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
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The correct answer is (c) ++40 : ab40 : +a10 : +b10
As given in the question, a and b will show 20% recombination, this means that there will exist a 20% crossover and the progeny will result in a recombination of both parents.
The percentage of linkage here is 80% so there is a 40% and 40% chance of getting the progeny identical to both parents.
As given in the question, a and b will show 20% recombination, this means that there will exist a 20% crossover and the progeny will result in a recombination of both parents.
The percentage of linkage here is 80% so there is a 40% and 40% chance of getting the progeny identical to both parents.
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