An applied statistician does probability:

Transcription

1 An applied statistician does probability: It s not pretty Karl W Broman Department of Biostatistics & Medical Informatics University of Wisconsin Madison

2 An applied statistician does probability: Don t think do! Karl W Broman Department of Biostatistics & Medical Informatics University of Wisconsin Madison

3 An applied statistician does probability: Don t think simulate! Karl W Broman Department of Biostatistics & Medical Informatics University of Wisconsin Madison

4 Genotype probabilities in the pre-collaborative Cross An applied statistician Karl W Broman Department of Biostatistics & Medical Informatics University of Wisconsin Madison

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6 Human vs mouse 6

7 Intercross P P 2 F F F 2 7

8 The Collaborative Cross G 0 A B C D E F G H G A B C D E F G H G 2 ABCD EFGH G 2 F G 2 F 2 G 2 F 8

9 Recombination A B A B A B A B A B B A ( r) 2 ( r) 2 r 2 r 2 r is the "recombination fraction" 9

10 X in the CC G 0 A B C D E F G H G A B C Y D E F G Y H G 2 AB C EF Y H G 2 F G 2 F 2 G 2 F 0

11 In the CC Autosomes Pr(G = i) = /8 Pr(G 2 = j G = i) = r/( + 6r) for i j Pr(G 2 G ) = 7r/( + 6r) X chromosome Pr(G = A) = Pr(G = B) = Pr(G = E) = Pr(G = F) = /6 Pr(G = C) = /3 Pr(G 2 = B G = A) = r/( + r) Pr(G 2 = C G = A) = 2r/( + r) Pr(G 2 = A G = C) = r/( + r) Pr(G 2 G ) = (/3)r/( + r) Broman, Genetics 69:33 6, 2005

12 Computer simulations R R = 7r / (+6r) R = (/3)r / (+r) r 2

13 -way RIL G A B C D G 2 A B C D G 2 F G 2 F 2 G 2 F 3

14 8-way RIL G 0 A B C D E F G H G A B C D E F G H G 2 ABCD EFGH G 2 F G 2 F 2 G 2 F

15 The PreCC What happens at G 2 F k? Pr(g = i) Pr(g = i, g 2 = j) as a function of k as a function of k and the recombination fraction...ideally, the joint probabilities for the pair of individuals....but, at least, for a random individual. 5

16 locus, X chr G 2 AB C G 2 F G 2 F 2 AA A AA C AC A AC C 6

17 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C 7

18 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C 8

19 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C AB A AB C BC A BC C 9

20 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C AB A AB C BC A BC C AB B AB C AC B AC C 20

21 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C AB A AB C BC A BC C AB B AB C AC B AC C BB B BB C BC B BC C 2

22 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C AB A AB C BC A BC C AB B AB C AC B AC C BB B BB C BC B BC C 22

23 locus, X chr G 2 AB C G 2 F G 2 F 2 AC A AC B BC A BC B AA A AA C AC A AC C AB A AB C BC A BC C AB B AB C AC B AC C BB B BB C BC B BC C Females AA, BB /8 each AC, BC / each AB / each Males A, B / each C /2 23

24 Markov chains Sequence of random variables {X 0, X, X 2,... } satisfying Pr(X n+ X 0, X,..., X n ) = Pr(X n+ X n ) Here, X n = parental type at generation n. Transition probabilities P ij = Pr(X n+ = j X n = i) If π 0 is the starting distribution (at G 2 ), π k = π 0 P k is the distribution at G 2 F k 2

25 locus, X chr, male A B C G G 2 F /2 /2 0 G 2 F 2 / / /2 G 2 F 3 / G 2 F 3/8 G 2 F 5 5/6 G 2 F 6 /32 G 2 F 7 2/6 G 2 F 8 3/28 G 2 F 9 85/256 25

26 locus, X chr, male A B C G G 2 F /2 /2 0 G 2 F 2 / / /2 G 2 F 3 / G 2 F 3/8 G 2 F 5 5/6 G 2 F 6 /32 G 2 F 7 2/6 G 2 F 8 3/28 G 2 F 9 85/256 G 2 F k (for k ): Pr(C) = k i=0 ( ) i 2 26

27 AA, BB AC, BC AB CC G G 2 F 0 /2 0 0 G 2 F 2 /8 2/8 2/8 0 G 2 F 3 /8 2/8 /8 /8 G 2 F 3/6 3/6 2/6 2/6 G 2 F 5 3/6 0/6 6/6 2/6 G 2 F 6 5/6 8/6 5/6 3/6 G 2 F 7 32/28 3/28 8/28 30/28 G 2 F 8 37/52 2/52 26/52 28/52 G 2 F 9 3/52 3/52 2/52 37/52 locus, X chr, female 27

28 AA,BB,CC,DD AB,CD AC,AD,BC,BD G 2 0 /2 0 G 2 F 0 0 / G 2 F 2 /6 2/6 2/6 G 2 F 3 3/32 2/32 /32 G 2 F /32 2/32 3/32 G 2 F 5 9/28 6/28 0/28 G 2 F 6 3/256 0/256 6/256 G 2 F 7 7/256 8/256 3/256 G 2 F 8 20/02 26/02 2/02 G 2 F 9 23/208 2/208 68/208 locus, autosome, ind l 28

29 Summary locus 2 loc, hap 2 loc, gen X chr Male Female Pair Autosome Ind l Pair 29

30 Φ (k) = kinship coefficient of ind l w/ itself at G 2F k Φ (k) 2 = kinship coefficient of sibs at G 2F k Kinship coefficients Φ (0) = 2, Φ(0) 2 = 0 Φ (k+) = Φ(k) 2 Φ (k+) 2 = 2 Φ(k) + 2 Φ(k) 2 Φ (k+2) 2 = 2 Φ(k+) 2 + Φ(k) 2 + Φ (k+3) 2 = 3 2 Φ(k+2) 2 Φ(k+) 2 Φ(k) 2 30

31 Fibonacci numbers 0,,, 2, 3, 5, 8, 3, 2, 3, 55, 89,, 233, 377, 60, 987,... x k = x k + x k 2, where x 0 = 0, x = Characteristic polynomial: x 2 x Roots: ϕ = + 5 2, ϕ = 5 2 ϕ is the golden ratio x k = Aϕ k + B( ϕ) k =... = ϕk ( ϕ) k 5 3

32 AA, BB AC, BC AB CC G G 2 F 0 /2 0 0 G 2 F 2 /8 2/8 2/8 0 G 2 F 3 /8 2/8 /8 /8 G 2 F 3/6 3/6 2/6 2/6 G 2 F 5 3/6 0/6 6/6 2/6 G 2 F 6 5/6 8/6 5/6 3/6 G 2 F 7 32/28 3/28 8/28 30/28 G 2 F 8 37/52 2/52 26/52 28/52 G 2 F 9 3/52 3/52 2/52 37/52 locus, X chr, female 32

33 locus, X chr, female Pr(AC) satisfies a k+2 = 2 a k+ + a k with a 0 = 0, a = 2 [ a k = ( 5 + ) k ( 5 5 ) ] k 5 Pr(AB) satisfies a k+2 = 2 a k+ + a k with a 0 =, a = 0 ( ) ( ) k ( ) ( ) k a k = Pr(AA) a k = 3 + ( 6 ) ( ) ( k 2 5+ ) ( ) k ( 5 5 ) ( 5 20 ) k 5 Pr(CC) a k = 3 + ( 6 ) ( ) ( k 2 5+ ) ( ) k ( 5 5 ) ( 5 20 ) k 5 33

34 AA,BB,CC,DD AB,CD AC,AD,BC,BD G 2 0 /2 0 G 2 F 0 0 / G 2 F 2 /6 2/6 2/6 G 2 F 3 3/32 2/32 /32 G 2 F /32 2/32 3/32 G 2 F 5 9/28 6/28 0/28 G 2 F 6 3/256 0/256 6/256 G 2 F 7 7/256 8/256 3/256 G 2 F 8 20/02 26/02 2/02 G 2 F 9 23/208 2/208 68/208 locus, autosome, ind l 3

35 locus, autosome, ind l Pr(AC) satisfies a k+2 = 2 a k+ + a k with a 0 = 0, a = [ a k = ( 5 + ) k ( 5 0 ) ] k 5 Pr(AB) satisfies a k+2 = 2 a k+ + a k with a 0 = 2, a = 0 ( ) ( ) k ( ) ( ) k a k = Pr(AA) (5+2 a k = ) ( ) k ( ) ( 5 0 ) k 5 35

36 2-locus hap, X chr, male For Pr(CC): r = 0.0 a k+ = a k a k a k a k r = 0.02 a k+ = a k a k a k a k r = 0.03 a k+ = a k a k a k a k a k+ = ( 2 r r ) ak+3 + ( 2 r ) ak+2 ( 3 r 8 ) ak+ ( ) 2r 8 ak Pr(CC) = 2r r+3 ( ) ( k 2 + α r+ ) k ( r 2 0r+5 + β where α = 2r +(2r 3 r 2 +r) r 2 0r+5 9r 3 +5r r 35r 3 29r 2 +5r+5 β = 2r (2r 3 r 2 +r) r 2 0r+5 9r 3 +5r r 35r 3 29r 2 +5r+5 r ) k r 2 0r+5 Pr(AC), Pr(AB), Pr(AA) similarly obnoxious 36

37 Consider the two individuals at a single autosomal locus. Consider a particular allele (say A). Pool others into a single allele (B). Fixation probability We want Pr(X k = AA AA) [or really that] Transition matrix, P: AA BB AA AB AB AB AB BB AA AA BB BB AA BB AA AB 0 AB AB AB BB AA AA BB BB

38 -way RIL G A B C D G 2 A B C D G 2 F G 2 F 2 G 2 F 38

39 Fixation probability We seek π k = π 0 P k where π 0 = ( ) [start at AB BB]. Write P = VDV where D = diagonal matrix of eigenvalues, V = eigenvectors So P k = VD k V, and so π k = π 0 VD k V. Thus, the fixation probability for - and 8-way RIL is + ( ) k ( 2 ) ( k 5 9 ) ( 5 ) k ( ) ( + ) k 5 For 2-way RIL, we use π 0 = ( ) [start at AA BB], and obtain + ( ( 2 ) ( k 5) 7 3 ) ( 5 ) k ( ) ( 5 + ) k 5 0 [Caveat: meaning of generation k somewhat different in 2-, -, and 8-way RIL] 39

40 Fixation probability Probability fixed A, 2 way RIL A, or 8 way RIL X, or 8 way RIL Generation 0

41 Time of fixation Probability fixed at exactly this generation mean = 6 2/3 mean = 7 mean = 7 2/3 A, 2 way RIL A, or 8 way RIL X, or 8 way RIL Generation

42 Kimura (963) C k (xy) = Pr(random chr from gen k has x at locus and y at locus 2) S k (xy) = Pr(random chr from gen k has x at locus and opposite chr in that ind l has y at locus 2) T k (xy) = Pr(random chr from gen k has x at locus and random chr from other individual has y at locus 2) Kimura, Biometrics 9: 7, 963 2

43 Kimura (963) C k (xy) = Pr(random chr from gen k has x at locus and y at locus 2) S k (xy) = Pr(random chr from gen k has x at locus and opposite chr in that ind l has y at locus 2) T k (xy) = Pr(random chr from gen k has x at locus and random chr from other individual has y at locus 2) Kimura, Biometrics 9: 7, 963 3

44 Kimura (963) C k (xy) = Pr(random chr from gen k has x at locus and y at locus 2) S k (xy) = Pr(random chr from gen k has x at locus and opposite chr in that ind l has y at locus 2) T k (xy) = Pr(random chr from gen k has x at locus and random chr from other individual has y at locus 2) C k+ = ( r)c k + rs k S k+ = T k T k+ = C k + S k + 2 T k Kimura, Biometrics 9: 7, 963

45 2-locus hap, autosome Pr(AA) = Pr(AB) = Pr(AC) = 8 { +6r + [ ] [ 2r z k { 2r +6r [ ] [ 2r z k { r +6r + [ ] [ 2r z k ] 6r 2 7r+3rz (+6r)z ] 0r 2 r+rz (+6r)z ] r 2 +6r 2rz (+6r)z [ ] [ 2r+z k + [ ] [ 2r+z k [ ] [ 2r+z k ]} 6r 2 7r 3rz (+6r)z ]} 0r 2 r rz (+6r)z ]} r 2 +6r+2rz (+6r)z where z = ( 2r)(5 2r) 5

46 2-locus gen, autosome We seek to extend the idea from Kimura (963) to get probabilities at G 2 F k for patterns like A A A A, A A B B, A A C C

47 2-locus gen, autosome The transition matrix (really, the matrix that defines the recursion): r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) 2 r 2 +( r) ( r) 2 2r( r) r 2 2 r 2 +( r) 2 ( r) 2 2 r( r) r 2 2 ( r) 2 r 2 r( r) 3 r 2 r ( r) r 6 7 ( r) r 8 r r r 2 r r r r r

48 What are we doing? We have a Markov chain with many states (say n) Transition matrix P; starting distribution π 0 We want to calculate π k = π 0 P k Really, we just need π k b = π 0 P k b for some n vector b Suppose we can expand b to an n m matrix B in a way that PB = BA for some m m matrix A Then P k B = P k (PB) = P k (BA) = = BA k And so π 0 P k B = π 0 BA k work with A [m m] in place of P [n n] 8

49 Counting states Two loci, X chromosome 3 alleles: 3 6 = 729 types Swap A B, two hapl in female, 2 loci 6 types To get at Pr(AA AA) in females, consider just 2 alleles (2 6 = 6 types) After symmetries: 2 types Two loci, autosome alleles: 8 = 65, 536 types After symmetries: 700 types To get at Pr(AA AA), consider just 2 alleles (2 8 = 256 types) After symmetries: 3 types 9

50 2-locus gen, autosome Returning to that 3 3 matrix, we need the eigenvalues and eigenvectors. The first 7 eigenvalues: ± 5 2r± r 2 2r+5 2r± ( 2r)(9 2r) 8 The other 6 eigenvalues are the roots of 026x 6 28(r 2 8r + 7)x 5 6(8r 2r r 2 0r + 3)x + 8(6r 80r r 2 8r + 23)x 3 8(6r 5 56r + 72r 3 8r 2 + 8r 3)x 2 + (6r 5 0r + 50r 3 35r 2 + 3r 2)x + (6r 5 0r + r 3 26r 2 + 8r ) 50

51 2-locus gen, X chr

52 2-locus gen, X chr The first 6 eigenvalues:, 2 ± 5 r± r 2 0r+5 The other 7 eigenvalues are the roots of and 6x 3 + x 2 (6r + )x + 2r 32x + (6r 2)x 3 + (6r 3 32r r 8)x 2 (r 3 8r 2 + 0r )x (r 3 6r 2 + r ) 52

53 Summary locus 2 loc, hap 2 loc, gen X chr Male Female Pair Autosome Ind l Pair 53

54 Acknowledgments Timo Seppäläinen Bret Larget Mathematics, UW Madison Statistics & Botany, UW Madison 5