genetics and genetic prediction in plant breeding genetics...genetics and genetic prediction in...
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Genetics and
Genetic Prediction
in Plant Breeding
Which traits will be most responsive to selection?
What stage will be best to select for specific characters?
What environments are most suited to the best response to selection?
What level of selection will provide the most practical response?.
There must be some phenotypic variation within the crop.
A breeder has to select only a proportion of the phenotypes.
A portion of that variation must be genetic in nature.
The ratio of total variation to genetic variation is called heritability.
Response to Selection = ih2
Carry out particular crosses so that the
resulting data can be partitioned into
genetic and environmental components.
Compare the degree of resemblance
between off-spring and their parents.
Measure the response to a given selection
operation (later in selection).
-1> h2 >+1
Genetic Variation
Total Variation
Genetic Variation
Additive Genetic
Variation (A)
Dominant Genetic
Variation (D)
h2b = Total Genetic Variance
Total Variance
h2b = f(A + D)
Total Variance
h2n = Additive Genetic Variance
Total Variance
h2n = f(A)
Total Variance
Broad-sense
heritability
Narrow-sense
heritability
V(F2) = [(xi-)2]/n
F2 = ¼ P1 + ½ F1 + ¼ P2
(F2) = m +½ [d]
= ¼(m+a)+½(m+d)+¼(m-a)]
V(F2) = [(xi-)2]/n
F2 = ¼ P1 + ½ F1 + ¼ P2
(F2) = m +½ [d]
V(F2) = f(x-)2
F2 = ¼ P1 + ½ F1 + ¼ P2
V(F2) = f(x-)2
P1 = ¼[(m+a)-(m+½d)]2
= ¼ [m+a-m-½d]2
= ¼ [a-½d]2
= ¼ [a2+¼d2-ad]
= ¼a2+1/16d2–¼ad
F2 = ¼ P1 + ½ F1 + ¼ P2
V(F2) = f(x-)2
P2 = ¼[(m-a)-(m+½d)]2
= ¼ [m-a-m-½d]2
= ¼ [-a-½d]2
= ¼ [a2+¼d2+ad]
= ¼a2+1/16d2+¼ad
F2 = ¼ P1 + ½ F1 + ¼ P2
V(F2) = f(x-)2
F1 = ½[(m+d)-(m+½d)]2
= ½[m+d-m-½d]2
= ½[½d]2
= ½[¼d2]
= 1/8d2
V(F2) = f(x-)2
V(F2)=¼a2+1/16d2–¼ad+1/8d2+¼a2+1/16d2+¼ad X X
V(F2) = ½a2 + ¼d2
V(F2) = ½A + ¼D
V(F2) = ½A + ¼D + E
h2b = Total Genetic Variance
Total Variance
h2b = ½ A + ¼ D
½ A + ¼ D + E
Estimate the total variation and
estimate the error variation to
estimate the broad-sense heritability
Genetic variation in straw length in an F2
population of oat was 125 cm2. The
environmental (error) variance was found to be
25 cm2. That is the broad-sense heritability?
h2b = Total Genetic Variance
Total Variance
h2b = 125
125+25
h2b = 0.833
A cross was made between two parents (P1 & P2) and
F1 seed produced. F1 plants are grown and selfed to
produce F2 seed. Both parents, and the F1 and F2
progeny are grown in a properly designed field trial
and yield recorded on individual plants.
The following data were obtained from the experiment.
What is the broad-sense heritability?
V(P1) = 35.5; V(P2) = 29.7
V(F1) = 34.5; V(F2) = 97.2
The following data were obtained from the experiment.
What is the broad-sense heritability.
V(P1) = 35.5; V(P2) = 29.7
V(F1) = 34.5; V(F2) = 97.2
h2b = ½ A + ¼ D
½ A + ¼ D + E
E = [35.5+29.7+34.5]/3 = 32.2
h2b = 97.2 – 32.2 = 0.669
92.2
h2b = Total Genetic Variance
Total Variance
h2b = ½ A + ¼ D
½A + ¼D + E
h2n = Additive Genetic Variance
Total Variance
h2n = ½A
Broad-sense
heritability
Narrow-sense
heritability ½A + ¼D + E
h2b = Total Genetic Variance
Total Variance
h2b = ½ A + ¼ D
½A + ¼D + E
h2n = Additive Genetic Variance
Total Variance
h2n = ½A
Broad-sense
heritability
Narrow-sense
heritability ½A + ¼D + E
P1 = m + [a]
P2 = m – [a]
F1 = m + [d]
F2 = m + ½ [d]
B1 = m + ½ [a] + ½ [d]
B2 = m – ½ [a] + ½ [d]
P1 = m + [a]
P2 = m – [a]
F1 = m + [d]
F2 = m + ½ [d]
B1 = m + ½ [a] + ½ [d]
B2 = m – ½ [a] + ½ [d]
P1 = m + [a]
P2 = m – [a]
F1 = m + [d]
F2 = m + ½ [d]
B1 = m + ½ [a] + ½ [d]
B2 = m – ½ [a] + ½ [d]
V(B1) = [(xi-)2]/n
B1 = ½ P1 + ½ F1
(B1) = m + ½ [a] +½ [d]
V(B1) = f(x-)2
P1 = ½ [(m+a)-(m+½a+½d)]2
= ½ [½a–½d]2
B1 = ½ P1 + ½ F1
(B1) = m + ½ [a] +½ [d]
F1 = ½ [(m+d)-(m+½a+½d)]2
= ½ [-½a+½d]2
V(B1) = ½[½a–½d]2 + ½ [-½a+½d]2
B1 = ½ P1 + ½ F1
(B1) = m + ½ [a] +½ [d]
= ½[¼a2+¼ d2–½ad]+½[¼a2+¼d2–½ad]
= ¼a2 + ¼d2 – ½ad
P2 = ½ [(m-a)-(m-½a+½d)]2
= ½ [-½a–½d]2
B2 = ½ P2 + ½ F1
(B2) = m - ½ [a] +½ [d]
F1 = ½ [(m+d)-(m-½a+½d)]2
= ½ [½a+½d]2
V(B2) = ½[-½a–½d]2 + ½ [½a+½d]2
B2 = ½ P2 + ½ F1
(B2) = m - ½ [a] +½ [d]
= ½[¼a2+¼ d2+½ad]+½[¼a2+¼d2+½ad]
= ¼a2 + ¼d2 + ½ad
V(B1) = ¼ A + ¼ D – ½ [AD] + E
V(B2) = ¼ A + ¼ D + ½ [AD] + E
V(B1) + V(B2) = ½ A + ½ D + 2E
V(F2) = ½ A + ¼ D + E
D = 4[V(B1) + V(B2) – V(F2) – E]
A = 2[V(F2) – ¼D – E]
V(F1) = 21 g2; V(F2) = 65 g2;
V(B1) = 42 g2; V(B2) = 49 g2
D = 4[V(B1)+V(B2)-V(F2)-E]
= 4[42 + 49 + - 65 – 21]
= 20 g2
E = V(F1) = 21 g2
A = 2[V(F2) – ¼ D – E
= 2[65 – ( ¼ x 20) – 21]
= 78 g2
V(F2) = 65 g2; E = 21 g2;
D = 20 g2; A = 78 g2
h2n = ½A
½A + ¼D + E
h2n = 0.5 x 78
0.5x78 + 0.25x20 + 21
h2n = 0.60
Question.
A crossing design involving two homozygous pea cultivars is carried
out and both parents are grown in a properly designed field
experiment with the F2, B1 and B2 families. Given the following
standard deviations for both parents (P1 and P2), the F2, and both
backcross progeny (B1 and B2), determine the broad-sense
heritability and narrow-sense heritability for seed size in dry pea
Family Standard Deviation
P1 3.521
P2 3.317
F2 6.008
B1 5.450
B2 5.157
Answer
Family Standard Deviation
P1 3.521
P2 3.317
F2 6.008
B1 5.450
B2 5.157
VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6
Answer
h2b = Genetic variance
Total variance
h2b = 36.1 – 11.7
36.1
E = [VP1+VP2]/2 = 11.7
h2b = 0.67
VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6
Answer
D = 4[V(B1)+V(B2)-V(F2)-E]
4[29.7+26.6-36.1-11.7] = 8.5
A = 2[V(F2)-¼D-E] = 2[36.1-2.1-11.7] = 22.3
h2n = ½A/V(F2) = 11.15/36.1 = 0.31
VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6
E = [VP1+VP2]/2 = 11.7
Heritability
Parent v Offspring
Regression
19th Century - Charles Darwin
Francis Galton: In the “law of universal regression” “ each peculiarity in a man is shared by his kinsman, but on average in a less degree”
Karl Peterson & Andrew Lee (statisticians) survey 1000 fathers and sons height
Using this data set Galton, Peterson and Lee formulated regression analyses
0
20
40
60
80
100
120
140
0 20 40 60 80 100 120
Hei
gh
t of
son
Height of father
Y = bo + b1x bo
b1
b1 = [SP(x,y)/SS(x)]
Y = bo + b1x
SP(x,y) = (xi-x)(yi-y)
SP(x,y) = (xy) - [(x) (y)]/n
SS(x) = (xi-x)2
SS(x) = (x2) - [(x)]2/n
Y = bo + b1x
bo = mean(y) - b1 x mean(x)
se(b1) = {SS(y) – [b x SP(x,y)]}
(n-2) x SS(x)
b = [SP(x,y)/SS(x)]
F2 > F3
b = Covariance between F2 and F3
Variance of F2
b = [SP(x,y)/SS(x)]
F2 > F3
¼ P2 ½ F1 ¼ P1
¼ P2 1/16 P2 2/16 B2 1/16 F1
½ F1 2/16 B2 4/16 F2 2/16 B1
¼ P1 1/16 F1 2/16 B1 1/16 P1
1/16 P1; 1/16 P2; 2/16 F1;
4/16 F2; 4/16 B1; 4/16 B2
(xi-x)(yi-y)
x = m+½d; y = m+½d
P1 family types = 1/16 a2 + 1/64 d2 – 1/16 ad P2 family types = 1/16 a2 + 1/64 d2 + 1/16 ad
F1 family types = -1/32 d2
F2 family types = 0 B1 family types = 1/16 a2 B1 family types = 1/16 a2
¼A
b = Covariance between F2 and F3
Variance of F2
b = ¼ A
Variance of F2
b = ¼ A
½ A + ¼ D + E
h2n = ½ A
½ A + ¼ D + E
h2n = 2 x b
Regression of one parent
onto off-spring
h2n = 2 x b
Regression of two parents
onto off-spring
h2n = b
Male Parent Female
Parent
Average
(P1+P2)/2 Off-spring
950 1160 1055 1060
1040 990 1015 1040
1090 960 1025 1080
1120 1140 1130 1100
1050 1140 1090 1130
1070 1040 1050 1070
1340 1140 1240 1240
1150 980 1065 1020
1170 1280 1225 1200
1030 1130 1080 1080
Male Female Average
SP(x,y) 474 427 450
SS(x) 1002 912 554
SS(y) 437.6 438 438
Covariance 52.667 47.422 50.044
Variance 111.333 101.378 61.567
Male Female Average
b 0.473 0.468 0.813
se(b) 0.1632 0.1805 0.1269
Male h2n = 2 x b = 0.946 + 0.1632
Female h2n = 2 x b = 0.936 + 0.1805
Average h2n = b = 0.813 + 0.1269
Correlation
and
Heritability
History
Francis Galton
(1888)
Recorded height of adult males
and length of forearm
He said “The two measurements were co-related”
from where correlated is derived
r = SP(x1,x2)/[SS(x1).SS(x2)]
SP(x1,x2) = (x1ix2i)-[(x1i) (x2i)]/n
SS(x1) = (x1i2)-[(x1i)]
2/n
SS(x2) = (x2i2)-[(x2i)]
2/n
SP(x1,x2) = 39; SS(x1) = 74; SS(x1) = 66
r = 39/[74 x 66] = 0.553
Neither x1 nor x2 are dependant or independent
Correlation coefficient (r) is a pure number without units.
r values range in value from -1 to +1.
If r value is +’ive, r2 = h2
F2
(x1)
71 68 66 67 70 71 70 73 72 65 66
F3
(x2)
69 64 65 63 65 62 65 64 66 59 62
r = SP(x1,x2)/[SS(x1).SS(x2)]
SS(x1,x2) = 39; SS(x1) = 74; SS(x1) = 66
r = 39/[74 x 66] = 0.553 with n-2 df
h2 = r2 = 0.306
Diallels