1. The document discusses analyses of multiple location trials to evaluate genotype performance across environments. It describes factors to consider in determining the number and location of environments for trials and statistical analyses for multiple experiments. 2. Key analyses covered include the Bartlett test for homogeneity of error variances, analysis of variance models for sites and years, and joint regression analysis to evaluate genotype by environment interactions. 3. Joint regression analysis fits linear regressions between genotype performances in each environment and the mean performance across environments to identify which interactions are linear versus non-linear.

1. Genotype x Environment
Interactions
Analyses of Multiple Location Trials

2. Previous Class
Why do researchers conduct experiments
over multiple locations and multiple times?
What causes genotype x environment
interactions?
What is the difference between a ‘true’
interaction and a scalar interaction?
What environments can be considered to
be controlled, partially controlled or nor
controlled.

3. How many environments do I need?
Where should
they be?

4. Number of Environments
Availability of planting material.
Diversity of environmental conditions.
Magnitude of error variances and genetic
variances in any one year or location.
Availability of suitable cooperators
Cost of each trial ($’s and time).

5. Location of Environments
 Variability of environment
throughout the target region.
Proximity to research base.
 Availability of good cooperators.
 $$$’s.

6. Analyses of Multiple
Experiments

7. Points to Consider before Analyses
Normality.
Homoscalestisity
(homogeneity) of error
variance.
Additive.
Randomness.

8. Points to Consider before Analyses
Normality.
Homoscalestisity
(homogeneity) of error
variance.
Additive.
Randomness.

9. Bartlett Test
(same degrees of freedom)
M = df{nLn(S) - Ln2}
Where, S = 2/n
2
n-1 = M/C
C = 1 + (n+1)/3ndf
n = number of variances, df is the df
of each variance

10. Bartlett Test
(same degrees of freedom)
df 2
Ln(2
)
5 178 5.148
5 60 4.094
5 98 4.585
5 68 4.202
Total 404 18.081
S = 101.0; Ln(S) = 4.614

11. Bartlett Test
(same degrees of freedom)
df 2
Ln(2
)
5 178 5.148
5 60 4.094
5 98 4.585
5 68 4.202
Total 404 18.081
S = 100.0; Ln(S) = 4.614
M = (5)[(4)(4.614)-18.081] = 1.880, 3df
C = 1 + (5)/[(3)(4)(5)] = 1.083

12. Bartlett Test
(same degrees of freedom)
df 2
Ln(2
)
5 178 5.148
5 60 4.094
5 98 4.585
5 68 4.202
Total 404 18.081
S = 100.0; Ln(S) = 4.614
M = (5)[(4)(4.614)-18.081] = 1.880, 3df
C = 1 + (5)/[(3)(4)(5)] = 1.083
2
3df = 1.880/1.083 = 1.74 ns

13. Bartlett Test
(different degrees of freedom)
M = ( df)nLn(S) - dfLn2
Where, S = [df.2]/(df)
2
n-1 = M/C
C = 1+{(1)/[3(n-1)]}.[(1/df)-1/ (df)]
n = number of variances

14. Bartlett Test
(different degrees of freedom)
df 2
Ln(2
) 1/df
9 0.909 -0.095 0.111
7 0.497 -0.699 0.1429
9 0.076 -2.577 0.1111
7
5
0.103
0.146
-2.273
-1.942
0.1429
0.2000
37 0.7080
S = [df.2]/(df) = 13.79/37 = 0.3727
(df)Ln(S) = (37)(-0.9870) = -36.519

15. Bartlett Test
(different degrees of freedom)
df 2
Ln(2
) 1/df
9 0.909 -0.095 0.111
7 0.497 -0.699 0.1429
9 0.076 -2.577 0.1111
7
5
0.103
0.146
-2.273
-1.942
0.1429
0.2000
37 0.7080
M = (df)Ln(S) - dfLn 2 = -36.519 -(54.472) = 17.96
C = 1+[1/(3)(4)](0.7080 - 0.0270) = 1.057

16. Bartlett Test
(different degrees of freedom)
S = [df.2]/(df) = 13.79/37 = 0.3727
(df)Ln(S) = (37)(=0.9870) = -36.519
M = (df)Ln(S) - dfLn 2 = -36.519 -(54.472) = 17.96
C = 1+[1/(3)(4)](0.7080 - 0.0270) = 1.057
2
3df = 17.96/1.057 = 16.99 **, 3df

17. Heterogeneity of Error Variance
0
10
20
30
40
50
60
70
80
Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze
Seed
Yield

18. Significant Bartlett Test
“What can I do where there is
significant heterogeneity of error
variances?”
Transform the raw data:
Often  ~ 
cw Binomial Distribution
where  = np and  = npq
Transform to square roots

19. Heterogeneity of Error Variance
0
2
4
6
8
10
Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze
SQRT[Seed
Yield]

20. Significant Bartlett Test
“What else can I do where there is
significant heterogeneity of error
variances?”
Transform the raw data:
Homogeneity of error variance can always
be achieved by transforming each site’s data
to the Standardized Normal Distribution
[xi-]/

21. Significant Bartlett Test
“What can I do where there is
significant heterogeneity of error
variances?”
Transform the raw data
Use non-parametric statistics

22. Analyses of Variance

23. Model ~ Multiple sites
Yijk =  + gi + ej + geij + Eijk
i gi = j ej = ij geij
Environments and Replicate blocks are usually
considered to be Random effects. Genotypes are
usually considered to be Fixed effects.

24. Analysis of Variance over sites
Source d.f. EMSq
Sites (s)
Rep w Sites (r)
Genotypes (g)
Geno x Site
Replicate error

25. Source d.f. EMSq
Sites (s) s-1
Rep w Sites (r) s(1-r)
Genotypes (g) g-1
Geno x Site (g-1)(s-1)
Replicate error s(r-1)(g-1)
Analysis of Variance over sites

26. Source d.f. EMSq
Sites (s) s-1 2
e + g2
rws + rg2
s
Rep w Sites (r) s(1-r) 2
e + g2
rws
Genotypes (g) g-1 2
e + r2
gs+ rs2
g
Geno x Site (g-1)(s-1) 2
e + r2
gs
Replicate error s(r-1)(g-1) 2
e
Analysis of Variance over sites

27. Yijkl = +gi+sj+yk+gsij+gyik+syjk+gsyijk+Eijkl
igi=jsj=kyk= 0
ijgsij=ikgyik=jksyij = 0
ijkgsyijk = 0
Models ~ Years and sites

28. Analysis of Variance
Source d.f. EMSq
Years (y) y-1 2
e+gy2
rwswy+rg2
swy+rgs2
y
Sites w Years (s) y(s-1) 2
e + g2
rwswy + rg2
swy
Rep w Sites w year (r) ys(1-r) 2
e + g2
rwswy
Genotypes (g) g-1 2
e + r2
gswy + rs2
gy + rl2
g
Geno x year (y-1)(g-1) 2
e + r2
gswy + rs2
gy
Geno x Site w Year y(g-1)(s-1) 2
e + r2
gswy
Replicate error ys(r-1)(g-1) 2
e

29. Interpretation

30. Interpretation
Look at data: diagrams and graphs
Joint regression analysis
Variance comparison analyze
Probability analysis
Multivariate transformation of residuals:
Additive Main Effects and Multiplicative
Interactions (AMMI)

31. Multiple Experiment Interpretation
Visual Inspection
Inter-plant competition study
Four crop species: Pea, Lentil,
Canola, Mustard
Record plant height (cm) every week
after planting
Significant species x time interaction

32. Plant Biomas x Time after Planting
0
20
40
60
80
100
120
140
1 2 3 4 5 6 7 8 9 17
Weeks
Height
(cm)

33. Plant Biomas x Time after Planting
0
20
40
60
80
100
120
140
1 2 3 4 5 6 7 8 9 17
Weeks
Height
(cm)
Pea Lentil
Mustard
Canola

34. Plant Biomas x Time after Planting
0
20
40
60
80
100
120
140
1 2 3 4 5 6 7 8 9 17
Weeks
Height
(cm)
Legume
Brassica

35. Joint Regression

36. Regression Revision
Glasshouse study, relationship
between time and plant biomass.
Two species: B. napus and S.
alba.
Distructive sampled each week
up to 14 weeks.
Dry weight recorded.

37. Dry Weight Above Ground Biomass
Wk SA BN Wk SA BN Wk SA BN
1 0.011 0.019 6 1.752 0.987 11 30.114 26.142
2 0.210 0.050 7 3.895 1.814 12 35.264 22.314
3 0.312 0.115 8 7.536 4.799 13 43.115 29.444
4 0.642 0.245 9 14.292 6.941 14 45.675 32.321
5 0.936 0.462 10 24.741 11.639

38. Biomass Study
0
10
20
30
40
50
0 5 10 15
Weeks after planting
Dry
weight
(g)
S. alba
B. napus

39. Biomass Study
(Ln Transformation)
-6
-4
-2
0
2
4
6
0 5 10 15
Weeks after planting
Dry
weight
(g)
S. alba
B. napus

40. B. napus
Mean x = 7.5; Mean y = 0.936
SS(x)=227.5; SS(y)=61.66; SP(x,y)=114.30
Ln(Growth) = 0.5024 x Weeks - 2.8328
se(b)= 0.039361

41. B. napus
Mean x = 7.5; Mean y = 0.936
SS(x)=227.5; SS(y)=61.66; SP(x,y)=114.30
Ln(Growth) = 0.5024 x Weeks - 2.8328
se(b)= 0.039361
Source df SS MS
Regression 1 57.43 57.43 ***
Residual 12 4.23 0.35

42. S. alba
Mean x = 7.5; Mean y = 1.435
SS(x)=227.5; SS(y)=61.03; SP(x,y)=112.10
Ln(Growth) = 0.4828 x Weeks - 2.2608
se(b)= 0.046068

43. S. alba
Mean x = 7.5; Mean y = 1.435
SS(x)=227.5; SS(y)=61.03; SP(x,y)=112.10
Ln(Growth) = 0.4828 x Weeks - 2.2608
se(b)= 0.046068
Source df SS MS
Regression 1 55.24 55.24 ***
Residual 12 5.79 0.48

44. Comparison of Regression Slopes
t - Test
[b1 - b2]
[se(b1) + se(b2)/2]
0.4928 - 0.5024
[(0.0460 + 0.0394)/2]
0.0096
0.0427145
= 0.22 ns

45. Joint Regression Analyses

46. Joint Regression Analyses
Yijk =  + gi + ej + geij + Eijk
geij = iej + ij
Yijk =  + gi + (1+i)ej + ij + Eijk

47. Yield
Environments
a
b
c
d

48. Joint Regression Example
Class notes, Table15, Page 229.
20 canola (Brassica napus)
cultivars.
Nine locations, Seed yield.

49. Joint Regression Example
Source df SSq MSq F
Sites 8 1125.2 140.6 8.2 **
Reps w Sites 27 462.5 17.1 11.4 ***
Cultivars 19 358.8 20.3 6.7 *
S x C 152 459.4 3.0 2.0 *
Rep Error 513 771.3 1.5

50. Joint Regression Example
Mo Ge Te Gr Pe Co Ka Mc Bo
Westar 22.0 48.9 28.7 11.1 14.2 54.5 44.3 22.1 25.0
Mean 19.5 52.0 32.0 13.1 15.6 56.1 47.2 21.2 25.7
Source df SSq MSq
Regression 1 1899 1899 ***
Residual 7 22 3.2
Westar = 0.94 x Mean + 0.58

51. Joint Regression Example
Mo Ge Te Gr Pe Co Ka Mc Bo
Bounty 17.2 54.8 32.8 15.4 19.6 60.0 47.6 22.7 28.9
Mean 19.5 52.0 32.0 13.1 15.6 56.1 47.2 21.2 25.7
Source df SSq MSq
Regression 1 2247 2247 ***
Residual 7 31 4.0
Bounty = 1.12 x Mean + 1.12

52. Joint Regression Example
Source df SSq MSq F
Sites 8 1125.2 140.6 8.2 **
Reps w Sites 27 462.5 17.1 11.4 ***
Cultivars 19 358.8 20.3 6.7 *
S x C 152 459.4 3.0 2.0 *
Heter of Reg 19 208.9 11.0 5.8 ***
Residual 133 250.5 1.9 1.2 ns
Rep Error 513 771.3 1.5

53. Joint Regression ~ Example #2
Genotype Site 1 Site 2 Site 3 Site 4 Site 5 Site 6
A 9.9 12.1 15.0 17.7 18.2 22.6
B 12.3 13.7 14.9 16.5 17.1 17.4
C 8.4 12.0 15.7 19.5 19.9 27.2
Mean 10.2 12.6 15.2 17.9 18.4 22.4

54. Joint Regression
5
10
15
20
25
30
8 13 18 23 28
Site Means
Yield

55. Problems with Joint Regression
Non-independence - regression of
genotype values onto site means, which
are derived including the site values.
The x-axis values (site means) are
subject to errors, against the basic
regression assumption.
Sensitivity (-values) correlated with
genotype mean.

56. Problems with Joint Regression
Non-independence - regression of
genotype values onto site means, which
are derived including the site values.
Do not include genotype value in mean
for that regression.
Do regression onto other values other
than site means (i.e. control values).

57. Joint Regression ~ Example #2
Genotype Site 1 Site 2 Site 3 Site 4 Site 5 Site 6
A 9.9 12.1 15.0 17.7 18.2 22.6
B 12.3 13.7 14.9 16.5 17.1 17.4
C 8.4 12.0 15.7 19.5 19.9 27.2
Mean for
A
10.3 12.8 15.3 18.0 18.5 22.3

58. Joint Regression ~ Example #2
Genotype Site 1 Site 2 Site 3 Site 4 Site 5 Site 6
A 9.9 12.1 15.0 17.7 18.2 22.6
B 12.3 13.7 14.9 16.5 17.1 17.4
C 8.4 12.0 15.7 19.5 19.9 27.2
Control 11.0 13.4 16.3 19.0 20.5 24.3

59. Problems with Joint Regression
The x-axis values (site means) are
subject to errors, against the basic
regression assumption.
Sensitivity (-values) correlated
with genotype mean.

60. Addressing the Problems
Use genotype variance over sites
to indicate sensitivity rather than
regression coefficients.

61. Genotype Yield over Sites
0
10
20
30
40
50
60
Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze
Seed
Yield
‘Ark Royal’

62. Genotype Yield over Sites
0
10
20
30
40
50
60
Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze
Seed
Yield
‘Golden Promise’

63. Over Site Variance
Genotype Mean 2
g
A 20.0 (3) 24.13 (2)
B 22.0 (1) 8.11 (4)
C 21.5 (2) 19.24 (3)
D 18.0 (4) 26.05 (1)

64. Univariate Probability
Prediction

65. Over Site Variance
Genotype Mean 2
g
A 20.0 (3) 24.13 (2)
B 22.0 (1) 8.11 (4)
C 21.5 (2) 19.24 (3)
D 18.0 (4) 26.05 (1)

66. Univariate Probability Prediction
ƒ(µ¸A)
T

A
.

67. ƒ(µ¸A)


T
ƒ(AddA
T

A
.
Univariate Probability Prediction

68. -200
0
200
400
600
800
1000
1200
1400
500 510 520 530 540 550 560 570 580 590 600 610 620
Environmental Variation
 1  2 T

69. Use of Normal Distribution
Function Tables
|T – m|
g
to predict values greater than
the target (T)
|m – T|
g
to predict values less than
the target (T)

70. The mean (m) and environmental variance (g
2) of a
genotype is 12.0 t/ha and 16.02, respectively (so  =
4).
What is the probability that the yield of that given
genotype will exceed 14 t/ha when grown at any site
in the region chosen at random from all possible
sites.
Use of Normal Distribution
Function Tables

71. T – m
g
 =
14 – 12
4
=
Use of Normal Distribution
Function Tables
= 0.5
Using normal dist. tables we have the probability
from - to T is 0.6915. Actual answer is 1 – 0.6916 =
30.85 (or 38.85% of all sites in the region).

72. Use of Normal Distribution
Function Tables
The mean (m) and environmental variance (g
2) of a
genotype is 12.0 t/ha and 16.02, respectively (so  =
4).
What is the probability that the yield of that given
genotype will exceed 11 t/ha when grown at any site
in the region chosen at random from all possible
sites.

73. T – m
g
 =
11 – 12
4
=
Use of Normal Distribution
Function Tables
= -0.25
Using normal dist. tables we have (0.25) = 0.5987,
but because  is negative our answer is 1 – (1 –
0.5987) = 0.5987 or 60% of all sites in the region.

74. Exceed the target; and (T-m)/ positive,
then probability = 1 – table value.
Exceed the target; and (T-m)/ negative,
then probability = table value.
Less than the target; and (m-T)/ positive,
then probability = table value.
Less than target; and (m-T)/ negative, then
probability = 1 – table value.
Use of Normal Distribution
Function Tables

75. Univariate Probability
Genotype Mean 2
g g
A 20.0
(3)
24.13 4.91
B 22.0
(1)
8.11 2.85
C 21.5
(2)
19.24 4.38
D 18.0
(4)
26.05 5.10

76. Univariate Probability
Genotype Mean 2
g g T=21
A 20.0
(3)
24.13 4.91 0.43
(3)
B 22.0
(1)
8.11 2.85 0.64
(1)
C 21.5
(2)
19.24 4.38 0.54
(2)
D 18.0
(4)
26.05 5.10 0.28
(4)

77. Univariate Probability
Genotype Mean 2
g g T=21 T=24 T=26
A 20.0
(3)
24.13 4.91 0.43
(3)
0.21
(3)
0.11
(2)
B 22.0
(1)
8.11 2.85 0.64
(1)
0.24
(2)
0.08
(3)
C 21.5
(2)
19.24 4.38 0.54
(2)
0.28
(1)
0.15
(1)
D 18.0
(4)
26.05 5.10 0.28
(4)
0.12
(4)
0.06
(4)

78. Multivariate Probability Prediction
T1
-
T2
-
Tn
-
….
f(x1,x2,..., xn) dx1, dx2,..., dxn

79. Problems with Probability Technique
Setting suitable/appropriate target
values:
Control performance
Industry (or other) standard
Past experience
Experimental averages

80. Complexity of analytical
estimations where number of
variables are high:
Use of rank sums
Problems with Probability Technique

81. Additive Main Effects and
Multiplicative Interactions
AMMI
AMMI analysis partitions the residual
interaction effects using principal
components.
Inspection of scatter plot of first two
eigen values (PC1 and PC2) or first eigen
value onto the mean.

82. AMMI Analyses
Yijk =  + gi + ej + geij + Eijk

83. AMMI Analyses
Yijk-  - gi - ej - Eijk = geij

84. AMMI Analyses
Yijk-  - gi - ej - Eijk = geij
ge11 ge12 ge13 ….. ge1n
ge21 ge22 ge23 ….. ge2n
. . . ….. .
gei1 gei2 gei3 ….. gein
. . . ….. .
gek1 gek2 gek3 ….. gekn

85. AMMI Analysis Seed Yield
0
20
40
60
80
100
0 10 20 30 40 50
Site/Genotype Means
Eigen
value
1
G1
S4
S7 S3
S5
G4
S1
S6
G2
G3
S2

86. 0
20
40
60
80
100
0 10 20 30 40 50
Site/Genotype Means
Eigen
value
1
G1
S4
S7 S3
S5
G4
S1
S6
G2
G3
S2
AMMI Analysis Seed Yield

87. 0
20
40
60
80
100
0 10 20 30 40 50
Site/Genotype Means
Eigen
value
1
G1
S4
S7 S3
S5
G4
S1
S6
G2
G3
S2
AMMI Analysis Seed Yield

88. Time Square
Chi-Square