Question
Easy

The distance between the genes a, b, c, d in mapping unit are: a-d=3.5; b-c=1; a-b=6; c-d=1.5; a-c=5 Find out the sequence of arrangement of these genes

1
abcd
2
adcb
3
acdb
4
acbd
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: HTET_L3
Chapter:
Topic:
Correct Answer
Option B
Explanation

The correct option is (2): adcb In gene mapping, the distance between two linked genes is measured in map units (centimorgans, cM), which represents the percentage of recombination frequency between them. A key principle of this mapping is that gene distances are considered additive, meaning the distance between two non-adjacent genes is the sum of the distances of the intervening segments. To find the correct sequence, we must arrange the…Read More

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