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BIMM122: Microbial Genetics Final WI18Section 2: Free response questions (93pts)

1. Toxin/Antitoxin (T/AT) systems (10 pts)a. Diagram and describe the general characteristics of toxin/antitoxin systems including

their functions. (5 pts)b. Describe one toxin/ anti toxin system in detail including its function. Do NOT choose a

system that is not well understood! (5 pts)a

1. One operon, 2 adjacent genes , 2 small proteins.2. The antitoxin gene is always upstream of the toxin gene.3. The antitoxin inhibits the toxin by directly binding to it.4. A specific ATP-dependent protease degrades the antitoxin.5. Transcription is autoregulated by the antitoxin or antitoxin-toxin complex.6. Some chromosomal addiction modules are activated by phage infection by inactivating the antitoxin, thereby altruistically sacrificing the infected cell to protect the population as a whole.

b.

+ detailed mechanism

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2. Transcription (11 pts)a. Why does E.coli have several different sigma factors? (2 pts)b. You are examining an RNA polymerase you have isolated. In an  in vitro assay, you find

that the only products are short “abortive” transcripts, no longer than about 8 - 10 ribonucleotides.

i. Based on what you know about E.coli transcription, hypothesize why this would happen? Explain by drawing the cycle of transcription, and indicating which steps may be problematic. (6 pts)

ii. What experiments would you conduct to test your hypothesis? (3 pts)

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3. The gal operon of E. coli (12 pts)a. Diagram or explain how and why the E. coli gal operon (galETK) is regulated so

that catabolic (galactose degradation) and anabolic (galactose biosynthesis) processes occur appropriately (as needed by the cell)? (5 pts)

b. In this galETK operon, there are two overlapping promoters and two transcriptional start sites. Transcription from the S1 site normally gives rise to coordinate levels of the three protein products (ratios of 1:1:1), but transcription from the S2 site normally gives rise to polar expression of the genes (ratios of 4:2:1).

a. What conditions (factors) determine whether expression will be polar or coordinate (describe both conditions)? (5 pts)

b. Does regulation of this operon occur at the transcriptional or translational level? Provide evidence to justify your answer. (2 pts)

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4. Two component sensory transduction pathways (12 pts)a. How does a generalized two component sensory transduction (sensor kinase/response

regulator) system work? Please describe the general features. (5 pts)b. Diagram the Agrobacterial virulence sensory transduction pathway. (4 pts)c. Explain how and why tumor formation in the plant benefits the bacteria. (3 pts)

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5. List and briefly describe the seven stages of sporulation. You may choose to draw each step with a brief explanation instead. (10pts)

I Chromosome condensation: axial filamentII Asymmetric septation: Septum formationIII Engulfment: a 2-membrane layer IV Cortex formation by foresporeV Coat formation by mother cellVI Maturation: resistance to environmental stresses develops to completionVII Lysis: mother cell death; release of the mature spore

6. Briefly explain how sigma F, the sigma factor required for septum formation, is activated only in forespores, but not in mother cells, by a partner switching mechanism.(7pts) What’s the anti-sigma, anti-anti-sigma and anti-anti-anti sigma factor in this system?(3pts)

IIAB-σF +IIAA-P (MC)↔σF + IIAA-IIAB(FS)

Mother cell: high ATP.IIAA~P: IIAA will be phosphorylated, P is from ATP.σF-IIAB: sigmaF is inactive.(3pts)Forespore: low ATP, high ADP.IIE, which is a phosphatase only in forespore, will take off the phosphate from IIAA.IIAA-IIAB-ADP: Phosphate-free IIAA will bind IIAB with higher affinity. σF is now released from IIAB and active.(4pts)IIAA is the anti-anti sigma factor. IIAB is the anti and anti anti anti sigma factor.(3pts)

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7. Consider the trans-compartment signaling that is achieved for the activation of sigma K in the Bacillus sporulation process. Which sigma factor(s) initiate(s) the process? How are the signals transduced? Please also indicate all the proteases that participate in this signaling process, what their target proteins are, and how this results in trans-compartment signaling. (10pts)

8. In prokaryotes, about 90% of the genome usually codes for gene products, and many of these genes are in operons. Bacteria have developed many gene arrangements that allows more efficient control and coordination of metabolic functions. a. Please give an example of a selfish arrangement and explain why it is beneficial. (5pts)b. Also please explain how bacteria compact their chromosomes. (3pts)

For a: Examples includes Lac operon, macro-molecular synthesis operon, etc.. (3pts) Selfish arrangements help increase fitness of the organism while horizontal gene transferring, allow coordinate synthesis. (2pts)

For b: Supercoiling, DNA binding proteins, and macromolecular crowding.(1pt for each)

9. The genetic code makes sense if analyzed correctly from structural and thermodynamic standpoints. a. Please give 3 or 4 examples where thermodynamics provides a guide to an

understanding the genetic code (5 pts). b. Also, indicate 3 or 4 steps that may have occurred in the evolution of the genetic code.

(4 - 6 sentences) (5 pts)

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For a. For Stop codons, P1 is always U, and P2 and P3 are always purines, A being preferred over G. The initiation “Wobble” position is P1, P2 and P3 are invariant. Position 2 primarily determines when position 3 is important, usually correlating (see rule 9) with H-bond strength. It must make a difference which nucleotide is in the mRNA and which is in the tRNA. T:A (mRNA:tRNA) must be stronger than A:T, and C:G must be stronger than G:C.

For b. Position 2 evolved first as a single codon, there were only 4 amino acids or 3 types of amino acids in the first place, 1 hydrophobic, 1 hydrophilic, 2 semi polar. To increase specificity and complexity of possible amino acids, position 1 evolved, then position 3 evolved.

Or: The codons started as triplets, then gradually gain specificity for each amino acid.

10. Diagram the leader region in front of the E. coli tryptophan biosynthetic operon, revealing how this leader regulates expression of the structural genes in the operon by a termination/ anti-termination mechanism. (12)

• No tryptophan: cell wants to activate the operon and make tryptophan.

• Ribosome will stall at the UGG site in site 1, because Trp-tRNA is empty. Site 2 can now only bind site 3. This form is an anti-terminator. Whole operon is transcribed, and the EDCBA can be translated because they have other SD sequences and start codons.

• With tryptophan available: cell wants to save energy and don’t make Trp.

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• Ribosome won’t interfere transcription. mRNA will form the most stable conformation, which is 1 binds 2 and 3 binds 4. 3&4 is terminator, no transcription, operon will be off. ECDBA is not even transcribed.