Scientific Communication and Laboratory Skills LabReport Abstract Bradfordprotein assay and direct ultraviolet absorption (UV light) were performed onunknown samples. These two methods were performed to determinate theconcentration of the unknown samples. The Bradfordassay is colorimetric assay depends on the reaction between Coomassie brilliantblue and the aromatic residues in the protein. After the dye react with theseresidues it increases the absorbance from 470nm to 595nm. In general, absorbanceof a serious of known concentrations of standard protein were measured andcreate a standard curve. use the created standard curve to calculate theconcentration of unknown sample based on its absorbance. The method is highlysensitive, and it is very fast but small materials can interfere with it andhighly effect the results. The UV absorptioncan be applied on protein that contains aromatic residues such as tyrosine,tryptophan.
There aromatic residues are able to absorb the UV light at 280 nmwithout adding colouring reagent. uv spectrometer is used to measure theabsorbance and calculate the concentration of protein by using beer’s law. Extinctioncoefficient for the protein is required to carry on the calculation. Thetechnique is quick but the successful of the method depends on having theaccurate extinction coefficient for the protein.
The aim ofusing the both methods is to compare between its results and show if there anydifferent between them. introduction protein assaysare one of the most widely used methods in life science research. Knowingprotein concentration is required in protein purification, cell biology,electrophoresis and other research applications, it is important to use morethan once method during experiment to obtain the require accurate results.there are wide variety of methods such as dye binding assays (Bradford), copperion based assay (lowry and BCA), and spectrophotometry based on UV absorption.Each proteinassay has it is advantage and limitations and it shows the reliability of themethod. The natural of the sample is the most important consideration to choosethe right assay method.
If the sample is in solid form it can be easilysolubilized in protein assay but most of the protein samples are complexsolutions which include non-protein materials. There is many issues effect thereliability of the methods such as interfering agents, sample preparation,assay sensitivity, sample size and time consideration. Proteins arecomplex polymers of amino acids with huge range of modification and different structureswhich required numerous numbers of chemical agents to stabilise it. Non-proteinagents can be a challenge for protein assays, protein solution containssurfactant can interfere with dye based assay as Bradford assay. The betterreliability will be possible with removing non-protein assays or by usingmethod which overcome the interfering effect of it.The size ofthe sample becomes a critical consideration, as most of the methods required atleast 0.5 µg of proteins for a reliable result, the method that requiresthe smallest amount of protein will offer an advantage over the others such asdotMETRIC assay.
The amount of the time which taken to complete the experimentwill depend on the complexity of the sample and the assay method. The methodswhich use standard plots and curves are faster while the protein samplecontains non-protein agents are consuming more time which needed to remove thenon-protein agents. The assays that doesnot required standard plots can be faster and cheaper such as protein dotMETRICprotein assay and dye binding CB-X.
The most reliable protein estimation is the one whichperformed using protein standard. The protein standard should have similarproperties to the sample that being tested. It is very difficult to findprotein standard with similar properties to the sample. Bovine serum albuminand gamma globulin are accepted to use as reference protein.As in thisexperiment, bovine serum albumin was used to create a serious of standardsolutions with known concentrations. Two spectroscopic methods were using toobtain the absorbance of the standard solutions. The absorbance and theconcentration of the standard solution were used to create calibration curve todetect the concentration of the unknown samples. The result of the twospectroscopic methods will be compared to discover if there any differencebetween them.
Materials and methods Bradford essay By using 20?lfrom each different solution mix with 1ml of the Coomassie reagent in an Eppendorftube, mix it on a vortex and leave it at room temperature for 5 minutes. A blue colour should start to develop once the proteinsolution will be mixed with the reagent. The power of the colour will berelated to the protein concentration in the sample. The unknown samples will be treated in the same way theother sample reagent with the same method. In addition of a blank has preparedto calibrate the spectrophotometer at the wavenumber 595nm. Record all the results in a table. Ultraviolet light In this experiment will not be added any reagent to generatethe colour needed because the protein can absorb the ultraviolet radiation at certainwavelength and that is because of the aromatic amino acid, such asphenylalanine, tyrosine and tryptophan.
Set the spectrophotometer will be set at 280nm, it will becalibrated with distilled water. Cuvettes that are compatible with UV lightwere used in this method. After the calibration, the solution will be measured and recorded.that experiment must be repeated 3 times and the results can be presented in atable. Materials used in theexperimentsProtein (Bovine serum albumin) stock solution 1 mg/mlUnknown solution W, X, Y, and ZCoomassie (Bradford) Protein Assay Reagent (ThermoScientific)Distilled waterEppendorf microtubesVisible light cuvettesPhosphoric acid Safety assessment and precaution: this experiment was conducted in level twolaboratory. Bradford assay contain methanol and low concentration phosphoricacid which handle in container in hood with away from eye, skin or any materialcan be flammable.As the chemical can beirritating or toxic it has to be handle in a specific way gloves must be wearall the time so as safety glasses and lab coat. Results and discussion preparation ofprotein standard solutions.
Bovine serum albumin in water 0.5%It means 0.5g in each 100 ml = 0.005g/ml 5000 ?g/mlTotal volume =protein stock solution + distilled water = 2000 µlConvert µl to ml dividing by 10002000 µl/ 1000= 2mlUse the following equation to find the concentration of thestandard solution C1 V1 = C2 V2C2 = C1 V1 / V2 Stander solution 1Protein stock solution 50 µl = 0.05 mlC2 = 5000x 0.05 / 2 = 125 ?g/ml Stander solution 2Protein stock solution 100 µl = 0.
1 ml C2 = 5000×0.1 / 2 = 250 ?g/ml Stander solution 3Protein stock solution 200 µl = 0.2 mlC2 = 5000x 0.2 / 2 = 500 ?g/ml Stander solution 4Protein stock solution 300 µl = 0.3 mlC2 = 5000x 0.
3 / 2 = 750 ?g/ml Stander solution 5Protein stock solution 400 µl = 0.4 mlC2 = 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 Bradfordassay This table showsthe 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 standarddeviation for each standard solution and unknown samples using data given intable Solution 1Average = (0.069 + 0.065 + 0.072) / 3 = 0.
069Calculation 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 standarddeviation.
(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 = 0.0000125Standard deviation = ? 0.0000125 =0.0035Solution 2Average = (0.092 + 0.
095 +0.099) / 3 = 0.096Standard deviation =0.0035Solution 3Average =(0.232 + 0.243 + o.235) / 3 = 0.
236Standard deviation= 0.0051Solution 4Average = (0.320 + 0.326 + 0.
331) / 3 = 0.325Standard deviation = 0.0055Solution 5Average = (0.
410 + 0.405 + 0.411) / 3 = 0.409Standard deviation = 0.0032Solution XAverage = (0.214 + 0.215 + 0.218) / 3 = 0.
216Standard deviation = 0.0021Solution YAverage = (0.267 + 0.271 + 0.
275) / 3 = 0.271Standard deviation = 0.004Solution ZAverage =(0.352 + 0.349 + 0.359) / 3 = 0.
353Standard deviation = 0.005table thatshows the concentration of each standard solution and its corresponding averageabsorbance. 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 calibrationcurve of absorbance against concentration of the standard solutions. Using thecalibration curve and its regression equation, calculate the concentration ofthe unknown samples. Use the average absorbance for each unknown in yourcalculation and report your answers in ?g/ml (to 2 decimal places) as well asin w/v% and ppm. From thecalibration curve y = mc + bY = 0.0004X +0.
0144X = (Y –0.0144) / 0.0004 Solution X0.216 =0.0004X + 0.0144X = (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 1005.04 x102?g / 106 = 0.000504 g/mlw/v% = 0.
0504%ppm= 504 ppm Solution YX = (0.271 –0.0144) / 0.0004 = 64.15 x10 ?g/ml (to 2 decimal places)w/v% = 0.
06415%ppm = 641.5ppm solution ZX = (0.353 –0.0144) / 0.0004 = 86.45 x10 ?g/ml (to 2 decimal places)w/v% 0.
08645%ppm = 864.5ppm standard X Y Z Concentration ?g/ml 504 641.5 864.5 Average absorbance at 595 nm 0.216 0.271 0.353 Ultravioletlight Table showsabsorbance 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 averageabsorbance and standard deviation for each standard solution and unknownsamples using data given in table Solution 1Average = (0.112 + 0.
105 + 0.111) / 3 = 0.109Standard deviation = 0.0038 Solution 2Average = (0.205 + 0.211 +0.217) / 3 = 0.211Standard deviation =0.
006 Solution 3Average =(0.415 + 0.427 + 0.422) / 3 = 0.421Standard deviation= 0.
006 Solution 4Average = (0.612 + 0.628 + 0.625) / 3 = 0.621Standard deviation = 0.0085 Solution 5Average = (0.795 + 0.812 + 0.
829) / 3 = 0.812Standard deviation = 0.017 Solution XAverage = (0.421 + 0.409 + 0.414) / 3 = 0.
415Standard deviation = 0.006 Solution YAverage = (0.525 + 0.532 + 0.515) / 3 = 0.524Standard deviation = 0.0085 Solution ZAverage = (0.541 + 0.
522 + 0.536) = 0.533Standard deviation = 0.0098 table thatshows the concentration of each standard solution and its corresponding averageabsorbance. 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 calibrationcurve of absorbance against concentration of the standard solutions. Usingthe calibration curve and its regression equation, calculate the concentrationof the unknown samples. Use the average absorbance for each unknown in yourcalculation and report your answers in ?g/ml (to 2 decimal places) as well asin w/v% and ppm. From thecalibration curve y = mc + bY = 0.0008X +0.0144X = (Y –0.
0144) / 0.0008 Solution X0.415 =0.0008X + 0.0144X = (0.415 –0.0144) / 0.
0008 = 500.75 ?g/ml (to 2 decimal places)w/v% first convert?g to gram then multiple by 100500.75 ?g / 106= 0.00050075 g/mlw/v% = 0.050075%ppm= 500.
75ppm Solution YX = (0.524 –0.0144) / 0.0008 = 6.37 x102 ?g/ml (to 2 decimal places)w/v% = 0.0637%ppm = 637 ppm solution ZX = (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 Throughthe experiment, we will able to find out the concentration of the unknown samplesby using calibration curve of absorbance against concentration. From thestandard curve of Bradford assay, the unknown solutions (X,Y,Z) haveconcentration of 504 µg/ml , 641.5 µg/ml and 864.5µg/ml. while thestandard 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 Bradfordassay and UV light absorption show that the absorbance has direct relationshipwith the concentration. References · Bradford, M. M. (1976). A rapid and sensitive method for the quantitation of microgram quantitiesof protein utilizing the principle of protein-dye binding. Anal Biochem 72:248-254.· Kalb, V.
F. and Bernlohr, R. W.
(1977). A newspectrophotometric assay for protein in cell extracts. Analyt. Biochem. 82, 362–371.
· Kirschenbaum, D. M. (1975) Molar absorptivity and A1%/1 cmvalues for proteins at selected wavelengths of the ultraviolet and visibleregions. Analyt. Biochem. 68, 465–84.· Stoscheck, C. M.
(1990). Quantitation of protein. Methods Enzymol 182: 50-68.· Spector, T. (1978) Refinement of the Coomassie Blue methodof protein quantitation. A simple and linear spectrophotometric assay for<0.5 to 50 g of protein.
Anal. Biochem. 86, 142–146.· Stoscheck, C.
M. (1990) Increased uniformity in the responseof the Coomassie Blue protein assay to different proteins. Anal. Biochem.
184, 111–116.· Waddell, W. J. (1956) A simple UV spectrophotometric methodfor the determination of protein. J. Lab. Clin. Med.
48, 311–314.· Zuo, S.-S. and Lundahl, P. (2000) A micro-Bradford membraneprotein assay. Anal. Biochem.
284, 162–164. Discussion questions Question 1Calculate 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 DaWhile 1Da = 1g/mol, then molecular weight of BSA = 66500 g/molEach 100 ml contain 0.
5 g so 1 litter contains 5gNumber of moles = 5/66500 = 0.000075188Molarity = 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 2The relationship between molarity and molecular weight is indirectrelationship. Due to sample X has the smallest molecular weight, it will havethe highest molarity of all the samples Question 3Bradford assay method is themost precise as it shows the less standard error Column1 Bradford assay UV absorption Mean 0.246875 0.45575 Standard Error 0.042195733 0.078814146 Median 0.2535 0.4725 Mode #N/A #N/A Standard Deviation 0.119347557 0.222920069 Sample Variance 0.014243839 0.049693357 Kurtosis -0.885728735 -0.046227263 Skewness -0.3524454 -0.102443072 Range 0.34 0.703 Minimum 0.069 0.109 Maximum 0.409 0.812 Sum 1.975 3.646 Count 8 8 Question 4The required amount of HCL = 250mlPh of the required HCL = 3To calculate the concentration, use the equation ph = – log (concentration)Concentration = anti log (-ph)Concentration = anti log (-3) = 0.001MUse the equation (c1 v1 = c2 v2)to calculate the required amount of 0.1M HCL to prepare 250ml of 0.001M HCL0.01 x v1 = 0.001 x 250v1 = (0.001 x250) / 0.1 = 2.5mladd 2.5ml of 0.1M HCL to 247.5 ml of distilled water to prepare 250 mlof HCL with a ph value of 3