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Ch. 18 - Control of Gene Expression in Bacteria
Chapter 17, Problem 10

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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Step 1: To identify the colonies with functioning copies of ββ-galactosidase, look for the colonies that have turned blue. This is because the enzyme ββ-galactosidase breaks down X-gal into a product that turns blue.
Step 2: The colonies that have not turned blue might have mutations in the lacZ or lacY genes. These genes code for proteins that are part of the lactose metabolism pathway, including ββ-galactosidase. If these genes are mutated, the enzyme may not be produced or may not function correctly, and X-gal will not be broken down.
Step 3: If the protein-coding sequences of the lacZ and lacY genes are normal, then the mutation causing the lack of color change could be in the regulatory region of the lac operon. This region controls the expression of the lacZ and lacY genes.
Step 4: The regulatory region includes the promoter, where RNA polymerase binds to start transcription, and the operator, where a repressor protein can bind to prevent transcription. If there is a mutation in these areas, it could prevent the lacZ and lacY genes from being transcribed into mRNA and translated into protein, even if the genes themselves are normal.
Step 5: To confirm this, you could sequence the regulatory region of the lac operon in the mutant colonies and compare it to the sequence in the wild type colonies. If there are differences, this could explain the lack of ββ-galactosidase activity and the lack of color change in the presence of X-gal.

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Key Concepts

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

Lac Operon

The lac operon is a set of genes in E. coli that are involved in the metabolism of lactose. It includes genes such as lacZ, which encodes β-galactosidase, and lacY, which encodes a lactose permease. The operon is regulated by the presence or absence of lactose, allowing the bacteria to efficiently utilize lactose when available. Understanding the lac operon is crucial for analyzing mutations that affect lactose metabolism.
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Guided course
07:06
The Lac Operon

β-galactosidase

β-galactosidase is an enzyme produced by the lacZ gene that catalyzes the hydrolysis of lactose into glucose and galactose. It also cleaves X-gal, a synthetic substrate, producing a blue pigment as a byproduct. The presence of functional β-galactosidase is indicated by blue colonies on X-gal media, making it a useful marker for identifying E. coli that can metabolize lactose.

Mutations in Regulatory Regions

Mutations can occur not only in the coding regions of genes like lacZ and lacY but also in regulatory regions that control their expression. For example, mutations in the promoter or operator regions of the lac operon can prevent the transcription of these genes, leading to a lack of β-galactosidase or lactose permease even if the coding sequences are normal. Analyzing these regulatory regions is essential for understanding the mutant phenotype in the context of lactose metabolism.
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Related Practice
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

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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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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Textbook Question

LuxR is allosterically regulated by the inducer molecule secreted by V. fischeri. What does it mean that LuxR is allosterically regulated?

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