Lecture 10 - Quant Gen II

# Lecture 10 - Quant Gen II - 1 Lecture 10: Quantitative...

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1 Lecture 10: Quantitative Genetics, con't 2. Average effect of an allele Parents pass alleles to their offspring, not genotypes. To understand the phenotype in the context of a (randomly mating) sexual population we need to consider phenotypes in terms of alleles, not genotypes Define the "average effect of an allele" α i as "the mean deviation from the population mean (M) of individuals which received that allele from one parent, the other allele having come at random from the population" (this is technically defined as the "average excess". With random mating, the average excess and the average effect are the same). In other words, let a sample of gametes containing A1 unite with a random sample of gametes from the population. The mean of the resulting genotypes deviates from the population mean by an amount which is the average effect of the A1 allele. Recall the population mean, M= a (p-q) + 2pq d gamete genotypic value Mean g'typic Pop mean (M) average effect (in red) value to be subtracted α i freq (p, q) A1A1 A1A2 A2A2 a d -a ______________________________________________________________________ A1 p q pa + qd [ a(p-q) + 2pqd ] q [ a+d(q-p) ] A2 p q -qa + pd "" -p [ a+d(q-p) ] Call the average effect of the A1 allele α 1 . A randomly drawn A1 allele will combine with another A1 allele with probability p , producing an A1A1 genotype with value a , and it will combine with an A2 allele with probability q , producing an A1A2 genotype with value d . So, the average offspring deviates from the population mean by α 1 = ( pa + qd ) - [ a(p-q) + 2pqd ] = q [ a+d(q-p) ]

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2 α 2 = ( -qa + pd ) - M = -p [ a+d(q-p) ] An intuitive way of understanding the "average effect" is to think of it as the "average effect of an allele substitution", α . Imagine you could "mutate" A2 alleles to A1 at random. The resulting change would be the effect of an allele substitution. Pr(A2 in A1A2 A1A1) = p ; effect is d to a = ( a - d ) Pr(A2 in A2A2 A1A2) = q ; effect is - a to d = d - (- a ) = ( d + a ) Average change is p ( a-d ) + q ( d+a ) = α = a + d (q-p), i.e., α = α 1 - α 2 so α 1 = q α and α 2 =- p α This reasoning can be extended to multiple alleles, which we won't consider. B. Breeding value A = twice the mean deviation of an individual's progeny from the population mean. Defined in terms of average effects, the breeding value is the sum of the average effects of the alleles it carries. Genotype Breeding value, A A1A1 α 1 + α 1 = 2 α 1 A1A2 α 1 + α 2 A2A2 α 2 + α 2 = 2 α 2 - Expected breeding value of an individual is the average of the breeding values of that individual's two parents. E[A Offspr ] = (1/2)[ A mom + A dad ] Population mean breeding value A = 0; to prove, multiply the breeding value of each genotype by the frequency of the genotype, so: A = p 2 (2q α ) + 2pq[(q-p) α ] + q 2 (-2p α ) A = 2pq α (p+q-p-q) = 0 C. Dominance deviation -
3 Define the Genotypic value G (in caps) in terms of deviation from the population mean,

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## This note was uploaded on 03/01/2010 for the course PCB 4683 taught by Professor Williams,j during the Spring '08 term at University of Central Florida.

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Lecture 10 - Quant Gen II - 1 Lecture 10: Quantitative...

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