Review the following concepts if needed:
 Viruses for Genetics
 Bacteriophages for Genetics
 Eukaryotic Viruses for Genetics
 Transposable Elements for Genetics
 Cancer for Genetics
Viruses, Transposable Elements, and Cancer Practice Problems
Practice 1
A prepared solution of bacteriophage has an unknown titer (phage number). An aliquot, 0.1 ml, is used to inoculate 1 ml of a susceptible E. coli culture (This amount of E. coli is enough to grow into a lawn on the appropriate type of culture medium.) The inoculated culture is then mixed with molten top agar (agar that is poured on top of another type of culture medium) and poured on top of a solid culture medium in a petri plate. The top agar is allowed to harden and the plated bacteria are incubated for 12–18 hours. After incubation, a lawn of bacteria is observed to contain many plaques. The number of bacteriophage plaques is counted and found to equal 220. Determine the infective phage titer in the original phage solution.
Solution 1
Each plaque can be assumed to be the result of a single bacteriophage infecting a single E. coli cell. This assumption may not be valid if too many phage particles are used for the inoculation. However, let us assume that one plaque results from one phage infecting a single cell. Thus, each plaque represents one bacteriophage particle, known as a plaqueforming unit (pfu). In order to find the number of infective bacteriophage particles in the original culture, simply divide the counted pfu number by the volume of the phage solution used to inoculate the bacterial culture, as follows:
 = 2200 pfu/ml phage solution
If a dilution of the phage solution, such as 1 : 100, was used as an inoculum, this dilution would have to be taken into account in the above calculation, as follows:
 × 100 = 2:2 × 10^{5} pfu/ml phage solution
Practice 2
In an attempt to determine the amount of recombination between two mutations in the rII region of phage T4, strain B of E. coli is doubly infected with both kinds of mutants. A dilution of 1 : 10^{9} is made of the lysate and plated on strainB.Adilution of 1 : 10^{7} is also plated on strainK. Two plaques are found on K, 20 plaques on B. Calculate the amount of recombination.
Solution 2
In order to compare the numbers of plaques on B and K, the data must be corrected for the dilution factor. If 20 plaques are produced by a (10^{9}) dilution, the lesser dilution (10^{7}) would be expected to produce 100 times as many plaques.
Practice 3
Seven deletion mutants within the A gene of the rII region of phage T4 were tested in all pairwise combinations for wildtype recombinants. In the table below of results, + = recombination, 0 = no recombination. Construct a topological map for these deletions
Solution 3
If two deletions overlap to any extent, no wildtype recombinants can be formed.
Step 1. Deletion I overlaps with 3, 4, 6, and 7 but not with 2 or 5.
Step 2. Deletion 2 overlaps with 3, 4, and 7 but not with 1, 5, or 6.
Step 3. Deletion 3 overlaps with 1, 2, 4, and 7 but not with 5 or 6.
Step 4. Deletion 4 overlaps with 1, 2, 3, 6, and 7 but not with 5.
Step 5. Deletion 5 overlaps with 6 and 7 but not with 1, 2, 3, or 4. To satisfy these conditions wewill shift 5 to the left of 1 and extend 7 into part of region 5 so that the overlaps in step 4 have not changed. Now segment 5 can overlap 6 and 7 without overlapping 1, 2, 3, or 4.
Step 6. Deletion 6 overlaps with 1, 4, 5, and 7 but not with 2 or 3. No change is required
Step 7. Deletion 7 overlaps with all other regions, just as it was temporarily diagrammed in step 5. This completes the topological map. Although the overlaps satisfy the conditions in the table, we have no information on the actual lengths of the individual deletions.

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