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In maize, coloured endosprem (C) is dom...

In maize, coloured endosprem (C) is dominant over colourless (c) , and full endosperm (R) is dominant over shrunken (r). When a dihybrid of `F_(1)` generation was test crossed, it produced four phenotypes in the following percentage :
Coloured full - 48 % Coloured shrunken - 5 %
Colourless full - 7 % Colourless shrunken - 40 %
From this data, what will be the distance between two non-allelic genes ?

A

48 units

B

5 units

C

7 units

D

12 units

Text Solution

Verified by Experts

The correct Answer is:
D
Given that recombinat percentage is 7 % and 5 % therefore, total recombinants would be `7 + 5 = 12 %`. It is known that one map unit is the distance that yields `1 %` recombinant chromosomes. Hence distance between two non-allelic genes is 12 map units.
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