LSM 1101- Lab Report

(Practical 2: Quantitative Protein Estimation)

Group A- Bench 9

Members name:

Lee Xiang Yi Joycelyn (U074244N)

Lee Xin Yi (U074121N)

TA: Suguna

Dear Ms Suguna,

Due to some unexpected errors in our practical, the results obtained for both experiments 3.1 and 3.2 are inaccurate. Therefore, to make up for our mistakes, we used the results of one of our friend to complete the questions and discussions in our lab report. The first two pages are our own results for the experiment (but without answers to questions), where we discussed the possible reasons of our abnormal results. From the fourth page onwards, we used our friend’s results to complete our own discussion.

If there is a need, please penalise us accordingly. Thank you for your kind understanding.

Yours Sincerely,

Joycelyn

Xin Yi

Experiment 3.1 Ultraviolet absorbance of protein

Results and discussion:

We are to prepare 10 ml of 1.0 mg/ml BSA from the given BSA solution which has a concentration of 5 mg/ml. The dilution factor is 5X. Therefore, four parts of water should be added to every one part of BSA. To obtain 10 ml of 1.0 ml/mg of BSA, 8 ml of water is added to 2 ml of BSA.

Absorbance, A of samples

wavelength/nm

Blank (water)

1mg/ml BSA U40 U182250 0 0.326 0.090 0.011

260 0 0.404 0.088 0.011

270 0 0.572 0.090 0.011

280 0 0.646 0.090 0.009

290 0 0.291 0.086 0.002

300 0 0.055 0.085 -0.002

310 0 0.017 0.083 -0.004

320 0 0.010 0.084 -0.005

330 0 0.008 0.085 -0.004

340 0 0.007 0.083 -0.005

350 0 0.006 0.083 -0.005

360 0 0.006 0.085 -0.004

250 270 290 310 330 350

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

waterBSAU182U40

Possible reasons for the abnormal results

1) The inaccurate results are not due to dilution. Regardless of the concentration of the unknowns, different wavelengths passing through the samples should still give different absorbance values. The graphs obtained should still follow the shape of BSA.

2) Due to ignorance and carelessness, we labeled the cuvettes at their windows. This might affect the wavelength and the amount of light passing through, accounting for the inaccurate results.

Experiment 3.2 Dye binding method

Table 2 showing the absorbance of the respective sample at 595nm

  Absorbance at wavelength 595nmSample

Concentration of BSA (µg/ml)  2 0.0954 0.1846 0.2738 0.347

10 0.466Unknown 40 0.797Unknown 182 0.480

1 2 3 4 5 6 7 8 9 10 110

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Absorbance at wavelength 595nm

Absorbance at wavelength 595nmLinear (Absorbance at wave-length 595nm)

Reasons for the abnormal results

1) The high values of absorbance for the unknowns could be due to inaccurate dilution. We might have added too little water to the samples.

Experiment 3.1 Ultraviolet Absorbance of Proteins

Results and discussion

The unknown samples given are #25 and #179.

We are to prepare 10 ml of 1.0 mg/ml BSA from the given BSA solution which has a concentration of 5 mg/ml. The dilution factor is 5X. Therefore, four parts of water should be added to every one part of BSA. To obtain 10 ml of 1.0 ml/mg of BSA, 8 ml of water is added to 2 ml of BSA.

Absorbance, A of sampleswavelength/nm

Blank (water)

1mg/ml BSA U25 U179

250 0 0.165 0.216 0.945

260 0 0.218 0.256 1.066

270 0 0.418 0.444 1.089

280 0 0.508 0.529 0.924

290 0 0.191 0.244 0.372

300 0 -0.025 0.047 0.05

310 0 -0.059 0.013 -0.013

320 0 -0.068 0.003 -0.026

330 0 -0.074 -0.001 -0.031

340 0 -0.071 0.001 -0.028

350 0 -0.062 0.013 -0.021

360 0 -0.056 0.02 -0.012

250 260 270 280 290 300 310 320 330 340 350 360-0.5

0

0.5

1

1.5

2

2.5

Graph of Absorbance, A against wavelength/ nm

U179U251mg/ml BSABlank (water)

Wavelength/ nm

Abso

rban

ce, A

Answers to Questions

Question1

When E 1% = 10.0 , 1% = 1g in 100ml

= 1000mg in 100ml= 10mg/ml

Hence, absorbance of 10.0 = 10mg/ml

BSAA= 0.508 at 280nm UV raysTherefore, protein content= 0.508mg ml-1

Unknown #U25A= 0.529 at 280nm UV raysTherefore, protein content= 0.529mg ml-1

Unknown #U179A= 0.924 at 280nm UV raysTherefore, protein content= 0.924mg ml-1

When E 1% = 6.67 ,

1% = 1g in 100ml= 1000mg in 100ml= 10mg/ml

Hence, absorbance of 6.67 = 10mg/ml

BSAA= 0.508Therefore protein content= 10/6.67 x 0.508= 0.762mg ml-1 (3sf)

Unknown #U25A= 0.529Therefore protein content= 10/6.67 x 0.529= 0.793mg ml-1 (3sf)

Unknown #U179A= 0.924Therefore protein content= 10/ 6.67 x 0.924= 1.39mg ml-1 (3sf)

Question 2

As seen from the graph, both U25 and 1mg/ml of BSA showed similar spectrum shape. Thus, U25 is probably a sample of pure BSA. Both proteins have a maximum absorption at 280nm.

However, U179 displayed a totally different spectrum, so we may conclude that it is not a pure protein and may contain other substances. It has a maximum absorbance at 270nm but not 280nm. This probably signifies that it contains traces of DNA and BSA, since DNA registers a maximal absorbance at 260nm.

Experiment 2 Dye binding method

Results

Dilution was necessary to bring down the concentration to linear range. Since linear range of assay of BSA is 1.2 to 10ug/ml, samples of 2ml of concentrations 10ug/ml, 8ug/ml, 6ug/ml, 4ug/ml and 2ug/ml were prepared via dilution.

To get 10ug/ml from 1mg/ml of BSA,

100X dilution is carried out

To get 8ug/ml from 10ug/ml of BSA,

1.25X dilution is carried out.

To get 6ug/ml from 10ug/ml of BSA,

1.667X dilution is carried out

To get 4ug/ml from 10ug/ml of BSA,

2.5X dilution is carried out

To get 2ug/ml from 10ug/ml of BSA,

5X dilution is carried out

Concentration of BSA (ug/ml)

Volume of water/ ul

Volume of BSA/ ul

Absorbance at 595nm

2 800 200 0.5484 600 400 0.5566 400 600 0.6848 200 800 0.76810 9900 100 0.752

Unknown samples are diluted 100 times and 2 ml of samples were respectively prepared as follows:

Unknown Volume of sample extracted (ml)

Volume of deionised water added (ml)

25 0.02 1.98179 0.02 1.98

Sample Absorbance at 595nm

Blank 0U25 0.488U179 0.745

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Graph of Absorption at 595nm against Con-centration of BSA/ ug/ml

Concentration of BSA/ ug/ml

Abso

rptio

n, A

Answers to Questions

1) U25dil has absorption of 0.488 at 595nm. Based on the graph above, [U25dil]≈ 4.2ug/ml

[U25] = 4.2ug x 100= 420ug/ml= 0.420mg/ml

U179dil has absorption of 0.745 at 595nm. Based on the graph above, [U179dil]≈ 6.2ug/ml

[U179]= 6.2ug/ml x 100= 620ug/ml= 0.620mg/ml

3)

Unknowns

Protein Content of unknowns / mgml-1 by

Calculations usingGraphical MethodE1% =

10.0E1% = 6.67

25 0.529 0.793 0.420

179 0.924 1.39 0.620

The protein content of U25 based on experiment 1 is 0.793mg/ml and that based on the dye binding method is 0.420mg/ml.

The protein content if U179 based on experiment 1 is 1.39mg/ml and that based on the dye binding method is 0.620mg/ml.

In both unknown samples, the disparity between both experimental values is a significant one. This suggests that only one of the methodologies is a more accurate means in the determination of protein content.

Experiment one was based on the fact that maximum absorption occurs at 280nm, due to the presence of tryptophan and tyrosine amino acids. It is however to be noted that proteins contain varying amounts of both amino acids, thus, different proteins have different values for it at 280nm. Measurement of protein content based on the E1% of BSA would either under or over represent the protein content in the unknown samples. This is proven by the use of E1%= 10 which gave values that differ significantly from the protein content calculated using E1%= 6.67.

Experiment 2 on the other hand measures protein content using the dye binding technique. Dyes bind to proteins to create a colour change measured at a constant wavelength. There is a direct relationship between protein content and colour change measured. There is no problem of misrepresentation and hence, it is very clear that the protein content of the unknown samples is to be based on experiment 2.

4) According to the Beer Lambert Law,

Absorbance, A= Molar Extinction coefficient, E X (concentration of the compound in solution, C / MW, molecular weight) X path length of the cuvette in which the sample is contained, L

0.667= E X (1.0/ 66, 000) X 1

E= 44022 M-1cm-1