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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: 9780135564172당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 23a

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?

검증된 단계별 안내
1
Step 1: Understand the genetic basis of the problem. The synthesis of histidine involves multiple genes (hisA to hisM). Mutations in these genes can disrupt the pathway, leading to histidine auxotrophy (inability to synthesize histidine). The his1⁻ and his2⁻ mutants are defective in different genes within this pathway.
Step 2: Analyze the conjugation experiment setup. In Experiment 1, a his⁺ F' donor strain carrying the hisJ gene on a plasmid is mated with the his1⁻ recipient strain. In Experiment 2, the same his⁺ F' donor strain is mated with the his2⁻ recipient strain. The exconjugants are tested for their ability to grow on plates lacking histidine.
Step 3: Consider the role of the hisJ gene. The hisJ gene is one of the genes involved in histidine synthesis. If the his1⁻ mutant has a defect in a gene other than hisJ, the introduction of the hisJ gene via the plasmid will not restore the ability to synthesize histidine, and growth will not occur. This explains the lack of growth in Experiment 1.
Step 4: Examine the outcome of Experiment 2. If the his2⁻ mutant has a defect specifically in the hisJ gene, the introduction of the hisJ gene via the plasmid will complement the mutation, restoring the ability to synthesize histidine. This explains the observed growth in Experiment 2.
Step 5: Conclude the reasoning. Growth is observed in Experiment 2 because the hisJ gene provided by the plasmid complements the mutation in the his2⁻ strain. In Experiment 1, the hisJ gene does not complement the mutation in the his1⁻ strain, as the defect lies in a different gene within the histidine synthesis pathway.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Gene Function and Pathway

The synthesis of histidine in E. coli involves a multistep anabolic pathway regulated by multiple genes. Each gene, from hisA to hisM, encodes enzymes that catalyze specific reactions in the pathway. Understanding how these genes interact and contribute to histidine production is crucial for analyzing the mutants' growth in the experiments.
추천 영상:
가이드 코스
08:26
Functional Genomics

Conjugation in Bacteria

Conjugation is a process of horizontal gene transfer in bacteria, where genetic material is transferred from a donor to a recipient cell through direct contact. In this scenario, the his⁺ F' donor strain provides the hisJ gene to the his1⁻ and his2⁻ mutants. The success of this transfer and the resulting phenotype depend on the compatibility of the genes involved in the recipient strains.
추천 영상:
가이드 코스
04:06
Conjugation Overview

Mutant Phenotypes

The his1⁻ and his2⁻ mutants represent different genetic alterations affecting histidine synthesis. The observation of growth in Experiment 2 but not in Experiment 1 suggests that his2⁻ has a mutation that allows it to utilize the hisJ gene from the donor, while his1⁻ likely has a mutation that disrupts this ability. Analyzing the specific mutations helps explain the differential growth outcomes.
추천 영상:
가이드 코스
10:48
Mutations and Phenotypes
관련 실천
교과서 질문

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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교과서 질문

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.

A new mutation, designated 9, fails to complement mutants 1, 3, 5, 7, and 8. Wild-type recombinants form between mutant 9 and mutations 3, 5, and 8; however, no wild-type recombinants form between mutant 9 and mutations 1 and 7. What kind of mutation is mutant 9? Explain your reasoning.

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교과서 질문

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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교과서 질문

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.

538
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교과서 질문

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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교과서 질문

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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