Review Question If two genes are 13 map units apart on a linkage map, what proportion of recombinant offspring will be s

1. Review Question If two genes are 13 map units apart on a linkage map, what proportion of recombinant offspring will be seen in a testcross? What proportion of meioses experienced crossing over between to the two genes?

2. Two X-linked genes Female Male w m w+ m+ —•——————————•————————— P:—•——————————•————/ w m Gametes: Egg (A B) Sperm (a b) w m w+ m+ —•———————— —•———————— —•———/

3. w+ m+ —•————————— F1:—•————————— w m w m —•————————— —•————/ X w+ m+ —•————————— w m —•————————— w m —•————————— Par Par —•————/ Rec Rec Offspring (male or female) 314 Wild type 314 white, miniature 186 miniature 186 white w+ m —•————————— w m+ —•—————————

4. ___%non-recombinant offspring,____% recombinant offspring Genes are how far apart? ___ map units

5. If two genes are 50 map units apart, what proportion of recombinant offspring would you observe? What would you conclude about these two genes?

7. Multiple crossovers are useful for mapping many genes at a time

8. A B C —•————————— —•————————— a b c A b C a B c Three-point mapping A BC —•————————— —•————————— a b c A B C a b c

9. 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

10. 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 known

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

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

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

14. Rule 1 • Two most-frequent gametes types are the __________types • Tells us the ___________________ 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

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

16. 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

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

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

19. 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

20. 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

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

22. 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 ---> recombination 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

23. L S G l s g

24. 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)

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

26. 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

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

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

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

30. 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

31. L G s 4 l g S 2 6 • 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

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

33. Quantifying Interference • Coefficient of coincidence = Obs DCO • Exp DCO • cc = _____ • Interference = (1 - cc) =

34. 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

35. 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:

36. 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.

37. Three-locus mapping