117451 Animal Breeding and Improvement

117451 /137451 Animal Breeding and Improvement (for Veterinary Medicine). ..

: Overview of Animal Breeding ... ? Animal Breeding : 1) (Selection)Phenotypic selection ()Breeding value selection ()2) (Mating)Inbreeding()Cross breeding(/)

Basic genetic model P = G + E P = Phenotype G = Genotype E = EnvironmentG = A + D + I

A = Additive gene D = Dominant gene I = Epitasis gene()()() P = G + E G = A + D + I

A = Additive gene D = Dominant gene I = Epitasis geneE = Ep + Et

Ep = Permanent evi. () Et = Temporary evi. ()Related Area in Animal BreedingAnimal BreedingClassical geneticMathematicPopulation geneticMolecular biologyStatistical genetic - Mendelian genetic - study of gene actions - probability - genetic engineering - gene transfer - gene marker - gene frequencyEstimation - - (Principle of genetics)Gene and ChromosomeDNA structureRegulation of gene expressionCell division

Gene and Chromosome DNA structureRegulation of gene expressionCell division

(cell and cell components)

(Gene and chromosome)

SpeciesChromosomeHuman46Cattle60Pig38Goat54House64Rabbit44Chicken78Dog78Cat38

(Structure of DNA)

(double-helix model) (anti-parallel) H-bond

1) 2-deoxyribose 5 2 (OH-group)

2) (nitrogenous base) - purine adenine, (A) guanine (G) - pyrimidine thymine (T) cytosine (C) A T, C G A+G T+C

3) (phosphate group) (gene structure) exon1exon2exon3exon4intron1intron2intron3terminatorpromoter Gene Subunit enzyme (Protein)mRNAProteinexon1exon2exon3exon4intron1intron2intron3mRNA5

33

5tRNA, rRNA155

33

5Transcription ()Translation ()m-RNAProtein

?

WHERE / WHEN / HOW MUCH18Functions:Storing informationFunctions:Copying information Functions:Transmitting information?: DNA : DNA : (DNA) DNA (the role of DNA) 2

( cell cycle) - (2n)

- (n)

(cell division)

1. ( interphase) 2. ( prophase) 3. ( metaphase)

4. ( anaphase) 5. ( telophase) ( daughter chromosome)

Mendelian genetics ...

Sex chromosomeAutosome

(genome) - human genome project- pig genome project

Locus

Homologous chromosomeLocusAllele AAllele a allele

LocusAAallele A

Allele aaa 1. Phenotype

2. Genotype AA, Aa,aa

3. Homologous ChromosomeBasic term :Genotype PhenotypeAA (homozygote dominance)

Aa (heterozygote)

aa (homozygote recessive) 4. Dominant gene heterozygous A5. Recessive gene heterozygous aMendelian Genetics Greger Johann Mendel

Law of segregation of gene () Law of independent assortment (/)

Law of segregation of gene () (gamete)

1

F1 F1 F2

- Phenotypic ratio 3 : 1 1Dominant Vs Recessive dominant recessive R r dominant genehomozygous genotype (RR, rr) heterozygous genotype (Rr)

35Monohybrid cross heterozygous 1 (complete dominant) dominant : recessive 3:1

36 dominant recessive B b recessive gene i, ii, iii..

iiiiii37

BBbbBb....Monohybrid cross 1 B b BB,Bb bb

parentBB x bb gamete B B b b F1 Bb

B>b 1 phenptype F1 (F1 ) Bb x BbGameteB b B b

F2BBBb Bb bb

Genotypic ratio 1BB :2Bb:1bbPhenotypic ratio 3():1 () 1

6 basic cross 1. AA xAAAA 2. AA xaaAa 3. aa xaaaa 4. Aa xAA Aa : AA 5. Aa xaa Aa : aa 6. Aa xAa AA : Aa : aa

2 2 : (Law of independent assortment) 2 2 (2 ) 1

Dihybrid cross 2 1 B = b =

2 P = p =

parentBBPP xbbppgamete BP bpF1 BbPp B>b P>p BbPp x BbPpParentPpBb x PpBb

Gamete ? gamete punnet (punnets square)

gamete punnets square (BbPp)

1 2 BbPpBPbPBpbp gamete BP, bP, Bp, bpBbPp x BbPp genotype Punnets squareGamete Gamete PBpBPbpbPBpBPbpbPPBBPpBBPPBbPpBbPpBBppBBPpBbppBbPPBbPpBbPPBbPpbbPpBbppBbPpbbppbb genotype ?2. Fork line PpBb x PpBb

Pp x Pp

1/4PP 2/4Pp 1/4ppBb x Bb

1/4BB 2/4Bb 1/4bb PP

2/4 Pp

pp BB2/4 Bb bb BB2/4 Bb bb BB2/4 Bb bb1/16 PPBB2/16 PPBb1/16 PPbb2/16 PpBB4/16 PpBb2/16 Ppbb1/16 ppBB2/16 ppBb1/16 ppbb1/16 2/16 1/16 2/16 4/16 2/16 1/16 2/16 1/16 3 BbSSHh x bbssHh -- --

1Bb x bb 1Bb : 1bb 2 SS x ss 1Ss 3 Hh x Hh 1HH : 1Hh : hh

BbSSHh x bbssHh-- --Fork lineLocus 1 Locus 2 Locus 3genotype phenotype1Bb 1Ss 1HH1BbSsHH 1 -- 2Hh2BbSsHh 2 -- 1hh 1BbSshh 1 --

1bb1Ss 1HH1bbSsHH 1 -- 2Hh 2bbSsHh 2 -- 1hh 1bbSshh 1 --

3 Hh x Hh 1HH : 1Hh : hh

... ?

? Test cross : (dominant phenotype) homozygous heterozygous

homozygous recessive

Hh x hh

1Hh : hh

heter. (Hh) 2 HH x hh

1Hh

homo. Ex. 10 1- (1/2)n n = ()

1- (1/2)10 = 0.99 99 % = 1- (1/2)n 99 % (0.99) 0.99= 1-(1/2)n (1/2)n = 1-0.99 Log(1/2)n = log0.01nlog(1/2) = log0.01 n = log0.01/log0.5 n = 6.64 Chi-square Test monohybrid 85 15

BBbbBb

Bb

851556 -

57

OEO-E(O-E)2(O-E)2/E8575-101001.331525-1010045.33 (observe)(expect) 3 : 1 4 100 75 : 2558 2 2 degree of freedom 2 2 1 -

99%0.0195%0.0590%0.1059 degree of freedomOEO-E(O-E)2(O-E)2/E8575-101001.331525-1010045.33 : 3 : 1 (hypothesis) 2 degree of freedom = - 1 = 2-1 = 160-

( ) 95% degree of freedom = 1 - 3.84161OEO-E(O-E)2(O-E)2/E8575-101001.331525-1010041001005.33- = 5.33- = 3.841

- > - 3 :1 62Chi-square Test (2 loci) dihybrid - 270 - 100 - 90 - 30

BBCCbbccBbCc

BbCc

270100

9030 ?63What we need? degree of freedom -

dihybrid 800 - 395 185 - 165 - 55

64 (B) (b) (C) (c)

Bb X Bb1BB : 2Bb : 1BB3 : 1 Cc X Cc1CC : 2Cc : 1cc3 : 1 2 3

1 3

1 1 23

1 9 -

3 -3 -

1 - 9:3:3:1

OEO-E(O-E)2(O-E)2/E-395450-5530256.7-1851503512258.2-165150152251.5-55505250.580080016.9 (observe)(expect) 9:3:3:1 16 800 1 800/16 = 50 450 : 150 : 150 : 5067 degree of freedom - : - : - : - 9 : 3 : 3 : 1 (hypothesis) 4 degree of freedom = - 1 = 4-1 = 368-

( ) 95% degree of freedom = 3 - 7.81869- = 16.9- = 7.818

- - 9:3:3:1OEO-E(O-E)2(O-E)2/E-395450-5530256.7-1851503512258.2-165150152251.5-55505250.580080016.970- 9:3:3:1 Back to dihybrid cross

BBCCbbccBbCc

BbCc

395185

16555 dihybrid 800 - 395 185 - 165 - 55

72- = 16.9- = 7.818

- - 9:3:3:1OEO-E(O-E)2(O-E)2/E-395450-5530256.7-1851503512258.2-165150152251.5-55505250.580080016.973 9:3:3:1 (linkage) ?

B C

B C genetic distance

BBCCbbccBbCc

BbCc

395185

16555 dihybrid 800 - 395 185 - 165 - 55

- - Genetic Distance = (185+165)/800= 0.3125 = 31.25%= 31.25 centriMorgan77Non-mendelian genetics ( phenotype 3 :1 9:3:3:1)

1. Incomplete dominant 2. Epistasis3. Multiple allele4. Poly gene5. Pleiotrophy 6. Lethal gene7. Modify gene1. (Incomplete dominant ) short horn R r R>r

Parent RR x rr F1 Rr

F1

Incomplete dominant : genotypic ration = phenotypic ratio2. Epistasis ( no allelic) .. 2 .

BbRrBbccRecessive epitasis homo. recessive homo. Recessive

AbBBDominant epitasis homo. dominant

e.g. White leghorn and White wyandotte chickens

Recessive epistasis

3. Multiple allele : locus allele 2 ABO

IAIA, I AIO = Type AIBIB, IBIO = Type BIAIB = Type ABIOIO = Type OIA, I B, I OIA, I B , I OExample: gene "Cagouti (C) > chinchilla (silver) cch > Himalayan (ch) > albino (ca)

agouti (C) > chinchilla (silver) cch > Himalayan (ch) > albino (ca)

4. Pleiotropic gene re re re , : 1 1

5. Lethal gene : 3 1. (prenatal dead)2. (still birth)3. ( postnatal dead)

lethal gene

Dominant lethal gene homo. Dominant

Recessive lethal gene homo. Recessive

Semi lethal gene homo. Dominant homo. Recessive LETHAL ALLELES :yellow coat color in mice (Y)

Yy x Yy

1YY : 2Yy : 1yy : : YYyyDominant lethal gene

(Manx cat) Tt x Tt

1TT : 2Tt : 1tt : : TT : TTtt Manx allele in cats (T)6. Modifier genes:

- modify gene

7. Poly gene : (poly = many)

P : AABB x aabb

F1 : AaBb (iner se mating)F2 :AaBb x AaBb

AABB = 1/16 blackAABb = 2/16 darkAAbb = 1/16 mediumAaBB= 2/16 darkAaBb= 4/16 mediumAabb= 2/16 lightaaBB= 1/16mediumaaBb= 2/16 lightaabb= 1/16 whiteMDLBW1/161/164/164/166/16(Inheritance in relation to sex)98 ...Sex- linked geneSex- limited geneSex- influenced gene

Sex-linked gene: X (X-chromosome) Z

Mammal XAXa XAYAvian ZAW ZAZA

(colour sexing)

P1 Rhod Island Red X Barred Plymouth Rock () (-)genotype ZbZb ZBW

gamete Zb Zb ZB W F1 ZbZB , ZbZB : ZbW , ZbW () ()

: 3-5 Ex. Sex-linked (feather sexing)

P1 Fast feather X Slow feather () ()genotype ZkZk ZKW

F1 ZkZK : ZkW () ()

http://www.ansi.okstate.edu/course/3443/study/ReproTech/Feathersex/index.htm Sex-influenced: (autosome)

1 genotype HI H HI > H Dorset horn Suffolk

P1 Dorset horn X Suffolkphenotype () ()

genotype HlHl HH

F1 HlH

(HlH) : (HlH) : Hl H H HI Sex-limited : (autosome) * 1 genotype * - - - GenotypePhenotypeFFFfff F f genotype (phenotype)

- sebright bantam

Genotypes FF

Genotypes FF, Ff ff genotype ?

Population genetic

; f(gene) ; f(genotype) (Evolution)

Evolution f(gene) f(genotype) Evolution

gene genotype co-dominant incomplete dominant R = , r = R>r Rr x Rr genotype 3 Phenotype Genotype RR 800 Rr 150 rr 50 Phenotype Genotype RR 800 Rr 150 rr 50 1000D = f(RR) = 800/1000 = 0.8 H = f(Rr) = 150/1000 = 0.15R = f(rr) = 50/1000 = 0.05 f(genotype) = D+H+R = 1 (0.8+0.15+0.05) D = f(RR) = 800/1000 = 0.8 H = f(Rr) = 150/1000 = 0.15R = f(rr) = 50/1000 = 0.05 D = Homozygous dominance H = Heterozygous R = Homozygous recessive f(R) = D+ H = p

f(r) = R + H = qGene frequency =

Hardy-Weinberg law ()Gene equilibrium ()

(random mating) (selection), (mutation), (migration) (genetic drift) genotype (equilibrium) Hardy-Weinberg genotype D = p2 H = 2pq R = q2 p q R (p) = 0.875 r (q) = 0.125 genotype F1 HW D= f(RR) = p2 = (0.875)2 = 0.766

H =f(Rr) = 2pq = (0.875)(0.125) = 0.218

R = f(rr) = q2 = (0.125)2 = 0.016f(R) = 0.766 + (0.218) = 0.875

f(r) = 0.016 + (0.218) = 0.125 F1 f(gene) F1 f(gene) D = d =

D>d 9 625 genotype gene (complete dominant) Population genetics 2: R r (co-dominant alleles) Rr 250 , 450 200 genotype genotype alleleRrRRRrrr250450200500450--4504009001,800genotype alleleRrRRRrrr250450200500450--4504009001,800 genotype:f(RR) = 250/900 = 0.278f(Rr) = 450/900 = 0.500f(rr) = 200/900 = 0.222 gene:f(R) = (500+450)/1,800 = 0.528f(r) = (400+450)/1,800 = 0.472

:f(RR) + f(Rr) + f(rr) = 1.0f(R) + f(r) = 1.0

genotype gene 1 genotype f(RR) f(Rr) f(rr) f(rr) = 1 f(RR) f(Rr)

1 f(R) f(r) f(r) = 1 f(R) ...- (2-test)

goodness of fits - (equilibrium population) p = A q = a p + q = 1 genotype binomial expansion (p+q)2 = (p+q)*(p+q) = p2+2pq+q2 = 1Ex. f(R) = 0.528 f(r) = 0.472 genotype

f(RR) = p2 = (0.528)2 = 0.279 f(Rr) = 2pq = 2(0.528)(0.472) = 0.498 f(rr) = r2 = (0.472)2 = 0.223

gene

f(R) = f(RR) + f(Rr) = 0.279 + (0.498) = 0.528 f(r) = f(rr) + f(Rr) = 0.223 + (0.498) = 0.472

Hardy-Weinberg genotype

d = D =

D>d 9 625 genotype gene (complete dominant) genotypeDD, DdDd6169625

d = D = D>d 9 625 genotype gene f(D) = p f(d) = q

f(dd) = 9/625 = 0.0144

Hardy-Weinberg f(dd) = q2f(d) = = = 0.12 p+q = 1f(D) = 1-0.12 = 0.88

0.12 0.88 DD Dd

f(DD) = p2 = (0.88)2 = 0.7744 f(Dd) = 2pq = 2(0.88)(0.12) = 0.2112

R r (co-dominant alleles) Rr 250 , 450 200 Rr : : 1:2:1 R r

Ho : : : = 1:2:1HA : : : 1:2:1

- (2-test)

2value= O = (observed number) E = (expected number) 2value < 2 (df)

Phenotype1212504502002254502254900900:1. = 3+1= 4= 900/4= 2252. = 1x225= 225= 2x225= 4502value = =

2(2) 0.05 df 2 5.99 2value < 2 (2) Ho (null hypothesis)

PhenotypeOf(genotype) EO-E(O-E)2(O-E)2/E2500.278250.2-0.20.040.00016 4500.5450000 2000.222199.80.20.040.0002 90019000.0004 Expected p2(N), 2pq(N), q2(N) RR, Rr rr H-W = p2(N) = (0.278)(900) = 250.2 = 2pq(N) = (0.5)(900) = 450 = q2(N) = (0.222)(900) = 199.8N = 2value < 2 (1) Ho (null hypothesis) H-W 2(1) 0.05 df 1 3.84 PhenotypeOf(genotype) EO-E(O-E)2(O-E)2/E2500.278250.2-0.20.040.00016 4500.5450000 2000.222199.80.20.040.0002 90019000.0004 : R r df = 1 Hardy-Weinberg gene force 2 Systematic processDispersive processNon-random matingMigrationMutationSelectionGenetic driftPositive assortive matingNegative assortive matingNon-recurrent mutationRecurrent mutationCullingFavouring (non-random mating) ..... positive assortive mating negative assortive mating genotype

147 (non-random mating) positive assortive mating

(non-random mating) negative assortive mating

(migration) f(a) = q1N= n1f(a)= q0N= n0f(a) = q1N = nm (migration)

/ / (, ) q0= q0 + m(q1- q0)q0= q0 = q1 = m = = migration rate

n0= nm=

Ex. 8,000 (q0) 0.2 (q1) 0.6 2,000 q0= q0 + m(q1-q0)= = 0.2 + 0.08= 0.28

q= q0 - q0= 0.28 - 0.2= 0.08

/

(mutation) DNA (A, T, C G) 10-4 10-5

(mutation)

(mutation) 2 1. (non-recurrent mutation) 2. (recurrent mutation) (mutation) (non-recurrent mutation) (one- way mutation) 0 1 :R r :p q f(R) , f(r) p0 , q0 u 1 f(R) = p1= p0 - up0= p0(1-u)f(r)= q1= q0 + up0 1 p1 p >> q f(R) up0 f(r) up0 (mutation) q q= q1 - q0= (q0 + up0) - q0= up0 (q0) (qt) (t) (q) q= up ; t = = u(1-q)

(mutation) (non-recurrent mutation) ( one way mutation)

0 1

(mutation)Ex. B (p) 0.8 (u) B b 10-4 B 0.1t = t = = 1,335.31

(mutation) (recurrent mutation) (two-way mutation) :B b

:pq u v 1 f(B) = p1= p0 - up0 + vq0 = p0(1-u) + vq0f(b)= q1= q0 + up0 - vq0VU (mutation) up0 > vq0 f(B) up0 < vq0 f(B) up0 = vq0 f(B)

(mutation)vqE= upEvqE= u(1-qE)vqE = u - uqEvqE + uqE= u(v+u)qE= u qE=

(mutation)

Ex. q0 = 0.2 (B->b) u 4.2 x 10-5 (b->B) v 2.1 x 10-5 1 q1 = q0 + up0 vq0

= 0.2 + (4.2x10-5)(0.8) - (2.1x10-5)(0.2) = 0.2+ (0.294x10-4)

qE =

= = 0.6667

Ex. b 10%

2 (selection)Natural selectionResistance to antibacterial soapGeneration 1: 1.00 not resistant 0.00 resistantNatural selectionGeneration 1: 1.00 not resistant 0.00 resistantResistance to antibacterial soapNatural selectionResistance to antibacterial soapmutation!Generation 1: 1.00 not resistant 0.00 resistantGeneration 2: 0.96 not resistant 0.04 resistantNatural selectionResistance to antibacterial soapGeneration 1: 1.00 not resistant 0.00 resistantGeneration 2: 0.96 not resistant 0.04 resistantGeneration 3: 0.76 not resistant 0.24 resistantNatural selectionResistance to antibacterial soapGeneration 1: 1.00 not resistant 0.00 resistantGeneration 2: 0.96 not resistant 0.04 resistantGeneration 3: 0.76 not resistant 0.24 resistantGeneration 4: 0.12 not resistant 0.88 resistant (selection) fitness adaptive value (selection)

1. recessive 2. dominant 3. recessive dominant (selection) (selection)1. recessive recessive (aa) dominant (AA, Aa) Genotyperelative fitnessAAAaaap022p0q0q02111-sp022p0q0(1-s)q02p02+2p0q0+q02p02+2p0q0+(1-s)q02 (selection)1. recessive

genotype genotype (selection)1. recessive genotype

(selection)1. recessive

(selection)Ex. aa 0.5 f(AA), f(Aa), f(aa) 0.36, 0.48, 0.16 1

(selection)

recessive (q0) (qt) (t)

(selection)Ex. homozygous recessive 20%

2. dominant dominant (AA, Aa) recessive (aa) Genotyperelative fitnessAAAaaap022p0q0q021-s1-s1(1-s)p02(1-s)2p0q0q02p02+2p0q0+q02(1-s)p02+(1-s)2p0q0+q022. dominant genotype

2. dominant

Ex. A 0.5 f(AA), f(Aa), f(aa) 0.36, 0.48, 0.16 1

dominant (q0) (qt) (t)

Ex. (feather shank) dominant 10% 1%

3. dominant recessive homozygous recessive dominant (aa, AA) heterozygous genotype (Aa) Genotyperelative fitnessAAAaaap022p0q0q021-s111-s2(1-s1)p022p0q0(1-s2)q02p02+2p0q0+q02(1-s1)p02+2p0q0+(1-s2)q023. dominant recessive genotype

3. dominant recessive

Ex. AA 0.5 aa 0.2 f(AA), f(Aa), f(aa) 0.36, 0.48, 0.16 1

Ex. (chestnut) genotype AA, (creamello) genotype aa, (palomino) genotype Aa palomino

(chance) 0 1

Genetic drift8 RR8 rrBefore:After:2 RR6 rr0.50 R0.50 r0.25 R0.75 r

/ (genetic variation) Why is genetic variation important?potential for change in genetic structure adaptation to environmental change- conservation divergence of populations- biodiversity

Why is genetic variation important?variationno variationEXTINCTION!!globalwarmingsurvivalWhy is genetic variation important?variationno variationWhy is genetic variation important?variationno variationdivergenceNO DIVERGENCE!!Quantitative genetics Vs Mendelian genetics: Mendelian Genetics 1-2

: Quantitative Genetics (polygenes) (genome) (variance)209 , , ADG, FCR, , , , (selection index, breeding program)

210P = PhenotypeG = Genetic or GenotypeE = EnvironmentGxE = Genetic - Environment interactionP = G + E + (GxE)P = G + E + (G x E)P = G = E = = ////G x E = G E212P = G + EG = A + D + Iadditive genedominant effectepistasis effectE = Ep +Etpermanent environmenttemporary environmentP = A + D + I + Ep + Et213Heritability VP = VG + VE

214 0 -1 - -

2152P = 2A+2D+2I+2Ep+2Et2A2P= (heritability, h2)2A2P= narrow sense heritability, h2N2G2P= board sense heritability, h2B12Broad sense VS Narrow sense VG VA + VD + VIVA (additive gene)VD (dominance)VI (epistatis)

218 = > 0.35 = 0.20 - 0.35 = < 0.20, selectionprogramheterosiscross breednoyesmanagementprogramno

VGVEVPVGVEVP80%20%90%10%A. sib analysisa.1 half-sib analysisa.2 full-sib analysisB. regression between parent and offspring 1. 2. (ANOVA) 3. Sib analysis1. ( 5 )A.1 half-sib analysis

sire 1sire 2sire 3

half-sib analysis

sire 1sire 2sire 32sire2offspring/sire = 2WYij = + sirei + offspring/sirej/i

2total = 2sire + 2W = 2PCOVhalf sib = 2A = 2sire 2A = 42sire

ANOVASOVdfSSMSE(MS)Sires-1SSSMSS2W+k2SOffspring/siren-sSSWMSW2WTotaln-1SSTYij = + sirei + offspring/sirej/is = k = SS = Sum squareMS = Mean square2S = 2W = ANOVASOVdfSSMSE(MS)Sires-1SSSMSS2W+k2SOffspring/siren-sSSWMSW2WTotaln-1SSTYij = + sirei + offspring/sirej/i2W = MSW2sire = (MSS MSW) k ANOVASOVdfSSMSE(MS)Sires-1SSSMSS2W+k2SOffspring/siren-sSSWMSW2WTotaln-1SSTYij = + sirei + offspring/sirej/i2W = 9002sire = (1,200 900) = 100 3 k = 32992028,80010,80018,0001,200900

A. 2 full-sib analysisYij = + sirei + dam/sirej/i+ offspring/dam/sirei/j/k

sire 1sire 22SireDam 1Dam 22Dam2Wfull-sib analysis

sire 1sire 22SireDam 1Dam 22Dam2W

42sire = 2A2(2sire+2Dam) = 2A+ 2D42Dam = 2A+2D

2A2Ph2 = _____ANOVASOVdfSSMSE(MS)Sires-1SSSMSS2W+k12D+k22SDam/sired-sSSDMSD2W+k12DOffspring/Dam/siren-dSSWMSW2WTotaln-1SST2W = MSW2sire = (MSS MSD) k2 k2 = Yij = + sirei + dam/sirej/i+ offspring/dam/sirei/j/kk1 = 2Dam = (MSD MSW) k1 ANOVASOVdfSSMSE(MS)Sires-1SSSMSS2W+k12D+k22SDam/sired-sSSDMSD2W+k12DOffspring/Dam/siren-dSSWMSW2WTotaln-1SSTYij = + sirei + dam/sirej/i+ offspring/dam/sirei/j/k799103,1276574907349601,980332W = 332sire = (73 49) 8 k2 = 8k1 = 42Dam = (49 33) 4 2Sire = 32Dam = 42A2Ph2 = _____

2W = 332Sire = 32Dam = 4B. regression between parent and offspring h2 1. 2. bXY = COV (XY) V (X)

Sire Dam Offspring Offspring One parents-offspring regressionMidparents-offspring regression

Offspring Sire Dam One parents-offspring regression

2A = bb = __________COV(,)V()Regression coefficient b = __________COV(,)V()= _________ 2A 2P= h2 = 2b h2 COV(,) = 100V() = 1000b = ______________b = 0.1h2 = 2b = 2(0.1) = 0.2

Dam Offspring Midparents-offspring regression

Regression coefficient b = __________COV(,,)V(,)= _________ 2A 2P= h2 b = h2Regression coefficient

2G2EP2P = 2A+2D+2I+2Ep+2Et2G+2Ep2P= (repeatability, t) 2 - lactation 1, 2, 3 2,400 2,560 2,610

2. -

2W2e2W = 2G + 2EP 2e = 2ET 2W2W+2et = __________ANOVASOVdfSSMSE(MS)between animala-1SSWMSW2e+k2Wwithin animaln-aSSEMSE2eTotaln-1SST2e = MSe2W = (MSW MSe) k k = SOVdfSSMSE(MS)animala-1SSWMSW2e+k2Werrorn-aSSeMSe2eTotaln-1SSTANOVA2e = 4502W = (2,200 450) = 583 3 k = 32992028,80019,8009,0002,200450

0 h2 10 t 1t = > 0.4 0.20 < t < 0.35t < 0.20 Genetic correlation (rG) (rG)1. Pleiotropy 1 2. genetic linkage

ADG FCR0+

ADG FCR0-XY 2X 2Yr = ____________-1 rG +1

Pleiotropic Effects
Pleiotropic gene - a gene that affects more than one phenotype

In this example the gene that causes yellowing of the coat also affects viability and is termed a pleiotropic gene.