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Chapter 6: Quantitative traits, breeding value and heritability • • • • • • Quantitative traits Phenotypic and genotypic values Breeding value Dominance deviation Additive variance Heritability Quantitative traits • Phenotype = Genotype + Environment P=G+E • Mean value (m) • Standard deviation (s) • QTL (Quantitative Trait Loci) Quantitative traits, breeding value and heritability Fat % in SDM: Mean value (m) = 4,3% Standard deviation (s) = 0,25% Fat % in Jersey: Mean value (m) = 6,4% Phenotype value (P) • Phenotypic value = own performance • Phenotypic value can be measured and is evaluated in relation to the mean value of the population • Phenotypic value is determined by the Genotype value (G) and Environmental effect (E) Genotype value (G) • Joint effect of all genes in all relevant loci • The phenotype mean value (Pg) of individuals with the same genotype Breeding value (A) A = 2(`Pg -`Ppop) `Pg Genotypic value and dominance deviation No Dominance deviation (D) heterozygote = the average of homozygotes Genotypic value and dominance deviation in a locus • In case of dominance for a locus, the genotypic value is determined as the breeding value plus type and size of the dominance deviation • G =A+ D • Dominance: Interaction within a locus Dominance types • No dominance : The heterozygote genotypic value is the average of the two homozygotes • Complete dominance : The heterozygote genotypic value is as one of the homozygotes • Over dominance : The heterozygote genotypic value is outside one of the two homozygotes Calculation of defined mean value of weight in mice P(A1)= 0,3 q(A2)= 0,7 Mouse weight for the genotypes: A2 A2 6 A2 A1 12 A 1 A1 14 gram Ppop = (genotype valuefrequency) 60.72 + 122 0.70.3 + 140.32 = = 9.24 Calculation of defined breeding value of an individual A1A1 Individual A1 A1 A1 A2 p(A1A1 offspring) = 1p = 10.3 p(A1A2 offspring) = 1q = 10,7 PA A = 140.3 + 120.7 = 12.6 1 1 Calculation of defined breeding value, continued PA A = 12,6 `Ppop = 9.24 AA A = 2(`PA A -`Ppop) = 2( 12.6 – 9.24) = 6,72 On phenotype scale: AA A = 2(`PA A -`Ppop) +`Ppop = 6.72 + 9.24 = 15.96 1 1 1 1 1 1 1 1 1 1 Genotype value, breeding value and dominance deviation • The effect on a quantitative trait of a single loci is difficult to identify • Solution: Ignore the individual loci and define the problem as quantitative! Calculation of mean value: Example Genotype: TT Tt tt Kg milk: 1882 1882 2082 Genotype frequencies: p2=0.45 2pq=0.44 q2=0.11 pT = 0.67 and qt = 0.33 Ppop = 0.45 1882 + 0.441882 + 0.112082 = 1904 kg Calculation of environmental effect: Example PTT = 1882 kg Mathilde: P = 1978 kg milk Maren: P = 1773 kg milk P=G+E E = +96 E = -109 Calculation of breeding value of the heterozygote • An animal’s breeding value is not necessarily the same as the genotypic value • The breeding value of a heterozygote is the average of the breeding values for the two homozygotes • ATt = (ATT + Att)/2 Calculation of breeding value and dominance deviation TT and Tt Mean value 1882 1904 -22 0 tt 2082 +178 A = 2(`Pg -`P ) p(T) = 0.67 q(t) = 0.33 ATT = 2((-22 0.67 + -22 0.33) - 0) = -44 Att = 2((-22 0.67 + 178 0.33) - 0) = 88 ATt = (ATT + Att)/2 = 22 Calculation of dominance deviation Genotype TT Tt tt G = A+D -22 = -44 + 22 -22 = 22 + (-44) 178 = 88 + 90 Additive variance (s2A) • The genetic variance (s2G) for a locus is due to differences in breeding values or in dominance deviations • s2A is calculated as the mean value of the additive genetic deviations squared • s2A is due to the differences in breeding values Additive variance • s2A = (genotype frequency (A - P) 2) • s2A = (-44-0)2 0.45 + (22-0)2 0.44 + (88-0)2 0.11 = 1926 Phenotypic variance s2p is estimated directly as the variance of the observed values Heritability • The proportion of the phenotypic variance, which is caused by the additive variance, is called the heritability • h2 = s2A / s2p Heritability and common environment • Common environment (c2) • Heritability is calculated as the correlation between half sibs, as they normally only have genes in common and not the environment Heritability estimation Selection response = R Selection difference = S • R = h2 S h2 = R/S • Heritability is the part of the parents’ phenotypic deviation, which can be transferred to their offspring Heritability estimation, continued • Heritability can be determined as the calculated correlation (r or t) between related individuals in relation to the coefficient of relationship (a) • h2 = r / a r = a h2 Estimation of common environment • The correlation between related individuals: t = a h 2 + c2 Weight offspring Weight mother Example: Estimation of heritability and common environment Half sib correlation: • t = 0.03 • 1/4 h2 + 0 = 0.03 Full sib correlation: • h2 = 0.12 t = 0.41 1/2 h2 + c2 = 0.41 c2 = 0.35 1/2 0.12 + c2 = 0.41