What is the difference between the mechanism of generalized transduction by phages P22 and P1?
ANSWER: Both phage use a headful packaging mechanism. P22 occasionally initiates packaging of chromosomal DNA from a pseudo-pac site. In contrast, P1 does not recognize pseudo-pac sites on the chromosome. P1 seems to begin packaging from random double-stranded breaks on the chromosome that are generated during phage infection.
Wild-type P22 cannot transduce plasmid pBR322 but P22 HT can transduce pBR322. Based upon the way these two phage package DNA, propose an explanation for this result.
ANSWER: P22 only transduces DNA with pac or good pseudopac sites while P22 HT recognizes and packages DNA fragments relatively nonspecifically. Thus, the results suggest that pBR322 does not have a pseudo-pac site so it cannot be transduced by P22, but because P22 HT can package the DNA "nonspecifically", pBR322 can be transduced by P22 HT.
To avoid obtaining lysogens that are resistant to further infection, a virulent mutant of phage P1 is often used for generalized transduction. However, this poses a second problem because any cells that are coinfected with both a transducing particle and a phage particle will lyse, preventing the isolation of transductants.
The F-plasmid is about 100 Kb. Because P22 HT only packages about 45 Kb of DNA, it cannot package the entire F plasmid. However, when a Tetracycline sensitive (TetS) recipient is infected with P22 HT grown on a strain with a F-plasmid marked with Tetracycline resistance (TetR), it is possible to obtain rare TetR transductants. Suggest a likely explanation for this result.
ANSWER: Recall that transduction of plasmids requires packaging of a linear concatemer of the plasmid DNA and recircularization of the plasmid DNA in the host cell. Thus, the simplest explanation is that the TetR colonies arise by transduction of rare, spontaneous deletions of the F plasmid. The deletion mutants would have to reduce the size of the F plasmid to less than 45 Kb (allowing sufficient terminal redundancy for recircularization by homologous recombination in the recipient), without removing the TetR marker or the genes required for vegetative replication.
Clark recently isolated a phage from Hong Kong sewage that grows on E. coli. The phage acts as a generalized transducing phage on some strains of E. coli but it acts as a specialized transducing phage on other strains of E. coli.
ANSWER: The phage is probably unable to integrate into the chromosome of strains where it does generalized transduction. This may be due to (i) lack of a host protein that is required (e.g. IHF) or (ii) lack of an att site or other homology required for recombination (possibly a cryptic prophage).
The mutation putA900 is a point mutation. The mutation put-557 is a small deletion within the put genes. The mutation put-544 is a large deletions that removes the entire put operon and extends into DNA on both sides of the put operon. P22 HT was grown on each of these mutant strains and used to transduce recipients with a pyrC or pyrD mutation, selecting for Pyr+. The results are shown below.
P1 transduction was done to determine the order of the fadD, gap, and pabB genes relative to each other. The donor was fadD gap+ pabB and the recipient was fadD+ gap pabB+. Gap+ transductants were selected. The coinheritance of the nonselected markers is shown in the table below. Based upon these results, what is the predicted order of the genes.
|fadD||pabB||Number of colonies|
ANSWER: The rarest class of recombinants was Gap+ FadD- PabB+. The recipient marker inherited in this strain is pabB+, suggesting that this is the middle marker (see figure below). Thus, the order of the genes is predicted to be: fadD pabB gap.
Note that the three-factor cross data also allows you to calculate the percent coinheritance between the selected marker and each of the unselected markers. When selecting for Gap+ the total number of colonies obtained was 180. The coinheritance of PabB- was 54 + 19 = 73 thus 73/180 = 41%. The coinheritance of FadD- was 54 + 9 = 63 thus 63/180 = 35%. From this data you could draw the map shown below.
In E. coli the genes for the biosynthesis of leucine (leu) and for the ability to ferment arabinose (ara) are approximately 50% cotransducible by phage P1. Four different leu auxotrophs were isolated. The mutations were mapped by P1 transductions. Donor strains were ara+ leu and recipients were ara leu. Transductants were selected on minimal medium containing glucose as a carbon source, and then replica plated to test for the ability to ferment arabinose. The table below shows the donor and recipient strains in use and the proportion of transductants that were Ara+ for each two-factor cross.
In a generalized transduction experiment, donor E. coli cells have the genotype X+ Y+ A- Z+ and recipient cells have the genotype X- Y- A+ Z-. P1-mediated transductants for X+ or Y+ were selected and their phenotypes were determined with the following results:
ANSWER: The trick to answering three-factor cross questions is to look for a rare class of recombinants. The rare class usually requires two pairs of crossover events, with the middle marker being inherited from the recipient and the two outside markers inherited from the donor. If there is no rare class of recombinants, then the middle marker is inherited from the donor because none of the recombinant classes will require more than one pair of crossovers.
ANSWER: The co-transduction frequency for X and A = 23 + 4 /101 = 27%. The co-transduction frequency for Y and Z = 5 + 24 /101 = 29%.
The general transducing phage P1 was grown on a wild-type bacterial strain (cys+ pyr+ trp+). The resulting phages were mixed with bacteria which require cysteine, uracil, and tryptophan (Cys- Pyr- Trp-). Recombinants were selected on medium containing uracil and tryptophan. The results are below.
|Phage and cells added||Number of colonies|
|Phage only control (108 phage)||0|
|Cells only control (108 cells)||4|
|Phage (108) + Cells (108)||200|
ANSWER: This is a control to prove that the phage were not contaminated with donor cells. If there are no bacteria contaminating the phage, no colonies will be obtained.
ANSWER: The selection was for Cys+ colonies. Given the number of cells plated, the four colonies that arose in the absence of phage were probably due to spontaneous revertants.
ANSWER: It is unlikely that the other mutations also reverted, so the phenotype is most likely: cys+ pyr- trp-.
ANSWER: First consider what type of colonies will grow on each medium.
Thus, only prototrophic colonies will grow on all the media in row (i). No prototrophic recombinants were obtained. Based upon these results, you know that none of the colonies that grow under the conditions shown in rows (ii), (iii), or (iv) have wild-type copies of all three genes. Thus, the 20 colonies that grow on the media shown in row (ii) must be Cys+ Trp+ recombinants. Likewise, the 30 colonies that grow on the media shown in row (iii) must be Cys+ Pyr+ recombinants. The 150 colonies corresponding to row (iv) must be Cys+ Pyr- Trp- recombinants.
ANSWER: 20 Cys+ Trp+/200 total = 10%
ANSWER: 30 Cys+ Pyr+/200 total = 15%
ANSWER: The results suggest that the gene order is trp cys pyr because cys is cotransducible with both trp and pyr, but neither trp or pyr is cotransducible with each other.
A series of P22 transductions were done in order to construct a map for the met, thi, trp, and his loci in S. typhimurium. The transduction data is shown in the table below.
|Donor||Recipient||Selected marker||Scored marker|
|met+ thi+||met- thi-||Met+||Thi+|
|met+ trp+||met- trp-||Met+||Trp+|
|met+ his+||met- his-||Met+||His+|
|his+ thi+||his- thi-||His+||Thi+|
|trp+ thi+||trp- thi-||Trp+||Thi+|
ANSWER: See figure below:
ANSWER: The cotransduction frequency between trp and his would provide more confidence in the two-factor cross data. This data could be obtained by growing phage on a trp+ his+ donor and tranducing a trp- his- recipient, selecting for Trp+ or His+. Alternatively, three factor crosses would be more give more definative results.
Yanofsky isolated two trpA mutants (trpA8 and trpA17). To determine the order of the two trpA mutants relative to the trpE gene, he did the P1 transduction experiment described below.
Donor: trpA17 trpE+Based upon these results, what is the most likely gene order? [i.e., Determine if the order is trpE trpA8 trpA17 or trpE trpA17 trpA8).]
Recipient: trpA8 trpE-
Selected marker: trpA+ 100 colonies
Scored markers: trpA+ trpE+ 10 colonies; trpA+ trpE- 90 colonies
ANSWER: The trick to answering this question is to note that in the donor strain the trpA8 allele is wild-type and in the recipient strain the trpA17 allele is wild-type. Any recombinants that become Trp+ must inherit the wild-type allele of both trp mutations. Thus, although it may look like two-factors this is a three-factor cross. Note that the rare class of recombinant inherits the trpA8+ and the trpE+ alleles from the donor, and the trpA17+ allele from the recipient.
When a generalized transducing particle enters a recipient cell, the DNA can recombine with the recipient chromosome. Alternatively, sometimes a phage protein binds to the ends of the transducing DNA, causing it to circularize and protecting it from nucleases. When this happens the transducing DNA is not a substrate for the RecBCD recombination pathway and thus remains in the cytoplasm as an "abortive transductant". Abortive transductants typically form tiny colonies that never grow to full size and usually fail to form colonies when picked and restreaked on a fresh plate.
For example, when a Salmonella del(proBA) recipient is transduced with phage P22 grown on a prototrophic strain of Salmonella selecting for growth on minimal medium without proline, some large colonies appear and some tiny colonies appear. The large colonies reproduce when restreaked on the same medium, but the tiny colonies fail to reproduce when restreaked. Suggest an explanation for these results.
ANSWER: The rationale for these results is shown in the following figure.
The large colonies are "true transductants" due to recombination between the pro+ alleles on the linear transducing DNA and the chromosomal DNA, resulting in repair of the auxotrophic mutation. The tiny colonies are due to abortive transduction. The DNA on an abortive transducing particle can be transcribed and translated, allowing complementation of the chromosomal mutation. However, because an abortive transducing particle does not have an origin of replication it cannot be replicated. Everytime a cell divides only one of the two daughter cells will get a copy of the abortive transducing particle that complements the chromosomal auxotrophy. The other daughter cell will retain a proportion of the complementing proteins made before cell division, but it will only be able to continue growing until the gene products are degraded or too dilute to satifify the auxotrophic requirement. Hence, at each cell division only one of the two cells will be able to continue to divide and produce daughter cells, so instead of the cells reproduces exponentally (i.e. # cells = 2[# generations]) the abortive transductant only reproduces geometrically (i.e. # cells = 1 + [2 x # generations]) and most of the cells in the colony cannot reproduce when restreaked.
For more examples of two-factor and three-factor crosses, see the mapping problems