1. Genetics and Genetic Prediction in Plant Breeding

2. Assuming an additive/dominance mode of inheritance for a polygenic trait, list expected values for P 1, P 2, and F 1 in terms of m, [a] and [d]. [3 points] Question 1a P 1 = m + a P 2 = m – a F 1 = m + d

3. From these expectations, what would be the expected values for F 2, B 1 and B 2 based on m, [a] and [d]. [3 points] Question 1b F 2 = m + ½d B 1 = m + ½a + ½d B 2 = m – ½a + ½d

4. Question 1c From a properly designed field trial that included P 1, P 2 and F 1 families, the following yield estimates were obtained. P 1 = 1928 Kg; P 2 = 1294 Kg; F 1 = 1767 Kg P 1 = 1928 Kg; P 2 = 1294 Kg; F 1 = 1767 Kg From these family means, estimate the expected value of F 2, B 1, B 2 and F α, based on the additive/dominance model of inheritance [3 points]. From these family means, estimate the expected value of F 2, B 1, B 2 and F α, based on the additive/dominance model of inheritance [3 points].

5. Question 1c F 2 = m + ½d = 1611 + 78 = 1689 B 1 = m+½a+½d = 1611+158.5+78 = 1397.5 B 2 = m–½a+½d = 1611-158.5+78 = 1530.5 P 1 = 1928 Kg; P 2 = 1294 Kg; F 1 = 1767 Kg a = [P 1 – P 2 ]/2 = [1928-1294]/2 = 317 m = [P 1 – a] = 1928 – 317 = 1611 d = [F 1 – m] = 1767 – 1611 = 156

6. Question 3a Two genetically different homozygous lines of canola (Brassica napus L.) were crossed to produce F 1 seed. Seed from the F 1 family was self pollinated to produce F 2 seed. A properly designed experiment was carried out involving both parents (P 1 and P 2, 10 plants each), the F 1 (10 plants) and the F 2 families (64 plants) was grown in the field and plant height of individual plants (inches) recorded. The following are family means, variances and number of plants observed for each family FamilyMeanVariance# Plants P 1 52 1.9710 P 2 41 2.6910 F 1 49 3.1410 F 2 4310.6934 Complete a statistical test to determine whether an additive/dominance model of inheritance is appropriate to adequately explain the inheritance of plant height in canola [7 points]. Complete a statistical test to determine whether an additive/dominance model of inheritance is appropriate to adequately explain the inheritance of plant height in canola [7 points].

7. Question 3a FamilyMeanVariance#Plants P1P1P1P1521.9710 P2P2P2P2412.6910 F1F1F1F1493.1410 F2F2F2F24310.6964 C-scaling test = 4F 2 – 2F 1 – P 1 – P 2 C = 172-98-52-41 = 19 V(C) = 16V(F 2 )+4V(F 1 )+V(P 1 )+V(P 2 ) V(C) = 171.04+12.56+1.97+2.69 = 188.26 se(C) =  188.26 = 13.72

8. Question 3a FamilyMeanVariance#Plants P1P1P1P1521.9710 P2P2P2P2412.6910 F1F1F1F1493.1410 F2F2F2F24310.6964 C = 172-98-52-41 = 19 se(C) =  188.26 = 13.72 t 90df = 19/13.72 = 1.38 ns Therefore, additive/dominance model is adaquate

9. Question 3a FamilyMeanVariance#Plants P1P1P1P1521.9710 P2P2P2P2412.6910 F1F1F1F1493.1410 F2F2F2F24310.6964 P 2 41 P 1 52 m 46.5 F 2 43 F 1 49 ?

10. Question 3b If the additive/dominance model is inadequate, list three factors which could cause the lack of fit of the model [3 points].  Abnormal chromosomal behavior: where the heterozygote does not contributes equal proportions of its various gametes to the gene pool.  Cytoplasmic inheritance: where the character is determined by non-nuclear genes.  Epistasis: where alleles at different loci are interacting.

11. Question 4a F 1, F 2, B 1, and B 2 families were evaluated for plant yield (kg/plot) from a cross between two homozygous spring wheat parents. The following variances from each family were found: σ 2 F1 = 123.7; σ 2 F2 = 496.2; σ 2 B1 = 357.2; σ 2 B2 = 324.7 Calculate the broad-sense (h 2 b ) and narrow-sense (h 2 n ) heritability for plant yield [10 points].

12. Question 4a σ 2 F1 = 123.7; σ 2 F2 = 496.2; σ 2 B1 = 357.2; σ 2 B2 = 324.7 h 2 b = Genetic variance Total variance h 2 b = 496.2 – 123.7 496.2 E = V(F 1 ) = 123.7 h 2 b = 0.75

13. Question 4a σ 2 F1 = 123.7; σ 2 F2 = 496.2; σ 2 B1 = 357.2; σ 2 B2 = 324.7 E = V(F 1 ) = 123.7 D = 4[V(B 1 )+V(B 2 )-V(F 2 )-E] 4[357.2+324.7-496.2-123.7] = 248 A = 2[V(F 2 )-¼D-E] = 2[496.2-62-123.7] = 621 h 2 n = ½A/V(F 2 ) = 310.5/496.2 = 0.63

14. Question 4b Given the heritability estimates you have obtained, would you recommend selection for yield at the F 3 in a wheat breeding program, and why? [2 points]. A narrow-sense heritability greater than 0.6 would indicate a high proportion of the total variance was additive in nature. Selection at F 3 is often advesly related to dominant genetic variation (caused by heterozygosity), here D is small compared to A. Additive genetic variance is constant over selfing generations and so selection would result in a good response.

15. Question 5a Four types of diallel crossing designs have been described by Griffing. Briefly outline the features of each Method 1, 2, 3, and 4 [4 points]. 1. Complete diallel with selfs, Method 1. 2. Half diallel with selfs, Method 2. 3. Complete diallel, without selfs, Method 3. 4. Half diallel, without selfs, Method 4.

16. Question 5a Why would you choose Method 3 over Method 1? [1 point]. In instances where it was not possible to produce selfed progeny (i.e. in cases of strong self-incompatibility in apple and rapeseed). Why would you choose Method 2 over Method 1? [1 point]. In cases where there is no recipricol or maternal effects.

17. Question 5b A full diallel, including selfs is carried involving five chick-pea parents (assumed to be chosen as fixed parents), and all families resulting are evaluated at the F 1 stage for seed yield. The following analysis of variance for general combining ability (GCA), specific combining ability (SCA) and reciprocal effects (Griffing analysis) is obtained: SourcedfMSq GCA530,769 SCA1010,934 Recipricol109,638 Error495,136

18. Question 5b SourcedfMSqF GCA530,7695.99 ** SCA1010,9342.12 ns Recipricol109,6381.87 ns Error495,136 GCA is significant at the 99% level, while SCA and reciprical differences were not significant. This indicates that a high proportion of phenotypic variation between progeny is additive rather than dominant or error.

19. Question 5b SourcedfMSqF GCA530,7692.81 ns SCA1010,9342.12 ns Recipricol109,6381.87 ns Error495,136 If a random model is chosen then the GCA term is tested using the SCA mean square. In this case the GCA term is not formally significant, and the indication overall would be that there is no significant variation between progeny in the diallel.

20. Question 5b Plant height was also recorded on the same diallel families and an additive/dominance model found to be adequate to explain the genetic variation in plant height. Array variances V i 's and non- recurrent parent covariances (W i 's) were calculated and are shown along-side the general combining ability (GCA) of each of the five parents, below: ViWiWi GCA Parent 1491.4436.8-0.76 Parent 2610.3664.2+12.92 Parent 3302.4234.8-14.32 Parent 4310.2226.9-15.77 Parent 5832.7769.4+17.93

21. Question 5b Visual inspection of V i and W i values would indicate a linear relationship with slope approximatly equal to 1, which would indicate a additive/dominance model of inheritance. Parents with lowest V i and W i values (those with greatest frequency of dominant alleles) have negative GCA values indicating short stature in chick pea is dominant over tall stature. ViWiWi GCA Parent 1491.4436.8-0.76 Parent 2610.3664.2+12.92 Parent 3302.4234.8-14.32 Parent 4310.2226.9-15.77 Parent 5832.7769.4+17.93

22. Bonus Question A 4x4 half diallel (with selfs) was carried out in cherry and the following fruit yield of each possible F 1 family observed. Sm. Red Sm. Reds12Big Yld 2736Jim’s D Jim’s D.213527Fellman 18272621 Calculate narrow-sense heritability.

23. Mid-ParentOff-spring x2x2x2x2  xy 2427576648 19.521380410 16.518272297 31.5359921102 28.527812770 2426576624 1441553,6093,850 SS(x) =  x 2 – (  x) 2 /n = 153.0 SP(x,y) =  xy – (  x  y)/n = 130.5 b = 130.5/153.0 = 0.8529 = h 2 n Bonus Question

24. The End Thank you all Good Luck on Friday