Back to Directory
NEET BIOLOGYHard

The linkage map of the X-chromosome of fruitfly 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:

A

60%

B

50%

C

≤ 50%

D

100%

Step-by-Step Solution

Alfred Sturtevant used the frequency of recombination between gene pairs to map their position on the chromosome . 1 map unit (centimorgan) equals 1% recombination frequency. However, the maximum recombination frequency between any two genes cannot exceed 50% (which indicates independent assortment). Even if the map distance (calculated by summing small intervals) is greater than 50 units (e.g., 66 units), the observable recombination frequency between the distant markers will saturate at 50%.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started