User:Puja Mody/Notebook/Chem 571: Gold Nanoparticles/2012/09/25: Difference between revisions
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* Todays steps were necessary in order to separate out the proteins from the cells and collect them. | * Todays steps were necessary in order to separate out the proteins from the cells and collect them. | ||
* sample 1 cells are still being tested separately due to the original discrepancy. filtration, sonication, all were performed separately for the main sample and sample 1. | * sample 1 cells are still being tested separately due to the original discrepancy. filtration, sonication, all were performed separately for the main sample and sample 1. | ||
* In repeating the same experiment as the previous class using varying concentrations of the tris buffer with the AuNP/BSa solution, no difference could be seen between the trials except for 5mM in the second trial. Either another trial needs to be run at this concentration, or a blank needs to be run to see if it is accountable in any way for the jump in absorbance. | * In repeating the same experiment as the previous class using varying concentrations of the tris buffer with the AuNP/BSa solution, no difference could be seen between the trials except for 5mM in the second trial. Either another trial needs to be run at this concentration, or a blank needs to be run to see if it is accountable in any way for the jump in absorbance. | ||
*'''[[User:Abigail E. Miller|Abigail E. Miller]] 11:05, 7 October 2012 (EDT)''': you have not provided enough information to know why a blank would help. also why didn't you run a blank? a blank should always be run. when do you plan to assess this experiment again? and will you test both conclusions? and oyu have not provided enough information to explain your conclusions? nothing drastic is different but without knowing what changed between the two you cannot draw conclusions about them | |||
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Revision as of 08:05, 7 October 2012
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Separate ProteinsGoals
Procedure
DATA/Calculations
to make a 25mL stock solution of the tris buffer at a pH of 10 1mol/L x .025= .025 mol .025 mol x 121.14 g/mol = 3.0285 g of tris
link for excel File:TrisvaryingmolarUVVis.xlsx
Conclusions
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