Scientific Communication and Laboratory Skills
Lab
Report
We Will Write a Custom Essay about Scientific absorbance and calculate the concentration of
For You For Only $13.90/page!
order now
Abstract
Bradford
protein assay and direct ultraviolet absorption (UV light) were performed on
unknown samples. These two methods were performed to determinate the concentration
of the unknown samples.
The Bradford
assay is colorimetric assay depends on the reaction between Coomassie brilliant
blue and the aromatic residues in the protein. After the dye react with these
residues it increases the absorbance from 470nm to 595nm. In general, absorbance
of a serious of known concentrations of standard protein were measured and
create a standard curve. use the created standard curve to calculate the
concentration of unknown sample based on its absorbance. The method is highly sensitive,
and it is very fast but small materials can interfere with it and highly effect
the results.
The UV absorption
can be applied on protein that contains aromatic residues such as tyrosine,
tryptophan. There aromatic residues are able to absorb the UV light at 280 nm
without adding colouring reagent. uv spectrometer is used to measure the
absorbance and calculate the concentration of protein by using beer’s law. Extinction
coefficient for the protein is required to carry on the calculation. The technique
is quick but the successful of the method depends on having the accurate extinction
coefficient for the protein.
The aim of
using the both methods is to compare between its results and show if there any
different between them.
introduction
protein assays
are one of the most widely used methods in life science research. Knowing
protein concentration is required in protein purification, cell biology,
electrophoresis and other research applications, it is important to use more
than once method during experiment to obtain the require accurate results.
there are wide variety of methods such as dye binding assays (Bradford), copper
ion based assay (lowry and BCA), and spectrophotometry based on UV absorption.
Each protein
assay has it is advantage and limitations and it shows the reliability of the
method. The natural of the sample is the most important consideration to choose
the right assay method. If the sample is in solid form it can be easily
solubilized in protein assay but most of the protein samples are complex
solutions which include non-protein materials. There is many issues effect the
reliability of the methods such as interfering agents, sample preparation,
assay sensitivity, sample size and time consideration.
Proteins are
complex polymers of amino acids with huge range of modification and different structures
which required numerous numbers of chemical agents to stabilise it. Non-protein
agents can be a challenge for protein assays, protein solution contains
surfactant can interfere with dye based assay as Bradford assay. The better
reliability will be possible with removing non-protein assays or by using
method which overcome the interfering effect of it.
The size of
the sample becomes a critical consideration, as most of the methods required at
least 0.5 µg of proteins for a reliable result, the method that requires
the smallest amount of protein will offer an advantage over the others such as
dotMETRIC assay. The amount of the time which taken to complete the experiment
will depend on the complexity of the sample and the assay method. The methods
which use standard plots and curves are faster while the protein sample
contains non-protein agents are consuming more time which needed to remove the
non-protein agents. The assays that does
not required standard plots can be faster and cheaper such as protein dotMETRIC
protein assay and dye binding CB-X.
The most reliable protein estimation is the one which
performed using protein standard. The protein standard should have similar
properties to the sample that being tested. It is very difficult to find
protein standard with similar properties to the sample. Bovine serum albumin
and gamma globulin are accepted to use as reference protein.
As in this experiment,
bovine serum albumin was used to create a serious of standard solutions with
known concentrations. Two spectroscopic methods were using to obtain the
absorbance of the standard solutions. The absorbance and the concentration of
the standard solution were used to create calibration curve to detect the
concentration of the unknown samples. The result of the two spectroscopic
methods will be compared to discover if there any difference between them.
Materials and methods
Results and discussion
preparation of
protein standard solutions.
Bovine serum albumin in water 0.5%
It means 0.5g in each 100 ml = 0.005g/ml
5000 ?g/ml
Total volume =
protein stock solution + distilled water = 2000 µl
Convert µl to ml dividing by 1000
2000 µl/ 1000
= 2ml
Use the following equation to find the concentration of the
standard solution
C1 V1 = C2 V2
C2 = C1 V1 / V2
Stander solution 1
Protein stock solution 50 µl = 0.05 ml
C2 = 5000x 0.05 / 2 = 125 ?g/ml
Stander solution 2
Protein stock solution 100 µl = 0.1 ml
C2 = 5000x
0.1 / 2 = 250 ?g/ml
Stander solution 3
Protein stock solution 200 µl = 0.2 ml
C2 = 5000x 0.2 / 2 = 500 ?g/ml
Stander solution 4
Protein stock solution 300 µl = 0.3 ml
C2 = 5000x 0.3 / 2 = 750 ?g/ml
Stander solution 5
Protein stock solution 400 µl = 0.4 ml
C2 = 5000x 0.4 / 2 = 1000 ?g/ml
Standard
1
2
3
4
5
solution
Protein
stock
50
100
200
300
400
solution (?l)
Distilled water
1950
1900
1800
1700
1600
?l
Concentration
?g/ml
125
250
500
750
1000
Bradford
assay
This table shows
the absorbance values for standard and unknown solutions in Bradford assay
solution
1
2
3
4
5
X
Y
Z
Absorbance
0.069
0.092
0.232
0.320
0.410
0.215
0.267
0.352
0.065
0.095
0.243
.0326
0.405
0.214
0.271
0.349
at 595 nm
0.072
0.099
0.235
0.331
0.411
0.218
0.275
0.359
Calculate the average absorbance and standard
deviation for each standard solution and unknown samples using data given in
table
Solution 1
Average = (0.069 + 0.065 + 0.072) / 3 = 0.069
Calculation of Stander deviation:
Calculate the mean (0.069).
For each number: subtract the mean. Square the result.
Add up all the squared results.
Divide this sum by one less than the number of data points (N –
1). This gives you the sample variance.
Take the square root of this value to obtain the sample standard
deviation.
(0.069 – 0.069)2 = 0
(0.065 – 0.069)2 = 0.00016
(0.072 – 0.069)2 = 0.000009
(0 + 0.00016 + 0.000009) /2 = 0.0000125
Standard deviation = ? 0.0000125 =0.0035
Solution 2
Average = (0.092 + 0.095 +0.099) / 3 = 0.096
Standard deviation =0.0035
Solution 3
Average =(0.232 + 0.243 + o.235) / 3 = 0.236
Standard deviation= 0.0051
Solution 4
Average = (0.320 + 0.326 + 0.331) / 3 = 0.325
Standard deviation = 0.0055
Solution 5
Average = (0.410 + 0.405 + 0.411) / 3 = 0.409
Standard deviation = 0.0032
Solution X
Average = (0.214 + 0.215 + 0.218) / 3 = 0.216
Standard deviation = 0.0021
Solution Y
Average = (0.267 + 0.271 + 0.275) / 3 = 0.271
Standard deviation = 0.004
Solution Z
Average =(0.352 + 0.349 + 0.359) / 3 = 0.353
Standard deviation = 0.005
table that
shows the concentration of each standard solution and its corresponding average
absorbance.
Standard
1
2
3
4
5
Concentration
125
250
500
750
1000
?g/ml
Average
0.069
0.096
0.236
0.325
0.409
Absorbance
at 595 nm
calibration
curve of absorbance against concentration of the standard solutions.
Using the
calibration curve and its regression equation, calculate the concentration of
the unknown samples. Use the average absorbance for each unknown in your
calculation and report your answers in ?g/ml (to 2 decimal places) as well as
in w/v% and ppm.
From the
calibration curve
y = mc + b
Y = 0.0004X +
0.0144
X = (Y –
0.0144) / 0.0004
Solution X
0.216 =
0.0004X + 0.0144
X = (0.216 –
0.0144) / 0.0004 = 5.04 x102 ?g/ml (to 2 decimal places)
w/v%
first convert ?g
to gram then multiple by 100
5.04 x102
?g / 106 = 0.000504 g/ml
w/v% = 0.0504%
ppm= 504 ppm
Solution Y
X = (0.271 –
0.0144) / 0.0004 = 64.15 x10 ?g/ml (to 2 decimal places)
w/v% = 0.06415%
ppm = 641.5
ppm
solution Z
X = (0.353 –
0.0144) / 0.0004 = 86.45 x10 ?g/ml (to 2 decimal places)
w/v% 0.08645%
ppm = 864.5
ppm
standard
X
Y
Z
Concentration
?g/ml
504
641.5
864.5
Average
absorbance at 595 nm
0.216
0.271
0.353
Ultraviolet
light
Table shows
absorbance values for standard and unknown solutions at 280 nm
solution
1
2
3
4
5
X
Y
Z
Absorbance
0.112
0.205
0.415
0.612
0.795
0.421
0.525
0.541
0.105
0.211
0.427
0.628
0.812
0.409
0.532
0.522
at 280 nm
0.111
0.217
0.422
0.625
0.829
0.414
0.515
0.536
the average
absorbance and standard deviation for each standard solution and unknown
samples using data given in table
Solution 1
Average = (0.112 + 0.105 + 0.111) / 3 = 0.109
Standard deviation = 0.0038
Solution 2
Average = (0.205 + 0.211 +0.217) / 3 = 0.211
Standard deviation =0.006
Solution 3
Average =(0.415 + 0.427 + 0.422) / 3 = 0.421
Standard deviation= 0.006
Solution 4
Average = (0.612 + 0.628 + 0.625) / 3 = 0.621
Standard deviation = 0.0085
Solution 5
Average = (0.795 + 0.812 + 0.829) / 3 = 0.812
Standard deviation = 0.017
Solution X
Average = (0.421 + 0.409 + 0.414) / 3 = 0.415
Standard deviation = 0.006
Solution Y
Average = (0.525 + 0.532 + 0.515) / 3 = 0.524
Standard deviation = 0.0085
Solution Z
Average = (0.541 + 0.522 + 0.536) = 0.533
Standard deviation = 0.0098
table
that shows the concentration of each standard solution and its corresponding
average absorbance. For example:
Standard
1
2
3
4
5
Concentration
125
250
500
750
1000
?g/ml
Average
0.109
0.211
0.421
0.621
0.812
Absorbance
at 280 nm
calibration
curve of absorbance against concentration of the standard solutions
Using the
calibration curve and its regression equation, calculate the concentration of
the unknown samples. Use the average absorbance for each unknown in your
calculation and report your answers in ?g/ml (to 2 decimal places) as well as
in w/v% and ppm.
From the
calibration curve
y = mc + b
Y = 0.0008X +
0.0144
X = (Y –
0.0144) / 0.0008
Solution X
0.415 =
0.0008X + 0.0144
X = (0.415 –
0.0144) / 0.0008 = 500.75 ?g/ml (to 2
decimal places)
w/v%
first convert ?g
to gram then multiple by 100
500.75 ?g / 106 = 0.00050075 g/ml
w/v% = 0.050075%
ppm= 500.75
ppm
Solution Y
X = (0.524 –
0.0144) / 0.0008 = 6.37 x102 ?g/ml (to 2 decimal places)
w/v% = 0.0637%
ppm = 637 ppm
solution Z
X = (0.533 –
0.0144) / 0.0008 = 648.25 ?g/ml (to 2 decimal places)
w/v% = 0.08645%
ppm = 652 ppm
standard
X
Y
Z
Concentration
?g/ml
500.75
637
648.25
Average
absorbance at 280 nm
0.415
0.524
0.533
Conclusion
Through
the experiment, we will able to find out the concentration of the unknown
samples by using calibration curve of absorbance against concentration. From the
standard curve of Bradford assay, the unknown solutions (X,Y,Z) have
concentration of 504 µg/ml , 641.5 µg/ml and 864.5µg/ml. while the
standard curve of UV light absorption shows a similar concentration for sample (X,Y)
500.75 µg/ml and 637 µg/ml but different concentration for sample (Z)
648.25 µg/ml and that can be because of interfering agents, sample preparation, and assay sensitivity.
Both Bradford assay
and UV light absorption show that the absorbance has direct relationship with
the concentration.
References
Discussion questions
Question 1
Calculate the molarity of the stock BSA solution (0.5%)
Convert molecular weight from kDa to Da (Da= KDa x 1000)
BSA 66.5 kDa = 66500 Da
While 1Da = 1g/mol, then molecular weight of BSA = 66500 g/mol
Each 100 ml contain 0.5 g so 1 litter contains 5g
Number of moles = 5/66500 = 0.000075188
Molarity = 752 x10-7 mol/L
solution
X
Y
Z
Concentration ?g/ml
504
641.5
864.5
Concentration g/l
0.504
0.6415
0.8645
Molecular weight in DA
42300
26500
19700
Molecular weight g/mol
42300
26500
19700
Molarity mol/l
119 x10-7
242 x10-7
439 x10-7
Question 2
