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By Garrett Birkhoff

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I−1 πi . . πj πj+1 . . 4 – Carte physique comparative homme-souris. 1. 2 12 Cybb Araf Camp2 .. 22 .. 1 6 Pdha1 .. 23 Zfx Ar Alas2 DXF34 ACAS2 5 .. .. 7 4 Cf8 Dmd ... LAMP2 COL4A5 AR 1 3 .. 17 ARAF .. CYBB DMD .. ZFX PDHA1 .. 5 – « Transformation » d’un chromosome X humain en un chromosome X de souris. π · ρ(i, j) = π1 . . πi−1 πj πj−1 . . πi+1 πi πj+1 . . πn . 5 présente un scénario de réarrangement qui décrit la transformation d’un chromosome X humain en un chromosome X de souris. Problème de la distance d’inversion Étant données des permutations π et σ, trouver une suite d’inversions ρ1 , ρ2 , .

I−1 πi . . πj πj+1 . . 4 – Carte physique comparative homme-souris. 1. 2 12 Cybb Araf Camp2 .. 22 .. 1 6 Pdha1 .. 23 Zfx Ar Alas2 DXF34 ACAS2 5 .. .. 7 4 Cf8 Dmd ... LAMP2 COL4A5 AR 1 3 .. 17 ARAF .. CYBB DMD .. ZFX PDHA1 .. 5 – « Transformation » d’un chromosome X humain en un chromosome X de souris. π · ρ(i, j) = π1 . . πi−1 πj πj−1 . . πi+1 πi πj+1 . . πn . 5 présente un scénario de réarrangement qui décrit la transformation d’un chromosome X humain en un chromosome X de souris. Problème de la distance d’inversion Étant données des permutations π et σ, trouver une suite d’inversions ρ1 , ρ2 , .

Tout gène de ce type sera sujet à un nouveau séquençage chez de nombreux individus pour vérifier cette hypothèse. On a trouvé dans le gène de la mucoviscidose une mutation (suppression de trois nucléotides, provoquant la suppression d’un acide aminé) commune à des personnes atteintes. C’était un fil conducteur et on a créé des amorces PCR pour passer au crible un grand nombre d’individus pour cette mutation. Celle-ci a été trouvée chez 70% des patients atteints de mucoviscidose, prouvant ainsi qu’elle était à l’origine de la maladie.

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