Lecture 4-Variance Components-ANOVA -2019

ANOVA

1. Variance Components





1
)(
2
i
ii
G
f
meangf
V
1
)(
2




n
meangf
V
ii
G





2
i
2
)(p
)(
meang
N
meangf
V
i
ii
G
for i
th
genotype
For entire F
2
population, use N instead of n -1

2. F2 variance
Frequency:
Value: a d -a
224
1
212
1
114
1
: : AAAAAA
2
4
12
2
1
daV
G

DAG VVV 
2
4
1
D
2
2
1
V ; daV
A


3. Expected variation in F
2 individuals:
EDAF
VVVV 
2
Variance of non-segregation generations
EPVV
1
EP
VV
2
EF
VV
1

4. Generation Mean Variance
P
1 69.44 59.73
P
2 59.04 65.71
F
1
83.44 51.81
F
2
74.36 100.75
BC
1.1
76.03 81.05
BC
1.2
71.28 90.83

5. Estimation of environmental variance
-Adequate randomization is required
-Individuals must be submitted to same
environmental conditions
-Expected independent estimate of V
E
-However, V
E
estimates expected not to differ
significantly for each other. (WHY?)

6. Average of non-segregating generations
P
1
= 59.73
P
2 = 65.71
F
1
= 51.81
V
E = (59.73+65.71+51.81)/3 = 59.08

7. If observations are different then use pooled
variance.
121
121
FPP
FPP
E
dfdfdf
ssssss
V




8. 1. Use Bartlett’s test or Levene's test
2. Use F-test (largest var/smallest var)
Test for
Ho: Var P
1 = Var P
2 = Var F
1
F-test showed Probability value of 0.6 in the data set shown
(Meaning?)
Therefore V
E
calculated can be used (59.08)

9. Why the need to estimate variance
components?
•Breeders want to know if a cross shows
significant variation
•How much of the variation is heritable?
•What type of gene effects are
significant?

10. 1.Comparison of variances (using figures
provided)
Segregating vs non-segregating
generations
F2
Var F
2/V
E = 100.75/59.08 = 1.71**
BC1.1
Var BC
1.1/V
E = 81.05/59.08 =1.37*
BC1.2
Var BC
1.2/V
E = 90.83/59.08 =1.54**

11. Generation Mean Variance
P
1 69.44 59.73
P
2 59.04 65.71
F
1
83.44 51.81
F
2
74.36 100.75
BC
1.1
76.03 81.05
BC
1.2
71.28 90.83

12. 2. Estimation of V
A, V
D and V
AD
2
4
1
D
2
2
1
V ; daV
A

2
1
adV
AD

)2(
2.11.12 BCBCFA VVVV 
)(
22.11.1 EFBCBCD VVVVV 
)(
1.12.12
1
BCBCAD
VVV 

13. EG
VVV
F

2
EDA
VVVV
F

2
2
4
12
2
1
2
daV
F 

14. EBC VaddaV 
2
12
4
12
4
1
1.1
EBC
VaddaV 
2
12
4
12
4
1
2.1
adV
AD 2
1
that Note 

15. EBCBC
VdaVV 2
2
2
1
2
2
1
2.11.1

ABCBCF VaVVV 
2
2
1
)(2
2.11.12

16. V
A = (2 x 100.75 – 81.05 – 90.83) = 29.62
V
D = (81.05 + 90.83 – 100.75 – 59.08)
=12.05
V
AD = (90.38 – 81.05)/2 = 4.89
V
E
= 59.08