一个数量遗传学题：如何计算育种值

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：

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：

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：

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)``

代码语言：javascript复制

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

代码语言：javascript复制

#构建模型：

ped <- dat[,c(1,4,5)]
ainv <- asreml.Ainverse(ped)$ginv
ainv_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);Y
u = matrix(c(1,2,1,2,1,2,3),,1);u
# x = as.matrix(dat[,c(2,3,6)])/u
X = 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);X
Z = diag(11);Z


tXX = t(X)%*%X;tXX
tXZ = t(X)%*%Z;tXZ

tZX = t(Z)%*%X;tZX
tZZk = t(Z)%*%Z   ainv_mat*1.5
dim(tZZk)

LHS = rbind(cbind(tXX,tXZ),cbind(tZX,tZZk))
dim(LHS)

tXY = t(X)%*%Y
tZY = t(Z)%*%Y
RHS = rbind(tXY,tZY)
dim(RHS)

library(MASS)
ab = ginv((LHS))%*%RHS
ab

R语言运行结果

解决方法2：asreml处理代码：

代码语言：javascript复制

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$gammas.table
t$Value <- c(2,3)
t$Constraint <- "F"
modb <- asreml(mm ~ Riliang   Sex   Chushe,random = ~ ped(ID),
               ginverse = list(ID = ainv) ,data=dat,
               G.param = t,R.param = t)
summary(modb)$varcomp
coef(modb)$fixed
dim(ab)
ab[8:18]
coef(modb)$random

asrmel运行

r语言

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