Question

An individual with cd genes was crossed with wild type $++$. On test crossing $F_1$, the progeny was $+c105,+d115,cd880$ and $+$ $+$ $900$. Distance between cd genes is?

• A. $11$ map units
• B. $5.5$ map units
• C. $44$ map units
• D. $88$ map units

Answer 1

The correct option is A $11$ map units

Since, recombination frequency = $\frac{No.ofrecombinations}{Totalno.ofproperty}\times 100$

$=\frac{115+108}{115+105+900+880}\times 100$

$=11$ map units