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Question

In experimental individual homozygous for ab genes were crossed with wild type (++). The F1 hybrid thus produced was test crossed and progenies produced in following ration ++548,ab 340,+a 44,+b 68

Calculate the distance between b and a genes.

A
46 m.u
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B
22.4 m.u
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C
44 m.u
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D
11.2 m.u
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Solution

The correct option is A 11.2 m.u
The recombination frequency for this crossbreed is

= number of recombinant progenytotal number of progeny ×100

= 44+68548+340+44+68 ×100

= 11.2 %

1% recombinant frequency is equivalent to 1 centimorgan or 1 map unit or m.u.
Therefore the distance between a and b genes is 11. 2 m.u.
Answer D is correct.

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