User:Matthew R Skorski/Notebook/471 - Exp BioChem/2015/11/04: Difference between revisions

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|style="background-color: #EEE"|[[Image:owwnotebook_icon.png|128px]]<span style="font-size:22px;"> Project name</span>
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##The supernatant was removed but the fibers were retained  
##The supernatant was removed but the fibers were retained  
##Protinase K tube 11 was mixed with 1 mL of 1 mL of 50 mM pH 8 phosphate buffer
##Protinase K tube 11 was mixed with 1 mL of 1 mL of 50 mM pH 8 phosphate buffer
###Protinase K concentration: (___)*(1mol/28,900g)*(1/0.001L)= 0.0000___ M Protinase K
###Protinase K concentration: (0.00135 g)*(1mol/28,900g)*(1/0.001L)= 0.0000467 M Protinase K
###Amount of Proteinase K solution needed for 1mL with 1μM concentration: M1*V1 = M2*V2 => (____ μM)*(V1) = (100 nM)*(1 mL) => V1 = 0.00__ mL
###Proteinase K dilution: (46713 nM)*(0.1 mL Proteinase K) = (x nM)*(10 mL total dilution) => x = 467 nM
###Amount of Buffer solution need to get to 1mL: (1mL total)-(0.____ mL Protinase K solution) = 0.____ mL buffer
###Proteinase K dilution: 0.1 mL of proteinase k solution was mixed with 9.9 mL of buffer
NEED TO GET DATA AS WELL AS DILUTION FROM SYDNEY
###Amount of Proteinase K solution needed for 1mL with 10 nM concentration: M1*V1 = M2*V2 => (467 nM)*(V1) = (10 nM)*(1 mL) => V1 = 0.0214 mL
###Amount of Buffer solution need to get to 1mL: (1mL total)-(0.0214 mL Protinase K solution) = 0.979 mL buffer  
#Incubating Samples
#Incubating Samples
##____ μL Protinase K and ____ μL buffer were mixed in the 7 fiber tubes as well as 7 blank eppendorf tubes
##21.4 μL Protinase K and 979 μL buffer were mixed in the fiber tubes as well as blank eppendorf tubes
##Eppendorf tubes were incubated on a shaker in a 37°C water bath for 25 minutes, 45 minutes, 75 minutes, 45 minutes, 105 minutes, 135 minutes, and 1440 minutes for one of the fiber tubes and one of the blanks.   
##Eppendorf tubes were incubated on a shaker in a 37°C water bath for 25 minutes, 45 minutes, 75 minutes, 45 minutes, 105 minutes, 135 minutes, and 1440 minutes for one of the fiber tubes and one of the blanks.   
#Running Samples
#Running Samples
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###Excitation at 390 nm
###Excitation at 390 nm
###Emission from 400 to 650 nm
###Emission from 400 to 650 nm
GET ADDITIONAL DETAILS FROM NICOLE TOMORROW
###Both slit widths set to 10 nm
###Scan Speed of 200


==Results==
==Results==
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[[Image:2015 11 4 Fluorescence Concentrations.png]]
[[Image:2015 11 4 Fluorescence Concentrations.png]]


The 135 minute sample should be re-run as it is far too high of a value relative to the others.  Also, more data points should be taken between 2 hours and 24 hours.  
The 135 minute sample should be re-run as it is far too high of a value relative to the others.  Also, more data points should be taken between 2 hours and 24 hours. Otherwise there is a general trend in increasing peptide concentration over time.   
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__NOTOC__
__NOTOC__

Latest revision as of 01:19, 27 September 2017

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Objective

To analyze the activity of proteinase k at 10 nM using fluorescence

Description

The guide for analyzing a protease with fluorescence can be found here at Dr. Hartings notebook

  1. Prepping AuNP Fiber Samples
    1. 7 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 11 was mixed with 1 mL of 1 mL of 50 mM pH 8 phosphate buffer
      1. Protinase K concentration: (0.00135 g)*(1mol/28,900g)*(1/0.001L)= 0.0000467 M Protinase K
      2. Proteinase K dilution: (46713 nM)*(0.1 mL Proteinase K) = (x nM)*(10 mL total dilution) => x = 467 nM
      3. Proteinase K dilution: 0.1 mL of proteinase k solution was mixed with 9.9 mL of buffer
      4. Amount of Proteinase K solution needed for 1mL with 10 nM concentration: M1*V1 = M2*V2 => (467 nM)*(V1) = (10 nM)*(1 mL) => V1 = 0.0214 mL
      5. Amount of Buffer solution need to get to 1mL: (1mL total)-(0.0214 mL Protinase K solution) = 0.979 mL buffer
  2. Incubating Samples
    1. 21.4 μL Protinase K and 979 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 25 minutes, 45 minutes, 75 minutes, 45 minutes, 105 minutes, 135 minutes, and 1440 minutes 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 12000 rpm for 1 minute to collect the remaining fibers
    2. From either the sample or blank, 20uL was taken and placed in a 600uL eppendorf tube
      1. Added 140uL of Assay Buffer
      2. Added 40uL of Assay Reagent
    3. Taking Measurement
      1. Excitation at 390 nm
      2. Emission from 400 to 650 nm
      3. Both slit widths set to 10 nm
      4. Scan Speed of 200

Results

The image below shows the intensity for the fluorescence of samples and blanks from 400 to 650 nm.


To find the concentrations of peptide released at the various time intervals the samples and blanks had the area of their intensity integrated from 420 nm to 650 nm. The equation (Wavelength2 - Wavelength1)/2*0.5 was used and the sums of each interval totaled. The samples were then corrected for by subtracting the total area of the blank from the total area of the sample. The resulting intensities were then used to calculate the concentrations based off the calibration curve made on 10/07/15. The graph below shows the concentrations of the peptides as a function of time.

The 135 minute sample should be re-run as it is far too high of a value relative to the others. Also, more data points should be taken between 2 hours and 24 hours. Otherwise there is a general trend in increasing peptide concentration over time.