Q. The linkage map of X-chromosomes 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
A) ≤ 50%
B) 100%
C) 66%
D) >50%
Explanation
The actual distance between two genes is said to be equivalent to the percentage of crossing over between these two genes. Since the two genes lie at the ends of the chromosome, there are 100% chances of their segregation during crossing over.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.