Genetics and Genetic Prediction in Plant Breeding

Transcription

1 Genetics and Genetic Prediction in Plant Breeding

2 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?.

3 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.

4 -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).

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

6 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

7 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]

8 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

9 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

10 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

11 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

12 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

13 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

14 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 = h 2 b = 0.833

15 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

16 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 = [ ]/3 = 32.2 h 2 b = =

17 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

18 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

19 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]

20 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]

21 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]

22 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

23 (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

24 (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

25 (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

26 (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

27 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]

28 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[ ] = 20 g 2 = 2[V(F 2 ) ¼ D E = 2[65 ( ¼ x 20) 21] = 78 g 2

29 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 x x h 2 n = 0.60

30 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 P F B B

31 Answer Family Standard Deviation P P F B B VP 1 =12.4; VP 2 =11.0; VF 2 =36.1; VB 1 =29.7; VB 2 =26.6

32 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 = h 2 b = 0.67

33 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[ ] = 8.5 A = 2[V(F 2 )-¼D-E] = 2[ ] = 22.3 h 2 n = ½A/V(F 2) = 11.15/36.1 = 0.31

34 Heritability Parent v Offspring Regression

35 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

36 Height of son b b o Y = b o + b 1 x Height of father

37 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

38 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)

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

40 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

41 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

42 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

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

44 Male Parent Female Parent Average (P 1 +P 2 )/2 Off-spring

45 Male Female Average SP(x,y) SS(x) SS(y) Covariance Variance

46 Male Female Average b se(b) Male h 2 n = 2 x b = Female h 2 n = 2 x b = Average h 2 n = b =

47 Correlation and Heritability

48 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

49 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

50 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

51 F 2 (x 1 ) F 3 (x 2 ) 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] = with n-2 df h 2 = r 2 = 0.306

52 Diallels