14-Chromogen Dilution Buffer Hydrogen Peroxide/Citrate-Phosphate
Buffered Saline
15-Washing Buffer (5x) Concentrated Phosphate Buffered Saline
16-Stop Solution 1M Phosphoric Acid
17-Standards 1-6 0.5, 2.0, 8.0, 20.0, 80.0, 200.0 ng/mL 8-OHdG (1 vial Each) 1 mL each
18-Plate Seal Adhesive sheet for covering plate
METHODS
Each sample had three replicates.
1.
Firstly primary antibody with the primary antibody dilution buffer was reconstituted.
2.
Then 50 ul of sample was added to each well. After that 50 ul of reconstituted primary antibody
was added to all wells expect blank.
3.
Plate was covered with a kind of cover and plate was shaked to mix fully and it was incubated
for 1 h at 37 C.
4.
At the end of incubation time, contents in plate was poured off. Then 250 ul diluted washing
buffer was added to each well. It was washed thoroughly by agitation. After that plate was
inverted and blot against clean paper towel to remove any remaining washing buffer. This step
was repeated twice.
5.
Secondary antibody with secondary antibody dilution buffer was reconstituted. 100 ul of
reconstituted secondary antibody was added to each well.
6.
Plate was covered with cover and after that plate was shaked from side to side to mix fully.
After shaking, it was incubated at 37C for 1 hour.
7.
Chromogen was diluted and 100 ul diluted chromagen added to each well. After chromogen
addition, plate was mixed and incubated at room temperature for 15 minutes.
8.
At last 100 ul stop solution was added and after 3 minutes, its absorbance was read at 450 nm.
4-RESULTS
Absorbance
Concentration
(ng/ml)
Blank
1,271
Sample 1
1,217
0,5
Sample 2
1,066
2
Sample 3
0,882
8
Sample 4
0,869
20
Sample 5
0,404
80
Sample 6
0,348
200
Equation
y=-0,151ln(x) + 1,1701
R2
0,938
Figure 4: Absorbance vs Log Scale of
Concentration
Figure 5: Absorbance vs Log Scale of Concentration
1,217
1,066
0,882 0,869
0,404
0,348
y = -0,151ln(x) + 1,1701
R² = 0,938
0
0,2
0,4
0,6
0,8
1
1,2
1,4
0,1
1
10
100
1000
Absorbance Absorbance-
1,1701=a
b=a/(-0,151)
Conc.(ng/ml)
Unknown1 0,85
-0,3201
2,11986755
8,330034
Unknown2 0,7
-0,4701
3,113245033
22,49392
Unknown3 1,207
0,0369
-0,244370861
0,783197
Figure 6: Absorbance and Standart
Deviation Calculations
1-For unknown 1 y=-0,151ln(x) + 1,1701
(0,85-1,1701)/(-0,151)=2,12
ln(x)=2,12 Concentration of Unknown 1: x= 8.33 ng/ml
2-For unknown 2 y=-0,151ln(x) + 1,1701
(0,7-1,1701)/(-0,151)=3,11
ln(x)=3,11 Concentration of Unknown 2: x= 22,49 ng/ml
3-For unknown 3 y=-0,151ln(x) + 1,1701
(1,207-1,1701)/(-0,151)=-0,24
ln(x)=-0,24 Concentration of Unknown 3: x= 0,78 ng/ml
5-DISCUSSION:
When we look the graph and after seeing the R2 value, it can be said that our data fits
the curve and the R2 is 0,938. This value should be more closer to 1 but even 0,938 can be
accepted as good one. In this experiment because of fluctuations between reading, the median
values were taken while drawind the graph. Also while drawing the graph, x axis was taken in
log scale and the logaritmic trendline was added to our graph. Thus we got an formula which
contains logarithm function. Our experiment proved that higher concentrations of 8-OhdG in
the sample solution lead to a reduced binding of the antibody to the 8-OhdG on the plate. When
the concentration of standard increased, the absorbance values decreased. When I calculated
the concentration of the unknowns, then I saw that there is a huge difference between our
results. For example the concentrations of unknowns are 8.33, 22.49 and 0.78 ng/ml. And the
expected/real concentration is the 8 ng/ml. So it can be concluded that there are some reasons
that cause such a kind of difference. The source or sources of such a kind of difference can be
because of various things:
Incomplete washing of wells
Contamination