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Question at position 15 The pairwise map distances for four linked genes (not the same as the quiz) are as follows: A↔B = 30 m.u., B↔C = 2 m.u., C↔D = 10 m.u., B↔D = 12 m.u., A↔D = 20 m.u., A↔C = 28 m.u. What is the order of these four genes?CBADBADCABDCBCDAADBCCADBABCD
Options
A.CBAD
B.BADC
C.ABDC
D.BCDA
E.ADBC
F.CADB
G.ABCD
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Step-by-Step Analysis
The prompt gives the pairwise map distances among four linked genes A, B, C, D: A↔B = 30, B↔C = 2, C↔D = 10, B↔D = 12, A↔D = 20, A↔C = 28. For a linear gene order g1–g2–g3–g4, the three adjacent distances are x = distance(g1,g2), y = distance(g2,g3), z = distance(g3,g4). Then the remaining pairwise distances must be sums along the path: g1–g3 = x+y, g2–g4 = y+z, g1–g4 = x+y+z. With four genes in a line, only four possible orders are distinct up to reversing the chain; the answer choices enumerate several of them. Below, I assess each option by trying to assign the three adjacent distances (the distances between consecutive letters in the option) to match the given pairwise data, and then checking whether the implied longer-distance pairs align with the provided values. I’ll break down each option with concrete checks and point out where the numbers align or clash.
Option 1: CBAD
- If the order is C–B–A–D, the three adjacent distances would be: C–B = 2, B–A = 30, A–D = 20.
- From these, predictions are: C......Login to view full explanationLog in for full answers
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