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?

Hello, everyone. Here we have a question. Science in the recombination testing. The number of resulting comins is three times 10 to the fifth milliliters. If the total number of progeny is 12 times 10 to the eighth milliliters, what is the frequency of recombination between the two mutants? So to figure this out, we're gonna take two times the number of our resulting recommence which is three times 10 to the fifth divided by the total number of progeny, which is 12 times 10 to the eighth. And that equals two times 0.25 times 10 to the negative third, which equals 0.5 times 10 to the negative three. And we want to take that out of scientific notation. So we have negative three. So we're going to move our decimal place, 123 places to the right or to the left. So that gives us 0.005. So our answer is B 0.005. Thank you for watching. Bye.