Lectures 29-30: ?Phase ?Lod scores Statistical Evaluation of Genetic Linkage 20 cM SSR12 SSR112 SSR31 SSR37 SSR5 genetic linkage mapping We genotype the six members of the family for SSRs scattered throughout the genome (which spans 3300 cM)— one SSR must be within 10 cM of the Huntington's gene: HD ? log 10 (0.024/0.0039) Same for families #2 and #3:  LOD 0.06 (families 1, 2, 3) = 3 x 0.796 = 2.388 Family #4: DDDDSSR37 HD+HDHDHDMaternal alleles P if linked at 0.06 = 1/2 (P if phase 1) + 1/2 (P if phase 2) = 1/2 LOD 0.06 (family 4) LOD 0.06 (family 1) = = log 10 (6.25) = 0.796 (0.47 = 0.0016x 0.47 x 0.47)x 0.03 + 1/2 (0.03 x 0.03 x 0.47 x 0.03) = log 10 (0.41)= log 10 (0.0016/0.0039) = - 0.387  LOD 0.06 (families 1, 2, 3, 4) = 2.388 – 0.387 = 2.001 Still not sufficient to publish. What to do? 1. It's tempting to ignore family 4 — But this would not be an acceptable solution. 2. Calculate LOD scores for other  values? to declare it to be irrelevant for some reason or another. -0.5 0 0.5 1 1.5 2 2.5 0 0.1 0.2 0.3 0.4 0.5 0.6 LOD  3. 4. Get more families — always a good idea Determine phase in affected parents In each of the four families, we were uncertain about phase, and our LOD calculations embodied those uncertainties. Family #4: D+ BHD Phase 2: B+ DHD Phase 1: SSR37HD two possible arrangements of alleles on mother's chromosomes Typing the maternal grandparents for SSR37:  D SSR37    E  C  B  A Now we can deduce the phase in the mother: Family #4: D+ BHD Phase 2: B+ DHD Phase 1: SSR37HDLocus: Here is a more realistic version of the genotypic information we might obtain:  D SSR37   E  C B  A Family #4: dead refused consent    or inferred Before we had written: P if linked at 0.06 = 1/2 (P if phase 1) + 1/2 (P if phase 2) = 1/2 (0.47 x 0.47 x 0.03 x 0.47) + 1/2 (0.03 x 0.03 x 0.47 x 0.03) = 0.0016 But we now know that P if linked at 0.06 = 1/2 (P if phase 1) + 1/2 (P if phase 2) = 1/2 (0.47 x 0.47 x 0.03 x 0.47) + 1/2 (0.03 x 0.03 x 0.47 x 0.03) = 0.0032 LOD 0.06 (family 4) We can sum the LOD 0.06 scores for all four families:  LOD 0.06 (family 1, 2, 3, 4) = phase 1 was correct: = log 10 (0.82)= log 10 (0.0032/0.0039) = - 0.086 – 0.0862.388 = 2.302 phase known Overall effect of determining phase in all four families:  LOD 0.06 (families 1,2,3,4: all phased) =  LOD 0.06 (families 1,2,3,4: unphased) + 4 log 10 (2) = 2.001 What if we had not been able to obtain samples from any grandparents? Try more markers Publish! Add increment of log 10 (2) = 0.301 to each family’s LOD score. = 3.205 + 4 (0.301) 20 cM SSR34 SSR35 SSR36 SSR37 SSR38 Chr 4 or HD Marker showing no recombination with HD  LOD 0 (families 1,2,3,4: unphased) = 4 x 0.903 Very strong conclusion!! = 3.609 Search for SSR marker showing no recombination with HD: Where to look?