Genotype x Environment Interactions

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) 2n-1 = M/C M = df{nLn(S) - Ln2} Where, S = 2/n 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 23df = 1.880/1.083 = 1.74 ns

13. Bartlett Test(different degrees of freedom) 2n-1 = M/C M = ( df)nLn(S) - dfLn2 Where, S = [df.2]/(df) 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 23df = 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 [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 igi = jej = ijgeij 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

25. Analysis of Variance over sites

26. Analysis of Variance over sites

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

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 InterpretationVisual 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. Plant Biomas x Time after Planting Mustard Canola Pea Lentil

34. Plant Biomas x Time after Planting Brassica Legume

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. 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. d c Yield b a Environments

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

49. Joint Regression Example

50. Joint Regression Example Westar = 0.94 x Mean + 0.58 Source df SSq MSq Regression 1 1899 1899 *** Residual 7 22 3.2