Mutation Frequency


When grown on MacConkey-Lactose plates, Lac+ cells from red colonies and Lac- cells form white colonies. How could you screen for strains with a mutator phenotype using MacConkey-Lactose plates? [Hint: you would want to start with a Lac- mutant. Why?]

ANSWER: "Mutator phenotype" indicates that the desired mutant strains exhibit an enhanced spontaneous rate of mutagenesis due to lesions in one or more of the DNA repair genes (ones that you should know about are: mutD which results in loss of proof-reading and mutSLH which results in loss of mismatch repair). The key to this question is realising that the Lac- colonies will appear white on the MacConkey lactose medium thus allowing rare red areas within a colony (papillae or sectors) to be easily identified. So what are these red areas and how do they get there? The mutation causing the Lac- phenotype is REVERTABLE to Lac+ at a VERY low frequency because the mutant allele (point mutation) requires a hit to a specific base pair so that the wild type gene function is restored. Lac+ colonies appear red on the MacConkey medium so if a strain exhibits a high level of spontaneous mutation (e.g. a mutator phenotype), it will accumulate mutations after beginning to grow on the medium. If a reversion occurs early in the life of the colony, the daughter cells within the colony will be numerous and form a sector, if the reversion occurs later, fewer daughter cells will arise before the colony stops growing and so a small spot or papillus arises. This idea is analogous to the "jackpot" concept in the Luria-Delbruck fluctuation test.


When mutD strains are growing fast (e.g., in rich medium) the mutation is much higher than when they are growing slowly (e.g., in minimal medium). Explain why.

ANSWER: More DNA replication occurs during fast growth than during slow growth conditions. During rapid growth, initiation of DNA replication occurs more often, resulting in an increased number of DNA replication forks. (DNA polymerase does not work any faster, there are just more DNA replication occuring simultaneously.) In a mutD strain, the epsilon subunit of DNA Pol III is no longer functional, preventing proof-reading by DNA polymerase and thus resulting in the accumulation of misincorporated nucleotides. However, the rate of misincorporation per nucleotide synthesized is the same under both fast and slow growth conditions.The different mutagenesis frequency between fast and slow growing mutD strains arises from the fact that any error correction in these mutants must occur as a POST-REPLICATIVE process via the mismatch repair system. During fast growth conditions mutD strains have so many more replication forks that the ability of the mismatch repair system to handle all of the error correction is saturated (i.e. a second round of DNA replication occurs before error correction can take place efficiently).


The dcm methylase adds a methyl group to the second cytosine in the sequence CCAGG in E. coli K-12. Cytosines in this sequence within the lacI gene are hot spots for transition mutations (Coulondre and Miller. 1977. J. Mol. Biol. 117: 577-606). Suggest two genetic ways to determine if the high mutation frequency at this position is due to the 5-methyl cytosine.

ANSWER:

  1. Assay the mutation frequency in a dcm mutant.
  2. Assay the frequency of lacI mutations in a mutant with an altered dcm recognition sequence (for example, mutated to CAAGG).


What is the difference between mutation rate and mutant frequency? [Give a general description of what experiments you would need to do to determine the two values.]

ANSWER: The mutation rate is the number of mutations per cell division. Because the cell population is so large, the number of cell divisions is approximately equal to the number of cells in the population (N). The value of h can be determined by a fluctuation test.

The mutant frequency is simply the ratio of mutants / total cells in the population. This can be determined by simply plating out aliquots of a culture and counting the number of mutants that arise and the number of cells plated. The mutant frequency is much easier to measure but may show large fluctuations depending upon when the first mutation appeared in the population.


Briefly explain the expected results for the Luria-Delbruck fluctuation test if mutants arise:

  1. spontaneously in a population of cells before exposure to a selective agent;
  2. due to adaption after exposure to the selective agent.

ANSWER: If the mutations arise spontaneously (as in "a") there will be wide fluctuation in mutants from cells grown in different tubes (i.e., independent samples) relative to cells grown in the same tube. That is, there will be great variation in the number of mutants in each individual tube relative to the mean number of mutants calculated from the sum of all of the tubes. In contrast, if the mutations arise by adaption (as in "b") there will be relatively little fluctuation in mutants from cells grown in different tubes (i.e., independent samples) relative to cells grown in the same tube. That is, in this case the number of mutants on each plate approximately equals the mean number of mutants.


Wild-type Salmonella typhimurium cells are killed by phage F0. A Luria-Delbruck fluctuation test was done to determine the rate of mutation to resistance to phage F0. Twenty tubes of medium were inoculated with S. typhimurium cells and the cultures were grown to 108 cells/ml. A 0.1 ml sample of each culture was then spread on plates with phage F0. The results are shown below. Using the equation for the Poisson distribution, determine the rate of mutation to F0R.


A Luria-Delbruck fluctuation test was done to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures grown to 109 cells / ml. A 0.1 ml sample of each culture was then plated on minimal medium to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate to AztR in mutations / cell division.

ANSWER:

Number of tubes with no AztR resistant mutants = 5 / 20 total
h = - ln [5 / 20] = 1.39 mutations per tube (average number of hits)
a = h / N, where a = mutation rate
N = number of cells = 109 cell / ml x 0.1 ml = 108 cells
a = 1.39 / 108 = 1.39 x 10-8 mutations / cell division.


Much of our early understanding of how genes encode proteins came from Seymour Benzer's elegant fine structure genetic analysis of the rII locus of phage T4. He observed that the frequency of mutations at some nucleotides ("hotspots") is much higher than at other sites. For example, he never observed mutations at 129 particular base pairs within the rII locus, but at one particular base pair he found 517 independent mutations. To determine if this observation was significant, he used the Poisson distribution. Benser's results are reconstructed in the following table [From Proc. Natl. Acad. Sci. USA 47: 403-415]:

  1. Based upon the above data for all of the sites, what is the average number of hits per site (that is, the value of h from the Poisson distribution)? [Hint - this problem is somewhat different than the calculation of mutation rates from Luria-Delbruck fluctuation experiments, but the idea is the same -- to determine whether or not an event is random. To calculate P0 you need to divide the number sites with zero hits by the total number of sites in the target (that is, the sum obtained by adding up the number of sites columns).

  2. Based upon the value of h calculated above, what is the predicted probability of having 517 hits in a site? [Hint: solve the Poisson equation for P517.] How does this compare with the observed results?


Wild-type E. coli is unable to grow on the fatty acid decanoate (Dec-). A Luria-Delbruck fluctuation test was done by plating 108 cells from 10 independent cultures onto decanoate plates, and at the same time plating 108 cells from a single culture onto 10 decanoate plates. The results obtained are shown below.

Based upon these results, would you conclude that the mutation to Dec+ is random or adaptive?

ANSWER: The variance between the number of Dec+ colonies from individual cultures equals that seen with multiple samples of a single culture. These results suggest that the phenotype is not likely to be due to a random, spontaneous mutation. (A random mutation which would show much greater variance between individual cultures.) Thus, the simplest interpretation of these results is that the mutation to Dec+ is an example of "adaptive mutagenesis".


A Luria-Delbruck fluctuation test was done to determine the rate of mutation to Dehydroproline resistance (a toxic proline analog) in E. coli. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures grown to 5 x 109 cells / ml. A 0.1 ml sample of each culture was then plated on minimal medium to detect DHPR mutants. The results are shown in the following table. Calculate the mutation rate of the DHPR mutants in mutations / cell division.

ANSWER:

Number of tubes with no DHPR mutants= 6/ 20 total.
h = - ln [6/20} = 1.2 mutations per tube (average number of hits)
a = h/N, where a= mutation rate
a = 1.2 / 108 = 1.2 x 10-8 mutations / cell division.


E. coli strain B is killed by phage T4. A Luria-Delbruck fluctuation test was done to determine the rate of mutation to resistance to phage T4. Twenty tubes of medium were inoculated with E. coli B and the cultures were grown to 109 cells/ml. A 0.1 ml sample of each culture was then spread on plates with phage T4. The results are shown below. Using the equation for the Poisson distribution, determine the rate of mutation to T4R.

ANSWER:

Number of tubes with no T4R mutants = 6 / 20 total
h = - ln [6/20} = 1.2 mutations per tube (average number of hits)
a = h/N, where a = mutation rate
a = 1.2 / 108 = 1.2 x 10-8 mutations / cell division


Wild-type E. coli cells are very sensitive to phage T1. Luria-Delbruck used the fluctuation test to determine the rate of mutation to resistance to phage T1 (Luria and Delbruck.1943. Genetics 28: 491). Twenty tubes of medium were inoculated with E. coli cells and the cultures were grown to 108 cells/ml. A 0.1 ml sample of each culture was then spread on plates with phage T1. The results are shown below.

Culture ## T1 resistant mutants
1 1
2 0
3 3
4 0
5 0
6 5
7 0
8 5
9 0
10 6
11 107
12 0
13 0
14 0
15 1
16 0
17 0
18 64
19 0
20 33

Using the equation for the Poisson distribution, determine the rate of mutation T1 resistance expressed as mutations/cell/division.

ANSWER:

Number of tubes with no T1 resistant mutants = 11 / 20 total
h = - ln [11 / 20] = 0.60 mutations per tube (average number of hits)
a = h / N, where a = mutation rate
N = number of cells = 108 cell / ml x 0.1 ml = 107 cells
a = 0.60 / 107 = 6 x 10-8 mutations / cell division.


When E. coli K-12 cells grown on long chain fatty acids (C18) as a sole carbon source are subcultured into the same medium, soon after subculturing the cells begin to double exponentially until they reach lag phase. In contrast, when these cells are subcultured into medium with a shorter fatty acid (C10) as a carbon source, there is a lag of about 40 hours before the cells begin to reproduce. When the cells that grew on C10 are then re-subcultured into medium with C10 as the carbon source, they begin to double without a long lag phase.

How could you determine if the cells that grew on C10 after the 40 hour lag phase are due to mutants that arose in the population or a slow physiological adaption required for cells to this carbon source?

ANSWER: If the cells that grew on C10 are rare mutants capable of using this carbon source that arose from the population, these mutants should retain the ability to grow on C10 after repeatedly subculturing in medium with long chain fatty acids as the carbon source. If the cells that grew on C10 are simply physiologically adapted to this carbon source, then after subculturing these cells in medium with long chain fatty acids as the carbon source, they would have to re-adapt to growth on C10 so there would be a long lag before the cells began to double (as originally observed).


A Luria-Delbruck fluctuation test was done to determine the reversion rate of a histidine auxotroph. Twenty tubes of rich medium were each inoculated with a few auxotrophic cells and the cultures grown to 108 cells/ml. A 0.1 ml sample of each culture was then plated on minimal medium to detect His+ revertants. The results are shown below. Based upon these results, calculate the reversion rate of the his mutants in reversion events/cell division.

ANSWER: 5 out of 20 tubes contained no revertants ("mutants" for our purposes). So, we can set up the equation to measure the reversion rate as if it were a calculation of the mutation rate. Thus:

h = -ln [5/20] = 1.386
a = h / N
N = 108 cells / ml x 0.1 ml = 107 cells = approximately 107 cell divisions
Reversion rate = 1.386 / 107 = 1.386 x 10-7 reversions / cell division


Proline transport mutants arise spontaneously in a culture of Salmonella at a frequency of 10 per cell. Azetidine resistant mutants arise spontaneously at a frequency of about 10-5 per cell.

  1. If these are two independent events, what is the probability that a single cell will acquire both mutations?

    ANSWER: 10-6 x 10-5 = 10-11

  2. If 100 proline transport mutants are tested and all of them are also Azetidine resistant, how can you explain the observed frequency of each type of mutant?

    ANSWER: If all of the proline transport mutants are also Azetidine resistant, then proline transport mutants account for 10-5 / 10-6 = 10% of the Azetidine resistant mutants. Thus, there must also be other types of mutations that can also result in Azetidine resistance.


The frequency of spontaneous mutants affecting a gene is typically about 10-6 per cell. Spontaneous mutants that are resistant to a high concentration of streptomycin are rare (about 10-9 per cell). What is a likely explanation for this result?

ANSWER: The low frequency of streptomycin resistant mutants suggests that the target for mutations with this phenotype is smaller than expected for a simple null mutation in a gene. However, the frequency is much higher than you would expect for a phenotype requiring two independent mutations (10-6 x 10-6 = 10-12). This suggests that streptomycin resistance requires a very specific mutation within a gene. [In fact, streptomycin resistance requires a specific base substitution within a gene for one of the ribosomal proteins.]


DHP is transported into the cell by proline permease and incorporated into proteins, resulting in defective proteins and subsequently cell death. Spectinomycin binds to ribosomes, inhibiting protein synthesis, and ultimately resulting in cell death. Following treatment with the intercalating agent ICR-191, the frequency of DHPr mutants is about 10-4 per cell but the frequency of Spc mutants is about 10-9 per cell. Suggest a reason for the difference in mutation rates to DHPr vs Sprr and propose a genetic experiment to test your idea.

ANSWER: ICR is an intercalating agent and thus causes frameshift mutations which usually result in a null phenotype. The relatively high frequency of DHPr mutants suggest that this gene is not essential because it is readily disrupted with frameshift mutations. In contrast, the low frequency of Spcr mutations suggest that ICR does not stimulate mutations in this gene, possibly because the gene is essential -- the few mutants obtained are probably due to spontaneous mutations that do not completely disrupt the gene. A test of this idea would be to compare the frequency of spontaneous Sprr mutations with the frequency of ICR generated Sprr mutations.


(The following question is courtesy of Kelly Hughes, Univ. Washington)

A picture of your house is shown above. You are the only person with a key to open the door. (The door is reinforced and cannnot be opened by force.)

  1. What is the selection that only allows you in?
  2. What screen(s) can you use to keep out the burglars and open the door only for your friends?
  3. Normally you are the only one who can get in because you are the only one with a key (that is, you are wild-type for ability to enter the house). Describe a mutant that might arise out of the population of burglars that would overcome this barrier.
  4. Would the mutation you described be dominant or recessive? Why?


Like other mutations, mutations that result in a mutator phenotype arise spontaneously in a population of bacteria. To determine the spontaneous frequency of mismatch-repair deficient mutants, Miller et al. devised a selection that demands the reversion of three independent frameshift mutations (lac, bgl, and met) -- an event that is extremely rare in nonmutator strains but detectable in mismatch-repair deficient mutator strains [Miller et al. 1999. Direct selection for mutators in Escherichia coli. J. Bacteriol. 181:1576-1584].

  1. To test for spontaneous mismatch-repair mutants,104 cells from 699 independent cultures were spread on selection plates that detect reversion of the three mutations. The triple revertants were observed on 44 of the 699 plates. Based upon these results, calculate the frequency of mismatch-repair deficient mutants using the Poisson distribution.

    ANSWER: P0 = fraction of total cultures with no mutants = 655/699 = 0.937 = e-h.
    Therefore, h = -ln 0.937 = 0.065 mutations per 104 cells = 6.5 x 10-6 mutations/cell.

  2. A nearly identical number of revertants of the single mutations were obtained in a nonmutator strain background, with or without treatment with the mutagen 2-aminopurine, and in a mutT mutator strain background. The number of revertants obtained in these conditions was 500-1000 fold lower than in a mismatch-repair deficient strain. Suggest a simple explanation for these results.

    ANSWER: The results imply that the type of mutations caused by 2-aminopurine or the mutT mutation do not allow reversion of the lac, bgl, and met mutations. In contrast the mismatch-repair deficient mutator strain must increase the type of mutation that allows reversion of the the lac, bgl, and met mutations. [The mutT mutator strains only show an increase in one particular type of mutation, AT to GC transversions, but the three mutant alleles are caused by frameshift mutations.]

  3. Upon selection for reversion of all three mutants, the frequency of reversion increased in mutT strains or when cells were treated with 2-aminopurine. How can you explain these results?

    ANSWER: Recall that reversion of the triple mutants is usually extremely rare unless the strain has a mutation that prevents mismatch repair. Hence, this result is probably due to the production of mismatch-repair mutants caused by the mutT or 2-aminopurine mutagenesis. Once a mismatch repair deficient mutant is produced, it can subsequently revert the three frameshift mutations.



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Last modified October 18, 2004