1. Genetics and Genetic Prediction in Plant Breeding

2. Question 5b.
Why are plant breeders interested in conducting scaling tests for quantitatively inherited characters of importance in the breeding scheme? [3 points]
Many of the important statistics that breeders are concerned with (i.e. response to selection) are based on the additive/dominance model. Scaling tests are used to determine if this model is adequate, and hence the predictions are accurate.

3. Question 5b. Family Mean Yield Variance of yield Number of plants P1
A properly designed experiment was carried out in canola where the high yielding parent (P1), is grown alongside F1 plants and B1 plants. The average yield of plants from each of the three families, the variance of each family and the number of plants evaluated from each family were:
Family
Mean Yield
Variance of yield
Number of plants P1
1923 74 16 F1
1629 69 B1
1985
132 31

4. V(A) = 4VB1 + VF1 + VP1 = 528+69+74 = 671 se(A) = V(A) = 29.90
Question 5b.
Family
Mean Yield
Variance of yield
Number of plants P1
1923 74 16 F1
1629 69 B1
1985
132 31
A = 2B1 – F1 – P1 = = 418
V(A) = 4VB1 + VF1 + VP1 = = 671 se(A) = V(A) = 29.90
t df = A/se(A) = 1.40 ns

5. Yijk =  + gi + gj + sij + eijk,
Question 6a.
The following is a model for the analysis of diallels (Griffing analysis).
Yijk =  + gi + gj + sij + eijk,
Explain what gi and sij represent in the model [4 points].
gi is the general combining ability of the ith parent, sij is the specific combing ability (not explained by GCA) between the ith and the jth parent.

6. Question 6a.
Briefly explain the difference between parents chosen at random (random parent effects) and parents specifically chosen (fixed parent effects). [4 points].
If parents are chosen at random it is assumed that inference from the analyses are to govern the situation of all possible parental cross combinations. When parental are fixed than it is assumed that the analysis is unique only to the parents in the design.

7. Question 6a.
A 5 x 5 full diallel (including selfs) design was conducted and yield of the parent self’s and the F1 progeny were obtained from a properly randomized and replicated experiment. A Griffing analysis of variance was carried out and sum of squares and degrees of freedom are shown below. Complete the analysis assuming that the parents are fixed and briefly outline your conclusions from the analysis. [4 points].
Source df
S.Sq
GCA 4
7,900
SCA 10
6,010
Reciprocal
5,100
Error 50
17,500
Total 74
35,710

8. Question 6a. Source df S.Sq M.Sq F GCA 4 7,900 1957 5.64 ** SCA 10
6,010
601
1.18 ns
Reciprocal
5,100
510
1.46 ns
Error 50
17,500
350
Total 74
35,710
Reciprocal effects were not significantly different from error so there were no maternal or cytoplasmic effects for yield. SCA was also non-significant while GCA was significant at the 99% level indicating a high proportion of additive genetic variance. From the analysis there would be good opportunity to determine progeny worth from GCA values of parents.

9. Question 6a.
What difference would you make if the parents in this analysis were chosen at random [2 points].
Source df
S.Sq
M.Sq F
GCA 4
7,900
1957
3.25 ns
SCA 10
6,010
601
1.18 ns
Reciprocal
5,100
510
1.46 ns
Error 50
17,500
350
Total 74
35,710
Use the SCA M.Sq to test the GCA term. In this case the GCA is not quite formally significant at the 5% level.

10. Question 6d. SP(Vi,Wi) = 165 – [26 x 22]/5 = 50.6
A Hayman & Jinks analysis was conducted and Vi and Wi values estimated for each parent. The sum of products between Vi and Wi ([Vi x Wi]) was found to be 165; the sum of Vi’s was 26; sum of Wi’s was 22; sum of squares of Vi ( [Vi2]) was 205; and the sum of squares of Wi ( [Wi2]) was Complete a regression analysis of Vi on to Wi. [4 points].
SP(Vi,Wi) = 165 – [26 x 22]/5 = 50.6
SS(Vi) = 205 – [262]/5 = 69.8
SS(Wi) = 118 – [222]/5 = 21.2

11. Question 6d. b1 = SP(Vi,Wi)/SS(Vi) = 50.6/69.8 = 0.72
SP(Vi,Wi) = 165 – [26 x 22]/5 = 50.6 SS(Vi) = 205 – [262]/5 = SS(Wi) = 118 – [222]/5 = 21.2
b = SP(Vi,Wi)/SS(Vi) = 50.6/69.8 = 0.72
se(b1) = {SS(Wi) – b1SP(Vi,Wi)}/(n-2)SS(Vi)
= {118 – 36}/209.4 = 0.39
b = 4.4 – 0.72 x 5.2 = 0.67
t3df = [1-0.72]/0.39 = 0.71 ns

12. Question 6d.
What can be determined from this analysis regarding the adequacy of the additive/dominance model and the importance of additive genetic variance (A) compared to dominant genetic variance (D). [4 points].
The regression slope (b1) is not significantly different from one, therefore, the additive/dominance model is adequate to explain the variation observed for yield in the diallel. The intercept (b0) is greater than zero so A is greater than D. However, b0 is almost equal to zero so A = D.

13. Question 7.
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 [10 points].
Family
Standard Deviation P1
3.521 P2
3.317 F2
6.008 B1
5.450 B2
5.157

14. Question 7. VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6 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

15. h2b = Genetic variance Total variance
Question 7.
VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6
h2b = Genetic variance Total variance
E = [VP1+VP2]/2 = 11.7
h2b = –
h2b = 0.67

16. D = 4[V(B1)+V(B2)-V(F2)-E]
Question 7.
VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6
E = [VP1+VP2]/2 = 11.7
D = 4[V(B1)+V(B2)-V(F2)-E]
4[ ] = 8.5
A = 2[V(F2)-¼D-E] = 2[ ] = 22.3
h2n = ½A/V(F2) = 11.15/36.1 = 0.31

17. Question 8a.
Assuming an additive/dominance mode of inheritance for a polygenic trait, list expected values for P1, P2, and F1 in terms of m, [a] and [d]. [3 points]
P1 = m + a
P2 = m – a
F1 = m + d

18. Question 8b.
From these expectations, what would be the expected values for F2, B1 and B2 based on m, [a] and [d]. [3 points]
F2 = m + ½d
B1 = m + ½a + ½d
B2 = m – ½a + ½d

19. Question 8c.
From a properly designed field trial that included P1, P2 and F1 families, the following yield estimates were obtained.
 B1 = 42.0 lb/a; B2 = 26.0 lb/a; F1 = 38.5lb/a
 From these family means, estimate the expected value of P1, P2 and F2, based on the additive/dominance model of inheritance [6 points].

20. Question 8c. 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]

21. Question 8c B1 = 42.0 lb/a; B2 = 26.0 lb/a; F1 = 38.5lb/a
B1 – B2 = m + ½a + ½d – m –(-½a) –½d = a B1 + B2 – F1 = 2m + d – m - d = m F1 – m = d
B1 – B2 = 42.0 – 26.0 = 36.0 = 16 = a B1 + B2 – F1 = 29.5 = m F1 – m = 9.0 = d

22. Question 8c.
B1 – B2 = 42.0 – 26.0 = 36.0 = 16 = a B1 + B2 – F1 = 29.5 = m F1 – m = 9.0 = d
P1 = m + a = = P2 = m – a = 29.5 – 16.0 = F2 = m + ½d = 29.5 – 4.5 = 25.0

23. Quantitative Genetics Models
P m P1

24. Bonus Question
It is important in quantative genetics to know whether the additive/dominance model based on m, [a], and [d] is appropriate to explain the variation observed in this study. Given that you have available progeny means and variances from both parents (P1 and P2), the B1, B2, and F3 families. Devise a suitable scaling test (hint as yet we have not talked about this one) involving these five families. [10 bonus points].

25. Bonus Question
P1= m + [a]; P2= m – [a]; F3 = m + ¼ [d]; B1= m + ½ [a] + ½ [d]; B2 = m – ½ [a] + ½ [d]
4F3 = 4m+d
B1+B2+P1+P2= 4m +d
4F3 – B1 – B2 – P1 – P2  0

26. City of the Dead
Moscow March 10, 2000