E. coli: The Lab Rat of Molecular Biology

1. E. coli: The Lab Rat of Molecular Biology Presented by: Verpooten & McQuade February 21st, 2010

2. Diversity of Prokaryotes • Today’s chapter focuses on the Genetics of Bacteria, and in particular the “true bacteria.” • Remember prokaryotes are classified into 2 Domains: Archaea and Bacteria • Archaeabacteria: Extremists • Eubacteria: Generalists • The structure & function of genes in archaebacteria is more similar to that of eukaryotes today we will focus on eubacteria

3. Bacteria Review • Unicellular • Lack a Nucleus • Have dsDNA in the cytoplasm (Nucleoid) • Also, have independently replicating circular plasmids.

4. Genome Details • E. coli has about 4.6 Mbps • Divide through Binary Fission (asexual) • Replication occurs in both directions from a single origin of replication • E. coli can cause disease, but it is primarily a beneficial commensalist that exists in our intestinal tract • Replication of Genome can occur in as little as 20 minutes (lab) or 12 hours (intestines)

5. Genetic Variation: It’s in the #s • 2 x 1010 E. coli in our gut • 1x 1010 pooped out daily • Mutations in any given gene occurs once in 10 million cell divisions. • Given the 4300 genes and 10000000000 new bacteria made everyday—that is 9 million new mutations per day per host.

6. Natural Selection in Action • We already learned genetic variation acts upon already existing genetic variants within a population. So, how big of a role is mutation in the following organisms? • E. coli vs. Humans

7. Answer: Mutation as Source of Genetic Variation. • Bacteria: Large part • They are rapidly dividing populations that normally reproduce asexually, so mutation is one of the major sources of variation. • Humans: Small part • In general, slowly reproducing populations are less affected by mutation, of course most of our genetic variation comes not from mutation but sexual reproduction (giving new combinations of genes from parents)

8. EXPERIMENT Can Bacteria Transfer Genes? Mixture Mutantstrainarg+trp– Mutantstrainargtrp+ This experiment was undergone to determine whether or not genetic recombination occurs in bacteria

9. Mixture Mutantstrainarg+trp– Mutantstrainarg–trp+ No colonies(control) No colonies(control) Bacterial samples streaked on minimal media Coloniesgrew RESULTS Only the samples from the mixed culture, contained cells that gave rise to colonies on minimal medium, which lacks amino acids.

10. CONCLUSION Conclusion “Because only cells that can make both arginine and tryptophan (arg+ trp+ cells) can grow into colonies on minimal medium, the lack of colonies on the two control plates showed that no further mutations had occurred restoring this ability to cells of the mutant strains. Thus, each cell from the mixture that formed a colony on the minimal medium must have acquired one or more genes from a cell of the other strain by genetic recombination.” Put into your own words……….

11. Mechanisms of Gene Transfer and Genetic Recombination in Bacteria • Three processes bring bacterial DNA from different individuals together • Transformation • Transduction • Conjugation

12. Transformation Is the alteration of a bacterial cell’s genotype and phenotype by the uptake of naked, foreign DNA from the surrounding environment (Could be antiR plasmids)

13. Transformation

14. Transduction 1 Phages carry bacterial genes from one host cell to another Phage DNA Phage infects bacterial cell that has alleles A+ and B+ B+ A+ Host DNA (brown) is fragmented, and phage DNA and proteins are made. This is the donor cell. A+ B+ Donorcell A bacterial DNA fragment (in this case a fragment withthe A+ allele) may be packaged in a phage capsid. A+ Phage with the A+ allele from the donor cell infects a recipient A–B– cell, and crossing over (recombination) between donor DNA (brown) and recipient DNA (green) occurs at two places (dotted lines). Crossingover A+ 4 A– B– Recipientcell The genotype of the resulting recombinant cell (A+B–) differs from the genotypes of both the donor (A+B+) and the recipient (A–B–). 5 A+ B– Recombinant cell

15. 1 m Sex pilus Conjugation and Plasmids • Conjugation • Is the direct transfer of genetic material between bacterial cells that are temporarily joined • Cells containing the F plasmid, designated F+ cells • Function as DNA donors during conjugation • Transfer plasmid DNA to an F recipient cell

16. 1 3 2 4 F Plasmid Bacterial chromosome F+ cell F+ cell Mating bridge F+ cell Bacterial chromosome F– cell A cell carrying an F plasmid(an F+ cell) can form amating bridge with an F– celland transfer its F plasmid. DNA replication occurs inboth donor and recipientcells, using the single parental strands of the F plasmid as templates to synthesize complementary strands. The plasmid in the recipient cell circularizes. Transfer and replication result in a compete F plasmid in each cell. Thus, both cells are now F+. A single strand of the F plasmid breaks at a specific point (tip of blue arrowhead) and begins tomove into the recipient cell. As transfer continues, the donor plasmid rotates(red arrow). (a) Conjugation and transfer of an F plasmid from an F+ donor to an F– recipient • Conjugation and transfer of an F plasmid from an F+ donor to an F recipient

17. The Situation Gets more Interesting……………… • Chromosomal genes can be transferred during conjugation • When the donor cell’s F factor is integrated into the chromosome • A cell with the F factor built into its chromosome • Is called an Hfr cell • The F factor of an Hfr cell • Brings some chromosomal DNA along with it when it is transferred to an F– cell

18. 2 1 Hfr cell F+ cell F factor The circular F plasmid in an F+ cellcan be integrated into the circularchromosome by a single crossoverevent (dotted line). The resulting cell is called an Hfr cell (for High frequency of recombination). B+ D+ C+ C+ A+ D+ A+ Hfr cell B+ D+ A+ C+ B+ A+ A+ D+ C+ B+ B+ C– C– C– B– F– cell D– D– A+ B– D– B– A– A– A– C– B+ D– B– A– 3 4 6 A+ 5 Since an Hfr cell has all the F-factor genes, it can form a mating bridge with an F– cell and transfer DNA. The mating bridgeusually breaks well before the entire chromosome andthe rest of the F factor are transferred. The location and orientation of the F factor in the donor chromosome determine the sequence of gene transfer during conjugation. In this example, the transfer sequence for four genes is A-B-C-D. A single strand of the F factorbreaks and begins to move through the bridge. DNA replication occurs in both donor and recipient cells, resulting in double-stranded DNA Temporary partial diploid C– C– B+ B– Recombinant F– bacterium D– D– B+ B– A– A– A+ A+ 7 8 The piece of DNA ending up outside thebacterial chromosome will eventually be degraded by the cell’s enzymes. The recipient cell now contains a new combination of genes but no F factor; it is a recombinant F– cell. Two crossovers can result in the exchange of similar (homologous) genes between the transferred chromosome fragment (brown) and the recipient cell’s chromosome (green).

19. Insertion Sequences • Insertion sequences can serve as an additional source of genetic variation. • They are simple transposable elements that are unique to bacteria. • They move around the genome and could potentially interrupt a gene or its promoter region leading to variations.

20. Insertion sequence A T C C G G T… A C C G G A T… 3 5 T A G G C C A … T G G C C T A … Transposase gene Inverted repeat Inverted repeat Insertion Sequences • An insertion sequence contains a single gene for transposase • An enzyme that catalyzes movement of the insertion sequence from one site to another within the genome 3 5

21. Transposon Antibioticresistance gene Insertion sequence Insertion sequence 5 3 5 3 Inverted repeats Transposase gene Transposons • Bacterial transposons • Also move about within the bacterial genome • Have additional genes, such as those for antibiotic resistance

22. Section 18.4 Regulation of Gene Expression • This would be a great topic for AP Exam emphasizing regulation. • We will look at some examples in E. coli using operons.

23. (a) Regulation of enzyme activity (b) Regulation of enzyme production Precursor Feedback inhibition Enzyme 1 Gene 1 Regulation of gene expression Enzyme 2 Gene 2 Gene 3 Enzyme 3 – Gene 4 Enzyme 4 – Gene 5 Enzyme 5 Tryptophan • This metabolic control occurs on two levels • Adjusting the activity of metabolic enzymes already present • Regulating the genes encoding the metabolic enzymes

24. Operon Model • Jacob and Monod (1961) - Prokaryotic model of gene control. • What is an operon? • A cluster of genes that carry out a common function. Operon Composition: 1. Regulatory Gene 2. Operon Area a. Promoter b. Operator c. Structural Genes

25. Components • Regulatory Gene: Makes Repressor Protein which may bind to the operator. • Promoter: Attachment sequence on the DNA for RNA polymerase. • Operator-The "Switch”, binding site for Repressor Protein. • If blocked, will not permit RNA polymerase to pass, preventing transcription. Structural Genes--Make the enzymes for the metabolic pathway.

26. Lac Operon • For digesting Lactose. • Inducible Operon - only works (on) when the substrate (lactose) is present. If no Lactose • Repressor binds to operator. • Operon is "off”, no transcription, no enzymes made

27. If Lactose is absent

28. If Lactose is present • Repressor binds to Lactose instead of operator. • Operon is "on”, transcription occurs, enzymes are made.

29. If Lactose is present

30. Enzymes • Digest Lactose. • When enough Lactose is digested, the Repressor can bind to the operator and switch the Operon "off”. Net Result • The cell only makes the Lactose digestive enzymes when the substrate is present, saving time and energy.

31. CAP - positive gene regulation • Catabolite Activator Protein • Accelerates the level of transcription by working with the RNA polymerase. • Uses cAMP as a secondary cell signal. CAP - Mechanism • Binds to cAMP. • Complex binds to the Promoter, helping RNA polymerase with transcription.

32. Note for understanding graphic on next slide • **Bacteria prefer to use glucose as their primary source of sugar. If glucose is scarce cAMP is high

34. Result • If the amount of glucose is low (as shown by cAMP) and lactose is present, the lac operon can kick into high gear. • So, the lac operon can undergo both positive and negative regulation.

35. Regulation Theme • Inducible enzymes • Usually function in catabolic pathways • Catabolism: the set of pathways that break down molecules into smaller units and release energy • Repressible enzymes • Usually function in anabolic pathways • Anabolism: is the set of metabolic pathways that construct molecules from smaller units. These reactions require energy Why does this make sense biologically, what if it were reversed?

36. The End