Question

If two genes 'a' and 'b' are linked and show 20% recombination, the proportion of gemetes produced in $F_1$ by a dihybrid ++/ab derived from a cross between ++/++ and ab/ab would be

• A. ++ 20% : ab 20% : + a 20% : +b 20%
• B. ++ 80% : ab 20%
• C. ++ 50% : ab 50%
• D. ++ 40% : ab 40% : +a 10% : +b 10%

Answer 1

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.

So the correct answer is ' D'.