Micro lab 7p BLU Lab - lab 7 PDF

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Aarati Sriram Brown B1 Data and Analysis For pBLU 1. Predict your results. Write “yes” or “no,” depending on whether you think the plate will show growth. Give the reason(s) for your predictions. 2. Observe the colonies through the petri plate lids. Do not open the plates. LB-plasmid (1) Prediction: Yes Reason: LB is a non-selective media that would allow for unfettered bacteria growth. Observed Result: >105 LB/amp-plasmid (3) Prediction: No Reason: Ampicillin is a selective media. As the antibiotic-resistant plasmid is not added to the negative plates, all the bacteria should die. Observed Result: 0 LB/amp/X-gal-plasmid (5) Prediction: No Reason: Ampicillin is a selective media. As the antibiotic-resistant plasmid is not added to the negative plates, all the bacteria should die.

LB+plasmid (2) Prediction: Yes Reason: LB is a non-selective media that would allow for unfettered bacteria growth. Observed Result: >105 LB/amp+plasmid (4) Prediction: Yes Reason: Ampicillin is a selective media. As the gene for antibiotic resistance is inserted, some bacteria should express the gene and survive. Observed Result: 324 LB/amp/X-gal+plasmid (6) Prediction: Yes Reason: Ampicillin is a selective media. As the gene for antibiotic resistance is inserted, some bacteria should express the gene and survive. Some bacteria should also express the X-gal gene. Observed Result: 69

3. Record your observed results in the spaces above. If your observed results different from your predictions, explain what you think may have occurred. The class results reflected the predictions. However, my group did not observe any bacteria on the LB/amp/X-gal+plasmid plate. The bacteria may have taken in the gene for antibiotic resistance but not expressed it. The tube also may not have been mixed thoroughly after adding the LB. 4. Count the number of individual colonies. 5. Compare and contrast the number of colonies on each of the following pairs or plates. What does each pair of results tell you about the experiment? a. LB+plasmid and LB-plasmid The cell growth was too dense to count in both plates. LB is a non-selective media that allows for unfettered bacteria growth. The plasmid is irrelevant

because no selective media was added. The bacteria are viable after transformation. b. LB/Amp-plasmid and LB-plasmid The ampicillin prevented the growth of bacteria that did not contain plasmid DNA. The LB/Amp-plasmid plate had no bacteria growth whereas the LBplasmid plate had >105 colonies. c. LB/Amp+plasmid and LB/Amp-plasmid 324 individual colonies grew on the +plasmid plates while none grew on the – plasmid plate. The gene for antibiotic resistance was taken in and expressed by the +plasmid plate. The –plasmid plate was a negative control because it did not receive the plasmid. d. LB/Amp+plasmid and LB+plasmid The LB/Amp+plasmid plate had significantly fewer colonies than the LB+plasmid plate because not all the bacteria cells expressed the gene for antibiotic resistance. Only a small fraction took up the plasmid. e. LB/Amp/X-gal-plasmid and LB-plasmid Bacteria that did not receive the plasmid did not grow in the presence of ampicillin or X-gal. However, LB did not prevent growth. f. LB/Amp/X-gal+plasmid and LB/Amp/X-gal-plasmid 69 individual colonies grew on the +plasmid plates while none grew on the – plasmid plate. The plasmid DNA was taken in by some of the bacteria on the + plate, making them ampicillin-resistant. They are blue in color because of X-gal. g. LB/Amp+plasmid and LB/Amp/X-gal+plasmid The ampicillin-resistant colonies only show the blue phenotype in the presence of X-gal. None of the colonies on the LB/Amp+plasmid plate were blue. h. LB/Amp/X-gal+plasmid and LB+plasmid Less colonies grew on the LB/Amp/X-gal+plasmid plate because only a small fraction took up the plasmid DNA. Not all the bacteria cells expressed the gene for antibiotic resistance. 6. What are you selecting for in this experiment? This experiment selects for resistance to ampicillin. 7. What does the phenotype of the transformed colonies tell you? The phenotype shows that the plasmid DNA has been taken up and expressed by the transformed bacteria cells. 8. What one plate would you first inspect to conclude that the transformation occurred successfully? The LB/amp/X-gal+plasmid plate would indicate the success of the transformation because antibiotic resistance and color change would be expressed.

9. Transformation efficiency is expressed as the number of antibiotic-resistant colonies per µg of plasmid DNA. a. Determine the total mass (in µg) of plasmid used. Total mass = volume x concentration 10µL x 0.005µg/µL = 0.05µg b. Calculate the total volume of cell suspension prepared. 250µL (LB) + 250µL (CaCl2) + 10µL (plasmid) = 510µL c. Calculate the fraction of the total cell suspension that was spread on the plate. Volume suspension spread/total volume suspension = fraction spread 30µL/1530µL = 1/51 ≈ 0.196 d. Determine the mass of the plasmid in the cell suspension spread. Total mass plasmid (a) x fraction spread (c) = mass plasmid DNA spread 0.05µg x (1/51) = 9.8x10-4µg e. Determine the number of colonies per µg plasmid DNA. Express your answer in scientific notation. Colonies observed/mass plasmid spread (d) = transformation efficiency 324/9.8x10-4 = 3.3x105 #colonies/µL LB/amp+plasmid 69/9.8x10-4 = 7.0x104 #colonies/µL LB/amp/X-gal+plasmid 10. What factors might influence transformation efficiency? Explain the effect of each factor you mention. - The amount of plasmid DNA would influence transformation efficiency. Oversaturation would decrease the transformation efficiency. - Mechanical mixing would increase transformation efficiency. - Healthy cells would be more likely to take in the plasmid DNA. - Incorrectly performing the heat shock would decrease the transformation efficiency....