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Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages
Klug - Concepts of Genetics 12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
All textbooksKlug 12th EditionCh. 6 - Genetic Analysis and Mapping in Bacteria and BacteriophagesProblem 22a
Chapter 6, Problem 22a

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.

Verified step by step guidance
1
Understand the problem: Recombination frequency is calculated as the number of recombinant plaques divided by the total number of plaques. Recombinant plaques are those that can grow on E. coli K12 (indicated by the '+' phenotype).
Identify the data provided: From the table, the number of plaques on E. coli K12 (recombinants) is 8, and the dilution factor for this strain is 10⁻². For E. coli B, the number of plaques is 4, and the dilution factor is 10⁻⁷.
Calculate the total number of plaques for each strain by multiplying the observed plaques by the dilution factor. For example, for E. coli K12, the total plaques = observed plaques × dilution factor.
Determine the total number of plaques by summing the plaques from both strains (E. coli B and E. coli K12).
Calculate the recombination frequency using the formula: Number of recombinant plaquesTotal number of plaques. This will give the recombination frequency as a proportion.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Recombination Frequency

Recombination frequency is a measure of the likelihood that two genes will be separated during meiosis due to crossing over. It is calculated as the number of recombinant offspring divided by the total number of offspring, often expressed as a percentage. This frequency helps in mapping the distance between genes on a chromosome, with higher frequencies indicating genes that are further apart.
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Recombination after Single Strand Breaks

Mutants and Phenotypes

In genetics, mutants refer to organisms that have undergone a change in their DNA sequence, resulting in a new phenotype, or observable characteristic. The phenotypes can be categorized as wild-type (normal) or mutant types, which can be used to study genetic variations and inheritance patterns. Understanding the phenotypes of the strains involved is crucial for interpreting recombination results.
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Mutations and Phenotypes

Plaque Assay

A plaque assay is a laboratory technique used to measure the number of viral particles or bacterial cells in a sample by observing the formation of plaques, which are clear zones on a bacterial lawn caused by viral lysis. In the context of the question, the number of plaques formed by different strains indicates the effectiveness of recombination and the presence of specific phenotypes, which are essential for calculating recombination frequency.
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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?

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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.

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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 PhenotypesE. coli B 10⁻⁷ 4 rE. 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.
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Textbook Question

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)?

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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).]

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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⁻ rifT (No map order is implied in the listing of the alleles; rifT 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?

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