BIOS 2011 Lab 4 Lac Operon

BIOS 2011 Lab 4 Lac Operon - Exercise 4 Lac Operon THE...

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Unformatted text preview: Exercise 4 Lac Operon THE LAC OPERON Genetic control of enzyme expression in Escherichia coli Learning Objectives: 1. To become familiar with the operon involved with lactose metabolism in E. coli. 2. To think about the mechanisms of gene regulation in general and how the lac operon provides a model for this process. 3. To run gel for the Ant PCR lab and gather results. Readings: Hillis, Sadava, Heller, and Price, Principles of Life, 2012. Pages 212 ­215. Griffiths AJF, Miller JH, Suzuki DT, et al. , An Introduction to Genetic Analysis. Catabolite repression of the lac operon: positive control: http://www.ncbi.nlm.nih.gov/books/NBK22065/ Assignment: Data and analysis writeup. See Appendix I. Worth 20 pts. Introduction A bacterial cell contains the genetic information to synthesize several thousand different enzymes. Regulatory mechanisms control the rate of synthesis of a number of these enzymes, enabling the cell to utilize its available resources most efficiently and to adapt rapidly to new environments. Such a regulatory mechanism is called an operon. In more precise language, an operon is a cluster of physically adjacent and functionally related genes whose transcription is turned on and off coordinately. The mRNA transcribed from an operon is often referred to as polycistronic, meaning that it carries the coded information for several protein molecules within a single large mRNA molecule. (The term 'cistron' is frequently used to describe the coding unit for a single peptide chain.) This gene cluster consists of a promoter, an operator, and one or more structural genes. Bear in mind that genes occupy certain sites, or loci (sing. locus), of lengths ranging from a few to several thousand base pairs long, and that they code for polypeptides or RNA. Structural genes are so ­called because they code for proteins that are needed by the cell for either structural or enzymatic functions. Regulator genes also code for proteins; these are used as "switches" in the control of structural gene expression. The promoter (P ) and operator (O ) sites are regions of attachment for regulatory molecules — they do not code for gene products. Three enzymes are involved in lactose metabolism in E. coli: (1) b galactosidase, an enzyme that splits lactose into glucose and galactose (the latter sugar can be converted to glucose by a different enzyme system), (2) galactosidase permease, an enzyme that specifically promotes uptake of ß ­galactosides, such as lactose, into the cell (indeed, without permease lactose 71 Exercise 4 Lac Operon cannot penetrate into the cell), and (3) b ­galactoside transacetylase, an enzyme that transfers an acetyl group from acetyl ­CoA to ß ­galactosides. (The acetylation reaction results in detoxification and excretion when bacteria are grown on certain substrates. This is the last mention that will be made of transacetylase since its production is peripheral to our understanding of the lac operon and it is beyond the concern of our experimental assay.) The structural genes directing the formation of the aforementioned enzymes are Z, Y, and A, respectively. A Z  mutant is incapable of synthesizing active b ­galactosidase. A Y  mutant cannot use lactose as a carbon source because it cannot transport lactose from the medium into the cell. Only traces of b ­galactosidase and permease are synthesized when the cell is grown in the absence of lactose; this residual volume is the result of a basal level of expression of the lactose operon. It makes sense for the bacterium to have a few copies of the enzymes on hand in case they are needed. Otherwise, how could the machinery of the operon get turned on at all without permease to allow substrate into the cell? When lactose is present as the only carbon source, the level of these enzymes rapidly increases approximately a thousand fold. This response is called enzyme induction, and wild type E. coli is inducible for b ­galactosidase and permease. In other words, the lac operon is an inducible system. Inducibility is controlled by the regulator gene, I . The I gene produces the repressor protein which binds to the operator site on the DNA. This prevents RNA polymerase, the enzyme that catalyzes transcription, from binding to the promoter, the site contiguous to the operator at which the transcription of the Z and Y genes is initiated. Allolactose acts as the inducer of the operon. This substance, an intermediate metabolite of lactose, can combine with the repressor protein. This binding causes an allosteric change in the repressor protein's configuration so that it is no longer able to bind to the operator. The result is the expression of the Z and Y genes; the repressor protein no longer prevents the binding of RNA polymerase to the promoter and transcription proceeds. Mutants of the I site (I  ­) can be isolated that synthesize the enzymes in maximal amounts whether or not lactose is present. In I  ­ mutants no active repressor is produced, and the Z and Y genes are always turned on. This is called constitutive expression (or constitutive transcription). b ­galactosidase and permease can also be induced by analogs of lactose that mimic its ability to inactivate the repressor. However, these analogs cannot be metabolized. One of these gratuitous inducers is TMG (thiomethly ­b ­galactoside) which has the additional property that, if present in sufficiently high concentrations, it can enter a permeaseless (Y  ­) cell. The lac operon is a glucose sensitive operon, meaning that the presence of glucose in the medium with lactose severely reduces the rate of synthesis of b ­galactosidase. Glucose will be used preferentially as the carbon source. Why? This phenomenon is known as catabolite 72 Exercise 4 Lac Operon repression, or the glucose effect. Cells actively catabolizing glucose have reduced levels of cyclic ­ AMP, a compound needed in conjunction with a protein called CAP (catabolite activator protein) for active transcription of the Z and Y genes. (CAP is made by another gene elsewhere on the bacterial chromosome; c ­AMP is an important and ubiquitous metabolite.) A low glucose concentration in the medium allows for a higher concentration of c AMP, which predisposes the cell for the metabolism of lactose. The cAMP ­CAP complex functions by binding to the regulatory region of the operon and thereby increases the rate of transcription by enhancing the affinity of RNA polymerase for the promoter. Take note that this mechanism exerts a positive control on the operon — it helps to turn on transcription. This is in contrast to the negative control exhibited by the action of the repressor protein, which keeps the system turned off unless its action is blocked by the presence of inducer molecules. NOTATION In this discussion we have been using the designations I , P, Z, Y , and A. There is an alternate and more conventional form of designation that you may encounter in your readings, that being lacI, lacP, lacO, lacZ, lacY, and lacA. As this is cumbersome and we know we are dealing with the lac region of the bacterial chromosome, we have utilized the simpler designations. The wild type condition of a gene may be described by the superscript (+) on the symbol for that gene, and a mutant with a defective or absent gene by the superscript ( ­). A full genetic description of an organism would include a mention of every gene but this is usually impossible and always unwieldy. Instead, it is usual to mention only the genes under consideration. Failure to specify the condition of a gene implies that it is normal. For example, the strain labeled I  ­ is in fact I  ­ Z +Y + and so on. PROTOCOL In this exercise you will study the regulation of b ­galactosidase synthesis in wild type (inducible) E. coli, in a constitutive I  ­ strain, in a Z  ­ strain which has a mutant structural gene so that it cannot synthesize active b ­galactosidase even in the presence of lactose, and in a Y  ­ strain which has a mutation in the structural gene for permease so that it cannot transport lactose into the cell. You will measure b ­galactosidase activity by following the hydrolysis (as mediated by the enzyme) of the indicator, ONPG (s ­nitrophenyl ­galactosidase), a colorless artificial substrate of the enzyme which, when hydrolyzed, yields s ­nitrophenyl. This is intensely yellow in alkali and can be readily measured by eyesight (although a photometric analysis would yield more quantitative results, it isn't necessary for our purposes). b ­galactosidase is the only enzyme in E. coli that splits ONPG so there is no need to purify the enzyme. To ensure that the rate of 73 Exercise 4 Lac Operon hydrolysis of ONPG is linearly dependent on the amount of enzyme present, the cells are pre ­ incubated with toluene to render the cell walls freely permeable to ONPG, a large excess of substrate is provided, and the reaction is run under controlled conditions for a comparatively short time. 1. Work in pairs. Collect 16 tubes: 4 each of Medium A, B, C, and D. The tubes are in colored racks on the front bench — refer to the table below for the key. LABEL THE TUBES AS SOON AS YOU GET THEM. (Use Sharpie markers and label the glass of the tubes themselves. Don't write on the caps or on the test tube racks.) Each tube contains 2 mls of minimal media (salts, buffer, and amino acids) supplemented with either glycerol, lactose, glycerol and TMG, or lactose and glucose, as described in the table below. Note that glycerol is a triose, a 3 carbon sugar that does not affect the regulation of the lac operon. MEDIUM COLOR CODE A RED B C D GREEN YELLOW BLUE INGREDIENTS glycerol (2 x 10 ­3 M ) lactose (2 x 10 ­2 M ) glycerol (2 x 10 ­3 M ); TMG (2 x 10 ­3 M ) lactose (2 x 10 ­2 M ); glucose (4 x 10 ­2 M ) 2. There are four strains of E. coli code ­labeled Strain 10, Strain 20, Strain 30, and Strain 40 growing in the pure glycerol medium (medium A). For each culture, place 5 drops of that culture into each of a set of 4 tubes containing the different media (A ­D). Note: None of these 4 strains is a pathogenic or biohazardous strain of E. coli. 3. Incubate the 16 tubes for one hour in a water bath at 37° C. Shake the tubes every 10 minutes. 4. After 1 hour, add 1 drop of toluene (in a hood, wear gloves) to each tube. Mix well. Incubate for an additional 10 minutes. Toluene does not lyse the cells but makes them accessible to the substrate. Safety note about toluene Toluene is a hazardous organic solvent chemical that you don’t want to breathe or get on your skin or clothing. Please be sure to take the following precautions: • Work with toluene in the hood, which basically keeps gases inside the hood from drifting outside where you’re standing. • Keep everything fully in the hood while you’re working with it. • WEAR GLOVES AT ALL TIMES when handling anything with toluene. • When you are done with the experiment, pour the all media with toluene into the hazardous liquid waste container in the hood. Do not pour any toluene containing solutions into the sink! 74 Exercise 4 Lac Operon 5. After 10 minutes, add 2 drops of the substrate, ONPG, to each tube. Mix well. Incubate for exactly 20 minutes. Remove the tubes from the water bath and record the results below. Worksheet for your color results data: Media Strain 10 Strain 20 Strain 30 Strain 40 A  ­ Glycerol B  ­ Lactose C  ­ TMG/Glycerol D  ­ Lactose/Glucose Notes: 75 Exercise 4 Lac Operon APPENDIX I. GUIDELINES FOR THE LAB WRITEUP As you know, the strains are coded. You will have to identify the strains by comparing their behavior in the different media. The various strains may not have been harvested at the same point in their growth curves, thus there may be no particular significance in subtle differences in color produced by different strains growing in the same medium. However, clearly visible differences are significant, as are comparisons of the enzyme levels of a single strain in different media. For the analysis, construct a table giving the results of your experiment. Remember to provide a table #, title, and row and column headings. Indicate the contents of each lettered media, the code # of the inoculating strain and the type or mutation (Z , etc.) you identify it to be. Record the final color of each tube (yellow, pale yellow, clear, etc.) In a caption explain what different color intensities mean in terms of b ­galactosidase activity (no activity, basal level, low, medium or high, etc.). In a series of separate paragraphs, answer the following questions for each strain: • What results did you obtain for the strain in each media? • Was active b ­galactosidase produced in each tube? • How does the pattern of color changes allow you to identify the mutation of each strain? • If applicable to the tube, explain the molecular mechanisms of catabolite repression of transcription. • Try to explain any results that don’t seem to fit the hypothetical results you might have expected. 76 Exercise 4 Lac Operon LAB WRITEUP CHECKLIST AND GRADESHEET: LAC OPERON Total Points – 20 Table – Table of the results of the Lac Operon experiment. Table should include:  ­ Table #  ­ Title  ­ Row and Column Headings  ­ The contents of each letter tube  ­ The code of the inoculating strain  ­ The type of mutation  ­ The final color  ­ A complete caption that:  ­ Answers the question: What level of b ­galactosidase activity is indicated by the various shades of yellow?  ­ Answers the question: What results did you obtain for the strain in each media?  ­ Answers the question: Was b ­galactosidase produced in each tube?  ­ Answers the question: How does the pattern of color changes allow you to identify the mutation of each strain?  ­ If applicable to the tube, explains the molecular mechanism of catabolite repression of transcription.  ­ Explains any anomalous results for your experiment. 77 ...
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