Discrete Probability Distributions: Poisson Applications to Genome Assembly

Transcription

1 Discrete Probability Distributions: Poisson Applications to Genome Assembly

2 Binomial Distribution Binomially distributed random variable X equals sum of n independent Bernoulli trials The probability mass function is: n x n x f x C p 1 p for x 0,1,... n (3-7) x Based on the binomial expansion: Sec 3=6 Binomial Distribution 2

3 Matlab exercise: Binomial distribution Generate a sample of size 100,000 for binomially distributed random variable X with n=100, p=0.2 Plot the approximation to the Probability Mass Function based on this sample Calculate the mean and variance of this sample and compare it to theoretical calculations: E[X]=n*p and V[X]=n*p*(1 p)

4 Matlab exercise: Uniform distribution n=100; p=0.2; Stats=100000; r1=rand(stats,n)<p; r2=sum(r1,2); mean(r2) var(r2) [a,b]=hist(r2, 0:n); p_b=a./sum(a); figure; stem(b,p_b); figure; semilogy(b,p_b,'ko ')

5 Poisson Distribution Limit of the binomial distribution when n, the number of attempts, is very large p, the probability of success is very small E(X)=np is just right From von Bortkiewicz 1898 Siméon Denis Poisson ( ) French mathematician and physicist

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7 Let x np E x, so p n P X x n p x 1 p x x n( n 1)...( n x 1) n 1 ~ ; x! n n x! n x! x x!. Normalization requires P X x 1. Thus P e X x x x x! e x n x x n x x 7

8 Poisson Mean & Variance If X is a Poisson random variable, then: Mean: μ = E(X) = λ Variance: σ 2 =V(X) = λ Standard deviation: σ=λ 1/2 Note: Variance = Mean Note: Standard deviation/mean = λ 1/2 decreases with λ Sec 2 8

9 Matlab exercise Find mean, variance, and plot the histogram of 100,000 Poisson distributed numbers with lambda=2 Use random('poisson',lambda) to generate Poisson distributed random numbers

10 Matlab: Poisson distributions Stats=100000; lambda=2; r2=random('poisson',lambda,stats,1); disp(mean(r2)); disp(var(r2)); disp(std(r2)); [a,b]=hist(r2, 0:max(r2)); p_p=a./sum(a); figure; plot(b,p_p,'ko ');

11 Discrete Poisson Distribution PMF: f(x)=exp( λ) λ x /x! Cumulative Distribution Function: A. exp( λ) B. 0 C. 1 exp( λ) D. 1 CDF(x)=P(X x)? What is CDF(1)? Get your i clickers 11

12 Discrete Poisson Distribution PMF: f(x)=exp( λ) λ x /x! Cumulative Distribution Function: A. exp( λ) B. 0 C. 1 exp( λ) D. 1 CDF(x)=P(X x)? What is CDF(1)? Get your i clickers 12

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16 Credit: XKCD comics

17 Poisson Distribution in Genome Assembly

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19 Poisson Example: Genome Assembly Goal: figure out the sequence of DNA nucleotides (ACTG) along the entire genome Problem: Sequencers generate random short reads Solution: assemble genome from short reads using computers. Whole Genome Shotgun Assembly.

20 MinION, a palm sized gene sequencer made by UK based Oxford Nanopore Technologies

21 Short Reads assemble into Contigs

22 Promise of Genomics I think I found the corner piece!

23 How many short reads do we need?

24 of K mers (5 mers shown here )

25 Where is the Poisson? G genome length (in bp) L short read average length N number of short read sequenced λ sequencing redundancy = LN/G x number of short reads covering a given site on the genome Ewens, Grant, Chapter 5.1 Poisson as a limit of Binomial. For a given site on the genome for each short read Prob(site covered): p=l/g is very small. Number of attempts (short reads): N is very large. Their product (sequencing redundancy): λ = NL/G is O(1).

26 What fraction of genome is covered? Coverage: λ=nl/g, X r.v. equal to the number of times a given site is covered Poisson: P(X=x)= λ x *exp( λ) /x! P(X=0)=exp( λ), P(X>0)=1 exp( λ) Total length covered: G*[1 exp( λ)] λ λ

27 How many contigs? Probability that a given short read is the right end of a contig = no left ends of other reads fall within it. Left ends of each of N 1 N other reads has Prob: p=(l 1)/G L/G to fall within given read. Expected number of short reads extending a given one is N*p=N*L/G= λ Probability that none will extend it = exp( λ): Number of contigs: N contigs =Ne λ :

28 Average length of a contig? Length of a genome covered: G covered =G P(X>0)=G (1 exp( λ)) Number of contigs N contigs =N e λ Average length of a contig = <L>=Σ i L i /N contigs =G covered /N contigs = G (1 exp( λ))/ N e λ =L (1 exp( λ))/ λ e λ λ

29 Estimate Human genome is 3x10 9 bp long Chromosome 1 spans about 250x10 6 bp Illumina generates short reads 100 bp long How many short reads are needed to completely assemble the 1 st chromosome?

30 The answer is N=44x10 6 short (100bp) reads or λ =17.6 fold redundant coverage. At 0.1$/Mb that means that the reads for de novo full assembly of human genome would cost (3 x10 9 x 17.6 /10 6 ) x 0.1$ =$5300 /genome In reality is cheaper as we don t need de novo assembly

31 What spoils these estimates? Try searching human genome for TTAGGGTTAGGGTTAGGG (18 bases) and you will find 100s of matches How many one expects by pure chance? 2 3x =0.08 << 1 match Genome has lots of repeats. DNA repeats make assembly difficult Side remark. If it was not for repeats 17 bases would be enough to specify unique position in the human genome. log(2 3x10 9 )/log(4)= That is why micrornas recognizing ~22 bases don t have a lot of off target cleavage

32 Stuttering human genome

33 Repeats are like sky puzzle pieces

34 How many repeats are in eukaryotic genomes?

35 The Little Prince, Antoine de Saint Exupéry, 1943

36 Repetitive DNA and next-generation sequencing: computational challenges and solutions Todd J. Treangen & Steven L. Salzberg Nature Reviews Genetics 13, (January 2012) doi: /nrg3117

37 How to assemble genome with repeats? Answer: longer reads Problem: cheap sequencing = short reads L=50 L=5000 L=1000

38 Credit: XKCD comics