Biometrical Genetics Shaun Purcell Twin Workshop, March 2004

Single locus model 1.Genetic effects → variance components 2.Genetic effects → familial covariances 3.Variance components → familial covariances

P T1 AD E P T2 ADE [0.25/1] [0.5/1] eadeda ADE Model for twin data

POPULATION MODEL –Allele & genotype frequencies TRANSMISSION MODEL –Mendelian segregation –Identity by descent & genetic relatedness PHENOTYPE MODEL –Biometrical model of quantitative traits –Additive & dominance components Some Components of a Genetic Theory

POPULATION MODEL –Allele & genotype frequencies TRANSMISSION MODEL –Mendelian segregation –Identity by descent & genetic relatedness PHENOTYPE MODEL –Biometrical model of quantitative traits –Additive & dominance components Some Components of a Genetic Theory G G G G G G G G Time G G G G G G G G G G G G G G GG PP

MENDELIAN GENETICS

Mendel’s Experiments Pure Lines AAaa F1Aa AAAa aa 3:1 Segregation Ratio Intercross

Mendel’s Experiments Aa aa Aaaa F1Pure line Back cross 1:1 Segregation ratio

Mendel’s Experiments Pure Lines AAaa F1Aa AA Aa aa 3:1 Segregation Ratio Intercross

Mendel’s Experiments Aa aa Aaaa F1Pure line Back cross 1:1 Segregation ratio

Mendel’s Law of Segregation A3A3 A4A4 Maternal ½½ A1A1 A2A2 Paternal ½ ½ ¼ ¼ ¼ ¼ A1A1 A2A2 A4A4 A4A4 A3A3 A3A3 A1A1 A2A2 Meiosis/Segregation Gametes

PHENOTYPE MODEL

Classical Mendelian Traits Dominant trait D, absence R –AA, Aa  D; aa  R Recessive trait R, absence D –AA, Aa  D; aa  R Co-dominant trait, X, Y, Z –AA  X; Aa  Y; aa  Z

Dominant Mendelian inheritance Dd Maternal ½½ D d Paternal ½ ½ D d d d D D D d

Recessive Mendelian inheritance Dd Maternal ½½ D d Paternal ½ ½ D d d d D D D d

Phenocopies Incomplete penetrance Dominant Mendelian inheritance Dd Maternal ½½ D d Paternal ½ ½ 60% 1% D d d d D D D d

AA Aa aa Quantitative traits

Biometrical Genetic Model m d +a-a Genotypic means AA Aa aa m + a m + d m – a P(X)P(X) X AA Aa aa

POPULATION MODEL

Population Frequencies A single locus, with two alleles –Biallelic / diallelic –Single nucleotide polymorphism, SNP Alleles A and a –Frequency of A is p –Frequency of a is q = 1 – p Every individual inherits two copies –A genotype is the combination of the two alleles –e.g. AA, aa (the homozygotes) or Aa (the heterozygote)

Genotype Frequencies (random mating) Aa Ap 2 pq p aqp q 2 q p q Hardy-Weinberg Equilibrium frequencies P(AA) = p 2 P(Aa) = 2pq P(aa) = q 2

Before we proceed, some basic statistical tools…

Means, Variances and Covariances

Biometrical Model for Single Locus GenotypeAAAaaa Frequencyp 2 2pqq 2 Effect (x)ad-a Residual var  2  2  2 Mean m = p 2 (a) + 2pq(d) + q 2 (-a) = a(p-q) + 2pqd

Biometrical Model for Single Locus GenotypeAAAaaa Frequencyp 2 2pqq 2 (x-m) 2 (a-m) 2 (d-m) 2 (-a-m) 2 Variance = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 = V G (Broad-sense) heritability at this loci = V G / V TOT (Broad-sense) heritability = Σ L V G / V TOT

Additive and dominance effects Additive effects are the main effects of individual alleles: ‘gene-dosage’ –Parents transmit alleles, not genotypes Dominance effects represent an interaction between the two alleles –i.e. if the heterozygote is not midway between the two homozygotes

Practical 1 H:\pshaun\biometric\sgene.exe 1.What determines additive genetic variance? 2.Under what conditions does V D > V A

Some conclusions 1.Additive genetic variance depends on allele frequency p & additive genetic value a as well as dominance deviation d 2.Additive genetic variance typically greater than dominance variance

Average allelic effect Average allelic effect is the deviation of the allelic mean from the population mean, a(p-q)+2pqd Of all the A alleles in the population: –A proportion (p) will be paired with another A –A proportion (q) will be paired with another a AAAaaaAllelic meanAverage effect ad-a Apqpa+qdq(a+d(q-p)) apqqa-pd-p(a+d(q-p))

Average allelic effect Denote the average allelic effects as α α A = q(a+d(q-p)) α a = -p(a+d(q-p)) If only two alleles exist, we can define the average effect of allele substitution α = α A – α a α = (q-(-p))(a+d(q-p)) = (a+d(q-p)) Therefore, α A = qα and α a = -pα

Additive genetic variance The variance of the average allelic effects Freq.Additive effect AAp 2 2α A = 2qα Aa 2pqα A + α a = (q-p)α aaq 2 2α a = -2pα V A = p 2 (2qα) 2 + 2pq((q-p)α) 2 + q 2 (-2pα) 2 = 2pqα 2 = 2pq(a+d(q-p)) 2

Additive genetic variance If there is no dominanceV A = 2pqa 2 If p = q V A = ½a 2

Additive and Dominance Variance aaAaAA m -a a d Total Variance = Regression Variance + Residual Variance = Additive Variance + Dominance Variance

Biometrical Model for Single Locus GenotypeAAAaaa Frequencyp 2 2pqq 2 (x-m) 2 (a-m) 2 (d-m) 2 (-a-m) 2 Variance = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 = 2pq[a+(q-p)d] 2 + (2pqd) 2 V G = V A + V D

VAVA

Allele frequency Additive genetic variance V A Dominance genetic variance V D a d +1 +1

a d AAAaaa Allele frequency V A > V D V A < V D

Cross-Products of Deviations for Pairs of Relatives AAAaaa AA(a-m) 2 Aa(a-m)(d-m)(d-m) 2 aa(a-m)(-a-m)(-a-m)(d-m)(-a-m) 2 The covariance between relatives of a certain class is the weighted average of these cross-products, where each cross-product is weighted by its frequency in that class:

Covariance of MZ Twins AAAaaa AAp 2 Aa02pq aa00q 2 Covariance = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 = 2pq[a+(q-p)d] 2 + (2pqd) 2 = V A + V D

Covariance for Parent-offspring (P-O) AAAaaa AA? Aa?? aa? ? ? Exercise 2 : to calculate frequencies of parent- offspring combinations, in terms of allele frequencies p and q.

Exercise 2 e.g. given an AA father, an AA offspring can come from either AA x AA or AA x Aa parental mating types AA x AA will occur p 2 × p 2 = p 4 and have AA offspring Prob()=1 AA x Aa will occur p 2 × 2pq = 2p 3 q and have AA offspring Prob()=0.5 and have Aa offspring Prob()=0.5 Therefore, P(AA father & AA offspring) = p 4 + p 3 q = p 3 (p+q) = p 3

Covariance for Parent-offspring (P-O) AAAaaa AAp 3 Aa?? aa? ? ? AA offspring from AA parents = p 4 +p 3 q = p 3 (p+q) = p 3

Parental mating types PatMatP(PxM)AAAaaa AA p4p4 p4p4 0? Aa2p 3 qp3qp3qp3qp3q? AAaap2q2p2q2 0p2q2p2q2 ? AaAA2p 3 q??? Aa 4p 2 q 2 ??? Aaaa2pq 3 ??? aaAAp2q2p2q2 ??? aaAa2pq 3 ??? aa q4q4 ???

Covariance for Parent-offspring (P-O) AAAaaa AAp 3 Aap 2 q? aa? ? ? AA offspring from AA parents = p 4 +p 3 q = p 3 (p+q) = p 3 Aa offspring from AA parents = p 3 q+p 2 q 2 = p 2 q(p+q) = p 2 q

Parental mating types PatMatP(PxM)AAAaaa AA p4p4 p4p4 Aa2p 3 qp3qp3qp3qp3q AAaap2q2p2q2 p2q2p2q2 AaAA2p 3 qp3qp3qp3qp3q Aa 4p 2 q 2 p2q2p2q2 2p 2 q 2 p2q2p2q2 Aaaa2pq 3 pq 3 aaAAp2q2p2q2 p2q2p2q2 aaAa2pq 3 pq 3 aa q4q4 q4q4

Covariance for Parent-offspring (P-O) AAAaaa AAp 3 Aap 2 qpq aa0 pq 2 q 3 Covariance = (a-m) 2 p 3 + (d-m) 2 pq + (-a-m) 2 q 3 + (a-m)(d-m)2p 2 q + (-a-m)(d-m)2pq 2 = pq[a+(q-p)d] 2 = V A / 2

Covariance for Unrelated Pairs (U) AAAaaa AAp 4 Aa2p 3 q4p 2 q 2 aap 2 q 2 2pq 3 q 4 Covariance = (a-m) 2 p 4 + (d-m) 2 4p 2 q 2 + (-a-m) 2 q 4 + (a-m)(d-m)4p 3 q + (-a-m)(d-m)4pq + (a-m)(-a-m)2p 2 q 2 = 0 ?

IDENTITY BY DESCENT

Identity by Descent (IBD) Two alleles are IBD if they are descended from and replicates of the same recent ancestral allele 7 AaAa AaAa AaAa A aa AA Aa

IBS  IBD A1A2A1A2 A1A3A1A3 A1A2A1A2 A1A3A1A3 IBS = 1 IBD = 0 IBS=Identity by State

IBD: MZ Twins ABABCDCD ACAC MZ twins always share 2 alleles IBD ACAC

IBD: Parent-Offspring ABABCDCD ACAC If the parents are unrelated, then parent-offspring pairs always share 1 allele IBD

IBD: Unrelated individuals ABABCDCD If two individuals are unrelated, they always share 0 allele IBD

IBD: Half Sibs ABABCDCD ACAC EE CE/DE IBD SharingProbability 0½ 1½

IBD: Full Sibs IBD of paternal alleles IBD of maternal alleles

IBD and Correlation IBD  perfect correlation of allelic effect Non IBD  zero correlation of allelic effect # alleles IBD Correlation at a locusAllelicDom. MZ211 P-O10.50 U000

Covariance between relatives Partition of variance  Partition of covariance Overall covariance = sum of covariances of all components Covariance of component between relatives = correlation of component  variance due to component

Average correlation in QTL effects MZ twinsP(IBD 0) = 0 P(IBD 1) = 0 P(IBD 2) = 1 Average correlation Additive component = 0×0 + 0×½ + 1×1 = 1 Dominance component = 0×0 + 0×0 + 1×1 = 1

Average correlation in QTL effects P-OP(IBD 0) = 0 P(IBD 1) = 1 P(IBD 2) = 0 Average correlation Additive component = 0×0 + 1×½ + 0×1 = ½ Dominance component = 0×0 + 1×0 + 0×1 = 0

Average correlation in QTL effects UnrelatedP(IBD 0) = 1 P(IBD 1) = 0 P(IBD 2) = 0 Average correlation Additive component = 1×0 + 0×½ + 0×1 = 0 Dominance component = 1×0 + 0×0 + 0×1 = 0

Mendel’s Law of Segregation A3A3 A4A4 Maternal ½½ A1A1 A2A2 A1A1 A2A2 Paternal ½ ½ ¼ ¼ ¼ ¼ A4A4 A4A4 A3A3 A3A3 A1A1 A2A2

IBD sharing for two sibs A 1 A 3 A 1 A 4 A 2 A 3 A 2 A 4 A1A3A1A4 A2A3A2A4A1A3A1A4 A2A3A2A4 A 1 A 3 A 2 A 4 A 1 A 4 A 2 A 3 A 2 A 3 A 1 A 4 A 2 A 4 A 1 A 3 A 1 A 3 A 1 A 4 A 1 A 3 A 2 A 3 A 1 A 4 A 1 A 3 A 1 A 4 A 2 A 4 A 2 A 3 A 1 A 3 A 2 A 3 A 2 A 4 A 2 A 4 A 1 A 4 A 2 A 4 A 2 A 3 A 1 A 3 A 1 A 4 A 2 A 3 A 2 A 4 Sib 2 Sib1Sib1

IBD sharing for two sibs A 1 A 3 A 1 A 4 A 2 A 3 A 2 A 4 A1A3A1A4 A2A3A2A4A1A3A1A4 A2A3A2A Pr(IBD=0) = 4 / 16 = 0.25 Pr(IBD=1) = 8 / 16 = 0.50 Pr(IBD=2) = 4 / 16 = 0.25 Expected IBD sharing = (2*0.25) + (1*0.5) + (0*0.25) = 1

Average correlation in QTL effects Sib pairsP(IBD 0) = ¼ P(IBD 1) = ½ P(IBD 2) = ¼ Average correlation Additive component = ¼×0 + ½×½ + ¼×1 = ½ Dominance component = ¼×0+ ½×0 + ¼×1 = ¼

Summary of shared genetic variance MZ = V A + V D P-O = ½ V A U = 0 DZ, FS = ¼(V A +V D ) + ½(V A /2) + ¼ (0) = ½V A + ¼V D These single locus results can be summed over all loci to give total genetic variance, heritability.

Figure 3. BIOLOGICAL PARENT - OFFSPRING ECADECAD d a ce d a c e 0.25 P MOTHER P FATHER 0.5 ECADDACE dhceechd P CHILD 2 P CHILD 1 R A 1 R A

Segregation variance If both parents transmit 1 allele A O = A P /2 + A M /2 which means Var(A O ) = ¼ Var(A P ) + ¼ Var(A M ) But… this violates Var(A O )=Var(A P )=Var(A M ) So this explains the extra path in the P-O model: as Var(A)=1, arbitrarily, then Var(A O ) = ¼ Var(A P ) + ¼ Var(A M ) + ½

Segregation variance Segregation variance (SV) represents the random variation among the gametes of an individual –i.e. an Aa parent could transmit either A or a A O = A P /2 + A M /2 + S P + S M Var (A O ) = Var(A P )/4 + Var(A M )/4 + Var(S P ) + Var(S M ) For homozygous locus, Var(S) = 0 For heterozygous locus, Var(S) =  2 /4 Given random mating, average SV = 2pq  2 /4 = V A /4 As V A is fixed to 1 in the path model, SV = ¼ + ¼ = ½