Haplotype Blocks

An Overview

A. Polanski

Department of Statistics

Rice University

Key Papers

1. N. Patil et al., (2001), Blocks of Limited Haplotype Diversity Revealed by High-Resolution Scanning of Human Chromosome 21, Science, vol. 294, pp. 1719-1723

2. N. Wang et al., (2002), Distribution of Recombination Crossovers and the Origin of Haplotype Blocks: The Interplay of Population History, Recombination and Mutation, Am. J. Hum. Genet., vol. 71, pp. 1227-1234.

3. K. Zhang et al., (2002), A Dynamic Programming Algorithm for Haplotype Block Partitioning, PNAS, vol. 99, pp. 7335-7339

Supplementary Papers

1. R. Hudson, N. Kaplan, (1985), Statistical Properties of the Number of Recombination Events in The History of a Sample of DNA sequences, Genetics, vol. 111, pp. 147-164

2. R. Hudson, 2002, Generating Samples under a Wright-Fisher Neutral Model of Genetic Variation, Bioinformatics, vol. 18, pp. 337-338

3. D. Reich et al., (2001), Linkage Disequilibrium in the Human Genome, Nature, vol. 411, pp. 199-204

What are Haplotype Blocks ?

Haplotype block = a sequence of contiguous markers on DNA, homogeneous according to some criterion

Markers = Single Nucleotide Polymorphisms (SNPs)

Data (Patil et al. 2001)

Chromosome 21

Physically separated the two copies of chromosome 21 using a rodent-human somatic cell hybrid technique

Sample of 20 copies of chromosome 21 (32397439 bases)

Found: 35989 SNPs

Fig. 2 from (Patil et al. 2001)

01000000000000000000100000000000000100001110000000001000000010010000000010010000000000000000000010000000011010000101010100000000010000000000010000000000100100001000000000000001011001001001010001001000000000010010001011000000001101010010101010000000000100010001011000101000000001010001100000000001010000000000010000010011000001110100100000011000011000100010001101000000000000001000100100010100000000101000110000000000101000000000001000001001100000111010010000001100001100010001000110100000000010000000000010000100000100100000000000000000001001001001001010001001000000000010010001011000000001100100000000000001000000010000100001001000000000001000001100000000001010000000010010011010001000000001000000100100000100111010000000000000000000100000000000100001001101001000000000000000000010010010010010100010010000000000100100010110000000011001000000000001000100000000000000001000001000101000000000000000001000000001001000001001001000000100000000100001000000001101010010101010000000000000100000001000000000000001000001100000000000000000100100000000100100000000000000000000100000000110100001010101000000000100000000000100001000001001000000000000000000010010011010010100010010000000000100100010110000000011001000000000001000100000000000000001000001000101000000000000000001000000001001000000001001000000000000000000001000010001101010010101010000010000000000010000100000000010100000000000000000000000000100101000000100100000000000000000000100000000110100001010101010001000000000000000010000010001010000000000000000010000000010010000010010010000001000000001000010000000011010100101010100000000100100000000010010000000000011000011010000000010100000010100100100100010010000010100001001000001001110100000000000100010000000001000000100000100010100000000000000000100000000100100000100100100000010000000010000100000000110101001010101000000000001000000000100100000000000100000110000000000101000000001001001001000100000000100000010010000010011101010000000010000000000100000000010010000000000010000011010000000010100000010100100100100010010000010000001001001001001110100000000000000100100001000000100010000000101000000001100111111000000011000000000000001001110101000000101010010000000000100000101111000001000000000001000010000000001010000000000000000000000000010010100000010010000000000000000000010000000011010000101010100001010000000000001000000000000010000010011101000010000000100000000000000010010001010000001000100100100000001000001011010

20 ……

i = 1, 2, …, 35989

SNP no i

Problems

How do we determine boundaries between blocks ?

1. Average value of standarized coefficient of linkage disequilibrium is greater than some threshold (Wang et al. 2002, Reich et al. 2001)

2. Infer sites in the sample of DNA sequences where recombination events happened in the past history (Wang et al. 2002, Hudson, 2002)

3. Chromosome coverage – minimum number of SNPs to account for majority of haplotypes (Patil et al. 2001, Zhang et al. 2002)

What evolutionary forces are responsible for haplotype blocks

formation ?

• Mutation

• Genetic drift

• Recombination

• Recombination hot spots

Methods

Method 1 (Wang et al. 2002)

Infer sites in the sample of DNA sequences where recombination events happened in the past history

Three gamete condition

Consider a pair of SNPs, SNP1 and SNP2. If there was no recombination between SNP1 and SNP2, they must satisfy three gamete condition

SNP1 SNP2SNP1 SNP2

AG

CC

G T

AG CTAC

GC

GT

Four gamete test (Hudson and Kaplan, 1985)

If we see all four gametes at SNP1 and SNP2

SNP1 SNP2

AG

CC

G T

A T

Then there must have been a recombination event between these sites in their past history

4GT

Array of pairwise 4GT test resultsHudson and Kaplan, 1985

D, dij=

0, if there are less then 4 gametes

1, if there are 4 gametes

What is the minimal number of recombinations that couldexplain observed data ?Statistics FR (Hudson and Kaplan, 1985)

Fig. 1 from Wang et al., 2002

D

Block 1 Block 2 Block 3

Wang et al., 2002 - Study

• R. Hudson’s program for simulating genealogies with mutation, drift and recombination under various demographic scenarios

• Study of dependence of average lengths of blocks on different factors

• Comparison of simulation results to data from Patil et al., 2002

Dependence of average lengths of blocks on recombination frequency

… on sample size

... on mutation intensity

Comparison to data from Patil et al. 2001

• Compute distribution of haplotype block lengths in the data from Patil et al. 2001

• Try to tune parameters and R to obtain similar distribution in the simulations

… Failed

Try a mixture of two different recombination frequencies - better

Method 2 (Patil, 2001)

Chromosome coverage – minimum number of SNPs to account for majority of haplotypes

Fig. 2 from (Patil et al. 2001)

Problem formulation

Define block boundaries to minimize the number of SNPs that distinguish at least percent of the haplotypes in each block

Common haplotypes

Those represented more than one in the block

Condition

Common haplotypes must constitute at least =80 percent of all haplotypes in the block

Blocks that do not satisfy this are not allowed

Fragment of Fig. 2 from Patil et al., 2001

Notation

• B – block defined as numbers of SNPs,

e.g., B = 45, 46,….50, or B = i, i+1,…, j

• L(B) length of the block (number of SNPs)

• f(B) – minimum number of SNP’s required to distinguish common haplotypes

Greedy 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

Start End

1. Increment end0. Fix Start =End

2. Compute ratio L(B)/f(B)

…….

3. Stop at max

4. Go to 0

Results

• 4563 representative SNPs (13%)

• 4135 blocks

Method 3 (Zhang et al. 2002)

Solves the same problem of 80% chromosome coverage, but using the better method of dynamic programming

Dynamic programming 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

……

Optimal partition of SNPs 1,2, … i

Assume that for all i=1, 2, …, j-1 we know optimal block partition,B1(i), B2(i), …, Bk(i) that minimizes:

i

K

kki iBfS

1

)]([

B1(i) B2(i) B3(i)

Bellman’s equation

)},...,1,({ 11,..1

min jiifSS iji

j

Results

• 3582 representative SNPs (compared to 4563 from greedy algorithm)

• 2575 blocks (compared to 4135 blocks from greedy algorithm)

Conclusions

• Studying haplotype block partitions is very important to

1. Constructing haplotype maps for genetic

traits

2. Understanding recombination in human

genome

To expect

• A lot of papers in this area appearing in scientific journals