假定表型值由均值+育种值+残差
y i = μ + a i + ϵ i y_i = \mu + a_i + \epsilon_i yi=μ+ai+ϵi
表型值 VS 育种值
他们之间的相关系数
c o r ( y , a ) = C o v ( y , a ) V a r ( y ) ∗ V a r ( a ) cor(y,a) = \frac{Cov(y,a)}{\sqrt{Var(y)*Var(a)}} cor(y,a)=Var(y)∗Var(a) Cov(y,a)
其中:
C o v ( y , a ) = C o v ( μ + a + ϵ , a ) = C o v ( a , a ) = V a r ( a ) Cov(y,a) = Cov(\mu + a + \epsilon, a) = Cov(a,a) = Var(a) Cov(y,a)=Cov(μ+a+ϵ,a)=Cov(a,a)=Var(a)
所以:
c o r ( y , a ) = V a r ( a ) V a r ( y ) ∗ V a r ( a ) = V a r ( a ) V a r ( y ) = h 2 cor(y,a) = \frac{Var(a)}{\sqrt{Var(y)*Var(a)}} = \sqrt{\frac{Var(a)}{Var(y)}} = \sqrt{h^2} cor(y,a)=Var(y)∗Var(a) Var(a)=Var(y)Var(a) =h2
回归系数 VS 相关系数 VS 遗传力
回归系数:
b a , y = C o v ( y , a ) V a r ( y ) = V a r ( a ) V a r ( y ) = h 2 b_{a,y} = \frac{Cov(y,a)}{Var(y)} = \frac{Var(a)}{Var(y)} = h^2 ba,y=Var(y)Cov(y,a)=Var(y)Var(a)=h2
回归系数和相关系数:
b a , y = c o r 2 ( a , y ) = h 2 b_{a,y} = cor^2(a,y) = h^2 ba,y=cor2(a,y)=h2
育种值 VS 表型值 VS 遗传力
a = h 2 ( y − μ ) a = h^2(y - \mu) a=h2(y−μ)
结论:
由此可以知道, 在模拟数据时, TBV和phenotype的关系, 以及与遗传力的关系为:
c o r ( T B V , p h e n o t y p e ) = h 2 cor(TBV, phenotype) = \sqrt{h^2} cor(TBV,phenotype)=h2
T B V = b + a ∗ p h e n o t y p e TBV = b + a*phenotype TBV=b+a∗phenotype
a 为回归系数:
a = h 2 a = h^2 a=h2
