User:Matthew R Skorski/Notebook/471 - Exp BioChem/2015/11/03: Difference between revisions

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|style="background-color: #EEE"|[[Image:owwnotebook_icon.png|128px]]<span style="font-size:22px;"> Project name</span>
|style="background-color: #EEE"|[[Image:owwnotebook_icon.png|128px]]<span style="font-size:22px;"> Project name</span>
|style="background-color: #F2F2F2" align="center"|<html><img src="/images/9/94/Report.png" border="0" /></html> [[{{#sub:{{FULLPAGENAME}}|0|-11}}|Main project page]]<br />{{#if:{{#lnpreventry:{{FULLPAGENAME}}}}|<html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>[[{{#lnpreventry:{{FULLPAGENAME}}}}{{!}}Previous entry]]<html>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</html>}}{{#if:{{#lnnextentry:{{FULLPAGENAME}}}}|[[{{#lnnextentry:{{FULLPAGENAME}}}}{{!}}Next entry]]<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html>}}
|style="background-color: #F2F2F2" align="center"|[[File:Report.png|frameless|link={{#sub:{{FULLPAGENAME}}|0|-11}}]][[{{#sub:{{FULLPAGENAME}}|0|-11}}|Main project page]]<br />{{#if:{{#lnpreventry:{{FULLPAGENAME}}}}|[[File:Resultset_previous.png|frameless|link={{#lnpreventry:{{FULLPAGENAME}}}}]][[{{#lnpreventry:{{FULLPAGENAME}}}}{{!}}Previous entry]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;}}{{#if:{{#lnnextentry:{{FULLPAGENAME}}}}|[[{{#lnnextentry:{{FULLPAGENAME}}}}{{!}}Next entry]][[File:Resultset_next.png|frameless|link={{#lnnextentry:{{FULLPAGENAME}}}}]]}}
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###Protinase K concentration: (0.00115g)*(1mol/28,900g)*(1/0.001L)= 0.0000397 M Protinase K
###Protinase K concentration: (0.00115g)*(1mol/28,900g)*(1/0.001L)= 0.0000397 M Protinase K
####Proteinase K dilution: (39792 nM)*(0.05 mL Proteinase K) = (x nM)*(10 mL total dilution) => x = 199 nM
####Proteinase K dilution: (39792 nM)*(0.05 mL Proteinase K) = (x nM)*(10 mL total dilution) => x = 199 nM
####Proteinase K dilution: 0.05 mL of proteinase k solution was mixed with 9.9 mL of buffer
####Proteinase K dilution: 0.05 mL of proteinase k solution was mixed with 9.95 mL of buffer
###Amount of Proteinase K solution needed for 1mL with 1 nM concentration: M1*V1 = M2*V2 => (199 nM)*(V1) = (1 nM)*(1 mL) => V1 = 0.005 mL
###Amount of Proteinase K solution needed for 1mL with 1 nM concentration: M1*V1 = M2*V2 => (199 nM)*(V1) = (1 nM)*(1 mL) => V1 = 0.005 mL
###Amount of Buffer solution need to get to 1mL: (1mL total)-(0.005 mL Protinase K solution) = 0.995 mL buffer
###Amount of Buffer solution need to get to 1mL: (1mL total)-(0.005 mL Protinase K solution) = 0.995 mL buffer
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The graph below shows the peak for the absorbance at 600 nm.  From 0 minutest to 75 minutes the aborbance rises, indicating more protein is free in solution.  Then it drops at 135 minutes before rising again at 26 hour.  More data points should be taken to ensure that the exact shape of the curve.   
The graph below shows the peak for the absorbance at 600 nm.  From 0 minutest to 75 minutes the aborbance rises, indicating more protein is free in solution.  Then it drops at 135 minutes before rising again at 26 hour.  More data points should be taken to ensure that the exact shape of the curve.   
[[Image:AMS_Bradford_1nM_Abs_600nm_vs_time.png|500px]]
[[Image:AMS_Bradford_1nM_Abs_600nm_vs_time.png|500px]]



Latest revision as of 01:21, 27 September 2017

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Objective

The objective of today was to run the Bradford Analysis of Protease Degradations for 1 nM Protinase K.

Description

The protocol is identical to the one used on 9/29/15 except the fibers were spun down at 300 rpm for 10 minutes, not 1500 rpm for 1 minute.

  1. Prepping AuNP Fiber Samples
    1. 6 AuNP fiber samples were spun at 300 RPM for 10 minutes
    2. The supernatant was removed but the fibers were retained
    3. Protinase K tube 1C was mixed with 1 mL of 50 mM Tris and 10 mM CaCl2 pH 8 buffer
      1. Protinase K concentration: (0.00115g)*(1mol/28,900g)*(1/0.001L)= 0.0000397 M Protinase K
        1. Proteinase K dilution: (39792 nM)*(0.05 mL Proteinase K) = (x nM)*(10 mL total dilution) => x = 199 nM
        2. Proteinase K dilution: 0.05 mL of proteinase k solution was mixed with 9.95 mL of buffer
      2. Amount of Proteinase K solution needed for 1mL with 1 nM concentration: M1*V1 = M2*V2 => (199 nM)*(V1) = (1 nM)*(1 mL) => V1 = 0.005 mL
      3. Amount of Buffer solution need to get to 1mL: (1mL total)-(0.005 mL Protinase K solution) = 0.995 mL buffer
  2. Incubating Samples
    1. 5 μL Protinase K and 995 μL buffer were mixed in the fiber tubes as well as blank eppendorf tubes
    2. Eppendorf tubes were incubated on a shaker in a 37°C water bath for 15 minutes, 45 minutes, 1 hrs, 1.25 hrs, 1.75 hours, 2.25 hours and 26 hrs for one of the fiber tubes and one of the blanks.
  3. Running Samples
    1. Once the time for a sample was up the tube was centrifuged at 300 rpm to collect the remaining fibers
    2. To a plastic cuvette was mixed in 600 uL of pre-mixed Bradford dilution, 750 uL of sample, and 1650 uL of buffer
    3. Samples were run from 400-800 nm on the UV-VIS

Results

The graph below shows the absorbance of the samples as function of the wavelength of incident light when corrected. The six samples were first corrected for by subtracting the absorbance of the superblank of buffer and Bradford reagents. Then the samples were corrected for again by subtracting the results of the first correction by the blanks for their corresponding time periods. Then, the isosbestic point at 535 nm from each samples was subtracted from all the wavelengths of that sample.

The graph below shows the peak for the absorbance at 600 nm. From 0 minutest to 75 minutes the aborbance rises, indicating more protein is free in solution. Then it drops at 135 minutes before rising again at 26 hour. More data points should be taken to ensure that the exact shape of the curve.