Linkage and Recombination Mendel Alleles segregate Alleles for

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

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

Mapping Genes on Chromosomes Alfred Sturtevant 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

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

The genes are on the same Sturtevant’s interpretation: chromosome & the parental alleles (mostly) co-segregate wild type tl+y + s + + o vg pr pr mvg gamete? x parental purple eyestype (same as gametes vestigial wings made prthat pr vg vg the parent) oc cro casi wild type phenotype ss ona gamete? recombinant (non ov l pr+ pr vg+ vg er -parental) type 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

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

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

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

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

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

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

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

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

Identifying the parental type Option 2. The two most abundant progeny types (assuming the genes are linked) Cross: pr+ pr vg+ vg x pr pr vg vg Progeny: 1287 1204 170 154 What were the gametes that made the heterozygous parent? pr+ vg pr vg+

A notation system that defines allele configuration linkage = Configuration (i. e. , cis or trans) ? ? ? pr+ vg+ e k impliesa m t ’ n s pr vg e o D n e s ! e s = w+ w ; pr+ pr separate chromosomes

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?

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

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

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

Map distance… example from last time Cross: pr+ pr vg+ vg x pr pr vg vg Progeny: 1287 1204 170 154 What is the map distance between these two genes?

Map distance… example (cont’d) Step 1. Identify the parental and non-parental types. 1287 parental types (most abundant) 1204 170 non-parental types (least abundant) 154 parental = pr+ vg and pr vg+ non-parental = pr+ vg+ and pr vg

Map distance… example (cont’d) Step 2. Calculate % recombinant products. 1287 parental types (most abundant) 1204 170 non-parental types (least abundant) 154 % recombinant = (170 + 154) x 100 = 11. 5% (170+154+1287+1204) Map distance = 11. 5 map units = 11. 5 centi. Morgan’s (c. M)

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

Another Example (recessive b mutation) 4615 red black 4707 purple tan 307 red tan Parental types pr + b pr 295 purple black Recombinant types b+ pr + b+ pr b Recombination Frequency = Number Recombinant Types Total Number of Progeny pr b Example: 307602 + 295 4615 + 4707 + 307 + 295 9924 = 0. 06 x 100 = 6. 0 %

Creating a Genetic Map Genetic loci: % Recombinants + + purple - vestigial - vg) b+ (pr pr vg 11. 5 % purple - black (pr - b) 6. 0 % Revealed vestigial - black (vg - b) from other 16. 5 % crosses b vg pr Alfred Sturtevant’s mental leap: % recombinants is directly proportional to distance He drew a genetic map of the second chromosome: 6 c. M 11. 5 c. M b pr vg Why 17. 5 c. M and not 16. 5 c. M?

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

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

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 1 2 Is the gamete that III-1 received from II-2 parental or non-parental? BUT FIRST… break down the question: II 1 2 Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem Step 1. III 1 = autosomal dominant Step 2. Step 3. = autosomal recessive = both traits etc.

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 1 2 Is the gamete that III-1 received from II-2 parental or non-parental? BUT FIRST… break down the question: II 1 2 Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem Step 1. Figure out all the genotypes! III 1 = autosomal dominant = autosomal recessive = both traits 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?

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

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

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

Example Suppose URA 1 and URA 2 are on the same chromosome… Diploid Spores Growth on ura plates? no yes no no Parental ditype? Non-parental ditype? Tetratype?

Exercise Suppose URA 1 and URA 2 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+ = URA 1 1+ 2 - 1+ 2 - 1 and 2 - 1 - 2 + 1+ 2 + PD NPD 1 - =ura 1 2+ =URA 2 2 - =ura 2

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

Using 2 analysis to explore linkage 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. It gives a precise Why test the null hypothesis? 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.

2 analysis • 2 = (Obs-Exp)2 Exp Out of 10, 000 testcross progeny. . . observed expected (o-e)2 long no speck 2929 vestigial speck 2921 vestigial no speck 2070 long speck 2080 (o-e)2 e

2 table P 0. 995 0. 975 0. 900 0. 500 0. 100 0. 050 0. 025 0. 010 0. 005 df 1 2 3 4 5 6 0. 000 0. 010 0. 072 0. 207 0. 412 0. 676 0. 000 0. 051 0. 216 0. 484 0. 831 1. 237 0. 016 0. 211 0. 584 1. 064 1. 610 2. 204 0. 455 1. 386 2. 366 3. 357 4. 351 5. 348 2. 706 4. 605 6. 251 7. 779 9. 236 10. 645 3. 841 5. 991 7. 815 9. 488 11. 070 12. 592 5. 024 7. 378 9. 348 11. 143 12. 832 14. 449 6. 635 9. 210 11. 345 13. 277 15. 086 16. 912 7. 879 10. 597 12. 838 14. 860 16. 750 18. 548 Degrees of freedom? E R What does this P value mean? What is the P value? T C JE NO WAY could these numbers be due to independent assortment!

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

Practice question What if Sturtevant had analyzed only 100 testcross progeny? Out of 100 testcross progeny. . . observed long no speck 29 vestigial no speck 21 long speck 21 What is the P value? What does this P value mean? expected (o-e)2 e

2 analysis The “null” hypothesis: The testcross data do not significantly differ from a • 2 = Exp 1: 1: 1: 1 ratio. Out of 100 testcross progeny. . . 2 2 (o-e) observed expected (o-e) e (Obs-Exp)2 29 25 42 = 0. 64 vestigial speck 29 25 42 = 0. 64 vestigial no speck 21 25 42 = 0. 64 long no speck long speck = 2. 56

2 table P 0. 995 0. 975 0. 900 0. 500 0. 100 0. 050 0. 025 0. 010 0. 005 df 1 2 3 4 5 6 0. 000 0. 010 0. 072 0. 207 0. 412 0. 676 0. 000 0. 051 0. 216 0. 484 0. 831 1. 237 0. 016 0. 211 0. 584 1. 064 1. 610 2. 204 0. 455 1. 386 2. 366 3. 357 4. 351 5. 348 2. 706 4. 605 6. 251 7. 779 9. 236 10. 645 3. 841 5. 991 7. 815 9. 488 11. 070 12. 592 5. 024 7. 378 9. 348 11. 143 12. 832 14. 449 6. 635 9. 210 11. 345 13. 277 15. 086 16. 912 7. 879 10. 597 12. 838 14. 860 16. 750 18. 548 Degrees of freedom? What does this P value mean? t c je e R ’t n a C Yes, these numbers could be due to independent assortment! Since Sturtevant DID analyze more flies, however, we are certain that vg and sp are linked.

Where does speck map? Genetic loci: % Recombinants purple - vestigial (pr - vg) 11. 5 % purple - black (pr - b) 6. 0 % vestigial - black (vg - b) 16. 5 % vestigial - speck (vg - sp) 41. 5 % 41. 5 c. M sp? 6 c. M 11. 5 c. M b pr 41. 5 c. M sp? vg How do we distinguish these possibilities? If to the left, fewer recombinants with pr (and b). If to the right, more recombinants with pr (and b).

P 1 cross: pr+ sp + pr+ sp+ F 1 : F 1 dihybrid X pr pr X sp sp testcross parent Testcross progeny: 2504 red no speck 2498 purple speck Parental types 2501 purple no speck 2497 red speck Recombinant types

Where shall we place speck? Genetic loci: % Recombinants 11. 5 % purple - vestigial (pr - vg) 6. 0 % purple - black (pr - b) 16. 5 % vestigial - black (vg - b) 41. 5 % vestigial - speck (vg - sp) 6 c. M 10. 7 c. M b pr 41. 5 c. M sp vg What is the map distance between b and sp? ~58. 2 c. M