The linkage map of the X-chromosome of fruit flies 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 linkage map of the X-chromosome of fruit flies 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 correct option is C ≤50%
Linkage/ Cross over/ Chromosome maps are a graphic representation of relative positions/ order and relative distances of genes in a chromosome in the form of a line like a linear road map depicting different places and their relative distances without giving exact mileage.
It is based on Morgan’s hypothesis (1911), which states that the frequency of crossing over/recombination between two linked genes is directly proportional to the physical distance between the two.
The maximum possible recombination frequency is 50% because the recombination may occur in the gene present between the given two genes. Putting that parameter, the possibility of recombination frequency higher than the maximum limit will not be there.
1 map unit or centimorgan is equivalent to 1% recombination between two genes.
Therefore, if the yellow body gene (y) and bobbed hair (b) gene are 66 units apart, the recombination frequency is unable to exceed the 50% mark. Either the recombination frequency between them will be 50% or lower than that.
Therefore, the correct option is ≤50%.
Linkage/ Cross over/ Chromosome maps are a graphic representation of relative positions/ order and relative distances of genes in a chromosome in the form of a line like a linear road map depicting different places and their relative distances without giving exact mileage.
It is based on Morgan’s hypothesis (1911), which states that the frequency of crossing over/recombination between two linked genes is directly proportional to the physical distance between the two.
The maximum possible recombination frequency is 50% because the recombination may occur in the gene present between the given two genes. Putting that parameter, the possibility of recombination frequency higher than the maximum limit will not be there.
1 map unit or centimorgan is equivalent to 1% recombination between two genes.
Therefore, if the yellow body gene (y) and bobbed hair (b) gene are 66 units apart, the recombination frequency is unable to exceed the 50% mark. Either the recombination frequency between them will be 50% or lower than that.
Therefore, the correct option is ≤50%.
