Question

Assume that genes a and b are linked and show 40% recombination. If ++ / ++ individual is crossed with ab / ab, then types and proportions of gametes in $F_1$ will be :

• A. ++ 20% : ab 20% : +b 20% : a + 40%
• B. ++ 50% : ab 50%
• C. ++ 25% : ab 25% : +b 25% : a + 25%
• D. ++ 30% : ab 30% : +b 20% : a + 20%

Answer 1

The correct option is D ++ 30% : ab 30% : +b 20% : a + 20%

Genes $aandb$ are incomplete in linkage.Here $40$ recombinants are reproduced through crossing over between segments of genes $aandb$.So the gametes of parental types $++andab$ should be produced in total frequency of $60$ and the total frequencies of recombinants such as $+b,b+or+a,a+$ should be 40%.Therefore F1 is ++ 30% : ab 30% : +b 20% : a + 20% 20%:a+20%