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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