1. Tardigrade
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  3. Biology
  4. The linkage map of X-chromosome of fruitfly 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:-

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Q. The linkage map of X-chromosome of fruitfly 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:-

AIPMTAIPMT 2003Molecular Basis of Inheritance

Solution:

The linkage map is a chromosome map which is determined by the recombination relations. The map distances are expressed by recombination frequencies and are given by recombination frequencies and are sometimes represented in map unit. X-chromosome has 66 crossover units with yellow and bobbed genes at two extreme ends of the map. Recombination frequency never exceed 50% between any two loci, but these 66 units will be actually obtained by making use of a mapping function.