Recombination & Genetic Analysis

1. Recombination & Genetic Analysis -The maximum recombination frequency -Quiz section 5 revisited -Genetic vs. Physical maps

2. 58.2 cM b+sp+ b sp 21 25 P 25 21 25 29 R 25 29 Test your understanding Predict the number of progeny for each type of offspring that result from the following cross. Assume 100 total offspring. Dihybrid female Testcross male X tan no speck Recombinants never exceed 50% black speck tan speck black no speck

3. No! b+ sp+ All four gamete types are equally frequent b+ sp + 1 b+ sp b+ 2 sp + b sp 3 b sp+ 4 b sp b sp Are b and sp linked? • These genes are so far apart, that they assort independently from one another. • b and spappear to be unlinked even though they are on the same chromosome! vg+ vg How do we know that b and sp are on the same chromosome? b is linked to vg and vg is linked to sp.

4. Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: Write down the parental types with respect to A/a and D/d:

5. Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: Write down the parental types with respect to A/a and D/d:

6. Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: A B D 1 A B D 2 a b d 3 a b d 4 After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: A B and a b Write down the parental types with respect to A/a and D/d: A D and a d

7. Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: A B D 1 A B D 2 a b d 3 a b d 4 After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: A B and a b Write down the parental types with respect to A/a and D/d: A D and a d

8. In-class experiment (cont’d) Looking first at just loci A/a and B/b… What are the genotypes of the products from your meiosis? 1. 3. 2. 4. Are these gametes all parental? All recombinant? 2 of each? Then look at just loci A/a and D/d… What are the genotypes of the products from your meiosis? 1. 3. 2. 4. Are these gametes all parental? All recombinant? 2 of each?

9. What must have happened to create these gametes? In-class experiment (cont’d) Class aggregate data: A-B Genotype P or R? Number? Number P gametes Number R gametes AB, ab 4 P 81 324 0 Ab, aB 4 R 2 0 8 AB, ab, Ab, aB 2 P, 2 R 43 86 86 # Recombinants X = % Rec? X Total # Gametes X

10. In-class experiment (cont’d) Class aggregate data: A-D Genotype P or R? Number? Number P gametes Number R gametes AD, ad 4 P 40 160 0 Ad, aD 4 R 19 0 76 AD, ad, Ad, aD 2 P, 2 R 61 122 122 # Recombinants 76+122 = % Rec? X Total # Gametes 76+122 +160 +122

11. Why do we observe IA? A B D A B D a b d a b d Everyone in the class drew crossovers somewhere between A/a and D/d, yet the overall % recombinants for the class was only ~50%. If we look at a large enough sample, even genes that are very far apart on the same chromosome cannot show more than 50% recombinant products. Need to look closer at meiosis itself to see why.

12. What is the maximum recombination frequency in any interval? Parental Reminder… one crossover gives 2 parental, 2 recombinant gametes Recombinant Recombinant Parental The range of possibilities: tightly linked  independent assort. Consider 100 cells undergoing meiosis… 2 recombinant out of 400 0.5% if one cell has a crossover   20 recombinant out of 400 5.0% if 10 cells have crossovers   200 recombinant out of 400 50% if all cells have crossovers   Maximum recombination frequency = 50% for single recombination events

13. 6 parental 6 non-par. What is the maximum recombination frequency in any interval? The effect of multiple crossovers: # xovers resulting gametes 2 (2 strands) 4 parental 2 (3 strands) 2 parental 2 non-par. 2 (4 strands) 4 non-par. = 50% recombinant and 50% parental. Also true for triple Xover, quadruple Xover, etc.

14. Human X-chromosome map… how could we get 180 cM?

15. R Y R Y X r y r y Now consider independent assortment …the “ultimate” in non-linkage Refer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round yellow): What were the parental types for the F1? What were the parental and recombinant gametes made by the F1 plants? What was the % recombinant gametes? RY and ry 1/4 RY 1/4 ry 1/4 Ry 1/4 rY 1/4 + 1/4 = 50%! So, even for independently assorting genes, the % recombinant products is only 50%

16. Conclusion For widely separated genes 1) An odd number of crossovers gives, on average, an equal number of parental and recombinant types. 2) An even number of crossovers gives, on average, an equal number of parental and recombinant types. 3) Alleles on two different chromosomes line up on the metaphase plate independently, giving on average equal numbers of parental and recombinant types. Thus, the maximum recombinant frequency =50% Loci can appear to be unlinked because: • They are on separate chromosomes • They are so far apart on the same chromosome that recombination always occurs

17. R f r F X r f r f Practice question • In a certain plant species… • flower fragrance (F) is dominant over unscented (f) • blue flower color (B) is dominant over white (b) • rounded leaves (R) is dominant over pointy (r); and • thorny stems (T) is dominant over smooth stems (t). • From the following crosses, can you determine whether the fragrance gene is linked to any of the other genes; if so, at what map distance? The parental and recombinant types are the same! Need to be heterozygous at both loci Bb Ff x bb ff Rr ff x rr Ff Tt Ff x tt ff 270 blue, fragrant 281 blue, non-fragrant 268 white, fragrant 275 white, non-fragrant 219 rounded, fragrant 222 rounded, non-fragrant 209 pointy, fragrant 216 pointy, non-fragrant 333 thorny, fragrant 36 thorny, non-fragrant 39 smooth, fragrant 342 smooth, non-fragrant F and T linked at 10 cM F not linked to B Can’t tell!

18. etc. QS7 revisited What were the main points of QS5? -To give you an opportunity to see actual data from a meiosis and to draw conclusions from the data based on your knowledge of this process. -To show what can be learned from looking at all four products of a single meiosis. The diagrams used in quiz section… …were designed to help set up specific predictions

19. Can’t distinguish Can distinguish! Setting up predictions for meiosis outcomes… But if we can’t see all of the products from a single meiosis we expect… then we expect… If… a parental ditype (PD) PD > T >> NPD mostly parental types either PD or non-parental ditype (NPD), with a 50:50 chance of each PD = NPD > T an equal proportion of parental and recombinant types mostly tetratypes (T), but also some PD, and NPD an equal proportion of parental and recombinant types

20. What did the data from quiz section tell you? Conclusion? Conclusion? Conclusion? Probably not linked Probably linked Probably not linked

21. Looking at the 10 tetrads in terms of LEU & TS How many PD? How many NPD? How many T? 4 5 1 So what do you conclude about the LEU and TS genes? they are independently assorting but each close to a centromere!

22. leu HIS ts LEU his ADE TS ade Our completed map Diploid genotype: MATaADEHIS leu ts URA1 ura2 MAT ade his LEU TS ura1 URA2

23. ADE2 HIS4 LEU2 CDC7 (TS) How well did we do? The actual gene names… ADE2 CDC7 (TS) HIS4 LEU2 Let’s look in SacchDB… Not bad!

24. U1 u2 U1 u2 u1 U2 u1 U2 What about URA? Know the parental types Look at spore phenotypes U1 u2 & u1 U2 Parental types? What spore genotypes would you expect in a PD tetrad? Phenotype on -ura plate? no growth So, given these parental types… 0/4 spores growing is diagnostic of PD no growth no growth no growth

25. U1U2 U1U2 u1 u2 u1 u2 What about URA? What spore genotypes would you expect in a NPD tetrad? U1 u2 & u1 U2 Parental types? Genotype? Growth phenotype on -ura? GROWTH GROWTH no growth no growth

26. U1 u2 U1U2 u1 U2 u1 u2 What about URA? What spore genotypes would you expect in a T tetrad? U1 u2 & u1 U2 Parental types? Genotype? Growth phenotype on -ura? no growth GROWTH no growth no growth

27. Looking at the 10 tetrads… How many PD? How many NPD? How many T? So what do you conclude about the ura genes?

28. Practice question B e Heterozygote genotype = b E Brown seed pods (B) in a plant species is dominant to green (b), and elongated pods (E) is dominant over squished (e). (a) A fully heterozygous plant has the dominant alleles linked in trans (i.e., dominant alleles not on the same homologue) at a map distance of 20 cM. What will be the genotypes of gametes produced by this plant, and in what frequencies (or percentages)? (b) If this plant is self-pollinated, what progeny phenotypes will you expect to see, and in what frequencies? Use a Punnett square to illustrate your answer.

29. 3-point testcrosses The problem with using two markers (like a and d below)… double crossovers can go undetected  underestimation of recombinant frequency Solution: include a third marker between the other two…  more DCOs revealed Plus… gene order revealed (more later)

30. + + a b c + b c a b c a 3-point testcross—predicting progeny from a known map Predict the progeny phenotypes and numbers from this cross: + = wild type, dominant Parent 1: b c a 3 cM 7 cM Map: Parent 2: Count 10,000 progeny Step 1. Determine the number of DCO products 3% = 0.03 Probability of recombinant product in (b-c) = Probability of recombinant product in (c-a) = 7% = 0.07 Probability of recombinant product in both = 0.03 x 0.07 = 0.0021

31. Predicting progeny from a known map (cont’d) Heterozygous parent: only one chromatid of each homologue shown on next slide

32. DCO: both together = SCO in b-c interval: Both together = SCO in c-a interval: Both together = NCO (non-crossover): Both together = Predicting progeny from a known map (cont’d) 10000 x 0.0021 = 21 (10000 x 0.03)-21 = 279 (10000 x 0.07)-21 = 679 10000-(SCO + DCO) = 9021

33. SCO DCO NCO NCO DCO SCO 3-point testcross—constructing a linkage map Construct a linkage map (gene order and map distance) for the following genes in Drosophila: Genes pr/+ (purple or red eyes) v/+ (vestigial or long wings) b/+ (black or tan body) Parents Female: pr/+ v/+ b/+ Male: pr/pr v/v b/b Step 1. Expand the shorthand Progeny phenotypes + + + 564 b 32 v 4125 pr 266 v b 272 pr b 4137 pr v 30 pr v b 574 pr+ v+ b+ pr+ b+ v+ pr+ v+ b+ Step 2. Identify the NCO and DCO classes SCO Step 3. Which gene is in the middle? ASK: Which order allows us to go from the NCO genotype to the DCO genotype. Total = 10000

34. order unknown b+ v pr+ b+ v+ pr+ b v+ pr b v pr X X X X b+ pr+ v b+ pr v b pr v+ b pr+ v+ Constructing a linkage map… Step 3 (cont’d) DCO (b+ v pr+) (b v+ pr) (b v+ pr+) (b+ v pr) We know: The process: • Try out the parental genotypes in the 3 possible orders • Do a “virtual double crossover” to see which one would give the correct DCO genotype. incorrect correct!

35. NCO: 4125+4137 = 8262 b+ pr v+ SCOb-pr: 266+272 = 538 b pr+ v b+ pr+ v+ SCOpr-v: 564+574 = 1138 b pr v b+ pr v DCO: 30+32 = 62 b pr v+ Constructing a linkage map (cont’d) Step 4. Calculate % recombinant products % recombinants in b-pr interval= (538+62)/10000 =600/10000 =6% % recombinants in pr-v interval= (1138+62)/10000 =1200/10000 =12%

36. or v pr b 12 cM 6 cM Constructing a linkage map (cont’d) Step 5. Draw the map b pr v 6 cM 12 cM

37. Interference and coefficient of coincidence (COC) Interference: Lower-than-expected frequency of DCO products - Chiasma at one one location blocks other chiasmata from forming nearby observed DCO COC = Interference = 1 - COC expected DCO In our example… expected DCO = 0.06 x 0.12 x 10000 = 72 Observed DCO = 62 COC = 62/72 = 0.86 Interference = 1 - 0.86 = 0.14

38. alleles! alleles! recombination: how frequent? Genetic vs. physical maps Genetic maps… based on recombinant frequencies between markers variation at location #2 variation at location #1 pr+ vg pr vg+ Alleles are detected as associated phenotypes New combination of phenotypes  new combination of alleles  recombination

39. Genetic vs. physical maps (cont’d) based on DNA sequence or landmarks in sequence Physical maps… For example: A B The number of chromosomal bands separating the known locations of genes. site1 site2 site3 site4 site5 The pattern of restriction sites in a DNA sequence. The number of bp of DNA.

40. Questions for Thought Which chromosome? Relation to centromere and telomere? pr+ vg+ pr vg Number of bp? How do we know where to place these genes relative to the centromere and telomeres? How do we know which physical entity (chromosome) our linkage group describes? “Linkage group” = chromosome What is the relationship between crossover frequency and gene order/distance (in bp of DNA) along the chromosomes?

41. Linking the Physical and Genetic Maps Novel Strain knob C Color is on chromosome IX Wx extra DNA Cri du chat is on chromosome V white is X-linked From lecture 6 From lecture 8

42. Linking the Physical and Genetic Maps * * * Metaphase chromosomes Double-stranded DNA segment from a particular location in the human genome Partially denature DNA * * * Fluorescently labeled single-stranded DNA -Fluorescence in situ hybridization (FISH)

43. genetic map (recombination) units = cM R-G colorblindness Fragile X Haemophilia 6 cM 10 cM telomere centromere Physical map (FISH) units = chrom. bands Haemophilia R-G colorblindness Fragile X ~10Xbp ~6Xbp physical map (DNA) units = bp What does the genetic map position tell us? -Order of genes is conserved in genetic and physical maps. -Distance separating markers in genetic and physical maps is ~proportional (but X varies in different organisms).