Two genes 'a' and 'b' show 20% recombination. | vedclass

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(C) The recombination frequency is given as $20\%$. This means the frequency of recombinant gametes ($a+$ and $+b$) is $20\%$ in total. Since there ar

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$++ \, 40\% : ab \, 40\% : a+ \, 10\% : +b \, 10\%$

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(C) The recombination frequency is given as $20\%$. This means the frequency of recombinant gametes ($a+$ and $+b$) is $20\%$ in total. Since there are two types of recombinant gametes,each will be $20\% / 2 = 10\%$. The parental gametes ($++$ and $ab$) will make up the remaining $100\% - 20\% = 80\%$. Since there are two types of parental gametes,each will be $80\% / 2 = 40\%$. Therefore,the proportions are: $++ = 40\%$,$ab = 40\%$,$a+ = 10\%$,and $+b = 10\%$. Thus,the correct option is $C$.

Two genes '$a$' and '$b$' show $20\%$ recombination. What will be the proportion of gametes produced by a dihybrid $F_1$ individual with the genotype $++/ab$,derived from a cross between $++/++$ and $ab/ab$?

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