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. $60%$ 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.
Complete Answer: - A map of genes is based on an analysis of the linkage on a chromosome. A linkage map does not indicate the actual differences between genes but rather their relative locations, as determined by how frequently they inherit two gene loci. The more related two genes are, the more likely they are shared together. - 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.
Now, let us find the recombination frequency between y and b genes. - 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 $\le 50%$.
Note: During the chromosomal fusion, two genetic markers which are physically similar to each other are unlikely to be split into different chromatids and are thus assumed to be more related than markers which are far apart.