Today…

1. Genome 351, 18 April 2014, Lecture 6 Today… • Linkage: genes close together on the same chromosome do not assort independently • Linkage can be used to map genes (i.e., determine what chromosome they are on and roughly where they are on the chromosome)

2. Independent assortment of genes on separate chromosomes Meiosis I CFTR+ CFTR- CFTR- CFTR+ What about genes on the same chromosome? i IA i IA Gamete types: CFTR+ CFTR- CFTR- CFTR+ 1/4 each IA i IA i Equally likely

3. Independent assortment of genes on separate chromosomes Lets imagine that CFTR & ABO genes both reside on chr 7 Meiosis I CFTR+ CFTR- CFTR- CFTR+ IA i i IA What about genes on the same chromosome? i IA i IA This is called “linkage” Alleles on the same chromosome (tend to) travel together!! Gamete types: Why do I say “tend to” CFTR+ CFTR- CFTR- CFTR+ 1/4 each 1/2 each IA IA i i IA i IA i same same Equally likely

4. Alleles on the same chromosome tend to travel together Meiosis I a a A A B b b B

5. Alleles on the same chromosome tend to travel together Meiosis I a a A A B b b B

6. Alleles on the same chromosome tend to travel together Meiosis II Meiosis I A A a a b b B B a B a B A b A b

7. …but crossing over is also part of meiosis… Meiosis I locations where crossing over occurs

8. …but crossing over is also part of meiosis… non-parental (recombinant) chromosomes Meiosis II parental (non-recombinant) chromosomes

9. …and crossing over can disrupt linkage Meiosis I a a A A b b B B

10. …and crossing over disrupts linkage Meiosis I a a A A b b B B

11. …and crossing over disrupts linkage Meiosis I a a A A b b B B

12. …and crossing over disrupts linkage Meiosis I Meiosis II A A parental recombinants parental a a b B b B a B a b A B A b

13. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10

14. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 sister chromatids homologs 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 First replication of DNA Then pairing of homologs

15. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Then crossing over (recombination)

16. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Then crossing over (recombination)

17. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Then crossing over (recombination)

18. Recombination at the gene level Regions of two homologs with 10 genes 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Then crossing over (recombination)

19. Recombination at the gene level After Meiosis II you end up with the 4 gametes Parental (non-recombinant) 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Non-parental (recombinants) 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Parental (non-recombinant)

20. Thinking about linkage… Inheritance patterns distinguish independent assortment from linkage Example: A CFTR+/- IA/i double heterozygote male has children with a CFTR-/-i/i double homozygote If the CFTR and ABO genes are on separate chromosomes: In equal proportions eggs: CFTR-i sperm: Non CF, A blood type CFTR-i CFTR+ IA CFTR+ IA CFTR-i CF, O blood type CFTR-i CFTR-i CF, A blood type CFTR-i CFTR- IA CFTR- IA CFTR-i Non CF, O blood type CFTR+i CFTR+i

21. Thinking about linkage… Inheritance patterns distinguish independent assortment from linkage Example: A CFTR+/- IA/i double heterozygote male has children with a CFTR-/-i/i double homozygote If the CFTR and ABO genes are on separate chromosomes: In equal proportions eggs: CFTR-i sperm: Non CF, A blood type CFTR-i CFTR+ IA CFTR+ IA CFTR-i CF, O blood type CFTR-i CFTR-i CF, A blood type CFTR-i CFTR- IA CFTR- IA CFTR-i Non CF, O blood type CFTR+i CFTR+i

22. Thinking about linkage… Inheritance patterns distinguish independent assortment from linkage Example: A CFTR+/- IA/i double heterozygote male has children with a CFTR-/-i/i double homozygote If the CFTR and ABO genes are on the same chromosome: Half of each type eggs: CFTR-i sperm: Non CF, A blood type CFTR-i CFTR+ IA CFTR+ IA CFTR-i CF, O blood type CFTR-i CFTR-i CF, O blood type CFTR-i CFTR-i CFTR-i CFTR-i Non CF, A blood type CFTR+ IA CFTR+ IA

23. Thinking about linkage… Inheritance patterns distinguish independent assortment from linkage Example: A CFTR+/- IA/i double heterozygote male has children with a CFTR-/-i/i double homozygote In summary: Independent assortment: Linkage: If the CFTR and ABO genes are on the same chromosome: Half of each type eggs: CFTR-i sperm: Non CF, A blood type CFTR-i CFTR+ IA CFTR+ IA CFTR-i CF, O blood type CFTR-i CFTR-i CF, O blood type CFTR-i CFTR-i CFTR-i CFTR-i Non CF, A blood type CFTR+ IA CFTR+ IA

24. Thinking about linkage… Inheritance patterns distinguish independent assortment from linkage Assumes no recombination between CFTR and ABO gene In summary: Independent assortment: Linkage: Non CF, A blood type Non CF, A blood type Equal proportions parental CF, O blood type CF, O blood type CF, A blood type CF, A blood type Generate mostly parental types recombinants Non CF, O blood type Non CF, O blood type What if recombination between CFTR & ABO gene happens rarely? What if it happens often?

25. What does the frequency of recombination between genes tell us? A hypothetical example: Imagine that: A = brown eyes a = blue eyes C = pigmented skin c = albinism Phenotype of this individual? C A c a Brown eyes Pigmented skin DNA replication C A c a meiosis I & II C A c A Gametes (sperm or eggs) C a c a

26. What does the frequency of recombination between genes tell us? Questions we will consider: 1. What does the frequency of recombination in the A-C interval tell us about the proximity of these genes to one another? 2. What effect would recombination events outside of the A-C interval have on the segregation of the eye color and skin pigmentation traits? 3. What effect would double recombination events within the A-C interval have on the segregation of the eye color and blood type traits? 4. What is the maximal recombination frequency that can be detected? A hypothetical example: Imagine that: A = brown eyes a = blue eyes C = pigmented skin c = albinism Phenotype of this individual? C A c a Brown eyes Pigmented skin DNA replication c a Phenotype of this individual? C A c a Blue eyes albino c a meiosis I & II meiosis I & II c C a A c c A a Gametes (sperm or eggs) C c a a c c a a

27. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a

28. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles frequency of recombination (A-C) Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a Meiosis I Meiosis II C A ?? c a

29. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles frequency of recombination (A-C) C B A c a b # of recomb. chromosomes = C B A 8 c a b # of non-recomb. chromosomes = C B A # of recomb. 12 c a b frequency of recomb. (A-C) = C B A 8 = 0.4 (8+12) c a b C = 40% recombination = 40 map units = 40 centiMorgans (cM) B A # of chrom. analyzed c a b

30. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles frequency of recombination (A-C) C B A c a b # of recomb. chromosomes = C B A 8 c a b # of non-recomb. chromosomes = C B A 12 c a b frequency of recomb. (A-C) = C B A 8 = 0.4 (8+12) c a b C = 40% recombination = 40 map units; 40 centiMorgans (cM) B A c a b

31. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles frequency of recombination (B-C) C B A c a b # of recomb. chromosomes = C B A 2 c a b # of non-recomb. chromosomes = C B A 18 c a b frequency of recomb. (B-C) = C B A 2 = 0.1 (2+18) c a b C = 10% recombination = 10 map units = 10 centiMorgans (cM) B A c a b

32. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles frequency of recombination (B-C) C B A c a b # of recomb. chromosomes = C B A 2 c a b # of non-recomb. chromosomes = C B A 18 c a b frequency of recomb. (B-C) = C B A 2 = 0.1 (2+18) c a b C = 10% recombination = 10 map units = 10 centiMorgans (cM) B A c a b

33. What the frequency of recombination between genes tells us The frequency of recombination is based on percentage of meiotic divisions that result in breakage of linkage between parental alleles General conclusions: C B A The frequency of recombination between two genes is proportional to the distance (in bp of DNA) between the genes Genes very close together… Low probability of crossover between them …very few recombinants “tight linkage” Genes further apart… more recombinants c a b C B A c a b C B A c a b C B A c a b C B A c a b

34. What is the maximum (detectable) frequency of recombination between genes? If you examined a population of meioses, what is the maximum recombination frequency you could see between two genes? frequency of recombination (A-C) C A c a C A # of recomb. chromosomes = 10 c a # of non-recomb. chromosomes = C A Can we exceed 50% with more recombinants? 10 c a frequency of recomb. (A-C) = C A 10 = 0.5 c a (10+10) = 50% recombination = 50 map units = 50 centiMorgans (cM) C A c a

35. What is the maximum (detectable) frequency of recombination between genes? If you examined a population of meioses, what is the maximum recombination frequency you could see between two genes? The frequency of recombination between genes far apart on the same chromosome will approach, but never exceed 50 map units frequency of recombination (A-C) C A c a C A # of recomb. chromosomes = 50% 8 10 c a % recomb. (obs) # of non-recomb. chromosomes = C A 10 12 c a frequency of recomb. (A-C) = distance apart (bp) C A 10 8 = 0.4 = 0.5 c a (10+10) (8+12) = 50% recombination = 50 map units = 50 centiMorgans (cM) = 40% recombination = 40 map units =40 centiMorgans (cM) C A c a

36. Summary • The frequency of recombination between two genes on the same chromosome is proportional to the distance (in bp of DNA) between the two genes; 1% recombination = 1 map unit = 1cM • Genes that are close together on the same chromosome (<50 map units apart) are said to be “linked” • Genes that are far apart on the same chromosome (~50 map units apart) behave like genes on different chromosomes and assort independently • Genes that assort independently are said to be “unlinked”

37. A practical example • The TAS2R38 gene encodes a protein involved in tasting phenylthiocarbamide (PTC) and comes in two versions (alleles): • One lets you taste PTC, extremely bitter (dominant). • The second allele doesn’t detect PTC (recessive). • TAS2R38resides on chromosome 7 starting at base 141,318,900. • A second gene MET controls cell division • One allele can predispose to renal (papillary) cancer (dominant) • The second allele has no increased risk of renal cancer (recessive). • MET lies on chromosome 7 starting at base 116,099,695. MET TAS2R38 Chr7: 110,000,000 120,000,000 130,000,000 140,000,000

38. A family history Mom can taste PTC, but dad can’t. Mom’s family has a history of renal cancer and she has had a bout of it herself. Dad’s family has no history and he is disease free. grandpa grandma MET TAS met tas Mom’s father could taste but her mother couldn’t. met tas ? ? ? ? ? ? ? ? ? ? MET TAS MET tas met tas ? ? ? ? ? ? ? ? ? ? met TAS met tas met tas dad mom ??? And mom’s father had renal cancer, but her mother didn’t.

39. ? TAS A family history Mom can taste PTC, but dad can’t. You can taste PTC. What are your chances of also having the MET allele? Mom’s family has a history of renal cancer and she has had a bout of it herself. Dad’s family has no history and he is disease free. grandpa grandma MET TAS met tas Mom’s father could taste but her mother couldn’t. met tas ? ? ? ? ? ? ? ? MET TAS met tas ? ? ? ? ? ? ? ? met tas met tas dad mom And mom’s father had renal cancer, but her mother didn’t.

40. Mom’s gametes sperm: eggs: met tas met tas PTC taster and RCa MET TAS MET TAS parental met tas Nontaster, no Ca met tas met tas met tas PTC taster, no Ca met TAS nonparental or recombinant met TAS Nontaster, RCa met tas MET tas MET tas

41. Recombination fraction nonparental recombinant fraction parental + nonparental recombinant fraction x 100 centiMorgans The frequency of recombination between two genes on the same chromosome is proportional to the distance (in bp of DNA) between the two genes In humans: 1cM ≈ 1,000,000bp DNA

42. Genetic map of TAS and MET TAS is at 141,318,900 on chromosome 7 MET is at 116,099,695 on chromosome 7 They recombine about 25% of the time. MET TAS met TAS nonparental or recombinant parental 75% 25% MET tas met tas ≈25cM MET TAS2R38 Chr7: 110,000,000 120,000,000 130,000,000 140,000,000

43. ? TAS A family history You can taste PTC. What are your chances of also having the MET allele? grandpa grandma MET TAS met tas met tas ? ? ? ? ? ? ? ? MET TAS met tas 75% ? ? ? ? ? ? ? ? met tas met tas dad mom

44. Genetic map of TAS and MET TAS is at 141,318,900 on chromosome 7 MET is at 116,099,695 on chromosome 7 They recombine about 25% of the time. CFTR is at 116,986,754 About how often does CFTR recombine with TAS? With MET? CFTR MET TAS2R38 Chr7: 110,000,000 120,000,000 130,000,000 140,000,000

45. Finding a gene for juvenile form of ALS (Lou Gehrig’s disease) Thomas Mattingly was founder of a Maryland based clan that includes 70 victims of a type of ALS. His great-great-great-great-great grandson developed a pedigree that spanned 11 generations (350 years) and obtained blood samples from 107 members of the clan. Analysis of DNA from these samples allowed Phillip Chance (at UW) to find the gene that causes the disease.

46. A gene bearing a disease-causing mutation Identifying a human “disease gene” = ALS = disease free How can we identify the ALS gene? The ALS gene Identify linked markers that lie progressively closer to the ALS gene; then look for defects in the genes near to linked markers Will markers here exhibit linkage to the ALS trait? Chromosome 9 = 140,273,252bp YES!!! Markers = alleles of a gene Will markers here exhibit linkage to the ALS trait? Will markers here exhibit linkage to the ALS trait?

47. A gene bearing a disease-causing mutation Identifying a human “disease gene” = ALS = disease free How can we identify the ALS gene? Identify linked markers that lie progressively closer to the ALS gene; then look for defects in the genes near to linked markers But need convenient markers to trace linkage

48. Markers used in human gene mapping Polymorphic DNA markers are typically used… A polymorphic marker is a location in the genome where at least two versions of the sequence exist in the population, each at a frequency of at least 1% e.g., UW student population— about 40,000  80,000 copies of (e.g.) chromosome 2 70,000 copies have A-T base pair 10,000 copies have C-G base pair each is at > 1% of total population, so this is a polymorphic site

49. ..TCTTGATC.. ..AGAACTAG.. ..TCTAGATC.. ..AGATCTAG.. TCAGCAACAACACAGTCATATTTAGCTCTACCACAGTGTG GCAACAACACAGTCATATCCATTCTCAATTAGCTCTACCACAGTGTG CGCACCGGCCGCAGCAGCAGCAGCAGATTCACTGAGCTTG CGCACCGGCCGCAGCAGCAGCAGCAGCAGCAGCAGCAGATTCACTGAGCTTG Types of polymorphisms -Correlated ~325,000 SNPs in 473 people to fitness increase during endurance training -Found 19 SNPs associated with a 3-fold increase in fitness -Biggest effect by SNPs in ACSL1, a gene that influences fat metabolism Single Nucleotide Polymorphisms (SNPs): INsertion/DELetion (INDEL) polymorphisms: Variable Number Tandem Repeat (VNTR) polymorphisms:

50. Using polymorphic DNA markers to map genes 1 allele with 5 copies: ….GTAGTCACACACACAGCTA…. 1 allele with 3 copies: ….GTAGTCACACAGCTA…. etc. 5/3 4/6 Number of repeats 5/6 5/4 size large Separates DNA fragments by size 6 5 4 3 small