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The linkage map of X chromosome of fruit fly has 66 units , with yellow body gene (y) at one end and bobbed hair gene (b) at the other end. This recombination frequency between these two genes (y and b) should be(a) 66% (b) > 50 % (c)

The linkage map of X chromosome of fruit fly has 66 units , with yellow body gene (y) at one end and bobbed hair gene (b) at the other end. This recombination frequency between these two genes (y and b) should be(a) 66% (b) > 50 % (c)

Grade:12th pass

2 Answers

Arun
25763 Points
3 years ago
  • The actual distance between two genes is said to be equivalent to the percentage of crossing over between these genes i.e. 66%. Crossing over chances between y and b genes suggest that these are to be placed on the chromosome at a distance of 66 units.
  •  
Aashna
79 Points
3 years ago
Your options are not complete but answer to this question is recombination frequency will be less than 50% because it`s the maximum value as recombinant can`t be more than 50%

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