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Biometrical Genetics. Shaun Purcell Twin Workshop, March 2004. Single locus model. Genetic effects → variance components Genetic effects → familial covariances Variance components → familial covariances. A DE Model for twin data. [0.25/1]. [0.5/1]. 1. 1. 1. 1. 1. 1. E. D. A.
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Biometrical Genetics Shaun Purcell Twin Workshop, March 2004
Single locus model • Genetic effects → variance components • Genetic effects → familial covariances • Variance components → familial covariances
ADE Model for twin data [0.25/1] [0.5/1] 1 1 1 1 1 1 E D A A D E e d a a d e PT1 PT2
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
Mendel’s Experiments AA aa Pure Lines F1 Aa Aa Intercross AA Aa Aa aa 3:1 Segregation Ratio
Mendel’s Experiments F1 Pure line Aa aa Back cross Aa aa 1:1 Segregation ratio
Mendel’s Experiments AA aa Pure Lines F1 Aa Aa Intercross Aa Aa aa AA 3:1 Segregation Ratio
Mendel’s Experiments F1 Pure line Aa aa Back cross Aa aa 1:1 Segregation ratio
Maternal A3 A4 ½ ½ A1 A1 A3 A4 ¼ ¼ A2 A2 A3 A4 ¼ ¼ Mendel’s Law of Segregation Gametes ½ A1 Paternal A2 ½ Meiosis/Segregation
Maternal D d ½ ½ D D D d 1 1 d d D d 1 0 Dominant Mendelian inheritance ½ D Paternal d ½
Maternal D d ½ ½ D D D d 1 0 d d D d 0 0 Recessive Mendelian inheritance ½ D Paternal d ½
Maternal D d ½ ½ D D D d Incomplete penetrance 60% 60% d d D d Phenocopies 60% 1% Dominant Mendelian inheritance ½ D Paternal d ½
AA Aa aa Quantitative traits
Biometrical Genetic Model P(X) Aa Genotypic means aa AA AA m + a X Aa m + d m aa m – a -a +a d
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 Ap2 pqp aqp q2q p q Hardy-Weinberg Equilibrium frequencies P(AA) = p2 P(Aa) = 2pq P(aa) = q2
Biometrical Model for Single Locus GenotypeAAAaaa Frequencyp2 2pq q2 Effect (x)a d -a Residual var2 2 2 Meanm = p2(a) + 2pq(d) + q2(-a) = a(p-q) + 2pqd
Biometrical Model for Single Locus GenotypeAA Aa aa Frequencyp2 2pq q2 (x-m)2(a-m)2 (d-m)2 (-a-m)2 Variance = (a-m)2p2 + (d-m)22pq + (-a-m)2q2 = VG (Broad-sense) heritability at this loci = VG / VTOT (Broad-sense) heritability = ΣLVG / VTOT
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 • What determines additive genetic variance? • Under what conditions does VD > VA
Some conclusions • Additive genetic variance depends on allele frequency p & additive genetic value a as well as dominance deviation d • 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
Average allelic effect • Denote the average allelic effects as α αA = q(a+d(q-p)) αa = -p(a+d(q-p)) • If only two alleles exist, we can define the average effect of allele substitution α = αA – αa α = (q-(-p))(a+d(q-p)) = (a+d(q-p)) • Therefore, αA = qα and αa = -pα
Additive genetic variance • The variance of the average allelic effects Freq.Additive effect AA p2 2αA = 2qα Aa 2pq αA +αa = (q-p)α aa q2 2αa = -2pα VA = p2(2qα)2 + 2pq((q-p)α)2 + q2(-2pα)2 = 2pqα2 = 2pq(a+d(q-p))2
Additive genetic variance • If there is no dominance VA = 2pqa2 • If p = q VA = ½a2
a d m -a Additive and Dominance Variance aa Aa AA Total Variance = Regression Variance + Residual Variance = Additive Variance + Dominance Variance
Biometrical Model for Single Locus GenotypeAA Aa aa Frequencyp2 2pq q2 (x-m)2(a-m)2 (d-m)2 (-a-m)2 Variance = (a-m)2p2 + (d-m)22pq + (-a-m)2q2 = 2pq[a+(q-p)d]2 + (2pqd)2 VG = VA + VD
Additive genetic variance VA -1 d -1 a +1 +1 Dominance genetic variance VD Allele frequency 0.01 0.05 0.1 0.2 0.3 0.5
AA Aa aa -1 0 +1 d -1 0 +1 a VA > VD VA < VD Allele frequency 0.01 0.05 0.1 0.2 0.3 0.5
Cross-Products of Deviations for Pairs of Relatives AA Aa aa 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 AA Aa aa AAp2 Aa 0 2pq aa 0 0 q2 Covariance = (a-m)2p2 + (d-m)22pq + (-a-m)2q2 = 2pq[a+(q-p)d]2 + (2pqd)2 = VA + VD
Covariance for Parent-offspring (P-O) AA Aa aa 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 AAfather, an AAoffspring can come from either AAx AAor AAx Aaparental mating types AAx AA will occur p2× p2 = p4 and have AA offspring Prob()=1 AAx Aa will occur p2× 2pq = 2p3q and have AA offspring Prob()=0.5 and have Aa offspring Prob()=0.5 Therefore, P(AA father & AAoffspring) = p4 + p3q = p3(p+q) = p3
Covariance for Parent-offspring (P-O) AA Aa aa AAp3 Aa ? ? aa ? ? ? • AA offspring from AAparents = p4+p3q = p3(p+q) = p3
Covariance for Parent-offspring (P-O) AA Aa aa AAp3 Aap2q ? aa ? ? ? • AA offspring from AAparents = p4+p3q = p3(p+q) = p3 • Aa offspring from AAparents = p3q+p2q2 = p2q(p+q) = p2q
Covariance for Parent-offspring (P-O) AA Aa aa AAp3 Aap2q pq aa 0 pq2 q3 Covariance = (a-m)2p3 + (d-m)2pq + (-a-m)2q3 + (a-m)(d-m)2p2q+ (-a-m)(d-m)2pq2 = pq[a+(q-p)d]2 = VA / 2
Covariance for Unrelated Pairs (U) AA Aa aa AAp4 Aa2p3q 4p2q2 aap2q2 2pq3 q4 Covariance = (a-m)2p4 + (d-m)24p2q2 + (-a-m)2q4 + (a-m)(d-m)4p3q+ (-a-m)(d-m)4pq + (a-m)(-a-m)2p2q2 = 0 ?
Identity by Descent (IBD) • Two alleles are IBD if they are descended from and replicates of the same recent ancestral allele 2 1 aa Aa 3 4 5 6 AA Aa Aa Aa 7 8 AA Aa
IBS IBD A1A2 A1A3 IBS = 1 IBD = 0 A1A2 A1A3 IBS=Identity by State
IBD: MZ Twins AB CD AC AC MZ twins always share 2 alleles IBD
IBD: Parent-Offspring AB CD AC If the parents are unrelated, then parent-offspring pairs always share 1 allele IBD