Day 2: Transformation Relevant Book Sections We will follow the protocols provided in various industry-standard kits, instead of the protocols described in these chapters, but the chapters provide good background and explanations.! Transformation: exercise 20! Expression from a T7 promotor: exercise 22! DNA and genetic elements: pgs 301-332. Likewise, protocols can be found on the web, there are now even journals and video journals devoted to these techniques and the tricks embedded in them. 64 Transformation: Theory Transformation means causing a bacterial cell line to take up foreign DNA. The term refers to the original observation that bacteria acquired new biological properties upon acquiring foreign DNA: they were transformed into a new strain more like the one from which the DNA was derived. For an intact plasmid with no toxic genes (puc or pbr322), 10 8 colonies are obtained per µg DNA used. (For a 4 kbp plasmid, at 600 g/mole bp, the MW is 2.4 x 10 6 g/mole = 2.4 x 10 12 µg/mole. 1 µg = 1 µg * (6 x 10 23 molec/mole) / (2.4 x 10 12 µg/ mole) 2.5 *10 11 molecules of plasmid. 10 8 colonies means we got 10 8 cells transformed. The transformation efficiency is therefore 1 cell / 2500 molecules of plasmid. Transformation will be less efficient when the plasmid is large or has incomplete ligation. It will not occur if a plasmid is not stable (it is not circular or does not contain an origin of replication). Our engineered plasmid should have a replication origin, but it will also likely retain nicks not sealed by ligase. Therefore, we can expect LOWER efficiency. 65
Transformation: Introduction of foreign DNA into cells! Cells have natural controls on access of foreign DNA.! We will use cells whose membranes have been permeabilized, by chemical treatment (CaCl2, DMSO and more). The resulting cells are called competent cells because they are competent to take up DNA. However they are fragile and must be handled with some care.! We will also heat-shock them.! DNA aggregated to exterior phospholipid head groups with Ca 2+ can be brought in.! Once the DNA gets in, it can be degraded, integrated, or independently replicated (requires a replication origin ). After-the-fact elimination of foreign DNA lowers the efficiency of transformation further.! We will transform with our engineered plasmid in parallel with a control plasmid that has not been manipulated. The positive control will be a plasmid carrying the WT GFP gene. 66 Day 2: Transformation and selection We will transform our plasmids directly into an expression strain of E. coli (T7 express) following the protocol that follows. T7 express has that name because it carries the gene for T7 RNA polymerase in its chromosome. Thus it can produce the RNA polymerase that would normally be produced by T7 virus. Genes behind a T7 promoter will be transcribed by this polymerase, and expressed. Hence the name of the cell line. Since only a few cells will actually take up the DNA and copy it (above), we need a good selective screen to identify the cells that do. Usually we choose a screen that kills all cells that are not transformed (zero background). Since a gene conferring ability to degrade kanamycin is on the plasmid, only cells with the plasmid might be expected to survive. (However, look carefully at your plates after they have been allowed to grow for several days.) 67
Day 2: Transformation Execution plan! Get the transformation started.! Pour plates of LB agar containing kanamycin (100 ug/ml). The agar will be provided in molten form (55 C, hot). Students will have a chance to try pouring. Kanamycin (Kn) should be added and agar should be poured when it is still molten but cool enough to touch.! We will be adding Kn from a stock solution that is 100 mg/ml, into 500 ml batches of molten agar. Calculate how much Kn to add.! Complete the transformation as per attached protocol.! Plate out cells on the LB-Kn plates (cartoons follow). T7 Express Iq Competent E.coli (High Efficiency) High Efficiency Transformation Protocol (C3016) Overview For C3016H, perform steps 1-7 in the tube provided. 1. C3016H: Thaw a tube of T7 Express I q Competent E. coli cells on ice for 10 minutes. C3016I: Thaw a tube of T7 Express I q Competent E. coli cells on ice until the last ice crystals disappear. Mix gently and carefully pipette 50 µl of cells into a transformation tube on ice. 2. Add 1-5 µl containing 1 pg-100 ng of plasmid DNA to the cell mixture. Carefully flick the tube 4-5 times to mix cells and DNA. Do not vortex. 3. Place the mixture on ice for 30 minutes. Do not mix. 4. Heat shock at exactly 42 C for exactly 30 seconds. Do not mix. ** 5. Place on ice for 5 minutes. Do not mix. 6. Pipette 950 µl of room temperature SOC into the mixture. 7. Place at 37 C for 60 minutes. Shake vigorously (250 rpm) or rotate. 8. Warm selection plates to 37 C. 9. Mix the cells thoroughly by flicking the tube and inverting, then perform several 10-fold serial dilutions in SOC. Day 2: Transformation Execution plan *** 10. Spread 50-100 µl of each dilution onto a selection plate and incubate overnight at 37 C. Alternatively, incubate at 30 C for 24-36 hours or at 25 C for 48 hours. * * Heat shock for 45 seconds (not 30) ** Add 800 µl SOC (not 950). *** We will plate all the cells. First plate 100 µl of cells. Then spin down remaining cells, decant all but 50 µl of supernatant, resuspend cells and plate those. 68 Selection will be accomplished by plating on agar containing 100 µg/ml ampicillin or kanamycin (engineered or control plasmid, respectively). 69
Sterile Technique Hygiene for you, and hygiene for your bacteria. Plating out bacteria. Swab the bench and your hands with 70% ethanol (wear gloves). Move only the items you need into your sterile field, light a burner to direct air currents up, work efficiently keeping tubes and plates covered whenever possible. At the end, extinguish flame, swab down area again including your hands. Day 3: Expression. Theory. Only cells carrying the plasmid will grow on Kn. However which of these plasmids contains the gene of interest? In order for T7 express cells to express genes cloned behind the T7 promotor, we must turn on production of T7 RNA polymerase. This is the only polymerase that will recognize the T7 promotor. (We should get very selective expression). To turn on (induce) expression of T7 RNA polymerase we will streak transformed cells onto an LB plate containing (provided). Our positive control transformation should provide a very visual measure of whether the plasmidborne gene is expressed because in that case the gene is for green fluorescent protein (GFP). We will be able to tell if it is expressed by illuminating the colonies with a black light. We will check for expression on Thursday. 71
Day 3: Expression. Execution plan: Protocol: Day 3: Phenotypic Screen.! Count colonies obtained from transformation using engineered plasmid, vs. control transformation. Also look at the T.A. s negative control transformation (no plasmid). Interpret these results in your report.! At this point we have screened colonies on the basis of possession of Kn resistance. This is another property encoded by our plasmid, but it is not our gene. We can now as, of the colonies that carry the plasmid (=are Kn resistant), how many can express our gene? To answer this question, we remove impediments to gene expression by comparing the colour of our colonies growing with vs. without.! Streak 3 colonies from each of the mutant and positive control transformations onto LB agar that contains Kn +.! For each colony you streak onto the LB Kn + plate also streak with the same colony onto an LB Kn plate, and number the streaks so you know that both come from the same cell. 1 2 3 m1 WT3 m2 WT1 WT2 m3 What you should have as a result of day 2 s work.! Obtain two LB/Kn plates. Spread 100 µl of stock solution on ONE of the plates and label that one LB/Kn,.! Divide up the area in each plate into six sectors as per the drawing, on the left.! From the plate of your bacteria transformed with mutated plasmid, locate three colonies that are isolated. On the bottom of the petry dish circle and number each one (see above, left). These are colonies m1, m2 and m3.! From your positive control LB/Kn plate, locate and label three unmutated (= wild type = WT) colonies. These are colonies WT1, WT2, WT3. Question: Why might you want a replicate culture that is not subjected to? (hint: think ahead to the topic of the next exercise). 72 73
Kn Protocol: Day 3: Phenotypic Screen.! From each of your labelled colonies, lightly touch the colony, taking up 1/3 the cells, and streak these on one sector of a LB/Kn plate: a plate in which the medium contains as well as Kn. (below, left).! Use a second third of the cells in each of your labelled colonies to streak sectors of an LB/Kn, plate (below, right). WT3 m1 m2 WT1 WT2 m3 Kn- WT3 m1 m2 WT1 WT2! The plates will grow overnight at 37 C. The next day, your TA will move them to the refrigerator. The T.A. will make up similar plates with negative control bacteria (not carrying a GFP plasmid). Question: What would be an ideal control? m3 74 Day 3: Carrying Cost. Objective: estimate the cost of carrying the plasmid and expressing the gene. In growth medium lacking Kn we expect that cells carrying the plasmid will have a disadvantage and become a smaller fraction of the total population. In growth medium lacking Kn and containing the penalty for carrying the plasmid should be even more severe. Mon: (T.A. will do this for you) Grow liquid cultures of the T7-express cells ± and ±Kn to cause bacteria to compete for resources with carrying the plasmid (or not) and expressing the gene (or not). Tues: Plate cells from liquid cultures onto different plates (± Kn) to evaluate retention of plasmid after growth requiring the plasmid (or not) and expressing the gene (or not). Thurs: count colonies from each liquid culture on each of the two plates. In your report: discuss the metabolic costs of expressing a gene, as well as the cost of carrying the plasmid. 75
Carrying cost: theory! Maintenance of the plasmid requires ongoing metabolic commitment.! When kanamycin is present, the commitment is required.! When is present, the cost of carrying the plasmid is augmented by the cost of expressing the gene.! When drug is not present, bacteria bearing plasmid have a growth disadvantage, due to the cost of carrying the plasmid. Thus in a liquid culture where they compete directly with all other bacteria, the plasmidbearing bacteria will tend to divide less frequently and dwindle as a fraction of the total population.! This disadvantage is accentuated when is present and they can be expected to dwindle even faster unless the gene they are expressing improves their ability to compete.! To learn what fraction of the total population is retains the plasmid, we plate 50 µl of liquid culture on LB agar where all bacteria will produce colonies, and also 50 µl on LB + Kn agar where only plasmid-carrying bacteria will produce colonies. The ratio of the colonies on the LB+Kn plate to those on the LB plate gives the fraction of colonies with the plasmid.! We do this for each type of liquid culture to assess the relative costs of carrying the plasmid vs. expressing the gene. 76 Carrying cost: Execution Plan The T.A. will do the following: Transfer cells carrying the GFP plasmid into 100 ml of LB liquid in sterile flasks, There will be four batches that are identical EXCEPT one is only LB medium. one is LB + Kn. one is LB +. one is LB + Kn +. All four will shake at 37 C overnight. LB only LB Kn LB LB Kn Question: In which culture do you expect the largest fraction of cells to retain the plasmid? Rank the cultures from the one that should retain the most plasmid-carrying cells (as fraction of total) to the one that should retain the least. 77
Carrying cost: Execution Plan Obtain 4 LB plates and 4 LB Kn plates. For the LB plates, label each on the bottom of the plate with the name of the overnight cultures from which the cells will come. Do the same for the LB-Kn plates. From the T.A., obtain 100 µl aliquots of cells from each of the four cultures. For each type of cell, spread 50 µl on the appropriately labelled LB plate and 50 µl on the appropriately labelled LB-Kn plate. Incubate the plates at 37 C overnight. (T.A. will then refrigerate plates.) Carrying cost: Hypothesis Which of the plates should have the most colonies? Consider the number of colonies on that plate the reference: 100%. Guess the relative number of colonies on each of the other plates, by writing the % you guess in each of the plates below. LB LB-Kn LB only LB LB Kn LB LB Kn LB-Kn Kn Kn Kn Kn Kn Kn Kn Kn 78 79
Aside: kanamycin! An aminoglycoside causes mistakes in protein production by interacting with the 30s subunit of ribosomes. (More precisely referred to as kanamycin A.)! kanamycin resistance is produced by placing a gene for a phosphotransferase enzyme on a plasmid.! The phosphotransferase phosphorylates kanamycin and renders it inactive. http://en.wikipedia.org/wiki/kanamycin 80