1. Genetic Heterogeneity Taken from: Advanced Topics in Linkage Analysis. Ch. 27 Presented by: Natalie Aizenberg Assaf Chen

2. Agenda Motivation General approach to genetic heterogeneity. –Allelic heterogeneity. –Non-allelic heterogeneity. Test for homogeneity given Linkage. Test for Linkage given heterogeneity. Using the HOMOG program. Exercise.

3. Motivation We will try to answer two important questions: 1.Given a positive Linkage test result, is there a significant evidence for a proportion of pedigrees segregating the linked gene and another proportion segregating a putative unlinked gene for the same disease ?

4. Motivation (Cont.) 2.Although not having a significant evidence for Linkage (assuming homogeneity of the disease gene), does allowing a certain percentage of pedigrees segregating an unlinked gene bring a significant evidence for linkage to a disease gene in a proportion of pedigrees?

5. General approach to genetic heterogeneity Number of genetic causes act differently to produce an identical disease phenotype. Allelic heterogeneity: Multiple separate disease alleles at the same locus can each cause the same disease phenotype. (CF) Non-allelic heterogeneity : Disease alleles at two or more independently acting loci could each cause the same disease phenotype.

6. General approach to genetic heterogeneity (Cont.) Heterogeneity is much easier to detect when there is a different mode of inheritance in some pedigrees from others. When 2 forms of the disease share a common mode of inheritance- the only way to determine that more than one genetic locus is involved is by using a special form of Linkage analysis that treats the heterogeneity as an additional parameter in the analysis.

7. Test for homogeneity given Linkage Given a significant test result (i.e. lod-score>3), we will check whether there is a significant evidence to support the hypothesis that some of the pedigrees might be segregating a different unlinked gene for the same disease, and not the gene that is linked to a marker in question in this analysis. We will use the A-test method.

8. A-test Assumptions: 1.Two categories of families in the data:  = 0.5  =  (1) < 0.5 2.  =  (1) in Proportion  of families (segregating the linked gene), and  =0.5 in proportion (1-  ). 3.No a priori assignment of any family to one class or the other. We will check H(0): {  =1} versus H(1): {  <1}

9. A-test (Cont.) The likelihood for any given family: L( ,  ) =  L(  ) + (1-  )L(  = 0.5) The total likelihood for n families: Testing the null hypothesis (all families are of linked type) by forming a likelihood ratio L( ,  ) / L(  =1,  ) = Finding the Chi-square :

10. Test for Linkage given heterogeneity The null hypothesis of no linkage against the hypothesis of linkage and heterogeneity. The ratio test: L( ,  ) / L(  = 1,  = 0.5 ) = R There are two problems in interpretation: 1.Trying to declare a linkage significant with heterogeneity as a parameter. We need a significance level equivalent to lod-score=3 in straight linkage analysis. The lod score calculates: log10 [ R ]. There is an additional free parameter in the numerator. To allow this additional degree of freedom, we add log10(2) = 0.3 to the critical value -> lod-score=3.3.

11. Test for Linkage given heterogeneity (Cont.) 2.Under the hypothesis  = 0.5 in all families, the parameter  disappears: L( ,  ) =  L(  ) + (1-  )L(  = 0.5), with  = 0.5 : L( ,  =0.5 ) =  L(  =0.5) + (1-  )L(  = 0.5) = = L(  =0.5). Making  = 0, causes  to disappear as a parameter: L(  =0,  ) = 0L(  ) + (1- 0 )L(  = 0.5) = L(  =0.5). We have a completely degenerate situation under the null hypothesis, where : L(  =0,  ) = L( ,  =0.5) = L(  =0.5).

12. Test for Linkage given heterogeneity (Cont.) Thus, there is one parameter under H(0), whereas under H(1) there are two, leading to a problem with the asymptotic distribution  further major problems. In light of all this – we would prefer not to apply any asymptotic theory to this test statistics (  ≠0,  ≠ 0.5), and use a lod-score above 3.3 to determine significance of a test result (ratio above 2000:1).

13. Using HOMOG to perform test Phase1: run a linkage program on the pedigrees calculating likelihood Phase 2: run LODOS on the out put file to receive lod-scores: Z(  )=log 10 L(  )- log 10 L(  =0.5) for each value of  at each pedigree Phase 3:Run HOMOG on calculated lod scores for each  and for each family

14. HOMOG input file Line1: Title of the problem Line2: N STEPSIZE Where: N= number of values of  at which lod-scores are given STEPSIZE = the step size in which  should be incremented (0.05 by default) Note! Values should not be given for 0.5 where Z=0

15. HOMOG input file (Cont.) Line3: OUT ALOW LL Where: OUT = output options (0..3) ALOW =smallest value of  to be considered LL = line length in the output file (80 char by default)

16. HOMOG input file (Cont.) Line4:  1,  2,…,  N Where  I the values of  for which the lod scares are going to provided below Line5: NFAM Where NFAM is the number of families for which lod scores are provided

17. HOMOG input file (Cont.) Line6: Z(  1 ), Z(  2 ), …, Z(  N ) in family1 Line7: Z(  1 ), Z(  2 ), …, Z(  N ) in family2 … Line(5+NFAM): Z(  1 ), Z(  2 ), …, Z(  N ) in family NFAM

18. An input example The pedigrees:

19. Input example (Cont.) Notice: the best  estimate for the entire family set together is 0.5 while fam1&3 appear to have no recombinants q=0.4 50.40.350.30.250.20.150.10.050 family 10.330.630.911.171.411.631.842.042.232.41 family 20.37-0.78-1.24-1.77-2.41-3.18-4.18-5.59-8-8-99- family 30.410.791.141.461.762.042.32.552.793.01 family 40.46--0.971.55-2.22-3.01-3.98-5.23-6.99-10-99- Resulting lod-scores

20. The Input file Ex 27 10 0.05 3 0 80 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 4 2.41 2.23 2.04 1.84 1.63 1.41 1.17 0.91 0.63 0.33 -99 -8.00 -5.59 -4.18 -3.18 -2.41 -1.77 -1.24 -0.78 -0.37 3.01 2.79 2.55 2.30 2.04 1.76 1.46 1.14 0.79 0.41 -99 -10.00 -6.99 -5.23 -3.98 -3.01 -2.22 -1.55 -0.97 -0.46

21. The output file Alpha| ln L(alpha,theta) 1.00 -99.00 -29.89 -18.40 -12.13 -8.04 -5.18 -3.13 -1.70 -0.76 -0.21 0.95 6.39 5.47 4.48 3.44 2.37 1.30 0.36 -0.20 -0.27 -0.11 0.90 7.66 6.74 5.75 4.72 3.65 2.53 1.45 0.57 0.09 -0.03 0.85 8.36 7.44 6.45 5.42 4.34 3.22 2.08 1.07 0.36 0.05 0.80 8.82 7.90 6.91 5.87 4.80 3.67 2.51 1.42 0.57 0.11 0.75 9.13 8.21 7.22 6.19 5.12 3.98 2.81 1.68 0.73 0.16 0.70 9.36 8.44 7.45 6.42 5.34 4.21 3.03 1.87 0.86 0.20 0.65 9.52 8.60 7.61 6.58 5.51 4.38 3.19 2.01 0.96 0.24 0.60 9.63 8.71 7.72 6.69 5.62 4.49 3.30 2.12 1.03 0.27 0.55 9.69 8.77 7.79 6.76 5.69 4.56 3.37 2.18 1.08 0.28 0.50 9.71 8.79 7.81 6.78 5.71 4.59 3.41 2.22 1.11 0.30 0.45 9.69 8.78 7.79 6.76 5.70 4.58 3.40 2.22 1.12 0.30 0.40 9.63 8.72 7.73 6.71 5.64 4.53 3.36 2.19 1.11 0.30 0.35 9.53 8.61 7.63 6.61 5.55 4.44 3.29 2.14 1.07 0.29 0.30 9.37 8.46 7.48 6.46 5.40 4.31 3.17 2.04 1.02 0.27 0.25 9.15 8.23 7.26 6.24 5.20 4.11 3.00 1.91 0.94 0.25 0.20 8.83 7.92 6.95 5.94 4.91 3.85 2.77 1.73 0.83 0.22 0.15 8.39 7.48 6.52 5.52 4.51 3.47 2.44 1.49 0.69 0.18 0.10 7.71 6.81 5.86 4.88 3.90 2.93 1.99 1.16 0.52 0.13 0.05 6.48 5.60 4.69 3.77 2.88 2.04 1.30 0.70 0.29 0.07 Theta 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45

22. The output file (Cont.) Table of conditional max. Ln(L) over alpha's, given theta or x theta or x alpha max Max.Ln(L) Lik. ratio 1 0.0000 0.5000 9.7123 1.652x10^4 2 0.0500 0.5000 8.7939 6593.7674 3 0.1000 0.5000 7.8082 2460.6301 4 0.1500 0.5000 6.7795 879.6603 5 0.2000 0.5000 5.7109 302.1472 6 0.2500 0.5000 4.5869 98.1858 7 0.3000 0.5000 3.4056 30.1313 8 0.3500 0.4500 2.2213 9.2196 9 0.4000 0.4500 1.1211 3.0683 10 0.4500 0.4500 0.3015 1.3519 This table shows the maximum natural log likelihood for each value of  over  normalized so that L(  =0.5)=0

23. The output file (Cont.) Estimates of Hypotheses Max.lnL Alpha Theta H2: Linkage, heterogeneity 9.71230.5000 0 linkage & heterogeneity H1: Linkage, homogeneity 0.0000 (1) 99 linkage & homogeneity H0: No linkage (0) (0) (99) neither linkage nor heterogeneity Components of chi-square Sourcedf Chi-square L ratio H2 vs. H1 Heterogeneity 1 19.425 16520 Heterogeneity given linkage H1 vs. H0 Linkage 1 0.00 1.0000 linkage assuming homogeneity H2 vs. H0 Total 2 19.425 1.6520 linkage & heterogeneity Note:  =99 means  =0.5

24. The output file (Cont.) Since there is absolutely no evidence for linkage (H 1 ), the test of H 2 versus H 1 is meaningless (null hypothesis is not valid). Instead, we must consider the comparison of H 2 versus H 0 (the correct null hypothesis of no linkage)

25. Conclusion from HOMOG output The likelihood ratio is 16,520 for H 2 vr. H 0 (heterogeneity) which exceeds the minimum ratio (2000) for significant evidence for linkage and heterogeneity.  although there is no evidence for linkage whatsoever under the assumption of locus homogeneity, as soon as we allow for heterogeneity, we have significant evidence for linkage (in at least some of the families