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Population Genetics, Recombination Histories & Global Pedigrees Finding Minimal Recombination Histories 1 23 4 1 23 4 1 2 3 4 Global Pedigrees Finding.

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Presentation on theme: "Population Genetics, Recombination Histories & Global Pedigrees Finding Minimal Recombination Histories 1 23 4 1 23 4 1 2 3 4 Global Pedigrees Finding."— Presentation transcript:

1 Population Genetics, Recombination Histories & Global Pedigrees Finding Minimal Recombination Histories 1 23 4 1 23 4 1 2 3 4 Global Pedigrees Finding Common Ancestors NOW Population Genetics and Genealogies NOW Forward in time Backward in time

2 Wright-Fisher Model of Population Reproduction Haploid Model i. Individuals are made by sampling with replacement in the previous generation. ii. The probability that 2 alleles have same ancestor in previous generation is 1/2N Diploid Model Individuals are made by sampling a chromosome from the female and one from the male previous generation with replacement Assumptions 1.Constant population size 2.No geography 3.No Selection 4.No recombination

3 10 Alleles’ Ancestry for 15 generations Generate new generations according to the haploid model Sort generations forward in time, such that individual 1’s children in the next generation comes first, then the children of 2.. Pick individuals in the present and label edges to their ancestors back in time

4 Mean, E(X 2 ) = 2N. Ex.: 2N = 20.000, Generation time 30 years, E(X 2 ) = 600000 years. Waiting for most recent common ancestor - MRCA P(X 2 = j) = (1-(1/2N)) j-1 (1/2N) Distribution of time until 2 alleles had a common ancestor, X 2 ?: P(X 2 > j) = (1-(1/2N)) j P(X 2 > 1) = (2N-1)/2N = 1-(1/2N) 1 2N 1 1 1 2 j 1 1 2 j

5 P(k):=P{k alleles had k distinct parents} 1 2N 1 (2N) k k -> any 2N *(2N-1) *..* (2N-(k-1)) =: (2N) [k] k -> k Ancestor choices: k -> k-1 For k << 2N: k -> j S k,j - the number of ways to group k labelled objects into j groups.(Stirling Numbers of second kind.

6 Expected Height and Total Branch Length Expected Total height of tree: H k = 2(1-1/k) +..+ 1/(k-1)) ca= 2*ln(k-1) 1 2 3 k 2 1 2/(k-1) Branch Lengths 1/3 1 Time Epoch i.Infinitely many alleles finds 1 allele in finite time. ii. In takes less than twice as long for k alleles to find 1 ancestors as it does for 2 alleles. Expected Total branch length in tree, L k : 2*(1 + 1/2 + 1/3 +..+ 1/(k-1)) ca= 2*ln(k-1)

7 6 Realisations with 25 leaves Observations: Variation great close to root. Trees are unbalanced.

8 Sampling more sequences The probability that the ancestor of the sample of size n is in a sub-sample of size k is Letting n go to infinity gives (k-1)/(k+1), i.e. even for quite small samples it is quite large.

9 Three Models of Alleles and Mutations. i. Only identity, non- identity is determinable ii. A mutation creates a new type. i. Allele is represented by a line. ii. A mutation always hits a new position. i. Allele is represented by a sequence. ii. A mutation changes nucleotide at chosen position. Infinite Allele  Infinite Site  Finite Site acgtgctt acgtgcgt acctgcat tcctgcat acgtgctt acgtgcgt acctgcat tcctggct tcctgcat 

10 Final Aligned Data Set: Infinite Site Model

11 Recombination-Coalescence Illustration Copied from Hudson 1991 Intensities Coales. Recomb. 1 2  3 2  6 2  1 (1+b)  0  3 (2+b)  b Backward in time

12 Local Inference of Recombinations 00001110000111 00011010001101 00 10 01 11 Four combinations Incompatibility: Myers-Griffiths (2002): Number of Recombinations in a sample, N R, number of types, N T, number of mutations, N M obeys: 00110011 01010101 T... G T... C A... G A... C Recoding At most 1 mutation per column 0 ancestral state, 1 derived state

13 ”Observing” Recombinations: Hudson & Kaplan’s R M If you equate R M with expected number of recombinations, this could be used as an estimator. Unfortunately, R M is a gross underestimate of the real number of recombinations. 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1

14 Minimal Number of Recombinations Last Local Tree Algorithm: L 21 Data 2 n i-1i 1 Trees The Kreitman data (1983): 11 sequences, 3200bp, 43(28) recoded, 9 different How many neighbors? Bi-partitions How many local trees? Unrooted Coalescent

15 Metrics on Trees based on subtree transfers. Pretending the easy problem (unrooted) is the real problem (age ordered), causes violation of the triangle inequality: Tree topologies with age ordered internal nodes Rooted tree topologies Unrooted tree topologies Trees including branch lengths

16 Observe that the size of the unit-neighbourhood of a tree does not grow nearly as fast as the number of trees Allen & Steel (2001) Song (2003+) Due to Yun Song Tree Combinatorics and Neighborhoods

17 1 2 3 4 5 6 7 Due to Yun Song

18 minARGs: Recombination Events & Local Trees True ARG Reconstructed ARG 1 23 45 123 45 ((1,2),(1,2,3)) ((1,3),(1,2,3)) n=7,  =10,  =75 Minimal ARG True ARG Mutation information on only one side Mutation information on both sides 0 4 Mb Hudson- Kaplan Myers- Griiths Song-Hein n=8,  =40 n=8,  =15

19 Counting + Branch and Bound Algorithm ? Exact length Lower bound Upper Bound 0 3 1 91 2 1314 3 8618 4 30436 5 62794 6 78970 7 63049 8 32451 9 10467 10 1727 289920 k-recombinatination neighborhood k Song, Y.S., Lyngsø, R.B. & Hein, J. (2006) Counting all possible ancestral configurations of sample sequences in population genetics. IEEE/ACM Transactions on Computational Biology and Bioinformatics, 3(3), 239–251 Lyngsø, R.B., Song, Y.S. & Hein, J. (2008) Accurate computation of likelihoods in the coalescent with recombination via parsimony. Lecture Notes in Bioinformatics: Proceedings of RECOMB 2008 463–477

20 BB & Heuristic minimal ancestral recombination graphs

21 Time versus Spatial 1: Coalescent-Recombination (Griffiths, 1981; Hudson, 1983 - Wiuf & Hein, 1999) Temporal ProcessSpatial Process ii. The trees cannot be reduced to Topologies i. The process is non-Markovian * * =

22 Elston-Stewart (1971) -Temporal Peeling Algorithm: Lander-Green (1987) - Genotype Scanning Algorithm: Mother Father Condition on parental states Recombination and mutation are Markovian Mother Father Condition on paternal/maternal inheritance Recombination and mutation are Markovian Time versus Spatial 2: Pedigrees

23 Time versus Spatial 3: Phylogenetic Alignment Optimisation Algorithms indels of length 1 (David Sankoff, 1973) Spatial indels of length k (Bjarne Knudsen, 2003) Temporal Statistical Alignment Spatial: Temporal:

24 - recombination 27 ACs 0 1 2 3 4 5 6 7 8 1 1 4 2 5 3 1 5 5 The Griffiths-Ethier-Tavare Recursions No recombination: Infinite Site Assumption Ancestral State Known History Graph: Recursions Exists No cycles Possible Histories without Recombination for simple data example + recombination 9*10 8 ACs

25 1st 2nd Ancestral configurations to 2 sequences with 2 segregating sites mid-point heuristic M M R C R C C

26 1st 2nd 0-ARG 5 states 1-ARG +15 states 2-ARG +10 states ,  = 2, 0 ,  = 1, 1 ,  = 2, 2 0.148 0.037 0.032 0.148 0.082 0.074 0.148 0.090 0.085

27 Likelihood Calculations on the  -ARG 010 101 110 Example: 0-ARG 2-ARG1-ARG

28 For sites=7, sequences=6, then 10 -7 th of states visited Cut-off  -ARG,  =2  -ARG likelihood calculation

29 Combining Ancestral Individuals and the Coalescent Wiuf & Hein, 2000. Let T be the time, when somebody was everybody’s ancestor. Changs’ result: lim T*/log 2 (N) =1 prob. 1 Unify the two processes: I. Sample more individuals II. Let each have 2 parents with probabilty p. Result: A discontinuity at 1. For p<1 change log 2  log p Comment: Genetic Ancestors is a vanishing set within Genealogical Ancestors.

30 Reconstructing global pedigrees: Superpedigrees Steel and Hein, 2006 The gender-labeled pedigrees for all pairs defines global pedigree k Gender-unlabeled pedigrees don’t!! Steel, M. & Hein, J. (2006) Reconstructing pedigrees: a combinatorial perspective. Journal of Theoretical Biology, 240(3), 360–367

31 All embedded phylogenies are observable Do they determine the pedigree? Genomes with  and  --> infinity  recombination rate,  mutation rate Benevolent Mutation and Recombination Process Counter example: Embedded phylogenies:

32 Infinite Sequences: From ARG to Pedigree Given A/B/C/D/E above how much does that constrain set of pedigrees How many pedigrees are compatible with A/B/C/D/E varying over data? A. The ARG? What can you observe from data (infinite sequences)? C. Sequence of local trees? D. Set of local trees? E. Set/Sequences of local unrooted tree topologies? B. Sequence of neighbor pairs of local trees with recombination points Going to neighbor triples, quadruples,.., be more restrictive than pairs? F. Set/Sequences of local bipartitions? (neighbor pairs…

33 Infinite Sequences: From ARG to Pedigree A. The ARG? What can you observe from data (infinite sequences)? C. Sequence of local trees? D. Set of local trees? E. Set/Sequences of local unrooted tree topologies? B. Sequence of pairs of local trees with recombination points Going to triples, quadruples,.., be more restrictive than pairs? 1 23 4 1 23 4 1 2 3 4 F. Set/Sequences of bipartitions


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