使用limma 进行转录组差异分析

作者: Seurat_Satija | 来源:发表于2021-10-18 17:30 被阅读0次

    基因表达差异分析是我们做转录组最关键根本的一步,edgeR+limma是目前最为推荐的方式。本文结合示例数据,将对这个过程进行梳理,让你明白limma包的why,what,how。
    本文示例数据下载

    什么是limma?

    首先要明白,不管哪种差异分析,其本质都是广义线性模型。limma也是广义线性模型的一种,其对每个gene的表达量拟合一个线性方程。limma的分析过程包括ANOVA分析、线性回归等。
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    limma对每个gene拟合出这样一个方程,其中:
    [图片上传失败...(image-8978de-1634549398661)]

    可以是:

    • 一个连续变量:如pH,RIN值,年龄,体重,身高...

    • 一个分类变量:如性别、种族、与中位数比较的gene高低表达...
      [图片上传失败...(image-cc3bd4-1634549398661)]

      是limma将要求出的值
      [图片上传失败...(image-5fa95-1634549398661)]

      是假定在整个数据集中正态分布的残差(residual)

    数据解释

    本文数据有两个因素,均为分类变量

    • 品种:cultivar(C,I5/I8)
    • 时间:time(6,9)
      cols为样本编号,rows为基因表达。和我们平时用的数据一致。

    开始分析

    1. 准备数据

    library(edgeR) #edgeR将同时引入limma
    counts <- read.delim("all_counts.txt", row.names = 1)
    head(counts)
    
    d0 <- DGEList(counts)
    # 注意: calcNormFactors并不会标准化数据,只是计算标准化因子
    d0 <- calcNormFactors(d0)
    d0
    
    # 过滤低表达
    cutoff <- 1
    drop <- which(apply(cpm(d0), 1, max) < cutoff)
    d <- d0[-drop,] 
    dim(d) # number of genes left
    
    # sample names
    snames <- colnames(counts)  
    snames
    # 此数据有两个因素:cultivar(C,I5/I8)和time(6,9)
    cultivar <- substr(snames, 1, nchar(snames) - 2) 
    time <- substr(snames, nchar(snames) - 1, nchar(snames) - 1)
    cultivar
    time
    
    # Create a new variable “group” that combines cultivar and time
    group <- interaction(cultivar, time)
    group
    
    # Multidimensional scaling (MDS) plot
    plotMDS(d, col = as.numeric(group))
    
    

    我们首先构建了edgeR的DGEList对象,这个对象将来将会转化成limma中的EList对象。然后计算了标准化因子,过滤低表达基因。然后按照分组整理了列名,并进行初步的MDS plot,以便看到样品的大概分布

    image

    2. limma

    mm <- model.matrix(~0 + group)
    par(mfrow = c(1, 3))
    y <- voom(d, mm, plot = T)
    #voom的曲线应该很光滑,比较一下过滤低表达gene之前的图形:
    voom(d0, mm, plot = T)
    # lmFit fits a linear model using weighted least squares for each gene:
    fit <- lmFit(y, mm)
    head(coef(fit))
    
    #Comparisons between groups (log fold-changes) are obtained as contrasts of these fitted linear models:
    # Comparison between times 6 and 9 for cultivar I5
    # makeContrasts实际就是定义比较分组信息
    contr <- makeContrasts(groupI5.9 - groupI5.6, levels = colnames(coef(fit)))
    # 比较每个基因
    tmp <- contrasts.fit(fit, contr)
    # Empirical Bayes smoothing of standard errors , (shrinks standard errors that are much larger or smaller than those from other genes towards the average standard error) 
    # (see https://www.degruyter.com/doi/10.2202/1544-6115.1027)
    tmp <- eBayes(tmp)
    # 使用plotSA 绘制了log2残差标准差与log-CPM均值的关系。平均log2残差标准差由水平蓝线标出
    plotSA(tmp, main="Final model: Mean-variance trend")
    # topTable 列出差异显著基因
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    # logFC: log2 fold change of I5.9/I5.6
    # AveExpr: Average expression across all samples, in log2 CPM
    # t: logFC divided by its standard error
    # P.Value: Raw p-value (based on t) from test that logFC differs from 0
    # adj.P.Val: Benjamini-Hochberg false discovery rate adjusted p-value
    # B: log-odds that gene is DE (arguably less useful than the other columns)
    head(top.table, 20)
    
    # p值<0.05的基因有多少个?
    length(which(top.table$adj.P.Val < 0.05))
    
    #Write top.table to a file
    top.table$Gene <- rownames(top.table)
    top.table <- top.table[,c("Gene", names(top.table)[1:6])]
    write.table(top.table, file = "time9_v_time6_I5.txt", row.names = F, sep = "\t", quote = F)
    
    

    limma的核心步骤包括voom、fit、eBays等步骤,注释里都有详细说明。最后我们用topTable方法按照p值排序输出结果。
    下图是我为了说明绘制的,顺序反了。。。中间的是没有过滤低表达基因之前的,左边是过滤后的,最后是fit后的,可以明显的看出区别。

    image

    这时差异分析就有已经完成了,怎样,是不是很简单?

    使用limma进行双变量、多变量、连续变量分析

    ###################双变量分析(cultivar+time)########################
    mm <- model.matrix(~cultivar*time)
    y <- voom(d, mm, plot = F)
    fit <- lmFit(y, mm)
    head(coef(fit))
    # (Intercept)  cultivarI5 cultivarI8      time9 cultivarI5:time9 cultivarI8:time9
    # AT1G01010    4.837410  0.53644370  0.2279446 0.20580445      -0.05565729       0.09265044
    # AT1G01020    3.530869 -0.03152318 -0.3180096 0.15875297       0.06289715       0.36468449
    # AT1G01030    1.250817 -0.32143420  0.3084243 0.03477863      -0.48099113      -0.37842909
    # AT1G01040    5.676015  0.27097286  0.1028739 0.50635951      -0.58923660      -0.46975071
    # AT1G01050    6.598712 -0.09734846 -0.1347759 0.02052702       0.23139851       0.22730960
    # AT1G01060    7.807988 -0.34550979 -0.4172467 1.15805850      -0.34989810      -0.17267051
    # 这个表中显示的是coefficient(相关系数) 
    # cultivarI5 这一列表示cultivar I5 组均值 vs cultivar C(参考cultivar)的差异, for time 6 (the reference level for time)
    #  time9 这一列表示time9 组均值 vs time6 ,forcultivar C的差异
    # cultivarI5:time9 : the difference between times 9 and 6 of the differences between cultivars I5 and C (interaction effect)
    
    # 接下来我们可以定义fit中的coef参数,来进行组间fit
    # Let’s estimate the difference between cultivars I5 and C at time 6
    tmp <- contrasts.fit(fit, coef = 2) #  the difference in mean expression between cultivar I5 and the reference cultivar (cultivar C), for time 6 (the reference level for time)
    tmp <- eBayes(tmp)
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    head(top.table, 20)
    
    tmp <- contrasts.fit(fit, coef = 5) # Test cultivarI5:time9
    tmp <- eBayes(tmp)
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    head(top.table, 20)
    
    ####################多变量分析########################
    #让事情更复杂一点,我们加入批次信息
    batch <- factor(rep(rep(1:2, each = 2), 6))
    # 只需要重新定义model matrix,其余都一样
    mm <- model.matrix(~0 + group + batch)
    y <- voom(d, mm, plot = F)
    fit <- lmFit(y, mm)
    contr <- makeContrasts(groupI5.6 - groupC.6, levels = colnames(coef(fit)))
    tmp <- contrasts.fitit(fit, contr)
    tmp <- eBayes(tmp)
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    head(top.table, 20)
    
    # 加入连续变量
    # Generate example RIN data
    set.seed(99)
    RIN <- rnorm(n = 24, mean = 7.5, sd = 1)
    RIN
    mm <- model.matrix(~0 + group + RIN)
    y <- voom(d, mm, plot = F)
    fit <- lmFit(y, mm)
    contr <- makeContrasts(groupI5.6 - groupC.6, levels = colnames(coef(fit)))
    tmp <- contrasts.fit(fit, contr)
    tmp <- eBayes(tmp)
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    head(top.table, 20)
    
    # What if we want to look at the correlation of gene expression with a continuous variable like pH?
    # Generate example pH data
    set.seed(99)
    pH <- rnorm(n = 24, mean = 8, sd = 1.5)
    pH
    mm <- model.matrix(~pH)
    head(mm)
    y <- voom(d, mm, plot = F)
    fit <- lmFit(y, mm)
    tmp <- contrasts.fit(fit, coef = 2) # test "pH" coefficient
    tmp <- eBayes(tmp)
    top.table <- topTable(tmp, sort.by = "P", n = Inf)
    head(top.table, 20)
    
    

    上面,我们分别加入了额外的二分变量、连续变量进行limma分析,结果都很好。
    这就是有关limma分析的全部内容,注释写的很清楚,可以用这个流程分析任何转录组数据,进行差异表达分析。
    limma 差异分析透彻讲解

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