A test cross of F1 flies +a/ +b produced the following offspring ++ / ab = 9 ab / ab = 9 +b / ab = 41 a+ / ab = 41 What will be distance between linked gene:-
A
82 cM
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B
18 cM (cis)
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C
20 cM
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D
18 cM (trans)
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Solution
The correct option is A 18 cM (trans)
The
distance between linked genes can be calculated by calculating the
recombination frequency of the linked genes. Here, the recombination frequency
of ++/ab and ab/ab is = 9/100 x100 + 9//100 x 100 = 18 cM. So, the correct answer is '18 cM (trans)'.