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
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
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
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
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?
Data Survey of 48000 SNPs in dogs and
wolves(grey wolf) Typed from 912 dogs - 85 breeds 225
grey wolves 11 globally distributed population
http://www.sciencemag.org/content/276/5319/1687.full
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Dog Evolution
Bayesian clustering &Neighbor joining trees
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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
Neighbor Joining Trees
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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=30r(B) = 42r(C) = 32r(D) = 38r(E) = 34r(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
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
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
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 DogsSutter 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 speciesin 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 TThe 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 = 100Value of U(for tortoises) is 261
(10 tortoises beaten by 9 hares) + (9 tortoises beaten by 19 hares) = 90 + 171 = 261Consulting 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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