Phage lambda is unable to grow lytically on wild-type Salmonella typhimurium.
ANSWER: Either (i) lambda may not be able to adsorb to S. typhimurium or (ii) S. typhimurium may be missing some host function required for maturation of lambda (i.e. lambda would adsorb to the cells but would not lyse them).
If a 0.2 ml aliquot of a bacterial culture with 2 x 108 cells per ml is mixed with 0.2 ml of a lysate with 5 x 108 phage per ml, what proportion of the bacteria will be uninfected? What proportion of the bacteria will be infected with a single phage? What proportion of the bacteria will be infected with two or more phage?
In separate test tubes, 0.1 ml of a S. typhimurium culture with 5 x 109 cells per ml was mixed with 0.1 ml of dilutions of a P22 lysate with 5 x 1010 phage per ml. Three dilutions of phage were used: (i) undiluted, (ii) a 1:10 dilution, and (iii) a 1:50 dilution. Use the Poisson distribution to answer the following questions.
Calculations for (i):
MOI = mean # phage / cell = (0.1 x 5 x 1010) / (0.1 x 5 x 109) = 10
P(0) = (100) (e-10) / (0!) = (1) (e-10) / 1 = 4.5 x 10-5 = 0.0045%
P(1) = (101) (e-10) / (1!) = (10) (e-10) / 1 = 4.5 x 10-4 = 0.045%
Because the probability of all events = 1; P(>1) = 1 - [P(1) - P(0)] P(>1) = 1 - [P(1) + P(0)] = 1 - [0.000045 + 0.00045] = 0.9995 or 99.95%
Calculations for (ii):
MOI = mean # phage / cell = (0.1 x 5 x 109) / (0.1 x 5 x 109) = 1
P(0) = (10) (e-1) / (0!) = (1) (e-1) / 1 = 0.37 = 37%
P(1) = (11) (e-1) / (1!) = (1) (e-1) / 1 = 37%
P(>1) = 1 - [P(1) + P(0)] = 1 - [0.37 + 0.37] = 0.26 or 26%
Calculations for (iii):
MOI = mean # phage / cell = (0.1 x 1 x 109) / (0.1 x 5 x 109) = 0.2
P(0) = (0.20) (e-0.2) / (0!) = (1) (e-10) / 1 = 0.82 = 82%
P(1) = (0.21) (e-0.2) / (1!) = (0.2) (e-0.1) / 1 = 0.16 = 16%
P(>1) = 1 - [P(1) + P(0)] = 1 - [0.82 + 0.16] = 0.02 or 2%
F0 is a virulent phage (that is, it cannot lysogenize its host). When wild-type phage F0 phage is spotted on a lawn of S. typhimurium the cells are lysed (indicated by a white spot in the figure below). Three conditional mutants were isolated that prevent cell lysis. All three of these mutants affect the synthesis and assembly of the phage head, a complex structure that requires proper interactions between several different proteins. To determine if these mutations affect different genes, cells were coinfected with two different mutant phage under nonpermissive conditions as shown in the figure below.
ANSWER: Mutation 1 and 2 probably map in the same gene because they do NOT complement. (Failure to complement is unlikely to be due to polarity because these are both "conditional mutants" indicating that they are caused by missense mutations. Complementation would result in complete clearing (i.e., lysis) of the spot.
ANSWER: The small plaques are due to RARE recombinants between the two mutants that result in wild-type progeny phage. Note that the frequency of recombination is MUCH lower than the frequency of complementation seen with mutants 1 and 3. It is unlikely that the small plaques are revertants because no revertants are seen in the spots containing mutant 1 or mutant 2 alone.
ANSWER: No. Different complementation groups often mean that mutants affect different genes, but they do NOT prove it -- intragenic complementation can also occur, especially in a case like this which requires many protein-protein interactions.
Phage P22 efficiently lyses Salmonella typhimurium but is unable to grow lytically on the closely related bacterium Salmonella typhi (Zahrt et al. 1994. J. Bacteriol. 176: 1527-1529).
ANSWER: One possible reason is that the phage cannot enter the cell (the one-step growth curve would look like the results observed for mutants of S. typhimurium that lack the P22 receptor site, like the LPS- mutant shown in the figure below) and another possibility is that the phage can enter the cell but cannot develop (the one-step growth curve would look like the results shown for the wild-type S. typhi in the figure below).
ANSWER: see the diagram below.
Packaging of P22 DNA requires a pac site while packaging of lambda DNA requires a cos site. Briefly describe:
Below is a genetic map of the phage T4 rII region. Mutants #1, 2, 3 are revertible mutants and mutants 101, 102, 103 are deletions. (The genetic material deleted in each is indicated by .) Unfortunately, the tubes containing each of these phage mutants were so old that it was impossible to read the strain numbers on the label. How could you identify each mutant by a few genetic experiments?
Soon after phage were discovered, it was proposed that they would be an effective tool for killing pathogenic bacteria in humans. ("Arrowsmith" by Sinclar Lewis was a best selling novel that popularized this idea.) With the development of effective antibiotics with a broad host range, most scientists gave up on this idea. However, with the growing problem of antibiotic resistance, it has resurfaced and some newer studies using phage as antibacterial agents in animals are very promising. (For example, see http://www.evergreen.edu/user/T4/PhageTherapy/phagethea.html). Suggest two different reasons why phage might seem to be inferior to antibiotics.
ANSWER: (1) Cells can develop resistance to phage at high frequency by simple mutations that alter cell surface receptors the phage bind to, and (2) because phage have antigenic capsids the host may readily develop immunity to the phage.