Linkage map of X -chromosome of fruit fly has 66 units with yellow body gene (y) at one end and bobb

English

Gujarati

Hindi

4.Principles of Inheritance and Variation

medium

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

$100\%$

$66\%$

$> 50\%$

$5.50\%$

(AIPMT-2003)

Solution

(b) 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.

Standard 12

Biology