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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
Chapter 6, Problem 22c

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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Step 1: Understand the problem. The goal is to determine the lost values (dilution and plaque numbers) for the recombination frequency between mutant 6 and mutant 1 on E. coli K12. The problem states that recombination was ten times more frequent than the recombination frequency between mutants 1 and 2.
Step 2: Calculate the recombination frequency for mutants 1 and 2 on E. coli K12. Use the formula for recombination frequency: \( \text{Recombination Frequency} = \frac{\text{Number of Recombinant Plaques}}{\text{Total Plaques}} \). From the table, the number of plaques for mutants 1 and 2 on K12 is 8, and the dilution is \( 10^{-2} \).
Step 3: Multiply the recombination frequency of mutants 1 and 2 by 10 to find the recombination frequency for mutants 6 and 1 on K12. This is because the problem states that recombination was ten times more frequent for mutants 6 and 1.
Step 4: Use the new recombination frequency to calculate the number of plaques for mutants 6 and 1 on K12. Rearrange the recombination frequency formula to solve for the number of plaques: \( \text{Number of Plaques} = \text{Recombination Frequency} \times \text{Total Plaques} \). Assume the same dilution factor (\( 10^{-2} \)) unless otherwise specified.
Step 5: Verify the results by ensuring that the calculated recombination frequency for mutants 6 and 1 is indeed ten times the recombination frequency for mutants 1 and 2. This step ensures consistency and correctness in the calculations.

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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 refers to the rate at which genetic recombination occurs between two loci during meiosis. It is often expressed as a percentage or a ratio, indicating how often offspring inherit a combination of alleles different from their parents. In this context, understanding recombination frequency is crucial for interpreting the results of mutant interactions and predicting the outcomes of genetic crosses.
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Mutant Phenotypes

Mutant phenotypes are observable traits resulting from genetic mutations that alter the normal function of genes. In the context of the question, the phenotypes of the E. coli strains (r and +) indicate different genetic characteristics that can affect recombination rates. Recognizing how these phenotypes relate to the underlying genetics helps in analyzing the data and understanding the implications of the recombination results.
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Dilution and Plaque Assays

Dilution and plaque assays are techniques used to quantify the number of viral particles or bacterial colonies in a sample. In this scenario, the dilution factor indicates how much the original sample was diluted, while the number of plaques represents the number of successful infections or colonies formed. These measurements are essential for calculating recombination frequencies and understanding the relationship between the mutants and their phenotypes.
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Related Practice
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.

Calculate the recombination frequency.

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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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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.)

Based on these data, determine the approximate location on the chromosome of the a, b, and c genes relative to one another and to the F factor.

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