题目
设动物个体效应为随机遗传效应（a），日粮、性别和畜舍为固定环境效应（b），背膘厚的遗传力为0.4，请完成以下工作：
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1，建立背膘厚的线性模型 -
2，写出模型的一般形式和矩阵形式 -
3，写出混合线性模型方程组的各组分成分 -
4，获得的估计值具有哪些特点 -
5，不同日粮和性别的效应值是多少 -
6，个体育种值是多少，是否和表型值排序一致？说明理由
处理思路
线性模型已经很清楚：
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固定因子：日粮，性别，畜舍 -
随机因子：加性效应 -
观测值：背膘厚
矩阵形式：在R语言中构建即可
方差组分形式：因为遗传力为0.4，可以假定加性Va=2，Ve=3，则遗传力为：2/(2+3) = 0.4
问题4，问题5，问题6需要根据结果来解答
解决方案1：R语言
构建数据：
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根据公式建立混合方程组，确定固定因子矩阵Z，随机因子矩阵X，亲缘关系逆矩阵`solve(A)``
ID <- c("A1","A2","DA","M1","C2","G","M2","CA","S","D","X")Riliang <- c(1,1,1,1,1,1,2,2,2,2,2)Sex <- c(2,1,2,2,1,1,2,2,1,2,1)Sire <- c(0,0,0,0,0,"A2","A2","C2","G","A2","S")Dam <- c(0,0,"A1","A1","A1","DA","M1",0,"M2","CA","D")Chushe <- c(1,2,1,3,3,1,3,2,2,2,3)mm <- c(17,20,15,30,18,12,17,16,23,19,17)dat <- data.frame(ID,Riliang,Sex,Sire,Dam,Chushe,mm)dat
#构建模型：ped <- dat[,c(1,4,5)]ainv <- asreml.Ainverse(ped)$ginvainv_mat <- asreml.sparse2mat(ainv)row.names(ainv_mat) = colnames(ainv_mat) <- attr(ainv,"rowNames")round(ainv_mat,3)dim(ainv_mat)Y = matrix(dat$mm,,1);Yu = matrix(c(1,2,1,2,1,2,3),,1);u# x = as.matrix(dat[,c(2,3,6)])/uX = matrix(c(1,0,0,1,1,0,0,1,0,1,0,0,1,0,1,0,0,1,1,0,0,1,0,0,1,0,0,1,1,0,1,0,0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,1,0,1,0,1,0,1,0,0,1,1,0,0,1,0,0,1,0,1,0,1,0,0,1,1,0,0,0,1),11,7,byrow = T);XZ = diag(11);ZtXX = t(X)%*%X;tXXtXZ = t(X)%*%Z;tXZtZX = t(Z)%*%X;tZXtZZk = t(Z)%*%Z + ainv_mat*1.5dim(tZZk)LHS = rbind(cbind(tXX,tXZ),cbind(tZX,tZZk))dim(LHS)tXY = t(X)%*%YtZY = t(Z)%*%YRHS = rbind(tXY,tZY)dim(RHS)library(MASS)ab = ginv((LHS))%*%RHSab
R语言运行结果
解决方法2：asreml处理代码：
for(i in 1:6) dat[,i] <- as.factor(dat[,i])moda <- asreml(mm ~ Riliang + Sex + Chushe,random = ~ ped(ID),ginverse = list(ID = ainv) ,data=dat,start.values = T)t <- moda$gammas.tablet$Value <- c(2,3)t$Constraint <- "F"modb <- asreml(mm ~ Riliang + Sex + Chushe,random = ~ ped(ID),ginverse = list(ID = ainv) ,data=dat,G.param = t,R.param = t)summary(modb)$varcompcoef(modb)$fixeddim(ab)ab[8:18]coef(modb)$random
asrmel运行
