Types of Mutations


Four independent mutations were obtained that affect the synthesis or assembly of fimbriae. The properties of the mutations are described in the table below (where + indicates that fimbriae were produced and - indicates that no fimbriae were produced).

  1. Based upon the above results, indicate whether each of the single mutations is a temperature sensitive (Ts) or cold sensitive (Cs) mutation.

    ANSWER: fim-1 and fim-3 are Cs, fim-2 and fim-4 are Ts.

  2. Based upon the above results, what is the predicted order of the mutant gene products in the pathway of fimbriae synthesis and assembly?

    ANSWER: First consider three pairs of double mutants separately.

    Combining these three results leads to the conclusion that the gene products probably act in the pathway in the following order: fim-2 - fim-1 - fim-4


Wild-type Salmonella typhimurium grows on minimal medium plates without addition of fatty acids. Fatty acid auxotrophs require supplementation with a fatty acid (for example, oleate) for growth. Fatty acid auxotrophs are normally rare (less than 10-6 fatty acid auxotrophs are found in a population of S. typhimurium cells).

  1. How would you distinguish fatty acid auxotrophs from wild-type prototrophic cells? [Specify the phenotype of both types of cells.]

    ANSWER: Replica plating on minimal medium + or - fatty acids. Fatty acid auxotrophs will not be able to grow without fatty acids in this screen.

  2. How could you could enrich for fatty acid auxotrophs? [Indicate the relative proportion of auxotrophs in the population at each step.]

    ANSWER: One approach would be to do a penicillin enrichment with growth in minimal medium + fatty acids followed by growth in minimal medium + penicillin without fatty acids, repeated multiple times. Each cycle of penicillin enrichment should result in approximately 100-fold enrichment for the mutant cells relative to the wild-type cells. [Note that any other auxotrophs that accumulated would not be able to grow in minimal medium without fatty acids either -- this is one good reason why you wouldn't want to use rich medium for growth between penicillin steps.]

  3. What is a potential problem with this method and how you could avoid this problem?

    ANSWER: A major problem with this method is that each enrichment culture is likely to contain siblings. This problem can be avoided by doing multiple, independent enrichments and only choosing one colony from each enrichment. [As described in the text, a second problem may be that lysis of cells in the cultures with penicillin could feed any penicillin auxotrophs, allowing their growth (and subsequent death) if the enrichment is not done properly.]


In order to identify new targets for antibiotics, a large collection of temperature-sensitive lethal mutations were isolated in Staphylococcus aureus.

  1. How could you isolate a collection of temperature-sensitive lethal mutations in Staphylococcus aureus (that is, mutants that survive at 30 C but die at 43 C)?

    ANSWER: Chemical mutagenesis (to increase the frequency of point mutations) followed by diluting and plating the cells on rich medium at 30 C (to obtain isolated colonies), and subsequent replica plating on rich medium at 43 C. The desired mutants would grow at 30 C but not 43 C. By using rich medium you can avoid most temperature sensitive auxotrophs.

  2. Is this approach a selection or a screen?

    ANSWER: A screen. [The mutants die at 43 C -- you would have to look through many replica plates to find the desired temperature sensitive mutants.]

  3. Why might these mutants be particularly useful targets for antibiotics?

    ANSWER: Temperature sensitive lethal mutations define essential genes -- that is, genes whose products are required for the bacteria to reproduce. Once you have identified such essential genes, you may be able to design an antibiotic that disrupts the target gene product and would thus prevent the bacteria from multiplying.


Three different proline auxotrophs were streaked on minimal plates without proline and incubated at 22 C, 30 C, or 42 C for two days. The results are shown in the table below:

Double mutant		Growth on minimal medium at:             
			22 C	30 C	42 C  
        
pro-1 pro-2		-	-	-             
pro-1 pro-3		-	+	-             
pro-2 pro-3		-	-	-             

For each strain, indicate if the Pro- mutation is a null mutation, a temperature sensitive mutation, or a cold sensitive mutation.

ANSWER: The null mutant will be inactive at all temperatures. The temperature sensitive mutation will be inactive at 42 C but active at lower temperatures. The cold sensitive mutation will be inactive at 22 C but active at higher temperatures. Thus, when the temperature sensitive and cold sensitive mutations are combined the phenotype strain will grow at 30 C but not at 22 C or 42 C.

pro-1 = temperature sensitive
pro-2 = null (non-conditional)
pro-3 = cold sensitive


Missense mutations in the phoA gene (which encodes alkaline phosphatase) are CRM+.

  1. Indicate whether each of the following types of mutants would produce CRM. Explain your answer. [Hint - CRM is an abbreviation for "cross-reacting material" -- see the discussion of CRM under Mutants and Mutations on the WWW supplement.]

    ANSWER: Missense mutants often result in a protein that is folded much like the wild-type protein and thus often is relatively stable. In contrast, misfolded proteins (due to premature termination or insertion of a block of incorrect amino acids from any of the above types of mutations) are typically degraded rapidly by cellular proteases.

  2. Monoclonal antibodies can also be used to test for CRM. What is a possible disadvantage of using monoclonal antibodies for this purpose instead of polyclonal antibodies? [Hint - A polyclonal antibody preparation contains a population of antibodies where each antibody recognizes a different part or "epitope" of the protein. A monoclonal antibody preparation contains a population of identical antibodies that recognize the same epitope of the protein.]

    ANSWER: A missense mutation may alter a particular amino acid of a protein in such a way that it destroys a single antibody recognition site ("epitope"), possibly preventing recognition by a monoclonal antibody. In contrast, polyclonal antibodies recognize many different epitopes of a protein so changing any single epitope will not prevent recognition by the polyclonal antibodies.


You have two strains of E. coli. One strain contains temperature sensitive trpA mutation and the other contains temperature sensitive RNA polymerase (RNAP) mutation. How could you distinguish between the two strains using a simple test?

ANSWER: Grow the strains on rich medium at 30 C, then replicate onto two plates each of the following agar medium: (1) minimal medium without tryptophan, (2) minimal medium + tryptophan, (3) rich medium (as a control for the efficiency of replica plating). A RNAP(Ts) mutant will grow on all three types of media at 30 C but will not grow on any of the media at 42 C. A trpA(Ts) mutant will grow on all of the media at 30 C, and will grow on the minimal medium + tryptophan and the rich medium at 42 C, but will not grow on the minimal medium without tryptophan at 42 C.


Suggest two methods to isolate a collection of cold-sensitive mutants in Salmonella typhimurium -- that is, mutants that grow at 42 C but do not grow at 30 C.

ANSWER: Although conditional mutations may not grow at the nonpermissive temperature, they often survive short exposure to the nonpermissive temperature. With this hint, consider the three basic approaches for isolating mutants: selections, screens, and enrichments. There is no obvious way of selecting for the desired mutants because the desired mutation is unable to grow under the nonpermissive conditions. It would be straightforward to screen for the desired mutants -- for example, (i) you could plate colonies at 30 C, replica plate the colonies to 30 C and 42 C, then look for colonies that grow at 42 C but not at 30 C, or (ii) you could plate the cells at the nonpermissive temperature for a short time then shift the plates to the permissive temperature -- the mutants usually form smaller colonies due to the effect of the temporary incubation at the nonpermissive temperature. An enrichment would also be a straightforward way to obtain the desired mutants -- for example, by doing a brief penicillin enrichment at the nonpermissive temperature, then washing out the penicillin and growing the cells at the permissive temperature. (This procedure would probably need to be repeated several times to adequately enrich for the mutants, and you would need to do multiple independent enrichments to avoid siblings). Would these approaches work for isolation of cold-sensitive lethal mutants?


Most nonsense mutations in a structural gene for a protein have a null phenotype.

  1. How many different single base substitution mutations in a protein coding sequence can produce a nonsense mutation?

    ANSWER: 19 codons. The possible single base substitutions that can produce a nonsense mutation are show in the table below. Note that stop codons are not found within a protein coding region, so although one stop codon could produce another stop codon by a single base substitution, such changes are unlikely to result in a nonsense mutation within a gene. In addition, some codons may produce either an amber or ochre codon by a single base substitution (for example, UAC may mutate to UAG or UAA).

  2. How many different amino acids are encoded by the codons that can be changed to a nonsense codon by a single base substitution?

    ANSWER: 9 amino acids: Lys, Gln, Glu, Ser, Trp, Leu, Tyr, Arg, and Cys.

Single base substitutions that can yield nonsense mutations

Nonsense mutationOriginal CodonAmino Acid
UAG (Amber)AAGLys
CAGGln
GAGGlu
UCGSer
UGGTrp
UUGLeu
UAAStop1
UACTyr
UAUTyr
UAA (Ochre)AAALys
CAAGln
GAAGlu
UCASer
UGAStop1
UUALeu
UACTyr
UAGStop1
UAUTyr
UGA (Opal)AGAArg
CGAArg
GGAGlu
UAAStop1
UCASer
UUALeu
UGCSer
UGGTrp
UGUCys


The DNA and corresponding amino acid sequence of a portion of the wild-type trpA gene is shown below. The DNA sequence of three different trpA mutations is shown directly below the corresponding region in the wild-type sequence. [The DNA sequence of the coding strand is shown -- i.e., TTG = UUG = Leu.]

209210211212213214215216217218 219 220 221
TTGCAGGGATTTGGTATTTCCGCCCCGGAT CAG GTA AAA
LeuGlnGlyPheGlyIleSerAlaProAsp GlnValLys

*
TAGCAGGGATTTGGTATTTCCGCCCCGGAT CAGGTAAAA
*
TTGCAGGGATTTGGTGTTTCCGCCCCGGAT CAGGTAAAA

del
TTGCAGGGATTTGGTATTTCCGCCCCGATC AGGTAAAA-

Using the codon table, show the amino acid sequence for the protein made in each of the three trpA mutants.

ANSWER: see figure below.

209210211212213214215216217218 219220221
*
TAGCAGGGATTTGGTATTTCCGCCCCGGAT CAGGTAAAA
AMBER

*
TTGCAGGGATTTGGTGTTTCCGCCCCGGAT CAGGTAAAA
LeuGlnGlyPheGlyVALSerAlaProAsp GlnValLys

del
TTGCAGGGATTTGGTATTTCCGCCCCGATC AGGTAAAA-
LeuGlnGlyPheGlyIleSerAlaProILE ARGOCHRE

What is the type of mutation in each of the three trpA mutants in the table?

ANSWER:


Mutagenesis with UV light causes a variety of types of mutations, but the mutagen ICR181 specifically causes frameshift mutations. A large number of Klebsiella aerogenes putP mutations were obtained with both mutagens. About 5% of the putP mutants obtained following UV mutagenesis had detectable CRM, while none of the mutants obtained following ICR181 mutagenesis had detectable CRM.

  1. What does the presence of CRM indicate?

    ANSWER: The presence of cross-reacting material suggests that the mutant phenotype is caused by a missense mutation (i.e. the protein is made and can be detected by antibodies but it is inactive).

  2. Suggest a likely explanation for these results.

    ANSWER: Frameshift mutations cause premature termination of putP and the truncated polypeptide is rapidly degraded by cellular proteases (hence no CRM). In contrast, a small percentage of the mutations that arise spontaneously are base substitution mutations that result in missense mutations (hence are CRM+).


Given the sequence of the beginning of the putP gene shown below, indicate a potential hot spot for frameshift mutations.

ATGGCTATTAGCACACCGATGTTGGTGACATTCTGTGTCTATATTTTTGGCATGATATTGATGGGTTTATCGCCTGGCGCT 
CAACCAAAAACTTTGATGACTATATTCTTGGCGGTCGCAGTCTGGGGCCGTTTGTTACGGCTTTATCAGCCGGCGCGTCGG 
ATATGAGCGGCTGGCTGTTAATGGGGCTGCCTGGCGCTATCTTTCTGTCGGGGATCTCTGAAAGCTGGATCGCCATTGGCC 
TGACGTTAGGCGCATGGATTAACTGGAAGCTGGTGGCCGGGCGCCTGCGCGTGCATACCGAATTTAACAATAACGCGCTCA 
CGCTGCCGGACTATTTTACCGGTCGGTTTGAGGATAAGAGCCGAGTCCTGCGTATTATTTCCGCGCTGGTCATTCTGTTGT 
TTTTCACT

ANSWER: Any run of identical base pairs is a potential hot spot for frameshift mutations due to strand slippage during DNA replication. A few examples are underlined in the sequence below.

ATGGCTATTAGCACACCGATGTTGGTGACATTCTGTGTCTATATTTTTGGCATGATATTGATGGGTTTATCGCCTGGCGCT
CAACCAAAAACTTTGATGACTATATTCTTGGCGGTCGCAGTCTGGGGCCGTTTGTTACGGCTTTATCAGCCGGCGCGTCGG
ATATGAGCGGCTGGCTGTTAATGGGGCTGCCTGGCGCTATCTTTCTGTCGGGGATCTCTGAAAGCTGGATCGCCATTGGCC
TGACGTTAGGCGCATGGATTAACTGGAAGCTGGTGGCCGGGCGCCTGCGCGTGCATACCGAATTTAACAATAACGCGCTCA 
CGCTGCCGGACTATTTTACCGGTCGGTTTGAGGATAAGAGCCGAGTCCTGCGTATTATTTCCGCGCTGGTCATTCTGTTGT
TTTTCACT


The aceA gene product is required to grow on acetate as a sole carbon source. Temperature sensitive (Ts) mutants that affect this gene were desired, but it is not possible to directly select for aceA mutants. Therefore, revertants of an aceA mutant that are AceA+ at 30 C but AceA- at 42 C were isolated.

How would you obtain the Ts revertants? (Note that the original mutation is not Ts -- you are looking for revertants that have a Ts phenotype!)

ANSWER:

  1. Select for AceA+ revertants by plating on minimal acetate plates at 30 C.
  2. Replicate the colonies onto minimal acetate plates incubated at 30 C and minimal acetate plates incubated at 42 C.
  3. Pick any colonies that grow on minimal acetate plates at 30 C but not 42 C and recheck their phenotype. Note an essential control is to make sure these are not simply temperature sensitive lethal mutations by checking for growth on some other carbon source at 42 C.

Why is this approach easier than simply screening for aceA(Ts) mutants?

ANSWER: Intragenic pseudorevertants are often temperature sensitive. Since there is a direct selection for revertants, it is possible to obtain the revertants without heavy mutagenesis, so the cells are not likely to have secondary mutations (which might have an unrelated temperature sensitive lethal phenotype). Thus, by looking for revertants first then scoring for temperature sensitivity, many fewer colonies will need to be screened to find the desired mutants.

The Ts revertants were grown at a permissive temperature, then isocitrate lyase activity (the aceA gene product) was assayed at a non-permissive temperature. The cells retained high levels of isocitrate lyase activity. Suggest a potential explanation for this result?

ANSWER: It is likely that the mutation is temperature sensitive for synthesis (Tss), but once the gene is translated and the protein folds it remains active at the nonpermissive temperature.


The biosynthesis of an amino acid in Serratia marcescens requires the synthesis of five enzymes encoded by genes V, W. X, Y, Z. You have isolated five mutants which cannot synthesize the amino acid. Next, you assay the enzymatic activities of the V, W, X, Y, Z in extracts made from each mutant. Table 1 shows the results of your assays. You also treat each of the mutants with a mutagen (or grow the cells in the absence of a mutagen as a control for spontaneous mutants) to determine the reversion properties of each mutant. The results are shown in Table 2.

  1. What is the probable order of the structural genes V, W, X, Y, Z? What is the direction of transcription?
  2. What can you say about the specific nature of the mutation present in each mutant (to a high degree of certainty) and the mechanism by which it affects the expression of the genes?
  3. You have just isolated a new mutant. All five enzymatic activities are at 0.1 to 0.2. What is the nature of the new mutation? Where would you predict this mutation would map?

Table 1. Enzymatic activity in wild-type and mutants.

MutantIntracellular Level of
VWXYZ
Wild type1010101010
10.10.210010
210100010
3101010100
40.10.100.110
500.1101010

Table 2. Effects of mutagens on reversion frequency.

MutantNumber of Revertants/108 cells
2APICRUVEMSNONE
17351752
200000
350220386
423164
500000


Which of the following types of mutations usually result in a null phenotype and why?

ANSWER: Nonsense, frameshift, deletion, and insertion mutations usually result in null mutants because the mutant gene product is severely disrupted and mutant protein products are often degraded by proteases. Missense mutations can also result in null mutations -- for example, a missense mutation that causes an amino acid substitution in a critical residue of the active site of a protein may result in complete loss of enzymatic activity. However, missense mutations may also result in a partially active ("leaky") gene product, a conditional phenotype, or a "silent" change that does not disrupt protein function.


What is the difference between a genetic selection and screen?

ANSWER: In a genetic selection only a cells with a specific phenotype will grow, cells that lack this phenotype will be unable to form colonies on the selective medium. Thus a genetic selection makes it possible to isolate a rare mutant from a large population of cells -- for example, rare His+ cells (prototrophs) can be selected from a population of His- mutants (auxotrophs) by demanding growth on minimal medium without histidine, rare streptomycin resistant cells can be selected from a population of streptomycin sensitive cells by demanding growth on rich medium with streptomycin, rare phage T1 resistant cells can be selected from a population of T1 sensitive cells by demanding growth on medium with phage T1, etc. In a genetic screen all cells can grow but it is possible to distinguish cells with one phenotype from cells that lack the phenotype. Because both types of cells grow in a genetic screen, it is much harder to identify rare mutants using this approach -- for example, Lac- cells can be distinguished from Lac+ cells because the Lac- cells form white colonies on indicator plates containing X-gal while Lac+ cells from blue colonies on this medium, His- cells can be distinguished from His+ cells because the His- cells cannot grow on minimal medium without histidine but can grow on minimal medium with histidine, while the His+ cells grow on both media.


Compare and contrast genotype vs phenotype.


Using the codon table, determine how many different types of mutations could arise from a single base substitution in a proline codon?

ANSWER: Seven different amino acids:



Please send comments, suggestions, or questions to smaloy@sciences.sdsu.edu
Last modified October 18, 2004