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Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
Chapter 6, Problem 23b

Synthesis of the amino acid histidine is a multistep anabolic pathway that uses the products of 13 genes (hisA to hisM) in E. coli. Two independently isolated his- E. coli mutants, designated his1⁻ and his2⁻ are studied in a conjugation experiment. A his⁺ F' donor strain that carries a copy of the hisJ gene on the plasmid is mated with a his1⁻ recipient strain in Experiment 1 and with a his2⁻ recipient in Experiment 2. The exconjugants are grown on plates lacking histidine. Growth is observed among the exconjugants of Experiment 2 but not among those of Experiment 1.
What is the genotype of exconjugants in Experiment 2?

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Step 1: Understand the problem. The synthesis of histidine involves multiple genes (hisA to hisM). The mutants his1⁻ and his2⁻ are defective in histidine synthesis. A his⁺ F' donor strain carrying the hisJ gene on a plasmid is used to complement these mutants in conjugation experiments.
Step 2: Analyze Experiment 1. In this experiment, the his⁺ F' donor strain is mated with the his1⁻ recipient strain. Since no growth is observed on plates lacking histidine, this suggests that the hisJ gene provided by the plasmid does not complement the defect in the his1⁻ strain. Therefore, the mutation in his1⁻ likely affects a gene other than hisJ.
Step 3: Analyze Experiment 2. In this experiment, the his⁺ F' donor strain is mated with the his2⁻ recipient strain. Growth is observed on plates lacking histidine, indicating that the hisJ gene provided by the plasmid complements the defect in the his2⁻ strain. This suggests that the his2⁻ strain has a mutation in the hisJ gene.
Step 4: Determine the genotype of the exconjugants in Experiment 2. After conjugation, the exconjugants inherit the hisJ gene from the plasmid, which complements the his2⁻ mutation. Therefore, the exconjugants are his2⁻ for chromosomal genes but hisJ⁺ due to the plasmid.
Step 5: Conclude the genotype. The genotype of the exconjugants in Experiment 2 is his2⁻ (chromosomal mutation) and hisJ⁺ (plasmid-borne gene). This combination allows them to synthesize histidine and grow on plates lacking histidine.

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Key Concepts

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

Gene Function and Pathways

In genetics, genes encode proteins that perform specific functions in biological pathways. The synthesis of amino acids, such as histidine, involves multiple genes working together in a coordinated manner. Understanding how these genes interact and contribute to metabolic pathways is crucial for analyzing genetic mutations and their effects on organismal growth.
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Functional Genomics

Conjugation in Bacteria

Conjugation is a process of horizontal gene transfer in bacteria, where genetic material is transferred from one bacterium to another through direct contact. In this experiment, the his⁺ F' donor strain provides the hisJ gene to the his1⁻ and his2⁻ mutants. The ability of the exconjugants to grow in the absence of histidine indicates successful gene transfer and functional complementation of the mutated pathways.
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Conjugation Overview

Mutations and Complementation

Mutations can lead to loss of function in genes, affecting metabolic pathways. Complementation occurs when a mutant strain receives a functional copy of a gene from another source, restoring the pathway's function. In this case, the growth of exconjugants in Experiment 2 suggests that the his2⁻ mutant can utilize the hisJ gene from the donor, while the his1⁻ mutant in Experiment 1 cannot, indicating a difference in the nature of the mutations.
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Related Practice
Textbook Question

An attribute of growth behavior of eight bacteriophage mutants (1 to 8) is investigated in experiments that establish coinfection by pairs of mutants. The experiments determine whether the mutants complement one another (+) or fail to complement (-). These eight mutants are known to result from point mutation. The results of the complementation tests are shown below.

New mutation 10 fails to complement mutants 1, 4, 5, 6, 8, and 9. Mutant 10 forms wild-type recombinants with mutants 1, 5, and 6, but not with mutants 4 and 8. Mutant 9 and mutant 10 form wild-type recombinants. What kind of mutation is mutant 10? Explain your reasoning.

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Textbook Question

An attribute of growth behavior of eight bacteriophage mutants (1 to 8) is investigated in experiments that establish coinfection by pairs of mutants. The experiments determine whether the mutants complement one another (+) or fail to complement (-). These eight mutants are known to result from point mutation. The results of the complementation tests are shown below.

Gene-mapping information identifies mutations 2 and 3 as the flanking markers in this group of genes. Assuming these mutations are on opposite ends of the gene map, determine the order of mutations in the region of the chromosome.

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Textbook Question

Synthesis of the amino acid histidine is a multistep anabolic pathway that uses the products of 13 genes (hisA to hisM) in E. coli. Two independently isolated his- E. coli mutants, designated his1⁻ and his2⁻ are studied in a conjugation experiment. A his⁺ F' donor strain that carries a copy of the hisJ gene on the plasmid is mated with a his1⁻ recipient strain in Experiment 1 and with a his2⁻ recipient in Experiment 2. The exconjugants are grown on plates lacking histidine. Growth is observed among the exconjugants of Experiment 2 but not among those of Experiment 1.

Why is growth observed in Experiment 2 but not in Experiment 1?

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Textbook Question

The phage P1 is used as a generalized transducing phage in an experiment combining a donor strain of E. coli of genotype leu⁺ phe⁺ ala⁺ and a recipient strain that is leu⁻ phe⁻ ala⁻. In separate experiments, transductants are selected for leu⁺ (Experiment A), for ala⁺ (Experiment B), and for phe⁺ (Experiment C). Following selection, transductant genotypes for the unselected markers are identified. The selection experiment results below show the frequency of each genotype.

What compound or compounds are added to the minimal medium to select for transductants in Experiments A, B, and C?

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Textbook Question

The phage P1 is used as a generalized transducing phage in an experiment combining a donor strain of E. coli of genotype leu⁺ phe⁺ ala⁺ and a recipient strain that is leu⁻ phe⁻ ala⁻. In separate experiments, transductants are selected for leu⁺ (Experiment A), for ala⁺ (Experiment B), and for phe⁺ (Experiment C). Following selection, transductant genotypes for the unselected markers are identified. The selection experiment results below show the frequency of each genotype.

Determine the order of genes on the donor chromosome.

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Textbook Question

The phage P1 is used as a generalized transducing phage in an experiment combining a donor strain of E. coli of genotype leu⁺ phe⁺ ala⁺ and a recipient strain that is leu⁻ phe⁻ ala⁻. In separate experiments, transductants are selected for leu⁺ (Experiment A), for ala⁺ (Experiment B), and for phe⁺ (Experiment C). Following selection, transductant genotypes for the unselected markers are identified. The selection experiment results below show the frequency of each genotype.

Diagram the crossover events that form each of the transductants in Experiment A.

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