1. Genetics and Genetic Prediction in Plant Breeding

2. Class Test #2, March, 1998 Eight Questions worth 100 points total Bonus Point worth 10 points Show all calculations 50 minutes

3. 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

4. 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

5. 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].

6. 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

7. Question 2 A spring barley breeding program has major emphasis in developing cultivars which are short in stature and with yellow stripe resistance. It is known that the inheritance of short plants is controlled by a single completely recessive gene (tt) over tall plants (TT), and that yellow stripe rust resistance is controlled by a single completely dominant gene (YY), over a recessive susceptible gene (yy). The tall gene locus and yellow rust gene locus are located on different chromosomes. Given that a tall resistant plant (TTYY) is crossed to a short susceptible plant (ttyy), both parents being homozygous, what would be the expected proportion of genotypes and phenotypes in the F 1 and F 2 families [12 points].

8. Tall, Res. (TTYY) x Short, Susc. (ttyy) F 1 TtYyTall/Resistant

9. Tall, Res. (TTYY) x Short, Susc. (ttyy) Gametes from female parent Gametes from male parent TYTytYty TYTTYYTTYyTtYYTtYy TyTTYyTTyyTtYyTtyy tYTtYYTtYyttYYttYy tyTtYyTtyyttYyttyy 1 TTYY:2 TTYy:1 TTyy:2 TtYY:4 TsYy:2 Ttyy:1 ttYY:2 ttYy:1 ttyy

10. Tall, Res. (TTYY) x Short, Susc. (ttyy) Gametes from female parent Gametes from male parent TYTytYty TYTTYYTTYyTtYYTtYy TyTTYyTTyyTtYyTtyy tYTtYYTtYyttYYttYy tyTtYyTtyyttYyttyy 9 T_Y_ : 3 T_yy : 3 ttY_ : 1 ttyy

11. Question 2 How many F 3 plants would need to be assessed to ensure, with 99% certainty, that at least one plant would exist that was short and homozygous yellow rust resistant (i.e. ttYY) [8 points]. ttYY = 9/64 = 0.1406 # = Ln[1-0.99] = Ln(0.01) Ln[1-0.1406]Ln(0.8594) = 30.39, need 31 F 3 plants

12. 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].

13. 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

14. 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

15. 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 ?

16. 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.

17. 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].

18. 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

19. 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

20. 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.

21. 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.

22. 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.

23. 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

24. 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.

25. 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.

26. 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

27. 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

28. Question 6a Two homozygous barley parents were crossed to produce an F 1 family. One parent was tall with awns and the other was short and awnless. Tall plants are controlled by a single dominant gene and awned plants are also controlled by a single dominant gene. The F 1 family was crossed to a plant which was short and awnless and the following number of phenotypes observed: Phenotype# observed Tall, awned954 Short, awned259 Tall, awnless221 Short, awned966

29. Question 6a Phenotype# observed Tall, awned954 Short, awned259 Tall, awnless221 Short, awned966 % Recombination [259+221]/2400 = 0.20

30. Gametes from female parent Gametes from male parent TA-0.4 Ta-0.1 tA-0.1 ta-0.4 TA – 0.4 TTAA 0.16 TTAa 0.04 TtAA 0.04 TtAa 0.16 Ta – 0.1 TTAa 0.04 TTrr 0.01 TtAa 0.01 Ttaa 0.04 tA – 0.1 TtAA 0.04 TtAa 0.01 ttAA 0.01 ttAa 0.04 ta – 0.4 TtRr 0.16 Ttaa 0.04 ttAa 0.04 ttaa 0.16 16 TTAA:8 TTAa:1 TTaa:8 TtAA:34 TtAa:8 Ttaa:1 ttAA:8 ttAa:16 ttaa 66 T_A_ : 9 T_aa : 9 ttA_ : 16 ttaa

31. Question 6a What is the difference between linkage and pleiotropy? [2 points]. Linkage is when alleles at two loci do not segregate independantly and hence there is linkage disequilibrium in segregating populations. The cause is that the two loci are located on the same chromosome. Pleiotropy is where two characters are controlled by alleles at a single locus.

32. Question 7a Two homozygous squash plants were hybridized and an F 1 family produced. One parent was long and green fruit (LLGG) and the other was round and yellow fruit (llgg). 1600 F 2 progeny were examined from selfing the F 1 's and the following number of phenotypes observed: L_G_L_ggllG_llgg 8913120397 Explain what may have caused this departuure from a 9:3:3:1 expected frequency of phenotypes [4 points].

33. Question 7a L_G_L_ggllG_llgg 8913120397 Explain what may have caused this departuure from a 9:3:3:1 expected frequency of phenotypes [4 points]. This departure from a 9:3:3:1 ratio could be caused by recessive epistasis, where ll is epistatic to G, so llG_ and llgg have the same phenotype.

34. Question 7a L_G_L_ggllG_llgg Observed8913120397 Expected900300-400 Difference912-3 D 2 /exp0.090.48-0.02  2 2df = 0.59 ns A appropriate test to use would be a chi-square test.

35. 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.

36. 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

37. The End Thank you all Good Luck on Friday