麻省理工大学：遗传学（Genetics）讲稿_lecture29_30

Lectures 29-30 Statistical evaluation of Genetic Linkage . Phase Lod scores

Lectures 29-30: •Phase •Lod scores Statistical Evaluation of Genetic Linkage

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 SSR12 SSR112 SSR31 SSR37 SSR5 20 CM

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 ?

LODoos(family1)=log0(0.024/0039)=0g10(6.25)=0796 Same for families #2 and#3 2LoDo6( families1,2,3)=3x0.796=2388 Family #4 Maternal HD HDHD HD alleles SSR37 D D D D P if linked at 0.06=1/2(P if phase 1)+1/2(P if phase 2) =12(0.47X0.47X0.03X0.47)+1/2(0.03×0.03X0.47X003)=0.0016 ODo6eamy4)=log10(0.00160.0039)=0g10(0.41)=-0387

log10 (0.024/0.0039) Same for families #2 and #3:
LOD0.06 (families 1, 2, 3) = 3 x 0.796 = 2.388 Family #4: SSR37 D D D D Maternal HD HD HD + HD alleles P if linked at 0.06 = 1/2 (P if phase 1) + 1/2 (P if phase 2) = 1/2 LOD0.06 (family 4) LOD0.06(family 1) = = log10 (6.25) = 0.796 (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 = log10 = log (0.41) 10 (0.0016/0.0039) = - 0.387

2LoD6ames1,2,3,4)=2388-0.387=2001 Still not sufficient to publish. What to do? 1. It's tempting to ignore family 4- to declare it to be irrelevant for some reason or another But this would not be an acceptable solution 2. Calculate LOD scores for other 0 values? 1.5 LOD 0.2 0.3 0.5 -0.5 6

LOD0.06 (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. Get more families always a good idea 4. 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 HD SSR37 Phase HD 1: B two possible arrangements of alleles on mother's chromosomes Phase HD BD

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 Phase HD B 2: + B Phase HD D 1: HD SSR37 two possible arrangements of alleles on mother's chromosomes

Typing the maternal grandparents for SSR37 Family #4 SSR37 ABCDE Locus HD HD Now we can deduce the Phase 1 B phase in the mother HD Phase 2: D

Typing the maternal grandparents for SSR37: D





SSR37


E C


B
A


Now we can deduce the phase in the mother: Family #4: + D HD B Phase 2: + B HD D Phase 1: Locus: HD SSR37

Here is a more realistic version of the genotypic information we might obtain dead Family #4 refused consent A SSR37 D inferred

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) =12(047X047X003X047)+12(038X0.03X047X003)=00016 But we now know that phase 1 was correct P if linked at 0.06 =Xe(p if phase 1)+1/2 APif phase 2) (0.47X047×003X047)+1/2(003×1e03X047×003)=0002 LODo o6(family4)=log10(0.00320.0039)=10g10(0.82)=-0.086 We can sum the lodo scores for all four families ∑ LODo o6(family12.34=2388-0086=2302 phase known

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 LOD0.06(family 4) We can sum the LOD0.06 scores for all four families:
LOD0.06(family 1, 2, 3, 4) = phase 1 was correct: = log10 = log (0.82) 10 (0.0032/0.0039) = - 0.086 2.388 – 0.086= 2.302 phase known

Overall effect of determining phase in all four families Add increment of logo(2)=0.301 to each familys LOD score X LOD.oG(families 1, 2, 3, 4: all phased) X LODo.06(families 1, 2, 3, 4:unphased)+4 log10(2) =2001+40301)=3205 Publish What if we had not been able to obtain samples from any grandparents Try more markers

Overall effect of determining phase in all four families:
LOD0.06 (families 1,2,3,4: all phased) =
LOD0.06 (families 1,2,3,4: unphased) + 4 log10 (2) = 2.001 What if we had not been able to obtain samples from any grandparents? Try more markers Publish! Add increment of log10(2) = 0.301 to each family’s LOD score. + 4 (0.301) = 3.205

Search for SSr marker showing no recombination with HD Where to look? SSR34 SSR35 SSR36 SSR37 SSR38 Chr 4 20 CM or Marker showing no recombination with HD HD X LODo(families 1, 2, 3, 4: unphased)=4 X0.903=3.609 Very strong conclusion!

20 cM SSR34 SSR35 SSR36 SSR37 SSR38 Chr 4 or HD Marker showing no recombination with HD
LOD0 (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?