In studies of recombination between mutants 1 and 2 from Problem 21, the results shown in the following table were obtained.

Calculate the recombination frequency.

An Hfr strain is used to map three genes in an interrupted mating experiment. The cross is Hfr/a⁺b⁺c⁺ rif x F⁻/a⁻b⁻c⁻ rif^T (No map order is implied in the listing of the alleles; rif^T is resistance to the antibiotic rifampicin.) The a⁺ gene is required for the biosynthesis of nutrient A, the b⁺ gene for nutrient B, and the c⁺ gene for nutrient C. The minus alleles are auxotrophs for these nutrients. The cross is initiated at time = 0, and at various times, the mating mixture is plated on three types of medium. Each plate contains minimal medium (MM) plus rifampicin plus specific supplements that are indicated in the following table. (The results for each time interval are shown as the number of colonies growing on each plate.)

What is the purpose of rifampicin in the experiment?

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?

In studies of recombination between mutants 1 and 2 from Problem 21, the results shown in the following table were obtained.

When mutant 6 was tested for recombination with mutant 1, the data were the same as those shown above for strain B, but not for K12. The researcher lost the K12 data but remembered that recombination was ten times more frequent than when mutants 1 and 2 were tested. What were the lost values (dilution and plaque numbers)?

During the analysis of seven rII mutations in phage T4, mutants 1, 2, and 6 were in cistron A, while mutants 3, 4, and 5 were in cistron B. Of these, mutant 4 was a deletion overlapping mutant 5. The remainder were point mutations. Nothing was known about mutant 7. Predict the results of complementation (+ or -) between 1 and 2; 1 and 3; 2 and 4; and 4 and 5.

In Bacillus subtilis, linkage analysis of two mutant genes affecting the synthesis of two amino acids, tryptophan (trp₂⁻) and tyrosine (trp₁⁻), was performed using transformation. Examine the following data and draw all possible conclusions regarding linkage. What is the purpose of Part B of the experiment? [Reference: E. Nester, M. Schafer, and J. Lederberg (1963).]