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John 15:4

John 15:4 4 Abide 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. Linkage, Crossing Over And Mapping In Eukaryotes. Timothy G. Standish, Ph. D. Mother. Father. Telophase I. Prophase I Crossing Over. e. E.

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John 15:4

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  1. John 15:4 4 Abide 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. Linkage, Crossing Over And Mapping In Eukaryotes Timothy G. Standish, Ph. D.

  3. Mother Father Telophase I Prophase I Crossing Over e E Replication e e E E e E e e E E n N N n n N n n N N e e E E n N N N n n Telophase II Chromosomal Theory of Inheritance

  4. Eggs E E e e N n N n EN En eN en Sperm EN EENN EENn EeNN EeNn E N En EENn EEnn EeNn Eenn E n eN EeNN EeNn eeNN eeNn e N en EeNn Eenn eeNn eenn e n Independent Assortment As long as genes are on different chromosomes, they will assort independently

  5. Mother Father Telophase I Replication e e E E e e e E E E n n N N Telophase II n n n N N N e E e E n n N N Two Genes On One Chromosome Genes linked together on the same chromosome should segregate together during meiosis.

  6. Eggs E e N n EN En eN en Sperm EN EENN EENn EeNN EeNn E N En EENn EEnn EeNn Eenn eN EeNN EeNn eeNN eeNn en EeNn Eenn eeNn eenn e n Linked Genes Because genes co-segregate, a 3:1 ratio results instead of the expected 9:3:3:1

  7. Mother Father Prophase I Telophase I Replication e e E E e e e E E E e E n n N N Telophase II n N n n n N N N e E e E n N n n N N Two Genes On One Chromosome With Crossing Over 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

  8. Eggs E E e e N n N n EN En eN en Sperm EN EENN EENn EeNN EeNn E N En EENn EEnn EeNn Eenn E n eN EeNN EeNn eeNN eeNn e N en EeNn Eenn eeNn eenn e n With Crossing Over As long as crossing over occurs between genes they will assort independently

  9. A B Tightly linked A C Closer to independent assortment 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 cross over event is between them

  10. e+e mah+mah e+mah+ e+mah e+e mah mah emah+ e emah+mah e emah mah emah • Imagine a situation where true breeding ebony body (e) mahogany eyes (mah) D. menanogaster are crossed with wild type flies: Linkage In A Test Cross • All F1 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 mahXe+e+mah+mah+ • Expected F1 would be: e+e mah+mah e e mah mahXe+emah+mah 1:1:1:1 emah

  11. e mah Chromosome 17 mu Linkage In A Test Cross ebony mahogany 41 Wild type 42 ebony 8 mahogany 9 • If the actual numbers were: • 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

  12. e+cu+ro+ e+cu+ro e+curo+ e+curo ecuro+ ecu+ ro ecu+ro+ e curo e+ecu+cu cd+cd e+ecu+cu cd cd e+ecucu cd+cd e+ecucu cd cd eecucu cd+cd eecu+cu cd cd eecu+cu cd+cd ee cucu cd cd 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+ecu+cu cd+cd X eecucu cdcd • Expected F1 would be: ecuro

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

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

  15. Phenotype Observed ebony curled cardinal 378 Wild type 375 ebony cardinal 103 curled 101 cardinal 19 ebony curled 18 curled cardinal 4 ebony 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 Mapping A Three Point Cross

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

  17. Phenotype Observed ebony curled cardinal 378 Wild type 375 ebony cardinal 103 curled 101 cardinal 19 ebony curled 18 curled cardinal 4 ebony 2 e e cu cd cu cd + + + Mapping A Three Point Cross

  18. Phenotype Observed ebony curled cardinal 378 Wild type 375 ebony cardinal 103 curled 101 cardinal 19 ebony curled 18 curled cardinal 4 ebony 2 e cu cd Mapping A Three Point Cross + cu cd + + e

  19. Phenotype Observed ebony curled cardinal 378 Wild type 375 ebony cardinal 103 curled 101 cardinal 19 ebony curled 18 curled cardinal 4 ebony 2 e cd cu Mapping A Three Point Cross 6 + cu cd + + e

  20. Phenotype Observed ebony curled cardinal 378 Wild type 375 ebony cardinal 103 curled 101 cardinal 19 ebony curled 18 curled cardinal 4 ebony 2 e cu cd Mapping A Three Point Cross 6 + + cu + e cd

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

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

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

  24. 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

  25. a Zygote Aa A a A Meiosis In An Ascus

  26. a a a a A A a A A A Meiosis In An Ascus

  27. a a a a a A A A a A A A Meiosis In An Ascus

  28. a a a a a a a A A A A a A A A A Meiosis In An Ascus

  29. a a a A a A A A Meiosis In An Ascus:If Crossing Over Occurs

  30. a a a Aa a A A a A a A A Aa A Meiosis In An Ascus:If Crossing Over Occurs

  31. a a a A A a A A Meiosis In An Ascus:If Crossing Over Occurs a a A A a A a A

  32. 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

  33. 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 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) a

  34. The End

  35. 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 • What was the phenotype of the F1 generation? • What F2 phenotypic ratio would you expect? • 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?

  36. Solution 1 • What was the phenotype of the F1 generation? VVMM makes VM gametes vvmm makes vm gametes Thus the F1 must be VvMm • 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)

  37. v+ m+ 0.49 v+ m+ v+ m v+ m 0.01 v m+ v m+ 0.01 v m v m 0.49 Solution 1 Continued • 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? F1 Gametes

  38. 0.49 v+m+ 0.01 v+m 0.01 vm+ 0.49 vm 0.49 v+m+ 0.2401 v+v+m+m+ 0.0049 v+v+m+m 0.0049 v+vm+m+ 0.2401 v+vm+m v+ m+ 0.49 v+ m+ v+ m v+ m 0.01 v+m 0.0049 v+v+m+m 0.0001 v+v+mm 0.0001 v+vm+m 0.0049 v+vmm 0.01 v m+ v m+ 0.01 v m 0.01 vm+ 0.0049 v+vm+m+ 0.0001 v+vm+m 0.0001 vvm+m+ 0.0049 vvm+m v m 0.49 0.49 vm 0.2401 v+vm+m 0.0049 v+vmm 0.0049 vvm+m Solution 1 Continued 0.74 v+_m+_ (0.74*200=148) 0.01 v+_mm (0.01*200=2) 0.01 vvm+_ (0.01*200=2) 0.2401 vvmm 0.24 vvmm (0.24*200=48)

  39. v m 1cM 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

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