Seminar

1. Seminar Wednesday, May 15, Biology 212, 4 pm "Genomic Consequences of Allopolyploidization" Luca Comai, Ph.D., Department of Botany, University of Washington 5 points for attendance, 1-5 points each person for good questions during the Q/A period after the talk.

2. Mutation in Bacteria …the ultimate source of variation in bacteria is spontaneous mutation, • generally errors in DNA replication, …mutations occur in specific genes at a rate of 1 in 106 to 1 in 107 cells, …adaptive mutations are quickly replicated and adaptive colonies predominate.

3. Spontaneous Mutations • DNA replication in E. coli occurs with an error every ~ 109 bases. • The E. coli genome is 4.6 x 106 bases. • An error occurs once per ~ 2000 replications. • If a single colony has 107 bacteria, • 5,000 cells carry a mutation, • or, one mutation every ~ 1,000 bases (across a colony), • or, a mutation in about every gene.

4. Conjugation … F+ cells donate genetic material via the F+ plasmid, …Hfr cells; strains with a chromosome integrated F factor that is able to mobilize and transfer part of the chromosome to the F- cell.

5. F’Cells • an F factor from an Hfr cell excises out of the bacterial genome and returns to plasmid form, • often carries one or more bacterial genes along, • F’cells behave like an F+ cells, • merizygote: partially diploid for genes copied on the F’plasmid, • F’plasmids can be easily constructed using molecular biology techniques.

7. Selective Media • wild-type bacteria grow on minimal media, • media supplemented with selected compounds supports growth of mutant strains, • minimal media + leucine supports leu- cells, • minimal media + leucine + arginine supports leu- arg- • etc. • Selective Media: a media in which only the desired strain will grow.

8. Selection ...the process that establishes conditions in which only the desired mutant will grow.

9. Problem You want to create cells that are only methionine auxitrophs.

10. The Wrong Answer Strain 1 x Strain 3 You have a recombinant that is met- only. How do you get rid of Strain 1?

11. Correct Answer Strain 2 x Strain 3 Grow on Minimal Media Plus Methionine Strain 2 dies because there is no arginine. Strain 3 dies because there is no threonine or thimine. The new exconjugate lives.

12. High Frequency of Recombination(Hfr) ...bacteria exhibiting a high frequency of recombination, …the F factor is integrated into the chromosomal genome.

13. F factor and Chromosomal DNA are Transferred

14. Double Crossover Recombination Requires Crossing over

15. Incomplete Transfer of DNA • Interrupted Mating: a break in the pilus during conjugation stops the transfer of DNA, • Transfer occurs at a constant rate, • provides a means to map bacterial genes.

16. How Do You Interrupt Bacterial Mating spread on agar mate for specified time frappe

17. Hfr and Mapping HfrH strs(sensitive to streptomycin) thr+ (able to synthesize the amino acid threonine) azir(resistant to sodium azide) tonr(resistant to bacteriophage T1) lac+ (able to grow with lactose as sole source of carbon) gal+ (able to grow with galactose as sole source of carbon) F- strr(resistant to streptomycin thr- (threonine auxotroph) azis (sensitive to sodium azide) tons (sensitive to phage T1) lac- (unable to grow on lactose) gal- (unable to grow on galactose)

18. Hfr and Mapping HfrH strs(sensitive to streptomycin) thr+ (able to synthesize the amino acid threonine) F- strr(resistant to streptomycin) thr- (threonine auxotroph) Streptomycin kills the HfrH cells in the mating mix. No threonine kills the F-cells in the mating mix.

19. Hfr and Mapping HfrH azir(resistant to sodium azide) tonr(resistant to bacteriophage T1) lac+ (able to grow with lactose as sole source of carbon) gal+ (able to grow with galactose as sole source of carbon) F- azis (sensitive to sodium azide) tons (sensitive to phage T1) lac- (unable to grow on lactose) gal- (unable to grow on galactose)

20. Interrupting Bacterial Mating spread on selective media mate 9 min blend

21. Replica Plating After 9 minutes, only azide resistant cells grow.

22. 10 Minutes Azide, and bacteriophage resistant cells grow.

23. 15 Minutes Azide, and bacteriophage resistant cells, and lactose utilizing cells.

24. 18 Minutes All recombinants grow.

25. % cells with markers

27. Bacterial Map Distances units = minutes

30. F factor inserts in different regions of the bacterial chromosome, Also inserts in different orientations.

31. Origin of Replication Hfr Order of transfer strain H thr azi ton lac pur gal his gly thi 1 thr thi gly his gal pur lac ton azi 2 lac pur gal his gly thi thr azi ton 3 gal pur lac tonazi thr thi gly his

32. Indicates direction of transfer. F factor A A a Hfr F-

33. A A Hfr F- F factor A transfers first. A A Hfr F- A transfers last. Leading Gene: the first gene transferred, it is determined empirically.

34. Hfr Order of transfer strain H thr azi ton lac pur gal his gly thi 1 thr thi gly his gal pur lac ton azi 2 lac pur gal his gly thi thr azi ton 3 gal pur lac tonazi thr thi gly his

35. E. coli Map • 0 minutes is at the threonine, • 100 minutes is required to transfer complete genome,

36. Typical Problem

38. combine

40. combine

41. +

42. Join Maps 11.5 minutes 26 minutes Refer to partial maps for map distances.

43. Practice • Insights and Solutions, #2, • Problem 8.17, 8.18, 8.19.

44. Transformation • heritable exchange brought about by the incorporation of exogenous DNA, • usually DNA from same, or similar species.

45. Not all cells are competent to receive DNA. Donor and Recipient

46. Competence …a transient state or condition in which a cell can bind and internalize exogenous DNA molecules, …often a result of severe conditions, • heat/cold, • starvation, etc.

47. Competent Cell Genes are expressed that produce proteins that, in turn, span the cell membrane.

48. Exogenous DNA Binds Receptor

49. ...one strand of the exogenous DNA is degraded also. Complementary Strand Degraded

50. Heteroduplex Exogenous DNA Incorporated