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Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.

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Presentation on theme: "Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2."— Presentation transcript:

1 Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2

2 Mapping a disease locus Fig. 11.A A1D A2d A1d d D A2

3 Mapping a disease locus Fig. 11.A A1d d A2D A1D A2d (sperm) A1 A2

4 LOD scores Odds = P(pedigree | r) P(pedigree | r = 0.5) r = genetic distance between marker and disease locus Odds = (1-r) n r k 0.5 (total # meioses) Odds = 0.7 7 0.3 1 0.5 8 = 6.325 Data >6 times more likely under LINKED hypothesis than under UNLINKED hypothesis. k = 1 recomb, n = 7 non-recomb. A1 A2

5 Just a point estimate observed recombination fraction = 1/8 = 12.5 cM Disease- causing mutation Restriction fragment length polymorphism True distance 30 cM this is our observation

6 LOD scores rodds 0.112.244 0.210.737 0.36.325 0.42.867 0.5?? Odds = P(pedigree | r) P(pedigree | r = 0.5) Odds = (1-r) n r k 0.5 (total # meioses) k = 1 recomb, n = 7 non-recomb.

7 How to get an overall estimate of probability of linkage? A.Multiply odds together B.Add odds together C.Take the largest odds D.Take the average odds Given r Odds 1 Given r Odds 2 Given r Odds 3 1,22,3 1,21,3 2,31,3 1,22,3 1,2 1,3 2,31,3 2,3 1,31,22,3 2,2 Combining families

8 More realistic situation: in dad, phase of alleles unknown A1d d D A2d A1 A2 or A1d A2D

9 More realistic situation: in dad, phase of alleles unknown Odds = 1/2[(1-r) n r k ]P(pedigree|r) A1 A2 A1D A2d + 1/2[(1-r) n r k ] assume one phase for dad 7 non-recomb, 1 recomb (k = # recomb, n = # non-recomb) A1d A2D assume the other phase for dad 1 non-recomb, 7 recomb

10 In real life this correction does matter… best r = 0.2873best r = 0.2771 Accounting for both phases Using only one phase family 1: 10 meioses, 1 (or 9) apparent recombinants family 2: 10 meioses, 4 (or 6) apparent recombinants family 3: 10 meioses, 3 (or 7) apparent recombinants family 4: 10 meioses, 3 (or 7) apparent recombinants total LOD = LOD(family 1) + LOD(family 2) + LOD(family 3) + LOD(family 4)

11 Modern genetic scans (single family)

12 Age of onset in breast cancer age of onset

13 Coins Odds = P(your flips | r) P(your flips | r = 0.5) r = intrinsic probability of coming up heads (bias) Odds = (1-r) n r k 0.5 (total # flips)

14 Coins 3 heads2 heads1 heads rodds 00 0.11.1664 0.21.6384 0.31.6464 0.41.3824 0.51 0.60.6144 0.70.3024 0.80.1024 0.90.0144 10 0 heads rodds 016 0.110.498 0.26.5536 0.33.8416 0.42.0736 0.51 0.60.4096 0.70.1296 0.80.0256 0.90.0016 10 4 heads rodds 00 0.10.0016 0.20.0256 0.30.1296 0.40.4096 0.51 0.62.0736 0.73.8416 0.86.5536 0.910.498 116 rodds 00 0.10.1296 0.20.4096 0.30.7056 0.40.9216 0.51 0.60.9216 0.70.7056 0.80.4096 0.90.1296 10 rodds 00 0.10.0144 0.20.1024 0.30.3024 0.40.6144 0.51 0.61.3824 0.71.6464 0.81.6384 0.91.1664 10 r = intrinsic probability of coming up heads (bias)

15 Coins By chance, can get good LOD score for just about anything. The more students you have flipping coins, the more likely you are to see this “unlikely” combination. The multiple testing problem

16 Coins Probability of one student observing 0 heads and 4 tails: 1/16 Estimated number of students out of 70 observing 4 tails: 70*(1/16) = 4 Probability of one student observing 1 head and 3 tails: 4/16 Estimated number of students out of 70 observing 1 heads and 3 tails: 70*(4/16)= 17.5 Probability of one student observing 2 heads and 2 tails: 6/16 Estimated number of students out of 70 observing 2 heads and 2 tails: 70*(6/16) = 26.3 … We would need to see >4 students get 0 heads and 4 tails before we believe any coins are biased.

17 Simulation/theory Expect 0.09 of a locus to reach LOD=3 by chance.

18 Simulation/theory But this would change in a different organism, with different number of markers, etc. So in practice, everyone does their own simulation specific to their own study.

19 Candidate gene approach Hypothesize that causal variant will be in known pigment gene or regulator. NOT randomly chosen markers genome-wide.

20 Candidate gene approach Red progeny have RFLP pattern like red parent

21 Affected sib pair method 2,22,3 2,2 4,41,3 1,4 … Sib pairsObserved Expected under null Same allele 2(1/2)*2 Different allele 0(1/2)*2  2 =  (O - E) 2 E Test for significant allele sharing. Total # families

22 Qualitative but polygenic Fig. 3.12 Two loci. Need one dominant allele at each locus to get phenotype.

23 “A weak locus”: need lots of data AAbb aaBB AaBb Flower color inter-mate Two loci. Need one dominant allele at each locus to get phenotype. AABb AaBb aaBbAaBB aaBB Aabb Genotype at marker close to A locus purplewhite Top allele 31 Bottom allele 23

24 More generally (one locus): AA x BB AB (F1) AB x AB AA AB BA BB (F2)

25 25% 50%25% “Effect of having a B” AA AB BA BB AA AB BA BB Effect of a B allele is the same regardless of genotype: additive 1 locus, incomplete dominance

26 1 locus, complete dominance 75%25% AB BA AA BB Dominance is a kind of epistasis: nonadditive

27 A real example (F2’s) CC x SS CS CS x CS CC SS CS SC (F2’s)

28 Quantitative trait linkage test (F2’s) Not counting recombinants. Statistical test for goodness of fit.

29 Locus effect vs. parents C3H parent F2’s, C/C at marker F2’s, C/S at marker F2’s, S/S at marker SWR parent Homozygotes do not look like parent. What do you infer? A single varying locus does not explain the data

30 >1 locus controlling trait (One mouse family)

31 A weak locus C3H parent F2’s, C/C at marker F2’s, C/S at marker F2’s, S/S at marker SWR parent Most loci underlying human disease look like this. “Effect of having an S allele”

32 Heritability in exptal organisms Genetic variance = total var - “environmental var” Heritability H 2 = ee tt  g  =  t  -  e  g/tg/t

33 Heritability in humans: MZ twins http://www.sciam.com/media/inline/15DD5B0E-AB41- 23B8-2B1E53E8573428C5_1.jpg http://www.twinsinsurance.net/images/twins.jpg http://www.twinsrealm.com/ot hrpics/sarahandsandra.jpg http://www.twinsrealm.com/othrpics/twins16.jpg Each individual = z ij Total mean sq =  (z ij - z) 2 T Mean each pair = z i Within pairs mean sq =  (z ij - z i ) 2 N Between pairs mean sq =  (z i - z) 2 N-1 =  b 2 =  w 2 =  t 2 h 2 =  b 2  w 2 t2t2

34 Linkage mapping (quantitative) intoleranttolerant

35 Transgenic test

36 Fine-mapping: new markers Best marker Position of true causal variant Because you have to hunt through by hand to find the causal gene, and test experimentally. The smaller the region, the better.

37 Fine-mapping: new markers Position of true causal variant Increased marker density

38 Two loci, incomplete dominance 0.511.52

39 2-locus interaction Effect of J at locus 2 Locus 2 is epistatic to locus 1: effects of locus 1 are masked in individuals with JJ or JL,LJ at locus 2 Locus 2 follows a dominance model: JJ and JL,LJ have the same phenotype, LL differs “The dominant allele of locus 2 does the masking”

40 NO progeny as extreme as diploid hybrid

41 Three mutant genes From pathogenic strain Alleles from the same strain at different genes/loci can have different effects.

42 Linked mutations of opposite effect Path Lab Very unlikely

43 Fine-mapping Inject into golden larvae Golden uninjected WT uninjected

44 No truncation in humans, but… No other species have the Thr allele: what does this mean? Could be deleterious, just an accidental mutation. Could be advantageous for some humans, no other species.

45 Correlates with human differences AA AG GG Allele is rare Perhaps explains phenotypic variation among people of African ancestry Thr Thr, Ala Ala

46 Association mapping (qualitative) Fig. 11.26 Blue alleles at markers are on the same haplotype as the M allele of the disease locus

47 Association scan, qualitative osteoarthritiscontrols C’s141797 G’s47433  2 test

48 Fine-mapping -log(  2 p-value) rs377472

49 Beginnings of molecular confirmation coding polymorphisms

50 Association scan, quantitative

51 Association vs. linkage Strong, easy to detect, but rare in population; may not be reflective of common disease. Also, hard to collect family data. Common but weak effects; need 1000’s of samples to detect. If no common cause, can fail. Unrelated individuals Related individuals

52 diabetescontrol Gm23270 no Gm13433284 Association mapping causal loci “Gm is protective against diabetes?”

53 Association and admixture these are all the Caucasians…

54 Association and admixture Cases Controls = = Don’t believe any one locus is causative!

55 Genotyping by array Fig. 11.8

56 Coding sequence array Fig. 1.13

57 Marker is linked to polymorphism in expression regulation cascade ORF TF G kinase TF G G

58 Marker is linked to polymorphism in expression regulation cascade ORF TF G kinase TF mRNA level shows linkage to locus of polymorphic regulator(s).

59 Clinical applications Colored curves = fat mass at different body locations

60 Association of human transcripts linkage (families) assoc (unrelated)

61 Protein inheritance

62 PSI+ phenotypes 50°C Genetically distinct S. cerevisiae strains


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