If two genes 'a' and 'b' are linked and show 20% recombination, the proportion of gemetes produced in F1 by a dihybrid ++/ab derived from a cross between ++/++ and ab/ab would be
A
++ 20% : ab 20% : + a 20% : +b 20%
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B
++ 80% : ab 20%
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C
++ 50% : ab 50%
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D
++ 40% : ab 40% : +a 10% : +b 10%
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Solution
The correct option is D ++ 40% : ab 40% : +a 10% : +b 10% As per the question, 20% recombination is present in genes 'a' and 'b' , so this shows 20% crossover and thus resultant progeny will be recombinants of both the parents. The linkage percentage here is 80% that means there will be 40% chances for progeny to be identical to both the parents.