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Quantitative Genetics Plant Breeding Populations

are created by breeders to serve as sources of

new cultivars. These populations are describable

by their genotypic and allelic frequencies (note

that as plant breeders we describe by their

phenotypic frequencies many times but need to

understand the underlying genetics) Recommend

Breeding for Quantitative Traits in Plants

Rex Bernardo (Dep. Agronomy Plt. Genetics,

Univ.of Minn.) published by Stemma

Press, Woodbury , MN

Quantitative Genetics Plant Breeding Consider a

diploid species with 2 alleles A1 and

A2 Genotype n Geno. freq. (f)

A1A1 240 P110.4 A1A2 240 P120.4

A2A2 120 P220.2 600

Consider A1 as the allele and A2 as the

allele Gene Action

Quantitative Genetics Plant Breeding Consider a

diploid species with 2 alleles A1 and

A2 Genotype n Geno. freq. (f)

A1A1 240 P110.4 A1A2 240 P120.4

A2A2 120 P220.2 600

Consider A1 as the allele and A2 as the

allele Gene Action

Dominance Genotype Additive Partial Complete Ov

erdom. A1A1 5 5 5

5 A1A2 3 4 5

6 A2A2 1 1 1 1

Quantitative Genetics Plant Breeding Consider a

diploid species with 2 alleles A1 and

A2 Genotype n Geno. freq. (f)

A1A1 240 P110.4 A1A2 240 P120.4

A2A2 120 P220.2 600 Genotypic freq

Allelic frequencies

Consider A1 as the allele and A2 as the

allele Gene Action

P11 240 / 600 .4 P12 240 / 600 .4 P22

120 / 600 .2 Sums to 1.0

p A1 P11 (.5)(P12) / 1200 .6 (240

240 240) / 1200 .6 q A2 P22 (.5)(P12)

/ 1200 .4 (120 120 240) / 1200 .4

Quantitative Genetics Plant Breeding Hardy-Weinb

erg Equilibrium (HWE) states that a populations

of a diploid species, considering 2 alleles or 1

allele versus all other alleles at that locus,

can be described by the expansion of (a b)2 or

in our terms (p q)2 p2 2pq q2, thus we

can add Consider a diploid species with 2

alleles A1 and A2 Genotype n Geno.freq.

(f) f after random mating A1A1 240 P110.4

A1A2 240 P120.4 A2A2 120 P220.2 Assumption

s and empirical proof of HWE (nexts)

p2P11 (.5)(P12)/1200 .62 .36 2pq 2.6.4

0.48 q2 P22 (.5)(P12)/1200 .42 .16

- Quantitative Genetics Plant Breeding
- HWE states that genotypic and allelic freq. will

not change in a cross pollinating species under

the following assumptions - random mating
- no selection is practiced (naturally or

otherwise) - no differential migration
- mutation rates are equal, i.e. A?a a?A
- species in question is diploid
- usually considering 1 gene and 2 alleles or 1

allele of an allelic series against all other

alleles in the series

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by crossing AA with aa AA x aa F1 Aa

(self) F2 .25 AA .5 Aa .25 aa

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by crossing AA with aa AA x aa F1 Aa

(self) F2 .25 AA .5 Aa .25 aa

But what if the species is X-pollinated? Then we

put AA and aa in a crossing block and allow

pollination to occur and HWE states that in the

next generation, analogous to the F2 ( skips the

F1) in the selfed scenario Freq. of A p .5

a 1-p .5 p2 2p(1-p) (1-p)2 .52

2(.5)(.5) .52 .25 AA .5 Aa .25 aa

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by crossing AA with aa

But what if the species is X-pollinated? Then we

put AA and aa in a crossing block and allow

pollination to occur and HWE states that in the

next generation, analogous to the F2 in the

selfed scenario Freq. of A p .5 a 1-p

.5 p2 2p(1-p) (1-p)2 .52 2(.5)(.5) .52

.25 AA .5 Aa .25 aa

Why? AA Aa aa AA x AA

4 AA x aa 4 aa x AA

4 aa x aa

4 4 8 4

.25 .5 .25

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by intercrossing P1 BB P2 BB P3

Bb P4 Bb

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by intercrossing P1 BB P2 BB P3

Bb P4 Bb

What is the freq. of p B and (1-p) b? 6/8

.75 B p 2/8 .25 b (1-p) HWE .752

2(.75)(.25) .252 .5675 BB .375 Bb .0625 bb

Quantitative Genetics Plant Breeding Empirical

proof of concept of HWE Consider a population

derived by intercrossing P1 BB P2 BB P3

Bb P4 Bb

What is the freq. of p B and (1-p) b? 6/8

.75 B p 2/8 .25 b (1-p) HWE .752

2(.75)(.25) .252 .5675 BB .375 Bb .0625 bb

Proof were in HWE p B .5675 .5 (.375) /

1 .75

Empirical proof of HWE for this population of 4

parents BB Bb bb BB1 x BB1 4 BB1 x BB2 4 BB1

x Bb3 2 2 BB1 x Bb4 2 2 BB2 x BB1 4 BB2 x

BB2 4 BB2 x Bb3 2 2 BB2 x Bb4 2 2 Bb3 x BB1

2 2 Bb3 x BB2 2 2 Bb3 x Bb3 1 2 1 Bb3 x

Bb4 1 2 1 Bb4 x BB1 2 2 Bb4 x BB2 2

2 Bb4 x Bb3 1 2 1 Bb4 x Bb4 1 2 1

Totals 36 24 4 64 .5625 .375 .0625

Quantitative Genetics Plant Breeding Lets go

back to a previouswhere we had a

population involving a diploid species with 2

alleles, A1 and A2, and plants in the following

freq. Genotype n freq. (f)

A1A1 240 P110.4 A1A2 240 P120.4

A2A2 120 P220.2

HWE p2 2p(1-p) (1-p)2 .36 A1A1 .48 A1A2

.16 A2A2

Can we devise empirical proof for this population?

Set up the table as below each individual has an

equal opportunity to be fertilized by any other

individual with each cross producing 4 possible

genotypes, i.e. a perfect population for each

plant to plant cross as shown earlier

p22p(1-p)(1-p)2 .62 (2)(.6).4) .42

.36 A1A1.48A1A2.16A2A2

Quantitative Genetics and Plant

Breeding Extension of HWE Recall the last

assumption as I listed the assumptions for HW to

be in effect was considering 1 gene and 2

alleles or 1 allele of an allelic series against

all other alleles in the series. However, HW can

be extended to describe the allelic and genotypic

frequencies encountered in a population with an

allelic series at a particular locus. What is an

allelic series? Consider a locus with 3 alleles

A1, A2, and A3

Quantitative Genetics and Plant Breeding HWE with

an allelic series Consider a locus with 3

alleles A1, A2, and A3, such that A1A1 A2A2

A3A3 A1A2 A1A3 A2A3 Obs. freq. 0.15 0.00 0.20 0

.25 0.35 0.05 If Pij freq. of the AiAj genotype,

then the freq. of the ith allele pi (i.e.

the freq. of Ai) or, pi freq. of AiAi

(1/2) (freq. of all heteroz. with Ai)

or, pi Pii ½ ?iltj Pij

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, if pi

Pii ½ ?iltj Pij then for this population,

allelic freq. are A1 p1 P11 ½ (P12

P13) .15 ½ (.25 .35) .45 A2 p2 P22

½ (P12 P23) 0 ½ (.25 .05) .15 A3

p3 P33 ½ (P13 P23) .2 ½ (.35 .05)

.40 Note sum to 1.0

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes are .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes is .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1 (2)(.45)(.15)

A1A2

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes is .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1 (2)(.45)(.15)

A1A2 (2)(.45)(.4) A1A3

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes is .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1 (2)(.45)(.15)

A1A2 (2)(.45)(.4) A1A3 .152 A2A2

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes is .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1 (2)(.45)(.15)

A1A2 (2)(.45)(.4) A1A3 .152 A2A2

(2)(.15)(.4) A2A3

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic

freq. across these genotypes is .45 A1 .15 A2

.40 A3 and at HWE, AFTER ONE GENERATION OF

RANDOM MATING, genotypic freq. will be described

by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22

2 p2p3 p32 .452 A1A1 (2)(.45)(.15)

A1A2 (2)(.45)(.4) A1A3 .152 A2A2

(2)(.15)(.4) A2A3 .42 A3A3

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 .45 A1 .15

A2 .40 A3 and .452 A1A1

(2)(.45)(.15) A1A2 (2)(.45)(.4) A1A3 .152

A2A2 (2)(.15)(.4) A2A3 .42 A3A3 .2025A1A1

.135A1A2 .36A1A3 .0225A2A2 .12A2A3

.16A3A3 Is this population in HWE? Recalculate

allelic freq. A1 p1 P11 ½ (P12 P13)

.2025 ½ (.135 .36) .45 A2 p2 P22 ½

(P12 P23) .0225 ½ (.135 .12) .15 A3

p3 P33 ½ (P13 P23) .16 ½ (.36 .12)

.40

YES

Quantitative Genetics and Plant Breeding Consider

a locus with 3 alleles A1, A2, and A3, such

that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.

freq. 0.15 0.00 0.20 0.25 0.35 0.05 And we could

do this with any number of alleles in a series at

a single locus, e.g. 5 alleles (p1 p2 p3

p4 p5)2 Note that we are determining HW for a

given locus or gene and not for every possible

gene or trait. Demonstrates however what occurs

in an idealized population for every gene

Quantitative Genetics and Plant Breeding In this

idealized HW population, we can describe the

genotypic mean for a given gene Consider a

single locus, A, with alleles A1 and

A2 Genotypic mean can be expressed as MMP

a(p-q) 2dpq where Mpopulation

mean MPmidparent value or mean of

homozygotes avalue of the homozygote

A1A1 - A2A2 p freq. of A1 q

freq. of A2 d value of the heterozygote

Quantitative Genetics and Plant Breeding In this

idealized HW population, we can describe the

genotypic mean for a given gene Consider a

single locus, A, with alleles A1 and

A2 Genotypic mean can be expressed as MMP

a(p-q) 2dpq or graphically A1A1 A1A2 A2A2

a d -a also described

as MPa MPd MP-a

Quantitative Genetics and Plant Breeding IF

aperformance of either homozygous genotype (

for A1A1 and for A2A2) minus the MP LET the

performance of A1A1 20 and A2A2 14 then a

20 ((2014)/2) 20 17 3 NOW d value of

the heterozygote MP d and if we rearrange to

MP d then d degree of dominance SUCH THAT

d A1A2 17 3 THEREFORE in this example A1

is completely dominant to A2 SO IF d a

then gene action is dominant d lt a then gene

action is partial dominance (incldg additive)

d gt a then gene action is overdominant

Quantitative Genetics and Plant Breeding Simple

example to demonstrate MMP a(p-q) 2dpq LET p

q 0.5 MMP a(p-q) 2dpq 17 3 (.5 - .5)

(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or

M for contribution of the gene only 18.5 MP

1.5

Quantitative Genetics and Plant Breeding Simple

example to demonstrate MMP a(p-q) 2dpq LET p

q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)

(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or

M for contribution of the gene only 18.5 MP

1.5 HW proof (recall HWp2 2pq q2) .52

(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2

.25 A2A2 AND SINCE A1A1 3 .25 (3) A1A1 .5

(3) A1A2 .25 (-3) A2A2 .75 A1A1 1.5 A1A2

(-.75) A2A2 ? 1.5

Quantitative Genetics and Plant Breeding Simple

example to demonstrate MMP a(p-q) 2dpq LET p

q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)

(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or

M for contribution of the gene only 18.5 MP

1.5 HW proof (recall HWp2 2pq q2) .52

(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2

.25 A2A2 AND SINCE A1A1 3 .25 (3) A1A1 .5

(3) A1A2 .25 (-3) A2A2 .75 A1A1 1.5 A1A2

(-.75) A2A2 ? 1.5

Quantitative Genetics and Plant Breeding Simple

example to demonstrate MMP a(p-q) 2dpq LET p

q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)

(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or

M for contribution of the gene only 18.5 MP

1.5 HW proof (recall HWp2 2pq q2) .52

(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2

.25 A2A2 AND SINCE M for A1A1 20 .25 (20)

A1A1 .5 (20) A1A2 .25 (14) A2A2 5 A1A1 10

A1A2 3.5 A2A2 ? 18.5

Quantitative Genetics and Plant Breeding Simple

example to demonstrate MMP a(p-q) 2dpq LET p

q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)

(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or

M for contribution of the gene only 18.5 MP

1.5 HW proof (recall HWp2 2pq q2) .52

(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2

.25 A2A2 AND SINCE M for A1A1 20 .25 (20)

A1A1 .5 (20) A1A2 .25 (14) A2A2 .75 A1A1

1.5 A1A2 (-.75) A2A2 E 18.5

Quantitative Genetics and Plant Breeding For a

polygenic trait without epistasis M ? MP

a(p-q) 2dpq or ? Mpi ? ai(pi qi) 2

??dipiqi AND there is no way to measure a, d,

p,or q for individual loci of a polygenic trait.

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