Module 1, Day 3: Difference between revisions
Crystal Mao (talk | contribs) |
Crystal Mao (talk | contribs) No edit summary |
||
Line 17: | Line 17: | ||
===Plaque Assay Titer=== | ===Plaque Assay Titer=== | ||
<math> Titer = (\frac{PFU}{Dilution * Volumes}) </math> | <math> Titer = (\frac{PFU}{Dilution * Volumes}) </math> | ||
*K01 | *K01: | ||
<math> Titer = \frac {880}{10^{-6}*10ul*\frac{1ml}{1000ul}} = 8.8 x 10^{10} pfu/ml</math> | <math> Titer = \frac {880}{10^{-6}*10ul*\frac{1ml}{1000ul}} = 8.8 x 10^{10} pfu/ml</math> | ||
*E4: | |||
As detailed in my lab notebook, we failed to observe E4 pfu colonies on the lb-plates, most likely due to | |||
uneven spreading and bubbles in the agar. As a result, I do not have the data available to calculate the | |||
titer for E4. | |||
<BR> | <BR> | ||
It makes sense that K07 would form less PFUs than E4--as a helper phage to other strains of M13, it is probably less effective at infecting bacteria with its own genome. Curiously however, our K07 colonies proliferated in similar number and morphology to other groups' E4 colonies. | |||
= | ===Refactoring Bacteriophage T7=== | ||
Read | |||
Latest revision as of 07:14, 21 September 2007
Transformation Efficiency
(a) [math]\displaystyle{ (10^9 cfu/ug) (\frac{1ug}{1000ng}) (\frac{1}{1000}) = 1000 colony forming units }[/math]
(b) [math]\displaystyle{ (50cfu)(\frac{1ug}{10^9cfu}) (\frac{100}{1}) (\frac{1}{2ul}) = 2.5 x 10^{-6} ug/ul }[/math]
Transformation Outcomes
Outcome 1: No bacteria were able to grow on the +kan plates, suggesting that none of the plasmids, including plate 1 with the positive control, were successfully transformed.
Outcome 2: The plates was either not treated with kanamycin, or a contamination occurred.
Outcome 3: The kill-cut mechanism failed, and the sticky ends of the backbones transformed into plate 3 were able to come together despite our efforts.
Plasmid Map
See Attached Sheet
Plaque Assay Titer
[math]\displaystyle{ Titer = (\frac{PFU}{Dilution * Volumes}) }[/math]
- K01:
[math]\displaystyle{ Titer = \frac {880}{10^{-6}*10ul*\frac{1ml}{1000ul}} = 8.8 x 10^{10} pfu/ml }[/math]
- E4:
As detailed in my lab notebook, we failed to observe E4 pfu colonies on the lb-plates, most likely due to uneven spreading and bubbles in the agar. As a result, I do not have the data available to calculate the titer for E4.
It makes sense that K07 would form less PFUs than E4--as a helper phage to other strains of M13, it is probably less effective at infecting bacteria with its own genome. Curiously however, our K07 colonies proliferated in similar number and morphology to other groups' E4 colonies.
Refactoring Bacteriophage T7
Read