Skip to main content
Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages

Chapter 6, Problem 22

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 + 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)?

Verified Solution
Video duration:
1m
This video solution was recommended by our tutors as helpful for the problem above.
231
views
Was this helpful?

Video transcript

Everyone. Let's take a look at this question. Together in mapping experiments, what does the frequency of recombination imply? Let's recall what we've learned about mapping experiments and what the frequency of recombination implies. So, we know in genetics that re combination frequency has to do with that frequency that a single almost almost crossover the curves and this is between two genes, meaning what is that percentage that that crossover between the two genes will occur during mitosis. But when we're dealing with genetic mapping here, we can see a gene that has two mutations here and here and so the frequency of recombination provides us with an estimate of this distance here between those two mutations, which means that answer, Choice B is the correct answer. Because during those mapping experiments, where we can see a gene that has two mutations, we can figure out an estimate of the distance between the two genes to figure out that frequency of recombination. Because we know that recombination frequency is the probability that a single chromosome will crossover will occur between two genes. So answer Choice B is the correct answer. I hope you found this video to be helpful. Thank you and goodbye
Related Practice
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?
232
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.
372
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.
262
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 + Mutant 7 (Problem 21) failed to complement any of the other mutants (1–6). Define the nature of mutant 7.
261
views
Textbook Question
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).] Donor DNA Recipient Cell Transformants No. trp⁺ tyr⁻ 196 A. trp₂⁺ tyr₁⁺ trp₂⁻ tyr₁⁻ trp⁻ tyr⁺ 328 trp⁺ tyr⁺ 367 trp₂⁺ tyr₁⁻ trp⁺ tyr⁻ 190 B. and trp₂⁻ tyr₁⁻ trp⁻ tyr⁺ 256 trp₂⁻ tyr₁⁺ trp⁺ tyr⁺ 2
310
views
Textbook Question
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 c⁺ 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.) Time of Interruption _ 5 min 10 min 15 min 20 min Nutrients A and B 0 0 4 21 Nutrients B and C 0 5 23 40 Nutrients A and C 4 25 60 82 What is the purpose of rifampicin in the experiment?
1091
views
1
rank