Question

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

• A. $++80:ab:20$
• B. $++50:ab:50$
• C. $++40:ab40:+a10+b:10$
• D. $++30:ab30:+a20:+b:20$

Answer 1

The correct option is C $++40:ab40:+a10+b:10$

The combined linkage frequency between the parental combinations ++ and ab is 80%

The combined recombination frequency between the recombinant combination +a and +b is 20%

The individual frequencies would be half of the combined.

So, the correct option is ' ++40 : ab40 : +a10 : +b10'.