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Two linked genes a and b show 20 % recombination. The individuals of a dihybrid cross between + + / + + crossed with ab / ab shall show gametes
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The correct answer is:
+ + 40 : ab 40 : + a 10 : + b 10
Two linked genes a and b show 20 % recombination, thus linkage is 80%. The given phenotype of parents is ++/++ and ab/ab.
The parental genotype will be ++ and ab, while the recombinant type will be +a and +b.
Since % of recombination i.e. cross over frequency is 20%, the recombinant phenotype will be 10% each.
+a: 10%
+b: 10%
Similarly, since linkage is 80%, parental types will be 80% or 40% each for ++ and ab respectively.
The parental genotype will be ++ and ab, while the recombinant type will be +a and +b.
Since % of recombination i.e. cross over frequency is 20%, the recombinant phenotype will be 10% each.
+a: 10%
+b: 10%
Similarly, since linkage is 80%, parental types will be 80% or 40% each for ++ and ab respectively.
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