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Genome-wide Studies: Linkage Analysis. Family Studies. Familial Relationships Twins Siblings Parents/offspring Phenotype information Affected/Unaffected (Prostate Cancer) Quantitative measure (Blood Pressure) No Genotype information required. Why do Family Studies?.
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Family Studies • Familial Relationships • Twins • Siblings • Parents/offspring • Phenotype information • Affected/Unaffected (Prostate Cancer) • Quantitative measure (Blood Pressure) • No Genotype information required
Why do Family Studies? • Is the trait genetic? • What is the mode of transmission? • Dominant • Recessive • Additive • Polygenic (Multiple genes involved)
First PTC Family Study L. H. Snyder Science 1931
PTC Population Genetics Wooding Genetics 2006, adapted from Cavalli-Sforza 1994
Genetic Epidemiology • Adding Genotype data • Candidate Gene study • Linkage Analysis study • Linkage Disequilibrium fine mapping
Linkage Analysis • Narrow down position of disease gene • No biological knowledge needed • Genetic markers (not disease gene) • Recombination
Recombination a A a a b B b b A a a a A a a a b b b b B b B b
Recombination a A a a b B b b R NR NR R A a a a A a a a b b b b B b B b
Independent Assortment a A a a b B b b 25% 25% 25% 25% A a a a A a a a b b b b B b B b
No recombination a A a a b B b b 0% 50% 50% 0% A a a a A a a a b b b b B b B b
Recombination Fraction a A a a b B b b 61 420 442 77 A a a a A a a a b b b b B b B b
Recombination Fraction Recombination Fraction q = Recombinants / Total = 61 + 77 / 61 + 77 + 442 + 420 = 138 / 1000 = 13.8% 61 420 442 77 A a a a A a a a b b b b B b B b
Linkage • Recombination fraction q < 50% • Two traits: PTC and KELL blood group • Two genetic markers • One trait and one genetic marker Linkage Analysis
Dominant Trait D d d d 1 2 3 3 D d D d d d 1 3 2 3 2 3
Linkage Analysis • LOD score based on recombination • LOD (q) = log (q)R (1 - q)NR ____________________ (q = 1/2) R + NR
Linkage Analysis 1 2 3 3 R NR NR 1 3 2 3 2 3
a a A A b B B b a A a a b B b b Phase known
Phase Known LOD (q) = log (q)1 (1 - q)2 ____________________ (q = 1/2) 1 + 2 = 0.07 for q = 1/3
Phase Unknown LOD (q) = ½ log (q)1 (1 - q)2 ____________________ (q = 1/2) 1 + 2 + ½ log (q)1 (1 - q)2 ____________________ (q = 1/2) 1+ 2 = -0.02 for q = 0.44
IBD • Identity by descent • Allele Sharing methods • Often used for affected sib pairs
Identity By Descent a A a A 25% 25% 25% 25% A A a A A a a a
Identity By Descent Sibling 1 A A Alleles shared IBD 2 1 1 0 A A a A A a a a
Identity By Descent Sibling 1 A A 2 1 1 0 A A a A A a a a
Identity By Descent Parent 1 a A Alleles shared IBD 1 1 1 1 A A a A A a a a
Identity By Descent Parent 1 a A 2 1 1 0 A A a A A a a a
Identity By Descent • IBD can be used for linkage analysis • Expect 50% alleles shared between siblings • Look for IBD > 50% for concordant pairs • Look for IBD < 50% for discordant pairs
Fine Mapping Linkage markers Genes Kim et al. Science 2003
Linkage Disequilibrium a A a a b B b b A a a a A a a a b b b b B b B b
Linkage Disequilibrium A a a a A a a a b b b b B b B b
Linkage Disequilibrium A a b b
Linkage Disequilibrium A a a a b b b b A a A a a a a a b b b b b b b b
Linkage Disequilibrium a A a A A a B b b b b b Time
Linkage Disequilibrium A G C T A T C G
Linkage Disequilibrium r2 = 1 D’ = 1 A G 50% C T 50% A T 0% C G 0%
Linkage Disequilibrium r2 < 1 D’ = 1 A G 34% C T 33% A T 33% C G 0%
Linkage Disequilibrium r2 = 0 D’ = 0 A G 25% C T 25% A T 25% C G 25%
Genomewide Linkage Analysis Genetic Markers q = 10% on average Genes
Linkage Disequilibrium Mapping Genetic Markers Genes
Gene Identification • Substitute disease status • Peak in LD should be near disease variant • Random SNPs • 50,000 basepairs apart on average
PTC Linkage Disequilibrium Mapping Kim et al. Science 2003
Candidate Gene Genetic Markers Gene of Interest
PTC Candidate Genes • 150 Genes in the region • Sequenced 16 genes • 9 TAS2R bitter taste receptors • 7 OR odorant like receptors • Strongest Association in TAS2R38
TAS2R38 Receptor Structure Kim et al. J Dent Res 2004