Answer Key. Bacterial Genetics, BIO 4443/6443 Spring Semester 2001 Exam I. Name. Student ID#

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1 Name Student ID# Answer Key Bacterial Genetics, BIO 4443/6443 Spring Semester 2001 Exam I 1.) Describe two general mechanisms by which transcription can terminate in bacteria. (5pts) Factor dependant termination occurs when the RNA polymerase encounters a particular sequence during transcription if a protein factor- such a Rho, Tau, or NusA. Factor independant terination occurs when a particular sequence (usually an inverted repeat followed by a string of Us in the RNA) that cause the RNA polymerase to terminate. 2.) What is unique about f met? When or where is it found? (5pts) a transformylase adds a formal group onto methionine. It is the first amino acid of every polypeptide synthesized in most bacteria and is used during translation initiation 3.) Define the following. (6pts) auxotrophic mutant: cannot make or utilize a nutrient that the wild-type organism uses for growth. transversion mutations: mutation that changes a purine to a pyrimidine or visa versa. missense mutations: base pair mutation in a coding region that changes an amino acid. nonsense mutations: base pair mutation in a coding region that produces a stop codon (terminates translation) nonsense suppressors: a mutation that allows an amino acid to be inserted at a nonsense codon during translation (often the mutation occurs in a trna gene) true reversion: a mutation that restores the wild-type sequence. 4.) Name three mechanisms by which the cell may repair UV-induced DNA lesions. How is the lesion repaired in each case? Write whether each mechanism is specific for UV-induced DNA lesions or is it a general repair mechanism that can recognize multiple types of DNA lesions? (9pts) 1.) Photolyase, specific repair of UV-photopoducts. Repairs by direct reversal of the lesion without excison. 2.) Nucleotide excision repair, a general mechanism that repairs many lesion types. Multiprotein subunit excises a twelve base pair gap around the DNA lesion. The gap is then resynthesized by PolI. 3.) Bypass polymerases, a specific (general OK) repair of DNA lesions. hese polmerases can incorporate bases opposite to DNA lesions that block replication forks. Thus the lesions are really tolerated rather than repaired. 4.) Base excison repair, a specific repair mechanism for UV-induced lesions. Some phage and organisms have glycosylases that excise the base to leave an AP site. An AP endonuclease then cleaves the DNA backbone and PolI (and other enzymes) remove and resynthesize the base that was damaged. 5.) Recombinational repair, a general repair pathway. Lesions are skipped over by the replication machinery leaving gaps in the daughter strands which are then filled in by using homologous undamaged, regions of the chromosome. Againthe lesions are tolerated (left in the DNA) rather than directly repaired. 6.) Mismatch repair (1/2 credit), a general repair mechanism. No evidence that the mismatch repair system can recognize UV-induced lesions...but worth an experiment to see. 5.) Mismatched base pairs which slip through the replication machinery contain two perfectly good bases that are paired incorrectly (A paired with G for example). Repairing one base restores the original sequence. However repairing the other base would result in a mutation. How do we think that the mismatch repair proteins recognize which of the mismatched bases is the correct one which should be repaired? (5pts) The Dam methylase methylates the adenine base in GATC sequences. Immediately replication, the newly synthesized strand has not yet had time to be methylated- the DNA is hemimethylated. The mismatch repair proteins (MutS,MutL, and MutH) bind to the mismatched base pairs and then this protein complex is thought to track along the DNA to determine which strand is unmethylated. The nonmethylated strand is then incised and repaired.

2 6.) You have mapped the origin of replication on the plasmid shown to the right. If the origin is unidirectional, draw where one would expect the termination of replication to occur on the plasmid? (4pts) oriv terminates at the origin replicating aroung the molecule 7.) ColE1 plasmids maintain a copy number of about 16 copies/cell. How would the copy number be effected by a mutation in the promoter region of the following genes or transcripts? Assume that the mutation prevents any transcription from occurring at each gene. Explain your answer. (6pts) RNAI RNAII oriv rop RNAI: The copy number would increase because the RNA I transcript hybridizes with RNAII, a transcript required for initiation, and prevents it from forming a functional initiator at the origin DNA region. RNAII: The copy number would decrease (The plasmid probably could not replicate) since the RNAII is needed for nreplication initiation. The transcript hybridizes to the origin region DNA, get processed by RnaseH, and serves as a primer to initiate replication. rop: The copy number would increase because the Rop protein promotes the hybridization of the inhibitor RNA I transcript to the RNAII transcript, preventing initiation from occurring. 8.) The single copy F plasmid replicates only once per cell generation. Thus at the time of cell division, there are only two copies of the plasmid per cell. Assuming that the plasmid is sorted randomly, calculate (or draw) how often the plasmid should be lost by chance each generation. (5pts) each cell division creates 2 cells, there s a 1/2 chance that each plasmid will go into a cell, in 2 of the 4 possibilities the compounded by the number OR plasmid is lost. of plasmids... 2(1/2) 2 = 1/2 the time 1/2 9.) By measuring the rate that the F plasmid is cured, you find that only one in every ten thousand cells actually loses the plasmid. Name two general mechanisms or processes that could account for the discrepancy between the observed cure rate and your calculated cure rate. (5pts) The plasmid may have 1.) a partitioning mechanism that ensures each cell receives one plasmid or 2.) plasmid addiction mechanism that kill cells when they lose the plasmid. Others... 3.) selection in the media kills cells which lose the plasmid 4.) integration of the plasmid into the chromosome keeps it stable.

3 Ten years down the road. The first round-trip space mission to Mars has just returned with large samples of soil and ice from the polar regions of Mars. The big news in all of the journals is that scientists found an organism that resembles bacteria on this planet and have cultured it out from one of the soil samples! The bacteria-like organisms are able to grow under culturing conditions very similar to the E. coli but seem to divide almost twice as fast. At this point, scientists have examined the chemical make up the organism and found that its DNA is identical in structure and composition to that which is found on earth. However, nothing yet is known about how they replicate or divide. NASA scientists have recently sent you a sample of the bacteria so that you can determine if the DNA is replicated by a similar mechanism to that which occurs on earth. You remember the Meselson-Stahl experiment and decide that repeating their experiment on these new bacteria may be the best way to begin an analysis of replication mechanisms. You grow three separate cultures of the bacteria. A.) For the first culture, you grow the bacteria in media containing heavy isotopes of nitrogen ( 14 N) and carbon ( 13 C) to density label the DNA during growth. B.) For the second culture, you grow the bacteria in normal media. C.) For the third culture, you begin by growing the culture in the media containing heavy isotopes. Then, several generations of growth in this media, you transfer the bacteria into normal media and takes samples at various times ( I. immediately before transfer, II. one doubling time in the normal media, and III. two doubling times in the normal media). You lyse the cells, isolate the DNA and centrifuge each sample to equilibrium in neutral CsCl gradients. Following centrifugation, you examine where the DNA ran in each gradient an observe the following results shown below: A B C immediately before transfer 1 doubling time 2 doubling times 10.) NASA needs to know. Do these bacteria replicate conservatively, semiconservatively, or distributively? (5pts) Conservatively

4 11.) What would you expect the results look like if the bacteria replicated by one of the alternative mechanisms? Circle one of the following replication mechanisms and then draw in where you would expect the bands to run. Explain your answer below. (5pts) Conservative. Semiconservative. Distributive. A B C immediately before transfer 1 doubling time 2 doubling times Explain: After on generation in normal media, all DNA would contain one heavy and one light strand. After a second generation, replication of the light strand of the hybrid creates a completely light molecule, while replication of the heavy strand would remain a hybrid density. Various size pieces of DNA that replicated either conservatively or semiconservatively would create a distribution of DNA that ranges in composition but gets gradually becomes less dense on average as it replicates in normal media. In the course of your studies, you notice that the bacteria are able to grow on plates that contain only toluene as the sole carbon source. Realising the potential commercial value of this ability, you decide to try and learn more about the genes involved in this pathway by screening for spontaneous mutants that cannot grow on toluene. 12.) Briefly describe how you would conduct your screen. (6pts) (May wish to mutagenize the bacteria first to enrich for mutants) 1 st Plate the bacteria on normal (nonselective or glucose) media: Everybody grows. 2 nd Replica plate the bacterial colonies onto minimal media plates that contain only toluene as a carbon source: tol mutants will not grow. 3 rd determine which colonies didn t grow on the toluene plates and isolate these from your original nonselective plates: These are your mutants. 13.) Does this screen involve a positive selection or a negative selection? (4pts) Negative selection Through your screen, you isolated 26 mutants out of the 100,000 cells which you plated that were could not grow on toluene. Using, the twenty-sixth mutant that you obtained (tol26), you screen to isolate repressor mutants. 14.) Briefly describe how you would conduct your screen for revertants. (6pts) (May wish to mutagenize the bacteria first to enrich for mutants) 1 st Plate many of the tol26 bacteria onto minimal media plates that contain only toluene as a carbon source. (Also plate an equal amount of cells on nonselective media so that you know how many cells were plated). 2 nd Any supressor mutants or revertants should be able to grow on toluene plates again so these are your mutants. 15.) Does the revertant screen involve a positive selection or a negative selection? (4pts) Positive selection

5 You plate more than 10 million tol26 cells and obtain only three revertants that are able to grow once again on toluene plates. 16.) Based upon the observed reversion rate, which of the following mutation types are not likely to have produced the original tol26 mutation? Why? (6pts) Basepair change Frameshift Deletions almost never produce revertions because large amounts of genetic information (multiple Deletion bases) have been lost entirely. 17.) Why would you expect the reversion mutation rate (tol26 changing to wild type) to be so much lower than the forward mutation rate (wild type changing to tol-)? Give one possible reason. (4pts) Toluene metabolism is likely to be controlled by many genes, so a mutation anywhere in this large target of genetic information could produce an inability to metabolize toluene. The tol26 mutation probably occured at a unique site in one toluene metabolism gene. So the chance that this mutation, or a compensating one, will occur again is much, much smaller since the target nuber of bases that could mutate to give you supression is so small. Health concerns abound about the possibility that a foreign and potentially contagious bacteria could accidentally be released on earth. Data involving its susceptibility to various antibiotics and its mutation rates are important for the safety of the general public. You find that the Mars bacteria is killed by treatment with rifampicin, a promising result. Not surprisingly however, you notice that if you plate enough bacteria, rifampicin resistant bacteria can be isolated. You decide to perform a fluctuation test to determine the mutation rate to rifampicin resistance. You grow 30 individual, 1 ml cultures of bacteria and spread each on a plate containing rifampicin. By counting the bacteria from a separate culture, you determine that there were 1 * 10 8 bacteria/ml at the time you plated them on rifampicin. After spreading each 1ml culture on separate plate, you let them grow overnight. The next day, you count the following number of rifampicin resistant colonies that grew on each plate: PLATE # of rif R colonies PLATE # of rif R colonies PLATE # of rif R colonies #1 1 #11 22 #21 2 #2 4 #12 16 #22 4 #3 0 #13 0 #23 0 #4 0 #14 0 #24 0 #5 0 #15 0 #25 0 #6 0 #16 0 #26 0 #7 6 #17 1 #27 9 #8 6 #18 2 #28 1 #9 1 #19 1 #29 1 #10 4 #20 1 #30 4 Assuming that the mutations occur at random and that the mutation rate, a = m/n. 18.) Estimate the mutation rate for rif R mutations occurring in the Martian bacteria by using the Poisson expression where the probability of having i mutations per culture is represented by P i = ( m i e -m )/ i!. Show your work. (10pts)

6 12 out of 30 cultures had 0 mutations. So the probability of having zero mutations (P i ) is 12/30, and i= 0 mutational events per culture in this situation. P i (12/30) = = ( m i e -m )/ i! ( m 0 e -m )/ 0! 0.4 = ( 1 e -m )/ = e -m -ln(0.4) = m.916 = m.916 mutational events per culture and each culture has 1 x 10 8 cells a = m/n a =.916 mutational events/1 x 10 8 cells a = 9.16 x 10-9 rif R mutational events per cell division