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III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage

III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT?. AaBb x aabb. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT?

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III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage

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  1. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? AaBb x aabb

  2. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? We map the genes using the knowledge that crossing-over is rare, and the frequency of crossing-over correlates with the distance between genes. AaBb x aabb

  3. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage OK… so we conclude the genes are linked… NOW WHAT? 1. Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. AaBb x aabb

  4. III. Linkage • A. ‘Complete’ Linkage • B. ‘Incomplete’ Linkage • OK… so we conclude the genes are linked… NOW WHAT? • Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. • This tells us the original arrangement of alleles in the heterozygous parent: AaBb x aabb A a B b

  5. III. Linkage • A. ‘Complete’ Linkage • B. ‘Incomplete’ Linkage • OK… so we conclude the genes are linked… NOW WHAT? • Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. • This tells us the original arrangement of alleles in the heterozygous parent: • Segregation without crossing over produces lots of AB and ab gametes. AaBb x aabb A a B b

  6. III. Linkage • A. ‘Complete’ Linkage • B. ‘Incomplete’ Linkage • OK… so we conclude the genes are linked… NOW WHAT? • Crossing-over is rare; so the RARE COMBINATIONS must be the products of Crossing-Over. The OTHERS, the MOST COMMON products, represent the PARENTAL TYPES. • 2. The frequency of crossing-over is used as an index of the distance between genes: • The other progeny are the products of crossing over, and they occurred 20 times in 100 progeny, for a frequency of 0.2. Multiply that by 100 to free the decimal, and this becomes 20 map units (CentiMorgans). AaBb x aabb A a B b 20 map units

  7. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage So, we used the chi-square test to identify pairs of genes on the same chromosome. Then, we mapped the distance between genes on the chromosome; all by looking at the frequency of phenotypes in the offspring of a test cross. AaBb x aabb A a B b 20 map units

  8. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Suppose you conduct two-point mapping and find: A - C are 23 centiMorgans apart C – B are 10 centiMorgans apart: A C B Well, it is not necessarily arranged like this!

  9. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Suppose you conduct two-point mapping and find: A - C are 23 centiMorgans apart C – B are 10 centiMorgans apart: A C B It could be either of these arrangements… A C B

  10. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc To determine the sequence of three genes, we need to do three-point mapping… crossing a triple heterozygote in a test cross, so we can see the frequency of crossing over between all genes…

  11. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 79 aBc = 39 abc = 27 abC = 5 So, suppose we do the cross and observe these results in the phenotypes of the offspring (which, of course, represent the frequencies that the heterozygote produced these types of gametes, too…)

  12. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - combine complementary sets Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC= 25 ABc = 3 Abc = 42 AbC = 85 aBC = 39…fixed here; error in audio ppt. aBc = 79 abc= 27 abC = 5 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 Abc = 42 aBC = 39 81 Since the heterozygote is heterozygous at all 3 loci, if ABC is on one chromosome, the ‘a’ and the ‘b’ and the ‘c’ must be on the other chromosome… these types (ABC, abc) are ‘complementary’. As are ‘Abc’ and ‘aBC’, etc. AbC = 85 aBc = 79 164

  13. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 39 aBc = 79 abc = 27 abC = 5 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 Abc = 42 aBC = 39 81 The parental types tell us the combination of alleles that were on each chromosome in the parent… but NOT the order of those alleles on the chromosomes = AbC, aBc AbC = 85 aBc = 79 164 MOST ABUNDANT ARE PARENTAL TYPES

  14. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping - the least abundant set are products of the rarest event – a double crossover (‘DCO’). Three Point Test Cross AaBbCc x aabbcc Phenotypic Ratio: ABC = 25 ABc = 3 Abc = 42 AbC = 85 aBC = 39 aBc = 79 abc = 27 abC = 5 ABC = 25 abc = 27 52 DCO Makes: D e F D E F ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” D E F Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 d e f MOST ABUNDANT ARE PARENTAL TYPES DCO’s = very rare, and gene in middle switches chromosomes

  15. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc Compare Parentals and DCO’s to determine which gene is in the middle… Parentals: AbC aBc DCO’s: abC ABc ‘B’ and ‘C’ alleles stay together; the ‘A’ switches. ‘A’ is in middle. REMAKE YOUR DRAWING TO DEPICT ALLELES AND ORDER: ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES

  16. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping ToMap distance from ‘B’ to ‘A’, need to know the frequency of crossing over in this region: Three Point Test Cross AaBbCc x aabbcc So, a “single cross-over” (SCO) between the B and A locus will produce the gametes/phenotypes: BAC and bac – find their frequency: 52… plus add DCO’s (8) = 60/305 = 0.197 x 100 = 19.7 cM ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES 19.7 cM

  17. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping ToMap distance from ‘A’ to ‘C’, need to know the frequency of crossing over in this region: Three Point Test Cross AaBbCc x aabbcc So, a “single cross-over” (SCO) between the A and C locus will produce the gametes/phenotypes: Abc and aBC – find their frequency: 81… plus add DCO’s (8) = 98/305 = 0.292 x 100 = 29.2 cM ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES 19.7 cM 29.2 cM

  18. III. Linkage A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage C. Three-point Mapping Three Point Test Cross AaBbCc x aabbcc “Coefficient of Coincidence” – are SCO’s occurring independently? If so, then DCO’s = product of SCO’s (0.197) x (0.292) x 305 = 17 We only observed 8. C.O.C. = obs/exp = 8/17 = 0.47 Interference = 1 – c.o.c. = 0.53 ABC = 25 abc = 27 52 ABc = 3 abC = 5 = 8 LEAST ABUNDANT ARE “DOUBLE CROSSOVERS” b A C Abc = 42 aBC = 39 81 AbC = 85 aBc = 79 164 B a c MOST ABUNDANT ARE PARENTAL TYPES 19.7 cM 29.2 cM

  19. Bacterial Genetics

  20. Bacterial Genetics • Overview - Domains of Life

  21. Bacterial Genetics • Overview - Domains of Life

  22. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission

  23. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission

  24. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • The rapid production of new organisms creates genetic diversity by mutation, alone; even though the rates of mutation are low for any given gene.

  25. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation

  26. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation

  27. Lederberg and Tatum – 1946 - certain strains of bacteria are able to donate genes to other strains – they have a “fertility factor” (F+). Other strains lack this factor (F-).

  28. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation Davis demonstrated that cell-cell contact was required…

  29. Figure 8-6 And Cavalli Sforza isolated a strain that would cause genetic change at a very high rate: Hfr (High frequency recombination). He recognized that the acquisition of traits was related to the duration of the conjugation event.

  30. He hypothesized that the time between transfer to the recipient cell was related to the distance between genes. As such, if he interrupted mating at specific intervals, he could use time between trait acquisition as an index of distance between genes.

  31. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation He isolated different strains that transferred genes in different order, suggesting that the transfer process could begin at different places.

  32. Figure 8-8b He isolated different strains that transferred genes in different order, suggesting that the transfer process could begin at different places.

  33. Figure 8-9 part 1

  34. Figure 8-9 part 2

  35. Figure 8-9 part 3

  36. Figure 8-9 part 4

  37. Figure 8-9 part 5

  38. An integrated Hfr plasmid can revert to a free F+ plasmid, and take chromosomal genes along, too.

  39. An integrated Hfr plasmid can revert to a free F+ plasmid, and take chromosomal genes along, too. It is now an F’ plasmid. Conjugation can now occur.

  40. Figure 8-10 part 4

  41. Figure 8-10 part 5

  42. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation • 2. transformation – absorption of DNA • from the environment.

  43. Figure 8-12 part 1

  44. Figure 8-12 part 2

  45. Figure 8-12 part 3

  46. Figure 8-12 part 4

  47. Figure 8-12 part 5

  48. Bacterial Genetics • Overview - Domains of Life • II. Prokaryotic Reproduction • A. fission • B. “sex” – genetic recombination • 1. conjugation • 2. transformation – absorption of DNA • from the environment. • 3. viral transduction

  49. Figure 8-14

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