2. Linkage Genes linked on the same chromosome may segregate together.

3. A b Independent Assortment 2 chromosomes a A A B B b a B a b P: AABB x aabb Parental Gametes: AB & ab F1 P P RR

4. Meiosis One Chromosome No Cross Over Parent Cell Daughter Cells Have Parental Chromosomes Aa Bb A B a b a b A B 2n = 2

5. Meiosis One Chromosome With Cross Over Parent Cell Daughter Cells Have Recombinant Chromosomes Aa Bb A B A b a b a B 2n = 2

6. Meiosis Prophase I If genes are linked, crossing over must occur for there to be recombination. AaAa BBbb aA

7. Linkage? P: AABB x aabb F1: AaBb Test cross: AaBb x aabb 1/4: AaBb 1/4: Aabb 1/4: aaBb 1/4: aabb P: AABB x aabb F1: AaBb Test cross: AaBb x aabb ?: AaBb ?: Aabb ?: aaBb ?: aabb no linkagelinkage

8. Recombination Frequency …or Linkage Ratio: the percentage of recombinant types, –if 50%, then the genes are not linked, –if less than 50%, then linkage is observed.

9. Linkage Genes located on the same chromosome do not recombine, –unless crossing over occurs, The recombination frequency gives an estimate of the distance between the genes.

10. Recombination Frequencies Genes that are adjacent have a recombination frequency near 0%, Genes that are very far apart on a chromosome have a linkage ratio of 50%, The relative distance between linked genes influences the amount of recombination observed.

11. AC In this example, there is a 4/10 chance of recombination. AB In this example, there is a 2/10 chance of recombination. ab ac

12. Linkage Ratio P GGWW x ggww Testcross F1: GgWw x ggww recombinant total progeny GWGwgWgw ???? = Linkage Ratio

13. Linkage Ratio Units % = mu (map units) - or - % = cm (centimorgan)

14. AC In this example, there is a 4/10 chance of recombination. AB In this example, there is a 2/10 chance of recombination. ab ac

15. cis “coupling”

16. trans “repulsion”

17. Fly Crosses (white eyes, minature, yellow body) In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu, In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu, When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu. Study Figs 4.2, 4.3, and 4.5

18. Simple Mapping white eyes x miniature = 36.9 mu, white eyes x yellow body = 0.5 mu, miniature x yellow body = 38 mu, my 38 mu 36.9 mu w 0.5 mu

19. Do We have to Learn More Mapping Techniques? Yes, –three point mapping, Why, –Certainty of Gene Order, – Double crossovers.

20. Gene Order It is often difficult to assign the order of genes based on two-point crosses due to uncertainty derived from sampling error. A x B = 37.8 mu, A x C = 0.5 mu, B x C = 37.6 mu,

21. Double Crossovers More than one crossover event can occur in a single tetrad between non-sister chromatids, –if recombination occurs between genes A and B 30% of the time, (p = 0.3), then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or nearly 10 map units. If there is a double cross over, does recombination occur? –how does it affect our estimation of distance between genes?

22. Why Me? Why Map? Over 4000 human diseases have a genetic component, –knowing the protein produced at specific loci facilitates the treatment and testing, Facilitates both classical and molecular analysis of organisms.

23. Classical Mapping Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. target What recombination frequency do you expect beteen the target and HY2? What recombination frequency do you expect beteen the target and TT2?

24. Three Point Testcross Triple Heterozygous (AaBbCc ) x Triple Homozygous Recessive (aabbcc)

25. Three Point Mapping Requirements The genotype of the organism producing the gametes must be heterozygous at all three loci, You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring, You must look at enough offspring so that all crossover classes are represented.

26. Representing linked genes... W G D w g d x w g d P Testcross = WwGgDd = wwggdd

27. Representing linked genes... + + + w g d x w g d P Testcross = WwGgDd = wwggdd

28. Phenotypic Classes W- ww G- gg G- gg D- dd D- dd D- dd D- dd W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- wwggdd

29. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- Crossovers 0 W G D w g d 1 1 2 # 179 52 46 4 22 2 wwggdd1730 1 1 2

30. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d III

31. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d I Total = 500 Region I: 46 + 52 + 2 + 4 500 x 100 = 20.8 mu

32. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d II Total = 500 Region II: 22 + 22 + 2 + 4 500 x 100 = 10.0 mu 20.8 mu

33. W G D w g d W-gg-D wwG-dd 4 2 Recombinants, double crossover Total = 500 10.0 mu20.8 mu 0.1 x 0.208 = 0.0208 6/500 = 0.012 NO GOOD!

34. Interference …the affect a crossing over event has on a second crossing over event in an adjacent region of the chromatid, –(positive) interference: decreases the probability of a second crossing over, most common in eukaryotes, –negative interference: increases the probability of a second crossing over.

35. Gene Order in Three Point Crosses Find either double cross-over phenotype, based on the recombination frequencies, Two parental alleles, and one cross over allele will be present, The cross over allele fits in the middle... –

36. # 2001 52 46 589 990 887 600 1786 Which one is the odd one? A C B a c b III A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- aabbcc

37. A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region I A C B a c b I 990 + 887 + 52 + 46 6951 x 100 = 28.4 mu

38. A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region II A C B a c b 28.4 mu 600 + 589 + 52 + 46 6951 x 100 = 18.5 mu II 18.5 mu

39. Today Coefficient of Confidence, Gene mapping in humans, Problems, problems, problems, –Be sure to at least try them before Friday.