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Evolution in dogs CS-374 Abhinay Nagpal abhinay@stanford.edu

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Papers that are discussed Evolution in dogs: A single IGF1 allele is a major determinant of small size in dogs Genome-wide SNP and haplotype analyses reveal a rich history underlying dog domestication

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Why are we interested in evolution of dogs? Dogs show greatest variation in size in vertebrates. It has been attributed to domestication Dogs have high similarity of multi locus haplotypes present in wolves in middle east

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A Single IGF1 Allele Is a Major Determinant of Small Size in Dogs

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The Experiment

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Identifying QTL Quantitative trait loci (QTLs)-> stretches of DNA linked to the genes that tie to a phentoype trait 2 radiographic skeletal measurements for size and shape-> two QTL (FH2017 at 37.9 Mb and FH2295 at 43.5 Mb) strongly associated with body size

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Relationships of skeletal size, SNP markers, IGF1 haplotype, and serum levels of the IGF1 protein in PWDs A Single IGF1 Allele Is a Major Determinant of Small Size in Dogs

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Mixed model for Portuguese water dog fine-mapping Y is the vector of the skeletal size trait; α is a vector of fixed effect, the SNP effect we are testing; u is a vector of random effect reflecting the polygenetic background; X and Z are known incidence matrices relating the observations to fixed and random effects, respectively. The variance in the model can be expressed as

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IGF1 Influences Size of dogs Average heterozygosity in small dogs near IGF1 is only 25% of that in large dogs A narrow precise genomic region holds the variant responsible for small size.

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Evidence of Association

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FISHER’S EXACT TEST Computes directly the probability of observing a particular set of frequencies in a 2 x 2 table Returns inflated p values

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Consider a hare and tortoise race in which the outcomes are as follows: H H H H H H H H H T T T T T T T T T T H H H H H H H H H H T T T T T T T T T Median tortoise here comes in at position 19 Median hare comes in at position 20. However, the value of U (for hares) is 100 Value of U(for tortoises) is 261 http://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U Mann Whitney

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Cumulative distribution function for Fisher’s exact test and Mann-Whitney U statistic calculated from 83 genomic control SNPs genotyped in small and giant dogs Mann Whitney V.S Fisher’s Test

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Association of body size and frequency of the SNP 5 A allele

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Fixation index Measure of the diversity of randomly chosen alleles within the same sub- population relative to that found in the entire population.

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Findings IGF1 haplotype substantially contributes to size Size diversity was present early in the history of domestication Ancestral small dog IGF1 haplotype was spread over a large geographic area by trade and human migration

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Genome-wide SNP and haplotype analyses reveal a rich history underlying dog domestication

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Highlights The dog is a “striking example” of variation under domestication Evolutionary processes poorly understood Did dogs first evolve in East Asia?

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Data Survey of 48000 SNPs in dogs and wolves(grey wolf) Typed from 912 dogs - 85 breeds 225 grey wolves 11 globally distributed population

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http://www.sciencemag.org/content/276/5319/1687.full Dog Evolution

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Bayesian clustering & Neighbor joining trees An example tree with 4 data points. The clusterings (1 2 3)(4) and (1 2)(3)(4) are tree-consistent partitions The clustering (1)(2 3)(4) is not a treeconsistent partition http://www.gatsby.ucl.ac.uk/~heller/bhcnew.pdf

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Bayesian Hierarchical Clustering Algorithm

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Neighbor Joining Trees http://www.icp.ucl.ac.be/~opperd/private/neighbor.html distance matrix:

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We have in total 6 elements (N=6). Step 1: We calculate the net divergence r (i) for each element from all other elements r(A) = 5+4+7+6+8=30 r(B) = 42 r(C) = 32 r(D) = 38 r(E) = 34 r(F) = 44 Step 2: Now we calculate a new distance matrix using for each pair M(ij)=d(ij) - [r(i) + r(j)]/(N-2) or in the case of the pair A,B: M(AB)=d(AB) -[(r(A) + r(B)]/(N-2) = -13

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Step 3: Choose as neighbors pairs for which Mij is the smallest. ==> A and B and D and E. Let's take A and B as neighbors and we form a new node called U. Calculate the branch length from the internal node U to A and B. S(AU) =d(AB) / 2 + [r(A)-r(B)] / 2(N-2) = 1 S(BU) =d(AB) -S(AU) = 4 Step 4: Now we define new distances from U to each other terminal node: d(CU) = d(AC) + d(BC) - d(AB) / 2 = 3 d(DU) = d(AD) + d(BD) - d(AB) / 2 = 6 d(EU) = d(AE) + d(BE) - d(AB) / 2 = 5 d(FU) = d(AF) + d(BF) - d(AB) / 2 = 7

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Neighbour-joining trees of domestic dogs and grey wolves.

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Variation within breed : 65% of total variation/diversity Variation within breed grouping: 31% of total variation/diversity Variation between breed groupings: 3.8% of total variation/diversity Analysis of molecular variance (AMOVA)

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Principal component analysis (PCA) of 48,036 SNPs

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For 5-SNP haplotype windows: haplotype sharing higher between modern dog breeds and Middle Eastern wolves For 15-SNP windows : the majority of breeds show the most sharing with Middle Eastern wolves This has dog breeds of diverse geographic origins Only two east Asian breeds (Akita and chow chow) had higher sharing with Chinese wolves Observations

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Haplotype sharing higher in modern dog breeds and Middle Eastern wolves Eg: basenji, chihuahua, basset hound and borzoi Neighbour-joining trees excellent for breed history & diversity Breed groupings mirror breed classification based on form and function Findings

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http://pritch.bsd.uchicago.edu/publications/structure.pdf References http://www.nature.com/nature/journal/v464/n7290/extref/nature08837-s1.pdf A Single IGF1 Allele Is a Major Determinant of Small Size in Dogs Sutter et al. Science,2007 Genome-wide SNP and haplotype analyses reveal a rich history underlying dog domestication. vonHoldt et al. nature,2010. http://en.wikipedia.org/wiki/Fisher's_exact_test

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Consider another hare and tortoise race, with 19 participants of each species in which the outcomes are as follows: H H H H H H H H H T T T T T T T T T T H H H H H H H H H H T T T T T T T T T The median tortoise here comes in at position 19, and thus actually beats the median hare which comes in at position 20. However, the value of U (for hares) is 100 (9 Hares beaten by (x) 0 tortoises) + (10 hares beaten by (x) 10 tortoises) = 0 + 100 = 100 Value of U(for tortoises) is 261 (10 tortoises beaten by 9 hares) + (9 tortoises beaten by 19 hares) = 90 + 171 = 261 Consulting tables, or using the approximation below shows that this U value gives significant evidence that hares tend to do better than tortoises (p < 0.05, two-tailed). Backup slide mann whitney

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