笔记 Bioinformatics Algorithms Chapter1
Chapter1 WHERE IN THE GENOME DOES DNA REPLICATION BEGIN
一、
·聚合酶启动结构域会结合上游序列的一些位点,这些位点有多个,且特异,并且分布在两条链上。通过计算,找到出现频率最高的k-mer可能为为聚合酶结合位点:dnaA BOX。
但是如何定位Ori的大概位置呢?
·DNA链复制的不对称性,其导致突变速率的不对称,使得有(forward链C->T,脱氨基)的趋势。由此,依据skew增的处于forward链,skew减的处于reverse链。(skew = G - C ,逢G+1 逢C-1)图中最低点代表Ori区域。
由此可以大致推测出ori的位置,然后在此位置内(100bp),寻找出现频率大的pattern,作为可能的dnaA box。
·由于k-mer之间,会有碱基的若干差异,故应使用能容错的计数方法。
二、
提出问题:The Clump Finding Problem
Find every k-mer that forms a clump in the genome.
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ComputingFrequencies(Text, k) #一种遍历一次计算频率的‘桶’方法 for i ← 0 to 4k − 1 FrequencyArray(i) ← 0 for i ← 0 to |Text| − k Pattern ← Text(i, k) j ← PatternToNumber(Pattern) #hash FrequencyArray(j) ← FrequencyArray(j) + 1 return FrequencyArray
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FindingFrequentWordsBySorting(Text , k) #排序法 FrequentPatterns ← an empty set for i ← 0 to |Text| − k Pattern ← Text(i, k) Index(i) ← PatternToNumber(Pattern) Count(i) ← 1 SortedIndex ← Sort(Index) for i ← 1 to |Text| − k if SortedIndex(i) = SortedIndex(i − 1) Count(i) = Count(i − 1) + 1 maxCount ← maximum value in the array Count for i ← 0 to |Text| − k if Count(i) = maxCount Pattern ← NumberToPattern(SortedIndex(i), k) add Pattern to the set FrequentPatterns return FrequentPatterns
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ClumpFinding(Genome, k, L, t)
FrequentPatterns ← an empty set
for i ← 0 to 4k − 1
Clump(i) ← 0
for i ← 0 to |Genome| − L
Text ← the string of length L starting at position i in Genome
FrequencyArray ← ComputingFrequencies(Text, k)
for index ← 0 to 4k − 1
if FrequencyArray(index) ≥ t
Clump(index) ← 1
for i ← 0 to 4k − 1
if Clump(i) = 1
Pattern ← NumberToPattern(i, k)
add Pattern to the set FrequentPatterns
return FrequentPatterns
我们不用每次挪动一位搜寻窗就重新计算kmer频率,搜寻窗每挪一位,原来的第一个kmer将少一个,结尾后一个kmer将多一个
BetterClumpFinding(Genome, k, t, L)
FrequentPatterns ← an empty set
for i ← 0 to 4k − 1
Clump(i) ← 0
Text ← Genome(0, L)
FrequencyArray ← ComputingFrequencies(Text, k)
for i ← 0 to 4k − 1
if FrequencyArray(i) ≥ t
Clump(i) ← 1
for i ← 1 to |Genome| − L
FirstPattern ← Genome(i − 1, k)
index ← PatternToNumber(FirstPattern)
FrequencyArray(index) ← FrequencyArray(index) − 1
LastPattern ← Genome(i + L − k, k)
index ← PatternToNumber(LastPattern)
FrequencyArray(index) ← FrequencyArray(index) + 1
if FrequencyArray(index) ≥ t
Clump(index) ← 1
for i ← 0 to 4k − 1
if Clump(i) = 1
Pattern ← NumberToPattern(i, k)
add Pattern to the set FrequentPatterns
return FrequentPatterns
posted on 2018-09-21 00:19 iojafekniewg 阅读(461) 评论(0) 编辑 收藏 举报