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NEET AIPMT SOLVED PAPER 2003

  • question_answer
                    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)                                                                                                                                                           \[\le \,50%\]

    B)                 100%     

    C)                 66%                       

    D)                 > 50%

    Correct Answer: B

    Solution :

                    The actual distance between two genes is safe 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.


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