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) -  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

Bartlett Test (same degrees of freedom) S = 101.0; Ln(S) = 4.614

Bartlett Test (same degrees of freedom) S = 100.0; Ln(S) = M = (5)[(4)(4.614) ] = 1.880, 3df C = 1 + (5)/[(3)(4)(5)] = 1.083

Bartlett Test (same degrees of freedom) S = 100.0; Ln(S) = M = (5)[(4)(4.614) ] = 1.880, 3df C = 1 + (5)/[(3)(4)(5)] =  2 3df = 1.880/1.083 = 1.74 ns

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

Bartlett Test (different degrees of freedom) S = [  df.  2 ]/(  df) = 13.79/37 = (  df)Ln(S) = (37)( ) =

Bartlett Test (different degrees of freedom) M = (  df)Ln(S) -  dfLn  2 = (54.472) = C = 1+[1/(3)(4)]( ) = 1.057

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

Heterogeneity of Error Variance

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

Heterogeneity of Error Variance

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 -  ]/ 

Significant Bartlett Test  “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 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.

Analysis of Variance over sites

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

Analysis of Variance

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

PeaLentil Mustard Canola

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

Biomass Study S. alba B. napus

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

Mean x = 7.5; Mean y = SS(x)=227.5; SS(y)=61.66; SP(x,y)= Ln(Growth) = x Weeks se(b)=

B. napus Mean x = 7.5; Mean y = SS(x)=227.5; SS(y)=61.66; SP(x,y)= Ln(Growth) = x Weeks se(b)= Source df SS MS Regression *** Residual

S. alba Mean x = 7.5; Mean y = SS(x)=227.5; SS(y)=61.03; SP(x,y)= Ln(Growth) = x Weeks se(b)=

S. alba Mean x = 7.5; Mean y = SS(x)=227.5; SS(y)=61.03; SP(x,y)= Ln(Growth) = x Weeks se(b)= Source df SS MS Regression *** Residual

Comparison of Regression Slopes t - Test [b 1 - b 2 ] [se(b 1 ) + se(b 2 )/2] [( )/2] = 0.22 ns

Joint Regression Analyses

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

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 SSqMSq Regression *** Residual Westar = 0.94 x Mean

Joint Regression Example Source df SSqMSq Regression *** Residual Bounty = 1.12 x Mean

Joint Regression Example

Joint Regression ~ Example #2

Joint Regression

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

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 ‘Ark Royal’

Genotype Yield over Sites ‘Golden Promise’

Over Site Variance

Univariate Probability Prediction

Over Site Variance

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

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

Environmental Variation  1 1 1 1  2 2 2 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 , 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 Actual answer is 1 – = (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 , 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 = Using normal dist. tables we have  (0.25) = , but because  is negative our answer is 1 – (1 – ) = 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

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

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

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

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

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

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

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

Time Square Chi-Square