Skip to main content
Pearson+ LogoPearson+ Logo
Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages
Back
Previous problem
Next problem
All textbooksKlug 12th EditionCh. 6 - Genetic Analysis and Mapping in Bacteria and BacteriophagesProblem 20

Chapter 6, Problem 20

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. What is the recombination frequency between the two mutants? Strain Plated Dilution Plaques E. coli B 10⁻⁵ 2 E. coli K12 10⁻¹ 5

Verified Solution
Video duration:
35s
This video solution was recommended by our tutors as helpful for the problem above.
236
views
Was this helpful?

Video transcript

Hello everyone. And here we have a question asking what is the main purpose of the serial dilution technique? So the serial dilution technique is used when the number of colonies is too big to count. And in this process, a series of sequential dilution czar performed to convert a dense solution into a more usable concentration. So our answer is c to reduce the colony number to the point where it can be accurately counted. Thank you for watching. Bye.
Previous problemNext problem
Related Practice
Textbook Question
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?
543
views
Textbook Question
In an analysis of rII mutants, complementation testing yielded the following results: Mutants Results (+/- lysis) 1, 2 + 1, 3 + 1, 4 - 1, 5 - Predict the results of testing 2 and 3, 2 and 4, and 3 and 4 together.
634
views
Textbook Question
If further testing of the mutations in Problem 18 yielded the following results, what would you conclude about mutant 5? Mutants Results 2, 5 - 3, 5 - 4, 5 -
266
views
Textbook Question
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?
231
views
Textbook Question
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.
370
views
Textbook Question
In studies of recombination between mutants 1 and 2 from Problem 21, the results shown in the following table were obtained. Strain Dilution Plaques Phenotypes E. coli B 10⁻⁷ 4 r E. coli K12 10⁻² 8 + Calculate the recombination frequency.
261
views