Extra Credit Are the b and vg genes on the same chromosome? If so, how far apart are they?

1. Extra CreditAre the b and vg genes on the same chromosome? If so, how far apart are they?

2. A B C —•————————— —•————————— a b c A b C a B c Multiple crossovers can be useful for mapping three genes at a time! A BC —•————————— —•————————— a b c A B C a b c

3. 3 recessive phenotypes in maize (corn), coded by three linked genes l l lazy or prostrate growth g g glossy leaves s s sugary endosperm

4. To map the genes, mate a triple heterozygote to triple recessive homozygote Ll Gg Ss x ll gg ss Gene order is not known, so the order shown here is arbitrary. Linkage phase is not know

5. How many different kinds of gametes do you get from triple heterozygote? Ll Gg Ss L or l G or g S or s 2 * 2 * 2

6. Recomb. Wildtpe for all lazy, gloss, sugary L G S / l g s x l g s / l g s

7. Where to begin? Parental types will constitute ≥ 50% of all progeny, so…

8. Rule 1 • Two most-frequent gametes types are the parentaltypes • Tells us the linkage phase of heterozygous parent: • L G S or L g S or l g S or L g s • l g s l G s L G s l G S

9. L G S // l g s x l g s // l g s

10. Linkage phase in heterozygous parent? • L G S or L g S or l g S or L g s • l g s l G s L G s l G S

11. Rule 2 • The double-recombinant gametes will be the two least frequent types A B C a b c

12. L G S / l g s x l g s / l g s

13. Rule 3 • Effect of double crossovers is to interchange the members of the middle pair of alleles between the chromosomes A B C AbC a b c a B c

14. Parental types: Double-crossover types: L G S and l g s • L G s and l g S Which gene is in the middle? LSG Ls G l sg l S g

15. Now you know linkage phase of heterozygous parent and gene order • L S G • l s g How far apart are the genes?

16. L S G l s g Count the crossovers between adjacent genes • In parents, L allele on same homolog as S and l on same homolog as s. So if these get broken up ---> crossover between L and S loci • In parents, S on same homolog as G and s on same homolog as g. If these get broken up --> recombination between S and G loci

17. L S G l s g

18. Rule 4: Reciprocal products expected to occur in approximately equal numbers • LGS ≈ lgs (286 ≈ 272) • LgS ≈ lGs(59 ≈ 44) • Lgs≈ lGS(40 ≈ 33) • LGs ≈ lgS(4 ≈ 2)

19. Rule 5 • Don't forget to include the double recombinants when calculating recombination frequency!

20. l G S 33 • L g s 40 • L G s 4 • l g S 2 • 79 Rec Freq S-G Rec Freq L-S L g S 59 l G s 44 L G s 4 l g S 2 109

21. Rec Freq L-S 79/740 or 10.7% of gametes recombinant between L & S. So, map distance between L & S = 10.7 map units l G S 33 L g s 40 L G s 4 l g S 2 79 Rec Freq S-G 109/740 or 14.8 % of gametes recombinant between S & G. So, map distance between S & G=14.8 map units L g S 59 l G s 44 L G s 4 l g S 2 109

22. Genetic Map 10.7 mu 14.8 mu _____________________________ L S G

23. Interference • Assuming independence, expected probability of double crossovers is the probability of recombination in one region times the probability of recombination in other (product rule).

24. Maize example • Probability of recombination between L and S is 10.7% • Probability of recombination between S and G is 14.8% • If crossovers independent, probability of double crossover should then be • 0.107 * 0.148 = 0.0158 • In 740 events, the double crossover class should occur • 0.0158 * 740 = 12 times

25. Expected DCO = 12 • Observed DCO = 6 • Typical Result: O < E • Conclusion: Crossing over in one region reduces probability of crossing over in adjacent regions • This is Interference

26. Why? • Physical constraints that prevent two chiasmata in close proximity during meiosis?

27. Quantifying Interference • Coefficient of coincidence = Obs DCO • Exp DCO • cc = 6/12 • Interference = (1 - cc) = 1 - 0.5 = 0.5

28. Recombination is not independent at small distances • If distance between genes is small (<10 map units in Drosophila) no double crossovers occur (interference is complete, I=1) • At large distances (> 45 map units, Interference disappears, Obs = Exp and I=0

29. In Drosophila, the allele b gives black body (wild type is tan); at a separate gene, the allele wx gives waxy wings (nonwaxy is wild type); and at a third gene, the allele cn gives cinnabar eyes (red is wild type). A female that is heterozygous for these three genes is testcrossed, and 980 progeny are classified as follows for body color, wing phenotype, and eye color:

30. Extra Credit Question for Next Time

31. a) What is the linkage phase of the heterozygous female parent?b) What is the order of the three genes?c) Construct a linkage map with the genes in their correct order and indicate the map distances between the genes. d) Calculate the Interference.

32. Where to begin? 50% Rec Recombinant genotypes of all sorts will constitute <= 50% of all progeny 0% Rec 50% Rec 50% Rec 100% Rec

33. Three-locus mapping

34. To map the genes, mate a triple heterozygote to triple recessive homozygote L G S x l g s l g s l g s Gene order is not known, so the order shown here is arbitrary

35. Multiple crossovers can lead to inaccuracy A C —•————————— —•————————— a c B b