Explain the purpose of the main steps in a DNA extraction procedure
Requirements: 1
1 Lab 3 In today’s lab you will extract a DNA plasmid from a bacterial culture. In this case, you will use the stock solutions you prepared in the previous labs to make the more complex solutions needed for this one. Along the way, you will also get some practice using some of the tools commonly used in various Biology disciplines. The skills you develop here will be useful to you, not just in this course, but in other courses and in many potential research positions. Learning Objectives: Students will: • Explain the purpose of the main steps in a DNA extraction procedure • Explain the importance of the various reagents being used in the DNA extraction procedure • Determine the amount and purity of their DNA samples Pre-Lab Questions: Please view the video linked here, and read the material on this page. Also, feel free to use any other resources to help you answer the following: • Explain the reasons why each of the 3 components are present in our Solution A. • The author of the video does not mention using Phenol/Chloroform, but we will be. What does this solution actually do? Why are we using it? • The author of the video made one fairly big mistake in his explanation. What was it? Important Note: This week’s lab procedure is fairly long. If you come unprepared, you may not be able to complete the procedure before the lab ends. It’s very important that: 1. You will receive an email from JOVE. Make sure to watch the video and do the questions. 2. you come in with your calculations completed (same as post-lab 2 questions) 3. you check what you calculated against what other students calculated
2 Introduction to Plasmid Isolation The DNA you will extract today is an extrachromosomal DNA molecule that is commonly used in molecular biology. Plasmids are small, circular, naturally occurring molecules that have been modified for use in laboratories. In nature, these molecules often provide genes which increase the host cell’s survival. For example, a plasmid might contain a gene which allows it to use a new macromolecule as a source of food. Plasmid DNA extraction is a crucial step in many molecular biology experiments because it provides researchers with a source of pure and concentrated plasmid DNA. This isolated plasmid DNA can then be further analyzed, manipulated, or used in various downstream applications. These plasmid molecules are useful in the lab because they can be used by scientists to insert new genes into bacterial cells (host cells) like E. coli and allow them to study how the new gene affects the bacterium. Or they can be used to produce the protein encoded by the inserted gene. Most commonly, however, plasmids are inserted into bacteria to amplify (make many copies of) a particular DNA sequence. A typical plasmid used for research purposes has been engineered to have the following three important features: an origin of replication, a selectable marker gene (ie. antibiotic resistance), and a multiple cloning site (MCS) where a gene of interest can be inserted. As you will see later this semester, inserting plasmids into bacterial cells is very inefficient – only a small percentage of cells will actually take up a plasmid into their cytoplasm. Thus, we need some way of identifying those cells, and this is accomplished through the use of the “selectable marker gene”. The selectable marker gene is usually a gene that provides the host cell with resistance to an antibiotic, and it is this resistance that allows researchers to “select” only the cells that contain the plasmid. Only the cells with a plasmid will be able to grow on the antibiotic
3 and cells that don’t have the plasmid will not grow. Thus we are “selecting” for a particular trait which “marks” these cells as having the plasmid that we want. The origin of replication determines how many copies of the plasmid will be present in each cell. “High-copy-number” plasmids will replicate themselves until there are hundreds of copies in a single cell. This, along with the ability of cells like E. coli to double in number every 20-30min (ie. 1 million cells becomes 2 million cells in about 20min) means that a DNA sequence inserted into a plasmid can be significantly amplified very quickly and easily. Once such a plasmid has been sufficiently amplified, scientists extract it from the bacteria and purify it. Today’s lab will introduce you to one of the most common techniques of purifying plasmid DNA from E. coli – this is called the Alkaline Lysis Mini Prep. Key Concepts Behind the Alkaline Lysis Procedure The alkaline lysis procedure is a widely used method for extracting plasmid DNA from bacterial cells. It relies on a series of chemical reactions to disrupt the bacterial cell wall, denature cellular proteins, and isolate plasmid DNA. The Alkaline Lysis procedure relies heavily on the small size of plasmids and the fact that they normally exist in a supercoiled conformation in the host cells. The supercoiling is important because it prevents the DNA strands from separating too much during the denaturation step – only short regions of the DNA will be able to separate because the physical strain introduced on the rest of the DNA double helix will prevent further separation. The small size is important because it will help keep the plasmids soluble while other macromolecules are precipitating out. We won’t go into the details of what each of the steps does here (please check the webpage linked in the pre-lab exercise), but the key steps to this protocol are the addition of Solution B (NaOH/SDS) and Solution C (Acetate solution). Solution B is responsible for the cell lysis and denaturation of proteins and DNA, while Solution C promotes renaturation and precipitation of insoluble molecules. The denaturation step affects plasmid and genomic DNA in similar ways – both of them will have their DNA strands separate, but during renaturation (after Solution C is added) the small size of the plasmid allows complementary regions to more easily find each other and stay soluble in the solution (please see this video). The genomic DNA, due to its much larger size, is unable to efficiently renature and so starts to clump together and becomes insoluble. Determining DNA Concentration & Purity There are a few different methods for determining the concentration of DNA in your sample. Some require a UV spectrophotomter and some just rely on you having a gel and a DNA sample whose concentration you already know. In general, if you have the necessary tools, use the spectrophotometric methods – they’re more accurate and allow you to get some idea of your sample’s purity. One of the most common ways to determine the concentration of RNA or DNA samples is by the use of a UV spectrophotometer. Both RNA and DNA absorb UV light very efficiently making it possible to detect and quantify either. In practical terms, the Absorbance of a sample is measured at 260nm and 280nm, and the absorbance at 260 can be used to find the concentration using this formula: 50µg/ml x A260 of the sample = concentration of your sample (µg/ml)
4 This formula is based on knowledge that 50µg/ml of pure DNA has an OD of 1. Similarly, 40µg/ml of pure RNA has an OD of 1 Purity at A260/A280 ratio To assess the purity of the extracted DNA, you can use the A260/A280 ratio, which is calculated by dividing the A260 value by the A280 value. This is because 260nm is maximally absorbed by nucleic acids (DNA as well as RNA), while light at 280nm is well absorbed by things like peptide bonds (ie. protein contamination) as well as Phenol and other reagents that are commonly used in DNA extraction protocols. Because we’re looking at the ratio of Nucleic Acids contaminants, the lower the A280 reading is, the higher the ratio, and the higher the purity the nucleic acids in your sample. Conversely, the more contaminants there are (high A280 reading), the lower the ratio will be. As a rule, a A260/A280 ratio in the range of 1.8-2.0 is considered pure. Usually pure DNA will give a ratio close to 1.8, while pure RNA will be closer to 2.0. If the A260/A280 ratio is significantly lower than 1.8, it indicates the presence of protein contamination in the DNA sample. A ratio higher than 2.0 can indicate the presence of RNA contamination in the DNA sample. RNA has a higher absorbance at 260 nm compared to DNA, leading to an elevated A260 value. Purity at A260/A230 ratio In some cases, researchers might also check the A260/A230 ratio. Using this ratio might be useful for those extracting from plants – carbohydrates (like cellulose) tend to absorb fairly strongly at 230nm. Also, a few extraction protocols use reagents (ie. TRIzol) which may be still present in the final sample and can be detected at 230nm. Knowing about such contaminants can be critical in cases where their presence could interfere with experiments you plan to do in the future (ie RT-PCR). Gel Method Alternatively, gel electrophoresis can be used to determine an approximate concentration of DNA in a sample. This is done by cutting the DNA with a restriction enzyme, running it on an agarose gel. The DNA is then visualized and the thickness/intensity of your sample can then be compared to a DNA standard (a sample of known concentration) as seen in the figure below.
5 While this method will tell you nothing about the purity of your sample, not all procedures require a DNA sample that is really clean. If your DNA sample will not be used for anything like that, running it out on a gel can be a simpler way of estimating your DNA sample concentration. Exercises In this lab, we will be using E. coli cells (a strain known as “TOP10”), which contain the pUC18 plasmid. The cultures you will be receiving were grown overnight at 37C and their growth was tested using spectrophotometry. Once the culture was determined to be at “mid-log” (OD600 = 1.0), it was harvested for this experiment. Materials • 3ml of E. coli cells (TOP10) • Solution A: 50ml of 25mM Tris-HCl; 50mM Glucose; 1mM EDTA-Na2; pH8.0 • Solution B: 5ml of 0.2M NaOH; 1% SDS • Solution C: 50ml of 5M Potassium Acetate/Acetic Acid • Solution D: Phenol:Chloroform:Isoamyl Alcohol (25:24:1) – in the fumehood • Solution E: 5ml of TE (10X): 100mM Tris-HCl ; pH: 8.0; 10mM EDTA-Na 2 • Solution F: TER: 1ml TE (1X) + 2μg RNase. – provided by lab tech • 5ml of 100% Ethanol • 5ml of 70% Ethanol 1. Solution Preparation 1. Use the stock solutions prepared in the previous week to prepare solution A, B, C, and E. 2. As soon as you finish them, place A, C and E on ice, and leave B at room temp. 2. Alkaline Lysis Mini-prep 1. Add 1.5ml of culture to two eppendorf tubes 2. Centrifuge for 1min at maximum speed 3. For each tube, discard the supernatant, then invert the tubes over a paper towel and tap gently to remove any remaining drops of media 4. Add 100μl of ice-cold Solution A to each tube and use a vortex to resuspend the pellets ◦ The solution should become cloudy 5. Add 200μl of Solution B, and mix by inverting each tube quickly about 5 times and place the tubes on ice for 5min. ◦ The solution should become clear ◦ Do NOT vortex at this step! This is a very sensitive stage of the protocol. Vortexing here will break up the genomic DNA into small pieces and will cause it to contaminate your plasmid DNA. 6. Add 150μl of ice-cold Solution C and mix for 10sec using the vortex. Place on ice for 5min ◦ You should see lots of white precipitate after vortexing – this is all the protein and genomic DNA along with the Potassium and SDS precipitating out of solution.
6 7. Centrifuge for 5min at maximum speed to pellet the precipitate. 8. Transfer 400μl of each supernatant into a new eppi tubes. 9. Add 400μl of Solution D, cap the tubes tightly, and shake them well for 1min ◦ Caution: Solution D is corrosive, please be very careful to not spill it, and dispose of any waste in a designated container in the fumehood ◦ This step removes most proteins and some RNA 10. Centrifuge for 1min at maximum speed. 11. Take the tubes out of the centrifuge very carefully and keep them at the same angle at which they were in the rotor. ◦ You should see something like the image on the right, but the difference in colour may not be as obvious. ◦ Do not allow the layers to mix 12. Carefully transfer 350μl of the aqueous phase into fresh eppi tubes. ◦ Use a P200 to do this because the tips are smaller and displace less liquid. 13. Add 700μl (2 volumes) of absolute ethanol. 14. Mix by inverting, and incubate for 2min at room temperature 15. Centrifuge for 5min at maximum speed, and discard the supernatant 16. Wash the pellet with 1ml 70% ethanol. Centrifuge for 1min and discard the supernatant 17. Invert the tubes and tap them on a paper towel to shake out as much of the ethanol as possible 18. Keep the eppendorf tubes open to air dry the DNA at room temperature for 10min to remove the remaining ethanol ◦ this can be done in the fumehood with the sash nearly closed ◦ You can check if the pellet is dry by checking if you can smell any ethanol. 19. While your DNA samples are drying, prepare 10ml of 1xTE from your 10x stock 20. Once the pellets are dry: 1. add 50μl of Solution F to one of your tubes (tube 1) and incubate at 37C for 30min 2. add 50μl of 1xTE to your second tube (tube 2) 21. While you’re waiting for tube 1 to incubate with the RNase , you can start working on Exercise 3 with tube 2. 3. Determination of Concentration and Purity of DNA (This part is tentative. It depends on the time available and it can done in lab 5 instead) Before storing your DNA samples (at -20C), you will determine their concentration and purity based on the information you were given in the introductory material to this lab. Because we will
7 be using UV light to determine DNA concentration and purity, we will need to use special quartz cuvettes (plastic cuvettes absorb UV light) – please be sure handle the quartz cuvettes carefully in order to avoid introducing any scratches. 1. For each sample, prepare a 250-fold dilution with a total volume of 2ml 2. Using 1xTE as a blank, measure the absorbance of each of the diluted samples and enter the results I the table below 3. Calculate the DNA concentration (using the formula given in the background section) and the A260/A280 ratio to get determine the purity of your samples. Data Table: Tube 1 DNA conc. (μg/ml) A260/A280 Tube 2 DNA conc. (μg/ml) A260/A280 A260 A280 A260 A280 1/250 dilution Post-Lab Questions: Take a look at your results and answer the following in your notebook: • What can you conclude about the purity of your samples (tube 1 and tube 2)? • Did using solution F have much of an effect on the amount of nucleic acids you extracted? Did it make a big difference in terms of the purity values? • While the plasmid concentration you calculated for the table is in μg/ml, it can be more useful to express the concentration in μg or ng/μl. Please try to make that conversion in your notes.
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