1. ©2000 Timothy G. Standish John 15:4 4Abide in me, and I in you. As the branch cannot bear fruit of itself, except it abide in the vine; no more can ye, except ye abide in me

2. ©2000 Timothy G. Standish Linkage, Crossing Over And Mapping In Eukaryotes Timothy G. Standish, Ph. D.

3. ©2000 Timothy G. Standish Chromosomal Theory of Inheritance Father Mother e N E n n E e N e n E N Telophase II E n e NN e E n Replication e N E n Prophase I Crossing Over N E n ee n E N Telophase I

4. ©2000 Timothy G. Standish EnEn eNeN enen ENENe ne ne Ne NE nE nE NE N Independent Assortment Sperm n E e N e n N E EeNnEeNnEeNNEENnEENN EennEeNnEeNnEEnnEENn eeNneeNNEeNnEeNnEeNN eenneeNnEennEeNnEeNn As long as genes are on different chromosomes, they will assort independently Eggs n E e n e N N E

5. ©2000 Timothy G. Standish Two Genes On One Chromosome Father Mother e n E N E N n e Replication e n E N Genes linked together on the same chromosome should segregate together during meiosis. Telophase I n e n e NN EE Telophase II E NN E n e n e

6. ©2000 Timothy G. Standish EnEn eNeN enen ENENe ne ne Ne NE nE nE NE N Linked Genes Sperm e n N E Eggs e n N E EeNnEeNnEeNNEENnEENN EennEeNnEeNnEEnnEENn eeNneeNNEeNnEeNnEeNN eenneeNnEennEeNnEeNn Because genes co- segregate, a 3:1 ratio results instead of the expected 9:3:3:1

7. ©2000 Timothy G. Standish Two Genes On One Chromosome With Crossing Over Father Mother e n E N Telophase II E NN e E nn e E N n e Replication e n E N Prophase I E N N n e n As long as genes on the same chromosome are located a long distance apart, they will assort independently due to crossing over during Prophase I of meiosis Telophase I N e E N E n e n

8. ©2000 Timothy G. Standish EnEn eNeN enen ENENe ne ne Ne NE nE nE NE N With Crossing Over EeNnEeNnEeNNEENnEENN EennEeNnEeNnEEnnEENn eeNneeNNEeNnEeNnEeNN eenneeNnEennEeNnEeNn As long as crossing over occurs between genes they will assort independently Sperm e n N E e N n E Eggs e n N E e N E n

9. ©2000 Timothy G. Standish Still Not 9:3:3:1 Linked genes, unless they are far apart on the chromosome, still do not exhibit the phenotypic ratio expected from independent assortment Deviation from the expected ratio allows mapping of genes The further apart two genes are, the more probable a crossover event is between them CA Closer to independent assortment BA Tightly linked

10. ©2000 Timothy G. Standish Linkage In A Test Cross All F 1 progeny should be phenotypically wild type If a test cross was done crossing the F1 flies with homozygous ebony mahogany flies, the expected outcome would be: e + e mah + mah Imagine a situation where true breeding ebony body (e) mahogany eyes (mah) D. menanogaster are crossed with wild type flies: e e mah mah X e + e mah + mah Expected F 1 would be: e e mah mah X e + e + mah + mah + e mah e e mah mahe e mah + mahe + e mah mahe + e mah + mah e mahe mah + e + mahe + mah + 1:1:1:1

11. ©2000 Timothy G. Standish Linkage In A Test Cross If the actual numbers were: ebony mahogany 41 Wild type 42 ebony 8 mahogany 9 Linkage would naturally be suspected The ebony and mahogany flies must have resulted from crossing over As there are a total of 100 flies in the sample and 17 represent crossover events, these genes are said to be 17 map units (or centimorgans) apart mah 17 mu e Chromosome

12. ©2000 Timothy G. Standish Mapping A Three Point Cross Two point crosses do not tell us much about gene order on chromosomes Three point crosses allow determination of both gene sequence and distances between genes Imagine the following cross between an ebony bodied, curly winged, cardinal eyed male and a female heterozygous for the same traits: e + e cu + cu cd + cd X ee cucu cdcd Expected F 1 would be: e cu ro e cu + ro + e + cu + roe + cu + ro + e + cu roe + cu ro + e cu + roe cu ro + ee cucu cd ee cu + cu cd + cd e + ecu + cu cd e + ecu + cu cd + cd e + ecucu cd e + ecucu cd + cd ee cu + cu cd ee cucu cd + cd

13. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 102 17 5 101 19 375 Observed 3 125 Expected 125

14. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 Rearrange in descending order of observance

15. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 Because crossing over is an infrequent event (and because we know the parents genotypes) the most commonly appearing class represent no crossing over Double crossovers would be expected much less frequently than single crossovers. Thus the least frequently observed class must represent crossover between the gene in the middle and the two flanking genes Intermediate classes represent single crossover events

16. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu Which is exactly the same as: ecdcu Knowing that ebony is in the middle allows construction of a tentative map

17. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu ecdcu + ++

18. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu e cdcu + + +

19. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 6 ecdcu e cdcu + + +

20. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu ecd cu + + + 6

21. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu ecd cu + + + 204 6 = 21 Double cross Total + Single cross = MU 1,000 (204 + 6) x 100 21

22. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu e cd cu + + 6 + Double cross Total + Single cross = MU 21

23. ©2000 Timothy G. Standish Mapping A Three Point Cross ebony curled cardinal ebony cardinal ebony curled curled cardinal curled cardinal Wild type Phenotype ebony 378 103 18 4 101 19 375 Observed 2 ecdcu e cd cu + + 37 + 6 Double cross Total + Single cross = MU 1,000 (37 + 6) x 100 4.3 = 4.3 21

24. ©2000 Timothy G. Standish Mapping In Haploid Organisms Haploid organisms are, in some ways, easier to work with because all genes impact phenotype On the other hand, they tend to be ugly smelly things that provide many other challenges Fungi in the phylum Ascomycota are easiest to work with as they show the order of division in their asci

26. ©2000 Timothy G. Standish Zygote Aa Meiosis In An Ascus a A a A

27. ©2000 Timothy G. Standish A a Meiosis In An Ascus a A a A a A A a

28. ©2000 Timothy G. Standish A a a A Meiosis In An Ascus a A a A a A A a

29. ©2000 Timothy G. Standish A A A A a a a a Meiosis In An Ascus a A a A a A A a

30. ©2000 Timothy G. Standish Meiosis In An Ascus: If Crossing Over Occurs A A a a A a A a

31. ©2000 Timothy G. Standish Aa Meiosis In An Ascus: If Crossing Over Occurs A A a a A a A a a A A a

32. ©2000 Timothy G. Standish A a A a Meiosis In An Ascus: If Crossing Over Occurs A A a a A a A a a A A a

33. ©2000 Timothy G. Standish A A a a A A a a Meiosis In An Ascus: If Crossing Over Occurs A A a a A a A a a A A a

34. ©2000 Timothy G. Standish A A a a A A a a Meiosis In An Ascus: If Crossing Over Occurs A A a a A a a A A a A a A 4:4 pattern in asci indicates no crossing over has occurred (First- Division Segregation) A 2:4:2 or 2:2:2:2 pattern indicates crossing over (Second- Division Segregation)

35. ©2000 Timothy G. Standish

36. Problem 1 In Drosophila, vermilion (v) is recessive to red (V) eyes and miniature (m) wings are recessive to normal (M) wings. The following cross was made: Male VVMM x vvmm Female A What was the phenotype of the F 1 generation? B What F 2 phenotypic ratio would you expect? C If the actual F 2 phenotypic numbers were: – 147 red-eyed normal winged –49 vermilion-eyed miniature winged, – 2 red-eyed miniature winged, –2 vermilion-eyed normal winged, How would you explain this?

37. ©2000 Timothy G. Standish Solution 1 A What was the phenotype of the F 1 generation? VVMM makes VM gametes vvmm makes vm gametes Thus the F 1 must be VvMm B What F2 phenotypic ratio would you expect? 9 red-eyed normal winged (V_M_) 3 red-eyed miniature winged (V_mm) 3 vermilion-eyed normal winged (vvM_) 1 vermilion-eyed miniature winged (vvmm)

38. ©2000 Timothy G. Standish Solution 1 Continued C If the actual F2 phenotypic numbers were: – 147 red-eyed normal winged –49 vermilion-eyed miniature winged, – 2 red-eyed miniature winged, –2 vermilion-eyed normal winged, How would you explain this? mv m+m+ v+v+ 0.49 0.01 0.49 0.01 mv+v+ m+m+ v mv m+m+ v+v+ mv+v+ m+m+ v F1 Gametes

39. ©2000 Timothy G. Standish Solution 1 Continued 0.0001 vvm + m + 0.0049 vvm + m 0.0049 vvm + m 0.2401 vvmm 0.01 vm + 0.49 vm 0.49 v + m + 0.01 v + m 0.24 vvmm (0.24*200=48) 0.01 vm + 0.49 vm 0.49 v + m + 0.01 v + m 0.0049 v + vmm 0.0049 v + vmm 0.0001 v + v + mm 0.0049 v + vm + m + 0.2401 v + vm + m 0.0001 v + vm + m 0.0049 v + vm + m + 0.0001 v + vm + m 0.2401 v + v + m + m + 0.0049 v + v + m + m 0.0049 v + v + m + m 0.2401 v + vm + m 0.01 vvm + _ (0.01*200=2) 0.01 v + _mm (0.01*200=2) 0.74 v + _m + _ (0.74*200=148) mv m+m+ v+v+ 0.49 0.01 0.49 0.01 mv+v+ m+m+ v mv m+m+ v+v+ mv+v+ m+m+ v

40. ©2000 Timothy G. Standish Solution 1 Continued Vermillion and miniature winged are closely linked genes on the same chromosome The distance between vermilion and miniature is 1 centimorgan The reason numbers in the cross do not fit the prediction of 1 centimorgan exactly is that the numbers are the result of chance and thus would not be expected to fit the predicted ratio perfectly mv 1cM