Becker's World of the Cell (9th Edition)
Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 20, Problem 20.1CC

Suppose you are analyzing a haploid E. coli strain and discover that the genes of the lac operon are never transcribed and that the bacteria cannot grow in lactose-containing medium. Provide two possible genotypes for your bacteria that would explain this phenotype.

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Summary Introduction

To determine: Two possible genotypes for a strain of haploid Escherichia coli bacteria in which the genes of the lac operon are never transcribed and the bacteria cannot grow in the lactose-containing medium.

Introduction: E. coli has a lactose operon, also called the lac operon. It is an inducible operon. This operon helps in the digestion and transportation of lactose. By genetic analysis of the lac operon of E. coli of various mutants, a lot of information has been retrieved.

Explanation of Solution

The lac operon under normal conditions can be represented by the genotype I+P+O+Z+ where the Z product will be expressed in the presence of lactose. The I gene represents the repressor gene, P represents the lac promoter gene, O represents the lac operator gene and the Z gene represents the protein product that is formed.

Two possible genotypes for a strain of haploid E. coli bacteria in which the genes of the lac operon are never transcribed and the bacteria cannot grow in the lactose-containing medium are as follows:

  • If the I gene has a mutation such that it shows the suppressor genotype, a repressor will be formed in every case. This repressor will bind to the operator in every case, that is whether a lactose inducer is present or absent. Thus, the transcription of the lac genes are suppressed and will never be transcribed. The genotype of such a strain of bacteria can be represented as IsP+O+Z+.
  • If the P gene has a mutation, the promoter will be non-functional. When the promoter is non-functional, the RNA polymerase enzyme will not be able to identify the promoter. Thus, the transcription of the lac genes are suppressed and will never be transcribed. The genotype of such a strain of bacteria can be represented as I+P-O+Z+.

Thus, two possible genotypes for a strain of haploid E. coli bacteria in which the genes of the lac operon are never transcribed and the bacteria cannot grow in the lactose-containing medium are IsP+O+Z+ and I+P-O+Z+.

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A number of mutations affect the expression of the lac operon in E. coli. The genotypes of several E. coli strains are shown below. ("+" indicates a wild-type gene with normal function and "-" indicates a loss-of-function allele.) Please predict which of the following strains would have the highest beta-galactosidase enzyme activity, when grown in the lactose medium. O CAP+ r* p* o* z O CAP* I P* o* z* O CAP* r* P O* z* O CAP I P* O z*
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Suppose you have six strains of E. coli. One is wildtype, and each of the other five has a single one of thefollowing mutations: lacZ−, lacY−, lacI−, oc, andlacIS. For each of these six strains, describe thephenotype you would observe using the following assays. [Notes: (1) IPTG is a colorless synthetic molecule that acts as an inducer of lac operon expressionbut cannot serve as a carbon source for bacterialgrowth because it cannot be cleaved byβ-galactosidase; (2) X-gal cannot serve as a carbonsource for growth; (3) E. coli requires active lactosepermease (the product of lacY) to allow lactose,X-gal, or IPTG into the cells.] Colony color in medium containing glycerol as theonly carbon source and X-gal, but no IPTG.d. Colony color in medium containing high levels ofglucose as the only carbon source, X-gal, andIPTG.e. Colony color in medium containing high levels ofglucose as the only carbon source and X-gal, butno IPTG
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