Home
Class 12
BIOLOGY
The linkage map of X - chrommosomes of f...

The linkage map of X - chrommosomes of fruitfly 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

`gt 50%`

C

`le 50%`

D

`100%`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the recombination frequency between the yellow body gene (y) and the bobbed hair gene (b) on the X chromosome of the fruit fly, we can follow these steps: ### Step 1: Understand Linkage and Recombination Linkage refers to the tendency of genes located close to each other on the same chromosome to be inherited together. Recombination occurs during meiosis, where chromosomes exchange segments, leading to new combinations of alleles. **Hint:** Remember that genes that are far apart on a chromosome have a higher chance of recombination compared to those that are close together. ### Step 2: Analyze the Distance on the Linkage Map The problem states that the distance between the yellow body gene (y) and the bobbed hair gene (b) is 66 map units. In genetics, 1 map unit corresponds to a 1% chance of recombination. **Hint:** The distance in map units directly correlates to the percentage of recombination frequency. ### Step 3: Calculate the Recombination Frequency Since the distance between the two genes is 66 map units, this suggests a recombination frequency of 66%. However, in practice, the maximum recombination frequency observed between two genes on the same chromosome cannot exceed 50%. This is because if genes assort independently, the maximum frequency of recombination is limited to 50%. **Hint:** Always remember that while the map distance can suggest a higher recombination frequency, the biological limit is 50% for genes on the same chromosome. ### Step 4: Conclusion Given that the genes are located at opposite ends of the chromosome and the maximum recombination frequency cannot exceed 50%, the recombination frequency between the yellow body gene (y) and the bobbed hair gene (b) should be 50%. **Final Answer:** The recombination frequency between the yellow body gene (y) and the bobbed hair gene (b) is 50%.
Promotional Banner

Similar Questions

Explore conceptually related problems

the linkage map of X -chroosomes of fruitfly has 66 units with yellow body gene (y) at one end and bobbed hair (b) gene at the other end the recomnation frequency between these two genes (y and b) should be :

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 he other end. The recombination frequency between these two genes Y and B should be

A 2.00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

If (a,b) be an end of a diagonal of a square and the other diagonal has the equation x-y=a, then another vertex of the square can be

Pipe A is 0.50 m long and open at both ends. Pipe B is open at one end and closed at the other end. Determine the length of pipe B so that it has the same fundamental frequency as A.

A scientist performed the gene mapping experiments in maize. He mapped the gene on chromosomes on the basis of % crossing over between different genes. One map unit corresponds to one % crossing over or recombination. Te genes showing more than 50% recombination were not supposed to be linked on same chromosome. In crossing over studies on maize, scientist observed the following % crossing over between A B, C, d- between. A and D 10% between A and C 3% between genes C and D 7% between genes A and B 5% , and between genes C and B 8% on the basis of above observation find out the correct sequence of tenes Alt B, C, and D on chromosomes :-