Genetics and

Genetic Prediction

in Plant Breeding

Qualitative Genetics

Segregation, linkage, epistasis,

tetraploid, 2

Quantitative Genetics

Continuous variation, genetic

models, model testing

QTL’s

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500

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3500

25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Fre

qu

ency

Plant height (cm)

Height of 4,000 F2 wheat plants

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25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105

Fre

qu

ency

Plant height (cm)

Height of 4,000 F2 wheat plants

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400

500

600

0.4 0.7 1 1.3 1.6 1.9 2.2 2.7 3 3.3 3.6 3.9 4.2 4.5 4.8 5.1 5.4

Fre

qu

ency

Plant Yield (kg)

Tuber yield of 4,000 genotypes

P = G + E + GE + 2e

Phenotype

Genotype

Environment

Geno x Environ

Error

Additive v other

Quantitative Genetics

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100

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500

600

0.4 0.7 1 1.3 1.6 1.9 2.2 2.7 3 3.3 3.6 3.9 4.2 4.5 4.8 5.1 5.4

Fre

qu

ency

Plant Yield (kg)

Tuber yield of 4,000 genotypes

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Histogram

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Normal Distribution

Normal Distribution Function

Relationship to plant genetics

Consider two homozygous parents of

canola (Brassica napus).

Parent 1 has yield of 620 lbs/acre.

Parent 2 has yield of 500 lb/acre.

The F1 has 560 lb/acre yield.

Parent 2 x Parent 1

aa AA

500+0+0=500 500+60+60=620

Parent 2 x Parent 1

aa AA

Parent 2 x Parent 1

aa AA

aA

Aa aa AA

500 560 620

Parent 2 x Parent 1

aabb AABB

500+0+0+0+0=500 500+30+30+30+30=620

Parent 2 x Parent 1

aabb AABB

Parent 2 x Parent 1

aabb AABB

Aabb

AaBb

aABb

AabB

aAbB

aaBB

aabB

aaBb

aAbb

Aabb

AABb

AAbB

aABB

AaBB aabb AABB

500 530 560 590 620

Parent 2 x Parent 1

aabbcc AABBCC

500+0+0+0+0+0+0=500 500+20+20+20+20+20+20=620

aaBbCC

aAbBcC

aAbBCc

aAbbCC

aAbbCC

AabBcC

AabBcC

AaBbcC

AaBbCc

AaBBcc

aabBCC

aaBBcC

aaBBCc

aABbcC

aABbCc

aABBcc

AAbbCc

AAbbcC

AAbBcc

AABbcc

aabbCC

aabBcC

aabBCc

aaBbcC

aaBbCc

aaBBcc

AabbcC

aAbbCc

aAbBcc

aABbcc

AabbcC

aAbbCc

AabBcc

AaBbcc

AAbbcc

AABBcc

AABbCc

AABbcC

AAbBCc

AAbBcC

AAbbCC

AaBBCc

AaBBcC

AaBbCC

AabBCC

aABBCc

aABBcC

aABbCC

aAbBCC

aaBBCC

aabbcC

aabbCc

aabBcc

aaBbcc

aAbbcc

Aabbcc

AABBcC

AABBCc

AABbCC

AAbBCC

aABBCC

AaBBCC aabbcc AABBCC

500 520 540 560 580 600 620

Relating Genetics to Normal Distribution

Parent 2 x Parent 1

aa AA

aA

Aa aa AA

500 560 620

Parent 2 x Parent 1

aabb AABB

Aabb

AaBb

aABb

AabB

aAbB

aaBB

aabB

aaBb

aAbb

Aabb

AABb

AAbB

aABB

AaBB aabb AABB

500 530 560 590 620

aaBbCC

aAbBcC

aAbBCc

aAbbCC

aAbbCC

AabBcC

AabBcC

AaBbcC

AaBbCc

AaBBcc

aabBCC

aaBBcC

aaBBCc

aABbcC

aABbCc

aABBcc

AAbbCc

AAbbcC

AAbBcc

AABbcc

aabbCC

aabBcC

aabBCc

aaBbcC

aaBbCc

aaBBcc

AabbcC

aAbbCc

aAbBcc

aABbcc

AabbcC

aAbbCc

AabBcc

AaBbcc

AAbbcc

AABBcc

AABbCc

AABbcC

AAbBCc

AAbBcC

AAbbCC

AaBBCc

AaBBcC

AaBbCC

AabBCC

aABBCc

aABBcC

aABbCC

aAbBCC

aaBBCC

aabbcC

aabbCc

aabBcc

aaBbcc

aAbbcc

Aabbcc

AABBcC

AABBCc

AABbCC

AAbBCC

aABBCC

AaBBCC aabbcc AABBCC

500 520 540 560 580 600 620

Relating Genetics to Normal Distribution

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1200

Yield

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1000

Yield

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Yield

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Yield

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500 510 520 530 540 550 560 570 580 590 600 610 620

Normal Distribution

= x = (x1 + x2 + x3 + …. + xn)

n

= x = in (xi)/n

0

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500 510 520 530 540 550 560 570 580 590 600 610 620

Normal Distribution

= [in (xi - )2/n]

^ S = [i

n (xi - x)2/(n-1)]

= {[in xi

2 – [(in xi)

2/n]}/n

^ S = {[i

n xi2 – [(i

n xi)2/n-1]}/(n-1)

P = G + E + GE + 2e

Quantitative Genetics

P = G + E + GE + e

2P =

2G +

2E +

2GE +

2e

2 = in (xi - )2/n

^ S2 = i

n (xi - x)2/(n-1)

2 = {in xi

2 – (in xi)

2/n}/n

^ S2 = {i

n xi2 – (i

n xi)2/n}/(n-1)

0

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500 510 520 530 540 550 560 570 580 590 600 610 620

Variance Partition

2P

2G

2GE

2e

.

.

.

.

Quantitative Genetics

Models

Consider two homozygous parents of

canola (Brassica napus).

Parent 1 has yield of 620 lbs/acre.

Parent 2 has yield of 500 lb/acre.

The F1 has 560 lb/acre yield.

P2 F1 P1

500 560 620 m

[a] [a]

[a] = (P1 - P2)/2

m = P1 – [a] or P2 + [a]

P1 = m + [a] : P2 = m – [a]

P2 F1 P1

500 560 620 m

[a] [a]

[a] = (620-500)/2 = 60

m = 620–60 or 500+60 = 560

P1 = 560 + 60 : P2 = 560 – 60

Parent 2 x Parent 1

aa AA

500+0=500 500+120=620

F1

Aa

500+120 = 620

Parent 2 x Parent 1

aa AA

500 620

aA

Aa

AA aa

F2 = 590

Parent 2 x Parent 1

aabb AABB

500+0+0+0=500 500+60+30+30=620

F1

AaBb

500+60+30+0 = 590

AABb

AAbB

aAbB

AabB

aABb

AaBb

AAbb

Aabb

aAbb

aaBB

AABB

aABB

AaBB

aabB

aaBb aabb 500 530 560 590 620

F2 = 575

Parent 2 x Parent 1

aabbcc AABBCC

500+0+0+0+0+0=500 500+40+20+20+20+20=620

F1

AaBbCc

500+40+20+20 = 580

500 520 540 560 580 600 620

F2 = 570

Quantitative Dominance AABBcc

AABbcC

AABbCc

AAbBCc

AAbBcC

AAbbCC

aAbBcC

aAbBCc

aAbbCC

AabbCC

AabBcC

AabBCc

AaBbcC

AaBbCc

AaBBcc

aABbcC

aABbCc

aABBcc

aaBBCC

aaBbCC

aabBCC

aaBBcC

aaBBCc

AaBbCc

AAbbcC

AAbbCc

AAbBcc

AABbcc

aAbbcC

aAbbCc

aAbBcc

aABbcc

AabbcC

AabbCc

AabBcc

AaBbcc

AABBcC

AABBCc

AAbBCC

AABbCC

AaBbCc

AaBbcC

AaBbCC

AabBCC

aABBCc

aABBcC

aABbCC

AabBCC

aabbCC

aaBbcC

aaBbcC

aabBCc

aaBbCc

aaBBcc

Aabbcc

aAbbcc

Aabbcc

aabbcC

aabbCc

aabBcc

aaBbcc

AABBCC

aABBCC

AaBBCC aabbcc

0

200

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500 510 520 530 540 550 560 570 580 590 600 610 620

Fre

quen

cy o

f G

enoty

pes

Yield

All additive

Quantitative Dominance

0

200

400

600

800

1000

500 510 520 530 540 550 560 570 580 590 600 610 620

Fre

quen

cy o

f G

enoty

pes

Yield

All

additive

1 Dominant,

5 additive

3 Dominant,

3 additive

5 Dominant,

1 additive

Quantitative Dominance

The performance of the F1 compared to

the mid-parent (m) will be dependant on

the proportion of dominant loci.

There is a relationship between m, the F1

performance, the F2 performance and

dominance (d).

P2 m F1 P1

500 560 580 620

[d]

m

[a] [a]

P1 = m + [a]

P2 = m – [a]

F1 = m + [d]

F1

B2 F2 B1

x P2 x self x P1

F2 = ¼ P1 + ½ F1 + ¼ P2

= ¼ (m+[a]) + ½ (m+[d]) + ¼ (m-[a])

= ¼m + ¼ [a] + ½m + ½[d] + ¼m – ¼ [a]

= m + ½ [d]

B1 = ½ P1 + ½ F1

= ½ (m+[a]) + ½ (m+[d])

= m + ½ [a] + ½ [d]

B2 = ½ P2 + ½ F1

= ½ (m-[a]) + ½ (m+[d])

= m - ½ [a] + ½ [d]

P1 = m + [a]

P2 = m – [a]

F1 = m + [d]

B1 = m + ½ [a] + ½ [d]

B2 = m – ½ [a] + ½ [d]

P1 = m + [a]

P2 = m – [a]

F1 = m + [d]

B1 = m + ½ [a] + ½ [d]

B2 = m – ½ [a] + ½ [d]

P2 m F1 P1

B2 B1

P2 = 110; P1 = 170; F1 = 160

[a] = (P1+P2)/n = (110+170)/2 = 30

m = P1 – [a] = 170 – 30 = 140

[d] = F1 – m = 160 – 140 = +20

P2 = 110; P1 = 170; F1 = 160

[a] = 30; m = 140; [d] = +20

F2 = m + ½ [d] = 140 + 10 = 150

B1 = m + ½[a] + ½[d] = 140+15+10 = 165

B1 = m - ½[a] + ½[d] = 140-15+10 = 135

P1 = 110; P2 = 170

m = 140; F1 = 160

B1 = 165; B2 = 135

P1 = 110; P2 = 170

m = 140; F1 = 160

B1 = 165; B2 = 135

P2 m F1 P1

135 165

110 140 160 170

B2 B1

Testing the

additive/dominance

inheritance model