Single Choice

In maize, coloured endosperm (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 = 45%, Coloured-Shrunken = 5%, Colourless-Full = 4% Colourless-Shrunken = 46% From these data what would be the distance between the two non-allelic genes?

A48 unit
B9 unit
Correct Answer
C4 unit
D12 unit

Solution

The percentage of recombinants formed by F$$_1$$ individuals is the measure of distance between genes under study. The higher percentage of recombinants for a pair of traits means those two loci are present far apart. The recombinants percentage is taken as the distance in centimorgans (cM); 1% recombinants means the genes are present 1cM apart. Since, the recombinants percentage in question is 5% (coloured-shrunken) + 4% (colourless-full) = 9% , the genes are 9.0 cM apart. Thus, the correct option is B.

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