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Rediscovery of Mendelism

In experiment...

Question

In experimental individual homozygous for $a b$ genes were crossed with wild type $(+ +)$ . The $F_{1}$ hybrid thus produced was test crossed and progenies produced in following ration $+ + 548, a b 340, + a 44, + b 68$
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

= $\frac{n u m b e r o f r e c o m b i n a n t p r o g e n y}{t o t a l n u m b e r o f p r o g e n y}$ $\times 100$

= $\frac{44 + 68}{548 + 340 + 44 + 68}$ $\times 100$

= $11.2$ %

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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