Home
Class 12
BIOLOGY
The map distance between genes A and B i...

The map distance between genes A and B is 3 units, between B and C is 10 units and between C and A is 7 units. Which of the following shows the correct order of the genes in a linkage map on the above data ?

A

A,B,C

B

A,C,B

C

B,C,A

D

C,A,B

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the correct order of genes A, B, and C based on the given map distances, we can follow these steps: ### Step 1: Identify the Distances We have the following distances: - Distance between A and B = 3 map units - Distance between B and C = 10 map units - Distance between C and A = 7 map units ### Step 2: Determine the Farthest Genes From the distances provided, we can see that the greatest distance is between B and C (10 map units). This suggests that B and C are the farthest apart in the linkage map. ### Step 3: Position the Farthest Genes We will place gene B on one end and gene C on the other end of our linkage map: - Position: B — 10 map units — C ### Step 4: Determine the Position of Gene A Next, we need to find the position of gene A. We know that the distance between A and B is 3 map units. Therefore, we can place gene A 3 map units from gene B: - Position: A — 3 map units — B — 10 map units — C ### Step 5: Verify the Distance Between A and C Now, we need to check the distance between A and C. The distance between C and A is given as 7 map units. If we add the distances: - From A to B = 3 map units - From B to C = 10 map units The total distance from A to C through B would be: 3 (A to B) + 10 (B to C) = 13 map units, which does not match the given distance of 7 map units. ### Step 6: Adjust the Position of Gene A Since the distance from A to C is only 7 map units, we need to adjust the position of gene A. Thus, we can place A between B and C: - Position: B — 3 map units — A — 4 map units — C This configuration satisfies all the distances: - Distance between A and B = 3 map units - Distance between A and C = 7 map units (3 + 4) - Distance between B and C = 10 map units (3 + 4 + 3) ### Conclusion The correct order of the genes in the linkage map is: - B — A — C ### Final Answer The correct order of the genes is B, A, C. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

If map distance between genes P and Q is 4 units, between P and R is 11 units, and between Q and R is 7 units, the order of genes on the linkage map can be traced as follows

The map distance in a certain organism between genes A and B is 8 units, between B and C is 4 units and between C and D is 12 units. Which one of these gene pairs will show more recombination frequency? Give reason in just one single sentence in support of your answer.

Unit of distance between genes in a chromosome is known as

If distance between gene on chromosome is more, then gene shows :-

Distance between two linked genes is measured in map units that depict

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 :-

The distance between two directrices of a rectangular hyperbola is 10 units. Find the distance between its foci.

The distance between two directrices of a rectangular hyperbola is 10 units. Find the distance between its foci.

Distance between the points (2,5,3) and (4,3,1) is _____ units