评估测序文库复杂度 Library Complexity

在制备测序文库时，经常会增加PCR的步骤来扩增DNA片段。如何评估PCR的效果和影响，本文主要分享ENCODE中针对ChIP-Seq和ATAC-Seq标准来说明。

ENCODE中主要通过三个参数来反应Library Complexity：PCB1，PBC2和NRF。以下分别介绍各自的定义：

PCR Bottlenecking Coefficient 1 (PBC1)

• PBC1=M1/M_DISTINCT where
  • M1: number of genomic locations where exactly one read maps uniquely
  • M_DISTINCT: number of distinct genomic locations to which some read maps uniquely

 

PCR Bottlenecking Coefficient 2 (PBC2)

• PBC2= M1/M2 where
  • M1: number of genomic locations where only one read maps uniquely
  • M2: number of genomic locations where two reads map uniquely

 

Non-Redundant Fraction (NRF) - Number of distinct uniquely mapping reads (i.e. after removing duplicates) / Total number of reads.

如何计算这三个数值呢，在我们拿到比对的结果后（SAM或者BAM文件）

对于single end测序，以align.bam为例：

bedtools bamtobed -i align.bam | \
	awk \'BEGIN{OFS="\t"}{print $1,$2,$3,$6}\' | \
	grep -v \'chrM\' | sort | uniq -c | \
	awk \'BEGIN{mt=0;m0=0;m1=0;m2=0} ($1==1){m1=m1 1} ($1==2){m2=m2 1} {m0=m0 1} {mt=mt $1} END{m1_m2=-1.0; if(m2>0) m1_m2=m1/m2; printf "%d\t%d\t%d\t%d\t%f\t%f\t%f\n",mt,m0,m1,m2,m0/mt,m1/m0,m1_m2}\' > pbc_qc.txt

对于pair-end测序，以align.bam为例：

bedtools bamtobed -bedpe -i align.bam | \
	awk \'BEGIN{OFS="\t"}{print $1,$2,$4,$6,$9,$10}\' | \
	grep -v \'chrM\' | sort | uniq -c | \
	awk \'BEGIN{mt=0;m0=0;m1=0;m2=0} ($1==1){m1=m1 1} ($1==2){m2=m2 1} {m0=m0 1} {mt=mt $1} END{m1_m2=-1.0; if(m2>0) m1_m2=m1/m2; printf "%d\t%d\t%d\t%d\t%f\t%f\t%f\n",mt,m0,m1,m2,m0/mt,m1/m0,m1_m2}\' > pbc_qc.txt

最后文件中包含7列，分别为：

1）TotalReadPairs

2）DistinctReadPairs

3）OneReadPair

4）TwoReadPairs

5）NRF=Distinct/Total

6）PBC1=OnePair/Distinct

7）PBC2=OnePair/TwoPair

对于ChIP-seq结果解读:

对于ATAC-Seq结果解读: