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  • The linkage map of the 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

The linkage map of the 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

Option: 1

\leq 50 \%


Option: 2

100 \%


Option: 3

66 \%


Option: 4

>50 \%


Answers (1)

best_answer

The actual distance between two genes is said to be equivalent to the percentage of crossing over between these two genes. Since, the two genes lie at the ends of the chromosome, there are 100% chances of their segregation during crossing over.Hence, the correct answer is option 2.

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vinayak

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