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Introduction to Morgan's Experiments

A test cross ...

Question

A test cross of $F_{1}$ 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)'.

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