Presentation is loading. Please wait.

Presentation is loading. Please wait.

Tutorial #5 by Ma’ayan Fishelson

Similar presentations


Presentation on theme: "Tutorial #5 by Ma’ayan Fishelson"— Presentation transcript:

1 Tutorial #5 by Ma’ayan Fishelson
The Elston-Stewart Algorithm References: Kenneth Lange “Mathematical and Statistical Methods for Genetic Analysis” Jurg Ott “Analysis of Human Genetic Linkage” Tutorial #5 by Ma’ayan Fishelson

2 The Given Problem Input: A pedigree + phenotype information about some of the people. These people are called typed. Output: the probability of the observed data, given some probability model for the transmission of alleles. founder leaf 1/2 typed

3 Q: What is the probability of the
observed data composed of ? A: There are three types of probability functions: founder probabilities, penetrance probabilities, and transmission probabilities.

4 Founder Probabilities – One Locus
Founders – individuals whose parents are not in the pedigree. We need to assign probabilities to their genotypes. This is done by assuming Hardy-Weinberg equilibrium. Suppose the gene frequency of d is 0.05, then: P(d/h) = 2 * 0.05 * 0.95 1 d/h d-mutant allele h-normal allele Pr(d/h, h/h) = Pr(d/h) * Pr(h/h) = (2 * 0.05 * 0.95)*(0.95)2 1 d/h 2 h/h Genotypes of different founders are treated as independent:

5 Founder Probabilities – Multiple Loci
According to linkage equilibrium, the probability of the multi-locus genotype of founder k is: Pr(xk) = Pr(xk1) *…* Pr(xkn) 1 d/h 1/2 Pr(d/h, 1/2) = Pr(d/h) * Pr(1/2) = Pr(d) * Pr(h) * Pr(1) * Pr(2) Linkage equilibrium Hardy-Weinberg Example:

6 Penetrance Probabilities
Penetrance: the probability of the phenotype, given the genotype. E.g.,dominant disease, complete penetrance: E.g., recessive disease, incomplete penetrance: d/d Pr(affected |d/d) = 1.0 d/h Pr(affected | d/h) = 1.0 Pr(affected | h/h) = 0 Can be, for example, sex-dependent, age-dependent, environment-dependent. d/d Pr(affected | d/d) = 0.7

7 Transmission Probabilities
Transmission probability: the probability of a child having a certain genotype given the parents’ genotypes. Pr(xc| xm, xf). If we split the ordered genotype xc into the maternal allele xcm and the paternal allele xcf, we get: Pr(xc| xm, xf) = Pr(xcm|xm)Pr(xcf|xf) The inheritance from each parent is independent.

8 Transmission Probabilities – One locus
The transmission is according to the 1st law of Mendel. Pr(Xc=d/h | Xm=h/h, Xf=d/h) = Pr(Xcm=h | Xm=h/h)*Pr(Xcf=d | Xf=d/h) = 1 * ½ = ½ 1 d/h 2 h/h 3 We also need to add the inheritance probability of the other phase, but we can see that it’s zero !

9 Transmission Probabilities – One locus
Different children are independent given the genotypes of their parents. Pr(X3=d/h, X4=h/h, x5=d/h | X1=d/h, X2=h/h) = = (1 * ½) * (1 * ½) * (1 * ½) 1 d/h 2 h/h 3 4 5

10 Transmission Probabilities – Multiple Loci
Let’s look at paternal inheritance for example. We generate all possible recombination sequences (s1,s2,…,sn), where sl = 1 or sl = -1. (2n sequences for n loci). Each sequence determines a selection of paternal alleles p1,p2,…,pn where: and therefore its probability of inheritance is: We need to sum the probabilities of all 2n recombination sequences.

11 Calculating the Likelihood of Family Data - Summary
The likelihood of the data is the probability of the observed data (the known phenotypes), given certain values for the unknown recombination fractions. Each person i has an ordered multi-locus genotype xi = (xi1, xi2, …,xin), and a multi-locus phenotype gi. For a pedigree with m people: where x=(x1,…,xm) and g=(g1,…,gm).

12 Computational Problem
Performing a multiple sum over all possible genotype combinations for all members of the pedigree. Solution: the Elston-Stewart algorithm provides a means for evaluating the multiple sum in a streamlined fashion, for simple pedigrees. This results in a more efficient computation.

13 Simple Pedigree No consanguineous marriages, marriages of blood-related individuals ( no loops in the pedigree). There is one pair of founders from which the whole pedigree is generated.

14 the genotypes of i’s parents
“Peeling” Order Assume that the individuals in the pedigree are ordered such that parents precede their children, then the pedigree likelihood can be represented as: where is: P(gi), if i is a founder, or , otherwise. In this way, we first sum over all possible genotypes of the children and only then on the possible genotypes for the parents. the genotypes of i’s parents

15 An Example for “Peeling” Order
1 2 3 4 5 7 6 h(gi) = P(xi|gi) P(gi) h(gm,gf,gc) = P(xc|gc) P(gc|gm,gf) According to the Elston-Stewart algorithm:

16 Elston-Stewart “Peeling” Order
As can be seen, this “peeling” order, “clips off” branches (sibships) of the pedigree, one after the other, in a bottom-up order. 1 2 1 1 2 3 4 1 2 3 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 7 6

17 Elston-Stewart – Computational Complexity
The computational complexity of the algorithm is linear in the number of people but exponential in the number of loci.

18 locus 1 variables locus 2 variables Sc[1,m] Sc[2,m] Ga[1,m] Ga[1,p]
Sc[1,p] Gc[1,p] Pa[1] Gb[1,m] Gb[1,p] Gc[1,m] Pb[1] Pc[1] Ga[2,m] Ga[2,p] Sc[2,p] Gc[2,p] Pa[2] Gb[2,m] Gb[2,p] Gc[2,m] Pb[2] Pc[2] locus 1 variables locus 2 variables Let’s look at an example for the model which describes our likelihood function. We have 3 types of variables in this model: genetic loci variables (the orange nodes) which represent genotypes. For each individual i and locus a we have 2 genetic loci variables that represent the alleles inherited from i’s father and mother at locus a. We have phenotype variables (yellow nodes). For each individual i and phenotype a we have a node that represents the value of the phenotype for individual a. We also have selector variables (green nodes), which denote the inheritance pattern in the pedigree. For each individual i and locus a we have 2 selector variables that indicate the haplotype of the parent from which i got his alleles. For example, the paternal selector of person c at locus 1 (Sc[1,p]) indicates whether c got his paternal allele at locus 1 from his father’s paternal haplotype or from his father’s maternal haplotype. Here we can see a fragment of a network that describes parents-child interaction in a 2-loci analysis. On the top we can see all the nodes that describe the 1st locus, and on the bottom all the nodes that describe the 2nd locus. Each node is associated with a probability function of the node given its parents. All the probability functions are logical functions, except for those of selector variables, which represent the recombination model. In the same way, we have the variable Sc[1,p] . If theta1 is the recombination fraction between these 2 loci, then the probability that these 2 selectors have different values is the probability that a recombination occurred, because it means that the 2 alleles came from different haplotypes.

19 Variation on the Elston-Stewart Algorithm in Fastlink
The pedigree traversal order in Fastlink is some modification of the Elston-Stewart algorithm. Assume no multiple marriages… Nuclear family graph: Vertices: each nuclear family is a vertex. Edges: if some individual is a child in nuclear family x and a parent in nuclear family y, then x and y are connected by and edge x-y which is called a “down” edge w.r.t. x and an “up” edge w.r.t. y.

20 Traversal Order One individual A is chosen to be a “proband”.
For each genotype g, the probability is computed that A has genotype g conditioned on the known phenotypes for the rest of the pedigree and the assumed recombination fractions. The first family that is visited is a family containing the proband, preferably, a family in which he is a child. Visit(w) { While w has an unvisited neighbor x reachable via an up edge: Visit(x); While w has an unvisited neighbor y reachable via a down edge: Visit(y); Update w; }

21 Traversal Order - Updates
If nuclear family w is reached via a down edge from z, the parent in w that nuclear families w and z share, is updated. If nuclear family w is reached via an up edge from z, then the child that w and z share is updated.

22 Example 1 An example pedigree: The corresponding nuclear family graph:
100 101 204 205 202 203 200 201 304 302 303 300 301 400 An example pedigree: The corresponding nuclear family graph:

23 Example 2 An example pedigree: The corresponding nuclear family graph:
100 101 304 305 202 203 405 404 302 303 403 201 102 103 300 301 400 401 402


Download ppt "Tutorial #5 by Ma’ayan Fishelson"

Similar presentations


Ads by Google