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Chapter 6: Quantitative traits, breeding value and heritability

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

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Chapter 6: Quantitative traits, breeding value and heritability

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  1. Chapter 6: Quantitative traits, breeding value and heritability • Quantitative traits • Phenotypic and genotypic values • Breeding value • Dominance deviation • Additive variance • Heritability

  2. Quantitative traits • Phenotype = Genotype + Environment P = G + E • Mean value (m) • Standard deviation (s) • QTL (Quantitative Trait Loci)

  3. 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%

  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)

  5. Genotype value (G) • Joint effect of all genes in all relevant loci • The phenotype mean value (Pg) of individuals with the same genotype

  6. Breeding value (A) A = 2(`Pg -`Ppop) `Pg

  7. Genotypic value and dominance deviation No Dominance deviation (D) heterozygote = the average of homozygotes 

  8. 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

  9. 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

  10. Calculation of defined mean value of weight in mice P(A1)= 0,3 q(A2)= 0,7 Mouse weight for the genotypes: A2 A2 A2 A1 A1 A1 6 12 14 gram Ppop= (genotype valuefrequency) = 60.72 + 122 0.70.3 + 140.32 = 9.24

  11. Calculation of defined breeding value of an individual A1A1 Individual A1A1 Population A1 A2 p(A1A1 offspring) = 1p = 10.3 p(A1A2 offspring) = 1q = 10,7  PA1A1= 140.3 + 120.7 = 12.6

  12. Calculation of defined breeding value, continued PA1A1 = 12,6 `Ppop = 9.24 AA1A1 = 2(`PA1A1 -`Ppop) = 2( 12.6 – 9.24) = 6,72 On phenotype scale: AA1A1 = 2(`PA1A1 -`Ppop) +`Ppop = 6.72 + 9.24 = 15.96

  13. 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!

  14. Calculation of mean value: Example Genotype: TT Tt tt Kg milk: 1882 18822082 Genotypefrequencies: p2=0.45 2pq=0.44 q2=0.11 pT = 0.67 and qt = 0.33 Ppop = 0.451882 + 0.441882 + 0.112082 = 1904 kg

  15. Calculation of environmental effect: Example PTT = 1882 kg Mathilde: P = 1978 kg milk E = +96 Maren: P = 1773 kg milk E = -109 P = G + E

  16. 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

  17. Calculation of breeding value and dominance deviation 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 TT and Tt Mean value tt 1882 1904 2082 -22 0 +178

  18. Calculation of dominance deviation Genotype G = A + D TT -22 = -44 + 22 Tt -22 = 22 + (-44) tt 178 = 88 + 90

  19. Additive variance (2A) • The genetic variance (2G) for a locus is due to differences in breeding values or in dominance deviations • 2A is calculated as the mean value of the additive genetic deviations squared • 2A is due to the differences in breeding values

  20. Additive variance • 2A = (genotype frequency  (A - P) 2) • 2A = (-44-0)2 0.45 + (22-0)2  0.44 + (88-0)2  0.11 = 1926

  21. Phenotypic variance 2pis estimated directly as the variance of the observed values

  22. Heritability • The proportion of the phenotypic variance, which is caused by the additive variance, is called the heritability • h2 = 2A/ 2p

  23. 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

  24. 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

  25. 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

  26. Estimation of common environment • The correlation between related individuals: t = a  h2 + c2 Weight offspring Weight mother

  27. Example: Estimation of heritability and common environment Half sib correlation: • t = 0.03 • 1/4  h2 + 0 = 0.03 h2 = 0.12 Full sib correlation: • t = 0.41 1/2  h2 + c2 = 0.41 1/2  0.12 + c2 = 0.41 c2 = 0.35

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