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

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

Transposition Summary
Mutagenesis in vitro
Effects of Mutations
Plasmids and

F Factor


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

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

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Sample problems part 2

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

  1. Question:

    You have isolated a large number of independent NTG-induced Arg- mutants and you examine those that are mutated in the argB region where there is known to be only a single gene involved in arginine biosynthesis. You perform complementation analysis with these mutations and find that they all fail to complement one another except for a single pair (argB17 and argB56) that, in a Rec- merodiploid, give an Arg+ phenotype. The controls all work: argB17/argB17 and argB56/argB56 are both Arg- and argB17/arg+ and argB56/arg+are Arg+.

    a. Propose a simple model for these results, making sure to mention what the result says about the structure of the protein product.

    b. If you isolated a large number of Tn5-induced argB mutations, approximately what fraction of these would you expect to complement either argB17 or argB56 and why?


    a. Since the problem states that you examined a "large number" of independently derived mutants, it is clear that this intragenic complementation is rare, suggesting that it must involve very specific protein-protein interaction, rather than a case of two different functional domains that happen to be linked on the same peptide chain. The strong suggestion is that the gene product functions as a multimer of (at least some) identical subunits.

    b. Few if any, since the transposon-induced mutations ought to produce a severely damaged product, not the sort of item that would be able to rescue either of these mutations by protein-protein interactions. If intragenic complementation had been more frequently observed in the original analysis, suggesting that you were dealing with a gene product with two distinct functional domains, then it might be possible that most or all transposon mutations in the downstream domain were capable of complementing point mutations in the upstream domain.

  2. Question:

    You are given two E. coli strains, UW1 and UW2, that have the following genotypes: UW1 is wild-type except for a recA1 mutation while UW2 contains the recA1 mutation as well as a particular F' that carries (in addition to all the F factor genes) about 10% of the E. coli chromosome as shown below.

    [Macro error: Can't locate an image object named "recA1".]

    You expose each strain to a chemical that inhibits an essential cell function and find that UW1 gives no spontaneously resistant colonies when 108 cells are plated. UW2, however, yields 5 colonies resistant to the chemical when similarly tested. You also notice that the mutation causing resistance always maps to the region made diploid by the F'.

    a. Assuming the results are repeatable and statistically valid, propose a simple model that explains why UW2 yields resistant colonies and UW1 does not. Remember that they are genotypically identical except that UW2 is merodiploid for a portion of the genome.

    b. Upon further analysis UW1 is found to yield resistant colonies, but at a frequency of 3 x 10-11. Propose a mechanism for the generation of these resistant mutants.

    c. Briefly state whether any of the following mutagens would likely stimulate the frequency of resistant mutants (using UW2) and why: 2-aminopurine, ICR-191, and Tn5.


    a. The fact that a particular class of revertants only appear in a merodiploid background is consistent with a number of schemes, but the simplest (and the one strongly suggested by the fact that "the mutation causing resistance always maps to the region made merodiploid") is that resistance is "recessively lethal". This implies that when you alter the target gene product to cause resistance (perhaps producing an altered protein that degrades the chemical), your little friends are dead because you lack the wild-type version of the gene product. Only when there are two copies of the region can one copy mutate to resistance without death (the mutation is recessive to wild-type allele in its lethality and dominant to the wild-type allele in its "resistance"). An alternative, but unlikely, explanation posits that, to be resistant, you need an interaction of both mutant and wild-type gene products, thus only a merodiploid strain will give rise to resistant colonies.

    b. This frequency is roughly 10-3 of that seen in the merodiploid, a clue that duplications might be involved. By either argument in "a", these mutants can only survive when diploid for the appropriate region and duplications satisfy that requirement. The number of 10-11comes from 10-3 of the population mutating to resistance at about 5 x 10-8, the same as the F' strain.

    c. Regardless of your arguments for "a" and "b", the frequency of resistance is appropriate for the alteration, not destruction, of the activity of the affected gene product. Therefore 2-AP and not ICR or Tn5 would likely stimulate appearance of the resistant mutants.

  3. Question:

    Briefly tell why the following planned experiments are impossible (and not just bad ideas):

    a. Fred plans to make a map of all his argH mutations using the technique of 2-point mapping. He intends to use no other mutations of any kind in his analysis.

    b. Fred also has an F-prime carrying that part of the E. coli chromosome encoding leuA+ and trpC+. He intends to mate this prime into another strain of E. coli that has point mutations in both leuA and trpC and determine the linkage of the two mutations by 2-point mapping.

    c. Enrico knows that the trpC and leuX genes are very close together on the E. coli chromosome and he has a specialized transducing phage that carries that region (and therefore the wild-type version of each of these genes). He wants to determine the 2-point linkage between the trpC3 and leuX47 mutations in another strain, so he infects that (doubly auxotrophic) strain with his specialized phage, selecting Trp+ and scoring Leu+.


    a. Two-point mapping requires dissimilar phenotypes for the two mutations being tested. This is nontrivial in the case of two Arg- mutants. Of course, if it was known that some responded to intermediates in the arginine biosynthetic pathway, then you might be able to map some versus others. For example, consider a case where a strain with arg-1 could use argininosuccinate as an arginine source while arg-2-containing strain could not. Transducing phage grown on an arg-1-containing strains could be used to transduce a strain with arg-2 to growth on minimal medium plus argininosuccinate. Such a cross selects for the wild-type allele of the gene affected by arg-2, but the presence or absence of arg-1 is unselected. The linkage of arg-1 and -2 would be indicated by the frequency with which the resultant colonies failed to grow on minimal medium alone, reflecting their acquisition of the arg-1 mutation. Notice that the cross cannot be done in the other direction (arg-2 phage into an arg-1 recipient), since the recipient would grow: There is no selection for gene transfer or recombination.

    b. You will have no selection for recombination since both versions of the region are already associated with a replicon. Because of that, they will complement and you will have no way of seeing if recombination even occurs.

    c. Since the specialized phage can integrate into the chromosome by its own int system, there is no selection for recombination in the trpC-leuX region. The integrated genes will complement the mutations originally in the recipient so that linkage will appear to be 100%.

  4. Question:

    Below is a map of a given gene showing the known deletion mutations. (The map is not to scale since it is based only on the "+" and "-" responses from deletion mapping.)

    delmap1 picture

    The deletion mutations divide the gene into 7 regions (I - VII). It is found that all of the 10 known missense mutations map in region III.

    a. Suggest two unrelated reasons that might explain this observation.

    b. No nonsense mutations (of any kind) and no frameshift mutations are found in region III although thirty of each type are distributed elsewhere in the gene. How does this help you decide between your answers above?


    a. (i) Region III might encode a critical region of the protein (in fact, THE critical region of the protein) so that only in this region would a high percentage of amino acid substitutions be detected by causing a mutant phenotype. (ii) Since it is a genetic map based on the circular "use of point mutations to define the ends of deletions and deletions to define the position of points," coupled with the possible non-randomness of deletion endpoints, the relative physical size of each deletion "group" is unclear: It could therefore be that region three is huge relative to the others:


    delmap2 picture

    In this case it would be statistically likely that most missense mutations map in region III.

    b. Since the detection of nonsense and frameshift mutations is efficient (i.e. they have such a deleterious effect on protein structure that they will nearly always be detectable as mutants), the position of these mutations can be an indication (if you have enough of them) of the physical size of the region because they will largely occur with a frequency appropriate to the target size and will not be limited to critical regions of the protein. Thus, if they are rare in region III, then that region is likely to be physically small (or at least not huge) and explanation (i) seems likely.

  5. Question:

    Eight arginine-requiring mutants have been isolated. All map in the same general region of the chromosome. Using the generalized transduction data below, construct a fine-structure map of the region. In the table a "+" indicates that recombinants were seen and a "-" indicates that they were not. You also know that mutations arg-1 through -6 were spontaneously derived while arg-7 and -8 are due to Tn10 insertions.

    4-+- ++

    a. Draw a map, indicate ambiguities with parentheses. (HINT: Find the deletions first! These are the mutations that fail to recombine with two or more other mutations that do show recombination between each other.)

    b. Which mutations might cause a temperature-sensitive (ts) phenotype? (i.e., might be missense mutations?)

    c. Which mutations are definitely not deletions?

    d. How many genes, involved in arginine biosynthesis, are encoded in this region?



    argmap picture

    b. In general, deletions and insertions would not be expected to cause ts gene products, so that arg-2, -3 and -4 are candidates. (If you argue that the loss of the gene product confers a ts phenotype because the product is necessary for growth at higher temperature, then all the mutations would cause a ts phenotype. If the question asked "which mutations might cause a ts gene product", the latter answer does not work. Such an argument is also a bit far-fetched in the case of a gene involved in arginine biosynthesis.)

    c. Data such as this can indicate that a mutation is a deletion, not that it is not. You therefore do not know that arg-2, -3, and -4 are point mutations, you simply know they "have not been shown to be deletions". Only the Tn-induced mutations are definitely not deletions. (OK, OK, even these might have deletions associated with them in some way.)

    d. You cannot say more than "at least one". This is mapping data, and you require complementation data to discern complementation groups which are, to a first approximation, genes.

  6. Question:

    You have isolated a collection of mutants affected in the biosynthesis of a vitamin. You carry out both recombination and complementation tests on all possible pairwise combinations of these mutants. The recombination data are presented below:

    4-+- ++++

    "+" indicates recombinants are detected for the cross

    "-" indicates no recombinants are seen

    a. Using these data alone, draw the best genetic map you can. Indicate uncertainties for which the data are insufficient to be more precise.

    The complementation results are indicated below:

    2-+-- +++-+

    b. Using this additional data draw the best possible genetic map. Taken together, the recombination and complementation data permit a unique map to be drawn.

    c. What can you say about:

    -the minimum/maximum number of genes?

    -the minimum/maximum number of cistrons?

    -the number and direction of transcripts?

    d. How would the answers to the above 3 questions change if mutation 9 failed to complement 3,6,7, and 8 (in addition to 1,2,4,5 and itself)?


    a. Find the deletion first: mutations 1,4, and 7 are clearly deletions by the criteria of failing to recombine with a number of mutations that themselves recombine with each other. Since 1 and 4 fail to recombine, they overlap. Now just put the points under the appropriate interval. Note that there are a number of ambiguities: 5 could be anywhere except under the deletions, 7,8,10 are of uncertain order with respect to anything else, the entire orientation of the region covered by 1 and 4 is unclear. Uncertainties are indicated by parentheses.

    map22a picture

    In the above drawing, the position and orientation of each of the regions in the square brackets is undefinable with the information considered so far. Similarly, the order of the mutations in the two small parentheses is unclear (with respect to other mutations).

    b. The complementation data suggests three complementation groups. This is easiest to see when you consider just the "point mutations" (these are not necessarily point mutations, it is just that there is no indication that they are deletions or anything else!): 5,9, and 2 fail to complement; 6,3 and 8 fail to complement; and 10 is by itself. Since each of the deletions extends into two complementation groups, you can order the mutations (i.e. resolve the ambiguities of the deletion mapping) to the following:

    argcomp picture

    where the heavy vertical lines separate the three complementation groups. Note, there is no necessary correlation between the apparent size of any of these regions and physical reality, e.g. the distance between 2 and 6 could be very large or very small.

    c. You cannot say anything with certainty about the number of genes. It is likely (barring intragenic complementation or non-essential genes etc.) to be the same as the number of complementation group. The minimum number of cistrons or complementation groups is three. There is no maximum because it could be argued that you have not looked at enough mutations and have thus missed a group at either end or even, albeit unlikely, in the middle. Nothing can be said about transcripts.

    d. With the altered information that 9 fails to complement 3,6,7 and 9: you still cannot address the number of genes with certainty, but since 9 fails to complement the left two groups but succeeds with the third, a minimum of two is fairly assured. The number of cistrons does not change, because there are still two mutations (2 and 5) defining the left group and three (6,3, and 8) defining the middle. It is almost inescapable that there is transcription from the left proceeding through the middle group and it is likely that the right group is separately transcribed in an unknown direction:

    argtran picture

    (Note: You would never approach a genetic system this way, i.e. use complementation data to "solve" problems or uncertainties in mapping data. You would really want the mapping analysis to stand independently.)

  7. Question:

    You have isolated the dnaZ gene, whose product is involved in DNA replication and want to overproduce the protein product. The dnaZ gene is on a 2.0-kb BamHI fragment and you have an expression vector with a single BamHI cloning site just downstream of a high level promoter (see figure). You mix the dnaZ fragment with the cut vector, ligate and transform (selecting drug-resistance on the plasmid). You get a large number of transformants all having the 2.0-kb insert. However, they are all oriented the "wrong way" -- the dnaZ is not under control of the promoter. What do you suppose is happening and how might you solve the problem?

    dnazclone picture


    Either you are very unlucky, or there is a selection against the desired class. A likely explanation is that the overexpression of dnaZ is a lethal event (perhaps due to interference with normal replication). A solution is a "tunable" promoter where the promoter allows low expression (and therefore does not kill) until an external signal is supplied. Such a signal might involve a small molecule that "induces" the removal of a repressor or high temperature in the case of a temperature-sensitive repressor. With such a system, the desired strain could be constructed and grown to high cell density and then synthesis of the desired, lethal product could be triggered.

  8. Question:

    You mutagenize E. coli with nitrosoguanidine and find a mutant lacking detectable activity of enzyme X. Biochemical characterization of the enzyme X from the mutant shows that it differs from wild-type protein by a single charge-change, suggesting to you that the mutation is a missense mutation (adding or subtracting a charged amino acid to the primary sequence of the protein). What model can you suggest when you then isolate several Tn10-containing mutants with the following properties: (i) They also lack enzyme X activity; (ii) the Tn10 mutations are highly (greater than 98%) linked to the NTG-induced mutation (by P1 phage) and fail to complement it for enzyme X activity; (iii) the enzyme X proteins from these Tn10-containing mutants are identical in charge to that of the NTG-induced mutant; (iv) the Tn10 mutations are capable of recombining at low frequency to produce a wild-type strain.


    Since all of the mutations fail to complement and have identical effects on enzyme X, it is reasonable to assume that they affect the same gene. That the affected gene is not the structural gene encoding enzyme X is shown by the fact that the Tn10 mutation has so little effect on the gene product. The fact that the Tn10 mutations are not genetically identical (they recombine), yet have the same effect on the protein rules out the obscure possibility that the insertions were all very near the 3' end of the gene, yielding the slightly altered product. Most reasonably, both mutations eliminate an enzyme that makes a post-translational modification of enzyme X that is necessary for its activity and that alters its charge (a phosphorylation, for example). By this model, the transposon and point mutations are either in or very polar onto the gene encoding the modification function.

  9. Question:

    You want to study the regulation of the groA gene (whose gene product function is unknown). You have a strain with a deletion of the chromosomal groA gene and containing a low copy- number plasmid with a groA:lacZ transcriptional fusion (the strain therefore lacks a functional groA gene). While b-galactosidase activity in this strain is low, spontaneous mutants appear with high activity. In these mutants, you replace the fusion plasmid with a similar one carrying groA+ (using an incompatible plasmid with a different, selectable drug-resistance), and analyze the level of groA product. Rationalize the following possible results.

    • Result a. The groA product is present at a very high level.

    • Result b. You detect approximately wild-type levels of the groA gene product.

    • Result c. Drug-resistant transformants of the second plasmid are hard to get and, when analyzed, all make little or no groA+ product.


    • a. This is the expected result. You have chosen mutants with altered regulation of the groA fusion, so it is not surprising that the groA+ gene shows the same altered regulation. This result is only obtained if the regulatory mutation occurs on the chromosome (for it to affect both plasmids).

    • b. The mutation causing high expression was on the groA::lacZ plasmid. Thus, replacing this plasmid with a groA+ one also removes the "regulatory" mutation.

    • c. Apparently the high-level expression of the groA+ product makes the cell sick (dead?). The introduction of the second (groA+) plasmid is therefore a double selection: drug-resistance and sub- lethal expression of groA+. This is most easily accomplished by rearrangements of the plasmid, affecting the groA gene. Such altered plasmids will be somewhat rare, which is why the transformants are rare.

  10. Question:

    A strain containing an amber mutation in leuA reverts to prototrophy at 30° at about 10-6. Upon further characterization, some of the revertants have curious properties at 42°:

    a. What would you say if some were Leu- auxotrophs at 42°? Are they internal or external suppressors (i.e., does the compensatory mutation map internal to the originally mutated leu gene)?

    b. What if some were dead at 42° even on rich media? Are they internal or external suppressors?


    a. You do not know. All you know is that you have made a temperature-sensitive protein almost certainly because you have corrected the nonsense mutation but caused the wrong amino acid to be put in. This can occur either by a mutation in the codon of the amber mutation or by generating an informational suppressor that inserts a non-wild-type amino acid.

    b. They are external since now you have a temperature-sensitive lethal (i.e. the mutation causes death even on rich media) and therefore it is probably due to something outside of the leu system: almost certainly an altered tRNA that becomes a problem for the cell at high temperature (for any number of reasons).

  11. Question:

    You have isolated a strain containing an extremely efficient (greater than 90%) amber suppressor and you store it away in a nutrient agar stab for later use. A while later, when you recover the strain from storage, you find the suppressor is still present (as detected by the successful plaquing of phage containing characterized amber mutations), but it is much less efficient than before (roughly 10% efficient). What happened?


    A suppressor that is that efficient would make the cell sick by fouling up proper protein termination at a reasonable frequency. Even as the strain sits in the nutrient agar stab, there is slow growth and a selection for a "happier" phenotype. In this case that meant a decrease, but apparently not the elimination, of the suppressor activity. You are therefore doing a selection for, and getting, secondary mutations that reduce the amount or effectiveness of the suppressor.

  12. Question:

    You have an E. coli strain with an amber mutation in proA. You "shotgun clone" (clone random DNA fragments) DNA from a Bacillus strain into a pBR322 derivative (multi-copy, capable of replication in E. coli, and confers Ampr) and transform the E. coli strain, selecting Pro+.

    Result a. You find a number of Pro+ colonies that fall into three classes: class I and II are Ampr, but the cloned chromosomal DNA of each class fail to hybridize with the other; class III colonies are Amps. Propose an explanation for the three classes.

    Result b. You find no Pro+ Ampr colonies, though you do succeed in getting prototrophic transformants for several other E. coli auxotrophs tested with the same set of clones. Suggest an explanation for the lack of Pro+ clones.


    a. Both class I and II have Bacillus DNA that causes a Pro+ phenotype but it is not likely that you have cloned two versions of the proA gene of Bacillus or the two classes would hybridize. If one class actually is proA the other class probably encodes a suppressor of the E. coli proA amber mutation. Class III strains contain no plasmid (Amps) and probably are spontaneous revertants of the amber mutation.

    b. While there are many possible technical reasons for such an experiment to fail, the success with other markers rules out most of these. Among the possibilities: (i) the normal promoter for the Bacillus proA gene may not function in E. coli. Such a result is not unusual, especially when the cloned DNA is from a microorganism unrelated to the transformed strain. This is sometimes overcome by cloning into an "expression vector" where promoters are so positioned on the vector that transcription proceeds into the cloned DNA. (ii) If the library was generated by restriction enzyme digestion of the Bacillus chromosome, multiple sites for that enzyme in the proA gene would result in the cloning of only part of the gene. A solution to this problem is the use of another enzyme.

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