Genetics and Genetic Prediction in Plant Breeding

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?.

Response to Selection = i h 2 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.

-1> h 2 >+1 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).

Genetic Variation Total Variation Genetic Variation Additive Genetic Variation (A) Dominant Genetic Variation (D)

h 2 b = Total Genetic Variance Total Variance h 2 b = f(a + D) Total Variance h 2 n = Additive Genetic Variance Total Variance h 2 n = f(a) Total Variance Broad-sense heritability Narrow-sense heritability

V(F 2 ) = [ (x i - ) 2 ]/n F 2 = ¼ P 1 + ½ F 1 + ¼ P 2 = ¼(m+a)+½(m+d)+¼(m-a)] (F 2 ) = m +½ [d]

V(F 2 ) = [ (x i - ) 2 ]/n (F 2 ) = m +½ [d] F 2 = ¼ P 1 + ½ F 1 + ¼ P 2 V(F 2 ) = f(x- ) 2

F 2 = ¼ P 1 + ½ F 1 + ¼ P 2 V(F 2 ) = f(x- ) 2 P 1 = ¼[(m+a)-(m+½d)] 2 = ¼ [m+a-m-½d] 2 = ¼ [a-½d] 2 = ¼ [a 2 +¼d 2 -ad] = ¼a 2 +1/16d 2 ¼ad

F 2 = ¼ P 1 + ½ F 1 + ¼ P 2 V(F 2 ) = f(x- ) 2 P 2 = ¼[(m-a)-(m+½d)] 2 = ¼ [m-a-m-½d] 2 = ¼ [-a-½d] 2 = ¼ [a 2 +¼d 2 +ad] = ¼a 2 +1/16d 2 +¼ad

F 2 = ¼ P 1 + ½ F 1 + ¼ P 2 V(F 2 ) = f(x- ) 2 F 1 = ½[(m+d)-(m+½d)] 2 = ½[m+d-m-½d] 2 = ½[½d] 2 = ½[¼d 2 ] = 1/8d 2

V(F 2 ) = f(x- ) 2 V(F 2 )=¼a 2 +1/16d 2 ¼ad+1/8d X 2 +¼a 2 +1/16d 2 +¼ad X V(F 2 ) = ½a 2 + ¼d 2 V(F 2 ) = ½A + ¼D V(F 2 ) = ½A + ¼D + E

h 2 b = Total Genetic Variance Total Variance h 2 b = 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 F 2 population of oat was 125 cm 2. The environmental (error) variance was found to be 25 cm 2. That is the broad-sense heritability? h 2 b = Total Genetic Variance Total Variance h 2 b = 125 125+25 h 2 b = 0.833

A cross was made between two parents (P 1 & P 2 ) and F 1 seed produced. F 1 plants are grown and selfed to produce F 2 seed. Both parents, and the F 1 and F 2 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(P 1 ) = 35.5; V(P 2 ) = 29.7 V(F 1 ) = 34.5; V(F 2 ) = 97.2

The following data were obtained from the experiment. What is the broad-sense heritability. V(P 1 ) = 35.5; V(P 2 ) = 29.7 V(F 1 ) = 34.5; V(F 2 ) = 97.2 h 2 b = ½ A + ¼ D ½ A + ¼ D + E E = [35.5+29.7+34.5]/3 = 32.2 h 2 b = 97.2 32.2 = 0.669 92.2

h 2 b = Total Genetic Variance Total Variance h 2 b = ½ A + ¼ D ½A + ¼D + E h 2 n = Additive Genetic Variance Total Variance h 2 n = ½A ½A + ¼D + E Broad-sense heritability Narrow-sense heritability

h 2 b = Total Genetic Variance Total Variance h 2 b = ½ A + ¼ D ½A + ¼D + E h 2 n = Additive Genetic Variance Total Variance h 2 n = ½A ½A + ¼D + E Broad-sense heritability Narrow-sense heritability

P 1 = m + [a] P 2 = m [a] F 1 = m + [d] F 2 = m + ½ [d] B 1 = m + ½ [a] + ½ [d] B 2 = m ½ [a] + ½ [d]

P 1 = m + [a] P 2 = m [a] F 1 = m + [d] F 2 = m + ½ [d] B 1 = m + ½ [a] + ½ [d] B 2 = m ½ [a] + ½ [d]

P 1 = m + [a] P 2 = m [a] F 1 = m + [d] F 2 = m + ½ [d] B 1 = m + ½ [a] + ½ [d] B 2 = m ½ [a] + ½ [d]

V(B 1 ) = [ (x i - ) 2 ]/n B 1 = ½ P 1 + ½ F 1 (B 1 ) = m + ½ [a] +½ [d] V(B 1 ) = f(x- ) 2

(B 1 ) = m + ½ [a] +½ [d] B 1 = ½ P 1 + ½ F 1 P 1 = ½ [(m+a)-(m+½a+½d)] 2 = ½ [½a ½d] 2 F 1 = ½ [(m+d)-(m+½a+½d)] 2 = ½ [-½a+½d] 2

(B 1 ) = m + ½ [a] +½ [d] B 1 = ½ P 1 + ½ F 1 V(B 1 ) = ½[½a ½d] 2 + ½ [-½a+½d] 2 = ½[¼a 2 +¼ d 2 ½ad]+½[¼a 2 +¼d 2 ½ad] = ¼a 2 + ¼d 2 ½ad

(B 2 ) = m - ½ [a] +½ [d] B 2 = ½ P 2 + ½ F 1 P 2 = ½ [(m-a)-(m-½a+½d)] 2 = ½ [-½a ½d] 2 F 1 = ½ [(m+d)-(m-½a+½d)] 2 = ½ [½a+½d] 2

(B 2 ) = m - ½ [a] +½ [d] B 2 = ½ P 2 + ½ F 1 V(B 2 ) = ½[-½a ½d] 2 + ½ [½a+½d] 2 = ½[¼a 2 +¼ d 2 +½ad]+½[¼a 2 +¼d 2 +½ad] = ¼a 2 + ¼d 2 + ½ad

V(B 1 ) = ¼ A + ¼ D ½ [AD] + E V(B 2 ) = ¼ A + ¼ D + ½ [AD] + E V(B 1 ) + V(B 2 ) = ½ A + ½ D + 2E V(F 2 ) = ½ A + ¼ D + E D = 4[V(B 1 ) + V(B 2 ) V(F 2 ) E] A = 2[V(F 2 ) ¼D E]

D A V(F 1 ) = 21 g 2 ; V(F 2 ) = 65 g 2 ; V(B 1 ) = 42 g 2 ; V(B 2 ) = 49 g 2 E = V(F 1 ) = 21 g 2 = 4[V(B 1 )+V(B 2 )-V(F 2 )-E] = 4[42 + 49 + - 65 21] = 20 g 2 = 2[V(F 2 ) ¼ D E = 2[65 ( ¼ x 20) 21] = 78 g 2

V(F 2 ) = 65 g 2 ; E = 21 g 2 ; D = 20 g 2 ; A = 78 g 2 h 2 n = ½A ½A + ¼D + E h 2 n = 0.5 x 78 0.5x78 + 0.25x20 + 21 h 2 n = 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 F 2, B 1 and B 2 families. Given the following standard deviations for both parents (P 1 and P 2 ), the F 2, and both backcross progeny (B 1 and B 2 ), determine the broad-sense heritability and narrow-sense heritability for seed size in dry pea Family Standard Deviation P 1 3.521 P 2 3.317 F 2 6.008 B 1 5.450 B 2 5.157

Answer Family Standard Deviation P 1 3.521 P 2 3.317 F 2 6.008 B 1 5.450 B 2 5.157 VP 1 =12.4; VP 2 =11.0; VF 2 =36.1; VB 1 =29.7; VB 2 =26.6

Answer VP 1 =12.4; VP 2 =11.0; VF 2 =36.1; VB 1 =29.7; VB 2 =26.6 h 2 b = Genetic variance Total variance E = [VP 1 +VP 2 ]/2 = 11.7 h 2 b = 36.1 11.7 36.1 h 2 b = 0.67

Answer VP 1 =12.4; VP 2 =11.0; VF 2 =36.1; VB 1 =29.7; VB 2 =26.6 E = [VP 1 +VP 2 ]/2 = 11.7 D = 4[V(B 1 )+V(B 2 )-V(F 2 )-E] 4[29.7+26.6-36.1-11.7] = 8.5 A = 2[V(F 2 )-¼D-E] = 2[36.1-2.1-11.7] = 22.3 h 2 n = ½A/V(F 2) = 11.15/36.1 = 0.31

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

Height of son 140 120 100 80 b 1 60 40 20 b o Y = b o + b 1 x 0 0 20 40 60 80 100 120 Height of father

Y = b o + b 1 x b 1 = [SP(x,y)/SS(x)] SP(x,y) = (x i -x)(y i -y) SP(x,y) = (xy) - [ (x) (y)]/n SS(x) = (x i -x) 2 SS(x) = (x 2 ) - [ (x)] 2 /n

Y = b o + b 1 x b o = mean(y) - b 1 x mean(x) se(b 1 ) = {SS(y) [b x SP(x,y)]} (n-2) x SS(x)

b = [SP(x,y)/SS(x)] F 2 > F 3 b = Covariance between F 2 and F 3 Variance of F 2

b = [SP(x,y)/SS(x)] F 2 > F 3 ¼ P 2 ½ F 1 ¼ P 1 ¼ P 2 1/16 P 2 2/16 B 2 1/16 F 1 ½ F 1 2/16 B 2 4/16 F 2 2/16 B 1 ¼ P 1 1/16 F 1 2/16 B 1 1/16 P 1

1/16 P 1 ; 1/16 P 2 ; 2/16 F 1 ; 4/16 F 2 ; 4/16 B 1 ; 4/16 B 2 (x i - x )(y i - y ) x = m+½d; y = m+½d P 1 family types = 1/16 a 2 + 1/64 d 2 1/16 ad P 2 family types = 1/16 a 2 + 1/64 d 2 + 1/16 ad F 1 family types = -1/32 d 2 F 2 family types = 0 B 1 family types = 1/16 a 2 B 1 family types = 1/16 a 2 ¼A

b = Covariance between F 2 and F 3 Variance of F 2 b = b = h 2 n = ¼ A Variance of F 2 ¼ A ½ A + ¼ D + E ½ A ½ A + ¼ D + E h 2 n = 2 x b

Regression of one parent onto off-spring h 2 n = 2 x b Regression of two parents onto off-spring h 2 n = b

Male Parent Female Parent Average (P 1 +P 2 )/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 h 2 n = 2 x b = 0.946 + 0.1632 Female h 2 n = 2 x b = 0.936 + 0.1805 Average h 2 n = 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(x 1,x 2 )/ [SS(x 1 ).SS(x 2 )] SP(x 1,x 2 ) = (x 1i x 2i )-[ (x 1i ) (x 2i )]/n SS(x 1 ) = (x 1i2 )-[ (x 1i )] 2 /n SS(x 2 ) = (x 2i2 )-[ (x 2i )] 2 /n SP(x 1,x 2 ) = 39; SS(x 1 ) = 74; SS(x 1 ) = 66 r = 39/ [74 x 66] = 0.553

Neither x 1 nor x 2 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, r 2 = h 2

F 2 (x 1 ) F 3 (x 2 ) 71 68 66 67 70 71 70 73 72 65 66 69 64 65 63 65 62 65 64 66 59 62 r = SP(x 1,x 2 )/ [SS(x 1 ).SS(x 2 )] SS(x 1,x 2 ) = 39; SS(x 1 ) = 74; SS(x 1 ) = 66 r = 39/ [74 x 66] = 0.553 with n-2 df h 2 = r 2 = 0.306

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