2. Linkage and Recombination Genes are on chromosomes Alleles segregate Alleles for different genes assort independently Pea plants have 7 pairs of chromosomes. Flies have 4 pairs. Humans have 23 pairs. Mendel Morgan and Bridges Bateson et al.

3. What is the inheritance pattern for two traits determined by genes on the same chromosome? “Linked” genes are those that do NOT assort independently. (They reside on the same chromosome.) We will see that... DMD SCID XIST XLP Fragile X Haemophilia R-G Colorblindness Human X chromosome RP3 HPRT PGK The corollary isn’t always true! Just because genes are on the same chromosome doesn’t necessarily mean they are linked.

4. As a student in Morgan’s lab, Sturtevant was the first person to correctly interpret linkage; he created the first genetic maps. Sturtevant in his lab at Caltech Alfred Sturtevant Mapping Genes on Chromosomes

6. Segregation of genes on the same chromosome wild type pr + pr + vg + vg + purple eyes vestigial wings pr pr vg vg gamete? wild type phenotype pr + pr vg + vg x testcross [ ] testcross parent genotype? pr + vg + pr + vgpr vg + pr vg phenotypes: For I.A., expect:1:1:1:1 ratio observed:1339:151:154:1195 pr pr vg vg

7. Sturtevant ’s interpretation: parental type (same as gametes that made the parent) mostly recombinant (non-parental) type occasional crossover wild type pr + pr + vg + vg + purple eyes vestigial wings pr pr vg vg gamete? wild type phenotype pr + pr vg + vg x The genes are on the same chromosome & the parental alleles (mostly) co-segregate How do we know the genotypes of the gametes? Testcross parent’s gametes… only recessive alleles present So testcross progeny phenotype allows us to deduce the heterozygous parent’s gamete genotype

8. sister chromatids (each is a double helix) homologues cohesins synaptonemal complex Recombination… a brief review [see lecture 2 for details] Single- strand nicks strand exchange Holliday junction -No deletions are caused by recombination. -No mutations are caused by recombination.

9. Recombination… a brief review (cont) ALWAYS pull from the centromeres Segregation of recombinant chromosomes during Meiosis I 3 4 pr + pr vg 1 2 pr pr + vg + 1 4 3 2 Label the chromatids

10. Telophase IGametes Parental Recombinant Telophase II One recombination event: 2 recombinant and 2 non-recombinant products Recombination… a brief review (cont)

11. Which chromatid goes where? An example with a double crossover: Test your Understanding Draw these chromosomes in anaphase I. Label all alleles appropriately.

12. ALWAYS pull from the centromeres 3 4 pr + pr vg vg + 1 2 pr pr + vg + vg

13. Other types of crossovers # xoversresulting gametes 0parental parental! 2 (2 strands) 2 parental 2 non-par. 2 (3 strands) 4 non-par. 2 (4 strands) Crossing over must occur for faithful segregation at meiosis I

14. Sturtevant’s findings—summary Genes on the same chromosome can show linkage instead of independent assortment Gametes (mostly) have the same allele combinations as the homologs in the parent Recombination can give rise to gametes with non-parental (=recombinant) allele combinations Two parental types are more abundant and occur at roughly equal frequency to one another Non-parental types are less abundant and occur at roughly equal frequency to each other

15. Identifying the parental type Option 1. Know the gametes that made the heterozygous parent pr + pr + vg + vg + pr pr vg vg pr + vg + pr vg gametes: pr + pr vg + vg pr + vgpr vg + gametes: dominant alleles together on the same chromosome = “cis” configuration dominant alleles on different chromosomes = “trans” configuration define the parental type X pr + pr + vg vgpr pr vg + vg + X pr + vg + pr vg pr + vg pr vg + The cross from our previous example: A different cross from our example:

16. Identifying the parental type Option 2. The two most abundant progeny types (assuming the genes are linked) 12871204 170154 Cross: pr + pr vg + vgxpr pr vg vg Progeny: What were the gametes that made the heterozygous parent? pr + vgpr vg +

17. =  = pr + pr vg + vg implies Doesn’t make sense! ??? A notation system that defines allele configuration linkage Configuration (i.e., cis or trans) separate chromosomes pr + pr ; w+w+ w

18. Sturtevant’s interpretation of linkage -Recombination involves the physical exchange of chromosomal segments between homologs. -The frequency of recombinant types indicates the distance between linked genes. What is the evidence in support of these hypotheses?

19. Test of Sturtevant’s hypothesis  Harriet Creighton and Barbara McClintock, maize  Curt Stern, Drosophila The problem: homologous chromosomes look alike… how to tell if they really exchanged segments? found genetically marked chromosome with structurally (visually) distinct homologues Prediction: Chromosome from genetically recombinant plants should show structural rearrangement

20. Creighton/McClintock test Sturtevant’s hypothesis C = coloredWx = starchy c = colorlesswx = waxy “knob”translocation x Progeny phenotypes colored, waxy colorless, starchy colored, starchy colorless, waxy Conclusion: genetic recombination  exchange of chromosome segments Structural markers:Genetic markers: recombinant genotypes and chromosomes!

21. Mapping genes using recombination Alfred Sturtevant’s major insight: If crossovers occur at random Probability of crossover between two genes is proportional to the distance between them Crossover between A and B much more likely than between B and D

22. Map distance… example from last time 12871204 170154 Cross: pr + pr vg + vgxpr pr vg vg Progeny: What is the map distance between these two genes?

23. non-parental types (least abundant) parental types (most abundant) 12871204 170154 Step 1. Identify the parental and non-parental types. parental = non-parental = pr + vg and pr vg + pr + vg + and pr vg Map distance… example (cont’d)

24. non-parental types (least abundant) parental types (most abundant) 12871204 170154 Step 2. Calculate % recombinant products. % recombinant = (170 + 154) (170+154+1287+1204) x100 = 11.5% Map distance = 11.5 map units = 11.5 centiMorgan’s (cM) Map distance… example (cont’d)

25. Number Recombinant Types Total Number of Progeny Another Example (recessive b mutation) Which are the parental types? and the recombinant types? Recombination Frequency = 4615 red black 4707 purple tan 307 red tan 295 purple black pr b Example:

26. Recombination Frequency = Total Number of Progeny pr b Example: 307 + 295 4615 + 4707 + 307 + 295 = 0.06 Parental typesRecombinant types pr + bpr b + pr + b + pr b = 6.0 %x 100 602 9924 4615 red black 4707 purple tan 307 red tan 295 purple black Number Recombinant Types Another Example (recessive b mutation)

27. Genetic loci:% Recombinants Alfred Sturtevant’s mental leap: 11.5 % % recombinants is directly proportional to distance purple - vestigial (pr - vg) 6.0 % purple - black (pr - b) 16.5 % vestigial - black (vg - b) He drew a genetic map of the second chromosome: vgpr 11.5 cM b 6 cM Creating a Genetic Map Revealed from other crosses Why 17.5cM and not 16.5cM? b+b+ bvg vg + pr pr +

28. Point to ponder: If you examined a population of meioses, what is the maximum recombination frequency you could see between two genes? Genes very close together… Low probability of crossover between them …very few recombinants “tight linkage” Genes further apart… more recombinants

29. * Crossing-over creates new combinations of traits. * The frequency of recombinant types indicates the distance between linked genes. * Two Parental types in ≈ frequencies. Two Recombinant types ≈ frequencies. * If genes are linked, Parental types > recombinant types. Summary

30. Practice question The pedigree shows segregation of two disorders one is autosomal dominant (A= disease, a = not) one is autosomal recessive (b = disease, B = not). I II III 12 12 1 = autosomal dominant = autosomal recessive = both traits Is the gamete that III-1 received from II-2 parental or non-parental? BUT FIRST… break down the question: Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem Step 1. Step 2. Step 3. etc.

31. The pedigree shows segregation of two disorders one is autosomal dominant (A= disease, a = not) one is autosomal recessive (b = disease, B = not). I II III 12 12 1 = autosomal dominant = autosomal recessive = both traits Is the gamete that III-1 received from II-2 parental or non-parental? BUT FIRST… break down the question: Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem Step 1. Figure out all the genotypes! Step 2. What are the gametes that made II-2? Step 3. What is the gamete that II-2 made? Step 4. Does the gamete that II-2 made have a different genotype than the gamete(s) that made him? Practice question

32. I II III 12 12 1 Is the gamete that III-1 received from II-2 parental or non-parental? a ba b a b A BA B ? AB A BA B a b A bA b a Ba B Ab ab gamete Gametes that made II-2= Gamete that II-2 gave to III-1= AB and ab Ab = autosomal dominant = autosomal recessive = both traits The pedigree shows segregation of two disorders one is autosomal dominant (A= disease, a = not) one is autosomal recessive (b = disease, B = not).

33. Quiz Section this week: yeast tetrad analysis (genetic mapping) An introduction to the yeast tetrad analysis terminology...

34. In yeast… Tetrad of spores… can examine products of individual meioses! Tetrad with only the 2 parental types = “parental ditype” (PD). Tetrad with only the 2 recomb. types = “non-parental ditype” (NPD). Tetrad with all four combinations = “tetratype” (T). Looking at whole tetrads (PD/NPD/T) is informative. non-parental 4 haploid spores 1n meiosis 2n diploid

35. Example Suppose URA1 and URA2 are on the same chromosome… DiploidSpores Growth on - ura plates? no yes Parental ditype? Non-parental ditype? Tetratype?

36. Exercise Suppose URA1 and URA2 are on separate chromosomes, but each tightly linked to their respective centromeres… what kinds of tetrads (growth on -ura plates) would this diploid produce? 1 + = URA1 1 - =ura1 2 + =URA2 2 - =ura2 1 + 2 - 1 - 2 + 1 - 2 - 1 + 2 + and PDNPD

37. Using  2 analysis to explore linkage P 1 cross: F 1 : All long, no speck X long, no speck vestigial, speck Test cross Long, no speck: 2929 Vestigial, speck: 2921 Vestigial, no speck: 2070 Long, speck: 2080 Can these results be explained by chance deviation from independent assortment?

38. Is this really a 1:1:1:1 ratio as we would expect for independent assortment? Is the deviation from independent assortment due to chance?  2 analysis: Test the “ null ” hypothesis — that the observed deviation from 1:1:1:1 segregation is due to chance variation. Why test the null hypothesis? It gives a precise expectation (1:1:1:1) We cannot test directly for linkage, because (assuming they are linked) we do not know the map distance separating these genes. Using  2 analysis to explore linkage

39. Out of 10,000 testcross progeny... expectedobserved long no speck 2929 (o-e) 2 e  (Obs-Exp) 2 Exp 2 =2 =  2 analysis vestigial speck 2921 vestigial no speck 2070 long speck 2080

40. P df 1 2 3 4 5 6 0.995 0.000 0.010 0.072 0.207 0.412 0.676 0.975 0.000 0.051 0.216 0.484 0.831 1.237 0.900 0.016 0.211 0.584 1.064 1.610 2.204 0.500 0.455 1.386 2.366 3.357 4.351 5.348 0.100 2.706 4.605 6.251 7.779 9.236 10.645 0.050 3.841 5.991 7.815 9.488 11.070 12.592 0.025 5.024 7.378 9.348 11.143 12.832 14.449 0.010 6.635 9.210 11.345 13.277 15.086 16.912 0.005 7.879 10.597 12.838 14.860 16.750 18.548  2 table What does this P value mean? REJECT NO WAY could these numbers be due to independent assortment! What is the P value? Degrees of freedom?

41. P 1 cross: F 1 : All long, no speck X long, no speck vestigial, speck Test whether the data truly show linkage by doing a Chi-square analysis. What is the null hypothesis? speck Draw the chromosomes of the F1 hybrid in prophase of Meiosis I such that the gametes will produce all four progeny types. Homework Practice question

42. 1: vg + sp + 3: vg sp + 2: vg + sp 4: vg sp sp + vg + sp + vg + spvg spvg 1 2 3 4

43. Out of 100 testcross progeny... expected observed long no speck29 (o-e) 2 e vestigial speck29 vestigial no speck21 long speck21 What does this P value mean? What is the P value? What if Sturtevant had analyzed only 100 testcross progeny? Practice question

44. Out of 100 testcross progeny... expectedobserved long no speck 29 (o-e) 2 e  = 2.56  (Obs-Exp) 2 Exp 2 =2 = vestigial speck 29 vestigial no speck 21 long speck 21 25 4242 4242 4242 4242 = 0.64  2 analysis The testcross data do not significantly differ from a 1:1:1:1 ratio. The “null” hypothesis:

45. P df 1 2 3 4 5 6 0.995 0.000 0.010 0.072 0.207 0.412 0.676 0.975 0.000 0.051 0.216 0.484 0.831 1.237 0.900 0.016 0.211 0.584 1.064 1.610 2.204 0.500 0.455 1.386 2.366 3.357 4.351 5.348 0.100 2.706 4.605 6.251 7.779 9.236 10.645 0.050 3.841 5.991 7.815 9.488 11.070 12.592 0.025 5.024 7.378 9.348 11.143 12.832 14.449 0.010 6.635 9.210 11.345 13.277 15.086 16.912 0.005 7.879 10.597 12.838 14.860 16.750 18.548  2 table What does this P value mean? Can’t Reject Yes, these numbers could be due to independent assortment! Degrees of freedom? Since Sturtevant DID analyze more flies, however, we are certain that vg and sp are linked.

46. 41.5 % Genetic loci:% Recombinants 11.5 % purple - vestigial (pr - vg) vestigial - speck (vg - sp) 6.0 % purple - black (pr - b) 16.5 % vestigial - black (vg - b) vgpr 11.5 cM sp? 41.5 cM b 6 cM sp? 41.5 cM If to the left, fewer recombinants with pr (and b). If to the right,more recombinants with pr (and b). Where does speck map? How do we distinguish these possibilities?

47. P 1 cross: X 2504 F 1 : F1 dihybrid 249825012497 Parental typesRecombinant types testcross parent X Testcross progeny: red no speck purple speck purple no speck red speck pr + sp + pr sp

48. Genetic loci:% Recombinants 11.5 % 41.5 % purple - vestigial (pr - vg) vestigial - speck (vg - sp) 6.0 % purple - black (pr - b) 16.5 % vestigial - black (vg - b) vgpr 10.7 cM sp41.5 cM b 6 cM Where shall we place speck? What is the map distance between b and sp?~58.2 cM