Principles of Inheritance and Variation Question 65
65. 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
(a) $100 \%$
(b) $66 \%$
(c) $>50 \%$
(d) $<50 \%$
Show Answer
Answer : d
Hints & Solution
(d) Mapping of genes on chromosomes is based on the assumption that genes are arranged on the chromosome and that the percentage of crossing over (recombination frequency) between two genes is an index of their distance apart. Distances between genes is expressed as map units, where one map unit is defined as 1 per cent recombination. The recombination frequency will be $50 \%$ when two genes are located on different chromosomes or widely separated on the same chromosome.
