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Ch. 18 - Control of Gene Expression in Bacteria
Chapter 17, Problem 9

In a mutant that lacks adenylyl cyclase, the enzyme that synthesizes cAMP, predict which of the following conditions of extracellular lactose and glucose would cause regulation of the lac operon to differ from that of wild-type cells. a. no lactose, no glucose b. no lactose, abundant glucose c. abundant lactose, no glucose d. abundant lactose, abundant glucose

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1
Understand the role of adenylyl cyclase: Adenylyl cyclase is responsible for converting ATP to cyclic AMP (cAMP). In the context of the lac operon, cAMP binds to the catabolite activator protein (CAP), which then binds to the promoter region of the lac operon to enhance transcription when glucose levels are low.
Analyze the effect of no adenylyl cyclase on cAMP levels: Without adenylyl cyclase, cAMP cannot be synthesized. This means that CAP cannot be activated, leading to reduced transcription of the lac operon regardless of lactose presence when glucose is low.
Evaluate condition a (no lactose, no glucose): In both mutant and wild-type cells, the lac operon would be off due to the absence of lactose, the inducer. The lack of adenylyl cyclase and thus cAMP does not change the regulation under this condition.
Evaluate condition b (no lactose, abundant glucose): Similar to condition a, the lac operon would be off in both mutant and wild-type cells due to the absence of lactose. High glucose levels would normally inhibit lac operon activity through catabolite repression, but the absence of cAMP in the mutant does not affect this since the operon is already off.
Evaluate condition c (abundant lactose, no glucose): In wild-type cells, low glucose would lead to high cAMP levels, activating CAP and thus enhancing lac operon transcription. In mutant cells, the absence of cAMP means CAP cannot be activated, leading to lower expression of the lac operon despite the presence of lactose. This condition shows a difference in regulation between mutant and wild-type cells.

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Key Concepts

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

Lac Operon Regulation

The lac operon is a set of genes in E. coli that are responsible for the metabolism of lactose. Its regulation is influenced by the presence of lactose and glucose. When lactose is present, it binds to the repressor protein, allowing transcription of the operon. Conversely, high glucose levels inhibit the operon through catabolite repression, preventing cAMP production and thus reducing the activity of the operon.
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cAMP and Adenylyl Cyclase

Cyclic AMP (cAMP) is a secondary messenger important for signal transduction in cells. It is synthesized from ATP by the enzyme adenylyl cyclase. In the context of the lac operon, cAMP levels are inversely related to glucose concentration; low glucose leads to high cAMP, which activates the CAP (catabolite activator protein) to enhance transcription of the lac operon. A mutant lacking adenylyl cyclase would therefore have low cAMP levels, affecting operon regulation.
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Catabolite Repression

Catabolite repression is a regulatory mechanism that ensures bacteria preferentially utilize the most efficient energy source. In the case of the lac operon, when glucose is abundant, it inhibits the expression of genes involved in lactose metabolism, even if lactose is present. This is mediated by low cAMP levels, which prevent CAP from binding to the promoter, thus reducing transcription of the lac operon genes.
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Related Practice
Textbook Question

Predict what would happen to regulation of the lac operon if the lacI gene were moved 50,000 nucleotides upstream of its normal location.

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Textbook Question

Explain why it makes sense for the lexA regulatory gene of the SOS regulon to be expressed constitutively.

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Textbook Question

IPTG is a molecule with a structure much like lactose. IPTG can be transported into cells by galactoside permease and can bind to the lac repressor protein. However, unlike lactose, IPTG is not broken down by ββ-galactosidase. Predict what would occur to lac operon regulation if IPTG were added to E. coli growth medium containing no glucose or lactose.

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Textbook Question

X-gal is a colorless, lactose-like molecule that can be split into two fragments by ββ-galactosidase. One of these product molecules creates a blue color. The photograph here shows E. coli colonies growing in a medium that contains X-gal. Find three colonies whose cells have functioning copies of ββ-galactosidase. Find three colonies whose cells might have mutations in the lacZ or the lacY genes. Suppose you analyze the protein-coding sequence of the lacZ and lacY genes of cells from the three mutant colonies and find that these sequences are wild type (normal). What other region of the lac operon might be altered to account for the mutant phenotype of these colonies?

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Textbook Question

The light-producing genes of V. fischeri are organized in an operon that is under positive control by an activator protein called LuxR. Would you expect the genes of this operon to be transcribed when LuxR is bound or not bound to a DNA regulatory sequence? Explain.

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Textbook Question

The diagram shown here is a model of the gene regulatory circuit for light production by V. fischeri cells. The lux operon contains genes for luminescence (luxCDABE) and a gene, luxI, that encodes an enzyme that catalyzes the production of an inducer. This inducer easily moves back and forth across the plasma membrane and acts as a signaling molecule. The lux operon is never completely turned off. The luxR gene codes for the activator LuxR. The inducer can bind to LuxR, and when it does, the LuxR–inducer complex can bind to a regulatory site to activate transcription of the lux operon and inhibit transcription of luxR. Explain how this gene regulatory circuit accounts for bacteria emitting light only when they reach a high cell density.

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