1. FCH4 key concepts Flinkage Frecombination Flinkage maps F3-point test cross F  2 test for linkage FCH4 problems... so far LECTURE 06: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION I

2. Fcan new combinations of “linked genes” be inherited? Fby what mechanism? Fis the frequency of new combinations related to their distance apart on the chromosome? Fhow do we know if genes are linked? LECTURE 06: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION I

3. MENDELIAN ANALYSIS: 1  n GENES... a quick review from CHAPTER 2 F3 methods of working out expected outcomes of controlled breeding experiments: 1. Punnet square 2. tree method (long) – genotypes 3. tree method (short) – phenotypes

4. MENDELIAN ANALYSIS: PUNNET SQUARE Fall pairings of  and  gametes F= probabilities of all pairings Fsome pairings occur >1 F  different P s for different genotypes & phenotypes F1 gene, 2 x 2 = 4 cells F2 genes, 4 x 4 = 16 cells F3 genes, 8 x 8 = 64 cells... Ftoo much work !

5. MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a 1/4 A/A 1/2 A/a 1/4 a/a GENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a 

6. MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a 1/4 A/A 1/2 A/a 1/4 a/a GENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a  PHENOTYPE 3/4 A 1/4 a

7. MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a ; B, b... 1/4 B/B 1/4 A/A 1/2 B/b 1/4 b/b 1/4 B/B 1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B 1/4 a/a 1/2 B/b 1/4 b/b GENOTYPE 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b 

8. MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a ; B, b... 1/4 B/B 1/4 A/A 1/2 B/b 1/4 b/b 1/4 B/B 1/2 A/a 1/2 B/b 1/4 b/b 1/4 B/B 1/4 a/a 1/2 B/b 1/4 b/b GENOTYPE 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b  PHENOTYPE 9/16 AB 3/16 Ab 3/16 aB 1/16 ab

9. MENDELIAN ANALYSIS: SHORT TREE 2 genes, alleles A, a ; B, b... ¾ B ¾ A ¼ b ¾ B ¼ a ¼ b much easier... can extend to >2 genes  PHENOTYPE 9/16 AB 3/16 Ab 1/16 ab

10. CHAPTER 4: KEY CONCEPTS Fgenes close together on a chromosome do not assort independently at meiosis Frecombination  genotypes with new combinations of parental alleles Fhomologous chromosomes can exchange segments by crossing-over Frecombination results from either independent assortment or crossing-over

11. CHAPTER 4: MORE KEY CONCEPTS Fgenes can be mapped by measuring frequencies of recombinants produced by crossing-over Finterlocus map distances based on recombination measurements are ~ additive Fone crossover can influence the occurrence of a second one in an adjacent region

12. LINKAGE Fobserve deviations from F9 : 3 : 3 : 1 ratios derived from dihybrid crosses F1 : 1 : 1 : 1 ratios derived from test crosses Ftoo many Ftoo few

13. LINKAGE SYMBOLISM & TERMINOLOGY Falleles on the same homologue have no punctuation between them... a b Falleles on different homologues separated by a slash... A B / a b Falleles written in same order on each homologue... A d B c / a D b C Funlinked genes separated by semicolon... A / a ; B / b Fgenes of unknown linkage separated by dot... A / a B / b

14. LINKAGE Funlinked genes, different chromosomes Fwrite as... Aa; Bb or A/a; B/b Fillustrate as... Fsegregate and assort independently Ftest cross to a/a; b/b  4 progeny types... 1 AB : 1 Ab : 1 aB : 1 ab

15. LINKAGE Flinked genes, same chromosomes Fwrite as... Aa Bb or AB/ab or Ab/aB F2 possible dihybrids, illustrate as... Fsegregate and assort dependently ~ distance Ftest cross to ab/ab  4 progeny types... in CIS: 0 Ab : > 0 aB : < 1 ab in TRANS: > 0 AB : 0 ab

16. LINKAGE Flinked genes, same chromosomes

17. LINKAGE Fchiasmata: visible manifestations of crossing over

18. LINKAGE Flinked genes (in cis here), same chromosomes also called non-parental or recombinant chromosomes

19. LINKAGE Fcrossing over when in meiosis? (tetrad analysis) if 

20. LINKAGE Fhow many chromatids involved? (tetrad analysis)  

21. RECOMBINATION Fproducts of meiotic recombination linkage unknown ati

22. RECOMBINATION Fdetection of recombination using a test cross

23. RECOMBINATION Ftest cross Funlinked genes Findependent assortment always  “recombination” frequency of 50%

24. RECOMBINATION Frecombinants arise when non-sister chromatids cross over between genes under study

25. RECOMBINATION Ftest cross Flinked genes Findependent assortment  recombination frequency < 50%

26.  2 TEST FOR LINKAGE A O E a O E  B 140 / 127.5115 / 127.5255 b 110 / 122.5135 / 122.5245  250 500

27.  2 TEST FOR LINKAGE A O E a O E  B 140 / 127.5 115 / 127.5 255 b 110 / 122.5135 / 122.5245  250 500

28.  2 TEST FOR LINKAGE

29. F express this as: 0.05 > P (  2 c = 5.02) > 0.01 F the data derived from a test cross deviate significantly from a 1:1:1:1 ratio F we reject our H 0 (alternatively we could not reject ) F “genes A and B are linked” F the probability of deviation by chance from the linked genes model is between 1 and 5% ( i.e., deviation from the 1:1:1:1 ration occurs because linkage is not supported by data)

30. RECOMBINATION PF1F2PF1F2 cn bw + cn + bw cn bw cn bw + cn bw + cn + bw cn bw cn + bw  x    x   score progeny Autosome Linkage X -Chromosome Linkage  no recombination in Drosophila  s y w + y + w y + w y w + Y y + w Y y + w  x   x   score  progeny   X -Chromosome Linkage must use  s only

31. RECOMBINATION PF1F2PF1F2 cn bw + cn + bw cn bw cn bw + cn bw + cn + bw cn bw cn + bw  x    x   score progeny Autosome Linkage X -Chromosome Linkage  y w + y + w y + w y w + Y y + w Y y + w  x   x   score  progeny   X -Chromosome Linkage tester  chromosomes

32. LINKAGE MAPS Fsites of recombination are ~ random Fprobability of recombination increases ~ distance

33. LINKAGE MAPS Fcomplications: 1. recombination rates  position on chromosome RECOMBINATION FREQUENCY Telomere centromere telomere 2. linked genes >50 cM apart look like unlinked genes 3. double recombination events are usually not detected 4. multiple strand (>2) exchanges 5. tester for multiple autosomal genes not always available 6. interference – reduced probability of adjacent events (later)

34. F2 possibilities for map Fadd the shorter distances LINKAGE MAPS

35. 3-POINT TEST CROSS P v + cv ct x v cv + ct + v + cv ct Y  F 1 v + cv ct x v cv + ct + v cv + ct + Y F 2

36. 3-POINT TEST CROSS Fdouble crossover involving 2 chromatids Fouter alleles look like parental types Frecombination cryptic without middle gene

37. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS

38. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS

39. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS Base your decision on... (1) similar numbers (2) reciprocal genotypes

40. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS

41. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS Draw all 3 possible linear arrangements (next), what do double crossovers between both PT chromosomes give...

42. Ftest all 3 possible orders of PT allele combinations  3-POINT TEST CROSS 3535 ct + 94 89 45 40

43. Ftest all 3 possible orders of PT allele combinations  3-POINT TEST CROSS PT chromosomes with genes in the order v ct cv (NOT the order shown in the table... it usually won’t be) give you the two F 2 groups with the smallest numbers with double crossovers between the genes. 3535

44. 3-POINT TEST CROSS REAL GENE ORDER DOES NOT HAVE TO BE AS LISTED IN THE DATA TABLE. Consider genes v & ct. PT are v ct + or v + ct in TRANS. All single crossovers between them yield reciprocal v ct & v + ct + CIS products as shown.

45. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time; e.g. : v  ct = [ ( 89 + 94 + 3 + 5 ) / 1448 ] x 100 4.construct map 5. calculate interference 3-POINT TEST CROSS

46. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map... 5. calculate interference 3-POINT TEST CROSS

47. v ct cv 18.5 cM 13.2 cM6.4 cM or... 13.2 + 6.4 = 19.6 cM Fdistances between outer genes is less accurate because double recombinants are cryptic Fadd distances for accurate map

48. Forder of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3.calculate distances: 2 genes at a time 4.construct map 5. calculate interference 3-POINT TEST CROSS

49. Finterference = 1 – COEFFICIENT OF COINCIDENCE = 1 – [ OBSERVED # DCO / EXPECTED # DCO ] = 1 – [ 8 / ( 0.132 x 0.064 x 1448 ) ] = 1 – [ 8 / 12 ] = 1 / 3 = 33 % 3-POINT TEST CROSS

50. LINKAGE MAPS why is mapping important ? $ fundamental aspect of genetic analysis, concerns: $ gene function $ gene evolution $ gene isolation

51. LINKAGE MAPS gene function... $ location influences function... “position effect” $ relative to heterochromatin $ other genes $ genes of related function often clustered $ operons in Prokaryotes $ homeotic genes in Eukaryotes

52. LINKAGE MAPS gene evolution... $ relative positions of genes in related organisms infer history of change

53. LINKAGE MAPS gene isolation... $ mapping 1 st step $ Drosophila...

54. LINKAGE MAPS gene isolation... $ mapping 1 st step $ tomato... prophase nucleus chromosome numbering map (ca. 1952)

55. LINKAGE MAPS - HUMANS why has this been difficult ? $ controlled breeding programs not possible (not ethical) $ sample sizes small (relatively) $ human genome is LARGE... distances are large

56. EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION: PROBLEMS F in Griffiths chapter 4, beginning on page 141, you should be able to do questions #1-30 F begin with the solved problems on page 138 if you are having difficulty F look at the way Schaum’s Outline discusses linkage and mapping for alternative explanations - especially tetrad analyses F try Schaum’s Outline questions in chapter 4, beginning on page 208

57. TUTORIAL #2: T.9.26 or R.9.28 F go to the TUTORIAL page on the genetics web site to download the file tut2-06F.pdf F complete all six questions before attending tutorial #2 next week F during tutorial, you will review solutions to these questions with your group and prepare to present solutions to all six questions F one member of your group will be selected by the TA to present a solution to one of these questions F presentations will be 10 min each + 5 min for questions and discussion