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Inbreeding

Individuals h...

Question

Individuals homozygous for CD genes were crossed with wild type (AA).
The F1 dihybrid thus produced was test crossed and it produced offsprings in the following ratio:
• AA – 950
• CD – 800
• AD – 130
• AC – 110
What is the distance between C and D genes?

A

88 cM

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B

12 cM

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C

44 cM

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D

56 cM

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Solution

The correct option is B 12 cM

Here offsprings with parental combinations
950 + 800 = 1750.

Offsprings with non-parental combinations and with recombinations
130 + 110 = 240.

Total offsprings = 1750 + 240 = 1990.

∴ Percentage of non-parental combination or recombination frequency = 240/1990 × 100
= 12%

As recombination frequency is 12%, hence distance between genes c and d is 12 m.u. or cM.

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