Evolution in dogs CS-374 Abhinay Nagpal [email protected].

Evolution in dogs

CS-374

Abhinay Nagpal

[email protected]

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



The Experiment



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



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





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.



Evidence of Association



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



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



Fixation index Measure of the diversity of randomly

chosen alleles within the same sub-population relative to that found in the entire population.



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



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



Dog Evolution

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



Bayesian Hierarchical Clustering Algorithm

Neighbor Joining Trees



http://www.icp.ucl.ac.be/~opperd/private/neighbor.html

distance matrix:



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





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





Neighbour-joining trees of domestic dogs and grey wolves.



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)



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

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

Evolution in dogs CS-374 Abhinay Nagpal [email protected].

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