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) S = 101.0; Ln(S) = 4.614

11. Bartlett Test (same degrees of freedom) 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) 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) S = [  df.  2 ]/(  df) = 13.79/37 = 0.3727 (  df)Ln(S) = (37)(-0.9870) = -36.519

15. Bartlett Test (different degrees of freedom) 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

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

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 [x i -  ]/ 

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 Y ijk =  + g i + e j + ge ij + E ijk  i g i =  j e j =  ij ge ij 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

27. Y ijkl =  +g i +s j +y k +gs ij +gy ik +sy jk +gsy ijk +E ijkl  i g i =  j s j =  k y k = 0  ij gs ij =  ik gy ik =  jk sy ij = 0  ijk gsy ijk = 0 Models ~ Years and sites

28. Analysis of Variance

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

33. PeaLentil Mustard Canola

34. 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

38. Biomass Study S. alba B. napus

39. Biomass Study (Ln Transformation) S. alba B. napus

40. 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 [b 1 - b 2 ] [se(b 1 ) + se(b 2 )/2] 0.4928 - 0.5024 [(0.0460 + 0.0394)/2] 0.00960.0427145 = 0.22 ns

45. Joint Regression Analyses

46. Y ijk =  + g i + e j + ge ij + E ijk ge ij =  i e j +  ij Y ijk =  + g i + (1+  i )e j +  ij + E ijk

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

50. Source df SSqMSq Regression 118991899 *** Residual 7 22 3.2 Westar = 0.94 x Mean + 0.58

51. Joint Regression Example Source df SSqMSq Regression 122472247 *** Residual 7 31 4.0 Bounty = 1.12 x Mean + 1.12

52. Joint Regression Example

53. Joint Regression ~ Example #2

54. Joint Regression

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

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 ‘Ark Royal’

62. Genotype Yield over Sites ‘Golden Promise’

63. Over Site Variance

64. Univariate Probability Prediction

65. Over Site Variance

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

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

68. Environmental Variation  1 1 1 1  2 2 2 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 16.0 2, 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 16.0 2, 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

78. Multivariate Probability Prediction T1T1--T1T1-- T2T2--T2T2-- TnTn--TnTn-- …. f (x 1,x 2,..., x n ) dx 1, dx 2,..., dx n

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 Y ijk =  + g i + e j + ge ij + E ijk

83. AMMI Analyses Y ijk -  - g i - e j - E ijk = ge ij

84. AMMI Analyses Y ijk -  - g i - e j - E ijk = ge ij ge 11 ge 12 ge 13 ….. ge 1n ge 21 ge 22 ge 23 ….. ge 2n... …...... …... ge i1 ge i2 ge i3 ….. ge in... …...... …... ge k1 ge k2 ge k3 ….. ge kn

85. AMMI Analysis Seed Yield G1 S4 S7 S3 S5 G4 S1 S6 G2 G3 S2

86. G1 S4 S7 S3 S5 G4 S1 S6 G2 G3 S2 AMMI Analysis Seed Yield

87. G1 S4 S7 S3 S5 G4 S1 S6 G2 G3 S2 AMMI Analysis Seed Yield

88. Time Square Chi-Square