1

Genetic Principles of Plant Breeding Quantitative Genetics

Models to compose and decompose the ‘genotypic value’

Wolfgang Link, University of Göttingen, Germany

Dedicated to H.H. Geiger, Hohenheim

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MODELS FOR THE GENOTYPIC VALUE In chapter 2 we did focus on genotypic effects, gi being the sum of many single effects, effects on the expression of a quantitative trait. In this chapter we shall study different linear models, describing the combined action of genes. These models belong to two categories:

metric models statistical models.

In both cases we mostly assume that the genes at different loci act independently (no epistasis) and that their effects are not correlated (no linkage). This implies that the effects (and contributions to the variance) of the several loci and combinations of loci behave purely additive and can be summarized to a genotypic value. Hence, it is valid to construct the models with only one of the pertinent loci, it being an valid example for all loci. Interactions between genes at different loci (epistasis) may involve 2, 3, 4, … loci. Any generalization of 1-locus-models to these situations requires a great number of additional parameters. We therefore restrict ourselves to the easiest case, i.e. 2-loci-interactions (digenic epistasis). This is the most important case of epistatic effects and the principle of the epistatic influence on the genotypic value is quite well elucidated with this case. Linkage between loci is mostly neglected, because the models would else become too cumbersome and because only for a small part of pairs of loci linkage is to be expected (this part being < 1/x, where x equals the number of chromosomes in the haploid genome).

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Metric model of the genotypic values

c + a

c + d

c

c - a

0 aa

Genotypic value

Aa AA

Genotype

d

a

a

aa AAAa

= metric domi-nance effect

= metric additive effect

= metric additive effect

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Metric model The parameters of the metric model describe the differences between the genotypic value of the geno-types under study independently from the gene- and genotype frequencies of the respective populations. They characterize the gene action; e.g. the degree of dominance or type and size of epistatic effects of genes. General asssumption: 2 alleles per locus a) 1 locus: Genotype Genotypic value

AA

c

+

a Aa c + d Aa c - a

Interpretation of the parameters:

c = (AA + aa)/2 Mean of the homozygotes

a = (AA – aa)/2 Half the difference between

homozygotes

d = Aa – (AA +

aa)/2

Deviation of the heterozygote from the mean of the

homozygotes AA, Aa & aa are taken here to specify gentypic values: instead of the more cumbersome expressions GAA, GAa and Gaa; (mostly we use AA etc. for both, genotype & genotypic value) .

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Degree of dominance = d/a

No dominance: d/a = 0Partial dominance: 0 < d/a < 1Complete dominance:d/a = 1

Overdominance:d/a > 1

The reference level in the ‘metric model’ is the level “c”. This c is the average genotypic value of all possible homozygotes (nota bene: not the population mean !). Full homozygosity is reached only after a number of n = ∞ generations of selfing, hence, the models were termed F∞-model or better, F∞-metric.

(In some texts, the reference level is chosen as the average genotypic value of a F2-equilibrium population, leading to a somewhat different metric. In this case the metric is analogously termed F2-metric.)

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Schön, C.C., 1993

AA-genotypes

aa-genotypes

+a-a

High ............Resistance............Low

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Schön, C.C., 1993

a d

8

Schön, C.C., 1993

9

Schön, C

.C., 1993

10

UMC33 UMC128

Schön, C.C., 1993

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Genetic Principles of Plant Breeding 27.08.2009 Examples of the concept of “Effect = (Main Effect)i + (Main Effect)j + (Interaction)ij”

Dominant – recessive gene action A a

A 20 20 Δ 0 ≠ 6 a 20 14

Epistasis

BB bb AA 20 20 Δ 0 ≠ 20 aa 20 0

Effect of genotype + environment on phenotypic value

Env. 1 Env. 2 Gentoype .1 20 20

Δ 0 ≠ 6 Genotype 2 20 14

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Genetic Principles of Plant Breeding 27.08.2009 It makes a difference whether there is epistasis involved or not! Example: Start with a HWE- population! Three genotypes: P(AA/BB) = 36%, H(Aa/Bb) = 48%, Q(aa/bb) = 16%. As the population continues to propagate by random mating (no selection …), “nothing should change”!

Genotypic frequencies Genotypic values Population

perfor-mance

First generation (epistasis ?) BB Bb bb BB Bb bb

AA 0.36 AA 20 Aa 0.48 Aa 20 aa 0.16 aa 10

18.40

No epistasis: Generation ∞ BB Bb bb Σ BB Bb bb

AA 0.1296 0.1728 0.0576 0.36 AA 20 20 16 Aa 0.1728 0.2304 0.0768 0.48 Aa 20 20 16 aa 0.0576 0.0768 0.0256 0.16 aa 14 14 10

Σ 0.36 0.48 0.16

18.40

With epistasis*: Generation ∞ BB Bb bb Σ BB Bb bb

AA 0.1296 0.1728 0.0576 0.36 AA 20 20 20 Aa 0.1728 0.2304 0.0768 0.48 Aa 20 20 20 aa 0.0576 0.0768 0.0256 0.16 aa 20 20 10

Σ 0.36 0.48 0.16

19.744

*(aa)12=(ad)12=(da(12)=(dd)12=-2,5, see further below!

It is of grate importance whether a population (the performance of which is e.g. 18.40) changes its performance without selection, mutation, drift etc., it means without any ‘reason’. This is in contradictory with DUS critera (distinctness, uniformity, stability; the ‘reason here is EPISTASIS)

It is of great importance whether a population (the per-formance of which is e.g. 18.40) changes its performance without (selection, mu-tation, drift etc., it means without) any ‘reason’. This is contradictory to the DUS critera (distinct-ness, uniformity, sta-bility; the ‘reason’ here is EPISTASIS)

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Albina-Locus

Xantha-Locus

Albina-LocusX

antha-Locus

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Modes of Gene action

Supplementary gene action: different alleles of one locus code for different effect that may compensate each other

Complementary gene action: for a full expression of the trait, two loci have to be favourably equipped with pertinent alleles.

Example: Resistance

Genotype of host

Resistance of host when attacked by

1 locus Race ‘R’ Race ‘Q’ Races ‘R’ & ‘Q’

R1R1 Yes No No

R1R2 Yes Yes Yes

R2R2 No Yes No

Mode of gene action

Dominant R1>R2

Dominant R2>R1

Supplementary Gene Action

2 loci* Race ‘R’ Race ‘Q’ Races ‘R’ & ‘Q’

R1R1/Q1Q1 Yes No No

R1R2/Q1Q2 Yes Yes Yes

R2R2/Q2Q2 No Yes No

Mode of gene action

Dominant R1>R2

Dominant Q1>Q2

Complementary Epistasis

*if linked, then only these genotypes may occur

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2 loci:

To fully describe the genotypic value based on 2 loci, we (additionally to the 1-locus-parameters a & d) need the following digenic epistasis parameters:

(aa)12 Additive x additive-interaction (ad)12 Additive x dominance-interaction (da)12 Dominance x additive-interaction (dd)12 Dominance x dominance-interaction

between the genes at locus 1 and 2

Genotypic value (c 0): Locus 2 Locus

1 BB Bb bb

AA

a1 + a2 + (aa)12

a1 + d2 + (ad)12

a1 - a2 - (aa)12

Aa d1 + a2 + (da)12 d1 + d2 + (dd)12 d1 - a2 - (da)12

aa -a1 + a2 - (aa)12 - a1 + d2 - (ad)12 - a1 - a2 + (aa)12

Interpretation:

c = 1/4 (AABB + AAbb + aaBB + aabb)

a1 =

½ [½ (AABB – aaBB) + ½ (AAbb – aabb)] Half difference between the two homozygotes at the first locus, averaged across the homozygous

phases (situations) at the second locus.

a2 =

½ [½ (AABB – AAbb) + ½ (aaBB – aabb)]

Half difference between the two homozygotes at the second locus, averaged across the homo-zygous phases (situations) at the first locus.

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d1 = ½{[AaBB–½(AABB+aaBB)] + [Aabb–½(AAbb+aabb)]}

Deviation of the heterozygote from the mean of the homozygotes at the first locus averaged across the homozygous phases at the second locus. ½

d2 = ½{[AABb–½(AABB+AAbb)] + [aaBb–½(aaBB+aabb)]}

(aa)12 = ½ [½(AABB–aaBB) – ½(AAbb–aabb)] = ½ [½(AABB–AAbb) – ½(aaBB–aabb)]

Dependency of the difference between the homozygotes at one locus from the homozygous phase at the other locus;

(ad)12 =

½{[AABb–(AABB+AAbb)/2] – [aaBb–(aaBB+aabb)/2]} Dependency of the dominance effects at the 2nd locus from the homozygous phase at the first locus

= ½{[AABb–aaBb] – ½[(AABB–aaBB) + (AAbb–aabb)]} Dependency of the homozygotes’ difference at 1st locus from the hetero/homozygosity at 2nd locus.

(da)12 = Analogous to (ad)12

(dd)12 = AaBb–½(AABb+aaBb) – ½{[AaBB–½(AABB+aaBB)] + [Aabb–½(AAbb+aabb)]} Dependency of the dominance effect at the first locus from the hetero/homozygosity an the second locus and vice versa.

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Y = vector of genotypic values, X = matrix of coefficients, = vector of parameters. Solution of the linear system of equations for by matrix inversion leads to:

1-X Y

Contributions of the parameters of the metric 2-loci-model to the genotypic values of the 9 genotypes that are possible with 2 alleles per locus (cf. Tab. at page 5)

Parameter

Genotype c a1 a2 d1 d2 (aa)12 (ad)12 (da)12 (dd)12

AABB 1 1 1 0 0 1 0 0 0

AABb 1 1 0 0 1 0 1 0 0

AAbb 1 1 -1 0 0 -1 0 0 0

AaBB 1 0 1 1 0 0 0 1 0

AaBb 1 0 0 1 1 0 0 0 1

Aabb 1 0 -1 1 0 0 0 -1 0

aaBB 1 -1 1 0 0 -1 0 0 0

aaBb 1 -1 0 0 1 0 -1 0 0

aabb 1 -1 -1 0 0 1 0 0 0

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How to “compose” parameters from genotypes (coefficients of the inverted system of equations1))

Genotype

Para-meter

AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb

c 1 0 1 0 0 0 1 0 1

a1 1 0 1 0 0 0 -1 0 -1

a2 1 0 -1 0 0 0 1 0 -1

d1 -1 0 -1 2 0 2 -1 0 -1

d2 -1 2 -1 0 0 0 -1 2 -1

(aa)12 1 0 -1 0 0 0 -1 0 1

(ad)12 -1 2 -1 0 0 0 1 -2 1

(da)12 -1 0 1 2 0 -2 -1 0 1

(dd)12 1 -2 1 -2 4 -2 1 -2 1 1)All matrix values must be multiplied by ¼, e.g.: c = ¼ [1 x G(AABB) + 1 x G(Aabb) + 1 x G(aaBB) + 1 x G(aabb)] etc.

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2–loci-Model für n loci:

n....54321Locus

....EeddCCBbAAAny genotype

Genotypic value:

- (aa)14 + (ad)15 + ... + (da)23 - (da)24 + (dd)25 + ...

+ a1 + d2 + a3 – a4 + d5 + ... + (ad)12 + (aa)13cGi =

From the single parameters a, d, (aa), (ad), (da) and (dd), a summation parameter can be built by simple addition. Here, we will elucidate the parameter system, the metric, based on several numerical examples and by experimental data sets. The genotypic values are ordered in the standard matrix form:

aabbaaBbaaBB

AabbAaBbAaBB

AAbbAABbAABB

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Examples for trait expression with 2 loci and 2 alleles each, allowing for different modes of gene action. 1) Intermediate 2) Partial dominance

20 18 16 20 19 16 17 15 13 19 18 15 14 12 10 14 13 10

3) Complete domin. 4) Overdominance

20 20 16 20 21 16 20 20 16 22 23 18 14 14 10 14 15 10

5) Complementary 6) Duplicate gene action

20 20 10 20 20 20 20 20 10 20 20 20 10 10 10 20 20 10

7) “Complex“ epistasis

38,7 6,6 3,2 Callus induction rate

(%) 4,0 2,0 1,3 Corn anther culture 2,7 3,4 1,6 (Cowen et al. 1992)

Exercise: Calculate the parameters a1, a2, d1, d2, (aa)11, (ad)12, (da)12, (dd)12 and the degree of dominance (a/d) for the two loci. Check whether you can „re-compose“ the genotypic values from the calculated parameters.

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Solutions

Interme-diate

Partial domininance

Complete dominance

Over- domin.

Comple-mentary Duplicatee Complex

c

15

15

15

15

12,5

17,5

11,55

a1

3

3

3

3

2,5

2,5

9,40

a2 2 2 2 2 2,5 2,5 9,15

d1

2

3

5

2,5

2,5

-8,90

d2 1 2 3 2,5 2,5 -6,55

(aa)12

2,5

-2,5

8,60

(ad)12 2,5 -2,5 -7,80

(da)12 2,5 -2,5 -7,80

(dd)12 2,5 -2,5 5,90

d1/a1 0,67 1 1,67 1 1 -0,95

d2/a2 0,5 1 1,5 1 1 -0,72

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Example for epistasis:

Inheritance of restoration of pollen fertility in a F2-population of winter rye (Secale cereale)

Two way tableau of marker class means for pollen fertility (1-9)a; marker loci are two unlinked RFLP-marker loci Ma (MWG59) und Mb (PSR371), originating from the female and male parent; Scores large enough for use in breeding are shaded in grey.

MWG59 Marker-genotype

MaMa Mama mama Mean

MbMb 7.9 7.1 1.5 5.5

Mbmb 7.8 8.0 2.0 5.9

PSR371 mbmb 4.6 5.6 3.5 4.6

Mean 6.8 6.9 2.3 5.3 a) 1 = totally male sterile, 9 = fully male fertile. (from: Dreyer, Ph.D. 2000) Exercise: Estimation of the parameters of the metric model:

c = a1 = (aa)12 = a2 = (ad)12 = d1 = (da)12 = d2 = (dd)12 =

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60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 Plant height [cm]

60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 Plant height [cm]

60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 Plant height [cm]

60 65 70 75 80 85 90 95 100 105 110 115 120 125 130

B1 B2

F 2

B

F 1

P2 P1 P

Heterosis

10

10

10

10

40

20

20

20

20

30

30

30

30

60

40

40

40

50

50

50

50

70

60

60

60

F

req

uen

cy [

%]

F

req

uen

cy [

%]

Plant height of winter rye

(from: Oetzel, 1977)

F

req

uen

cy [%

]

Fre

que

ncy

[%]

83,8

38.0

121,8

102,5 18,7

101,7 Plant height [cm]

71,1 96,5

95,7 106,8

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Schierholt, Antje, 2000: Hoher Ölsäuregehalt (C18:1) im Samenöl: genetische Charakterisierung von Mutan-ten im Winterraps (Brassica napus L.). Dissertaion, Universität Göttingen.

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Schierholt, Antje, 2000: Hoher Ölsäuregehalt (C18:1) im Samenöl: genetische Charakterisierung von Mutan-ten im Winterraps (Brassica napus L.). Dissertaion, Universität Göttingen.

ExampleF2- ½(B1+B2)= ¼(aa)12

i.e., 70.5 – ½ (65.1+72.7) = 1.6thus,

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COV 3.1.2 - 16 –

Genetic interpretation of generation means, 2-loci-model (F-metric), no linkage

b) P1 = AAbb, P2 = aaBB (Repulsion phase)

Gen.

c

a1

a2

d1

d2

(aa)12

(ad)12

(da)12

(dd)12

P1

1

1

-1

-1 P2 1 -1 1 -1 P 1 -1

F1

1

1

1

1 F2 1 ½ ½ ¼

F 1 0 0 0

B1

1 ½ -½ ½ ½ -¼ ¼ -¼ ¼

B2 1 -½ ½ ½ ½ -¼ -¼ ¼ ¼

B 1 ½ ½ -¼ ¼

Compare the cases „coupling“ and „repulsion“; the parameters a1, a2, (aa)12, (ad)12 and (da)12 enter with varyiing sign into the generation means. The sign depends on the type of association of the genes in the parents. This fact governs the above studied contrasts between mean. Heterosis: F1 – P = d1 + d2 + (aa)12 + (dd)12

F2

-

(F1 + P)/2

=

½ (aa)12 – ¼(dd)12

F2

-

B

=

¼(aa)12

(F∞-metric), no linkage

F∞

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0

1

2

3

4

5

6

7

8

9

10

Yie

ld p

erfo

rman

ce (

t/ha)

Ert

rags

leis

tung

(t/h

a)

0.000.250.500.751.00

Inbreeding coefficient, Inzuchtkoeffizient

Paren-tal mean; F∞

F1-hybridF

2- mean; B

C1-m

ean

F3-generation m

ean

Any difference of F∞

and the parental mean shows additiv-additiv-epistatic effects

Any deviation from this linearity is indicative for epistasis. The type(s) of epistasis depend(s) on the actual non-linearity.

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Statistical models The parameters of the statistical models characterize the average effects of genes and gene combinations in populations with given, in reality mostly unknown genotype frequencies like p(A) and q(a). In the following we will restrict ourselves to populations in the generalized HWE (this implies: N , random mating, no selection, no migration, no mutation).

Parameters of the statistical models are

- The population mean µ, - The additive or average effect of an allele, α, - The dominance effect (interaction of allelic

genes), δ, - The epistatic effects (interaction of non-allelic

genes), (αα) etc. Different from the parameters of the metric models, here the parameters are written in Greek letters - µ, α, δ, (αα), … - and the sums of effects of a given type are written in capital Latin letters (A, D, AA, ...).

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0

20

40

60

80

Gen

otyp

ic tr

ait v

alue

of o

ffspr

ing

fam

ilies

0 20 40 60 80 100

Genotypic trait value of parents

2=1.554=1.247

2=11.603=3.406

2=2.901=1.703 = 3.406/2

h²=0.50

h²=0.37

²G=11.603²A= 6.218A = 2.494²D= 5.385

100Gentoypic variance; 100 loci; a=d=0.5; p(A)=0.634Genotypic variance; 100 loci; a=0.5; d=0; p(A)=0.634

Random mating

„Value“ of AA = 1 „Value“ of aa = 0

WHY ?

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Population mean Definition: = E {Pij} = E {Gi} Assumption: Random mating, generalized HWE

a) 1 locus, 2 alleles:

Runner (i) Genotype Frequency (zi)

Gi

1

AA

p2

c + a 2 Aa 2pq c + d 3 aa q2 c – a

=3

i=1

ziGi = (p2 + 2pq + q2) c + (p2 – q2) a + 2pqd

=

c + (p - q)a + 2pqd

As clear from definition, homozygotes AND hetero-zygotes contribute to the population mean. b) n loci, 2 alleles each, no epistasis:

=

c +

n

1l

(pl – ql)al + 2

n

1l

plqldl

c = Mean of all 2n homozygotes

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COV 3.2 - 25 -

frequency (p) of the favourable allele when allowing for different degrees of dominance.

= c + (p - q)a + 2pqd

d = 0 intermediate gene effects

d = ½ a partial dominance

d = a23

overdominance

d = a complete dominance

p*=5/6

0 1

1 0

μ

c + a

c - a

c + a

c

0 1

c + a c + a

c - a

c

1 0

c

c + a

c - a

c - a

c + a c + a

c

Dependency of the population mean () on the

p

p

p

p

d ap*

2d

for d>a

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Analysis of genotypic value: AVERAGE EFFECT OF AN ALLELE refer to Falconer, D.S., 1981: I ntrodution to quantitave genetics. Longman.

Example: Corn (panmictic species), imagine an improvement of a population

via mass selection. Trait: Grain yield in saline soil Mean yield of genotypes with allelic equipment of locus A: A1A1 = 30 dt/ha A1A2 = 30 dt/ha A2A2 = 26 dt/ha Frequency of A1 = 10% Population mean: M = 0,12 x 30 + 2 x 0,1 x 0,9 x 30 + Frequency of A2 = 90% 0,92 x 26 = 26,76 dt/ha The favourable allele A1 is rather rare, and it is dominant. Mating of the rare gametes with the allele A1 to the gametes, as delivered by the population (imagine this as pollen pool), gives: A1 with A1 = 10%, A1 with A2 = 90% with a resulting mean of the offspring : 0,1 x 30 dt/ha + 0,9 x 30 dt/ha = 30 dt/ha. A2 with A1 = 10%, A2 with A2 = 90% With a resulting mean of the offspring : 0,1 x 30 dt/ha + 0,9 x 26 dt/ha = 26,4 dt/ha. Difference of these offsprings to population mean = average effect(1) The average effect of allele A1 ( 1) is:

Mean of the offspring - population mean = 1

30 dt/ha - 26,76 dt/ha = 3,24 dt/ha

The average effect of allele A2 ( 2) is:

26,4 dt/ha - 26,76 dt/ha = - 0,36 dt/ha Accordingly the breeding value A of the genotypes amounts to: A1A1 A1A2 A2A2

21 1 + 2 22 6,48 dt/ha 2,88 dt/ha - 0,72 dt/ha Average effects and breeding values depend on the frequency of the alleles and their effect. An individual has a high breeding value, if it is carrying many favourable and/or dominant alleles and if these alleles are rare in the population that has to be improved.

The variance of the breeding values of genotypes is called additive variance, σ²A .

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Analysis of genotypic value: AVERAGE EFFECT OF AN ALLELE refer to Falconer, D.S., 1981: I ntrodution to quantitave genetics. Longman.

Example: Corn (panmictic species), imagine an improvement of a population

via mass selection. Trait: Grain yield in saline soil Mean yield of genotypes with allelic equipment of locus A: A1A1 = 30 dt/ha A1A2 = 30 dt/ha A2A2 = 26 dt/ha Frequency of A1 = 10% Population mean: M = 0,12 x 30 + 2 x 0,1 x 0,9 x 30 + Frequency of A2 = 90% 0,92 x 26 = 26,76 dt/ha The favourable allele A1 is rather rare, and it is dominant. Mating of the rare gametes with the allele A1 to the gametes, as delivered by the population (imagine this as pollen pool), gives: A1 with A1 = 10%, A1 with A2 = 90% with a resulting mean of the offspring : 0,1 x 30 dt/ha + 0,9 x 30 dt/ha = 30 dt/ha. A2 with A1 = 10%, A2 with A2 = 90% With a resulting mean of the offspring : 0,1 x 30 dt/ha + 0,9 x 26 dt/ha = 26,4 dt/ha. Difference of these offsprings to population mean = average effect(1) The average effect of allele A1 ( 1) is:

Mean of the offspring - population mean = 1

30 dt/ha - 26,76 dt/ha = 3,24 dt/ha

The average effect of allele A2 ( 2) is:

26,4 dt/ha - 26,76 dt/ha = - 0,36 dt/ha Accordingly the breeding value A of the genotypes amounts to: A1A1 A1A2 A2A2

21 1 + 2 22 6,48 dt/ha 2,88 dt/ha - 0,72 dt/ha Average effects and breeding values depend on the frequency of the alleles and their effect. An individual has a high breeding value, if it is carrying many favourable and/or dominant alleles and if these alleles are rare in the population that has to be improved.

The variance of the breeding values of genotypes is called additive variance, σ²A .

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COV 3.2.2 - 26 -

3.2.2 Additive and dominance effects

Definitions: The additive effect (or average effect) of an allele Ai,

i, is defined as the average deviation in performance (plus or minus) of all those genotypes arising from mating Ai with the gametic array of the population. The deviation is measured from the population’s mean ():

1 = p · G(A1A1) + q · G(A1A2) -

= p(c + a) + q(c + d) - c + (p – q)a + 2pqd

= q a – (p – q)d

2 = p · G(A2A1) + q · G(A2A2) -

= p(c + d) + q(c – a) - c + (p – q)a + 2pqd

= -p a – (p-q)d

The dominance effect of the allelic gene combination

AiAj, ij, is defined as the deviation of the genotypic value of genotype AiAj from the sum of the population mean plus the two additive effects

i & j of the

corresponding two alleles Ai and Aj.

11 = G(A1A1) – { + 1 + 1}

= c + a – {c + (p – q)a + 2pqd + 2qa – (p-q)d}

= -2q2d

12 = G(A1A2) – { + 1 + 2}

= +2pqd

22 = G(A2A2) – { + 2 + 2}

= - 2p2d

α1- α 2= α =[a- (p-q)d]

α is sometimes called „average effect of a gene

substitution“

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COV 3.2.3 - 32 -

3.2.3 Epistasis effects

Here (statistical model), we consider only digenic epistasis – as we did in case of the metric model. Hence, the model for any genotypic value can be written in case of 2 loci (genotype AiAjBkBl):

In short:

Gijkl = Population mean

+ i + j + k + l Additive effects

+ ij + kl Dominance effects

Additive x additive–effects ++

()ik + ()il ()jk + ()jl

= interaction between 1 allele of each locus, respectively

Additive x dominance or dominance x additive–effects

++

()ikl + ()jkl ()ijk + ()ijl

= interaction between 1 gene of one locus and 2 genes at the other locus

Domin. x domin. –effects + ()ijkl = interaction between 2 genes

of each locus, respectively

G = μ + A + D + AA + AD + DD

39

40

COV 3.3 - 38 – 3.3 Summary The genotypic values of the individuals of a population can be described by

metric and statistical models.

Metric models characterize the gene effects, the type of action of genes, and hence are useful to analyse the trait differences between generation: especially if the generations trace back to homozygous lines and hence we know the allele frequencies. Important phenomena of gene action for application in breeding are:

heterosis and inbreding depression, deviation from the linearity between perfor-

mance and heterozygosity.

The parameters of the statistical models charac-terize the effects of genes and of combinations of genes on average, across all genotypes of a popu-lation, with unknown allele frequencies in the popu-lation. Based on statistical models, we can des-cribe and forecast (predict):

the suitability of a genotype for use in further breeding,

the effects of shifts in gene frequencies caused by selection.

41

COV 3.4 - 39 - 3.4 Exercises

(cf. FALCONER Exercise 7.1-7.8) 1 FOR a maize single hybrid cultivar and its parental homozygous inbred lines, the following yields were measured [t ha-1]:

What is the amount of heterosis of the hybrid ? Imagine we harvest the (open pollinated) F1-plants

and took the harvested seed to sow the resulting generation (= F2); neglect epistasis; what is the expected yield in F2?

Discuss the potential effects of epistasis on the performance of the performance of this F2.

2 ACHIEVE a thorough understanding of the following equation: it describes average genotypic value of an inbred population (F; 0 < F 1; epistasis is neglec-ted. Which factor(s) determine the difference in perfor-mance between a non inbred (F=0) and an inbred (F>0) population, i.e., O - F ?

F1: 6.80 P1: 1.65 P2: 2.35

Population mean under inbreeding, F; no epistasis:

F = c + (p² + Fpq)a + 2pq(1 – F)d + (q² + Fpq)(-a) F = c + (p – q)a + 2pq (1 – F)d

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COV 3.4 - 40 - 3.4 Exercises (ff) (cf. FALCONER Exercise 7.1-7.8)

3 IN THE following table you find data on yield and height for corn, for progenies with different degrees of inbreeding. Do graphically show the relationship between performance (y-axis) and inbreeding coefficient (x-axis). Choose differenct sympols for selfing and full-sib-mating like and . Do comment the graphic !

Inbreeding depression (yield, height) under selfing & fullsib mating in corn (Cornelius & Dudley 1974, Tab. 2)

Selfing Fullsib Mating Inbreeding Yield Height Yield Height

genera-tion

coeffi-cient F

(t/ha) (cm) F

(t/ha) (cm)

0 0,00 5,43 229 0,00 5,43 229

1 0,50 3,29 201 0,25 4,44 215

2 0,75 2,97 185 0,375 4,02 210

3 - - - 0,50 3,59 202

4 A GRASS breeder develops a synthetic cultivar from 5 inbres lines (= starting generation: Syn-0). The first Syn-generation (Syn-1) is created by diallel crossing. The seed of the (5·4)/2 = 10 crosses are mixed with equal proportions, and the second generation, Syn-2, is created by open pollination (random mating) of the crop stand comprising this equal-dose mixture. All further multiplications are as well by open pollination. Do make a scheme of the expected performance across the generations Syn-0 to Syn 4! Do neglect epistasis!

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COV 3.4 - 42 -

Solutions of the exercises to chapter 3

1 Heterosis = 6,80 – (1,65 + 2,35) / 2

= 4,80 t ha-1

F2 = F1 –½ heterosis = 6,80 – 2,40 = 4,40 t ha-1

Positive aa and/or dd effects would decrease the performance of the F2, negative effects would increase it.

2 Change of population mean due to inbreeding;

no epistasis:

O - F = c + (p - q)a + 2pqd - F

O - F = 2pqFd

We find a linear dependency of the population performance on its inbreeding coefficient F !

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2

3

4

5

6

Y

ield

(t

ha

-1)

1 8 0

1 9 0

2 0 0

2 1 0

2 2 0

2 3 0

2 4 0

Pla

nt

he

igh

t (c

m)

0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

0 .0 0 .2 0 .4 0 .6 0 .8 1 .0

In b re e d in g c o e ff ic ie n t (F )

In b re e d in g c o e ff ic ie n t F

3 Graphic Dependency of grain yield and plant height on the inbree-ding coefficient in corn (Cornelius & Dudley 1974) Basic population Selfing, Fullsib mating Comment: Nearly linear relation of performance & inbreeding coefficient F. In case of grain yield there is a hint on a deviation from linearity with the last inbreeding generation. This could be the effect of selection when choosing the parental individuals. In case of height, there is no such deviations. This trait most probably is not or nearly not affected by selection (natural or artificial), hence, a nearly-at- random-choice of parents in the experiment could be realized.

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t

0 1 2 3 4

Syn generation (t)

Expected performance (t) of a synthetic population in the first generations of multiplicaitons

4