Solve this:
Q.151. In maize, coloured endosperm (C) is dominant over colourless (C), and full endosperm (F) is Dominant over shrunken (f). When a dihybrid of generation was test crossed, It produced Four phenotypes in the following percentage :-
Coloured full - 41 %
Coloured shrunken - 9 %
Colourless full - 9 %
Colourless shrunken - 41 %
From this data, what will be the distance between two non-allelic genes?
(1) 48 units
(2) 5 units
(3) 7 units
(4) 18 cM
Dear Student,
Recombination frequencies for pairs of genes reveal the distances between them along a chromosome.
Different gene pairs show different linkage rates because the closer together two genes are the less likely they will be separated by a recombination event.
The recombination frequency could be used to determine the physical distance between two genes.
1% RF = 1 cM = 1 map unit.
Recombination frequency = (no of recombinants/total progeny) x 100.
Given,
Coloured full - 41%.
Coloured shrunken - 9%
Colourless full - 9%
Colourless shrunken - 41%
So, Recombination frequency = (9+9)/100=18%
Thus distance between two non-allelic genes = 18 cM = 18 map unit
Option (4) is correct
Regards,
Recombination frequencies for pairs of genes reveal the distances between them along a chromosome.
Different gene pairs show different linkage rates because the closer together two genes are the less likely they will be separated by a recombination event.
The recombination frequency could be used to determine the physical distance between two genes.
1% RF = 1 cM = 1 map unit.
Recombination frequency = (no of recombinants/total progeny) x 100.
Given,
Coloured full - 41%.
Coloured shrunken - 9%
Colourless full - 9%
Colourless shrunken - 41%
So, Recombination frequency = (9+9)/100=18%
Thus distance between two non-allelic genes = 18 cM = 18 map unit
Option (4) is correct
Regards,