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Biom-33-GxE II.ppt

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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.

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Genotype x Environment Interactions Analyses of Multiple Location Trials
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.
How many environments do I need? Where should they be?
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).
Location of Environments Variability of environment throughout the target region. Proximity to research base. Availability of good cooperators. $$$s.
Analyses of Multiple Experiments
Points to Consider before Analyses Normality. Homoscalestisity (homogeneity) of error variance. Additive. Randomness.
Points to Consider before Analyses Normality. Homoscalestisity (homogeneity) of error variance. Additive. Randomness.
Bartlett Test (same degrees of freedom) M = df{nLn(S) - Ln2} Where, S = ワ2/n 2 n-1 = M/C C = 1 + (n+1)/3ndf n = number of variances, df is the df of each variance
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
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
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
Bartlett Test (different degrees of freedom) M = ( df)nLn(S) - dfLn2 Where, S = [df.2]/(df) 2 n-1 = M/C C = 1+{(1)/[3(n-1)]}.[(1/df)-1/ (df)] n = number of variances
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
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
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
Heterogeneity of Error Variance 0 10 20 30 40 50 60 70 80 Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze Seed Yield
Significant Bartlett Test 金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
Heterogeneity of Error Variance 0 2 4 6 8 10 Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze SQRT[Seed Yield]
Significant Bartlett Test 金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 sites data to the Standardized Normal Distribution [xi-]/
Significant Bartlett Test 金What can I do where there is significant heterogeneity of error variances? Transform the raw data Use non-parametric statistics
Analyses of Variance
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.
Analysis of Variance over sites Source d.f. EMSq Sites (s) Rep w Sites (r) Genotypes (g) Geno x Site Replicate error
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
Source d.f. EMSq Sites (s) s-1 2 e + g2 rws + rg2 s Rep w Sites (r) s(1-r) 2 e + g2 rws Genotypes (g) g-1 2 e + r2 gs+ rs2 g Geno x Site (g-1)(s-1) 2 e + r2 gs Replicate error s(r-1)(g-1) 2 e Analysis of Variance over sites
Yijkl = +gi+sj+yk+gsij+gyik+syjk+gsyijk+Eijkl igi=jsj=kyk= 0 ijgsij=ikgyik=jksyij = 0 ijkgsyijk = 0 Models ~ Years and sites
Analysis of Variance Source d.f. EMSq Years (y) y-1 2 e+gy2 rwswy+rg2 swy+rgs2 y Sites w Years (s) y(s-1) 2 e + g2 rwswy + rg2 swy Rep w Sites w year (r) ys(1-r) 2 e + g2 rwswy Genotypes (g) g-1 2 e + r2 gswy + rs2 gy + rl2 g Geno x year (y-1)(g-1) 2 e + r2 gswy + rs2 gy Geno x Site w Year y(g-1)(s-1) 2 e + r2 gswy Replicate error ys(r-1)(g-1) 2 e
Interpretation
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)
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
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)
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
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
Joint Regression
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.
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
Biomass Study 0 10 20 30 40 50 0 5 10 15 Weeks after planting Dry weight (g) S. alba B. napus
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
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
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
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
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
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
Joint Regression Analyses
Joint Regression Analyses Yijk = + gi + ej + geij + Eijk geij = iej + ij Yijk = + gi + (1+i)ej + ij + Eijk
Yield Environments a b c d
Joint Regression Example Class notes, Table15, Page 229. 20 canola (Brassica napus) cultivars. Nine locations, Seed yield.
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
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
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
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
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
Joint Regression 5 10 15 20 25 30 8 13 18 23 28 Site Means Yield
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.
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).
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
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
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.
Addressing the Problems Use genotype variance over sites to indicate sensitivity rather than regression coefficients.
Genotype Yield over Sites 0 10 20 30 40 50 60 Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze Seed Yield Ark Royal
Genotype Yield over Sites 0 10 20 30 40 50 60 Mosc Gene Tens Gran Pend Colf Kalt Mocc Boze Seed Yield Golden Promise
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)
Univariate Probability Prediction
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)
Univariate Probability Prediction (袖存A) T A .
(袖存A) T (AddA T A . Univariate Probability Prediction
-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
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)
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
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).
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.
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.
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
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
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)
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)
Multivariate Probability Prediction T1 - T2 - Tn - . f(x1,x2,..., xn) dx1, dx2,..., dxn
Problems with Probability Technique Setting suitable/appropriate target values: Control performance Industry (or other) standard Past experience Experimental averages
Complexity of analytical estimations where number of variables are high: Use of rank sums Problems with Probability Technique
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.
AMMI Analyses Yijk = + gi + ej + geij + Eijk
AMMI Analyses Yijk- - gi - ej - Eijk = geij
AMMI Analyses Yijk- - gi - ej - Eijk = geij ge11 ge12 ge13 .. ge1n ge21 ge22 ge23 .. ge2n . . . .. . gei1 gei2 gei3 .. gein . . . .. . gek1 gek2 gek3 .. gekn
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
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
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
Time Square Chi-Square

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Biom-33-GxE II.ppt

  • 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.
  • 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) - Ln2} Where, S = ワ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) - dfLn2 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 金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 金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 sites data to the Standardized Normal Distribution [xi-]/
  • 21. Significant Bartlett Test 金What can I do where there is significant heterogeneity of error variances? Transform the raw data Use non-parametric statistics
  • 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 + g2 rws + rg2 s Rep w Sites (r) s(1-r) 2 e + g2 rws Genotypes (g) g-1 2 e + r2 gs+ rs2 g Geno x Site (g-1)(s-1) 2 e + r2 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+gy2 rwswy+rg2 swy+rgs2 y Sites w Years (s) y(s-1) 2 e + g2 rwswy + rg2 swy Rep w Sites w year (r) ys(1-r) 2 e + g2 rwswy Genotypes (g) g-1 2 e + r2 gswy + rs2 gy + rl2 g Geno x year (y-1)(g-1) 2 e + r2 gswy + rs2 gy Geno x Site w Year y(g-1)(s-1) 2 e + r2 gswy Replicate error ys(r-1)(g-1) 2 e
  • 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
  • 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
  • 46. Joint Regression Analyses Yijk = + gi + ej + geij + Eijk geij = iej + ij Yijk = + gi + (1+i)ej + ij + Eijk
  • 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)
  • 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)
  • 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)
  • 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