If a single bacteriophage infects one E. coli cell present on a lawn of bacteria and, upon lysis, yields 200 viable viruses, how many phages will exist in a single plaque if three more lytic cycles occur?

Two theoretical genetic strains of a virus (a⁻b⁻c⁻ and a⁺b⁺c⁺) were used to simultaneously infect a culture of host bacteria. Of 10,000 plaques scored, the following genotypes were observed. Determine the genetic map of these three genes on the viral chromosome. Decide whether the interference was positive or negative.

Seven deletion mutations (1 to 7 in the table below) are tested for their ability to form wild-type recombinants with five point mutations (a to e). The symbol "+" indicates that wild-type recombination occurs, and "-" indicates that wild types are not formed. Use the data to construct a genetic map of the order of point mutations, and indicate the segment deleted by each deletion mutation. 

The bacteriophage genome consists of many genes encoding proteins that make up the head, collar, tail, and tail fibers. When these genes are transcribed following phage infection, how are these proteins synthesized, since the phage genome lacks genes essential to ribosome structure?

If a single bacteriophage infects one E. coli cell present on a lawn of bacteria and, upon lysis, yields 200 viable viruses, how many phages will exist in a single plaque if three more lytic cycles occur?

In recombination studies of the rII locus in phage T4, what is the significance of the value determined by calculating phage growth in the K12 versus the B strains of E. coli following simultaneous infection in E. coli B? Which value is always greater?

In an analysis of rII mutants, complementation testing yielded the following results:

If further testing of the mutations in Problem 18 yielded the following results, what would you conclude about mutant 5?

Using mutants 2 and 3 from Problem 19, following mixed infection on E. coli B, progeny viruses were plated in a series of dilutions on both E. coli B and K12 with the following results.

Another mutation, 6, was tested in relation to mutations 1 through 5 from Problems 18–20. In initial testing, mutant 6 complemented mutants 2 and 3. In recombination testing with 1, 4, and 5, mutant 6 yielded recombinants with 1 and 5, but not with 4. What can you conclude about mutation 6?