1. Genetics and Genetic Prediction in Plant Breeding

3. Class Test #2, March 10, 2000 Eight Questions worth 100 points total
Bonus Point worth 10 points
Show all calculations
50 minutes

4. Question 1a.
A cross is made between two homozygous barley plants. One parent is tall and with short leaf margin hairs (TTss), and the other is short with long leaf margin hairs (ttSS). Single genes control both plant height and leaf margin hair length. Tall plants (T_) being completely dominant to short (tt), and long leaf margin hairs (S_) dominant to short (ss). If a sample of F1 plants from this cross were self pollinated, a large population of F2 plants grown, and their phenotype for height and margin hairs noted, what would be the expected ratio of phenotypes based on the following genetic situations. [8 points].

5. Question 1a. Genetic situation Plant Phenotype T_S_ T_ss ttS_ ttss
Complete dominance: T dominant to t, S dominant to s.
Duplicate recessive epistasis: tt epistasis to S and ss; ss epistatic to T and tt.
Recessive epistasis: tt epistatic to S and ss.
Duplicate dominant epistasis: T epistasis to S and ss; S epistasis to T and tt.

6. Question 1a. Genetic situation Plant Phenotype T_S_ T_ss ttS_ ttss 9 3
Complete dominance: T dominant to t, S dominant to s. 9 3 1
Duplicate recessive epistasis: tt epistasis to S and ss; ss epistatic to T and tt. 7
Recessive epistasis: tt epistatic to S and ss. 4
Duplicate dominant epistasis: T epistasis to S and ss; S epistasis to T and tt. 15

7. Question 1b
In the above experiment, a sample of F2 plants were self-pollinated and 900 F3 plants evaluated for plant height and leaf margin hair length. After evaluating the phenotypes it was found that there were: 335 tall plants with long leaf margin hairs; 220 tall plants with short margin hairs; and 345 short plants with short margin hairs.
Explain what could have caused the observed frequency of phenotypes and test your theory using a suitable statistical test. [6 points].

8. Question 1b. T_S_ T_ss ttS_ ttss 335 220 345
345
Explain what may have caused this departure from a 25:15:15:9 expected frequency of phenotypes [6 points].
This departure from a 25:15:15:9 ratio could be caused by recessive epistasis, where tt is epistatic to S, so ttS_ and ttss have the same phenotype.

9. A appropriate test to use would be a chi-square test.
Question 1b.
A appropriate test to use would be a chi-square test.
L_G_
L_gg
llG_
llgg
Observed
335
220
345
Expected
352
211 -
337
Difference
-17 9 8
D2/exp
0.821
0.384
0.190
2 2df = ns

10. Question 2.
A spring wheat breeding program aims to develop cultivars that are resistant to foot-rot, controlled by a single completely dominant gene (FF) which confers resistance, and that are resistant to yellow strip rust, which is controlled by a single dominant (YY) gene conferring resistance. A cross is made between two parents where one parent is resistant to foot rot and susceptible to yellow strip rust (FFyy) while the other is resistant to yellow strip rust but susceptible to foot rot (ffYY). A sample of F1 plants was self-pollinated, without selection, and a large sample of F2 plants were grown and allowed to self-pollinate. At harvest, only plants that were phenotypically resistant to both diseases (foot-rot and yellow strip rust) are retained and used to plant a large population of F3’s.

11. Question 2.
What proportion of these F3 plants would be expected to be:
Resistant to foot rot and yellow strip rust =
Resistant to foot rot but susceptible to yellow strip rust =
Susceptible to foot rot but resistant to yellow strip rust =
Susceptible to both foot rot and yellow strip rust =
[10 points].

12. (FFyy) x (ffYY) F3 F2 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyy
Total
1/16
2/16
4/16

13. (FFyy) x (ffYY) F3 F2 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyy
Total
4/64
8/64
16/64

14. (FFyy) x (ffYY) F3 F2 4 2 - 1 9 6 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY
Total
4/64
8/64
16/64 4 2 - 1 9 6

15. (FFyy) x (ffYY) F3 F2 4 2 1 9 - 6 3 FFYY FFYy FfYY Total 4/64 8/64
16/64 4 2 1 9 - 6
Ffyy 3
FfYy
ffYY
ffYy

16. Question 2.
What proportion of these F3 plants would be expected to be:
Resistant to foot rot and yellow strip rust = 25
 Resistant to foot rot but susceptible to yellow strip rust = 5
 Susceptible to foot rot but resistant to yellow strip rust = 5
 Susceptible to both foot rot and yellow strip rust = 1
 [10 points].

17. Question 3.
Assume the situation in question 3 (above) where a foot rot wheat parent (FFyy) is crossed to a yellow rust resistant parent (ffYY). The heterozygous F1 is crossed to a double susceptible homozygous line with genotype ffyy and 2,000 BC1 progeny evaluated for disease resistance. The following numbers of phenotypes were observed:
Resistant to both foot rot and yellow strip rust = 90 Resistant to foot rot but susceptible to yellow strip rust = 893 Susceptible to foot rot but resistant to yellow strip rust = 907 Susceptible to both foot rot and yellow strip rust = 110
Determine the percentage recombination between the foot rot and yellow strip rust loci. [4 points].

18. Question 3. Recombination (R) Number of Recombinants
Number of observations
= 0.10 = 10%

19. Question 3. Gametes from female parent Gametes from male parent
FY-0.05
Fy-0.45
fY-0.45
fy-0.05
FY – 0.05
FYFY
FFYy
FfYY
FfYf
Fy – 0.45
FFyy
FfYy
Ffyy
fY – 0.45
ffYY
ffYy
fy – 0.05
FfYy
ffyy

20. Question 3.
25 FFYY:450 FFYy:2025 FFyy: FfYY:4100 FfYy:450 Ffyy: ffYY:450 ffYy:25 ffyy
F_Y_ = 5,025 F_yy = 2,475 ffY_ = 2,475 ffyy =

21. n = 6.60 ~ need 7 F2 plants to be 99% sure of one.
Question 3.
Estimate the number of F2 plants that would need to be evaluated to be 99% sure of identifying one F2 plant that is resistant to both foot rot and yellow strip rust. [3 points].
n = Ln(1-p)/Ln(1-x)
n = Ln(1-0.99)/Ln( )
n = 6.60 ~ need 7 F2 plants to be 99% sure of one.

22. Question 4a.
Potato cyst nematode resistance is controlled by a single completely dominant allele (R). A cross is made between two auto-tetraploid potato cultivars where one parent is resistant to potato cyst nematode and the other is completely susceptible. Progeny from the cross are examined and it is found that 83.3% of the progeny are resistant to potato cyst nematode. What can be determined about the resistant parent in the cross? [4 points].
One parent must be nulliplex (rrrr), and one resistant (Rrrr, RRrr, RRRr, or RRRR). If RRRr or RRRR all progeny are resistant. If Rrrr 50% are resistant (1:1). If RRrr then 83% (5:1) are resistant. Answer parent is duplex (i.e.. RRrr x rrrr).

23. Question 4b.
A breeding program aims to produce potato parental lines that are either quadriplex or triplex for the potato cyst nematode resistant gene R. A cross is made between two parents, which are know to be duplex for the resistant gene R. What would be the expected genotype and phenotype of progeny from this diploid x diploid cross? [4 points].

24. Question 4b.
1 RR
4 Rr
1 rr
1 RRRR
4 RRRr
1 RRrr
16 RRrr
4 Rrrr
1 rrrr

25. 1 RRRR: 8 RRRr : 18 RRrr : 8 Rrrr : 1 rrrr
Question 4b.
1 RR
4 Rr
1 rr
1 RRRR
4 RRRr
1 RRrr
16 RRrr
4 Rrrr
1 rrrr
1 RRRR: 8 RRRr : 18 RRrr : 8 Rrrr : 1 rrrr

26. 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.

27. 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

28. 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

29. 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.

30. 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.

31. 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

32. 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.

33. 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.

34. 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

35. 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

36. 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.

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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].

44. 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]

45. 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

46. 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

47. Quantitative Genetics Models
P m P1

48. 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].

49. 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

50. City of the Dead
Moscow March 10, 2000