, suppose AABB and aabb are crossed, so the gametes produced will be AB and ab respectively, which will combine to form AaBb as F1 progeny.
Now, these F1 offsprings when crossed with one another, these will produce offspring with the following genotypes-
Question ImageThe phenotype of F1 progeny was, both genes are showing dominant character and its genotype is AaBb.
If we look at the  F2 generation progenies, 9 of the 16 plants produced show phenotypes that are similar to those of the parent plant, (both characters exhibit dominant traits). 4 of these plants share similar genotype with their parent plant.
In question, we have been given a total of 128 plants, so we will apply th evalues that are obtained to us to solve the question further.
Number of F2 progenies that are similar to F1 phenotype are -
 916×128=72
Number of F2 progenies that are similar to F1 genotype are -
 416×128=32
So, the right answers are 72 and 32 respectively.
Hence, option 2 is the correct answer.
 
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