斯坦福-计算生物学课程lecture

1、Evolution in dogsCS-374Abhinay NPapers that are discussedEvolution in dogs:A single IGF1 allele is a major determinant of small size in dogsGenome-wide SNP and haplotype analyses reveal a rich history underlying dog domesticationWhy are we interested in evolution of dogs?Dogs

2、 show greatest variation in size in vertebrates. It has been attributed to domesticationDogs have high similarity of multi locus haplotypes present in wolves in middle eastA Single IGF1 Allele Is a Major Determinant of Small Size in DogsThe ExperimentIdentifying QTLQuantitative trait loci (QTLs)- st

3、retches of DNA linked to the genes that tie to a phentoype trait2 radiographic skeletal measurements for size and shape- two QTL (FH2017 at 37.9 Mb and FH2295 at 43.5 Mb) strongly associated with body sizeRelationships of skeletal size, SNP markers, IGF1 haplotype, and serum levels of the IGF1 prote

4、in in PWDsA Single IGF1 Allele Is a Major Determinant of Small Size in DogsMixed model for Portuguese water dog fine-mapping Y is the vector of the skeletal size trait; is a vector of fixed effect, the SNP effect we are testing; u is a vector of random effect reflecting the polygenetic background; X

5、 and Z are known incidence matrices relating the observations to fixed and random effects, respectively. The variance in the model can be expressed asIGF1 Influences Size of dogsAverage heterozygosity in small dogs near IGF1 is only 25% of that in large dogsA narrow precise genomic region holds the

6、variant responsible for small size.Evidence of AssociationFISHERS EXACT TEST Computes directly the probability of observing a particular set of frequencies in a 2 x 2 table Returns inflated p valuesConsider a hare and tortoise race in which the es are as follows:H H H H H H H H H T T T T T T T T T T

7、 H H H H H H H H H H T T T T T T T T TMedian tortoise here comes in at position 19Median hare comes in at position 20.However, the value of U (for hares) is 100Value of U(for tortoises) is 261 Mann WhitneyCumulative distribution function for Fishers exact test and Mann-Whitney U statistic calculated

8、 from 83 genomic control SNPs genotyped in small and giant dogsMann Whitney V.S Fishers TestAssociation of body size and frequency of the SNP 5 A alleleFixation indexMeasure of the diversity of randomly chosen alleles within the same sub-population relative to that found in the entire population. Fi

9、ndings IGF1 haplotype substantially contributes to sizeSize diversity was present early in the history of domesticationAncestral small dog IGF1 haplotype was spread over a large geographic area by trade and human migrationGenome-wide SNP and haplotype analyses reveal a rich history underlying dog do

10、mesticationHighlightsThe dog is a “striking example” of variation under domestication Evolutionary processes poorly understoodDid dogs first evolve in East Asia?DataSurvey of 48000 SNPs in dogs and wolves(grey wolf)Typed from 912 dogs - 85 breeds 225 grey wolves11 globally distributed population Dog

11、 EvolutionBayesian clustering &Neighbor joining treesAn 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 Bayesian Hierarchical Clustering AlgorithmNeighbor Joining Trees distance mat

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

13、 of the pair A,B:M(AB)=d(AB) -(r(A) + r(B)/(N-2) = -13Step 3: Choose as neighbors pairs for which Mij is the smallest. = A and B and D and E. Lets 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) /

14、 2(N-2) = 1 S(BU) =d(AB) -S(AU) = 4Step 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 = 7Neighbour-joining trees of domestic dogs an

15、d grey wolves.Variation within breed : 65% of total variation/diversityVariation within breed grouping: 31% of total variation/diversityVariation between breed groupings: 3.8% of total variation/diversityAnalysis of molecular variance (AMOVA)Principal component analysis (PCA) of 48,036 SNPsFor 5-SNP

16、 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 originsOnly two east Asian breeds (Akita and chow chow) had higher s

17、haring with Chinese wolvesObservations Haplotype sharing higher in modern dog breeds and Middle Eastern wolvesEg: basenji, chihuahua, basset hound and borzoi Neighbour-joining trees excellent for breed history & diversity Breed groupings mirror breed classification based on form and functionFindings

18、 References A Single IGF1 Allele Is a Major Determinant of Small Size in DogsSutter et al. Science,2007Genome-wide SNP and haplotype analyses reveal a rich history underlying dog domestication. vonHoldt et al. nature,2010. s_exact_testConsider another hare and tortoise race, with 19 participants of each speciesin which the es are as follows:H H H H H H H H H T T T T T T T