Single Choice

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

AHigher number of the recombinant types
BSegregation in the expected 9:3:3:1 ratio
CSegregation in 3:1 ratio
DHigher number of the parental types
Correct Answer

Solution

The closely located genes show the tendency of linkage and they are transmitted together to the next generation. If two closely located genes show linkage, they do not show cross over and result in formation of higher number of the parental types. Option D is correct.


SIMILAR QUESTIONS

Genetics

What map unit(Centimorgan) is adopted in the construction of genetic maps?

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

Genetics

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?

Genetics

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

Genetics

The first attempt to show linkage in plants was done in

Genetics

How many linkage groups are there in nucleoid of bacteria?

Genetics

If distance between genes on chromosome is more, then gene shows

Genetics

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

Contact Details