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, 20:44, 11 March 2015
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| It is important that the normalization results in a unique representation of the variant. Before we begin the proof, intuitively, accept that any representation of a variant can be | | It is important that the normalization results in a unique representation of the variant. Before we begin the proof, intuitively, accept that any representation of a variant can be |
− | transformed to another representation by removing or adding nucleotides from the reference sequence. | + | transformed to another representation by adding nucleotides from the reference sequence to either ends of all the alleles at the same time or removing equivalent nucleotides from the ends of |
| + | all the alleles at the same time. |
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| Now suppose there are 2 normalized variants A and B. Suppose A is at a different position from B and B is to the right of A (without loss in generality), this is not possible because by the definition of a normalized variant, it is left aligned, | | Now suppose there are 2 normalized variants A and B. Suppose A is at a different position from B and B is to the right of A (without loss in generality), this is not possible because by the definition of a normalized variant, it is left aligned, |
| and if they were at different positions, that means B may be left aligned to A since they represent the same variants leading to a contradiction. So A and B must be at the same position. | | and if they were at different positions, that means B may be left aligned to A since they represent the same variants leading to a contradiction. So A and B must be at the same position. |
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− | Now, suppose that A and B are of different lengths where B is longer than A, then this is not possible as B is then not parsimonious, so B can be trimmed to the same length as A. | + | |
| + | Now, suppose that A and B are at the same position but are of different lengths where B is longer than A (without loss in generality), this is not possible as B is then not parsimonious, so B can be trimmed to the same length as A. |
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| Thus A and B have to be at the same position and have the same length and variant normalization is unique. | | Thus A and B have to be at the same position and have the same length and variant normalization is unique. |