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MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • 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  gene at the other end. The recombination frequency between                these two genes (y and b) should be

    A)  50%                                      

    B)  100%

    C)  66%                                      

    D)  50%

    Correct Answer: B

    Solution :

                     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 per cent chance of their segregation during crossing over.


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