Genotype x Environment Interactions

1. Genotype x Environment Interactions Analyses of Multiple Location Trials

2. Previous Class • How many locations are sufficient. • Where should sites be located. • Assumptions of over site analyses. • Homoscalestisity of error variance. • Bartlett Test – Same d.f., Different d.f. • Transforming the data. • Analyses of variance table and EMS.

3. Interpretation

4. 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)

5. 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

6. Plant Biomas x Time after Planting

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

8. Plant Biomas x Time after Planting Brassica Legume

9. Joint Regression

10. 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.

11. Dry Weight Above Ground Biomass

12. Biomass Study S. alba B. napus

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

14. 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

15. 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

16. 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

17. 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

18. 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

19. Joint Regression Analyses

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

21. d c Yield b a Environments

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

23. Joint Regression Example

24. Joint Regression Example Westar = 0.94 x Mean + 0.58 Source df SS Regression 1 b1sp(x,y)/ss(x) = [sp(x,y)]2/ss(x) Residual 12 Difference = SS(Res) Total 13 ss(y)

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

26. Joint Regression Example Bounty = 1.12 x Mean + 1.12 Source df SS Regression 1 b1sp(x,y)/ss(x) = [sp(x,y)]2/ss(x) Residual 12 Difference = SS(Res) Total 13 ss(y)

27. Joint Regression Example Bounty = 1.12 x Mean + 1.12 Source df SSq MSq Regression 1 2247 2247 *** Residual 7 31 4.0

28. Joint Regression Example Source SS Heter. Reg ∑[SP(x,yi)]2/SS(x)-[∑SP(x,yi)]2/[SS(x)]2 Residual ∑SS(Resi) G x E 459.4

29. Joint Regression Example

30. Joint Regression ~ Example #2

31. Joint Regression C A B

32. 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.

33. 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).

34. Joint Regression ~ Example #2

35. Joint Regression ~ Example #2

36. 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.

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

38. Genotype Yield over Sites ‘Ark Royal’

39. Genotype Yield over Sites ‘Golden Promise’

40. Over Site Variance

41. Univariate Probability Prediction

42. Over Site Variance

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

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

45. Environmental Variation  1  2 T

46. 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)

47. Use of Normal Distribution Function Tables The mean (m) and environmental variance (g2) 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.

48. Use of Normal Distribution Function Tables T – mg 14 – 12 4 = 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).

49. Use of Normal Distribution Function Tables The mean (m) and environmental variance (g2) 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.

50. Use of Normal Distribution Function Tables T – mg 11 – 12 4 = -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.