1. Genetic principles for linkage and association analyses Manuel Ferreira & Pak Sham Boulder, 2009

2. Gene mapping LOCALIZE and then IDENTIFY a locus that regulates a trait Linkage analysis Association analysis

3. If a locus regulates a trait, Trait Variance and Covariance between individuals will be influenced by this locus. Linkage: Association: If a locus regulates a trait, Trait Mean in the population will also be influenced by this locus.

4. Revisit common genetic parameters - such as allele frequencies, genetic effects, dominance, variance components, etc Use these parameters to construct a biometric genetic model Model that expresses the: (1)Mean (2)Variance (3)Covariance between individuals for a quantitative phenotype as a function of the genetic parameters of a given locus. See how the biometric model provides a useful framework for linkage and association methods.

5. Outline 1. Genetic concepts 2. Very basic statistical concepts 3. Biometrical model 4. Introduction to linkage analysis

6. 1. Genetic concepts

7. B. Population level C. Transmission level D. Phenotype level G G G G G G G G G G G G G G G G G G G G G G GG PP Allele and genotype frequencies Mendelian segregation Genetic relatedness Biometrical model Additive and dominance components A. DNA level DNA structure, organization recombination

8. A DNA molecule is a linear backbone of alternating sugar residues and phosphate groups Attached to carbon atom 1’ of each sugar is a nitrogenous base: A, C, G or T Two DNA molecules are held together in anti-parallel fashion by hydrogen bonds between bases [Watson-Crick rules] Antiparallel double helix Only one strand is read during gene transcription Nucleotide: 1 phosphate group + 1 sugar + 1 base A. DNA level A gene is a segment of DNA which is transcribed to give a protein or RNA product

9. DNA polymorphisms A B Microsatellites >100,000 Many alleles, eg. (CA) n repeats, very informative SNPs 14,708,752 (build 129, 03 Mar ‘09) Most with 2 alleles (up to 4), not very informative Copy Number Variants >5,000 Many alleles, just recently automated

10. Haploid gametes ♁ ♂ ♂ ♁ G 1 phase chr1 S phase Diploid zygote 1 cell M phase Diploid zygote >1 cell ♁ ♂ ♁ A - B - A - B - A - B - A - B - A - B - A - B - ♂ ♁ A - B - A - B - ♂ ♁ - A - B - A - B - A - B - A - B DNA organization Mitosis 22 + 12 (22 + 1)

11. Diploid gamete precursor cell (♂)(♁) (♂) (♁) Haploid gamete precursors Hap. gametes NR R R ♁ A - B - - A - B A - B - - A - B A - B - - A - B A - B - - A - B ♂ ♁ A - B - A - B - - A - B - A - B DNA recombination Meiosis 2 (22 + 1) 22 + 1 chr1

12. Diploid gamete precursor (♂)(♁) (♂) (♁) Haploid gamete precursors Hap. gametes NR ♁ A - B - - A - B A - B - - A - B A - B - - A - B A - B - - A - B ♂ ♁ A - B - A - B - - A - B - A - B DNA recombination between linked loci Meiosis 2 (22 + 1) 22 + 1

13. B. Population level 1. Allele frequencies A single locus, with two alleles - Biallelic - Single nucleotide polymorphism, SNP Alleles A and a - Frequency of A is p - Frequency of a is q = 1 – p A a A genotype is the combination of the two alleles A A A A A A A A A A A A A A a a a a a a a a a a a A a Aa

14. B. Population level 2. Genotype frequencies (Random mating) A (p)a (q) A (p) a (q) Allele 1 Allele 2 AA (p 2 ) aA (qp) Aa (pq) aa (q 2 ) Hardy-Weinberg Equilibrium frequencies P (AA) = p 2 P (Aa) = 2pq P (aa) = q 2 p 2 + 2pq + q 2 = 1

15. Segregation (Meiosis) Mendel’s law of segregation A3 (½)A3 (½)A4 (½)A4 (½) A1 (½)A1 (½) A2 (½)A2 (½) Mother (A 3 A 4 ) A1A3 (¼)A1A3 (¼) A2A3 (¼)A2A3 (¼) A1A4 (¼)A1A4 (¼) A2A4 (¼)A2A4 (¼) Gametes Father (A 1 A 2 ) C. Transmission level

16. D. Phenotype level 1. Classical Mendelian traits Dominant trait - AA, Aa1 - aa0 Recessive trait - AA1 - aa, Aa0 Cystic fibrosis 3 bp deletion exon 10 CFTR gene Huntington’s disease (CAG)n repeat, huntingtin gene

17. 2. Quantitative traits AA Aa aa D. Phenotype level e.g. cholesterol levels

18. P(X)P(X) X AA Aa aa AA Aaaa D. Phenotype level m

19. m d +a P(X)P(X) X AA Aa aa – a AA Aaaa Biometric Model Genotypic effect D. Phenotype level

20. 2. Very basic statistical concepts

21. Mean, variance, covariance 1. Mean (X)

22. 2. Variance (X) Mean, variance, covariance

23. 3. Covariance (X,Y) Mean, variance, covariance

24. 3. Biometrical model

25. Biometrical model for single biallelic QTL Biallelic locus - Genotypes: AA, Aa, aa - Genotype frequencies: p 2, 2pq, q 2 Alleles at this locus are transmitted from P-O according to Mendel’s law of segregation Genotypes for this locus influence the expression of a quantitative trait X (i.e. locus is a QTL) Biometrical genetic model that estimates the contribution of this QTL towards the (1) Mean, (2) Variance and (3) Covariance between individuals for this quantitative trait X

26. Biometrical model for single biallelic QTL 1. Contribution of the QTL to the Mean (X) aa Aa AA Genotypes Frequencies, f(x) Effect, x p2p2 2pqq2q2 a d-a = a(p 2 ) + d(2pq) – a(q 2 )Mean (X)= a(p-q) + 2pqd e.g. cholesterol levels in the population

27. Biometrical model for single biallelic QTL 2. Contribution of the QTL to the Variance (X) aa Aa AA Genotypes Frequencies, f(x) Effect, x p2p2 2pqq2q2 a d-a = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 Var (X) = V QTL Heritability of X at this locus = V QTL / V Total

28. Biometrical model for single biallelic QTL = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 Var (X) = 2pq[a+(q-p)d] 2 + (2pqd) 2 = V A QTL + V D QTL Additive effects: the main effects of individual alleles Dominance effects: represent the interaction between alleles m = a(p-q) + 2pqd

29. Biometrical model for single biallelic QTL m = 0 d+a – a A aAaAa d = 0 (no dominance) +a+a+a+a Additive model

30. Biometrical model for single biallelic QTL m = 0 d+a – a A aAaAa d > 0 (dominance) +a+d+a-d Dominant model

31. Biometrical model for single biallelic QTL m = 0 d+a – a A aAaAa d < 0 (dominance) +a-d+a+d Recessive model

32. aa Aa AA aa Aa AA +4 +0.7+0.4 log (x) No departure from additivity Significant departure from additivity Statistical definition of dominance is scale dependent

33. Biometrical model for single biallelic QTL Var (X) = Regression Variance + Residual Variance -a 0 a aa Aa AAaa Aa AAaa Aa AA aa Aa AA AdditiveDominant Recessive Genotypic mean = Additive Variance + Dominance Variance = V A QTL + V D QTL

34. Practical H:\manuel\biometric\sgene.exe

35. Practical Aim Visualize graphically how allele frequencies, genetic effects, dominance, etc, influence trait mean and variance Ex1 a=0, d=0, p=0.4, Residual Variance = 0.04, Scale = 2. Vary a from 0 to 1. Ex2 a=1, d=0, p=0.4, Residual Variance = 0.04, Scale = 2. Vary d from -1 to 1. Ex3 a=1, d=0, p=0.4, Residual Variance = 0.04, Scale = 2. Vary p from 0 to 1. Look at scatter-plot, histogram and variance components.

36. 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

37. Biometrical model for single biallelic QTL 3. Contribution of the QTL to the Covariance (X,Y) 2. Contribution of the QTL to the Variance (X) 1. Contribution of the QTL to the Mean (X)

38. Biometrical model for single biallelic QTL AA Aa aa AA Aaaa (a-m)(a-m) (d-m)(d-m) (-a-m) (a-m)(a-m) (d-m)(d-m) (a-m)2(a-m)2 (a-m)(a-m) (d-m)(d-m) (a-m)(a-m) (d-m)2(d-m)2 (d-m)(d-m) (-a-m) 2 3. Contribution of the QTL to the Cov (X,Y)

39. Biometrical model for single biallelic QTL AA Aa aa AA Aaaa (a-m)(a-m) (d-m)(d-m) (-a-m) (a-m)(a-m) (d-m)(d-m) (a-m)2(a-m)2 (a-m)(a-m) (d-m)(d-m) (a-m)(a-m) (d-m)2(d-m)2 (d-m)(d-m) (-a-m) 2 p2p2 0 0 2pq 0 q2q2 3A. Contribution of the QTL to the Cov (X,Y) – MZ twins = (a-m) 2 p 2 + (d-m) 2 2pq + (-a-m) 2 q 2 Cov(X,Y) = V A QTL + V D QTL = 2pq[a+(q-p)d] 2 + (2pqd) 2

40. Biometrical model for single biallelic QTL AA Aa aa AA Aaaa (a-m)(a-m) (d-m)(d-m) (-a-m) (a-m)(a-m) (d-m)(d-m) (a-m)2(a-m)2 (a-m)(a-m) (d-m)(d-m) (a-m)(a-m) (d-m)2(d-m)2 (d-m)(d-m) (-a-m) 2 p3p3 p2qp2q 0 pq pq 2 q3q3 3B. Contribution of the QTL to the Cov (X,Y) – Parent-Offspring

41. 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

42. Biometrical model for single biallelic QTL AA Aa aa AA Aaaa (a-m)(a-m) (d-m)(d-m) (-a-m) (a-m)(a-m) (d-m)(d-m) (a-m)2(a-m)2 (a-m)(a-m) (d-m)(d-m) (a-m)(a-m) (d-m)2(d-m)2 (d-m)(d-m) (-a-m) 2 p3p3 p2qp2q 0 pq pq 2 q3q3 = (a-m) 2 p 3 + … + (-a-m) 2 q 3 Cov (X,Y) = ½V A QTL = pq[a+(q-p)d] 2 3B. Contribution of the QTL to the Cov (X,Y) – Parent-Offspring

43. Biometrical model for single biallelic QTL AA Aa aa AA Aaaa (a-m)(a-m) (d-m)(d-m) (-a-m) (a-m)(a-m) (d-m)(d-m) (a-m)2(a-m)2 (a-m)(a-m) (d-m)(d-m) (a-m)(a-m) (d-m)2(d-m)2 (d-m)(d-m) (-a-m) 2 p4p4 2p 3 q p2q2p2q2 4p 2 q 2 2pq 3 q4q4 = (a-m) 2 p 4 + … + (-a-m) 2 q 4 Cov (X,Y) = 0 3C. Contribution of the QTL to the Cov (X,Y) – Unrelated individuals

44. Biometrical model for single biallelic QTL Cov (X,Y) 3D. Contribution of the QTL to the Cov (X,Y) – DZ twins and full sibs ¼ genome ¼ (2 alleles) + ½ (1 allele) + ¼ (0 alleles) MZ twins P-O Unrelateds ¼ genome # identical alleles inherited from parents 0 1 (mother) 1 (father) 2 = ¼ Cov(MZ) + ½ Cov(P-O) + ¼ Cov(Unrel) = ¼(V A QTL +V D QTL ) + ½ (½ V A QTL ) + ¼ (0) = ½ V A QTL + ¼V D QTL

45. Summary so far…

46. Biometrical model predicts contribution of a QTL to the mean, variance and covariances of a trait Var (X) = V A QTL + V D QTL Cov (MZ) = V A QTL + V D QTL Cov (DZ) = ½V A QTL + ¼V D QTL On average! 0 or 10, 1/2 or 1 IBD estimation / Linkage Mean (X) = a(p-q) + 2pqd Association analysis Linkage analysis For a given locus, do two sibs have 0, 1 or 2 alleles in common?

47. 4. Introduction to Linkage analysis

48. For a heritable trait... localize region of the genome where a QTL that regulates the trait is likely to be harboured identify a QTL that regulates the trait Linkage: Association: Family-specific phenomenon: Affected individuals in a family share the same ancestral predisposing DNA segment at a given QTL Population-specific phenomenon: Affected individuals in a population share the same predisposing DNA segment at a given QTL Can only detect very large effects Can detect weaker effects

49. No Linkage No Association Linkage No Association Linkage Association Families Cases Controls

50. Non-parametric linkage approach If a trait locus truly regulates the expression of a phenotype, then two relatives with similar phenotypes should have inherited from a common ancestor the same predisposing allele at a marker near the trait locus, and vice-versa. Interest: correlation between phenotypic similarity and genetic similarity at a locus Linkage tests co-segregation between a marker and a trait

51. Phenotypic similarity between relatives Squared trait differences Squared trait sums Trait cross-product Trait variance-covariance matrix Affection concordance T2T2 T1T1

52. Genotypic similarity between relatives IBS Alleles shared Identical By State “look the same”, may have the same DNA sequence but they are not necessarily derived from a known common ancestor IBD Alleles shared Identical By Descent are a copy of the same ancestor allele M1M1 Q1Q1 M2M2 Q2Q2 M3M3 Q3Q3 M3M3 Q4Q4 M1M1 Q1Q1 M3M3 Q3Q3 M1M1 Q1Q1 M3M3 Q4Q4 M1M1 Q1Q1 M2M2 Q2Q2 M3M3 Q3Q3 M3M3 Q4Q4 IBS IBD 2 1 Inheritance vector (M) 0 001 1

53. Genotypic similarity between relatives - M1M1 Q1Q1 M3M3 Q3Q3 M2M2 Q2Q2 M3M3 Q4Q4 Number of alleles shared IBD 0 M1M1 Q1Q1 M3M3 Q3Q3 M1M1 Q1Q1 M3M3 Q4Q4 1 M1M1 Q1Q1 M3M3 Q3Q3 M1M1 Q1Q1 M3M3 Q3Q3 2 Proportion of alleles shared IBD - 0 0.5 1 Inheritance vector (M) 0 011 0 001 0 000

54. Genotypic similarity between relatives - 2 2n ABC D

55. Var (X) = V A QTL + V D QTL Cov (MZ) = V A QTL + V D QTL Cov (DZ) = ½V A QTL + ¼V D QTL On average! Cov (DZ) = V A QTL + V D QTL For a given locus Cov (DZ) = V A QTL

57. Statistics that incorporate both phenotypic and genotypic similarities to test V QTL Regression-based methods Variance components methods Mx, MERLIN, SOLAR, GENEHUNTER Haseman-Elston, MERLIN-regress (X 1 -X 2 ) 2 = -2 * V A QTL

58. Should we still use linkage analysis? Given dense SNP data Rare genetic variant (not covered by the genotyping platform)... or allelic heterogeneity (multiple disease variants in the same gene) *AND* strong effect on phenotype... Linkage analysis can complement association and provide an additional approach to localise a disease locus (with no loss).