Question

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) would be
A. 100%
B. 66%
C. 50%
D. 5.50%

Answer 1

Hint: Recombination frequency (θ) is the frequency with which a single chromosomal crossover will take place between two genes during meiosis. A centimorgan (cM) is a unit that describes a recombination frequency of 1%. There are two types of genetic mapping they are physical mapping and genetic linkage mapping

Complete answer: In recombinant frequency, the one unit of distance is considered as one per cent. As per the given information in question, the distance is 66 units in between gene (y) at one end and bobbed hair(b) gene at the other end. The recombination frequency between these two genes (y and b) would be 66%. The genetic linkage map shows the relative positions of genetic markers on a chromosome. The positions of genetic markers are determined based on how the loci are inherited together rather than on the specific physical distance between them. It means that the more the two genetic markers are inherited together the farther apart they are present. In reverse, the lower frequency of being inherited together the smaller the physical distance is between them. There are two types of genetic mapping: physical mapping and genetic linkage mapping. This distance is measured in base pairs and recombination frequency.
Hence, the correct answer is option B.

Note: As an example of linkage, consider the classic experiment by William Bateson and Reginald Punnett. There are three types of linkages: external linkage, internal linkage, and no linkage. Anything internal to a function, its arguments, variables, and so on always has no linkage and so can only be accessed from inside the function itself.