Genetics of Bacteria Practice Problems (page 2)
Review the following concepts if needed:
- Genetics of Bacteria for Genetics
- Bacterial DNA Replication and Cell Division for Genetics
- Bacterial Transcription for Genetics
- Bacterial Translation for Genetics
- Genetic Recombination for Genetics
- Regulation of Bacterial Gene Activity for Genetics
- Mapping the Bacterial Chromosome for Genetics
Genetics of Bacteria Practice Problems
Under optimal conditions, some bacteria can divide every 20 min. Suppose each cell has a mass of 2 × 10–9 milligrams (mg). The mass of Earth is approximately 5:97 × 1027 grams (g). Determine the time (in hours) required for the progeny of a single cell dividing without restriction at the above rate to equal the weight of Earth.
At time zero we have one cell; 20 min later we have two cells; at 40 min there are four cells; at 60 min there are eight cells; etc. Thus, there are three cell divisions per hour. The number of cells at any hour, t, is 23t. The mass of each cell in grams is 2 × 10– 9 mg × 10–3 mg / g = 2 × 1010–12 g. The number of cells equivalent to the weight of Earth is
- (5.97 × 1027 g) / 2 × 10–12) = 2:98 × 10–39 = 23t hours(h)
from which 3t log 2 = log 2.98 + log 10–39,
The discipline of bacterial genetics began in 1943 when S. E. Luria and M. Delbrüuck published a paper entitled "Mutations of bacteria from virus sensitivity to virus resistance." Before this time, it was not known if the heredity of bacteria changed adaptively in specific ways as a consequence of exposure to specific environments, or whether specific mutants existed in the population prior to an environmental challenge. The former idea was Lamarckian, the latter was neo-Darwinian. Luria and Delbrück found that there was great variation from one trial to another in the number of E. coli that were resistant to lysis by phage T1. In order to determine which of the two hypotheses was correct, they devised the following "fluctuation test." Twenty 0.2-ml "individual cultures" and one 10-ml "bulk culture" of nutrient medium were incubated with about 103 E. coli cells per milliliter. The cultures were incubated until they contained about 108 cells per milliliter. The entire 0.2 ml of each individual culture was spread on a nutrient agar plate heavily seeded with T1 phages. Ten 0.2-ml samples from the bulk culture were also treated in similar fashion. After overnight incubation, the total number of T1-resistant (Tonr) bacterial cells was counted; the results are presented in the following table. What inferences can be drawn from this "fluctuation test"?
Variances for each experiment can be calculated from the square of formula (8.2) or (8.3); the individual cultures have a variance of 714.5, whereas the variance of samples from the bulk culture is 16.4.2 In a Poisson distribution, the mean and the variance are essentially identical; hence, the variance/mean ratio should be near unity (1.0). The variance/mean ratio for the bulk culture samples is 16.4/16.7 = 0.98 or nearly 1.0, as expected from a random distribution of rare events. The samples from the bulk culture collectively serve as a control for the individual cultures. The same ratio for the individual cultures, however, is 714.5/11.3 = 63.23, indicating that there are extremely wide fluctuations of the numbers of Tonr cells in each culture around the mean. If resistance to the T1 phages occurs with a given probability only after contact with the phages, then each culture from both the individual and batch experiments should contain approximately the same average number of resistant cells. On the other hand, if Tonr mutants occurred prior to contact with the phages, great variation around the mean is expected from one individual culture to another because some will incur a mutation early and others late (or not at all) during the incubation period. This experiment argues in favor of the mutation hypothesis and against the induced resistance hypothesis.
Certain mutations, such as that to phage resistance, are preadaptive in that their selective advantage only becomes manifest when phages are in the environment as a selective agent; in this case, T1-sensitive bacteria (Tons) are killed by T1 phages, allowing only the few Tonr cells to survive and multiply. Phage resistance depends upon altering the structure of the bacterial receptor sites to which T1 phages normally attach. Immunity to superinfection by a specific phage is based upon production of a repressor of phage replication by a lysogenic cell.
Two triple auxotrophic strains of E. coli are mixed in liquid medium and plated on complete medium, which then serves as a master for replica plating onto six kinds of media. From the position of the clones on the plates and the ingredients in the media, determine the genotype for each of the six clones. The gene order is as shown.
Clone 1 grows when supplemented with Thr and Leu or Thr and Thi, but not with Leu and Thi. Therefore, this colony is auxotrophic for Thr alone (thr– leu+ thi+ bio+ phe+ cys+).
Clone 2 appears only on the plate supplemented with Phe and Cys. This is a double auxotrophic colony of genotype thr+ leu+ thi+ bio+ phe– cys–.
Clone 3 appears on all replica plates and therefore must be prototrophic (thr+ leu+ thi+ bio+ phe+ cys+).
Clone 4 grows only when supplemented with Thr and Leu and therefore must be a double auxotroph of genotype thr– leu– thi+ bio+ phe+ cys+.
Clone 5 and clone 1 always appear together on the replica plates and therefore have the same genotype.
Clone 6 can grow in the presence of Phe and Cys or Bio and Cys. The common factor is Cys, for which this strain is singly auxotrophic (thr+ leu+ thi+ bio+ phe+ cys–).
A strain of E. coli unable to ferment the carbohydrate arabinose (ara–) and unable to synthesize the amino acids leucine (leu–) and threonine (thr–) is transduced by a wild-type strain (ara+ leu+ thr+). Recombinants for leucine are detected by plating on minimal medium supplemented with threonine. Colonies from the transduction plates were replicated or streaked onto plates containing arabinose. Out of 270 colonies that grewon the threonine-supplemented plates, 148 could also ferment arabinose. Calculate the amount of recombination between leu and ara.
In order for a transductant to be leu+ ara+, crossing over in regions (1) and (3) must occur; for leu+ ara– to arise, crossing over in regions (1) and (2) must occur.
Double crossovers are expected to be much more frequent than quadruple crossovers. The above scheme does not fit the data because the first mating was more frequent than the reciprocal mating. Our assumption must be wrong.
Let us assume that the order is lacZ1-lacZ2-ade. The first cross now requires a double crossover.
The reciprocal cross is expected to be much less frequent under this assumption and is in agreement with the observations.
Six mutations are known to belong to three genes. From the results of the complementation tests, determine which mutants are in the same gene.
Mutations 3 and 5 are in the same gene, since they fail to complement each other. Mutations 1 and 3 are in different genes, since they do complement each other. We will arbitrarily assign these to genes A and B.
Mutations 1 and 2 are in different genes, but we do not know whether 2 is in A or C. However, 5 and 2 complement and therefore 2 cannot be either in gene A or B and thus must be in C.
Mutations 3 and 4 complement; thus, 4 must be in either BorC. But 2 and 4 also complement; thus, 4 cannot be in Cand must reside in B.
Mutation 6 cannot be inAsince it complements with 5. Thus, 6 is either in BorC. Since 6 and 4 complement, they are in different genes. If 6 cannot be in A or B, it must be in C. The mutants are grouped into genes as shown below.
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