$F - 1$ individual of the cross AAbb X aaBB was test crossed .If both non-allergic genes are 24 map units apart then calculate the percentage of aaBb:

24 %

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12 %

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76 %

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38 %

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Solution

The correct option is B $12 %$
It is given that the both non-allelic genes are 24 map units apart. It means they have 24% recombination. Each recombinant, that is Aabb and aaBb will be 12%.
The parental recombination = 100 - recombinanat recombination = 100 - 24 = 76%. Each parental combination, that is AABB and aabb will be 38%.
Thus, the correct answer is option B.

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When AABB and aabb are crossed, in F $_{2}$ generation the ratio of AaBb will be

_{2}

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