genetics and genetic prediction in plant breeding...in plant breeding qualitative genetics...
TRANSCRIPT
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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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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500
1000
1500
2000
2500
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
0
100
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300
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
0
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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
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500 510 520 530 540 550 560 570 580 590 600 610 620
Histogram
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500 510 520 530 540 550 560 570 580 590 600 610 620
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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Yield
0
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Yield
0
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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
200
400
600
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1000
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
200
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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
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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