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 (  Standard deviation  QTL (Quantitative Trait Loci)

3. Quantitative traits, breeding value and heritability Fat % in SDM: Mean value  Standard deviation (  Fat %  in Jersey: Mean value 

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(  P g -  P pop ) Pg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(A 1 )= 0,3 q(A 2 )= 0,7 Mouse weight for the genotypes: A 2 A 2 A 2 A 1 A 1 A 1 6 12 14 gram P pop =  (genotype value  frequency) = 6  0.7 2 + 12  2  0.7  0.3 + 14  0.3 2 = 9.24

11. Calculation of defined breeding value of an individual A1A1 Individual  A 1 A 1  A 1 A 2 p(A 1 A 1 offspring) = 1  p = 1  0.3 p(A 1 A 2 offspring) = 1  q = 1  0,7 P A 1 A 1 = 14  0.3 + 12  0.7 = 12.6 Population 

12. Calculation of defined breeding value, continued P A 1 A 1 = 12,6  P pop = 9.24 A A 1 A 1 = 2(  P A 1 A 1 -  P pop ) = 2( 12.6 – 9.24) = 6,72 On phenotype scale: A A 1 A 1 = 2(  P A 1 A 1 -  P pop ) +  P pop = 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: TTTttt Kg milk: 1882 1882 2082 Genotype frequencies: p 2 =0.45 2pq=0.44 q 2 =0.11 p T = 0.67 and q t = 0.33 P pop = 0.45  1882 + 0.44  1882 + 0.11  2082 = 1904 kg

15. Calculation of environmental effect: Example P TT = 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 A Tt = (A TT + A tt )/2

17. Calculation of breeding value and dominance deviation A = 2(  P g -  P )p(T) = 0.67 q(t) = 0.33 A TT = 2((-22  0.67 + -22  0.33) - 0) = -44 A tt = 2((-22  0.67 + 178  0.33) - 0) = 88 A Tt = (A TT + A tt )/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 (  2 A ) The genetic variance (  2 G ) for a locus is due to differences in breeding values or in dominance deviations  2 A is calculated as the mean value of the additive genetic deviations squared  2 A is due to the differences in breeding values

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

21. Phenotypic variance  2 p is 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 h 2 =  2 A /  2 p

23. Heritability and common environment Common environment (c 2 ) 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 = h 2  Sh 2 = 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) h 2 = r / a r = a  h 2

26. Estimation of common environment The correlation between related individuals: t = a  h 2 + c 2 Weight mother Weight offspring

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