Genetics and Genetic Prediction in Plant Breeding genetics...Genetics and Genetic Prediction in Plant Breeding ... Measure the response to a given selection operation (later in selection).

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