Genetics and Genetic Prediction in Plant Breeding

Genetics and Genetics and Genetic Prediction Genetic Prediction in Plant Breedingin Plant Breeding

Class Test #2, March 10, 2000Class Test #2, March 10, 2000

Eight Questions worth 100 points totalEight Questions worth 100 points total

Bonus Point worth 10 pointsBonus Point worth 10 points

Show all calculationsShow all calculations

50 minutes50 minutes

Question 1a.Question 1a.

A cross is made between two homozygous barley plants. A cross is made between two homozygous barley plants. One parent is tall and with short leaf margin hairs One parent is tall and with short leaf margin hairs ((TTssTTss), and the other is short with long leaf margin hairs ), and the other is short with long leaf margin hairs ((ttSSttSS). Single genes control both plant height and leaf ). Single genes control both plant height and leaf margin hair length. Tall plants (margin hair length. Tall plants (T_T_) being completely ) being completely dominant to short (dominant to short (tttt), and long leaf margin hairs (), and long leaf margin hairs (S_S_) ) dominant to short (dominant to short (ssss). If a sample of F). If a sample of F

11 plants from this plants from this

cross were self pollinated, a large population of Fcross were self pollinated, a large population of F22 plants plants

grown, and their phenotype for height and margin hairs grown, and their phenotype for height and margin hairs noted, what would be the expected ratio of phenotypes noted, what would be the expected ratio of phenotypes based on the following genetic situations. based on the following genetic situations. [8 points].[8 points].


Genetic situationPlant Phenotype

T_S_ T_ss ttS_ ttss

Complete dominance: Complete dominance: TT dominant to dominant to t, S t, S dominant to dominant to ss..

Duplicate recessive epistasis: Duplicate recessive epistasis: tt tt epistasis to epistasis to S S and and ssss;; ss ss epistatic to epistatic to T T and and tt.tt.

Recessive epistasis: Recessive epistasis: tt tt epistatic to epistatic to S S and and ssss..

Duplicate dominant epistasis: Duplicate dominant epistasis: T T epistasis to epistasis to S S and and ssss; ; S S epistasis to epistasis to T T and and tttt..


Genetic situationPlant Phenotype

T_S_ T_ss ttS_ ttss

Complete dominance: Complete dominance: TT dominant to dominant to t, S t, S dominant to dominant to ss.. 99 33 33 11

Duplicate recessive epistasis: Duplicate recessive epistasis: tt tt epistasis to epistasis to S S and and ssss; ; ss ss epistatic to epistatic to T T and and tt.tt.

9 9 00 00 77

Recessive epistasis: Recessive epistasis: tt tt epistatic to epistatic to S S and and ssss.. 99 33 00 44

Duplicate dominant epistasis: Duplicate dominant epistasis: TT epistasis to epistasis to S S and and ssss; ; S S epistasis to epistasis to T T and and tttt..

1515 00 00 11

Question 1bQuestion 1b

In the above experiment, a sample of FIn the above experiment, a sample of F22 plants were plants were

self-pollinated and 900 Fself-pollinated and 900 F3 3 plants evaluated for plant plants evaluated for plant

height and leaf margin hair length. After evaluating the height and leaf margin hair length. After evaluating the phenotypes it was found that there were: 335 tall plants phenotypes it was found that there were: 335 tall plants with long leaf margin hairs; 220 tall plants with short with long leaf margin hairs; 220 tall plants with short margin hairs; and 345 short plants with short margin margin hairs; and 345 short plants with short margin hairs.hairs.

Explain what could have caused the observed Explain what could have caused the observed frequency of phenotypes and test your theory using a frequency of phenotypes and test your theory using a suitable statistical test. suitable statistical test. [6 points].[6 points].

Question 1b.Question 1b.

T_S_ T_ss ttS_ ttss

335 220 0 345

Explain what may have caused this departure from a Explain what may have caused this departure from a 25:15:15:9 expected frequency of phenotypes 25:15:15:9 expected frequency of phenotypes [6 points][6 points]..

This departure from a 25:15:15:9 ratio could be This departure from a 25:15:15:9 ratio could be caused by recessive epistasis, where caused by recessive epistasis, where tttt is epistatic to is epistatic to S, S, so ttS_so ttS_ and and ttss ttss have the same phenotype.have the same phenotype.


L_G_ L_gg llG_ llgg

Observed 335 220 0 345

Expected 352 211 - 337

Difference -17 9 - 8

D2/exp 0.821 0.384 - 0.190

2 2df = 1.395 ns

A appropriate test to use would be a chi-square test.A appropriate test to use would be a chi-square test.

Question 2.Question 2.

A spring wheat breeding program aims to develop cultivars that A spring wheat breeding program aims to develop cultivars that are resistant to foot-rot, controlled by a single completely are resistant to foot-rot, controlled by a single completely dominant gene (dominant gene (FFFF) which confers resistance, and that are ) which confers resistance, and that are resistant to yellow strip rust, which is controlled by a single resistant to yellow strip rust, which is controlled by a single dominant (dominant (YYYY) gene conferring resistance. A cross is made ) gene conferring resistance. A cross is made between two parents where one parent is resistant to foot rot and between two parents where one parent is resistant to foot rot and susceptible to yellow strip rust (susceptible to yellow strip rust (FFyyFFyy) while the other is resistant ) while the other is resistant to yellow strip rust but susceptible to foot rot (to yellow strip rust but susceptible to foot rot (ffYYffYY). A sample of ). A sample of FF11 plants was self-pollinated, without selection, and a large plants was self-pollinated, without selection, and a large

sample of Fsample of F22 plants were grown and allowed to self-pollinate. At plants were grown and allowed to self-pollinate. At

harvest, only plants that were phenotypically resistant to both harvest, only plants that were phenotypically resistant to both diseases (foot-rot and yellow strip rust) are retained and used to diseases (foot-rot and yellow strip rust) are retained and used to plant a large population of Fplant a large population of F

33’s. ’s.


What proportion of these FWhat proportion of these F33 plants would be expected to be: plants would be expected to be:

Resistant to foot rot and yellow strip rust Resistant to foot rot and yellow strip rust ==

Resistant to foot rot but susceptible to yellow strip rustResistant to foot rot but susceptible to yellow strip rust = =

Susceptible to foot rot but resistant to yellow strip rustSusceptible to foot rot but resistant to yellow strip rust ==

Susceptible to both foot rot and yellow strip rustSusceptible to both foot rot and yellow strip rust ==

[10 points].[10 points].

((FFyyFFyy) x () x (ffYYffYY))

F3

F2

FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyyTotal

1/16 2/16 1/16 2/16 4/16 2/16 1/16 2/16 1/16


F3

F2


4/64 8/64 4/64 8/64 16/64 8/64 4/64 8/64 4/64


F3

F2


4/64 8/64 4/64 8/64 16/64 8/64 4/64 8/64 4/64

FFyy 44 22 -- 22 11 -- -- -- -- 99

FFYy -- 44 -- -- 22 -- -- -- -- 66

FFyy -- 22 44 -- 11 22 -- -- -- 99

FfYY -- -- -- 44 22 -- -- -- -- 66

FfYy -- -- -- -- 44 -- -- -- -- 44

Ffyy -- -- -- -- 22 44 -- -- -- 66

ffYY -- -- -- 22 11 -- 44 22 -- 99

ffYy -- -- -- -- 22 -- -- 44 -- 66

ffyy -- -- -- -- 11 22 -- 22 44 99

F3

F2

FFYY FFYy FfYY FfYYTotal

4/64 8/64 8/64 16/64

FFYY 44 22 22 11 99

FFYy -- 44 -- 22 66

Ffyy -- 22 -- 11 33

FfYY -- -- 44 22 66

FfYy -- -- -- 44 44

Ffyy -- -- -- 22 22

ffYY -- -- 22 11 33

ffYy -- -- -- 22 22

Ffyy -- -- -- 11 11



What proportion of these FWhat proportion of these F33 plants would be expected to be: plants would be expected to be:

Resistant to foot rot and yellow strip rust Resistant to foot rot and yellow strip rust = 25= 25

Resistant to foot rot but susceptible to yellow strip rustResistant to foot rot but susceptible to yellow strip rust = 5= 5

Susceptible to foot rot but resistant to yellow strip rustSusceptible to foot rot but resistant to yellow strip rust = 5= 5

Susceptible to both foot rot and yellow strip rustSusceptible to both foot rot and yellow strip rust = 1= 1

[10 points].[10 points].


Assume the situation in question 3 (above) where a foot rot wheat Assume the situation in question 3 (above) where a foot rot wheat parent (parent (FFyyFFyy) is crossed to a yellow rust resistant parent () is crossed to a yellow rust resistant parent (ffYYffYY). ). The heterozygous FThe heterozygous F

11 is crossed to a double susceptible is crossed to a double susceptible

homozygous line with genotype homozygous line with genotype ffyy ffyy and 2,000 BCand 2,000 BC11 progeny progeny

evaluated for disease resistance. The following numbers of evaluated for disease resistance. The following numbers of phenotypes were observed:phenotypes were observed:

Resistant to both foot rot and yellow strip rust Resistant to both foot rot and yellow strip rust = 90 = 90 Resistant to foot rot but susceptible to yellow strip rustResistant to foot rot but susceptible to yellow strip rust = 893 = 893 Susceptible to foot rot but resistant to yellow strip rustSusceptible to foot rot but resistant to yellow strip rust = 907 = 907 Susceptible to both foot rot and yellow strip rustSusceptible to both foot rot and yellow strip rust = 110= 110

Determine the percentage recombination between the foot rot and Determine the percentage recombination between the foot rot and yellow strip rust loci. yellow strip rust loci. [4 points].[4 points].

Recombination (R)Recombination (R)

Number of RecombinantsNumber of RecombinantsNumber of observationsNumber of observations


90+11090+110 = 0.10 = 10% = 0.10 = 10%

2000 2000

Gametes from female parent

Gametes from male parent

FY-0.05 Fy-0.45 fY-0.45 fy-0.05

FY – 0.05FYFY FYFY

0.0025 0.0025 FFYy FFYy

0.02250.0225FfYY FfYY

0.02250.0225FfYf FfYf

0.00250.0025

Fy – 0.45FFYy FFYy

0.02250.0225FFyy FFyy

0.20250.2025FfYy FfYy

0.20250.2025Ffyy Ffyy

0.02250.0225

fY – 0.45FfYY FfYY

0.02250.0225FfYy FfYy

0.20250.2025ffYY ffYY

0.20250.2025ffYy ffYy

0.02250.0225

fy – 0.05FfYy FfYy

0.00250.0025Ffyy Ffyy

0.02250.0225ffYy ffYy

0.02250.0225ffyy ffyy

0.00250.0025



25 25 FFYYFFYY:450 :450 FFYyFFYy:2025 :2025 FFyyFFyy: : 450 450 FfYYFfYY:4100 :4100 FfYyFfYy:450:450 FfyyFfyy: :

20252025 ffYYffYY:450 :450 ffYyffYy:25 :25 ffyyffyy

F_Y_F_Y_ = 5,025 = 5,025 F_yy F_yy = 2,475 = 2,475 ffY_ ffY_ = 2,475 = 2,475 ffyy ffyy = 25 = 25


Estimate the number of FEstimate the number of F22 plants that would need to be evaluated plants that would need to be evaluated

to be 99% sure of identifying one Fto be 99% sure of identifying one F22 plant that is resistant to both plant that is resistant to both

foot rot and yellow strip rust. foot rot and yellow strip rust. [3 points].[3 points].

n = Ln(1-p)/Ln(1-x)

nn = Ln(1-0.99)/Ln(1- 0.5025) = Ln(1-0.99)/Ln(1- 0.5025)

nn = 6.60 ~ need 7 F = 6.60 ~ need 7 F2 2 plants to be 99% sure of one.plants to be 99% sure of one.

Question 4a.Question 4a.Potato cyst nematode resistance is controlled by a single Potato cyst nematode resistance is controlled by a single completely dominant allele (R). A cross is made between two completely dominant allele (R). A cross is made between two auto-tetraploid potato cultivars where one parent is resistant to auto-tetraploid potato cultivars where one parent is resistant to potato cyst nematode and the other is completely susceptible. potato cyst nematode and the other is completely susceptible. Progeny from the cross are examined and it is found that 83.3% of 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 the progeny are resistant to potato cyst nematode. What can be determined about the resistant parent in the cross? determined about the resistant parent in the cross? [4 points].[4 points].

One parent must be nulliplex (rrrr), and one resistant (Rrrr, One parent must be nulliplex (rrrr), and one resistant (Rrrr, RRrr, RRRr, or RRRR). If RRRr or RRRR all progeny are RRrr, RRRr, or RRRR). If RRRr or RRRR all progeny are resistant. If Rrrr 50% are resistant (1:1). If RRrr then 83% 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).(5:1) are resistant. Answer parent is duplex (i.e.. RRrr x rrrr).


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


1 RR 4 Rr 1 rr

1 RR 1 RRRR1 RRRR 4 RRRr4 RRRr 1 RRrr1 RRrr

4 Rr 4 RRRr4 RRRr 16 RRrr16 RRrr 4 Rrrr4 Rrrr

1 rr 1 RRrr1 RRrr 4 Rrrr4 Rrrr 1 rrrr1 rrrr


1 RR 4 Rr 1 rr

1 RR 1 RRRR1 RRRR 4 RRRr4 RRRr 1 RRrr1 RRrr

4 Rr 4 RRRr4 RRRr 16 RRrr16 RRrr 4 Rrrr4 Rrrr

1 rr 1 RRrr1 RRrr 4 Rrrr4 Rrrr 1 rrrr1 rrrr

1 RRRR: 8 RRRr : 18 RRrr : 8 Rrrr : 1 rrrr1 RRRR: 8 RRRr : 18 RRrr : 8 Rrrr : 1 rrrr


Why are plant breeders interested in conducting Why are plant breeders interested in conducting scaling tests for quantitatively inherited scaling tests for quantitatively inherited characters of importance in the breeding scheme? characters of importance in the breeding scheme? [3 points][3 points]

Many of the important statistics that breeders are Many of the important statistics that breeders are concerned with (concerned with (i.e. response to selectioni.e. response to selection) are ) are based on the additive/dominance model. Scaling based on the additive/dominance model. Scaling tests are used to determine if this model is tests are used to determine if this model is adequate, and hence the predictions are accurate.adequate, and hence the predictions are accurate.


A A properly designed experiment was carried out in canola where properly designed experiment was carried out in canola where the high yielding parent (Pthe high yielding parent (P

11), is grown alongside F), is grown alongside F11 plants and B plants and B

11

plants. The average yield of plants from each of the three plants. The average yield of plants from each of the three families, the variance of each family and the number of plants families, the variance of each family and the number of plants evaluated from each family were:evaluated from each family were:

FamilyMean Yield

Variance of yield

Number of plants

PP11 19231923 7474 1616

FF11 16291629 6969 1616

BB11 19851985 132132 3131


Family Mean YieldVariance of

yieldNumber of

plants

P1 1923 74 16

F1 1629 69 16

B1 1985 132 31

A = 2BA = 2B11 – F – F11 – P – P11 = 3970-1629-1923 = 418 = 3970-1629-1923 = 418

V(A) = 4VBV(A) = 4VB11 + VF + VF11 + VP + VP11 = 528+69+74 = 671 = 528+69+74 = 671

se(A) = se(A) = V(A) = 29.90 V(A) = 29.90

t 15+15+30 df = A/se(A) = 1.40 ns


The following is a model for the analysis of The following is a model for the analysis of diallels (Griffing analysis).diallels (Griffing analysis).

YYijkijk = = + g + gii + g + gjj + s + sijij + e + eijk,ijk,

Explain what gExplain what gii and s and sijij represent in the model represent in the model [4 points].[4 points].

ggii is the general combining ability of the i is the general combining ability of the i thth

parent, sparent, sij ij is the specific combing ability (not is the specific combing ability (not

explained by GCA) between the iexplained by GCA) between the i thth and the j and the jthth parent.parent.


Briefly explain the difference between parents chosen at Briefly explain the difference between parents chosen at random (random parent effects) and parents specifically random (random parent effects) and parents specifically chosen (fixed parent effects). chosen (fixed parent effects). [4 points].[4 points].

If parents are chosen at random it is assumed that If parents are chosen at random it is assumed that inference from the analyses are to govern the inference from the analyses are to govern the situation of all possible parental cross situation of all possible parental cross combinations. When parental are fixed than it is combinations. When parental are fixed than it is assumed that the analysis is unique only to the assumed that the analysis is unique only to the parents in the design.parents in the design.

Question 6a.Question 6a.A A 5 x 5 full diallel (including selfs) design was conducted and yield 5 x 5 full diallel (including selfs) design was conducted and yield of the parent self’s and the Fof the parent self’s and the F

11 progeny were obtained from a properly progeny were obtained from a properly

randomized and replicated experiment. A Griffing analysis of randomized and replicated experiment. A Griffing analysis of variance was carried out and sum of squares and degrees of freedom variance was carried out and sum of squares and degrees of freedom are shown below. Complete the analysis assuming that the parents are shown below. Complete the analysis assuming that the parents are fixed and briefly outline your conclusions from the analysis. are fixed and briefly outline your conclusions from the analysis. [4 [4 points].points].

Source df S.Sq

GCAGCA 44 7,9007,900

SCASCA 1010 6,0106,010

ReciprocalReciprocal 1010 5,1005,100

ErrorError 5050 17,50017,500

TotalTotal 7474 35,71035,710

Question 6a.Question 6a.Source df S.Sq M.Sq F

GCAGCA 44 7,9007,900 19571957 5.64 **5.64 **

SCASCA 1010 6,0106,010 601601 1.18 ns1.18 ns

ReciprocalReciprocal 1010 5,1005,100 510510 1.46 ns1.46 ns

ErrorError 5050 17,50017,500 350350

TotalTotal 7474 35,71035,710

Reciprocal effects were not significantly different from error so Reciprocal effects were not significantly different from error so there were no maternal or cytoplasmic effects for yield. SCA was there were no maternal or cytoplasmic effects for yield. SCA was also non-significant while GCA was significant at the 99% level also non-significant while GCA was significant at the 99% level indicating a high proportion of additive genetic variance. From the indicating a high proportion of additive genetic variance. From the analysis there would be good opportunity to determine progeny analysis there would be good opportunity to determine progeny worth from GCA values of parents.worth from GCA values of parents.


Source df S.Sq M.Sq F

GCAGCA 44 7,9007,900 19571957 3.25 ns3.25 ns

SCASCA 1010 6,0106,010 601601 1.18 ns1.18 ns

ReciprocalReciprocal 1010 5,1005,100 510510 1.46 ns1.46 ns

ErrorError 5050 17,50017,500 350350

TotalTotal 7474 35,71035,710

What difference would you make if the parents in this analysis were What difference would you make if the parents in this analysis were chosen at random chosen at random [2 points][2 points]..

Use the SCA M.Sq to test the GCA term. In this case the Use the SCA M.Sq to test the GCA term. In this case the GCA is not quite formally significant at the 5% level.GCA is not quite formally significant at the 5% level.

Question 6d.Question 6d.

A Hayman & Jinks analysis was conducted and VA Hayman & Jinks analysis was conducted and Vii and W and W

ii values values

estimated for each parent. The sum of products between Vestimated for each parent. The sum of products between Vii and W and W

ii

(([V[Vi i x Wx W

ii]) was found to be 165; the sum of V]) was found to be 165; the sum of Vii’s was 26; sum of ’s was 26; sum of

WWii’s was 22; sum of squares of V’s was 22; sum of squares of Vii ( ( [V [V

ii22]) was 205; and the sum of ]) was 205; and the sum of

squares of Wsquares of Wii ( ( [W [W

ii22]) was 118. Complete a regression analysis of ]) was 118. Complete a regression analysis of

Vi on to WVi on to Wii.. [4 points]. [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


SP(Vi,Wi) = 165 – [26 x 22]/5 = 50.6 SS(Vi) = 205 – [262]/5 = 69.8 SS(Wi) = 118 – [222]/5 = 21.2

b1 = 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

b0 = 4.4 – 0.72 x 5.2 = 0.67

tt3df3df = [1-0.72]/0.39 = 0.71 ns = [1-0.72]/0.39 = 0.71 ns


What can be determined from this analysis regarding the What can be determined from this analysis regarding the adequacy of the additive/dominance model and the adequacy of the additive/dominance model and the importance of additive genetic variance (A) compared to importance of additive genetic variance (A) compared to dominant genetic variance (D).dominant genetic variance (D). [4 points].[4 points].

The regression slope (bThe regression slope (b11) is not significantly different from ) is not significantly different from

one, therefore, the additive/dominance model is adequate to one, therefore, the additive/dominance model is adequate to explain the variation observed for yield in the diallel. The explain the variation observed for yield in the diallel. The intercept (bintercept (b00) is greater than zero so A is greater than D. ) is greater than zero so A is greater than D.

However, bHowever, b00 is almost equal to zero so A = D. is almost equal to zero so A = D.

Question 7.Question 7.A crossing design involving two homozygous pea cultivars is A crossing design involving two homozygous pea cultivars is carried out and both parents are grown in a properly designed carried out and both parents are grown in a properly designed field experiment with the Ffield experiment with the F

22, B, B11 and B and B

22 families. Given the families. Given the

following standard deviations for both parents (Pfollowing standard deviations for both parents (P11 and P and P

22), the F), the F 2 2, ,

and both backcross progeny (Band both backcross progeny (B11 and B and B

22), determine the broad-), determine the broad-

sense heritability and narrow-sense heritability for seed size in sense heritability and narrow-sense heritability for seed size in dry pea dry pea [10 points].[10 points].

Family Standard Deviation

PP11 3.5213.521

PP22 3.3173.317

FF22 6.0086.008

BB11 5.4505.450

BB22 5.1575.157


Family Standard Deviation

PP11 3.5213.521

PP22 3.3173.317

FF22 6.0086.008

BB11 5.4505.450

BB22 5.1575.157

VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6


hh22b b = = Genetic varianceGenetic variance

Total varianceTotal variance

hh22b b = = 36.1 – 11.736.1 – 11.7

36.136.1

E = [VPE = [VP11+VP+VP22]/2 = 11.7]/2 = 11.7

hh22b b = 0.67= 0.67

VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6


D = 4[V(B1)+V(B2)-V(F2)-E]

4[29.7+26.6-36.1-11.7] = 8.5

A = 2[V(FA = 2[V(F22)-¼D-E] = 2[36.1-2.1-11.7] = 22.3)-¼D-E] = 2[36.1-2.1-11.7] = 22.3

hh22nn = ½A/V(F = ½A/V(F22) = 11.15/36.1 = 0.31) = 11.15/36.1 = 0.31

VP1=12.4; VP2=11.0; VF2=36.1; VB1=29.7; VB2=26.6

E = [VPE = [VP11+VP+VP22]/2 = 11.7]/2 = 11.7

Assuming an additive/dominance mode of Assuming an additive/dominance mode of inheritance for a polygenic trait, list expected inheritance for a polygenic trait, list expected values for Pvalues for P

11, P, P22, and F, and F

11 in terms of m, [a] and [d]. in terms of m, [a] and [d].

[3 points][3 points]


PP11 = = m + am + a

PP22 = = m – am – a

FF11 = = m + dm + d

From these expectations, what would be the From these expectations, what would be the expected values for Fexpected values for F

22, B, B11 and B and B

22 based on m, [a] based on m, [a]

and [d]. and [d]. [3 points][3 points]


FF22 = = m + ½dm + ½d

BB11 = = m + ½a + ½dm + ½a + ½d

BB22 = = m – ½a + ½dm – ½a + ½d

Question 8c.Question 8c.

From a properly designed field trial that From a properly designed field trial that included Pincluded P

11, P, P22 and F and F

11 families, the following families, the following

yield estimates were obtained. yield estimates were obtained.

BB11 = 42.0 lb/a; = 42.0 lb/a; BB22 = 26.0 lb/a; = 26.0 lb/a; FF11 = 38.5lb/a = 38.5lb/a

From these family means, estimate the expected From these family means, estimate the expected value of Pvalue of P

11, P, P22 and F and F

22, based on the , based on the

additive/dominance model of inheritance additive/dominance model of inheritance [6 points][6 points]..

PP11 == m + [a]m + [a]

PP22 == m – [a]m – [a]

FF11 == m + [d]m + [d]

FF22 == m + ½ [d]m + ½ [d]

BB11 == m + ½ [a] + ½ [d]m + ½ [a] + ½ [d]

BB22 == m – ½ [a] + ½ [d] m – ½ [a] + ½ [d]


Question 8cQuestion 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


B1 – B2 = 42.0 – 26.0 = 36.0 = 16 = a B1 + B2 – F1 = 29.5 = m

F1 – m = 9.0 = d

PP11 = = m + a = m + a = 29.5 + 16.0 = 45.5 29.5 + 16.0 = 45.5

PP22 = = m – am – a = 29.5 – 16.0 = 13.5 = 29.5 – 16.0 = 13.5

FF22 = = m + ½dm + ½d = 29.5 – 4.5 = 25.0 = 29.5 – 4.5 = 25.0

Quantitative Genetics ModelsQuantitative Genetics Models

PP22 mm 38.5 P 38.5 P11

26.026.0 42.042.0

13.513.5 29.5 38.5 29.5 38.5 45.545.5

26.026.0 34.034.0 42.042.0

Bonus QuestionBonus Question

It is important in quantative genetics to know whether It is important in quantative genetics to know whether the additive/dominance model based on the additive/dominance model based on mm, , [a][a], and , and [d] [d] is appropriate to explain the variation observed in this is appropriate to explain the variation observed in this study. Given that you have available progeny means study. Given that you have available progeny means and variances from both parents (Pand variances from both parents (P

11 and P and P22), the B), the B

11, B, B22, ,

and Fand F33 families. Devise a suitable scaling test (hint as families. Devise a suitable scaling test (hint as

yet we have not talked about this one) involving these yet we have not talked about this one) involving these five families. five families. [10 bonus points].[10 bonus points].

PP11= = m + [a]m + [a]; P; P22= = m – [a]m – [a]; F; F33 = = m + ¼ [d]m + ¼ [d]; B; B11= =

m + ½ [a] + ½ [d]m + ½ [a] + ½ [d]; B; B2 2 = = m – ½ [a] + ½ [d] m – ½ [a] + ½ [d]

Bonus QuestionBonus Question

4F4F33 = 4 = 4m+dm+d

BB11+B+B22+P+P11+P+P22= 4= 4m +dm +d

4F3 – B1 – B2 – P1 – P2 0

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Moscow March 10, 2000Moscow March 10, 2000

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Genetics and Genetic Prediction in Plant Breeding