Lab Exercise: Transformation

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1 Lab Exercise: Transformation Background Genetic transformation is used in many areas of biotechnology, and, at its heart, requires two things: Donor DNA and recipient cells. Cells which receive the donor DNA are considered genetically recombined, that is, they have their original DNA (their chromosome) and new DNA (the plasmid) and whatever genes that plasmid carries. Before considering the details of recombination, we will consider each of these players individually. Plasmids were discovered as extra-chromosomal genetic material in the late 1960s. Like the bacterial chromosome, they are circular but they are much smaller (2,000 10,000 bp), and usually contain genes for one or more traits that may be beneficial to bacterial survival. In nature, bacteria can transfer plasmids back and forth allowing them to share these beneficial genes and adapt to new environments. For instance, the quick dissemination of bacterial resistance to antibiotics is due in part to the transmission of plasmids. These naturally occurring plasmids have been engineered to contain not only antibiotic resistance (which is used in the laboratory as a selective marker for successful transformation) but other genes of interest. If a plasmid is transformed into an E. coli cell that is sensitive to the antibiotic ampicillin, it will confer resistance to that antibiotic. Growing the transformants in the presence of ampicillin is an easy way to select for recombined E. coli cells. However, E. coli that have become antibiotic-resistant are not themselves useful in biotechnological applications. As such, plasmids must be engineered to include some other gene of interest. This gene of interest is spliced into the plasmid at its multiple cloning site- an area of the plasmid that contains many restriction enzyme recognition sites within a reporter gene. When this reporter gene does not contain the gene of interest its phenotype is easily recognized in the successfully transformed cells. Many plasmids are commercially available, and all of them contain four key components: 1. A selectable marker (i.e., a gene that encodes for antibiotic resistance); 2. An origin of replication (ori) for plasmid replication; 3. A multiple cloning site (MCS), containing the restriction enzyme recognition sites; 4. A reporter gene containing the MCS. pbluescript is a commonly used plasmid that contains the reporter gene lacz and the selection marker gene bla which codes for resistance to the antibiotic ampicillin (Figure 1). In vivo, the lacz gene produces an enzyme called -galactosidase which cleaves the disaccharide lactose. In vitro, using the inducer isopropyl- -D-thio-galactoside (IPTG) this enzyme can be made to catabolize a second substrate, X-gal, producing a colored product that allows transformed bacterial colonies to be easily distinguished from wild-type colonies by their dark blue color (Figure 2). However, when a gene of interest has been engineered into the MCS of the reporter gene lacz, these recombined plasmids will confer antibiotic resistance but will not result in blue-colored colonies (Figure 3). 1

2 Figure 1: Diagram of the pbluescript plasmid. Note the three major functional domains of interest in this experiment: the amp r gene which codes for resistance to the antibiotic ampicillin; the ori which allows plasmid replication and the lacz gene which encodes for -galactosidase and allows for the colorimetric detection of transformed cells. Figure 2: Chemical structure of X-gal and the insoluble blue dye that results from cleavage with - galactosidase. Figure 3: On left: Diagram of non-recombined plasmid (no gene of interest ) exhibiting blue colonies and a recombined plasmid (with a gene of interest) exhibiting white colonies due to the insertion of a gene within the reporter gene lacz. On right: Photograph of a plate containing X-gal, IPTG and ampicillin on which transformed cells have been grown. Note the blue color of cells containing pbluescript with an intact lac Z gene and white cells containing pbluescript with an interrupted lac Z gene (a recombined plasmid). 2

3 DNA is a hydrophilic molecule which will not normally pass through the plasma membrane of a bacterial cell. In order to accomplish transformation, the bacterial cells must first be made competent to take up the plasmid DNA. This is done by neutralizing the negative charge of the DNA molecule using CaCl 2. This is followed by a heat shock step that causes the plasma membrane to become permeable. Together the CaCl 2 treatment and heat shock will cause the cells to take up the plasmid DNA. Once the cells have had a chance to recover, they can be plated out on antibiotic-containing media to select for successful transformants. Introduction The DNA you will be transforming your bacteria with is pbluescript, but you will not be performing any manipulation of the plasmid to create recombined plasmids. Thus, the recombined cells should all exhibit the blue phenotype, provided the inducer IPTG is added to the media. The lac Z gene encoded in the pbluescript plasmid is part of the inducible lac operon, and is susceptible to induction with the chemical inducer IPTG. This way, the expression of the lac Z gene can be controlled at the transcriptional level. In media that lacks the inducer IPTG, the lac Z gene will not be transcribed, when IPTG is added, transcription occurs and -galactosidase is produced. This, in turn, cleaves the X-gal. This system produces recombinant cells that are easily recognizable from their non-transformed neighbors. In addition to the lac Z gene, the pbluescript plasmid encodes the bla gene which confers resistance to the antibiotic ampicillin. In this experiment we will be transforming a strain of E. coli called HB101. As part of the experiment, it is important to ensure that these cells do not exhibit the selective or reporter phenotypes without the plasmid genes. This can be confirmed by observing the original E. coli colonies. It will also be important, during the course of the experiment, to ensure that these E. coli are not naturally resistant to ampicillin. This transformation procedure involves three main steps. These steps are intended to introduce the plasmid DNA into the E. coli cells and provide an environment for the cells to express their newly acquired genes. To move the pbluescript plasmid DNA through the cell membrane you will: Use a transformation solution of CaCl 2 (calcium chloride). Carry out a procedure referred to as heat shock. For transformed cells to grow in the presence of ampicillin you must: Provide them with nutrients and a short incubation period to begin expressing their newly acquired genes For transformed cells to exhibit the blue phenotype you must: Provide them with the operon inducer IPTG and the substrate X-gal. 3

4 Lab Exercise Class supplies Table supplies Group supplies 42 C water bath E. coli starter plate LB, LB/Amp, LB/Amp/X-gal/IPTG plates 1 tube pbluescript plasmid labeled + 1 tube negative control (H 2O) labeled - 1 tube transformation solution (CaCl 2) 1 tube recovery broth (LB broth) P200 and P1000 pipets Pipet tips Spreader Ice bath Protocol 1. Collect your two eppendorf tubes: + (which contains 10µl of the pbluescript plasmid) and (which contains 10µl of sterile H 2 O). 2. Open the tubes and, using a sterile transfer pipet, transfer 250µl of transformation solution (CaCl 2 ) into each tube. 3. Place the tubes on ice. 4. Add 100 µl or 1 colony of the E. coli HB101 (your instructor will indicate which) to each of the tubes from step one. 5. Incubate the tubes on ice for 10 minutes, making sure to push the tubes all the way down in the rack so they make contact with the ice. 6. While the tubes are sitting on ice, label your LB nutrient agar plates on the bottom (not the lid) as follows: Label ½ of one LB/amp plate: + pbluescript Label the other ½ of the LB/amp plate: - pbluescript Label ½ of one LB/amp/IPTG/X-gal plate: + pbluescript Label ½ of one LB plate: - pbluescript 7. Transfer both the +pbluescript and -pbluescript tubes into a 42 C water bath, for exactly 50 seconds, making sure to push the tubes all the way down in the rack so the tubes make contact with the warm water. 8. Place both tubes back on ice for 2 minutes. 9. Remove the tubes from the ice and place them on the bench top. Using a new sterile pipet for each tube, add 250 µl of recovery broth to each tube. 10. Incubate the tubes for 10 minutes at room temperature. 11. Flick the closed tubes with your finger to mix the contents. Pipet 100µl of the transformation and control suspensions onto the appropriate portions of the nutrient agar plates. 12. Using your inoculating loop spread the suspensions evenly across the surface of the LB nutrient agar. 13. Incubate your plates for 24 hours at 37 C. 4

5 Prior to collecting your data, answer the following questions: 1. Why did you plate your experiment on the various plates? Which plates serve as a control? 2. What is meant by a control plate? What purpose does a control serve? 3. What do you expect to see growing on each plate (+ and/or cells)? Answer this question in terms of growth and phenotype (i.e., color). 4. Which plates should be compared to determine if any genetic transformation has occurred? Why? Data and observations 1. Observe and draw the results of your transformation for each of your 3 plates. Make sure to count the CFUs on each plate. LB plate LB/Amp plate CFUs CFUs 5

6 LB/Amp/Xgal/IPTG plate CFUs 2. If the genetically transformed cells have acquired the ability to live in the presence of the antibiotic ampicillin, then what might be inferred about the other genes on the plasmid that you used in your transformation procedure? 3. What evidence exists to show that the changes that occurred were due to this experiment? Discussion 1. Describe the evidence that indicates whether your attempt at performing a genetic transformation was successful. 2. Look again at your plates; do you observe some E. coli growing on the LB plate that does not contain ampicillin or IPTG & X-gal? From your results, can you tell if these bacteria are ampicillin resistant by looking at them on the LB plate? Explain your answer. 6

7 3. Very often an organism s traits are caused by a combination of its genes and its environment (multifactorial traits). Think about the blue color you saw in the genetically transformed bacteria: a. What two factors must be present in the bacteria s environment for you to see the blue color? b. What is each of the two factors you listed above doing to cause the bacteria to turn blue? c. Why is it advantageous for an organism to be able to turn particular genes on or off? 4. Calculate transformation efficiency Your next task in this experiment will be to learn how to determine the success of your transformation of the E. coli cells with the pbluescript plasmid DNA. This quantitative measurement is referred to as transformation efficiency. Transformation efficiency is a number which represents the total number of bacterial cells that express the blue protein, divided by the amount of DNA used in the experiment (i.e. it tells us the total number of bacterial cells transformed by one microgram of DNA). The transformation efficiency is calculated using the following formula: Before you can calculate the efficiency of your transformation, you will need two pieces of information: a. The total number of blue fluorescent colonies growing on your LB/amp/IPTG/Xgal plate (data previously collected). This is the total CFUs in the transformation efficiency equation above. b. The original concentration of pbluescript DNA, and more importantly, the resulting amount plated on the LB/amp/IPTG/X-gal plate. You can determine this by completing the calculations below. 7

8 In this experiment the pbluescript had a stock concentration of ng/µl (1µg = 1000ng). This means that each microliter of pbluescript solution has µg of pbluescript DNA. If you put µl of this pbluescript solution into your +pbluescript tube, your pbluescript tube should have µg of pbluescript DNA total. Now calculate the amount of DNA transferred from that +pbluescript tube to your LB/amp/IPTG/X-gal plate. You plated µl of the +pbluescript solution onto your LB/amp/IPTG/X-gal plate. To determine the percentage transferred of the original amount, divide the number above (amount plated) by the total reaction volume (transformation solution + pbluescript DNA + LB nutrient medium). This will give you the percentage transferred. Multiply this number by the total micrograms of pbluescript DNA in your +pbluescript tube, this will give you the DNA (micrograms) for your transformation efficiency equation. Biotechnologists are in general agreement that the transformation protocol that you have just completed generally has a transformation efficiency of between and transformants/µg DNA. How does your transformation efficiency compare with the above? How does your transformation efficiency compare with the other groups in class? References 1. Hanahan, Douglas. Studies on transformation of Escherichia coli with plasmids. J. Mol. Biol., 166, Hanahan, Douglas. Techniques for transformation of E. coli. In DNA Cloning: A Practical Approach (Ed. D. M. Glover), vol. 1. IRL Press, Oxford Schleif, Robert. Two positively regulated systems, ara and mal, In Escherichia coli and Salmonella, Cellular and Molecular Biology, Neidhardt. ASM Press, Washington, D.C Bio-Rad Laboratories. Student Manual pglo Transformation. 5. Madigan, Michael, John M. Martinko, Paul V. Dunlap and David P. Clark. Brock s Biology of Microorganisms, 11 th Edition. Benjamin Cummings Tortora, Gerard J., Berdell R. Funke and Christine L. Case. Microbiology An Introduction,9 th Edition. Addison-Wesley

9 Additional info (if you are so inclined ) The Lac Operon The genes involved in the transport and breakdown (catabolism) of food are good examples of highly regulated genes. For example, the sugar lactose is both a source of energy and a source of carbon. E. coli produce three enzymes (proteins) needed to digest lactose as a food source (Figure 4). Because of this, the lactose operon is considered inducible. In this specific example, LacI is found normally bound to the operon at the site of the promotermaking it impossible for RNA polymerase to begin the process of transcription. When lactose (or IPTG) is present, it directly binds the repressor and causes it to release the promoter region. The subsequent binding of RNA polymerase results in the transcription of the structural genes and the cell s ability to catabolize lactose (Figure 5). Figure 4: Diagram of the lac operon. Figure 5: Top: Inducible operon when no lactose (or IPTG) is present; Bottom: Induced operon when repressor (LacI) is bound by the inducer lactose (or IPTG) 9

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