HEREDITY
CROSSOVER
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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A, B, C
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A, C, B
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B, C, A
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B, A, C
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Detailed explanation-1: -Therefore, with the help of this data, the sequence of the genes can be as follows. Therefore, the correct answer is option iv, i.e., B, A, C.
Detailed explanation-2: -The map distances between genes A and B is 3 units, between B anc C is 10 units and between C and A is 7 units. The order of the genes in a linkage map constructed on the above data would perhaps be: Q.
Detailed explanation-3: -As shown in the next video, the map distance between loci B and E is determined by the number of recombinant offspring. Remember: The # of recombinant offspring / total # of offspring x 100% = recombination frequency. Recombination frequency = map units = centiMorgan (cM)
Detailed explanation-4: -Gene map distance is the distance between points on a chromosome which can be estimated by counting the number of crossovers between them. Therefore, the distance between two points on the genetic map of a chromosome is the average number of crossovers between them.
Detailed explanation-5: -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.