jellyfish K-mer analysis and genome size estimate
So, to estimate manually, take the sum of the counts of unique kmers under the first peak and multiply by 1; add the sum of the counts of unique kmers under the peak at 2x the depth of the first peak and multiply by 2; etc, for all peaks. This will give you the haploid genome size. So if your genome is tetraploid, the actual size will be 1/4 of your result, since the first peak will correspond to mutations present on only 1 ploidy (1/0/0/0 genotype).
You can make this more accurate by modelling the peaks as a sum of Gaussian curves, but that probably won't change the result much. Of course, this method is subjective because calling peaks is subjective.
Please note - I think 17-mers are too short for this kind of analysis. I prefer 31-mers because they are the longest computationally-efficient kmers. Also, FYI, BBNorm is faster than Jellyfish and can also generate kmer-frequency histograms:
khist.sh in=reads.fq hist=khist.txt
Outline
- count k-mer occurence using Jellyfish (jellyfish count)
- summarize as histogram (jellyfish histo)
- plot graph with R
- determine the total number of k-mer analyzed and the peak position
- compare the peak shape with poisson distribution
Count k-mer occurence
In this example we have 5 pair of fastq files in three different subdirectories. The file to process can be specified with "*/*.qf.fastq" and veriied with ls.
$ ls */*.qf.fastq run1/s_1_1_sequence.qf.fastq run2/s_2_2_sequence.qf.fastq run1/s_1_2_sequence.qf.fastq run3/s_1_1_sequence.qf.fastq run2/s_1_1_sequence.qf.fastq run3/s_1_2_sequence.qf.fastq run2/s_1_2_sequence.qf.fastq run3/s_2_1_sequence.qf.fastq run2/s_2_1_sequence.qf.fastq run3/s_2_2_sequence.qf.fastq
Next, we issue the jellyfish count command
jellyfish count -t 8 -C -m 25 -s 5G -o spec1_25mer --min-quality=20 --quality-start=33 */*.qf.fastq
- -t 8
- specifies the number of threads to be used. This value should be equal to the number of cores on the machine or the number of slots you reserved through job management system ($NSLOTS in SGE or UGE).
- -C
- specifies the both strands are considered. If you do not specify this, the apparent depth would be half, --- that is undesirable
- -m 25
- specified that now you are counting for 25 mer (i.e., k=25)
- -s 5G
- is some kind of magical number specification of hash size. This should be as high as the physical memory allows. The higher the faster, but exceeding the available memory leads to failure or extremely slow counting.
- -o spec1_25mer
- specifies the prefix of output file names.
- --quality-start=33
- specified that your fastq file have 33 based quality value string. Be careful on the dataformat. There are cases that your data are 64 based depending on the sequending system and software versions. This is relevant only when you specify --min-quality
- --min-quality=20
- specifies that nucleotide having qv lower than 20 should not included in the count. This selection reduces the k-mers derived from sequence errors and make the peak clearer.
- */*.qf.fastq
- will be expanded to the ten filenames explained above by the shell and passed to jellyfish as input files
summarize as histogram (jellyfish histo)
First confirm that you got the output file
$ ls spec1_25mer* spec1_25mer_0
now that there is a single file spec1_25mer_0
$ jellyfish histo -o spec1_25mer.histo spec1_25mer_0
Confirm that you got the output
$ ls spec1_25mer* spec1_25mer_0 spec1_25mer.histo
Examine the numbers by your eyes
$ head -25 spec1_25mer.histo 1 461938583 2 95606044 3 19280477 4 13836754 5 11018480 6 9555090 7 8557935 8 7863244 9 7319505 10 6920880 11 6589723 12 6321923 13 6148638 14 6036120 15 5972264 16 5962234 17 5987696 18 6051171 19 6154429 20 6297373 21 6485135 22 6700579 23 6932570 24 7217627 25 7533211
terminate called after throwing an instance of 'jellyfish::invertible_hash::ErrorAllocation'
what(): Failed to allocate 628292358736 bytes of memory
jellyfish K-mer analysis and genome size estimate的更多相关文章
- Evaluate|GC content|Phred|BAC|heterozygous single nucleotide polymorphisms|estimate genome size|
(Evaluate):检查reads,可使用比对软件:使用SOAPaligner重新排列:采用massively parallel next-generation sequencing technol ...
- Maximum Size Subarray Sum Equals k -- LeetCode
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If t ...
- 【LeetCode】325. Maximum Size Subarray Sum Equals k 解题报告 (C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 prefix Sum 日期 题目地址:https:// ...
- The sequence and de novo assembly of the giant panda genome.ppt
sequencing:使用二代测序原因:高通量,短序列 不用长序列原因: 1.算法错误率高 2.长序列测序将嵌合体基因错误积累.嵌合体基因:通过重组由来源与功能不同的基因序列剪接而形成的杂合基因 se ...
- [LeetCode] Longest Substring with At Least K Repeating Characters 至少有K个重复字符的最长子字符串
Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...
- [LeetCode] Kth Largest Element in an Array 数组中第k大的数字
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the so ...
- k近邻算法的Java实现
k近邻算法是机器学习算法中最简单的算法之一,工作原理是:存在一个样本数据集合,即训练样本集,并且样本集中的每个数据都存在标签,即我们知道样本集中每一数据和所属分类的对应关系.输入没有标签的新数据之后, ...
- 6.3Sum && 4Sum [ && K sum ] && 3Sum Closest
3Sum Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find a ...
- 剑指offer系列55---最小的k个数
[题目] 输入n个整数,找出其中最小的K个数.例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4,. *[思路]排序,去除k后的数. package com.exe11 ...
随机推荐
- CPU信息查询
cat /proc/cpuinfo |grep "physical id"|sort |uniq|wc -l //查看CPU的个数 cat /proc/cpuinfo |grep ...
- 一些qml资料
qml开发ios应用 http://www.seanyxie.com/qt-qml%E7%A7%BB%E5%8A%A8%E5%BC%80%E5%8F%91%E4%B9%8B%E5%9C%A8ios%E ...
- HTML元素,属性,基础标签
元素,属性 元素 html有父元素和子元素,被包含的叫子元素,如html是head的父元素,他们是父子关系,head和body是兄弟关系 <html> <head></h ...
- MySQL存储过程之事务管理
原文链接:http://hideto.iteye.com/blog/195275 MySQL存储过程之事务管理 ACID:Atomic.Consistent.Isolated.Durable 存储程序 ...
- [转载]java NIO详解
Java NIO(New IO)是从Java 1.4版本开始引入的一个新的IO API,可以替代标准的Java IO API.下面的文章写的很详细,还配有插图,有助于深入学习和理解java NIO 文 ...
- js 替换 当前URL 特定参数
js 替换 当前URL 特定参数 2012-12-24 20:45:53| 分类: JS&JQuery |举报 |字号 订阅 //替换指定传入参数的值,paramName为参数,repl ...
- (转载)Java基础知识总结
写代码: 1,明确需求.我要做什么? 2,分析思路.我要怎么做?1,2,3. 3,确定步骤.每一个思路部分用到哪些语句,方法,和对象. 4,代码实现.用具体的java语言代码把思路体现出来. 学习新技 ...
- arm-linux-gcc-4.3.2安装步骤
安装交叉编译工具链: 1.首先以root用户登入 2.复制arm-linux-gcc-4.3.2.tgz到根目录下tmp文件夹里 3.解压命令tar xvzf arm-linux-gcc-4. ...
- Hibernate中的多对多映射
1.需求 项目与开发员工 一个项目,有多个开发人员 一个开发人员,参与多个项目 [多对多] 2.实体bean设计 Project: public class Project { private int ...
- HDU 1098 Ignatius's puzzle 费马小定理+扩展欧几里德算法
题目大意: 给定k,找到一个满足的a使任意的x都满足 f(x)=5*x^13+13*x^5+k*a*x 被65整除 推证: f(x) = (5*x^12 + 13 * x^4 + ak) * x 因为 ...