Two linked genes a and b show 20% recombination. The individuals of a dihybrid cross between ++/++ × ab/ab shall show gametes (a) ++80 : ab20 (b) ++50 : ab50 (c) ++40 : ab40 : +a10 : +b10 (d) ++30 : ab30 : +a20 : +b20. Plzz explain me in detail
Dear Student.
Please find below the solution to the asked query.
The correct answer is:
(c) ++ 40 : ab 40 : + a 10 : + b : 10
As it is mentioned a and b shows 20% recombination, this suggests there will be 20% crossing over & the progeny will be a recombination of both the parents.
The percentage of linkage is 80%, so for both parents, there will be 40% and 40% chance of getting the progeny identical to the two parents.
Hope this information will clear your doubts about the topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards.
Please find below the solution to the asked query.
The correct answer is:
(c) ++ 40 : ab 40 : + a 10 : + b : 10
As it is mentioned a and b shows 20% recombination, this suggests there will be 20% crossing over & the progeny will be a recombination of both the parents.
The percentage of linkage is 80%, so for both parents, there will be 40% and 40% chance of getting the progeny identical to the two parents.
Hope this information will clear your doubts about the topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards.