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What is the probability for this phenotype to 'contain' a positive or a negative mean inter-action effect with the two environments of the experiment?

Lecture on the concept of heritability in Plant Breeding (wlink@gwdg.de, March 2012)Lecture on the concept of heritability in Plant Breeding (wlink@gwdg.de, March 2012)

2

Heritability is a shrinkage coefficient: 0<h²<1It shrinks the phenotypic variance σ²P to the genetic variance σ²G: σ²G / σ²P = h²h² is not the square of heritability, it is heritability itself.

If you do not have honest environments (like seasons, years, agro-ecological conditions, soils, rotation etc.) at your disposal, then you may not talk about heritability but perhaps about mere repeatability. This discussion leads as well to the item of (1) mega-environments versus years x location-combinations and (2) fixed versus random effects.

There is heritability in broad sense and in narrow sense. The latter is focussing on breeding values and additive variance, items that are dealt with in quantitative genetics. Here we talk about heritability in broad sense.

3

10 11 12 13 14 15 16 17 18 19 20

10 11 12 13 14 15 16 17 18 19 20

10 11 12 13 14 15 16 17 18 19 20

Var = 11.667

Var = 2.333

Var = 3.889

Top: six ‘components‘, each with its result. µ=15, N-weighted variance is 11.67Middle: 20 ‘mixtures‘ of three each. µ=15, N-weighted variance is 2.33Bottom: 216 ‘mixtures‘ of three each. µ=15, N-weighted variance is 3.89

What is the difference between ‘Middle‘ and ‘Bottom‘?

[1] On variance between single values (samples) and means of samples

4

σ²x = σ²x / nσ²x = [σ²x/n]λ

λ=1-[(n-1)/(N-1)]

Components 1. 2. 3.

Means

1 2 3 12.00 1 2 4 12.67 1 2 5 13.33 1 2 6 14 1 3 4 13.33 ... ... ... ... 3 5 6 17.33 4 5 6 18.00

N 20 Mean 15.00

Variance 2.333

Components 1. 2. 3.

Means

1 1 1 10.00 1 1 2 10.67 1 1 3 11.33 1 1 4 12.00 1 1 5 12.67 1 1 6 13.33 1 2 1 10.67 1 2 2 11.33 .. .. .. ... 6 6 3 18.00 6 6 4 18.67 6 6 5 19.33 6 6 6 20.00

N 216 Mean 15.00

Variance 3.889

Component Performance 1 10 2 12 3 14 4 16 5 18 6 20 N 6

Mean 15 Variance 11.667

Sampling without replacement

(11.667/3)[1 - (3-1)/(6-1)] = (3.889)(3/5) = 2.333 ►►voilà

Sampling with replacement

11.667/3 = 3.889

N very large; then the question of replacement vanishes.

Important:The variance among means (n values per mean) is 1/n of the variance among the single values.

[1]

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[2] The variance between means is smaller than the mean variance

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µ=15; σ²=5.6

µ=17; σ²=14

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µ(σ²)=11.2

µ(µ)=15; σ²(µ)=10.8

As soon as there is G x E-interaction, the variance between genotypes in a single environments is ON AVERAGE larger than the variance of the genotypic means across environments;

local/regional breeding > ‚wide-area‘-breeding

Env.3

[2]

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The data (mainly) used here is invented data. There are 12 „environments“ with means of environments from 8.38 up to 11.36; and with means of genotypes from 7.58 to 12.27. The actually highest performance of a genotype at an environment was 13.47, the lowest was 5.77.The pure genetic, pure environmental and the GxE variance was – according to intention – taken to be 1 and normally distributed. The general mean was taken as 10.

[2]

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Environments Genotypes

1 2 3 4 5 .. 12 Mean Gi 1 13.23 13.68 14.00 12.23 13.24 10.76 12.27 2.337 2 13.47 12.17 13.30 12.83 13.21 9.12 11.63 1.699 3 13.41 11.39 12.61 11.11 11.30 10.89 11.44 1.513 4 12.66 12.54 12.41 10.95 12.88 10.23 11.54 1.613 5 11.43 13.55 12.28 11.42 11.04 8.03 10.98 1.048 … ... … … … 34 11.57 8.58 9.58 9.01 10.89 7.64 8.77 -1.161 35 8.87 7.27 9.65 8.49 9.21 8.20 8.11 -1.815 36 9.29 8.60 9.00 6.59 8.27 5.77 7.58 -2.346

Mean 11.36 10.82 10.69 10.55 10.37 8.38 Ej 1.433 0.892 0.758 0.623 0.445 -1.548

Variance►► 1.807 2.464 2.350 1.685 2.466

1.355 Mean►► Variance

2.0268

µ=9.9295

Single result minus µ minus Gi minus Ej = Gi x Ei interaction effect13.23 - 9.9295 - 2.337 - 1.433 = -0.4696 -0.4696 = G x E-interaction between Gen.1 and Env.1‘Best’ phenotypic result of genotype 1 is G1 = 12.27 (its mean across 12 env’s.)The true genotypic value of genotype 1 is unknown!Similarly, the true value of E1 and G1xE1 etc. is unknown. What we have here are just estimates of truth from the data.

[3] GxE

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Single result minus µ minus Gi minus Ej = Gi x Ei interaction effect13.23 - 9.9295 - 2.337 - 1.433 = -0.4696 -0.4696 = G x E-interaction between Gen.1 and Env.1‘Best’ phenotypic result of genotype 1 is G1 = 12.27 (its mean across 12 env’s.)The true value of genotype 1 is unknown!Similarly, the true value of E1 and G1xE1 etc. is unknown. What we have here are just estimates of truth from the data.

Environments Gen.

1 2 3 4 5 … 12 Mean Gi 1 -0.47 0.53 0.98 -0.65 0.53 0.04 ? 2.337 2 0.41 -0.35 0.92 0.58 1.14 -0.96 1.699 3 0.53 -0.94 0.40 -0.96 -0.59 1.00 1.513 4 -0.32 0.11 0.11 -1.21 0.90 0.23 1.613 5 -0.98 1.68 0.54 -0.18 -0.39 -1.40 1.048 … 34 1.37 -1.08 0.06 -0.38 1.67 0.42 -1.161 35 -0.67 -1.73 0.78 -0.25 0.65 1.63 -1.815 36 0.27 0.13 0.66 -1.62 0.24 -0.26 -2.346

Mean ? Ej 1.433 0.892 0.758 0.623 0.445 -1.548

Variance►► 0.848 0.790 1.432 0.717 1.405

0.540

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I:\Wegen Erblichkeit\ANOVA-GR1.spf

Main effects of envs.

Mean = 9.92947Variance = 0.81853SD=0.90473

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C:\Dokumente und Einstellungen\Wolfgang Link\Desktop\Wegen Erblichkeit\ANOVA-GR2.spf

Main effects of genotypes

Mean = 9.92947Variance = 1.06449SD=1.03174

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[3]

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All 12 x 36 combinations36 GxE-effects at the first env.Var.=0.8478

Var.=0.9378

[3]

15

Each single result such as the 13.23 is (thought, is modelled to be …) composed additively from the following effects: Gij = µ + Gi + Ej + (GE)ij

The mean of genotypes 1, 2 etc. across the 12 environments is composed additively from the following effects: 12.27 = µ + (12 G1)/12 + (∑E(1 to12))/12 + (∑G1 x E(1 to12)) /12 11.63 = µ + (12 G2)/12 + (∑E(1 to12))/12 + (∑G2 x E(2 to12)) /12 etc.

So: what makes these genotypes’ means to vary? The variance between them is (of course) caused by the Gi (the genotype’s main

effects) which are unique and different from each other (this causes this variance to mainly contain the true genotypic variance).

The Ej effects do not contribute. They contribute the same, constant item to each genotype’s mean. Tricky: when estimating from the data, then ∑E(1 to12)) is zero. The estimated environments’ main effects are the deviation from the grand mean, their sum = 0. BUT: These 12 envs. are not all envs. that exist in the basic pool (of envs. in the target region) => their true mean Ej will not be exactly zero.

There is variance between genotypes’ results due to GxE! This is because ∑G1 x E(1 to12)) is independent and different from ∑G2 x E(1 to12)) etc. (is not a constant item). Taking the estimates for the GxE from the data, ∑G1 x E(1 to12)) is zero. But we KNOW that we sample only 12 out of ∞ available true GxE effects, so two samples of 12 are only expected to add up to the same (zero); yet in real cases such samples will very probably differ from each other.

Environments Genotypes

1 2 3 4 5 .. 12 Mean Gi 1 13.23 13.68 14.00 12.23 13.24 10.76 12.27 2.337 2 13.47 12.17 13.30 12.83 13.21 9.12 11.63 1.699 … ... … … …

Mean 11.36 10.82 10.69 10.55 10.37 8.38 9.9295 0.00 Ej 1.433 0.892 0.758 0.623 0.445 -1.548

Variance ►►► 1.06449

[3]

16

Each single result such as the 13.23 is (thought, is modelled to be …) composed additively from the following effects: Gij = µ + Gi + Ej + (GE)ij

The mean of genotypes 1, 2 etc. across the 12 environments is composed additively from the following effects: 12.27 = µ + (12 G1)/12 + (∑E(1 to12))/12 + (∑G1 x E(1 to12)) /12 11.63 = µ + (12 G2)/12 + (∑E(1 to12))/12 + (∑G2 x E(2 to12)) /12 etc.

So: what makes these genotypes’ means to vary? The figure 12.27 contains the mean of 12 independent true GxE-effects. The figure 11.63 contains the mean of 12 other, again independent true GxE-effects. All these GxE-effects are (thought to be) sampled from a joint, common basic pool of GxE-effects with mean value of zero and variance of σ²GE (which is not truly known to us); they are (in the basic pool) uncorrelated with G and E main effects (yet, few of them such as 12 may well be correlated to G or E, just by chance). So; each of the 36 genotype means contains such a mean-of-12 GxE effects. How big is the variance between these 36 mean-of-12 GxE effects? It is expected to be 1/12 of the σ²GE variance (σ²µ[x] = σ²x/n). We expect the variance between the 36 means of genotypes to be larger than their pure genetic variance σ²G (which is caused by their different true Gi values); this increment is expected to be 1/12 of the σ²GE (E=12 environments).

Environments Genotypes

1 2 3 4 5 .. 12 Mean Gi 1 13.23 13.68 14.00 12.23 13.24 10.76 12.27 2.337 2 13.47 12.17 13.30 12.83 13.21 9.12 11.63 1.699 … ... … … …

Mean 11.36 10.82 10.69 10.55 10.37 8.38 9.9295 0.00 Ej 1.433 0.892 0.758 0.623 0.445 -1.548

Variance ►► 1.06449

I am sure that I did not as thoroughly as this would be done by an honest statistician distinguish between ‚true value‘ and ‚estimate of the true value‘! Alas.

[3]

17

Phenotypic* estimate of Gi = Ĝi Ĝi = Gi + “constant mean Ej” + “Gi-specific mean GixEj” Phenotypic variance between the mean values of genotypes taken from current data:

σ²Ĝ = σ²P = σ²G + (σ²GE)/E with E being the number of environments. *Estimates are always ‘phenotypic’, I mean ‘based on experimental data’, or could there be anything else?

[4] h² from ANOVA

18

ANOVA across 12 environments and 36 genotypes

Source of var. DF MS Var.comp. F-value LSD(5%)

Environments 12-1 29.4672 0.7894 28.07** 0.47

Genotypes 36-1 12.7738 0.9770 12.17** 0.82

G x E 385 1.0498 1.0498

Phenotypic variance = σ²G + σ²GE /12 = 0.9770 + 1.0498 / 12 = 1.0645 MS(Genot.) = 12 σ²P Heritability h² = genot. variance / phenot. variance

h² = 0.977 / [(1/12)12.7738] = 0.9178 h² = 0.977 / [(0.977 + (1/12)1.0498] = 0.9178 h² = Var(G) / [(Var(G)+ (1/12)Var(GxE)]

It looks rather ‘easy’ to write: ‘The mean square of genotypes is twelve times the phenotypic variance of the genotypes; and it includes one time genotypes x environments interaction variance’. Mean Square(Genot.) = 12 σ²P = 12 σ²G + σ²GE Honestly speaking there is much to say to these “=” signs The event here is what is literally called: ‘equate the calculated mean square to its expected composition’

E{MS(G)} = 12σ²G + σ²GE Searle, SR (1991). C. R. Henderson, the Statistician and his Contributions to Variance Components Estimation. J Dairy Sci 74:4035-4044.

[4]

…distinguish between ‚true value‘ and ‚estimate of the true

value‘! Alas.

19

ANOVA across 12 environments and 36 genotypes

Source of var. DF MS Var.comp. F-value LSD(5%)

Environments 12-1 29.4672 0.7894 28.07** 0.47

Genotypes 36-1 12.7738 0.9770 12.17** 0.82

G x E 385 1.0498 1.0498

Heritability h² = genot. variance / phenot. variance h² = 0.977 / [(1/12)12.7738] = 0.9178 h² = 0.977 / [(0.977 + (1/12)1.0498] = 0.9178 h² = Var(G) / [(Var(G)+ (1/12)Var(GxE)]

Variance between the 432 GxE-effects (12 environments x 36 genotypes): Var(GE)ij = 0.93776; This is NOT what is calculated in the ANOVA! For the ANOVA, the 432 GxE-effects are considered as 12 groups (envs.) of values and 36 values (genots.) each (or as 36 groups à 12 values each, leading to the same final result). For example, the variance of the GxE-effects in the first environment, Var(GxE)i1 = 0.84778; And, in the second environment, this variance is Var(GxE)i2 = 0.79007; and so on. For the ANOVA, the 12 estimates of Var(GxE)ij from the 12 envs. are pooled (averaged). Therefore, the sums of squares SS(GxE)ij when calculating the variance is not divided by [(12 x 36)-1], but it is divided by only (12-1)(36-1) = 385. In this example, the sums of squares of the GxE-interaction effects amounts to SS(GxE) = 404.1746; (N.B: SS(GxE)=Var(GixEj)x(432-1) = 0.93776 x 431 = 404.1746); Divide this SS(GxE) by (12-1)(36-1), i.e. by 385 degrees of freedom; Var(GixEj) = 404.1746 / 385 = 1.0498; which is the figure given in the ANOVA.

[4]

20

Imagine you have data of two locations.

By mistake, yield data of the second location include (throughout) the weight of the bucket that was used for weighing. How will this mistake influence heritability of the data?

Or:

By mistake, yield of the second location (where plot size was double than plot size at location 1) was not divided by 2. How will this mistake influecen heritability of your data?

Check supplementary files, and tell me about this asap.

21

ANOVA across 12 environments and 36 genotypes

Source DF MS Var.cp. F

Environments 11 29.4672 0.7894 28.07**

Genotypes 35 12.7738 0.9770 12.17**

G x E 385 1.0498 1.0498

ANOVA across environments E2 and E7 and across 36 genotypes

Source DF MS Var.cp. F

Environments 1 15.4089 0.4062 19.60**

Genotypes 35 4.5043 1.8591 5.73**

G x E 35 0.7860 0.7860

4.5043/2=2.2522

[4]

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Variance at the 12 environments (I mean, single by single!)

Var. of means at E2 & E7

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Env. 2 Var.=2.4639

Envs. 2&7 Var.=2.2522

So what ?

"There is no variance between genotypes left" ... Cannot be true

[4]

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Environments

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Environments

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h²= σ²G / σ²P

h²=σ²G / [σ²G+ (1/Y)σ²GY + (1/L)σ²GL+ (1/YL)σ²GYL + (1/YLR)σ²R:GYL ] σ²R:GYL = σ²e

G = (number of) genotypes; L=(number of) locations; Y = (number of) yearsIf you increase Y, then you decrease σ²GY and σ²GYL and σ²R:GYL If you increase L, then you decrease σ²GL and σ²GYL and σ²R:GYL If you increase R, then you only decrease σ²R:GYL

Thus, if seed or budget for plots is limited, then

you better increase Y or L (with small R) instead of R (with small Y,L)!

Var(P) = Var(G) + Var(GxE) / EE ∞ => Var(P) = Var(G)E ∞ => h² 1.0

This here is on of the most prominent findings from statistics for applied plant breeding

[4]

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Mean = 1.5019

Mean = 1.0645

Mean = 2.0268

Lower Limit = 0.977

Var(P) at the 12 environments

Var(P) for all 66 combinations of 2 environments

Predictions of Var(P) from all 66 2-env.-combinations for a 12- env.-situation

σG2

[4]

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r=0.8334**

From E2 & E7

From E8 & E10

[4]

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(39.5-36.0) = h² (56.0-49.0)

S

G

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Heritability ‘h²’ The superiority of the selected genotypes ‘S’ is only partly (0<h²<1) composed of genetic gain ‘G’. This is true as long as there is genotype x envi-ronment interaction. If offspring of selected geno-types are genetically different from them, e.g., due to cross-fertilization, h² is even smaller, because non-inherited gene x gene interactions (domi-nance, epistasis) cause further ‘slippage and shrinkage’.

This here is from a true data set on faba bean, yield (dt/ha) of 36 lines across E=12.

Very different from deducing h² from ANOVA across environment, heritability is here not taken from ’within‘ such a data set (what is, honestly speaking, similar to pulling oneself up by ones bootstraps; sich an den eigenen Haaren aus dem Sumpf ziehen). The approach to verify h² is here quite different: it is based on the true genotypic values as firm anchor points (alas, I know, such ‚firm anchor points‘ are normally unknown).

[5] h² from regression of G on P

30

0 10 20 30 40 50 60 70 80 90 100

Oil content of a bulk of 1200 entries

Bulk of 1200 "events" from 4 initial genotypesInitial Genotype 1Initial Genotype 2Initial Genotype 3Initial Genotype 4

Hightest phenotypic value among 1200 "events"

Imagine you have 1200 DH-lines taken in equal-dose from 4 initial parental genotypes; you analyse newly acquired diversity;Or you have 1200 transgenic events taken in equal-dose from …Or you have 1200 mutation events taken in equal-dose from …

310 10 20 30 40 50 60 70 80 90 100

Oil content of a bulk of 1200 entries

Bulk of 1200 "events" from 4 initial genotypesInitial Genotype 1Initial Genotype 2Initial Genotype 3Initial Genotype 4

Hightest phenotypic value among 1200 "events"

Assumption: there is zero genetic variance, all variance is just error.The (error-)variance among the 1200 is (10%)²; one plant per entry is analysed. The four initial genotypes are each a random sample of plants from the same basic pool [N(40,10)]; but their values are means of 100 plants each. Therefore, the (error-) variance between these initial genotypes is (1/100)(10%)²=(1%)².

It is necessary to conduct a test of significance in such a situation (before e.g. em-barking in a pro-gram to se-quence the ge-nome of this very “promising” highest entry (that looks like having nearly 80% of oil.

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s.e.=0.15245 DH-lines:h² = 0.64LSD(5%) = 1.84%

[%]

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The regression of the true, genotypic values (G) on the phenotypic values is ßG,P= ωGP / σ²P. Greek letter such ß, , ω indicate ‘true’ regression and correlation coefficient and true covariance (ω), whereas Latin (b, r, cov) indicate their estimates. It is ~ difficult to always properly contemplate this. The Gi-values are supposed to be uncorrelated with all effects that are enclosed in phenotypic values such as GxE (except G itself; mind: P=µ+G+GE; or P=µ+G+GY+GL+GYL+RGYL). Therefore the ωGP between the (1) genotypic values and their (2) phenotypic values equals the covariance between the (1) genotypic values and the (2) genotypic values within the phenotypic values (the only reason for the phenotypic data to be ‘similar’ with the true genotypic data is that there is some truth in the phenotypic data). Thus, ωGP = ωGG = σ²G (covariance between x and x is the variance of x; [(xi-µx)(xi-µx)=(xi-µx)²]. ßG,P= ωGP / σ²P => ßG,P = σ²G / σ²P = h²; the regression of G on P estimates h². And: GP = ωGP /(σG · σP) = σ²G /(σG · σP) = σG / σP. Consequently: GP = h The correlation coefficient between the G and the P values equals the square root of h². You can estimate a genotypic effect as Gi = µ + h² ( Pi - µ); Gi as effect, deviation from mean. THREE values that in a way estimate heritability h²: Regression coefficient of the Gi on their Pi (you need to know the Gi) Square of correlation coefficient between the Gi on their Pi (need to know the Gi) Ratio σ²G / σ²P = h² as taken from ANOVA (you need not know Gi).

[5]

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What is the probability for this phenotype to 'contain' a positive or a negative mean inter-action effect with the two environments of the experiment?

Pi = µ + Gi + (GixE1+ GixE2)/2

positive? negative?

[5]

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What is the probability for this phenotype to 'contain' a positive or a negative mean inter-action effect with the two environments of the experiment?

This specific genotype showed an inferior perfor-mance: probably because it is honestly an inferior genotype. Yet, the proba-bility that a

(A) negative GxE inter-action contributed to its low performance is larger than the probability that

(B) it is genotypically even more inferior and a positive GxE interaction improved its actual performance.

Why? Because these ‘even more inferior genotypes’ occur less frequent!

[5]

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What is the probability for this phenotype to 'contain' a positive or a negative mean inter-action effect with the two environments of the experiment?

Take the ‘Tour de France’. It is more probable that the winners were doped than the others. Correspondingly, it is more probable that the losers were sick (or suffered from any other specific disadvantage) than the others. Clear?

[5]

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r = 0.888**m=0.5789**h²=0.8255

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InteractionGxE

This specific genotype showed a inferior perfor-mance: probably because it is honestly an inferior genotype. Yet, the proba-bility that a

(A) negative GxE inter-action contributed to its low performance is larger than the probability that

(B) it is genotypically even more inferior and a positive GxE interaction improved its actual performance.

Why? Because these ‘even more inferior genotypes’ occur less frequent!

[5]

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cro s

s 3 6

En v

s .

6 7 8 9 10 11 12 13

Means of genotypes at two Environments

r = 0.888**m=0.5789**h²=0.8255

µ

µ

gj

gj

InteractionGxE

This specific genotype showed a inferior perfor-mance: probably because it is honestly an inferior genotype. Yet, the proba-bility that a

(A) negative GxE inter-action contributed to its low performance is larger than the probability that

(B) it is genotypically even more inferior and a positive GxE interaction improved its actual performance.

Why? Because these ‘even more inferior genotypes’ occur less frequent!

[5]

39

7

8

9

10

11

12

13

Mea

ns o

f gen

toy p

es a

c ro s

s 36

En v

s.

7 8 9 10 11 12 13

Means of genotypes at two Envs.

"E8&E10""E2&E7"

r = 0.867**m=0.8595**h²=0.2243

r = 0.888**m=0.5789**h²=0.8255

[5]

40

0

10

20

30

40

50

60

70

80

90H

erita

b ilit

y h²

from

AN

OV

A

0.4 0.5 0.6 0.7 0.8 0.9Coeff. of Determination R² between G and P

U2&U7

U8&U10

r=0.1280ns

h² as taken from ANOVA versus R² of G versus P in all 66 pairs of two environments each.

µ=0.6596

µ=0.

6412

[5]

41

0

10

20

30

40

50

60

70

80

90H

erita

b ilit

y h²

from

AN

OV

A

0.4 0.5 0.6 0.7 0.8 0.9Coeff. of regression of G on P

h² as taken from ANOVA versus regression of G on P in all 66 pairs of two environments each.

U2&U7

U8&U10

r=-0.3487**

µ=0.6596

µ=0.

6412

[5]

42

0.4

0.5

0.6

0.7

0.8

0.9C

oeff .

of D

e te r

mi n

a tio

n R

² be

twe e

n G

an d

P

0.4 0.5 0.6 0.7 0.8 0.9Coeff. of regression of G on P

U2&U7

U8&U10

r=0.4898**

R² of G vs. P versusregression of G on P in all 66 pairs of two environments each.

µ=0.6596

µ=0.

6727

[5]

43

6

7

8

9

10

11

12

13

14R

e su l

t s o

f ge n

o typ

e s a

t en v

. 7

6 7 8 9 10 11 12 13 14

Results of genotypes at env. 2

r=0.705**

[13.846; 9.825]

G11+(GE)11.2

Significant correlation between results at two locations.

r<1 because of GE-interactions

Significance of r tells whether r<0, not whether r<1!

The larger the variance due to GE, the smaller is r.

I do not know any equation that connects these two parameters.

Diagonal (bisecting line) not really useful here

[5]

44

-4

-3

-2

-1

0

1

2

3P

h en o

t. d e

v . fr

o m m

ean

a t e

nv. 7

-4 -3 -2 -1 0 1 2 3

Phenot. dev. from mean at env. 2

r=0.705**

[+3.0254;-0.0709]

Effects!

G11+(GE)11.2

Diagonal useful

[5]

45

6

7

8

9

10

11

12

13

14G

e not

ype s

' me a

ns a

c ro s

s e

nvs .

2 &

7

6 7 8 9 10 11 12 13 14

Results of genotypes at env. 2

R²=0.843**m=0.877

[13.846; 11.84]

h²(ANOVA, only 1env.)=0.703

Var

ianc

e=2.

252

Variance=2.463

[5]

46

-4

-3

-2

-1

0

1

2

3M

e an

of E

2&E

7: P

h en o

t. d e

v . fr

om m

e an

-4 -3 -2 -1 0 1 2 3

Phenot. dev. from mean at env. 2

[+3.0254;1.4814]

R²=0.843**m=0.877

Var

ianc

e=2.

252

Variance=2.463

Effects!

(GE)11.2

[5]

47

6

7

8

9

10

11

12

13

14G

e not

ype s

' me a

ns a

c ro s

s e

nvs .

2 &

7

6 7 8 9 10 11 12 13 14

Results of genotypes at env. 2 & 7

R²=0.843**m=0.877

[13.846; 11.84][9.82487; 11.84]

R² = 0.862**m = 0.829

h²(ANOVA, only 1env.)=0.703

Var

ianc

e=2.

252

Effects!

With just two environments, (GE)11.2 + (GE)11.7 = 0

[5]

48

6

7

8

9

10

11

12

13

Mea

ns o

f ge

ntoy

pes

acro

ss 1

2 E

nvs.

6 7 8 9 10 11 12 13

Means of genotypes at two Envs.

"E2&E7"

r = 0.888**m=0.5789**h²=0.8255

[5]

49

6

7

8

9

10

11

12

13

Tru

e g e

noty

p ic

v alu

es

6 7 8 9 10 11 12 13

Means of genotypes at two Envs.

"E2&E7"

r = 0.888**m=0.5789**

Variance=2.463

Var

ianc

e=0.

9574

The true genotypic values and their estimates from averaging all 12 environments are nearly the same here.

The genetic variance is 0.9574

[5]

50

-4

-3

-2

-1

0

1

2

3

4T

rue

g ent

oyp i

c e f

f ec t

s a r

e un

k now

n !

-4 -3 -2 -1 0 1 2 3 4

Means of genotypes' effects (E2&E7)

m=h²=0.8255 (A

NOVA)

Variance=2.2523

Var

ianc

e=0.

9574

Variance=

1.53476

TRUE genetic variance=0.9574Var.cp(G)=1.8591h²=0.8255h² * 1.8591 = 1.5348Var(BLUPs)=1.5348

Predicted (!) genotypic effects

[5]

51

I could not find any of the 66 bi-environment data sets to obey sufficiently to these requirements. Thus, I fully artificially constructed such a data set.

-20

-10

0

10

20

GxE

at

1rst

env

. as

est

imat

ed f

rom

dat

a

7 8 9 10 11 12 13

True Genotypes' performance

r = -0.1051 nsTheoretically and in ideal case we expect the correlation between true G-values and GxE interaction effects with any environment to be zero; and the GxE at one environment to be uncorrelated with the GxE at any other environment.

Now, in an actual (non-ideal) data set that will not be true. This is the reason for h² from ANOVA, b of G on P and R² of G with P to differ in such real data sets.

[5]

52

This data set was constructed to have a true genotypic vari-ance of 1.000, a GxE-variance of 1.000, zero correlations bet-ween G vs. GxE at both en-vironments and zero correla-tions between the GxE of the two environments.

Moreover, the true mean and the mean of the two environ-ments is 10; it is the same, so the difference between the mean of a genotype across both environments and its true value does not contain any E-effect (it shows just the average of its two GxE-effects).

[5]

53

-2

-1

0

1

2G

xE-e

ffec t

s a t

En v

i r.1

-2 -1 0 1 2

1

4

9

12

18 20

Slope of regression = 0.0

Effects of G (g i)

[5]

54

-2

-1

0

1

2G

xE-e

ff ect

s a t

Env

ir .1

-2 -1 0 1 2

1

4

9

12

18 20

Slope of regression = 0.0

Effects of G (g i)

-2

-1

0

1

2

Gx E

-eff e

cts

a t E

nvir .

1

-2 -1 0 1 2

1 4

9 12

18 20

Slope of regression = 0.0

Effects of G (g i)

Gxe

-eff

ect

s a

t E

nvi

r. 2

-2

-1

0

1

2

GxE

-effe

cts

at E

nvir.

2

-2 -1 0 1 2

1 4

9 12

1820

Slope of regression = 0.0

GxE-effects at Envir.1

[5]

556

7

8

9

10

11

12

13

14

Tru

e va

lues

of G

7 8 9 10 11 12 13 14 15

Values of G at Envir.1

Slope of regression = 0.50Correl. Coeff. r = 0.7071**Coeff. of Det. R² = 0.50

Source DF MS Var.cp F LSD5 U 1 72.0 1.9722 72.00** 0.48 G 35 3.0 1.0000 3.00** 2.03 GU 35 1.0 1.0000 Total 71

h² = 1.0/(3.0/2) = 1/1,5 = 0,667

[5]

56

6

7

8

9

10

11

12

13

14T

rue

va

lue

s o

f G

6 7 8 9 10 11 12 13 14P-values, means across Env1 & Env2

Slope of regression = 0.6667Coeff. of det. R² = 0.6667

Source DF MS Varcp U 1 72.0 1.972 G 35 3.0 1.000 GU 35 1.0 1.000 h²= 1/(3/2) =1/1.5= 0.6667

In this ‚ideal‘ data set, we find (ANOVA) h²=0.667. Variance component (genotypes) is 1.00 Variance for GxE-interactions is 1.00. Regression of G on P is b=0.667. Coefficient of determination R² (G vs. P) = 0.667.

P-data from one environment: phenoytpic variance is 2.0; h²=0.5, and b = R² = 0.5.

Variance between the true genotypic values is larger than variance between the genotypic values on the regression line (their BLUPs)!

Var(G)=Var(BLUP)+Var(G-BLUP) 1.0 = 0.6667 + 0.3333 >>> 2/3 is explained by the regression (R²=2/3)

Var(P)=Var(BLUP)+VAR(G-BLUP)+Var(GE) 1.5 = 0.6667 + 0.3333 + 0.5000

[5]

57

-4

-3

-2

-1

0

1

2

3

4T

rue

g ent

oyp i

c e f

f ec t

s

-4 -3 -2 -1 0 1 2 3 4

Means of genotypes' effects (E2&E7)

m=h²=0.8255 (A

NOVA)

Variance=2.2523

Var

ianc

e=0.

9574

Variance=

1.53476

TRUE genetic variance=0.9574Var.cp(G)=1.8591h²=0.8255h² * 1.8591 = 1.5348Var(BLUPs)=1.5348

Predicted (!) genotypic effects

r=0.8885**m=0.5789

[5]

58

R = i h² σP h² = σ²G / σ²P

R = i h σG

Why do we need to shrink with ‚h‘ even if we go in steps of genetic standard deviation? I mean, it is ‘genetic’ standard deviation, not phenotypic, so why is this scale not valid?

We would like to make use of the true genetic variance. But we do not know it. We only estimate it as σ²G = h² σ²P from ANOVA.

When we select between our genotypes, then we select implicitly between the estimates-of-their-genetic-values, this is, between their BLUPs – these are the points on the regression line. Their variance is smaller than the true genetic variance. The variance of these BLUPs is that part of the genetic variance that is explained by its regression G on P. Thus, the variance of these BLUPs is R² of the genetic variance.

Since R² = h², variance to exploit in selection is

σ²BLUPs = h²(h² σ²P); σ²BLUPs = h²(σ²G); we go in steps of h (σG) !

[5]