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 Experiment 8 Bacterial Transformation VY NGUYEN 1 April 2016 ABSTRACT 1. Understand the concepts of plasmids and genetic transformation 2. Perform a bacterial transformation with appropriate laboratory techniques. 3. . Describe the function of a selectable marker in the isolation of transformants. 4. Analyze the results and calculate the transformation efficiency. Experiment 8: ​
Bacterial Transformation
Vy Nguyen 1 1) Include a title page and present data in a table format (see below) and drawings (or photographs) with appropriate labeling. Plates Characteristics Plasmid added Colony number Colony Fluorescence LB/Amp Yes 194 No LB/AmpP/Ara Yes 232 Yes LB/Amp (negative control) No 0 No LB (positive control) No >300 No Experiment 8: ​
Bacterial Transformation
Vy Nguyen 2 2. Data Analysis: b) Calculate the transformation efficiency of your OWN experiment after you obtain the results the following week. Number of colonies on LB/amp plate = 194 Total amount of pGLO DNA used = 0.250 μg Fraction of DNA used = 0.247 Micrograms of DNA spread on the plates = 0.0617 μg Transformation efficiency (1pts) = 3.14 * 10^3 transformants/μg (DNA in μg) = (concentration of DNA in μg/μl) x (volume of DNA in μl) = (0.05 μg/μl) (5 μl) = 0.25 μg Fraction of DNA used = ​
Volume spread on LB/amp plate (μl​
) Total sample volume in microfuge tube (μl) = ​
100 μl ​
= 0.2469 405 μl pGLO DNA spread = Total amount of DNA used in μg x fraction of DNA used pGLO DNA = 0.25 μg * 0.2469 = 0.0617 μg Transformation efficiency = ​
Total number of cells growing on the agar plate Amount of DNA spread on the agar plate = 194/0.0617 = 3.142 * 10^3 transformants/μg 3. Discussion and conclusions: (Include the questions in your report) 1) Based on the experimental design, on which plate(s) should the genetically transformed bacterial cells be located? Which plates should be compared to determine if successful genetic transformation has occurred? What should growth on each of these plates look like? The genetically transformed bacterial cells should be located in the LB/Amp and LB/AmpP/Ara plates. The positive (LB/Amp)​
and negative (LB/Amp) control plates should be compared to determine if successful genetic transformation has occurred. The negative control has no growth because of the absence of ampicillin­resistant bacteria, and the positive control has a the growth of bacteria because they grow in the Luria Broth. 2) What allows the transformed cells to grow on LB/Amp? How? r​
The ampicillin resistant gene (bla or Amp​
), which codes for beta­lactamase protein, allows the transformed cells to grow on LB/Amp. In other words, ​
the transformed cells contain genes for beta lactamase, an enzyme that provides resistance to the antibiotic, ampicillin. Thus, the transformed cells can grow and survive on the LB/Amp plate even though it contains ampicillin.​
This was done by inserting a plasmid that contains ampicillin­resistant gene into bacterias. Experiment 8: ​
Bacterial Transformation
Vy Nguyen 3 3) Why do transformed cells display fluorescence in the presence of arabinose? Explain the mechanism. The transformed cells contain the genes for GFP (green fluorescent protein) which allows the bacteria to display green fluorescence in the presence of arabinose and a UV light. GFP can be switched on and off. Arabinose, a biological molecule, allows GFP to turn on and, therefore, fluoresce. The araC protein binds pGLO DNA and blocks RNA polymerase from binding. Arabinose interacts with araC, changing its conformation that enables RNA polymerase to be bound, and once transcription begins, GFP is produced. 4) Based on the analysis of your results, describe the evidence that indicates your transformation experiment was successful or not successful? Offer a possible explanation if your result was unsuccessful? Explain the possible errors resulting in the data of group 12 on page 6. The experiment was successful because of our high transformation efficiency, ​
3.1x103​
transformants/μg. The bacterias were transferred without being killed, and a high number of cells were infected by the plasmid. Possible errors resulting in the data of group 12 on page 6 could be that they did not add pGLO to the tube, or that they did not follow the procedures correctly. 5)Which plate in your experiment is the negative control plate? What might cause growth on the negative control plate? The negative control plate was LB/Amp. If transformed bacteria was transferred to the plate, or if the ampicillin was not working properly, there could be growth on the negative control plate. 6) What purpose does spreading –pGLO on the LB plate serve? What might cause your cells to fail to grow on the LB plate? The purpose of spreading pGLO on the LB plate was to enable bacteria to grow, otherwise the ampicillin would kill them. If the pGLO is not spread, then the cells would be killed by the ampicillin. 7)If you spread 100 μl of the +pGLO transformation solution onto an LB plate, what should you see after incubating the plate at 37­degrees for a few days and why? If you used the velvet stamping method demonstrated in lab 3 to replica­plate the growth from this LB plate onto an LB/Amp plate, what would the LB/Amp plate look like after a few days of incubation at 37 degrees. If you spread 100μl of pGLO to LB plate, you should observe colonies of bacteria growing across the surface of the plate. If you used a stamp, then the colonies may not be as separated, but there would still be a large amount of growth. Experiment 8: ​
Bacterial Transformation
Vy Nguyen 4 REVIEW FOR FINAL EXAM 1. Define genetic transformation. The ability to introduce a specific gene into an organism and have it express that gene 2. Compare constitutive vs. inducible gene expression. Constitutive systems​
are always on, lack regulation. ex: the gene for the repressor protein in prokaryotes which sits at the operator for lac operon. Inducible systems​
are turned on in the presence of a molecule that needs to be broken down. Enzymes only produced when specific chemical substrates are present. Ex: lac operon on prokaryotes. Inducible system – Arabinose present in growth media • The gene for GFP will be switched on • Transformed cells will appear fluorescent green under UV light – No arabinose present in growth media • The gene for GFP will remain switched off • Transformed cells will appear white under UV light 3. What are some important characteristics of the strain and the plasmid in this experiment? We used E.Coli bacteria and placed a unique pGLO plasmid that codes for GFP and bla, which provides antibiotic resistance. pGLO can be used to control expression of fluorescent protein in transformed cells. E.Coli was used because it is a single celled organism that produces quickly and will not infect plants or animals and it cannot grow on plates with antibiotic (ampicillin) 4. Understand how bacterial transformation is set up in this lab. a)​
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Why do we use LB/amp media to select for transformants? The bacteria with the pGLO DNA also codes for beta lactamase, which allows for the bacteria to be resistant to ampicillin. The ones that did not transform would be killed. b)What was the purpose of the arabinose? Experiment 8: ​
Bacterial Transformation
Vy Nguyen 5 If Arabinose is present, it turns on the beta lactamase gene of the PGLO plasmid which is an antibiotic resistant gene. This phenomenon allows the transformed cells to grow even in the presence of ampicilin. c)​
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Which plate­­­LB/amp or LB/amp/arab­­­ produced glowing colonies? WHY? LB/amp/arab produced glowing colonies because the arabinose activated the gene so it could be expressed which means glow. d)What are the control plates? What are their purposes? the LB plate was used to ensure that we had live bacteria & they were not killed off during the process. LB plate with no plasmid (­pGLO) demonstrates Positive control with TMTC LB/AMP plate with no plasmid (­pGLO) demonstrates Negative control with no growth 5. Be able to determine the transformation efficiency given the number of transformants, the concentration and the volume of plasmid used. a. Total amount of pGLO DNA=concentration of DNA (ug/uL) * volume of DNA in uL b. Fraction of DNA used= Volume spread on plate/to sample in microfuge tube c. DNA spread = fraction of DNA * total DNA used d. Transformation efficiency = number of cells on plate / DNA spread Ok Four different treatments of bacterial cells that enhance transformation efficiency a. Incubation i. Cells mixed with CaCl2 + plasmid b. Heat shock i. Quick temperature changes (destabilizes cell membrane to create thermal imbalance) c. Recovery i. Short incubation to begin expressing new genes d. Cells of log phase