麻省理工学院：《动物遗传学 Genetics》课程教学资源（英文讲义）Lecture 29、30 Statistical Evaluation of Genetic Linkage

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 ?

LOD0.06(family1)=log1o(0.024/0.0039)=log1o(6.25)=0.796 Same for families #2 and #3: ΣL0Do.o6(families1,2,3)=3×0.796=2388 Family #4: Maternal HD HD HD HD alleles SSR37 D P if linked at 0.06 1/2(P if phase 1)+1/2 (P if phase 2) =1/2(0.47×0.47×0.03×0.47)+1/2(0.03×0.03×0.47×0.03)=0.0016 L0D0.06(family4)=log10(0.0016/0.0039)=1og10(0.41)=-0.387

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

ΣL0D0.o6(families1,2,3,4)=2.388-0.387=2.001 Still not sufficient to publish.What to do? 1.It's tempting to ignore family 4 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? 2.5 2 1.5 LOD 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 -0.5 8

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 B 2: + D

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: B SSR37 E Locus: HD SSR37 HD D Now we can deduce the Phase 1: + B phase in the mother: HD B 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 B SSR37 E or 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) =1/2(0.47×0.47×0.03×0.47)+1/2(0.03×0.03×0.47×0.03)=0.0016 But we now know that phase 1 was correct: Pit linked at.06i phase 1)2) 047x0.47x008x047)+1203x0x047X0.031= 0.0032 LOD0.o6(family4)=log1o(0.003210.0039)=1og1o(0.82)=-0.086 We can sum the LOD.06 scores for all four families: ΣL0Do.o6(family1,2,3,④=2.388-0.086=2.302 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 log1o(2)=0.301 to each family's LOD score. >LOD.6(families 1,2,3,4:all phased) LOD.06 (families 1,2,3,4:unphased)+4 log10(2) =2.001+4(0.301)=3.205 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 LOD(families 1,2,3,4:unphased)=4 x 0.903=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?