1.) Draw the structure of guanine. Indicate where the hydrogen bonds form with cytosine. (4pts)

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1 Name Student ID# p 1.) Draw the structure of guanine. Indicate where the hydrogen bonds form with cytosine. (4pts) hydrogen bonds form at these sites O HN HN H N N R Microbial Genetics BIO 410/510 Fall Quarter 2007 Exam I 2.) Thermosensitive mutations exist for several of the E. coli proteins associated with DNA replication. Describe the role that each of the following proteins play during semiconservative replication AND predict what would happen to replication if that protein were inactivated. EXPLAIN your rationale. (10pts) DnaB helicase: separates and unwinds the two strands DNA as replication progresses. Following DnaB inactivation, the replication fork would rapidly be arrested. DnaG primase: Repeatedly synthesizes a short RNA primer that allows the lagging strand polymerase to reinitiate on the lagging strand template. Following DnaG inactivation, the replication fork would either arrest or, if it continued, the lagging strand synthesis would not occur. DNA Ligase: Joins the Okazaki fragments on the nascent lagging strand following synthesis. Following inactivation of DNA Ligase, the synthesis on the nascent lagging strand would remain discontinuous (contain nicks). Pol I: Removes the RNA primer on each Okazaki fragment and resynthesizes it with DNA on the lagging strand. Following inactivation of Pol I, the synthesis on the nascent lagging strand would remain discontinuous (contain nicks). DNA Gyrase topoisomerase: Relieves the positive supercoiling induced by replication forks progressing along the DNA. Following inactivation of DNA Gyrase, replication would arrest as the supercoiling stress would prevent further DNA winding (replication fork progression) from occurring.

2 Name page 2 3.) Escherichia coli replication machinery makes a mistake about once every times it incorporates a nucleotide. E. coli O157 is a pathogenic strain that was isolated from hamburgers at Jack in the Box restaurants in the late 1990 s. It caused severe fevers and death in several customers. Its genome contains approximately 5 * 10 6 basepairs. How many changes (mutations) would be expected to occur in its genome each time it replicates? Show your work. (3pts) (5 * 10 6 bases / genome) * (1 error/ 1 * bases) = (5 * 10 6 bases / genome) * (1 error/ 1 * bases) = 5 * 10-4 errors / genome 4.) The actual error (mutation) rate per cell division in E. coli strain O157 was measured and it was found to make a mistake about once every 10 6 times it incorporates a nucleotide. How many changes (mutations) occur in its genome each time it replicates? Show your work. (2pts) (5 * 10 6 bases / genome) * (1 error/ 1 * 10 6 bases) = (5 * 10 6 bases / genome) * (1 error/ 1 * 10 6 bases) = 5 * errors / genome 5.) The muts gene of E. coli O157 is inactivated by a mutation. MutS is part of the methyl directed mismatch repair system. Describe how DNA methylation allows replication to correct errors that were made during replication. (4pts) Methylation occurs at GATC sites on the DNA. Since the methylation process takes some time to occur, the methyl-directed mismatch repair system is able to identify which strand is the newly replicated (unmethylated) strand of the DNA. Mismatched base pairs are then excised from the unmethylated, daughter strand and the region is then re-replicated to correct the error. 6.) The E. coli origin or replication contains multiple DnaA boxes and AT-rich 13-mers. Describe what role each of these plays in the initiation of replication? (5pts) Multiple DnaA proteins bind in a sequence specific manner to the DnaA boxes and associated with each other. The DnaA binding induces stress on the surrounding helical DNA regions and promote strand separation at the AT-rich regions. These single stranded regions then serve as the binding sites for the delivery and loading of the DnaB helicase by DnaC to initiate the subsequent steps of establishing the replication forks.

3 Name page 3 A soil bacteria that was isolated replicates in every 10 hours when grown in lab cultures. To examine whether replication occurs conservatively or semiconservatively in this bacteria, you decide to utilize a variation of the approach that Meselson-Stahl originally used to examine this question in E. coli. For your controls, you grow the bacteria in two separate culture media for several generations. C1.) One culture is grown in normal media C2.) The other culture is grown in media that contains 5-bromouracil, an analog of thymine that has a much higher buoyant density. For your experimental analysis, you inoculate a third culture of the bacteria in normal media and allow it to grow for 2 days. At this time, you then transfer the cells into media containing 5- bromouracil. To examine the mode of replication, you collect a portion of the cells E1) immediately after changing the media, E2) 10 hours after you changed the media, and E3) 20 hours after you changed the media. You then lyse the cells and load the cell lysates (and DNA) from each sample in neutral CsCl gradients before centrifuging them to equilibrium. The results of your controls, C1 and C2, are shown below. 7.) On the tubes below, 8.) On the tubes below, clearly indicate where the clearly indicate where the DNA would be expected to DNA would be expected to appear if the bacteria replicate appear if the bacteria replicate conservatively? (5pts) semi-conservatively? (5pts) C1 Normal media C2 5-bromouracil media 0 hrs 10 hrs 20 hrs 0 hrs 10 hrs 20 hrs E1 E2 E3 E1 E2 E3

4 Name page 4 9.) What DNA sequences are important for factor independent transcriptional termination? How are these thought to promote transcription termination? (3pts) Termination by this mechanism relies upon an inverted repeat sequence that is followed by a stretch of UUUUs in the RNA transcript. Transcription of the inverted repeats produces a hairpin in the RNA that destabilizes the RNA polymerase enough to dissociate it when followed by a string of UA base pairs in the transcription bubble. The UA base pairs are less stable than GC base pairs due to the lower number of hydrogen bonds formed between these base pairs. 10.) What DNA sequences are important for factor dependent transcriptional termination? How are these thought to promote transcription termination? (3pts) Factor dependant termination occurs when a protein factor, such a Rho, binds to a specific sequence on the RNA transcript. In the case of Rho, the rut sequence is bound. The Rho factor is an RNA helicase that, in effect, chases or follows the RNA polymerase and dislodges it at specific pause sites downstream on the transcript. Since Rho can only bind to rut when translation is not "hiding" the rut sequence, this also provides a mechanism to regulate the transcription of polycistronic messages. 11.) Briefly describe (or draw) the events involved during translation elongation (do not include events that are associated specifically with initiation or termination). Include the following terms, if appropriate (not all terms are appropriate). TATA Box, 30S Subunit, 50S Subunit, 70S Subunit, RNA Polymerase, Shine Delgarno, rho, ATP, GTP, formylmethionine, P site, A site, RF1, IF3 and IF1, IF2, EF-Tu, EF-G, t-rna, sigma factor. (6pts) A. EF-Tu loads a charged trna into the empty A site of the 30S ribosome subunit using GTP as an energy source. (EF-Ts then exchanges the GDP for GTP on Ef-Tu to "recharge" the enzyme for the next round. B. The polypeptide attached to the trna in the P site is transferred (bonded) to the amino acid-trna in the A site using a peptidyltransferase activity found on the 50S subunit. C. GTP charged EF-G powers the ribosome forward moving the trna containing the polypeptide into the P site. (The cycle then repeats.) You ve identified a small gene from a strain of Streptomyces whose expression prevents other bacteria from growing when Streptomyces is present. You are interested in this peptide because of its potential as a new antimicrobial agent. The entire open reading frame (ie. the TRANSLATED portion of the mrna transcript has following the sequence: 5' GUG AUC AUU AUA ACU ACC ACA ACG UAA 3' 12.) Using the table for the genetic code and your knowledge of how translation occurs, translate the given DNA sequence into its appropriate amino acids. (4pts) fmet-ile-ile-ile-thr-thr-thr-thr 13.) What s the absolute minimum number of different trna molecules that would be needed to translate this peptide? Why? (2pts) At least three, an initiation trna, an Ile-tRNA and a Thr-tRNA. The wobble position of an anticodon on a given trna may recognize multiple bases, meaning that the same trna may deliver a given amino acid to codon sequences that differ at the wobble position.

5 Name page 5 The gene below encodes Homo sapiens coagulation factor III, a cell surface glycoprotein that enables cells to initiate the blood coagulation cascade. This protein is the only one in the coagulation pathway for which a congenital deficiency has not been described. Your working for a biotech company that is interested in producing drugs to treat specific forms of hemophelia. Your task is to clone this gene into an expression vector as a potential mode of treatment. 1 gggggggggg cccccgcgca ccccctcgca ctccctctgg ccggcccagg gcgccttcag 61 cccaacctcc ccagccccac gggcgccacg gaacccgctc gatctcgccg ccaactggta 121 gacatggaga cccctgcctg gccccgggtc ccgcgccccg agaccgccgt cgctcggacg 181 ctcctgctcg gctgggtctt cgcccaggtg gccggcgctt caggcactac aaatactgtg 241 gcagcatata atttaacttg gaaatcaact aatttcaaga caattttgga gtgggaaccc 301 aaacccgtca atcaagtcta cactgttcaa ataagcacta agtcaggaga ttggaaaagc 361 aaatgctttt acacaacaga cacagagtgt gacctcaccg acgagattgt gaaggatgtg 421 aagcagacgt acttggcacg ggtcttctcc tacccggcag ggaatgtgga gagcaccggt 481 tctgctgggg agcctctgta tgagaactcc ccagagttca caccttacct ggagacaaac 541 ctcggacagc caacaattca gagttttgaa caggtgggaa caaaagtgaa tgtgaccgta 601 gaagatgaac ggactttagt cagaaggaac aacactttcc taagcctccg ggatgttttt 661 ggcaaggact taatttatac actttattat tggaaatctt caagttcagg aaagaaaaca 721 gccaaaacaa acactaatga gtttttgatt gatgtggata aaggagaaaa ctactgtttc 781 agtgttcaag cagtgattcc ctcccgaaca gttaaccgga agagtacaga cagcccggta 841 gagtgtatgg gccaggagaa aggggaattc agagaaatat tctacatcat tggagctgtg 901 gtatttgtgg tcatcatcct tgtcatcatc ctggctatat ctctacacaa gtgtagaaag 961 gcaggagtgg ggcagagctg gaaggagaac tccccactga atgtttcata aaggaagcac 1021 tgttggagct actgcaaatg ctatattgca ctgtgaccga gaacttttaa gaggatagaa 1081 tacatggaaa cgcaaatgag tatttcggag catgaagacc ctggagttca aaaaactctt 1141 gatatgacct gttattacca ttagcattct ggttttgaca tcagcattag tcactttgaa 1201 atgtaacgaa tggtactaca accaattcca agttttaatt tttaacacca tggcaccttt 1261 tgcacataac atgctttaga ttatatattc cgcacttaag gattaaccag gtcgtccaag 1321 caaaaacaaa tgggaaaatg tcttaaaaaa tcctgggtgg acttttgaaa agcttttttt 1381 tttttttttt tttgagacgg agtcttgctc tgttgcccag gctggagtgc agtagcacga 1441 tctcggctca cttgcaccct ccgtctctcg ggttcaagca attgtctgcc tcagcctccc 1501 gagtagctgg gattacaggt gcgcactacc acgccaagct aatttttgta ttttttagta 1561 gagatggggt ttcaccatct tggccaggct ggtcttgaat tcctgacctc agtgatccac 1621 ccaccttggc ctcccaaaga tgctagtatt atgggcgtga accaccatgc ccagccgaaa 1681 agcttttgag gggctgactt caatccatgt aggaaagtaa aatggaagga aattgggtgc 1741 atttctagga cttttctaac atatgtctat aatatagtgt ttaggttctt ttttttttca 1801 ggaatacatt tggaaattca aaacaattgg gcaaactttg tattaatgtg ttaagtgcag 1861 gagacattgg tattctgggc agcttcctaa tatgctttac aatctgcact ttaactgact 1921 taagtggcat taaacatttg agagctaact atatttttat aagactacta tacaaactac 1981 agagtttatg atttaaggta cttaaagctt ctatggttga cattgtatat ataatttttt 2041 aaaaaggttt ttctatatgg ggattttcta tttatgtagg taatattgtt ctatttgtat 2101 atattgagat aatttattta atatacttta aataaaggtg actggaaaaa tttttttttt 14.) Design two 15 base primers that you could use in a PCR reaction to amplify this gene from human cells? (4pts) 5 GGG GGG GGG GCC CCC 3 5 AAA AAA AAA ATT TTT 3 15.) You know that you will need to clone your PCR product into a BamHI restriction site, GGATCC, found on the expression vector. Modify your primers below such that you will be able to easily clone the PCR product into a BamHI site on an expression vector. (3pts) 5 NNN NNN GGA TCC- GGG GGG GGG GCC CCC 3 5 NNN NNN GGA TCC- AAA AAA AAA ATT TTT 3 16.) The human genome contains 3.3*10 9 basepairs. Is the 15 base sequence of your primer likely to be unique in the human genome? On average, how many times would your primer sequence be found on the human genome? Show your work (4pts) 4 possible bases at every position so a sequence 15 bases long would be found every 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 bases 4 15 bases = a specific15 base sequence would be found once in every 1,073,741,824 bases. With 3.3 * 10 9 bases in the human genome a specific 15 base sequence would appear, on average, 3.3 * 10 9 bases/ 1.07 * 10 9 bases = 3.08 times so its not long enough to be unique.

6 Name page 6 Your biotech company is using 4 different experimental cloning vectors called pexpress1, pexpress2, pexpress3, and pexpress4. All the vectors have similar constructions (shown to the right). However, the cloning region on each vector is slightly different. Each of the vector cloning regions is shown below. 18.) For each vector, determine if the Coagulation factor III product is likely to be expressed when your PCR product is cloned into the Bam HI site. Explain Why or Why not in each case. (8 pts) ampicillin resistance Cloning Region pexpress Vectors origin of replication BamHI site GGA TCC Promoter consensus Shine Delgarno sequence ATG Stop codons UAA UAG Cloning Region of pexpress 1 Transcription termination sequences pexpress1: Coagulation factor III product will not be expressed when the PCR product is cloned into the Bam HI site because the gene will be upstream of the promoter and ribosome binding site that are needed for transcription and translation to occur. Shine Delgarno sequence Promoter consensus ATG BamHI site GGA TCC Stop codons UAA UAG Cloning Region of pexpress 2 Transcription termination sequences pexpress2: Coagulation factor III product will not be expressed when the PCR product is cloned into the Bam HI site because the ribosome binding site that is needed for transcription is upstream of the transcriptional promoter. Thus the mrna containing the gene will not be translated because it does not contain the Shine Delgarno sequence. Promoter consensus Shine Delgarno sequence ATG BamHI site GGA TCC Stop codons UAA UAG Cloning Region of pexpress 3 Transcription termination sequences pexpress3: Coagulation factor III product will be expressed when the PCR product is cloned into the Bam HI site. The open reading frame, when cloned into the BamHI site, has a transcriptional promoter, the mrna have a ribosome binding site before the first start codon as well as additional stop codons prior to the end of the transcript. Promoter consensus Shine Delgarno sequence ATG BamHI site GGA TCC Transcription termination sequences Cloning Region of pexpress 4 Stop codons UAA UAG pexpress4: Coagulation factor III product will probably be expressed when the PCR product is cloned into the Bam HI site. If the open reading frame includes its own stop codon, the gene should be expressed normally.

7 Name page 7 The Luria-Delbruck experiment demonstrated that bacteria can become resistant to bacteriophage T1 infection through random mutations, rather than through a directed change. You decide to try and see if the random mutation hypothesis also applies to how bacteria become resistant to the antibiotic, rifampicin. You do the following experiment. You inoculate 200 mls of media with a dilute bacterial culture. Then, you split the culture A) growing 100 mls in a single flask, and B) growing the other half in 100 separate, 1 ml test tubes. After allowing the cultures to grow for several hours, you spread A) 1ml aliquots from the single flask on 100 plates that contain rifampicin, and B) the 100 1ml cultures onto 100 plates that contain rifampicin. You allow each set to incubate overnight and count the number of rifampicin resistant colonies on each plate in the morning. Five representative plates from A) the single 100 ml flask are shown below: 19.) If mutations arose primarily through a process of directed change, CLEARLY diagram how you would expect to five representative plates to look from B) the 100 individually grown 1ml cultures. Explain why. (4pts) The experiment would predict that each cell has an equal probability of generating resistance to the antibiotic WHEN exposed to it. Since all cells were exposed to the antibiotic at the same time, each plate should have roughly equal numbers of mutations. 20.) If mutations primarily arose randomly within growing populations, CLEARLY diagram how you would expect to five representative plates to look from B) the 100 individually grown 1ml cultures. Explain why. (4pts) The experiment would predict that mutations occur randomly within the population prior to any exposure to the antibiotic...some of these mutations confer resistance to the antibiotic and depending on when they first occur in each culture, a given culture may have many (if the mutation occurred early) or few to no (if the mutation occurred late to never) resistant mutants. The number on each plate should vary significantly. 21.) Does this experiment involve positive or negative selection? (2 pts) positive selection 22.) If 20 out of your 100 plates of 1ml cultures had no colonies on them, and the cultures had 10 8 cells/ml, calculate the mutation rate (a) for generating resistence to rifampicin using the Poisson expression, where the probability of having i mutations per culture is represented by P (i) = ( m i e -m )/ i! and a = m/n. (5pts). 20 out of 100 cultures had 0 mutations. So the probability of having zero mutations (P i ) is 20/100, and i= 0 mutational events per culture in this situation.

8 Name page 8 P i = ( m i e -m )/ i! (20/100) = ( m 0 e -m )/ 0! 0.2 = ( 1 e -m )/ = e -m -ln(0.2) = m.916 = m 1.6 mutational events per culture and each culture has 1 x 10 8 cells a = m/n a = 1.6 mutational events/1 x 10 8 cells a = 1.6 x 10-8 rif R mutational events per cell division

9 Name page 9 Primer design (10pts): Design primers for recombineering that would delete the recf gene of E.coli and replace it with the catrsacb cassette from plasmid pel4. >E. coli EG10828 recf: 1074 bp - Recombination and repair g ccagagcgcggcttatgttgtcatgccaatgagactgta 1 - atg tcc ctc acc cgc ttg ttg atc cgc gat 31 - ttc cgc aac att gaa acc gcg gat ctc gcc 61 - tta tct ccc ggc ttt aac ttt ctg gta ggt 91 - gcc aac ggc agt ggc aaa acc agc gtg ctg gaa gcc atc tat acg ctc ggc cat ggt cgg gcg ttt cgc agt ttg cag att ggt cgc gtc att cgc cat gag cag gag gcg ttt gtt ctc cac ggg cga tta cag ggc gaa gag cgc gag aca gcg att ggc tta acc aaa gac aaa cag ggc gac agc aaa gtc cgc atc gac ggt aca gac ggg cat aag gtc gcg gaa ctg gcg cac ctg atg cca atg cag ttg ata acg cca gaa ggg ttt act tta ctc aac ggc ggc ccc aaa tac aga aga gca ttc ctc gac tgg gga tgc ttt cac aac gaa ccc gga ttt ttc acc gcc tgg agc aat ctc aag cga ttg ctc aag cag cgc aat gcg gcg ctg cgc cag gtg aca cgt tac gaa cag cta cgc ccg tgg gat aaa gag ctg atc ccg ctg gcg gag caa atc agc acc tgg cgc gcg gag tat agc gcc ggt atc gcg gcc gat atg gct gat acc tgt aag caa ttt ctc cct gag ttt tct ctg act ttc tct ttc cag cgc ggc tgg gag aaa gag aca gaa tat gct gag gtg ctg gaa cgt aat ttt gaa cgc gat cgc cag cta acc tac acc gcg cac ggc ccg cac aaa gcg gac tta cgc att cgc gcc gac ggt gcg ccg gtg gaa gat acc tta tcg cgt ggg cag ctt aag ctg ttg atg tgc gcc tta cgt ctg gcg caa gga gag ttc ctc acc cgt gaa agc ggg cgg cgg tgt ctc tac ctg ata gat gat ttt gcc tct gag ctt gat gat gag cgt cgc ggg ctg ctt gcc agc cgc tta aaa gcg acg caa tca cag gtc ttt gtc agc gcg atc agt gct gaa cac gtt ata gac atg tcg gac gaa aat tcg aag atg ttt acc gtg gaa aag ggt aaa ata acg gat taa cccaagtataaatgagcgagaaacgttgatgtcgaattc t Primers used to PCR amplify the cat-sacb cassette: Primer-L: CCTGTGACGGAAGATCACTTCG Primer-R: CTGAGGTTCTTATGGCTCTTG Forward primer: 5 gccagagcgcggcttatgttgtcatgccaatgagactgtacctgtgacggaagatcacttcg Reverse primer: 5 agaattcgacatcaacgtttctcgctcatttatacttgggctgaggttcttatggctcttg

Microbial Genetics Bi 410/510 Fall 2009 Exam I

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