CHEM 2313 Lecture Notes - Asepsis, Plasmid

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lemonrat895

17 Feb 2014

School

Northeastern University

Department

Chemistry & Chemical Biology

Course

CHEM 2313

Professor

Stenburg

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Document Summary

For the pipet exercises our results were as expected. Variables that might contribute to pipet error are contamination if you touch the pipet tip to yourself or any other object. Also, if you depress the button to the second stop while you are drawing liquid into the pipet, you will probably damage the pipet as well as get an inaccurate amount. If drops of liquid are left on the end of your pipet, it could contribute to an inaccurate amount of liquid transferred. Also, if you do not depress the button far enough while drawing liquid into and releasing liquid from the pipet it could contribute to pipet error. For the aseptic technique with the liquid media, our results were as expected. However, with the streak plates, our results were not as expected. Plate b was not supposed to have growth and there was growth in one quadrant.

Related Questions

Please answer questions at the end.

In this lab, you will add different volumes of âcell-free supernatantâ (CFS) to fresh cultures of V. harveyi and observe bioluminescence. The CFS is prepared by culturing V. harveyi in liquid medium overnight, then separating the cells from the supernatant by centrifugation.

MATERIALS â¢ overnight culture of V. harveyi in AB medium â¢ fresh AB medium â¢ 1 microcentrifuge tube â¢ 3 glass culture tubes

PROCEDURE

1. This experiment will be performed with your lab group. Label the microcentrifuge tube and 3 glass cultures tubes with your group number. Also, label the 3 glass culture tubes as follows: âMedia only,â âCells+fresh media,â and âCells+CFS.â

2. Using aseptic technique, pipet different volumes of fresh AB medium into the glass tubes as follows: 1 mL in the âMedia onlyâ tube, 800 ul in the âCells+fresh mediaâ tube, and 500 ul in the âCells+CFSâ tube.

3. Pipet 500 ul of the overnight culture into a microcentrifuge tube and centrifuge at maximum speed (>13,000 x g) for 5 min to pellet the cells.

4. After centrifugation is complete, transfer 300 ul of the cell free supernatant (CFS) from the microfuge tube to the glass tube labeled âCells+CFSâ. Be careful to avoid the cell pellet.

5. Remove the remaining supernatant from the microfuge tube using a micropipette and discard as culture waste.

6. Add 500 ul of fresh AB medium to the cell pellet in the microfuge tube and pipet up and down to resuspend until no clumps are visible.

7. Add 200 ul of the cell suspension to the âCells+fresh mediaâ tube and 200 ul of the cell suspension to the âCells+CFSâ tube. Note the time.

8. Place the 2 glass tubes containing cells in the 30 degree incubator/shaker.

9. After 1 hour and 2 hours, observe your tubes next to the âMedia onlyâ tube in a dark room. You will need to let your eyes adjust for at least 30 seconds (count to 30 slowly in the dark) before you can evaluate whether the cultures are luminescent or not. Also, make sure you try swirling the tubes in the darkroom and observe the tubes for a minute after you have stopped swirling. After the observation at the 1 hour time point, put your tubes back in the incubator shaker until the 2 hour time point.

10. Note your observations. At the end of the lab, place your tubes in the tube waste.

Questions

1. What were the expected results? Explain the rationale of the experiment. If you didnât obtain the expected results, hypothesize why not and describe how the experiment could be improved.

2. What are differences between luminescence and fluorescence?

3. Why do V. harveyi cultures glow more brightly when swirled vigorously? Be specific.

4. Imagine you have a mutant V. harveyi strain that canât synthesize autoinducer and another mutant V. harveyi strain that makes a defective luciferase. If you streaked both strains near (but not touching) each other on the same plate, would you expect to see light produced? If so, by which strain, where, and why?

5. You have two identical tubes of wild-type V. harveyi (with the same cell concentration in each tube) in fresh media. To one tube, you add CFS from a light-producing V. fischeri culture, and to the other tube you add an equal volume of fresh media. If both tubes of V. harveyi have the same light production, what can you conclude? Explain.

6. Many biomedical research labs study quorum sensing in V. fischeri or V. harveyi. How is studying light production in marine bacteria relevant to human health? List at least 3 different examples.

7. Why do you think it is important for bacteria to be able to control the expression of some genes based on the number of individuals present?

graylion297

In a genetic experiment:

The use and limits of RNA interference as a reverse genetics technique for knocking down gene expression were tested. The wild-type and mutants worms for the strains were selected. The RNAi feeding strains and control bacterial were used for the the RNAi experiment. The eri-1 C. elegan strains (wild-type) were used for RNAi treatment and mutant strains that were used ( unc-22 and dpy-13) C. elegans. Both carbencillin and IPTG wered added to the RNAi feeding plates and E.coli containing the RNAi transgenes were spread on the plates. The antibodic carbacillin only allows growth of bacteria which contains the RNAI construct. IPTG induces the expression of the polymerase which then transcribes the siRNA in the bacteria. 6 NGM plates without bacteria were used and 20ul of 12.5 mg/ml carbencillin concentration were spread on the 6 of the plates. The carbencillin dried and then IPTG was spread (25ug/ml) on each plate. Then 6 plates with the RNAi bacterial strains were used. Theree of plates for strain one and the other three for strain the second strain. When IPTG dried , 150 ul of the appropraite bacteria was added onto each plate. 2NGM plates (without carbenicillin or IPTG) was spread with 150 ul of OP50 bacterial culture. OP50 is the regular E. coli strain for culturing C. elegans and does not contain the RNAi constructs. These plate were used as the control plates6. Then the plates were incubated for two days. Then the adult hermaphrodites from the RNAi feeding plates and the op50 plates were removed. Then from the eri-1 wildtype stock, 2 or 3 L4 hermaphrodites or
young adults were transferred onto each RNAi feeding plate. Then L4 worms were transferred to an intermediate plate first. After they have a minute to move away from each other and the eggs, and transfered to the RNAi plates. On the two control plates seeded with OP50, 2 L4 or young adult hermaphrodites were placed there. Plates were obtain as the reference mutants (unc-22, bli-1 or unc-13). Then the plates were incubated to allow the worm to develop and lay eggs in two days. Then the adult hermaphrodites from your RNAi feeding plates and the op50 plates were removed and then placed back into the incubator.

The op500 was the negative control plate wtih 80 and 300 progeny and unc-22 had around 50. (reference plates)

Mating in the RNAi plates

Dpy (60) x wildtype (15) and unc( 70) x wildtype ( 8) progeny

the phenotype was movement and it seemed similar phenotype within the progeny.

Can you help me understand the results and what is happening genetically in the experiment ? Are the for dyp-15 x unc-22 mating cross and unc-22 x eri-1 wildtype C. elegans. Based on the data:

Briefly discuss the results of the RNAi experiment. How closely did the RNAi-treated worms phenocopy the mutant worms?

Was the gene knockdown 100% effective?

Was the RNAi treatmentequally effective for the two genes? If not, discuss this result and why that could occured.

Please be specific and detailed

Genetics HELP ASAP

Genetics HELP ASAP In a genetic experiment: The use and limits of RNA interference as a reverse genetics technique for knocking down gene expression were tested. The wild-type and mutants worms for the strains were selected. The RNAi feeding strains and control bacterial were used for the the RNAi experiment. The eri-1 C. elegan strains (wild-type) were used for RNAi treatment and mutant strains that were used ( unc-22 and dpy-13) C. elegans. Both carbencillin and IPTG wered added to the RNAi feeding plates and E.coli containing the RNAi transgenes were spread on the plates. The antibodic carbacillin only allows growth of bacteria which contains the RNAI construct. IPTG induces the expression of the polymerase which then transcribes the siRNA in the bacteria. 6 NGM plates without bacteria were used and 20ul of 12.5 mg/ml carbencillin concentration were spread on the 6 of the plates. The carbencillin dried and then IPTG was spread (25ug/ml) on each plate. Then 6 plates with the RNAi bacterial strains were used. Theree of plates for strain one and the other three for strain the second strain. When IPTG dried , 150 ul of the appropraite bacteria was added onto each plate. 2NGM plates (without carbenicillin or IPTG) was spread with 150 ul of OP50 bacterial culture. OP50 is the regular E. coli strain for culturing C. elegans and does not contain the RNAi constructs. These plate were used as the control plates6. Then the plates were incubated for two days. Then the adult hermaphrodites from the RNAi feeding plates and the op50 plates were removed. Then from the eri-1 wildtype stock, 2 or 3 L4 hermaphrodites or young adults were transferred onto each RNAi feeding plate. Then L4 worms were transferred to an intermediate plate first. After they have a minute to move away from each other and the eggs, and transfered to the RNAi plates. On the two control plates seeded with OP50, 2 L4 or young adult hermaphrodites were placed there. Plates were obtain as the reference mutants (unc-22, bli-1 or unc-13). Then the plates were incubated to allow the worm to develop and lay eggs in two days. Then the adult hermaphrodites from your RNAi feeding plates and the op50 plates were removed and then placed back into the incubator. The op500 was the negative control plate wtih 80 and 300 progeny and unc-22 had around 50. (reference plates) Mating in the RNAi plates Dpy (60) x wildtype (15) and unc( 70) x wildtype ( 8) progeny the phenotype was movement and it seemed similar phenotype within the progeny. Can you help me understand the results and what is happening genetically in the experiment ? Are the for dyp-15 x unc-22 mating cross and unc-22 x eri-1 wildtype C. elegans. Based on the data: Briefly discuss the results of the RNAi experiment. How closely did the RNAi-treated worms phenocopy the mutant worms? Was the gene knockdown 100% effective? Was the RNAi treatmentequally effective for the two genes? If not, discuss this result and why that could occured. Please be specific and detailed

CHEM 2313 Lecture Notes - Asepsis, Plasmid

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