Question

Linkage map of the x- chromosome of fruit fly has 66 units with yellow body gene(y) at one end and bobbed hair (b) at the other. the recombination frequency between y and b gene would be:
A) 66%
B) >50%
C) <50%
D)100%

Answer 1

Hint: Drosophila melanogaster is a species of fly in the family Drosophilae. The species is known generally as the common fruit fly or vinegar fly.

Complete answer:
 Recombination frequencies are directly proportional to the distance between the genes and are used in lineage map preparation.

Distance between two genes is measured in a map unit. One map unit is equal to 1% recombination frequency and is also referred to as centimorgan.

This linear relationship holds true for lower values only as the recombination frequency increases beyond 50%, the linear relationship doesn’t hold true owning to double and multiple crossover and recombination frequency is always less then map distance and never exceeds than 50%.the yellow gene(y) and hair(b) gene are present 66 map unit apart which means that there is <50% chances of recombination between them.

Hence the correct answer is option ‘C’.

Note:Drosophila melanogaster is having relatively simple genetics with four pairs of chromosomes and large number of offspring per generation. Fruit fly was originally an African species.Wild type fruit flies are yellow-brown, with brick red eyes and transverse black rings across the abdomen. Females are about 2.5mm long, males are slightly smaller.