POPULATION GENETICS
Quantitative traits and evaluation of phenotypic
variance.
Heritability of quantitative traits
Ing. Radovan KASARDA, PhD.
Estimation of Heritability
• Method „ parent – offspring“ h2OP
–
–
–
–
similarity (kinship) between parents and offspring
according to Wright (1920) relationship coefficient is 0,5
comparison of relatives between groups
h2 is twice of rXY resp. regress coefficient bYX..
Principle of calculation:
Pair
X
Y
1
production of
mother
production of
daughter
ΣX
ΣY
X2
Y2
X.Y
ΣX2
ΣY2
ΣX.Y
n
To calculate rXY resp. bYX estimate semi-values A, B, C.
AX 
2
( X ) 2
variance of X
n
B  Y 
2
( Y ) 2
variance of Y
n
X Y

C   XY 
n
covariance between X and Y
Estimation of Heritability
• Method „ parent – offspring“ h2OP
than:
rXY 
C
 XY

A.B
( X2  Y2 )
C
bYX 
A
h 2 OP  2.rXY
h 2 OP  2.bYX
In case data are divided by fathers, calculate partial Ai, Bi, Ci separatelly for each
father and sumarize it.
Estimation of Heritability
• Method „ parent – offspring“ h2OP
With use of ANOVA:
Father
Variance σ2
Covariance
ni
ni .σ2i
ni.covXY
1
710
77
20
14200
1540
2
520
154
30
15600
4620
3
800
100
30
24000
3000
4
400
20
20
8000
400
1
640
128
40
25600
5120
2
700
70
10
7000
700
4
530
78
20
10600
1560
2
480
35
20
9600
700
3
550
90
30
16500
2700
4
640
64
30
19200
1920
Σ
250
150300
22260
bOP 
covOP
 P2
22260
2

 0,148; hOP
 2.0,148  0,296
150300
Method „half-sibs“ - ANOVA
• half-sibs have one common parent (father)
– use of genetic similarity between half-sibs (0,25)
– universal method
– Conditions of use:
•
•
•
•
groups of half-sibs in same conditions (min. 3 in group)
group based by father (min. 3 fathers) (p)
in group min. 3 measurements (ni)
i.e. min. 9 measurements in total (n)
ANOVA – initial informations
Father
i
Daughter
j
Prod. kg
Yij
1
1
5
25
20
400
σ2
2
2
g
30
1
30
900
2
20
400
p=3
σ2e
30
Σiyij = 80
1008,33
2133,33
900
900
1
30
900
2
25
625
3
Σ
Σiyij= 55
Y2ij
3
σ2g 3
3
σ2e
Yi.2/ni
Yi.
σ2e
ni=3
n=9
35
Σiyij = 90
Y..= 225
2700
Y2i.=5841,66
1225
Σi Σj Y2ij= 6275
ANOVA – One Way
σ2
S
DF
MS
S g  Yi.2 / ni  Y..2 / n
fg=p -1
MSg=Sg:fg
Se  
fe=n - p
MSe=Se:fe
2
2
Y

Y
 ij  i. / ni
i
 g2 
MS g  MSe
ni
 e2  MSe
j

ri 

2
g
2
p
h2=4.ri
σ2P =σ2g + σ2e = 84,259;
S
F
MS
σ2
216,666
2
108,333
12,037
433,33
6
72,222
72,222
ri = 0,143;
h2 = 0,57
ANOVA – „full-sibs“
• in populations of multiparous animals (pigs, poultry,..)
– high genetic similarity of full-sibs (0,5), whose should under equal
conditions show equal phenotype
– difficult for segmentation
• in group – variability of environment
• between groups – variability depended on diverse genetics of
mothers
• between classes – variability deperded on diverse genetics of fathers
• σ2P =σ2F + σ2M + σ2E
• two-factorial ANOVA is used
– i = father; j = mother; k = progeny
ANOVA – initial informations
Father
i
1
Mother
j
1
Daughter
k
Trait
Yijk
Yij.
Y2i../ni
Yi..
Yij.2/nij
Y2ijk
1
2
3
2
1
2
3
3
1
2
3
Σ
p
mi
nj
ΣY... ΣY2i../n Σi Σj Y2ij/nij Σi Σj Σk Y2ijk
i
ANOVA – two factorial
S
DF
MS
S F  Yi.2 / ni  Y...2 / n
fF=p -1
MSF=SF:fF
SM  
i
Se  
i
Yij2. / nij  Yi..2 / ni
j
2
kij
j
k

i
Y
2
ij.
j
/ nij
 F2 
k1.MSF  k2 .MSM  (k2  k1 ).MS E
k1.k3
MSM  MSE
k1
fM=m - p MSM=SM:fM
 M2 
fe=n - m MSe=Se:fe
 e2  MSe
i
 Y
σ2
k1 – average # of daughter per mother
k2 – average # of mother per father
k3 – average # of daugh. per father
ANOVA – two factorial
2
FM
h
 
2
2
P

h 4

2
F
2
F
2
F
2
P

h 4

2
M
2
M
2
P
2
M
ANOVA: Half-sibs
• Exercise: Calculate heritability
p=8
Yi.
Yi.2/ni
Y2ij
1
75,94
288,3430
288,6938
2
75,50
285,0112
285,7600
3
79,33
314,6611
315,2639
4
77,98
304,0430
304,6091
5
73,65
271,2151
271,7600
6
76,38
291,6943
292,1326
7
75,34
283,8047
284,1924
8
75,33
283,7295
284,1279
ni=20
ΣY..=
ΣY2i. /ni=
Σi Σj Y2ij=
ANOVA: Full-sibs
• Exercise:
p =10 m = 130
MSF=391,3
MSM=37,50
MSE=6,40
n = 1955
k1=15,00
k2=15,40
k3=193,00
• Calculate:
σ2 F=
h2 F=
σ2M=
h2M=
σ2E=
h2FM=
σ2P=
Non-parametric estimates of
heritability
• very good elaborated
• used if less information is known
– low number of observation, ranking evaluation
• could resuld in non-realistic estimates
• only informative results of h2
• analyses of influence of genotypes on traits
– Estimates by Young
– estimates using Spearman correlation coeficients
– estimates based on realised heritability
Non-parametric estimates of
heritability
– Estimates by Young:

D
X X
h 
2
(X

M

D

M
 X ).2

above-average mothers

below-average mothers

above-average daughters
XM
XM
XD

XD
below-average daughters
– Estimates using Spearmann rank-order correlation coefficient:
h2 = r S
rS  1 
6 d 2
n(n 2  1)
mother
rank
5,5
5,5
10
daughter
rank
9
8
4
d
3,5
2,5
6
d – absolute difference between mother and daughter ranking
• Calculate heritability:
Non-parametric estimates of
heritability
• estimates based on realised heritability
– by division of pop. on better (+) resp. worser (-) half
h 
2

O

O

P

P
Y Y
Y Y
O – offspring; P - parents
– according to realized selection
 Gr
h 
d
2
h 
2
- realized genetic gain
- selection difference (d=i.s)
XO X
XPX
- difference of selected offspring from average of off. generation
- difference of selected parents from average of parental generation