This preview shows pages 1–15. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 4: Phenotypes & Gene Interactions Interactions between alleles of the same gene (biochemical basis of dominance and recessiveness). Interactions between genes in pathways. Examining and inferring gene interactions. Complementation testing (test for allelism). Types of interactions among loci. Penetrance and expressivity. Wrapup of classical Mendelian and transmission genetics. Reading (9th edition) pp. 221249 ; 3133; 5761; 163164 The Chi Square Test in Linkage Analysis The Chi Square Test in Linkage Analysis 142 133 113 112 500 The Chi Square Test in Linkage Analysis P(B)= 254/500=.508 P(b)= 246/500=.492 P( A )= 255/500 =. 510 P( a )= 245/500 =.490 The Chi Square Test in Linkage Analysis P( B )= 254/500=.508 P( b )= 246/500=.492 P( A )= 255/500 =. 510 Exp( A B )= .510 X .508X 500 = 129.54 Exp( A b )= .510 X .492X 500 = 125.46 Exp( a B )= .490 X .508X 500 = 124.54 Exp( a b )= .490 X .492X 500 = 120.54 P( a )= 245/500 =.490 Expected number ( AB ) = P( A ) X P ( B ) X total The Chi Square Test in Linkage Analysis The Chi Square Test in Linkage Analysis ( Observed i Expected i ) 2 Expected i = i=1 n LOD scores: Does this pedigree show linkage? LOD scores: Does this pedigree show linkage? If the trait and polymorphism are not linked then the probability of getting this family is ... product rule P P R R If not linked Prob @ RF=0.5 Prob 0.5 (this family) = B X 1/4 X 1/4 X 1/4 X 1/4 X 1/4 X 1/4 = B X 0.00024 What about linked at 0.3 RF? Prob 0.3 (this family) = B X .35 X . 15 X . 35 X .35 X . 15 X .35 = B X 0.00034 LOD scores: Does this pedigree show linkage? The LOD score is the logarithm of the odds ratio Collect pedigree information from families of interest And pool information on numbers of parental type and recombinant progeny.(see next slide) LOD scores: Does this pedigree show linkage? LOD scores: Does this pedigree show linkage? The LOD score is the logarithm of the odds ratio A LOD score of 3 is required to support the data at a given RF A real case (HD) cM cM cM Lecture 4: Phenotypes & Gene Interactions Interactions between alleles of the same gene (biochemical basis of dominance and recessiveness). Interactions between genes in pathways. Examining and inferring gene interactions....
View Full
Document
 Summer '08
 D.SCHOEN
 Genetics

Click to edit the document details