Transcript Quantitative traits, breeding value and heritability.

Chapter 6:
Quantitative traits, breeding
value and heritability
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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 valuefrequency)
60.72 + 122 0.70.3 + 140.32
=
= 9.24
Calculation of defined breeding
value of an individual A1A1
Individual 
A1 A1 
A1
A2
p(A1A1 offspring) = 1p = 10.3
p(A1A2 offspring) = 1q = 10,7
PA A = 140.3 + 120.7 = 12.6
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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
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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 + 0.441882 + 0.112082
= 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:
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t = 0.03
• 1/4  h2 + 0 = 0.03
Full sib correlation:
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h2 = 0.12
t = 0.41
1/2  h2 + c2 = 0.41
c2 = 0.35
1/2  0.12 + c2 = 0.41