Genetics
Percentage of recombination between A and B is 9%, A and C is 17% and B and C is 26%. The arrangement of genes would be
What map unit(Centimorgan) is adopted in the construction of genetic maps?
$$1$$ map unit represent $$1\%$$ cross over.
Map unit is used to measure genetic distance. This genetic distance is based on average number of cross over frequency.
So, the correct answer is 'A unit of distance between genes on chromosomes, representing 1%
1% cross over'
Percentage of recombination between A and B is 9%, A and C is 17% and B and C is 26%. The arrangement of genes would be
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?
The linkage map of X chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be
The first attempt to show linkage in plants was done in
How many linkage groups are there in nucleoid of bacteria?
If distance between genes on chromosome is more, then gene shows
In a genetic map, distance between P and Q is 12.4 map unit, Q and R is 7.6 map unit and P and R is 18.2 map unit instead of 20 map unit. This is because of
Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybrid, the F$$_2$$ segregation will show