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Table of Contents
Why Do Genetics
Genetic Terms
More Terms
Basic Molelcular
Biology

More Basic Concepts
Screens
Selections
Mutation Frequency
Chemical Mutagenesis
Frameshift Mutation
DNA Repair
Mutation Summary
Detecting Mutants
Complex Mutation
Insertion Sequences
Compound Transposons
Complex Transposons
Models of
Transposition

Transposition Summary
Mutagenesis in vitro
Effects of Mutations
Complementation
Plasmids and
Conjugation

F Factor
Transformation
Transduction
Generalized
Transduction

Specialized
Transduction

Complementation
Mapping
Two Factor Crosses
Deletion Mapping
Other Mapping Methods
Strain Construction
Inverse Genetics
Gene Isolation
Characterization of
Clones

Sequence Data
General Approaches
Fusions
Supression
Final Summary
Problem Set 1
Problem Set 2


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Sample problems part 1

©2000 written by Gary Roberts, edited by Timothy Paustian, University of Wisconins-Madison

The following problems are not very "methodologically current", as few if any deal with cloning and sequencing and all their related technologies. Nevertheless, they do address more fundamental issues of the nature of genes, polarity, and most importantly, the interpretation of frequencies from any molecular genetic experiment. I don't believe that there at too many errors in here, but if you catch any, or things any f the wording herein is unnecessarily obscure, please Email me at groberts@bact.wisc.edu. I believe that I am the actual author of all of these (in the specific), but they are all some form of progeny of the fertile mind of John Roth, currently at the University of Utah, and reflect the approach he used in a molecular genetics course in the early '70's at Berkeley.

Gary Roberts

  1. Question:

    You have generated a Trp- mutant using the mutagen 2-aminopurine. It reverts spontaneously at a frequency of 10-8. Which of the following mutagens might you expect would increase this reversion frequency and briefly why? NTG, 5BU, ICR-191, EMS, proflavine.

    Answer:

    2-Aminopurine causes base substitution mutations and, typically, you would need another base substitution mutation in order to "revert" a base substitution mutation. NTG, 5BU, and EMS all cause base substitutions and could be expected to increase the reversion frequency. It would be quite unlikely for frameshift mutagens like ICR and proflavine to stimulate reversion of a base substitution mutation. The low frequency with which it reverts spontaneously suggests that a specific amino acid may be required at the site of the protein affected by the mutation. It is therefore conceivable that some of the base substitution mutagens (including 2-AP itself) might not be able to induce the particular base change required because of their mode of action.

  2. Question:

    You plate 108 cells of an E. coli mutant to minimal medium containing either histidine or leucine and then put a few crystals of ICR-191 and 2-AP on opposite edges of both plates as indicated in the figure. After several days incubation at 25°C, you see the following result:

    icr2ap picture

    a. What is the nature of the mutations in the original mutant in general terms (requirements, type of DNA alteration)?

    b. What would you expect to see if you plated the culture on medium containing both histidine and leucine?

    c. If you picked one colony from the histidine-containing plate, grew it up in liquid and plated 108 cells to an identical pair of plates as shown above (plus the mutagens) what result should you see?

    d. Being a careful scientist, you repeat the first experiment exactly to confirm the result. Unfortunately you make a little mistake and incubate these plates at 42°C (instead of 25°C) and find that one of the two plates now shows confluent growth (a dense lawn of growth). Which plate do you suppose might give that result and why wouldn't such a result be expected on the other plate? (You know that wild type does fine at 42°C.)

    Answer:

    a. Since the strain reverts to growth on each medium at a frequency appropriate for the reversion of a point mutation of some sort, there is only a single mutation preventing growth under each condition. If the reversion pattern (number and distribution of colonies on the plate) was the same on each plate, then the easiest explanation would be "there is a single mutation having nothing to do with histidine or leucine biosynthesis". However, since the plate appearances are very different, they cannot reflect the reversion of the same mutation. There are therefore 2 mutations in the strain, only one of which displays a mutant phenotype on each medium. It then follows that the supplied nutrient (histidine or leucine) must be satisfying the requirement caused by the other mutation. Therefore, there is a leu mutation that is due to a frameshift (since reversion to leucine independence is stimulated by ICR, a frameshift mutagen) and a his mutation due to a base substitution (reversion to His+ is stimulated by 2-AP, a mutagen causing base substitution mutations).

    b. Confluent growth; since the strain is His- Leu- .

    c. Such a colony would be Leu+ since it was selected on medium lacking leucine and His- since there has been no selection for His+ and the likelihood of an "unselected" colony being coincidentally His+ is 10-7. Therefore on the histidine-containing plate, you would see confluent growth. The leucine-containing plate would look identical to that in the figure on the original problem since the original his mutation will behave the same.

    d. Temperature-sensitive mutations typically involve the alteration, not elimination, of a gene product. Such an alteration would be due to a base substitution mutation (in this case, a missense), not a frameshift mutation. Therefore, you predict that the leucine-containing plate would be confluent, since it is the his mutation that is a base substitution mutation.

  3. Question:

    108 cells of a mutant bacterial strain are spread on each of several plates. The appearance of these plates after two days incubation is indicated below:

    Temperature
    Medium in plate40°30°
    MinimalNo growth50 colonies
    Minimal plus histidine10 coloniesConfluent growth
    Complete (rich) medium10 coloniesConfluent growth

    a. Suggest a reasonable genetic constitution of the strain that will account for this behavior.

    b. Describe a simply testable prediction based on your explanation.

    c. How would your story change if the data at 30° on minimal were changed to "no growth"?

    Answer:

    a. By the arguments in sample problem 2, there are more than one mutations in the strain, but there are conditions (30° or presence of histidine) where only 1 mutation displays a mutant phenotype. Confluent growth at 30° in the presence of histidine indicates that neither mutation has a mutant phenotype under those conditions, while the failure to see colonies on minimal medium at 40° indicates that both mutations have a mutant phenotype so that a vanishingly rare (10-7 x 5x10-7 = 5x10-14) double event in required for growth. We therefore have a temperature-sensitive lethal (it dies even on rich medium at 40°) that reverts at 10-7 and a his mutation that reverts at 5 x 10-7.

    b. Any of the "10 colonies" found on the 40° plate will, upon retest, prove to be His-; any of the "50 colonies" from minimal at 30° will retest as temperature-sensitive at 40° on all media, since each of these conditions only selects for the reversion of one of the mutations.

    c. It would simply indicate that the his mutation was stable (i.e., does not revert at a detectable frequency).

  4. TRICKY QUESTION!! THINK CLEARLY AND CAREFULLY!!!

    You plate 108 cells of strain UW375 onto minimal medium and put crystals of nutrients and mutagens on the edge of the plates as shown below, with the following result:

    hisleu picture

    a. Based on these results, how many mutations do you think UW375 has? What can you say about these mutations (reversion, phenotype)?

    b. What nutrients would you add to the 42° plate to make sure of getting confluent growth?

    c. If you pick one of the single colonies from the 30° plate, grow it up in rich liquid medium at 30°, and plate 108 cells onto the identical conditions as shown, what do you expect to see and why?

    Answer:

    a. At least 2: (1) a his mutation that reverts at 10-7 and (2) a temperature-sensitive mutation of unknown requirement and unknown reversion frequency. At 42°, apparently both mutations show their phenotype, so the lack of colonies does not necessarily mean that the temperature-sensitive mutation fails to revert. Alternatively, you could say the his mutation is cold-sensitive, displaying its phenotype only at 30° while the temperature-sensitive mutation does not revert (but is this likely??).

    b. You cannot know with certainty. The ts mutation could be in a biosynthetic pathway (like arg, trp, etc. or even his or leu) whereupon the strain would require that supplement plus histidine. However, the ts mutation could also be in an essential function so that the strain would die on any medium at the non-permissive temperature.

    c. The revertant strain will certainly give confluent growth on minimal at 30° since that is what it was picked from. Appearance of the 42° plate is unknowable: it depends on the reversion frequency of the ts mutation. Any stimulation of reversion by ICR and 2-AP again depends on the nature of the mutation causing the ts phenotype. (Since a ts phenotype is often the result of a protein with a single amino acid change, 2-AP would be expected to stimulate reversion. It is conceivable, on the other hand, that the strain is ts not because the affected gene product is ts but because the affected gene product is necessary for growth at 42°. In the latter case the ts mutation could be a frameshift, insertion, deletion, or anything else.)

  5. Question:

    You isolate a large number of mutants, by the procedures noted below, that are resistant to a given drug. In each case, assume it is known that all selected mutations lie in a single gene. Below are listed several possible situations that you might encounter. Indicate what you would conclude in each case about the nature of the gene product and about the means of achieving drug resistance.

    Possible Result A:

    Acridines (which induce single base additions and deletions) do not induce drug-resistant mutants. None of the drug-resistance mutations induced by other mutagens is an amber. Several base analogue-induced mutants are found that can grow at 30° in the presence of the drug, but fail to grow at 40° on any medium.

    Possible Result B:

    Among spontaneously derived drug-resistant isolates, those containing frameshift and nonsense mutations are very frequent but several with missense mutations are also found.

    Possible Result C:

    Drug-resistant mutations are induced by all mutagens tested, including acridines and base analogues. Some of the acridine-induced mutations cause a "leaky" (partially resistant) phenotype and others cause a temperature-sensitive (resistant at 40° but sensitive at 30°) one.

    Answer:

    Result A: Since frameshift and amber mutations are not found among resistant isolates, it argues that you cannot get a resistant phenotype by the "knockout" of a gene. That fact, coupled with the presence of ts-lethals, suggest that to be resistant you need to modify an essential function in the cell, without destroying it.

    Result B: The presence of frameshift and nonsense mutations indicates that it is the loss or destruction of a gene product function that causes the phenotype of interest. The paucity of missense mutations (that must occur more frequently but are apparently not being detected) suggests that they are not often tight enough (destructive enough of the product function) to give the required phenotype, whereas frameshift and nonsense mutations are.

    Result C: The ability of all manner of mutagens to stimulate the frequency of resistant colonies suggests that it is a loss of function that is demanded. The leaky phenotype due to frameshifts is curious and unusual: one possibility is that the gene product is an RNA (e.g., tRNA or rRNA) so that a frameshift might not be as completely destructive as it would be in the case of a protein product.

  6. Question:

    Assume that you know, from previous biochemical work, that there are 3 gene products necessary to make Blubb, a requirement for growth in the bacterium you are studying. For ease we will call these genes blbA, blbB, and blbC. Following NTG mutagenesis of a wild-type strain:

    a. You pick up 100 independent Blb- mutants that you subsequently find they assort as follows: 95 in blbA, 3 in blbB and 2 in blbC. Propose a reasonable explanation for this distribution of mutations.

    b. What would you propose if the results were 50 in blbA, 50 in blbB, and none in blbC?

    c. How about 60 in blbA, 20 in blbB and 20 in blbC?

    d. What if you got the results in a), but the isolates all came from a single mutagenized culture; i.e. they were not independent?

    Answer:

    a. NTG is not sufficiently specific in its choice of target sites to explain such a dramatically nonrandom result. Among the possibilities:

    -The gene product of blbA is limiting in the metabolism of the pathway; blbB and blbC gene products are significantly in excess. Thus most missense mutations in blbB and C will not have a sufficiently deleterious effect to be seen as Blb- .

    -Alternatively, blbB and C products are required for another pathway, so that their loss makes you "more" (i.e. a more striking phenotype) than just Blb-. Rare mutations might destroy only the Blb function of the products.

    -blbB and C are regulated such that they can be derepressed more if the organism is not making enough Blubb, whereas A expression is either fixed or at least does not respond nearly as much to this stimulus. Thus only very tight mutations are detected in blbB and C.

    b. (Here you need to explain the lack of blbC mutations; we will assume there is nothing about the equal number of blbA and B mutations that needs explaining.)

    -Again, BlbC might be an integral part of another pathway so that its loss leaves you with a phenotype not recognizable as Blb- .

    -The strain may contain a duplication of blbC, thus any mutation that eliminated one copy of the gene would not be detected because the strain would remain Blb+ .

    -There could be another protein in the cell that could perform the biochemical function of BlbC well enough so that the absence of BlbC would not be detected.

    c. Perhaps blbA is a little bigger, but this result is really pretty close to random. At least it is close enough that no firm or striking interpretations can be made.

    d. This result is essentially uninterpretable. Since you have not taken care to choose independent isolates, you may just be dealing with the same isolate again and again (siblings) that all resulted from a spontaneous mutation that occurred before the mutagenesis.

  7. Question:

    You have isolated a His- mutant with the following properties: it reverts to His+ spontaneously at 10-7 and this reversion frequency is increased by exposure to ICR-191, but not to EMS (an alkylator). When the reversion is performed at 25°, 90% of the His+ revertants turn out to be His- at 42°.

    a. Suggest a simple model for the nature of the original mutation and the curious revertants.

    b. What do you predict the spontaneous reversion frequency would be if the selection for His+ was done at 42°? How might you explain a frequency of 10-6?

    Answer:

    a. The original mutation is a frameshift since it is stimulated to revert with a frameshift mutagen. Some of the revertants have an apparently wild-type genotype (although this point is not necessarily true, nor crucial for the answer - more properly, they have an "apparently wild-type" phenotype). The funny revertants are certainly not due to a restoration of the wild-type genotype since they display a mutant phenotype, at least under some conditions. They are likely due to a compensatory frameshift such that the non-wild-type protein sequence encoded between the two frameshifts causes a temperature-sensitive protein product.

    b. The easy answer is 10-8 since 10% of the revertants found at a frequency of 10-7 grow at 42°. This is actually a minimum estimate, since it is also possible that a different "second site" exists (on the other side of the original mutation??) that generates a protein that only functions at high temperature, thus being missed in the reversion at 25°C. The 10-6 frequency would be explained by such a second-site reversion coupled with the added assumption that this other site (giving only cold- sensitive revertants) is a really good target for frameshifts, e.g., GGGGGGGGGGG, and therefore such frameshifts are frequent.

  8. Question:

    a. A new drug, Buffamycin, is toxic to wild- type E. coli at levels above 1 mg/ml. Mutants resistant to this level of Buffamycin (a phenotype abbreviated Bufr) appear spontaneously at a frequency of 10-8 (1 colony appears when 108 cells are plated) and the frequency of such mutants is increased by exposure to NTG but not by exposure to ICR-191. What sort of mutation do you expect you need to achieve a Bufr phenotype and why? Suggest both the sort of alteration you would require in the DNA as well as its effect on the protein product.

    b. Plating 109 unmutagenized, wild-type E. coli on 2 mg/ml Buffamycin yields NO resistant colonies. Given that result, it is not surprising that when a similar number of any of the Bufr mutants (from part "a" above; they are resistant to 1 mg/ml already) are plated onto 2 mg/ml Buffamycin, resistant colonies do appear very frequently (10-3 frequency) and these are called Bufr2. Why isn't it surprising that the Bufr cells fail to grow on 2 mg/ml Buffamycin?

    c. In what way do you suppose the Bufr2 strains differ from the Bufr strains? What easily observable property could you look for?

    Answer:

    a. The low frequency of spontaneous appearance and the stimulation by a mutagen (NTG) that causes predominately base substitution mutations both argue that Bufr is due to the alteration (in a very specific way) and not the destruction of a gene product. Since ICR does not stimulate reversion, the genetic change is likely a base substitution mutation.

    b. Since you see no colonies resistant to Buf at 2 mg/ml when you screen 109 (given in the first part of section "b"), then resistance to that level of Buffamycin clearly occurs at less than 10-9. In part "a" you only screened 108 cells for resistance to 1 mg/ml, so you did not screen enough to have found one resistant to the higher level. In other words, the experiment has already been done: resistance to 2 mg/ml clearly requires a different event than resistance to 1 mg/ml and the frequency of the former has been shown by a separate experiment to be less than 10-9.

    c. The frequency of Bufr2 arising from Bufr is too high for most mutation types. The numbers (10-3 frequency) strongly suggest that the Bufr2 class is arising by a duplication of a particular region of the genome of the Bufr cells. Apparently the regulation of the gene product necessary for Buffamycin resistance is such that a double dose of the gene yields a double dose of the gene product (this is by no means typical) leading to a doubling of the tolerated Buffamycin concentration. The easiest property to look for is instability: The Bufr2 cells should be unstable due to loss of one copy of the duplication resulting from homologous recombination, giving rise to Bufr cells at about a 1% frequency.

  9. Question:

    You are interested in an amino acid analogue that kills wild-type cells when added to the growth medium. It kills by being incorporated into protein and frequently causing those proteins to be nonfunctional. Mutants resistant to the analogue occur spontaneously at 10-5. You analyze 1000 independent, spontaneously derived analogue-resistant mutants and find 998 are due to mutations in gene anaA, while the other two have mutations in anaB, which is located in another part of the chromosome.

    a. Propose an explanation for the rarity of mutations in anaB relative to those in anaA. (i.e., What sort of alteration in each gene product is necessary to produce the desired phenotype?)

    b. Based on your answer in part "a" if you now generate 1000 independent analogue- resistant mutants following mutagenesis by a transposon (known to be random), what fraction of those mutants will fall in anaA? Why?

    Answer:

    a. anaA mutations are occurring frequently (10-5), which is appropriate for the knockout of the gene; apparently the loss of anaA product yields resistance. anaB mutations are about 103 times less frequent (2/1000 Ana- mutants are anaB-), appropriate for a very specific alteration, but not the knockout, of the gene product. Either this is because the anaB product is essential and therefore its loss is lethal (but its alteration is not) or because it is the altered product that is necessary for resistance (for example, maybe the altered anaB product degrades the analogue whereas the wild-type product does not).

    b. Insertion mutations will destroy the product of the affected gene and therefore you will see no analogue resistant mutations in anaB; they will all be found in anaA.

    (A more devious alternative can also be suggested: anaB exists in a tandem duplication and you need its loss to be analogue resistant. anaB mutants of the desired phenotype will only be found after the duplication has "fallen apart" (at a frequency of 10-2 - 10-3) and therefore be that much rarer than mutations detected in anaA. By this argument, part b would be answered: 998 in anaA and 2 in anaB (more or less) since you still need to get rid of the duplication of anaB in order to "see" anaB mutations.)

  10. Question:

    You are studying the following operon: a gene whose product is necessary for leucine biosynthesis, leuD, is transcriptionally upstream of a gene whose product is necessary for isoleucine biosynthesis, ileA.

    You mutagenize wild-type cells with NTG and find that 20% of your independent Leu- mutants are affected in leuD (the other 80% are affected in other leu genes). In a separate experiment, 25% of independent, NTG-induced Ile- mutants have mutations in ileA. Following Mu mutagenesis (an essentially random transposon), 25% of your Ile- are again found in ileA but surprisingly none of the Mu-induced Leu- are affected in leuD! (Assume you isolated 100 mutants in each hunt.)

    a. Propose a simple explanation for the lack of Leu- insertion mutations in leuD.

    b. What results would you expect following ICR mutagenesis? (i.e., Will you get Leu- mutations in leuD)?

    Answer:

    a. The failure to detect a particular class of mutations (e.g., Leu- insertions in leuD) will be due to one of two possibilities: They did not occur or they did not have the expected phenotype and therefore you missed them. In this case, you know that Mu is random (the problem states it!) and therefore mutations in leuD must have occurred with Mu mutagenesis. How might you have missed them? The suggestion that leuD is essential (and therefore Mu insertions in leuD are missed because they are lethal) is a nice try, but does not work because the NTG-induced mutations are common and this result is not appropriate for an essential gene. This assumes that the other leu genes do not have an essential function so that the comparable frequency of Leu- mutations in the various leu genes argues they are comparably able to give rise to Leu- auxotrophy by loss or alteration of product function. If leuD encoded an essential function, NTG-induced mutations, which altered the product enough for a Leu- phenotype but not enough for death, might be found there. The notion that insertions in leuD would be polar onto an essential function does not work since ileA is downstream and insertions in that gene are not lethal. Mu insertions in leuD are, however, polar onto ileA which means that they will be Leu- Ile- . This is not the phenotype that was sought (you looked for Leu- ) and therefore these mutants were missed.

    b. Most frameshift mutations in leuD will be polar onto ileA and therefore be missed for the reason described above. However, at least some will not be very polar and you will see a Leu- Ile+ phenotype. This will cause a reduced number of leuD mutations relative to those found in other leu genes (which do not have this funny polarity problem).

  11. Question:

    You isolate a spontaneous ochre (UAA) mutation in the leuB gene and find it reverts spontaneously at 10-8. Mutagens that cause frameshift or transversion mutations have no effect on that frequency while a base-analogue mutagen, which causes only transition mutations, greatly increases the reversion frequency. What amino acid do you expect to find at the affected position in the wild-type protein?

    Answer:

    Since the frequency of reversion is rather low, it suggests that only a very particular base change restores wild-type function. This fact suggests that the product requires a specific amino acid at the mutant site. Since the reversion frequency is stimulated only by mutagens that cause transition mutations, the acceptable (and therefore, in all probability, the wild-type) amino acid must be encoded by a codon that is a transition away from UAA. Since two of these, UAG and UGA, are also stop signals, they cannot be right; it must be CAA, which encodes glutamine.

  12. Question:

    You have isolated a phage- resistant mutant of a given strain by plating the bacteria onto a phage-seeded lawn (soft agar), picking survivors (appearing at 10-7), and retesting to verify phage resistance. After using the strain a bit for other matters (and without single colony isolation), you happen to plate the culture for single colonies and get two size classes, only the smaller of which remains phage resistant. What's going on?

    Answer:

    Apparently, the mutation that causes resistance to the phage coincidentally causes poorer cell growth. (This is not at all far fetched: Very often mutations to phage- or drug-resistance alter such a basic element in the cell that normal growth under the best of conditions is impaired.) When the strain was subsequently grown repeatedly in the absence of the selection for phage resistance, a different selection was being performed for normal growth and therefore for phage sensitivity. Thus the smaller colonies are the phage resistant slow growers and the faster growers have restored at least the wild-type phage sensitivity and growth properties presumably, but not certainly, by the restoration of the wild-type genotype.

  13. Question:

    You have a low copy number, conjugative plasmid carrying the argABC operon in an E. coli strain that is deleted for the entire argABC region as well as being Rec- . You want to isolate Tn10 insertions in the plasmid's argABC region and toward that end you mutagenize the strain with a nonreplicating phage vector carrying the transposon and generate tons of Tetr colonies (i.e., you get many Tn10 insertions into either the plasmid or the chromosome). How might you most easily look for the class you want: insertions in the argABC cluster on the plasmid? You can assume you have any other bacterial strains you want to help you look for the mutants.

    Answer:

    argsch12 picture

    (1) You have demanded transposition into a replicon somewhere in the recipient by infecting the Rec-, plasmid-containing strain with the transposon vector (a phage in this case) and demanding tetracycline resistance. This will, as the problem states, give you lots of drug-resistant colonies all of which are due to transposition of the Tn10 from the vector into a replicon in the recipient, but only some of which have transposons in the plasmid of interest.

    (2) This latter class is detected by pooling the tetracycline-resistant cells (scrape the plate of Tetr colonies and resuspend the cells in a liquid medium) and mating the plasmid into another recipient, namely one which is tetracycline-sensitive, resistant to another drug (to which the "donor" is sensitive, providing a counter-selective marker), and deleted for the argABC cluster (this latter property is not necessary here but will be useful in a later step). The mating is performed on rich medium containing tetracycline plus the other drug. The only colonies that appear will be those of the "recipient" that have received versions of the original plasmid carrying the Tn10. If your plasmid was not conjugatible but mobilizable by another plasmid, you could do the same experiment by introducing this latter plasmid to generate a strain capable of donating the plasmid of interest. If the plasmid was neither conjugatible or mobilizable, you could perform a plasmid preparation on the pooled Tetr cells and transform a Rec- recipient to Tetr. Only transposons on the plasmid will be detected here.

    (3) Only a fraction of these will have insertions in the argABC region. These can be detected by testing on minimal medium and minimal + arginine. Since the chromosomal arg genes are deleted, strains that grow on minimal must carry a nonmutated argABC region on their plasmid and these are discarded. The Arg- strains carry the desired plasmid, resistant to another drug (which the "donor" is sensitive to, to provide a counter-selective marker), and deleted for the argABC cluster (this latter property is not necessary here but will be useful in a later step). The mating is performed on rich medium containing tetracycline plus the other drug. The only colonies that appear will be those of the "recipient" that have received versions of the original plasmid carrying the Tn10. If your plasmid was not conjugatible but mobilizable by another plasmid, you could do the same experiment by introducing this latter plasmid to generate a strain capable of donating the plasmid of interest. If the plasmid was neither conjugatible or mobilizable, you could perform a plasmid preparation on the pooled Tetr cells and transform a Rec- recipient to Tetr. Only transposons on the plasmid will be detected here.

  14. Question:

    Where is the crucial technical flaw in the following experiment?

    George wishes to generate a large number of Tn5-induced auxotrophs (Tn5 encodes kanamycin resistance) in a rifampicin-resistant E. coli. He proceeds by mating this strain with a Rifs Rec- donor carrying the Tn5 on an F', selecting Rifr Kanr. He gets lots of colonies of that phenotype and he screens these for auxotrophs. (The donor is also Rec- and the gene conferring Rifr is not in the region made merodiploid by the F'.)

    Answer:

    As described, the donor will move the Tn5-containing F' to the recipient with high frequency, producing lots of RifR KanR cells. Only at low (10-5) frequency will Tn5 transpose into the chromosome and only about 1% of those will cause auxotrophy by random insertion. The problem is that the vector, the F', can replicate in the recipient so there is no selection for transposition. You want a vector that fails to either replicate or integrate (e.g., specialized phage are also unacceptable for that reason) so that the only KanR colonies seen must have arisen because of transposition events.

  15. Question:

    You receive a manuscript to review for the Journal of Bacteriology wherein the authors suggest that they have found a mutant of phage P22 that is capable of transducing all tested chromosomal markers (of Salmonella) at a frequency 10X that of P22HT. Do you accept the paper or not? If you accept, what step in transduction do you think they have "improved" in this mutant? If you reject, state why?

    Answer:

    The frequency of transduction is a function of 3 factors (which can be multiplied together to figure the final efficiency): Frequency of transducing particles among phage particles; frequency of transducing particles carrying a given region of interest among total transducing particles; and the frequency of a given region being recombined into the recipient's chromosome among total copies of that region carried by the transducing particles. The last factor cannot be improved by a phage mutant (since there are typically no phage products inside the recipient cell when a transducing particle "infects" a cell). The second factor, which, for a randomly packaged transducing phage like P22HT, is essentially the fraction of a cell's chromosome that the phage can carry, is not likely to be changed in a mutant because that would involve changing all sorts of protein-protein interactions involved in building a phage. The only factor that can be improved by a phage mutation is the first one, which is already 0.5 for P22HT. Since this cannot be increased by a factor of ten (a frequency approaching 1 is the maximum since you need some "real" phage for propagation of phage) then the proposed mutant is theoretically impossible. REJECT!!

  16. Question:

    Given the following linear array of mutations causing leucine auxotrophy (determined by mapping) where mutations leu-1, leu-3, leu-5 and leu-7 were induced by a base-substitution mutagen and leu-2, leu-4, leu-6 and leu-8 by a transposon.


    leu-1 leu-2 leu-3 leu-4 leu-5 leu-6 leu-7 leu-8

    You do all the pairwise combinations using F-primes and Rec- chromosomes and get this map

    plasmid
    leu+ leu-1leu-2(Tn)leu-3leu-4(Tn)leu-5leu-6(Tn)leu-7leu-8(Tn)
    chromosomeleu+ +++++++++
    leu-1+--+ +++++
    leu-2(Tn)+----++++
    leu-3++--- ++++
    leu-4(Tn)++---++++
    leu-5+++++----
    leu-6(Tn)+++++----
    leu-7+++++----
    leu-8(Tn)+++++----

    What are your conclusions about

    1. minimum/maximum number of complementation groups

    2. minimum/maximum number of genes

    3. minimum/maximum number of transcripts/organization of transcripts

    Draw the simplest model you can of the complementation groups and transcripts in the region. Can you draw a more complicated model?

    Answer:

    1. (at least) 3 complementation groups (leu-1, leu-2) (leu-3, leu-4) (leu-5, leu-6, leu-7, leu-8).

    2. You can make no conclusive statement about the number of genes.

    3. The complementation group with mutations (leu-1, leu-2) is transcriptionally upstream of the complementation group with mutations (leu-3, leu-4). The complementation group with mutations (leu-5, leu-6, leu-7, leu-8) is transcribed separately from the other two.

    4. Simplest model:

    leusimtrans picture

    A much more complicated but consistent model: (to emphasize that the data only gives minimum estimates).

    leucomtrans picture

    (Where complementation groups G,I & J encode non-essential functions. No mutations in E simply because you have not isolated enough mutations to have hit all the genes. Mutations leu-6,-7 & -8 must be polar.)

    Lessons:

    1. You can always make more complicated models.

    2. You need a map to make sense out of complementation data.

    3. You do not detect mutations in non-essential genes unless they are polar onto essential ones.

    4. You do not detect mutations in all essential genes if you have not isolated enough mutants.

  17. Question:

    You are working with the truAB region of the E. coli chromosome and you generate 4 tru mutations in separate strains. By genetic sleight of hand you also move the mutant tru region of each of these strains onto F primes carrying the truAB region. This allows you to construct all possible merodiploids (in a Rec- background, of course). Here are the results:

    Chromosomal genotype
    plasmidtru+ tru-1tru-2tru-3tru-4
    tru+ +++++
    tru-1+----
    tru-2+--+-
    tru-3+-+-+
    tru-4+--+-

    (assume all mutations are point mutations and that tru-2 is in truB)

    a. Assuming truA is transcriptionally upstream of truB and that these are the only genes of interest in the region, suggest a reasonable gene assignment for each mutation, noting whatever assumptions you need to make about the specific properties of the individual mutations.

    b. How would your answer change if the result of the tru-4 by tru-4 complementation analysis was "+" (!!?)?

    c. You find a new mutation in the region, tru-5, that fails to complement tru-1, -2, and -5 but is Tru+ in complementation with tru-3, tru-4, and tru+ . Provide an explanation.

    Answer:

    a. The data indicate that tru-2 and tru-4 complement tru-3 but not each other. Since the problem stated that tru-2 is in truB, then apparently so is tru-4 while tru-3 is in truA. tru-1 is a puzzle since it is negative with all the other mutations in complementation. It is not due to negative complementation since it is recessive to tru+, therefore it must be a polar mutation in the upstream gene that the problem defines to be truA.

    b. If the tru-4/ tru-4 merodiploid is Tru+, it suggests (but does not prove) that tru-4 is just barely Tru- so that 2 doses of that gene (and therefore its product) cause you to be Tru+. It also means that since one of the controls of the experiment failed, you should be cautious in your interpretation of any of the merodiploids involving tru-4.

    c. Either (i) tru-5 is in truB and shows intragenic complementation with truB4 or (ii) tru-5 lies in yet another gene, (truC??) in the same transcript downstream of truAB. In the latter case, tru-5 would complement tru-3 and -4 because they are in different cistrons and are apparently not polar onto truC. The failure of tru-5 to complement tru-1 and -2 would be because these two mutations are polar onto truC. (tru-1 is known to be polar from the above analysis; tru-2 was not known to be polar, but it was not known NOT to be polar either, since there was not a downstream function to look for.)

  18. Question:

    You have found 100 independent Arg- mutants following NTG mutagenesis and have shown they all map within the argABC operon. When these strains are made Rec- and an F' argABC+ plasmid is mated in, all the resulting merodiploids are Arg+ except one: the argB33/F'arg+ strain remains Arg- .

    a. What, if anything, can you say about the product of the mutant (argB33) gene?

    b. You know that the haploid (argB33) strain reverts spontaneously to Arg+ at 5 x 10-8. With approximately what frequency do you expect the argB33/F'arg+ merodiploid reverts to Arg+ and why?

    c. You manage to move the argB33 mutation to the plasmid used above (the plasmid is now genotypically argA+argB33argC+) and then make a merodiploid that is arg+ /F'argB33. Not surprisingly, this strain is Arg- . What frequency do you expect Arg+ revertants to appear and by what mechanism?

    Answer:

    a. Since the mutant version is trans-dominant to the wild-type allele, it must make a product that is interfering with the wild-type products to cause an Arg- phenotype. If the mutant product was interfering with some other pathway in the cell, it would not cause an Arg- phenotype.

    b. This merodiploid should revert at approximately 10-5 since the required event is the destruction of the "interfering" product of the argB33 allele.

    c. This merodiploid need only lose the plasmid carrying the trans-dominant argB33 allele to become Arg+ , therefore revertants should appear at about 10-2. Mutations in the argB33 gene will also occur as in "b", but these will be 103 less frequent and therefore not be noticed.

  19. Question:

    Two leu mutations are known that map in the same general area of the chromosome. These mutations cause a requirement for leucine. The following data is collected: (assume all strains are Rec- ).

    CellsDoubling time on minimal medium (no leucine)
    wild type40 minutes
    leu-1no growth
    leu-2no growth
    merodiploid leu-1/F'leu-2300 minutes

    a. List possible explanations for this observation. See if you can come up with three distinct explanations.

    b. Which of the explanations would you prefer if you were told that both mutations were induced by proflavin (a mutagen that causes single base additions and deletions)?

    Answer:

    a. (i) The mutations are in different co-transcribed genes and the one in the upstream gene is partially polar onto the downstream gene so that neither version of the region makes the amount of the downstream gene product necessary for a wild-type phenotype. (ii) The mutations are in the same gene. The gene product is apparently a "multimer" and the mutant products partially correct one or the other by protein-protein interaction (intragenic complementation). (iii) Both mutations are in the same gene and make slightly functional products. The merodiploid has therefore two "doses" of these slightly functional products, bringing the total activity up to a level sufficient for a "leaky" phenotype. Note that the appropriate control, the mutants versus themselves, was not performed. This model would predict leu-1/F'leu-1 would also show slow growth, for example. (iv) Both mutations are in different genes but one of the mutations may be rather trans-dominant. The appropriate controls, mutants versus wild-type, were not performed. This model predicts that one of these 2 controls would be Leu- .

    If trans-dominance is so difficult to have in bacteria, why is it so common in eukaryotes? I think it is because altered phenotypes are much more detectable in the latter organisms, because of the wealth of easily observable properties. As a consequence, relatively small degrees of trans-dominance are detectable in eukaryotes, but typically missed in prokaryotes.

    b. Choice (i). The other explanations demand that the mutant proteins possess some activity that is not likely for the product of a frameshift-mutated gene. The polarity posited by (i) is quite reasonable for frameshift mutations.

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