ACGTTTGACTGAGGAGTTTACGGGAGCAAAGCGGCGTCATTGCTATTCGTATCTGTTTAG Human Population Genomics

Transcription

1 ACGTTTGACTGAGGAGTTTACGGGAGCAAAGCGGCGTCATTGCTATTCGTATCTGTTTAG Human Population Genomics

2 Heritability & Environment Feasibility of identifying genetic variants by risk allele frequency and strength of genetic effect (odds ratio). TA Manolio et al. Nature 461, (2009) doi: /nature08494

3 Global Ancestry Inference G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} Nature November 6; 456(7218):

4 Modeling population haplotypes VLMC G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} G i = H i1 + H i2, where, H i = h ij1 h ijn ; h ijk {0, 1} Browning, 2006

5 Phasing Browning & Browning, 2007

6 Identity By Descent { {

7 IBD detection IBD = F IBD = T Parente Rodriguez et al FastIBD: sample haplotypes for each individual, check for IBD Browning & Browining 2011

8 Mexican Ancestry The genetics of Mexico recapitulates Native American substructure and affects biomedical traits, Moreno-Estrada et al. Science, 2014.

9 Population Sequencing C = 2-6x A/T C/C C/G 1,000 to A/A C/T G/G 1,000,000 A/A T/T C/G A/A C/T G/G

10 Population Sequencing C = 2-6x A/T C/C C/G 1,000 to A/A C/T G/G 1,000,000 A/A T/T C/G A/A C/T G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ]

11 Population Sequencing When C is high (>30x), 1,000 to A/T A/A C/C C/T C/G G/G 2-6x Prob(g ij = k data) ~ Prob(g ij = k reads mapping on (i, j)) fast & easy 1,000,00 0 A/A A/A T/T C/T C/G G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] When C is low, Prob(g ij = k data) needs to leverage LD: positions j j in all individuals in principle, intractable

12 Population Sequencing Summarization - Maximization: 1,000 to A/T A/A C/C C/T C/G G/G 2-6x 1. Identify candidate polymorphic sites 2. Initialize G (0) 3. Summarization: 1,000,00 0 A/A A/A T/T C/T C/G G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] p (n+1) ijk = Prob(g ij = K G(n), data) 4. Maximization: g (n+1) ijk = argmax p(n+1) ijk 5. Repeat until convergence

13 Modeling LD: Nearest Neighbors i j Sample 1 Sample 2 Sample 3 Sample l Let S i = { samples with >= one read covering minor allele } S i = {1, 2, 3, 10} S j = {1, 3, 4} Sample Then, Sim 1 (i, j) = (S i S j ) / (S i U S j ) = 2/4

14 Reveel Algorithm Summarization Maximization: 1,000 to A/T A/A C/C C/T C/G G/G 2-6x 1. Identify candidate polymorphic sites 2. Initialize G (0) 3. Summarization: 1,000,000 A/A T/T C/G A/A C/T G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] p (n+1) ijk = Prob(g ij = K G(n), data) 4. Maximization: g (n+1) ijk = argmax p(n+1) ijk 5. Repeat until convergence Candidate Polymorphic site Essentially, pos n j where some individuals have at least 2 reads with same minor allele

15 Reveel Algorithm Summarization Maximization: 1,000 to A/T A/A C/C C/T C/G G/G 2-6x 1. Identify candidate polymorphic sites 2. Initialize G (0) 3. Summarization: 1,000,000 A/A T/T C/G A/A C/T G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] p (n+1) ijk = Prob(g ij = K G(n), data) 4. Maximization: g (n+1) ijk = argmax p(n+1) ijk 5. Repeat until convergence At each position j, Use sum of read counts at j and its nearest neighbors

16 Reveel Algorithm: calculate P (n+1) p (n+1) ijk = P(g ij = k G(n), reads) 1,000 to A/T A/A C/C C/T C/G G/G 2-6x ~ P(g ij = k g knn, reads) = P(reads g ij = k) P(g ij = k g knn) ) 1,000,000 A/A T/T C/G A/A C/T G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] P(reads g ij = k) : easy P(g ij = k g knn ) = Let C 0, C 1, C 2 = # samples matching i in knn at (n), with j th genotype pos n = 0, 1, 2 Then, P(g ij = k g knn ) = C k / (C 0 + C 1 + C 2 )

17 Reveel Algorithm Summarization Maximization: 1,000 to A/T A/A C/C C/T C/G G/G 2-6x 1. Identify candidate polymorphic sites 2. Initialize G (0) 3. Summarization: 1,000,000 A/A T/T C/G A/A C/T G/G G 1,, G N ; G i = g i1 g in ; g ij {0, 1, 2} P 1,, P N ; P i : [ p ijk = Prob(g ij = k data) ] p (n+1) ijk = Prob(g ij = K G(n), data) 4. Maximization: g (n+1) ijk = argmax p(n+1) ijk 5. Repeat until convergence At each position j, Use sum of read counts at j and its nearest neighbors

18 Fixation, Positive & Negative Selection How can we detect negative selection? Negative Selection How can we detect positive selection? Neutral Drift Positive Selection

19 How can we detect positive selection? Ka/Ks ratio: Ratio of nonsynonymous to synonymous substitutions Very old, persistent, strong positive selection for a protein that keeps adapting Examples: immune response, spermatogenesis

20 How can we detect positive selection?

21 Positive Selection in Human Lineage

22 Positive Selection in Human Lineage

23 Long Haplotypes EHS, ihs tests Less time: Fewer mutations Fewer recombinations

24 Application: Malaria Study of genes known to be implicated in the resistance to malaria. Infectious disease caused by protozoan parasites of the genus Plasmodium Frequent in tropical and subtropical regions Transmitted by the Anopheles mosquito Slide Credits: Image source: wikipedia.org Marc Schaub

25 Application: Malaria Image source: Slide Credits: NIH - Marc Schaub

26 Application: Malaria Image source: CDC - malaria_risk_2003.gif Slide Credits: Marc Schaub

27 Results: G6PD Source: Sabeti et al. Nature Slide Credits: Marc Schaub

28 Results: TNFSF5 Source: Sabeti et al. Nature Slide Credits: Marc Schaub

29 Malaria and Sickle-cell Anemia Allison (1954): Sickle-cell anemia is limited to the region in Africa in which malaria is endemic. Distribution of malaria Distribution of sickle-cell anemia Slide Credits: Image source: wikipedia.org Marc Schaub

30 Malaria and Sickle-cell Anemia Single point mutation in the coding region of the Hemoglobin-B gene (glu val). Heterozygote advantage: Resistance to malaria Slight anemia. Image source: wikipedia.org Slide Credits: Marc Schaub

31 Lactose Intolerance Source: Ingram and Swallow. Population Genetics of Encyclopedia of Life Slide Credits: Sciences Marc Schaub

32 Lactose Intolerance LCT, 5 LCT, 3 Source: Bersaglieri et al. Am. J. Hum. Genet Slide Credits: Marc Schaub

33 Positive Selection in Human Lineage