1. . Basic Model For Genetic Linkage Analysis Lecture #5 Prepared by Dan Geiger

2. 2 Using the Maximum Likelihood Approach The probability of pedigree data Pr(data |  ) is a function of the known and unknown recombination fractions denoted collectively by . How can we construct this likelihood function ? The maximum likelihood approach is to seek the value of  which maximizes the likelihood function Pr(data |  ). This is the ML estimate.

3. 3 Constructing the Likelihood function L ijm = Maternal allele at locus i of person j. The values of this variables are the possible alleles l i at locus i. First, we need to determine the variables that describe the problem. There are many possible choices. Some variables we can observe and some we cannot. X ij = Unordered allele pair at locus i of person j. The values are pairs of i th -locus alleles (l i,l’ i ). L ijf = Paternal allele at locus i of person j. The values of this variables are the possible alleles l i at locus i (Same as for L ijm ). As a starting point, We assume that the data consists of an assignment to a subset of the variables {X ij }. In other words some (or all) persons are genotyped at some (or all) loci.

4. 4 What is the relationships among the variables for a specific individual ? L 11f L 11m X 11 Paternal allele at locus 1 of person 1 Unordered allele pair at locus 1 of person 1 = data Maternal allele at locus 1 of person 1 P(L 11m = a) is the frequency of allele a. We use lower case letters for states writing, in short, P(l 11m ). P(x 11 | l 11m, l 11f ) = 0 or 1 depending on consistency

5. 5 What is the relationships among the variables across individuals ? L 11f L 11m L 13m X 11 P(l 13m | l 11m, l 11f ) = 1/2 if l 13m = l 11m or l 13m = l 11f P(l 13m | l 11m, l 11f ) = 0 otherwise L 12f L 12m L 13f X 12 First attempt: correct but not efficient as we shall see. Mother Father Offspring

6. 6 Probabilistic model for two loci L 11f L 11m L 13m X 11 L 12f L 12m L 13f X 12 X 13 Model for locus 1 L 21f L 21m L 23m X 21 L 22f L 22m L 23f X 22 X 23 Model for locus 2 L 23m depends on whether L 13m got the value from L 11m or L 11f, whether a recombination occurred, and on the values of L 21m and L 21f. This is quite complex.

7. 7 Adding a selector variable L 11f L 11m L 13m X 11 S 13m Selector of maternal allele at locus 1 of person 3 Maternal allele at locus 1 of person 3 (offspring) Selector variables S ijm are 0 or 1 depending on whose allele is transmitted to offspring i at maternal locus j. P(s 13m ) = ½ P(l 13m | l 11m, l 11f,,S 13m =0) = 1 if l 13m = l 11m P(l 13m | l 11m, l 11f,,S 13m =1) = 1 if l 13m = l 11f P(l 13m | l 11m, l 11f,,s 13m ) = 0 otherwise

8. 8 Probabilistic model for two loci S 13m L 11f L 11m L 13m X 11 S 13f L 12f L 12m L 13f X 12 X 13 Model for locus 1 S 23m L 21f L 21m L 23m X 21 S 23f L 22f L 22m L 23f X 22 X 23 Model for locus 2

9. 9 Probabilistic Model for Recombination S 23m L 21f L 21m L 23m X 21 S 23f L 22f L 22m L 23f X 22 X 23 S 13m L 11f L 11m L 13m X 11 S 13f L 12f L 12m L 13f X 12 X 13  is the recombination fraction between loci 2 & 1.

10. 10 Constructing the likelihood function I S 13m L 11f L 11m L 13m X 11 P(l 11m, l 11f,, x 11, s 13m,l 13m ) = P(l 11m ) P(l 11f ) P(x 11 | l 11m, l 11f, ) P(s 13m ) P(l 13m | s 13m, l 11m, l 11f ) Joint probability Prob(data) = P(x 11 ) =  l11m  l11f  s13m  l13m P(l 11m, l 11f,, x 11, s 13m,l 13m ) Probability of data (sum over all states of all hidden variables) All other variables are not-observed (hidden) Observed variable

11. 11 Constructing the likelihood function II = P(l 11m ) P(l 11f ) P(x 11 | l 11m, l 11f, ) … P(s 13m ) P(s 13f ) P(s 23m | s 13m,  2 ) P(s 23m | s 13m,  2 ) P(l 11m,l 11f,x 11,l 12m,l 12f,x 12,l 13m,l 13f,x 13, l 21m,l 21f,x 21,l 22m,l 22f,x 22,l 23m,l 23f,x 23, s 13m,s 13f,s 23m,s 23f,  2 ) = Product over all local probability tables Prob(data|  2 ) = P(x 11, x 12, x 13, x 21, x 22, x 23 ) = Probability of data (sum over all states of all hidden variables) Prob(data|  2 ) = P(x 11, x 12, x 13, x 21, x 22, x 23 ) =  l11m, l11f … s23f [ P(l 11m ) P(l 11f ) P(x 11 | l 11m, l 11f, ) … P(s 13m ) P(s 13f ) P(s 23m | s 13m,  2 ) P(s 23m | s 13m,  2 ) ] The result is a function of the recombination fraction. The ML estimate is the  2 value that maximizes this function.

12. 12 Modeling Phenotypes I L 11f L 11m L 13m X 11 S 13m Phenotype variables Y ij are 0 or 1 depending on whether a phenotypic trait associated with locus i of person j is observed. E.g., sick versus healthy. For example model of perfect recessive disease yields the penetrance probabilities: P(y 11 = sick | X 11 = (a,a)) = 1 P(y 11 = sick | X 11 = (A,a)) = 0 P(y 11 = sick | X 11 = (A,A)) = 0 Y 11

13. 13 Modeling Phenotypes II L 11f L 11m L 13m X 11 S 13m Note that in this model we assume the phenotype depends only on the alleles of one locus. Also we did not model levels of sickness. We did not model continuous phenotypic observations either. Y 11

14. 14 Introducing a tentative disease Locus S 23m L 21f L 21m L 23m X 21 S 23f L 22f L 22m L 23f X 22 X 23 S 13m L 11f L 11m L 13m X 11 S 13f L 12f L 12m L 13f X 12 X 13 The recombination fraction  2 is unknown. Finding it can help determine whether a gene causing the disease lies in the vicinity of the marker locus. Disease locus: assume sick means x ij =(a,a) Marker locus

15. 15 Locus-by-Locus Summation order Sum over locus i vars before summing over locus i+1 vars Sum over orange vars (L ijt ) before summing selector vars (S ijt ). This order yields a Hidden Markov Model (HMM). S i3m L i1f L i1m L i3m X i1 S i3f L i2f L i2m L i3f X i2 X i3 1 2 3 4

16. 16 Hidden Markov Models in General Application in communication: message sent is (s 1,…,s m ) but we receive (r 1,…,r m ). Compute what is the most likely message sent ? Application in speech recognition: word said is (s 1,…,s m ) but we recorded (r 1,…,r m ). Compute what is the most likely word said ? Application in Genetic linkage analysis: to be discussed now. X1X1 X2X3Xi-1XiXi+1R1R1 R2R2 R3R3 R i-1 RiRi R i+1 X1X1 X2X3Xi-1XiXi+1S1S1 S2S2 S3S3 S i-1 SiSi S i+1 Which depicts the factorization:

17. 17 Hidden Markov Model In our case X1X1 X2X3Xi-1XiXi+1X1X1 X2X2 X3X3 X i-1 XiXi X i+1 X1X1 X2X3Xi-1XiXi+1S1S1 S2S2 S3S3 S i-1 SiSi S i+1 The compounded variable S i = (S i,1,m,…,S i,2n,f ) is called the inheritance vector. It has 2 2n states where n is the number of persons that have parents in the pedigree (non-founders). The compounded variable X i = (X i,1,m,…,X i,2n,f ) is the data regarding locus i. To specify the HMM we need to write down the transition matrices from S i-1 to S i and the matrices P(x i |S i ). Note that these quantities have already been implicitly defined.

18. 18 The transition matrix Recall that: Therefore, in our example, where we have one non-founder (n=1), the transition probability table size is 4  4 = 2 2n  2 2n, encoding four options of recombination/non-recombination for the two parental meiosis: (The Kronecker product) For n non-founders, the transition matrix is the n-fold Kronecker product:

19. 19 Probability of data in one locus given an inheritance vector S 23m L 21f L 21m L 23m X 21 S 23f L 22f L 22m L 23f X 22 X 23 Model for locus 2 P(x 21, x 22, x 23 |s 23m,s 23f ) = =  P(l 21m ) P(l 21f ) P(l 22m ) P(l 22f ) P(x 21 | l 21m, l 21f ) P(x 22 | l 22m, l 22f ) P(x 23 | l 23m, l 23f ) P(l 23m | l 21m, l 21f, S 23m ) P(l 23f | l 22m, l 22f, S 23f ) l 21m,l 21f,l 22m,l 22f l 22m,l 22f The five last terms are always zero-or-one, namely, indicator functions.

20. 20 Posterior decoding H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi The standard query for HMM is belief update (also called posterior decoding). 1. Compute the posteriori belief in H i (specific i) given the evidence {x 1,…,x L } for each of H i ’s values h i, namely, compute p(h i | x 1,…,x L ). 2. Do the same computation for every H i but without repeating the first task L times. The solution is called the forward-backward algorithm.

21. 21 Likelihood of evidence To compute the likelihood of evidence P(x 1,…,x L ), which depends on the recombination fractions in our case, we will use either the forward or the backward algorithm to be described now. H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi

22. 22 Decomposing the computation P(x 1,…,x L,h i ) = P(x 1,…,x i,h i ) P(x i+1,…,x L | x 1,…,x i,h i ) H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi Equality due to Ind({x i+1,…,x L }, {x 1,…,x i } | H i } = P(x 1,…,x i,h i ) P(x i+1,…,x L | h i )  f(h i ) b(h i ) Answer: P(h i | x 1,…,x L ) = (1/K) P(x 1,…,x L,h i ) where K=  hi P(x 1,…,x L,h i ).

23. 23 The forward algorithm P(x 1,x 2,h 2 ) =  P(x 1,h 1,h 2,x 2 ) {Second step} =  P(x 1,h 1 ) P(h 2 | x 1,h 1 ) P(x 2 | x 1,h 1,h 2 ) h1h1 h1h1 Last equality due to conditional independence =  P(x 1,h 1 ) P(h 2 | h 1 ) P(x 2 | h 2 ) h1h1 H1H1 H2H2 X1X1 X2X2 HiHi XiXi The task: Compute f(h i ) = P(x 1,…,x i,h i ) for i=1,…,L (namely, considering evidence up to time slot i). P(x 1, h 1 ) = P(h 1 ) P(x 1 |h 1 ) {Basis step} P(x 1,…,x i,h i ) =  P(x 1,…,x i-1, h i-1 ) P(h i | h i-1 ) P(x i | h i ) h i-1 {step i}

24. 24 The backward algorithm The task: Compute b(h i ) = P(x i+1,…,x L |h i ) for i=L-1,…,1 (namely, considering evidence after time slot i). H L-1 HLHL X L-1 XLXL HiHi H i+1 X i+1 P(x L | h L-1 ) =  P(x L,h L |h L-1 ) =  P(h L |h L-1 ) P(x L |h L-1,h L )= hLhL hLhL Last equality due to conditional independence =  P(h L |h L-1 ) P(x L |h L ) {first step} hLhL P(x i+1,…,x L |h i ) =  P(h i+1 | h i ) P(x i+1 | h i+1 ) P(x i+2,…,x L | h i+1 ) h i+1 {step i} =b(h i )= =b(h i+1 )=

25. 25 The combined answer 1. To Compute the posteriori belief in H i (specific i) given the evidence {x 1,…,x L } run the forward algorithm and compute f(h i ) = P(x 1,…,x i,h i ), run the backward algorithm to compute b(h i ) = P(x i+1,…,x L |h i ), the product f(h i )b(h i ) is the answer (for every possible value h i ). 2. To Compute the posteriori belief for every H i simply run the forward and backward algorithms once, storing f(h i ) and b(h i ) for every i (and value h i ). Compute f(h i )b(h i ) for every i. H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi

26. 26 Likelihood of evidence revisited 1.To compute the likelihood of evidence P(x 1,…,x L ), do one more step in the forward algorithm, namely,  f(h L ) =  P(x 1,…,x L,h L ) 2. Alternatively, do one more step in the backward algorithm, namely,  b(h 1 ) P(h 1 ) P(x 1 |h 1 ) =  P(x 2,…,x L |h 1 ) P(h 1 ) P(x 1 |h 1 ) H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi hLhL h1h1 hLhL h1h1

27. 27 Time and Space Complexity of the forward/backward algorithms Time complexity is linear in the length of the chain, provided the number of states of each variable is a constant. More precisely, time complexity is O(k 2 n) where k is the maximum domain size of each variable. H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi Space complexity is also O(k 2 n). In our case: O(k 2 n) is really O(2 4n L). Next class we will see how to save on these computations using the special matrices we have. The savings have been implemented in GeneHunter – a leading software of linkage.

28. 28 The Maximum APosteriori query H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi 1.Recall that the query asking likelihood of evidence is to compute P(x 1,…,x L ) =  P(x 1,…,x L, h 1,…,h L ) 2.Now we wish to compute a similar quantity: P * (x 1,…,x L ) = MAX P(x 1,…,x L, h 1,…,h L ) (h 1,…,h L ) And, of course, we wish to find a MAP assignment (h 1 *,…,h L * ) that brought about this maximum.

29. 29 Example: Revisiting likelihood of evidence H1H1 H2H2 X1X1 X2X2 H3H3 X3X3 P(x 1,x 2,x 3 ) =  P(h 1 )P(x 1 |h 1 )  P(h 2 |h 1 )P(x 2 |h 2 )  P(h 3 |h 2 )P(x 3 |h 3 ) h3h3 h2h2 h1h1 =  P(h 1 )P(x 1 |h 1 )  b(h 2 ) P(h 1 |h 2 )P(x 2 |h 2 ) h1h1 h2h2 =  b(h 1 ) P(h 1 )P(x 1 |h 1 ) h1h1

30. 30 Example: Computing the MAP assignment H1H1 H2H2 X1X1 X2X2 H3H3 X3X3 maximum = max P(h 1 )P(x 1 |h 1 ) max P(h 2 |h 1 )P(x 2 |h 2 ) max P(h 3 |h 2 )P(x 3 |h 3 ) h3h3 h2h2 h1h1 = max P(h 1 )P(x 1 |h 1 ) max b (h 2 ) P(h 1 |h 2 )P(x 2 |h 2 ) h1h1 h2h2 h3h3 Replace sums with taking maximum: = max b (h 1 ) P(h 1 )P(x 1 |h 1 ) h1h1 h2h2 {Finding the maximum} h 1 * = arg max b (h 1 ) P(h 1 )P(x 1 |h 1 ) h1h1 h2h2 {Finding the map assignment} h 2 * = x* (h 1 * ); h2h2 x* (h 2 ) h3h3 x* (h 1 ) h2h2 h 3 * = x* (h 2 * ) h3h3

31. 31 Viterbi’s algorithm For i=1 to L-1 do h 1 * = ARG MAX P(h 1 ) P(x 1 |h 1 ) b (h 1 ) h2h2 h2h2 h i+1 * = x* (h i *) h i+1 Forward phase (Tracing the MAP assignment) : H1H1 H2H2 H L-1 HLHL X1X1 X2X2 X L-1 XLXL HiHi XiXi x* (h i ) = ARGMAX P(h i+1 | h i ) P(x i+1 | h i+1 ) b (h i+1 ) For i=L-1 downto 1 do b (h i ) = MAX P(h i+1 | h i ) P(x i+1 | h i+1 ) b (h i+1 ) h i+1 h i+2 b (h L ) = 1 h L+1 h i+1 h i+2 Backward phase: (Storing the best value as a function of the parent’s values)