MCB 421 Exam #1 (A) Fall There are 9 questions and 1 supplement on last page. Answer all 9 questions. Be sure your name is on each page

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1 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

2 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 10 9 cells / ml. A 0.1 ml sample of each culture was then plated on minimal medium to detect DHP R mutants. The results are shown in the following table. Culture # # DHP R mutants Culture # # DHP R mutants 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.]

3 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 37 o 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 37 o 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 37 o C until colonies are formed. The colonies are then replica plated to minimal plates and grown at 37 o 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?

4 [Enrichment] G). Why were 50 tubes rather than 1 tube used for mutagenesis? [Prevent isolation of siblings.]

5

6 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 firstsite. (Interaction suppressors are usually allele specific).]

7 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 thr thr thr thr 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 transdomionant, the phenotype would be Thr-.]

8 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 C C 1 nad 2 nad nad nad nad nad nad nad-601 nad nad-601 nad nad-601 nad nad-601 nad 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 nonconditional) 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 nad-602 nad-603 nad-604 nad-606 nad-607 Ts, missense (Probably AA substitution that destabilized protein) Ts, missense Ts, missense Cs, missense Leaky, nonconditional (probably a missense mutation because gene product retains some activity) Cs, missense

9 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 # 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

10 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. Next, you decide to isolate a revertant that restores the activity of the protein encoded by the rii gene. C). (3 points). How would you isolate the revertant? Which E. coli strain would you use and how would you identify the revertants? Is this a selection or a screen? Why? [Revertants could be isolated by plating a pool of rii phage on K12 (λ+); only revertants will be able to form plaques so this is a selection. Looking for wild type plaques on B would require a lot more work since most of the plaques would be r type and wild type plaques would be exceedingly rare)] D). (4 points). What are the two most likely types of mutations that could restore activity?

11 [A true revertant or a frameshift suppressor (pseudorevertant). The original rii mutant must be a frameshift mutant since we used proflavin as the original mutagen. Thus, they can only be reverted by true reversion or a frameshift mutation at a second site.] NOTE: E was deleted from exam copy B so everyone got 4 points for E and F. E). (2 points). How could you distinguish the two types of revertants using a back cross? (ie. crossing your revertant with wild type T4 ). Draw diagrams showing the predicted results and phenotype for each type of cross. [A true revertant, when crossed with a wild type phage, will produce only wild type progeny. A revertant with an intragenic frameshift suppressor will produce some rii type progeny when crossed with wild type T4. Recombination between the sites of the 2 frameshift mutations will produce phage that contain a single frameshift each. One will have the original rii mutation and the other will carry the suppressor.] F). (2 [points). Which strain would you use for the cross and analyzing the progeny of the cross? Why? [E. coli B, because both phage types can grow on B and the genotypes distinguished on B]

12 7). (12 points). Van Way et al. [2000. J. Mol. Biol. 297: 7-24] took advantage of amber suppressors to determine the role of specific amino acid residues in the flagellar motor protein, MotB. Amber mutations were isolated at many sites throughout motb. The mutants were then tested for suppression in isogenic genetic backgrounds with either sup o, supe, or supfamber suppressors. The results are shown in the Table below. Allele Mobility relative to wild-type: sup o supe supf wild-type 100% 80% 60% Q4Am 0% 80% 100% Q57Am 0% 100% 100% Q94Am 0% 50% 80% Q100Am 0% 10% 10% Q112Am 0% 0% 0% Q124Am 0% 20% 30% Q145Am 0% 100% 100% Q271Am 80% 100% 100% Q281Am 100% 100% 100% This table was prepared from the data in Figure 6 of Van Way et al. [2000]. In the allele column, the mutations are shown as follows: Q, the one letter symbol for gln, indicates the wild-type codon that was mutated, the number indicates the position of that particular amino acid in the protein, and Am indicates the change to a stop codon. A). (6 points). What is the advantage and disadvantage of using suppressor mutations to make multiple amino acid substitutions at a particular position? ANSWER: 2 points Advantage = a single mutation will allow multiple amino acid substitutions any 2 2 points each Disadvantages = different suppressors have different suppression efficiencies, so the amount of substituted product may differ depending upon the sup trna; the suppression efficiency depends upon codon context so you cannot predict the level of suppression a priori; suppression is never 100% efficient so some truncated product will always be produced B). (2 points). Amber mutations after position 271 in the protein do not eliminate motility. Suggest a possible explanation for this result.

13 ANSWER: The simplest explanation of this result is that the C-terminus of the protein is not required for motility. C). (2 points). Some amber mutations, such as Q112Am, cannot be suppressed by supe or supf, and other amber mutations, such as Q94Am, are suppressed to different extents by supe or supf. Suggest a possible explanation for these result. ANSWER: Some positions in the protein may not tolerate either of the amino acid substitutions, and some positions may tolerate one amino acid substitution but not another. D). (2 points). Some amber mutations, such as Q57Am, are suppressed equally well by supe or supf. Suggest a possible explanation for this result. ANSWER: Some positions in the protein are tolerant of many different amino acid substitutions. Such positions probably do not play a direct role in the function of the protein.

14 8). (12 points). In E. coli dam mutants display a mutator phenotype. That is, they have a higher spontaneous mutation frequency relative to dam + strains. The dam gene encodes an enzyme that methylates the adenine residue in the DNA sequence 5 G-A-T-C-3. dam mutants fail to methylate the adenine residue in that sequence. In addition, dam mutants are killed by 2-aminopurine (2-AP). This is because this adenine analog incorporates into DNA, but is then recognized as a mismatch and is excised, somehow causing formation of lethal double strand breaks in the DNA of the dam mutants (but not in WT cells). A). (4 points). How could you show that the dam mutants have a mutator phenotype using a simple genetic test? [Check for spontaneous StrR, phage resistance, or other easily selectable resistance marker relative to the spontaneous resistance by an isogenic dam + strain.] Glickman and Radman [(1980) P. N. A. S. 77: ] carried out a suppressor analysis of the sensitivity to 2-AP exhibited by dam mutants. When they isolated suppressors, they found that all of the mutations mapped in several genes not linked to dam (don t worry about how this was done). They also found that when they moved the suppressor mutations to a new strain (dam + ) the new strains also had a mutator phenotype. B). (4 points). How did they likely isolate the suppressors of the original dam mutation? [Plate cells on plates supplemented with 2-AP so that resistant mutants are able to grow. (A mutagen is not necessary since this is a selection).] C). (4 points). What is the likely cellular function of the suppressors with respect to their mutator phenotypes? [The mutants likely encode enzymes that are involved in removing mismatches in DNA caused by 2-AP incorporation. Thus, the DNA is not repaired and the double strand breaks do not accumulate. (In fact most of the suppressors mapped in the muth, mutl and muts genes that are involved in mismatch correction. This helped define the proteins that are required for mismatch correction).]

15 9). (9 points). This question will require consultation with the Genetic Code Dictionary which is located on the last page of the Exam. The starting strain has a deletion of the prob gene and is suppressor-free (sup o ). A plasmid that carries the wild type prob + gene makes the strain proline independent. You treat your plasmid with a mutagen like 2 AP and isolated a prob mutant. DNA sequence analysis showed that codon 64, which is normally a tyrosine, is replaced by an amber codon. Next, you introduce a copy of the supd amber suppressor (inserts Serine) and find that the strain is still a proline auxotroph. A). (1 point). What is the likely explanation for this result? [Replacement of tyrosine with serine does not restore activity of the protein (or the suppressor is not efficient enough to produce enough copies of the protein to restore activity.] You next select a Pro + revertant using hydroxylamine (HA) as the mutagen. HA is specific for G:C to A:T transitions. After isolating such a Pro + revertant, you replace the copy of supd with the wild-type sup o allele. You now find that this strain becomes a proline auxotroph. B). (2 points). Could the reversion event affect codon 64? What are two acceptable molecular explanations for the reversion event that produced the HA induced revertant? (A G:C to A:T mutation is not a sufficient answer). [No, because the UAG codon cannot be mutated to a sense amino acid using HA because the UAG to UAA change makes an ochre codon. A second site suppressor in proa that restores activity of the ProA protein with the serine at position 64 (inserted by supd). must have occurred. (The results rule out intergenic suppressors because only the proa gene on the plasmid was mutagenized). Thus the mutation has to be elsewhere in proa. This could be allele specific or non allele specific from the data shown.] C). (4 points). What would you predict would happen of you changed codon 64 back to a tyrosine in the revertant (Hint: Two answers)?

16 [If the interaction between the suppressor amino acid residue and the Ser at codon 64 is allele specific, the cell would be a proline auxotroph when Tyr is prsent at codon 64 because the Tyr and second site suppressor could not interact. Alternatively, if the second site suppressor improved that activity of the protein the Tyr would be acceptable and the cells would be Pro +.] D). (2 points). How could you test your hypothesis by purely genetic methods? [Introduce other known amber suppressors that insert other amino acid residues at codon 64 to see if they are allele-specific or not. This could be tested by looking for growth in the absence of proline.]