What is the phenotype of a lambda int mutant?
ANSWER: An int mutant would be unable to integrate or excise but would be fully proficient for lytic growth.
Which of the following genes are required for efficient lysogenization following infection of a sensitive host by phage lambda: int, xis, N, cII, Q?
ANSWER: int, N, and cII are required for efficient lysogeny, xis is only required for excision. Q is required for expression of late genes during lysis. (Note that N is also required for expression of lysis genes.)
What is the phenotype of a lambda xis mutant?
ANSWER: An xis mutant would be able to integrate, but unable to excise and would remain fully proficient for lytic growth.
How could you isolate lambda int or xis mutants?
ANSWER: There are a variety of creative tricks for isolating int and xis mutants. However, the key concept for each trick is the inability to integrate or, more often, the inability to excise. One of the most commonly used approaches is to begin with a lambda lysogen inserted in the gal operon. Because the insertion prevents the synthesis of gene products required for the catabolism of galactose, this strain forms white colonies on MacConkey-Galactose plates. However, precise excision of the prophage yields Red colonies on MacConkey-Galactose plates. Thus, any int or xis mutants will be unable to form red colonies.
A lysogen of the double temperature sensitive mutant lambda cI(Ts) cII(Ts) was isolated at 30 C. What would the phenotype be if the lysogen was shifted to 42 C? Which mutation is epistatic?
ANSWER: When shifted to 42 C the cI repressor would be inactivated, the lysogen would be induced, the phage would enter the lytic cycle, and the cells would lyse. The cII protein is only required for establishment of lysogeny, not for maintence of lysogeny or for lytic growth, so the cI mutation would be epistatic to the cII mutation.
Most E. coli recA mutants do not prevent lambda lysogeny. However, a unique recA mutant was isolated that permits lysogeny of lambda at 30 C but not at 42 C. What is the probable mechanism for this observation?
ANSWER: When cells encounter DNA damage (e.g. by UV irradiation or treatment with mitomycin C) wild-type RecA becomes activated to a form called RecA*. RecA* interacts with the lambda cI repressor and induces autoproteolysis of the repressor. The resulting proteolytic fragments are not functional for repression. Hence, one explanation for the new recA missense mutant is that the mutation does not affect the conformation of RecA at 30 C, but it causes the RecA protein to assume the activated RecA* conformation at 42 C. Thus, at 30 C lambda would make cI and be able to lysogenize the mutant, but at 42 C the RecA* conformation would stimulate autoproteolysis of cI, preventing lysogeny.
After hydroxylamine mutagenesis of a wild-type lambda lysate, it is possible to isolate rare lysogens in the above recA mutant strain at 42 C. What is the probable explanation for this result? How could you test your hypothesis? Would you expect to obtain these lambda mutants using proflavin as a mutagen? Why?
ANSWER: Missense mutations in lambda cI that prevent the interaction of the repressor protein with RecA*. If wild-type cells were lysogenized with mutant phage with this phenotype, the resulting lysogens would not be induced by UV light or mitomycin C. This phenotype would be most likely to result from a missense mutation, so it would be unlikely that a frameshift mutagen like proflavin could produce this type of mutant.
The lysis-lysogeny decision of P22 is very similar to phage lambda. A S. typhimurium(P22 sieA) lysogen is wild-type for immunity and sensitive to superinfection with other P22 phage. When infected with wild-type P22, no plaques arise because the phage are homoimmune. Clear plaque mutants of phage P22 fall into two categories: (i) when these mutants infect the lysogen no plaques arise; (ii) when these mutants infect the lysogen, clear plaques are produced. What is a likely explanation for each of these classes of clear plaque mutants?
ANSWER: Class (i) mutants are probably due to mutations in the proteins required for establishment and maintance of lysogeny (c1, c2, c3). These mutations are complemented by the repressor protein made in the lysogen. Class (ii) mutants are probably due to mutations in the repressor binding site. These mutations are dominant. Because the repressor made by the lysogen cannot bind to these sites, the incoming phage will always enter the lytic pathway.
Matt Lawes isolated a new phage that forms turbid plaques on Klebsiella aerogenes. Describe a simple test to determine whether the turbid plaques are due to formation of lysogens.
ANSWER: The turbidity in a turbid plaque arises from the growth of surviving bacteria within the zone of lysis. These surviving bacteria can be isolated by picking from the plaque and streaking for isolation on an agar plate. Isolated colonies are restreaked out several times to purify them away from contaminating phage. The surviving bacteria may be phage resistant for one of two reasons. Either they are lysogens and prevent superinfecting phage from lysing due to repression of the lytic cycle of the entering phage by repressor produced by the resident prophage (a phenomenon called "superinfection immunity"). Alternatively they are rare receptor minus mutants to which the phage cannot adsorb. A simple to test determine if the isolated colonies are phage resistant is to streak a colony at 90¡ across a plate previously inoculated with a bead of the Boneyard phage. Sensitive bacteria will lyse upon meeting the phage whereas resistant bacteria will continue to grow beyond the point of contact with the phage. However, this test will not distinguish between a receptor mutant and a lysogen. To do that, isolated colonies must be grown up in broth, induced via UV or some other DNA damaging agent and then allowed to grow overnight. The next day the supernatant is tested for plaque forming units on a lawn of phage sensitive K. aerogenes. Receptor mutants will not release phage whereas the lysogens will.
When an Hfr carrying lambda prophage transfers its chromosome into a F- cell that does not contain lambda, the prophage is induced and the F- cell is killed. Why?
ANSWER: The lambda prophage in the Hfr donor expresses the lambda cI protein from PRM. The concentration of cI protein in the cell is maintained at a level sufficient to repress the lytic functions of the phage. However, there is no cI protein in the F- recipient, so when the lambda DNA enters the F- cell RNA polymerase can bind to PR and PL and express the genes required for lysis. [Note that because each recipient only gets one copy of the lambda DNA, the concentration of cIII protein will be low, favoring lytic growth.]
When the normal lambda attachment site on DNA is deleted by mutation, lambda can establish lysogeny only by inserting at a new attachment site, although it does so less efficiently than with the original attachment site near gal. The new attachment sites can lead to the production of novel lambda transducing phage that carry genes other than gal. Given an Hfr with the normal lambda attachment site deleted, how would you isolate a prophage inserted at a secondary attachment site?
E. coli C600 was coinfected with lambda imm434 and lambda c (where c indicates a clear mutation) and the offspring are plated on C600 (lambda+). What class of lambda clear mutants would produce plaques on this host strain?
If a lambda cI mutant infected a hfl mutant of E. coli would you expect the phage to grow lytically or to form a lysogen?
ANSWER: These two mutations have opposite effects -- a lambda cI mutant will grow lytically (because it lacks the repressor), but lambda forms lysogens at a high frequency on a hfl mutant host (because cII accumulates at high levels). When combined, the lambda cI mutant phenotype will be epistatic to the hfl mutant phenotype (because accumulation of cII does not affect the lysis/lysogeny decision in the absence of cI) -- that is, the phage will grow lytically.
Three new phages were isolated that can infect and lyse Klebsiella aerogenes. Phage PK1 and PK2 are temperate phage (forms turbid plaques) and phage PK3 is a lytic phage (forms clear plaques). During characterization of these phage, a clear-plaque mutant of phage PK1 (named PK1 c1) was also isolated.
K. aerogenes cells isolated from the center of the turbid plaques formed after infection with phage PK1 were designated K. aerogenes PK1. Similarly, K. aerogenes cells isolated from the center of the turbid plaques formed after infection with phage PK2 were designated K. aerogenes PK2. Growth of each of the phage on the wild-type K. aerogenes and the two K. aerogenes derivatives is shown in the following table. [Turbid indicates the phage forms turbid plaques, clear indicates the phage forms clear plaques, and none indicates the phage does not lyse the cells.]
|Strain||Lysis by phage:|
|K. aerogenes wild-type||turbid||clear||turbid||clear|
|K. aerogenes PK1||none||clear||none||clear|
|K. aerogenes PK2||none||none||none||clear|
ANSWER: PK3 is a lytic phage that is not homoimmune with PK1 or PK2 (that is, the repressor produced by PK1 or PK2 does not repress PK3)
ANSWER: Either (i) the repressor produced by the PK1 lysogen can also repress any superinfecting PK2 (that is, the two phage are homoimmune) or (ii) the PK1 lysogen produces a superinfection exclusion system that prevents growth of PK2. As expected, PK2 would be repressed by a PK2 lysogen.
ANSWER: Either (i) the repressor produced by the PK2 lysogen can also repress any superinfecting PK1 (that is, the two phage are homoimmune) or (ii) the PK2 lysogen produces a superinfection exclusion system that prevents growth of PK1. As expected, PK1 would be repressed by a PK1 lysogen.
ANSWER: Note that this mutant can plaque on a PK1 lysogen but not a PK2 lysogen, indicating that either there is a difference between phage entry, repression, or growth in the two lysogens. There are several possible reasons. For example, the PK2 lysogen may have a superinfection exclusion which prevents entry of the 0PK1 c1 phage; the c1 mutation may affect an operator site in such a way that the PK1 repressor cannot bind to the mutant operator but the PK2 repressor can still bind to the mutant operator; or the c1 mutation may affect a PK1 protein such that it inactivates the PK1 repressor (a dominant-negative phenotype) but cannot inactivate the PK2 repressor.
Phages L5, 29, and TM4 can infect and lyse Mycobacterium smegmatis. Phage L5 is a temperate phage (forms turbid plaques) and phage 29 is a lytic phage (forms clear plaques). Gene product 71 ("gp71") from phage L5 encodes the repressor for lysogeny (similar to the lambda cI protein). M. smegmatis strains that contain a L5 lysogen were infected with different phages as shown in thetable below. [+ indicates lysis and - indicates no lysis; * indicates mutant form of gene 71]
Answer the following questions based upon what you know about the lysis/lysogeny decision in phage lambda.
Lambda, 424, and 21 are all lambdoid phage. Lysogens of these phage are indicated by writing the phage in parenthesis after the bacterial strain. For example, an E. coli K-12 lambda lysogen is indicated K-12(lambda). HK22 is a newly isolated lambdoid phage isolated from pig sillage in Hong Kong. When testing the growth of phage HK22 on various E. coli K-12 lysogens, Max Gottesman and coworkers noted the following results.
ANSWER: Lambda is heteroimmune with phages 424 and 21 but homoimmune with lambda lysogens. When lambda infects a homoimmune lysogen, the repressor produced by the prophage will repress the incoming phage.
ANSWER: This cannot be a simple case of homoimmunity because the results indicate that lambda and 424 are heteroimmune (as in the answer to part a). The simplest explanation of these results is that the HK22 lysogen produces an alternative function that results in superinfection exclusion of both lambda and 424 but not 21. Two common ways a lysogen may do this are to produce a protein that (i) degrades the DNA from certain phages or (ii) modifies their receptors on the cell surface. In this case, the HK22 lysogen produces a protein called Nus that prevents antitermination by N-protein, so the late gene products required for lysis cannot be made.
The following question was posed by a scientist in the biotech industry, based upon a problem he encountered that posed a serious roadblock to his progress.
Upon cultivation of a Salmonella type B strain, phage were present in the culture supernatant. There was no obvious lysis of the culture.
Because phage recognize specific receptors on the cell surface of their bacterial host, testing for sensitivity to a battery of phage is a simple, effective way of identifying particular strains of bacteria. This method is called "phage typing". Anderson assembled a collection of phage that can distinguish strains of Salmonella. To determine if these phage shared were homo- or hetero-immune, three strains of bacteria were infected [Adapted from Schmieger, H. 1999. Molecular survey of the Salmonella phage typing system of Anderson. J. Bacteriol. 181:1630-1635.]:
The results for several of the phage are shown in the following table:
|A1||+||+||+||Heteroimmune with both P22 and ES18|
|A2||+||+||+||Heteroimmune with both P22 and ES18|
|A8||+||-||+||Homoimmune with P22 and|
|Heteroimmune with ES18|
How does the data support these conclusions?
ANSWER: All three phage lyse DB21, indicating that they can adsorb to S. typhimurium and complete phage replication and morphogenesis in this host. Phage A1 and A2 can lyse DB21(P22) or DB21(ES18) lysogens, indicating that these phage do not repress growth of A1 and A2 and, hence, A1 and A2 must be heteroimmune with phages P22 and ES18. This result indicates that P22 and ES18 are themselves heteroimmune. Phage A8 can lyse the DB21(ES18) lysogen indicating that it is heteroimmune with ES18, but it cannot lyse DB21(P22) indicating that it is homoimmune with P22. When the homoimmune phage infects the lysogen, the repressor produced in the lysogen binds to the incoming phage and prevents lytic growth.
Many natural isolates of bacteria are lysogenic for phage. To test for lysogens, the supernatent from several strains of Salmonella were assayed for spontaneous release of phage. The released phage were first tested for lysis of S. typhimurium DB21 which is a nonlysogen, then any phage obtained were tested for lysis of a DB21(P22) which is lysogenic for phage P22. The results are shown in the following table [Adapted from Schmieger, H. 1999. Molecular survey of the Salmonella phage typing system of Anderson. J. Bacteriol. 181:1630-1635.]:
|Parent strain||Lysis of DB21||Lysis of DB21(P22)|
Suggest an explanation for these results for each of the three bacterial strains.
ANSWER: Although lysogens are quite stable, lysis typically occurs at a frequency of about 10-6. Thus the supernatent of a culture with 109 lysogenic bacteria per ml will typically contain about 103 phage per ml.
Strain BA1 seems to be a nonlysogen because no spontaneous release of phage is detected from this culture. Strain BA7 seems to be a lysogen with a phage that is heteroimmune with phage P22 because this culture releases phage that can plaque on either DB21 or the DB21(P22) lysogen. Strain BA14 seems to be a lysogen with a phage that is homoimmune with P22 because this culture releases a phage that can plaque on DB21 but not on the DB21(P22) lysogen.
Last modified April 17, 2003