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Variance components and Estimation of genetic covariance between relatives

Variance components

The variance of a random variable is the expectation of the squared deviation between a random variable (X) and its expectation (E). The expectation of a discrete random variable is obtained by weighting the possible outcomes X with their probabilities (in the case of a dice the probability is 1/6 for each outcome).

The covariance between two random variables measures whether large values of X occur often together with large values of Y. The covariance depends on the scale of each variable. For getting an independent estimator the correlation of two traits is calculated as cor(x,y)=\frac{cov(x,y)}{\sqrt{\sigma _{x}^{2}\sigma _{y}^{2}}}

The mean square error of the genotype has to be large in comparison to the residual to get a significant difference within genotypes. The residual error is high if the measurement error, the environmental effect as well as soil in-homogeneity were large.

From the sum of squares and the expected sum of squares the variance components are derived.

Estimation of genetic covariance between relatives

The phenotypic variance can be decomposed into genotypic and environmental variance.

 \sigma _{p}^{2}= \sigma _{g}^{2}+\sigma _{e}^{2}+\sigma _{gxe}^{2} ,
g=genotype, e=environment, gxe = genotype by environment interaction

 \sigma _{g}^{2}= s\sigma _{a}^{2}+t\sigma _{d}^{2} ,
a= additive genetic variance, d= dominant genetic variance, s and t depend on the genetic covariance between relatives.


P=0 is the probability that all alleles differ
P=1 is the probability that 1 allele is in common
P=2 is the probability that 2 alleles are in common

In identical twins the probability of P=2 is 1. As both alleles are transferred from the parent to the sibling the full additive and dominance effects are expressed.

Between parent and child the probability of P1 is 1. As only one allele of each parent is present in the child, only half of the additive genetic variance is expressed and no dominance variance. Dominance variance can only be the result of a specific combination of two alleles.

In full sibs P0 and P1 is ¼ and P1 is ½. Dominance effects can be observed if the locus is heterozygote, and if both alleles expressing dominance are transmitted. The additive genetic variance is ¼ of P2 and ½ of P1, so ½. The dominance variance is equal to P2.

 Estimating dominance and additive variance

from half sib families: a is the family effect and e the error within families
\sigma _{\alpha }^{2}= (MS_{\alpha }-MS_{e})/n

\sigma _{A }^{2}= 4(MS_{\alpha }-MS_{e})/n


with a hierarchical design, m is the male, f is the female nested within males
\sigma _{A }^{2}= \frac{4}{1+F_{m}}\sigma _{m}^{2}

\sigma _{D }^{2}= \frac{4}{1+F_{m}}\ (\frac{\sigma _{f}^{2}}{1+F_{f}}-\frac{\sigma _{m}^{2}}{1+F_{m}})


with a factorial design
\sigma _{Am}^{2}= \frac{4}{1+F_{m}}\sigma_{m}^{2}

\sigma _{Af}^{2}= \frac{4}{1+F_{f}}\sigma_{f}^{2}

\sigma _{A}^{2}= \frac{\sigma_{Am}^{2}+\sigma_{Af}^{2}}{2}

\sigma _{D}^{2}= \frac{4\sigma _{mf}^{2}}{(1+F_{m})(1+F_{f})}

Bulmer, M.G. 1980. The Mathematical Theory of Quantitative Genetics. Clarendon, Oxford.
Searle, S.R., G. Casela, C.E. McCulloch. 1992. Variance Components. Wiley, New York.




November 26th, 2010
Topic: Crop Science, Plant breeding Tags: None

One Response to “Variance components and Estimation of genetic covariance between relatives”

  1. Heritability / Repeatability of a trait | Says:

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