MCB 421 Exam #1 (A) Fall 2006 There are 9 questions and 1 supplement on last page. Answer all 9 questions. Be sure your name is on each page 1). (8 points) A Luria-Delbruck fluctuation test was done to determine the rate of mutation to Dehydroproline resistance (a toxic proline analog) in E. coli. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures grown to 5 x 109 cells / ml. A 0.1 ml sample of each culture was then plated on minimal medium to detect DHPR mutants. The results are shown in the following table. Culture # # DHPR mutants Culture # # DHPR mutants 1 32 11 69 2 0 12 3 3 8 13 0 4 14 14 0 5 465 15 14 6 73 16 41 7 0 17 62 8 22 18 3 9 0 19 12 10 0 20 1 A). (2 points). From the data shown in the table is the resistance to dehydroproline due to induced or spontaneous mutation? B). (3 points). Why? [Spontaneous mutation. Even a superficial (non-statistical) analysis of the data shows that there is probably a high variance indicating that the appearance of mutations was random and the number is dependent upon when the mutations occurred during the life of the culture before exposure to the dehydroproline. If induced mutations were occurring, one would expect a similar number of colonies on all the plates.] C). (3 points). If you picked a colony from the plate supplemented with dehydroproline and grew it in LB broth for several generations, would the cells be resistant or sensitive to dehydroproline? [The cells would all (or almost all) be resistant to dehydroproline because the mutation is inherited by all the progeny.] 2). (12 points). Consider the following experiment. Your ultimate goal is to isolate tryptophan auxotrophs but you did not review my lecture notes carefully and did the following experiment. You mutagenize 50 cultures of wild type E. coli with a EMS, chemical mutagen. The cells are then grown for several generations in LB broth at 37o C to recover from the treatment with mutagen. Next the cells are centrifuged and resuspended in minimal medium supplemented with penicillin and grown several for several generations at 37o C. Next, the cells from each of the 20 cultures are resuspended in LB broth, grown for a few generations and the minimal medium-penicillin step is repeated. At the end of the second penicillin step, 100 cells from each of the 20 cultures are plated on LB plates and incubated overnight at 37o C until colonies are formed. The colonies are then replica plated to minimal plates and grown at 37o C. Approximately 95% of the colonies that grew on the master LB plates also grow on the Minimal plates while about 5% do not grow on minimal plates. A). (2 points). Why do most colonies grow on the minimal plate? [They come from cells that are still prototrophs that escaped killing by the penicillin.] B). (2 points). Why don’t some of the colonies grow on the minimal plate? [They are auxotrophs that require supplements included in LB broth. This would also include tryptophan auxotrophs.] C). (1 point). Would any of the colonies be tryptophan auxotrophs? [Probably because there are 50 x 5 = 250 auxotrophs represented in the experiment and several independent cultures.] D). (2 points). How could you identify tryptophan auxotrophs? [Replica plate or streak the auxotrophs from the LB plate onto min + trp plates. Trp auxotrophs will grow while other auxotrophs will not grow.] E). (2 points). How could you have modified the protocol to get only Trp auxotrophs? [Grow cells in Min + Trp at the steps where LB was used.] F). (1 point). Is the experiment a selection, enrichment or a screen? [Enrichment] G). Why were 50 tubes rather than 1 tube used for mutagenesis? [Prevent isolation of siblings.] 3). (9 points). A). (3 points). Would you expect an amber nonsense suppressor mutation to be dominant or recessive to the wild-type tRNA gene? Explain your answer with regard to the molecular mechanism involved. [The suppressor would be dominant (expressed) because it is expressed independently of the wild type tRNA. Thus both sense and amber codon can be decoded.] B). (3 points). Would you expect a bypass suppressor to be dominant or recessive to the wild-type gene? Explain your answer with regard to the molecular mechanism involved. [Dominant because it is expressed independently of the wild type gene.] C). (3 points). What does allele-specific mean? What does it tell you if a suppressor is allele-specific? [ANSWER: An allele-specific suppressor is a second-site mutation that repairs the mutant phenotype but only in strains with certain, specific mutations at the first- site. (Interaction suppressors are usually allele specific).] 4). (10 points). Complementation analysis was done on six mutants that lack theonine synthetase activity. The order of the mutational sites is not known. The results are shown below: thr-1 thr-2 thr-3 thr-4 thr-5 thr-6 thr-1 - + + - - - thr-2 - - + + - thr-3 - + + - thr-4 - - - thr-5 - - thr-6 - A). (2 points). How many complementation groups are represented? [At least two complementation groups: Group 1 = thr-1, thr-4, thr-5; Group 2 = thr-2, thr-3; Ungrouped = thr-6.] B). (6 points). Suggest two explanations for the results for thr-6. [ thr-6 fails to complement all of the other mutants. This could be either due to a trans-dominant negative phenotype of this mutant (e.g. due to a missense mutation that poisons threonine synthetase) or a cis-dominant negative phenotype caused by the mutation (e.g. due to an amber mutation that prevents expression of downstream genes). In either case, it is impossible to determine whether the thr-6 mutation is within one of the two complementation groups described by the other mutations or whether it is in a different complementation group.] C). (2 points). How could you distinguish between your explanations? [ Do a complementation analysis with the WT and thr-6. If the thr-6 mutation is cis-dominant, the phenotype would be wild type. If thr-6 is trans- domionant, the phenotype would be Thr-.] 5). (12 points). NADP is an essential cofactor for many cellular processes. Because it is not transported, exogenous NADP cannot supplement mutants unable to synthesize intracellular NADP. Five independent mutations were obtained that affect the synthesis of NADP. The properties of the mutations are described in the table below (where + indicates growth on rich medium, - indicates that no growth on rich medium, and -/+ indicates weak growth on rich medium). Row Mutation Growth temperature # 30°C 42°C 30 → 42°C 42 → 30°C 1 nad 2 nad-601 3 nad-602 4 nad-603 5 nad-604 6 nad-606 7 nad-607 8 nad-601 nad-602 9 nad-601 nad-603 10 nad-601 nad-604 11 nad-601 nad-607 12 nad-604 nad-607 A). (7 points). Note the properties of nad-601, nad-602, nad-603, nad-604, nad-606, and nad-607 in the above Table. Indicate both whether the mutant has a conditional phenotype (temperature sensitive, cold sensitive, or non- conditional) and whether the allele is likely to be due to a missense, nonsense, frameshift, deletion, or insertion mutation? Briefly explain your answers. ANSWER: nad-601 Ts, missense (Probably AA substitution that destabilized protein) nad-602 Ts, missense nad-603 Ts, missense nad-604 Cs, missense nad-606 Leaky, nonconditional (probably a missense mutation because gene product retains some activity) nad-607 Cs, missense All of these mutations are probably missense because Ts and Cs mutations usually arise due to single amino acid substitutions, and the leaky mutation retains some activity so it is clearly not due to a complete gene disruption. B). (5 points). Interpret the results for each pair of double mutants in rows # 8-12. If you are not able to determine the order of the reactions catalyzed by some of the gene products from the data given, suggest a likely reason for this result. ANSWER: 8:Cannot interpret gene order because both mutations are Ts 9:Cannot interpret gene order because both mutations are Ts 10:Mutation nad-604 (Cs) must act before nad-601 (Ts) 11:Mutation nad-601 (Ts) must act before nad-607 (Cs) 12:Cannot interpret gene order because both 6). (16 points). Phage T4 forms wild-type plaques on both E. coli K-12 (λ+) and E. coli B. Phage geneticists isolated mutants they designated “rII” for rapid lysis. rII mutants form large plaques (larger than wild-type plaques) on E. coli B but cannot grow on E. coli K-12 (λ+) so no plaques are formed. These facts are shown in the Table below. Phage genotype E. coli K12(λ+) E. coli B Wild type normal plaques normal plaques rII no plaques r-type plaques You decide to isolate an rII mutant using proflavin as a mutagen. Proflavin induces frameshift mutations when phage are grown in cells treated with proflavin. A). (2 points). Which E. coli strain would you use for the mutagenesis? Why? [E. coli B because rII mutants can grow in E. coli B but cannot grow in E. coli K-12 (λ+) and would be lost.] B). (3 points). How would you identify rII mutants? Is this a selection or screen? Why? [By plating mutagenized phage on a B strain and identifying plaques with rII morphology. This is a screen because all types may grow and the desired mutant can be identified by screening many plaques.] Assume you were successful in isolating an rII mutant.