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Question
The distance between the genes a, b, c and d in mapping units are a-d = 3.5; b-c = 1; a-b = 6; c-d = 1.5 and a-c = 5. Find out the sequence of arrangement of these genes.
The distance between the genes a, b, c and d in mapping units are a-d = 3.5; b-c = 1; a-b = 6; c-d = 1.5 and a-c = 5. Find out the sequence of arrangement of these genes.
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Solution
The correct option is E adcb
A linkage map is a genetic map of a species or experimental population that shows the position of its known genes or genetic markers relative to each other in terms of recombination frequency, rather than a specific physical distance along each chromosome. Linkage mapping is critical for identifying the location of genes that cause genetic diseases.The distance between a and b is longest (6), thus they would be at extreme ends, distance between a and c is 5 & distance between a and d is 3.5 therefore d will come before c. The final sequence hence is adcb.
a 3.5 d 1.5 c 1 b
A linkage map is a genetic map of a species or experimental population that shows the position of its known genes or genetic markers relative to each other in terms of recombination frequency, rather than a specific physical distance along each chromosome. Linkage mapping is critical for identifying the location of genes that cause genetic diseases.
The distance between a and b is longest (6), thus they would be at extreme ends, distance between a and c is 5 & distance between a and d is 3.5 therefore d will come before c. The final sequence hence is adcb.
a 3.5 d 1.5 c 1 b
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