1. Mapping and cloning Human Genes

2. Finding a gene based on phenotype 1. 100’s of DNA markers mapped onto each chromosome – high density linkage map. 2. identify markers linked to trait of interest by recombination analysis 3. Narrow region down to a managable length of DNA – for cloning and sequence comparison 4. Compare mutant and wild type sequences to find differences that could cause mutant phenotype 5. Prove that mutation is responsible for phenotype.

3. Identify polymorphic DNA markers Minisatellites Microsatellites (SSR or simple sequence repeats) SNP or single nucleotide polymorphisms

4. 1. 100’s of DNA markers mapped onto each chromosome – high density linkage map.

5. Mouse mapping Panels Backcross panel BXS cross to either B or S Progeny are BS/S or BS/B Each mouse represents one BS recombinant chromosome. Other chromosome from backcross parent always contributes same allele. If markers are linked they with be the same, either B or S, in most mice

6. Linkage Locus 1 50% B: 50% S Locus 2 50% BB: 50%BS; 50%SS and BS Unlinked: 50% parental : 50% recombinant Locus 3 linked: 38% parental : 12 % recombinant Use Chi-square test to determine if this data supports null hypothesis of unlinked genes

7. 2. Identify markers linked to trait of interest by recombination analysis Test degree of linkage: LOD score: Logarithm of odds of linkage MLS: Maximum likelihood LOD score 1.Establish trait is heritable: family studies, twin studies, adoption studies. 2. Identify linked markers: test many mapped polymorphic markers and look for cosegregation of markers with trait.

9. Test degree of linkage: odds of linkage

10. Lod score Probability gene and marker are linked at a certain map distance / Probability they are unlinked.

11. 1.Probability they are unlinked: Father is Pp and M1M2. Mother is pp and M1M1. Her alleles can be ignored here. Chance of each allele combination in children is:.25 PM1;.25 PM2;.25 pM1;.25 pM2 Probability of any genotype is.25 With 8 children of genotypes: pM2;PM1;pM2;PM1;pM2;pM1;pM2;PM2 p(unlinked)=.25 x.25 x.25 x.25 x.25 x.25 x.25 x.25 =.0000153

12. Probability they are linked at 10 map units: Chance of each allele combination in children is.45 PM1;.05 PM2;.05 pM1;.45 pM2 With 8 children of genotypes: pM2;PM1:pM2;PM1;pM2;pM1;pM2;PM2 p(linked at 10 cM) =.45 x.45 x.45 x.45 x.45 x.45 x.45 x.05 =.003736

13. BUT - formally you don’t know the phase of the two alleles of P and M: If the genes were linked so that P and M2 were on the same chromosome: Chance of each allele combination in children is.05 PM1;.45 PM2;.45 pM1;.05 pM2 With 8 children of genotypes: pM2;PM1;pM2;PM1;pM2;pM1;pM2;PM2 p(linked at 10 cM) =.05 x.05 x.05 x.05 x.05 x.05 x.05 x.45 =.000023

14. Odds of Linkage is L = p(.1)/p(.5) = [.5p(.1 in coupling) +.5p(.1 in repulsion)]/p(.5) p(.5 in coupling) = p(.5 in repulsion) In our case: L = [½(.003736) + ½(.000023)]/.0000153 L = 6.1 Log of L or LOD = 0.8 Maximum likelihood odds of linkage; Change estimated linkage distance p(.1) to get the best LOD score for the data.

15. To achieve significant LOD score: Combine odds of linkage for many families: p1(L)/p1(NL) x p2(L)/p2(NL) xp3(L)/p3(NL) In practice we combine the log of odds: LOD1 + LOD2 + LOD3. Continue until LOD > 3.0 before linkage is accepted Linkage distance is based on the linkage distance that gives the maximum value for the data.

16. If genes and markers are unlinked the p(L)/p(NL) will be <1.0 in some families and the LOD will be Negative. Therefore, as you add more families the LOD will only increase if the data of the majority of families supports linkage.

17. Summary 1.100’s of DNA markers mapped onto each chromosome – high density linkage map. the relative location of 100s of polymorphic DNA markers on chromosomes can be mapped using mapping panels. 2. identify markers linked to trait of interest by recombination analysis. Use LOD score to determine if markers are linked to gene in human families. The LOD score allows you to compare families in which marker and gene are either in repulsion or in coupling.

19. Fig. 11.17

20. Fig. 10.12

21. Fig. 10.10

23. Fig. 10.9a

24. Chose new markers within your large mapped interval and Repeat recombination analysis using those markers Continue until markers show no recombination Identify clones of DNA in ordered library that carry your Markers

25. Identify a BAC clone that must include your gene Find two flanking markers contained in a single BAC (large insert plasmid) clone. Look at GenBank entry for that BAC clone to identify candidate genes between your flanking markers –Open reading frames, –mRNA (cDNA) clone already identified, –Predicted gene regions

26. Expression pattern of genes in mapped interval can help choose best candidate gene

27. Final confirmation Sequence mutant and wild type – multiple mutant alleles needed to be convincing Complement mutation by making a transgenic with the wild type copy of the candidate gene.

28. Transformation to correct mutation using candidate gene

29. Take cloning slides from Ch. 10