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 (• Standard deviation• QTL (Quantitative Trait Loci)

Quantitative traits, breeding value and heritability

Fat % in SDM:

Mean value

Standard deviation (

Fat %in Jersey:

Mean value

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)= 0,7Mouse weight for the genotypes:

A2 A2 A2 A1 A1 A1

6 12 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

A1A1 A1 A2

p(A1A1 offspring) = 1p = 10.3 p(A1A2 offspring) = 1q = 10,7

PA1A1 = 140.3 + 120.7 = 12.6

Population

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

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 = 0.451882 + 0.441882 + 0.112082

= 1904 kg

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

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

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) = 88ATt = (ATT + Att)/2 = 22

TT and Tt Mean value tt 1882 1904 2082 -22 0 +178

Calculation of dominance deviation

Genotype G = A + D

TT -22 = -44 + 22 Tt -22 = 22 + (-44) tt 178 = 88 + 90

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

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

Phenotypic variance

2p 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 = 2A / 2

p

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 h2 + c2

Weight mother

Weight offspring

Example: Estimation of heritability and common

environmentHalf 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