Week2n3_AlignmentDynamicProgramming

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CAP5540/CIS4930 (Fall 2022) BIOINFORMATICS SEQUENCE ANALYSIS Week 2 & 3.1: Sequence Alignment (Dynamic Programming) Lecturer: Xian Mallory
Recap of last week We cannot sequence the DNA from the beginning to the end. DNA is broken into small fragments, whose order of A, C, G and T are to be determined by a sequencing machine. The resulting sequence of each fragment is called a read . In shotgun sequencing , a read loses its position on the DNA. In addition to the sequences of the small fragments of an individual’s DNA, we also have the human reference genome . To identify the differences between individual’s DNA and the human reference genome (for the purpose of detecting mutations that may cause genetic diseases etc), we need to align the reads to the reference genome.
Recap of last week (con’d): I/O of a mutation calling tool Input: 1) a reference genome, and 2) the sequences of an individual’s DNA, which is not exactly the same as the reference genome. Output: the “equivalent” regions on the reference for each read to maximize their similarity. Reference Individual’s Genome Fragments of individual’s genome
Recap of last week (con’d): scoring scheme in alignment It is important to have a scoring scheme to quantify the evaluation of the alignments. Alignment A ACGGGTTTT ACGTGTTTT Alignment B ACGGGTTTT AAGTGTCTT mismatches/substitutions 8 6 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 A C G T A C G T Reward 1 for each match.
Recap of last week (con’d): indels in alignment Three strategies to handle indels: Strategy #1: give zero points for indels; Strategy #2: penalize the indels. Penalty is linear to the number of nucleotides being inserted or deleted. Strategy #3: penalize the number of indels, as well as the number of nucleotides being inserted or deleted. Alignment C ACGGGTTTT ACTTTAAT deleted inserted Score = ?
Recap of last week (con’d): sequence length in alignment Suppose V is the matching score, m and p are the sequence lengths for X and Y . The final alignment score S is V – m – p. ACGGGTTTT ACGTGTTTT 8 ATGGGAAAATTTTAGCTG ACGTGAGACTCATCGGAG 9 v.s. V = S = 8 – 9 – 9 = -10 9 – 18 – 18 = -27 9 nucleotides 18 nucleotides 9 nucleotides 18 nucleotides >
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