Lecture 11 - Quant Gen II

# Lecture 11 - Quant Gen II - 1 Lecture 11: Quantitative...

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1 Lecture 11: Quantitative Genetics, con't Recall from last time: 1. Average effect of an allele, α i : "the mean deviation from the population mean (M) of an individual carrying allele i ". 2. Average effect of an allele substitution ( α ): the difference between the average effects of the (two) alleles; α = α 1 - α 2 . 3. Breeding value (A): (Twice) the average deviation of an individual's offspring from the population mean. A 11 = α 1 + α 1 =2 α 1 A 12 = α 1 + α 2 A 22 = α 2 + α 2 =2 α 2 4. Genotypic value (G ij ): the deviation of an individual genotype ij from the population mean (M) G 11 = a-M G 12 = d-M G 22 = -a-M 5. Dominance deviation (D): the difference between the Genotypic value and the Breeding value, i.e., G=A+D G'type (G ij ) Breeding Value (A ij ) Dominance Dev (D ij ) A1A1 2 α 1 = 2q α - 2 q 2 d A1A2 α 1 + α 2 = (q-p) α 2pgd A2A2 2 α 1 = -2p α - 2 p 2 d Important point: Dominance deviations depend on only on d and allele frequencies, not on a . If no dominance, Genotypic values = Breeding values. Show figure from Fisher II. Components of phenotypic variance Remember, our goal in quantitative genetics is to be able to provide a description of the phenotypic variation in a population , and how it relates to the genetic properties of the

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2 population. Further, understanding the variance in the population will allow us to understand how it will evolve, for example, in response to selection. Call the Phenotypic value of an individual P . Until now we've considered the phenotype of an individual as being entirely genetically determined, so P = G , which we can break down into P = A + D , i.e., an individual's phenotype is composed of its breeding value plus its dominance deviation. Let us further consider a contribution of the environment to the phenotype, call it E , so P = A + D + E (assuming no epistasis) Recall the variance = i i i x x p ) ( 2 , where p i is the frequency of the i th type Variance of a sum Z = X + Y is Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y) Covariance is a measure of association between two variables; if they are independent the covariance is zero. Cov(X,Y) = ) )( ( y y x x p j i j ij i IF covariance is zero, then Var(X+Y) + Var(X) + Var(Y) Thus, the phenotypic variance in a population, V P = V A + V D + V E IF the covariances between the terms are zero. i) It Can Be Shown that the covariance between the breeding value and the dominance deviation is always zero. Proof ** Below ii) We ASSUME the covariance between the genotypic component of variance V G = V A + V D and the environmental component of variance is zero, i.e., Cov(G,E) = 0. So, the phenotypic variance present in a population is composed of a component due to variation in breeding values, V A , variation in dominance deviations, V D , and variation due to the random effects of the environment, V E . Q:
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## This note was uploaded on 06/08/2011 for the course PCB 4674 taught by Professor Baer during the Fall '08 term at University of Florida.

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Lecture 11 - Quant Gen II - 1 Lecture 11: Quantitative...

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