Zoology - Principles of Inheritance and Variation Question with Solution | TestHub
ZoologyPrinciples of Inheritance and VariationMutationHard2 minPYQ_2003
ZoologyHardsingle choice
The linkage map of -chromosome of fruitfly has 66 units, with yellow body gene at one end and bobbed hair gene at the other end. The recombination frequency between these two genes and ) should be:NEET 2003
Options:
Answer:
C
Solution:
The yellow body gene (y) and bobbed hair (B) gene are present 66 map unit apart which means that there is chances of recombination between them (recombination frequency). Related Theory One map unit is equal to recombination frequency. This linear relationship holds true for lower values only; as the recombination frequency increases beyond , the linear relationship does not hold true owing to double and multiple cross overs and recombination frequency is always less than map distance and never exceeds than .
Stream:NEETSubject:ZoologyTopic:Principles of Inheritance and VariationSubtopic:Mutation
⏱ 2mℹ️ Source: PYQ_2003
Doubts & Discussion
Loading discussions...
