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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 (b) gene at the other end. The recombination frequency between these two genes (y and b) should beA. $60%$ B. $>50%$ C. $\le 50%$D. $100%$

Hint: Recombination frequencies are approximately equal to the difference between the genes and hence are used in the preparation of linkage maps.

- According to the question, the linkage mapping of X chromosome is $66$ units. ‘y’ gene represents ‘yellow body’ and ‘b’ gene represents ‘bobbed hair’ of the fruit fly.
- The recombination frequency is usually less than map distance and never exceeds 50 percent and here the map distance is $66$ units. Thus, $66%$ cannot be the correct answer.
- This linear relationship is valid only for lower values; when the recombination frequency rises above $50$ percent, the linear relationship is not accurate due to double and multiple crossover frequencies and recombination frequency is usually less than map distance and never exceeds $50$ percent.
- We know that the recombination frequency between two genes is equal to the distance between the two genes.The distance between the two genes is $66$ units. Thus, the recombination frequency is $66%$.
- It is said that the total difference between two genes is proportional to the rate of crossover between both genes, i.e. $66%$. Crossing the chances between y and b genes indicates that both should be positioned at a gap of $66$ units on the chromosome. $100%$ cannot be the correct answer.
Thus, the correct answer is $\le 50%$.