Principle Of Inheritance And Variation Question 212
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) should be [CBSE PMT 2003]
Options:
A) 100 %
B) 66 %
C) 50 %
D) 5.50 %
Show Answer
Answer:
Correct Answer: B
Solution:
The actual distance between two genes is said to be equivalent to the percentage of crossing over between these genes i.e.
66%.
Crossing over chances between y and b genes suggest that these are to be placed on the chromosome at a distance of 66 units.
BETA
