Question

Two linked genes a and b show $20\%$ recombination. The individuals of a dihybrid cross between $++/++$ crossed with $ab/ab$ shall show gametes -----------------------.

• A. + + 80 : ab : 20
• B. + + 50 : ab : 50
• C. + + 40 : ab 40 : + a 10 : + b 10
• D. + + 30 : ab : 30 : + a 20 : + b : 20

Answer 1

Answer: (C)

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, when the linkage is $80\%$ , parental types will be $80\%or40\%$ each for $++$ and $ab$ , respectively.