Hi Eman,

2 ul from stock give you 4 ug/ml concentration

0.2 ul (which is too small volume in terms of accuracy) from stock is suppose to be 0.4 ug/ml concentration. 

I suggest you use less stock concentration of BSA because If you want more accurate results you have to use 0.2 mg/ml BSA concentration which results in higher volumes which those you used. Or you can higher the concentrations of samples.

Another point is if you prepare Bradford samples, you can use the below procotol. It will give you more accurate results. 

900 ul Bradford solution + sample volume + (100 ul -sample volume) ddH2O

With this formulation, you can say, the protein in exactly in 1ml, so the final concentration will be (micro)ug/ml. In your case the total volume is 1020 ul, so you can just normalize your results with direct proportion. However lots of people just ignore that 20 ul which you used to prepare the protein sample. For your protein of interest, you add 20 ul from the stock, so you should divide the protein concentration/ 20 to get the concentration of 1 ul. 

After calculation ug/ul of the enzyme, you can calculate total protein.

I hope it helps

If your stock solution is 2 mg/mL and you use 20 µL of it, the sample contains m = c * V = 2 mg/mL * 0.02 mL = 0.04 mg of protein, or 40 µg.

If your sample gives the same absorption as that then you conclude that it contained 40 µg of protein. If the sample volume was, say,  200 µL, then the protein concentration is c = m / V = 40 µg / 200 µL = 0.2 µg/µL, which is equivalent to 0.2 mg/mL.

Apart from the maths there is another factor to consider: With the exception of Kjeldahl's nitrogen and the biuret method (which are very insensitive and hence rarely used) all methods for determination of protein concentration give different absorbtion for the same amount of different proteins, depending on their amino acid composition. I have tried this with Hsc70 once, depending on the method (Bradford, Lowry, BCA) and standard protein (BSA, IgG) the results varied by a factor of 20! However, this doesn't matter as long as you need only relative values and use the same method and standard protein throughout your study. But violation of any of these conditions can lead to seriously misleading results.

Many people find this confusing. Think about the amount of protein (in µg) added in each standard, rather than the volume or concentration.

For example, if you added 2 µl of a 2 mg/ml (= 2 µg/µl) BSA solution, then there was 4 µg of BSA. It is important that all the standards have the same volume (20 µl in your case), and that the sample also have that volume. If the volume is a constant, you can forget about it until the end.

Plot the standard curve of absorbance versus µg of BSA.

For your unknowns, read the µg of BSA-equivalent protein off the standard curve based on the absorbance measured.

Divide the µg by the sample volume, which was 20 µl, to get µg/µl (= mg/ml) of the sample used in the assay.

If the sample was diluted before it was used in the assay, calculate the concentration of the original sample using a proportionality. Done.

For the specific activity calculation, you can take the same approach of focusing on the amount of protein that was used in the enzyme assay rather than the volume. Specific activity is units of activity/mg of protein, so you need to calculate how many mg of protein was present in the enzyme reaction.

Thanks all for your valuable cooperation , I am really appreciate that

I will explain the results I have I hope that I am on the right way

I used  20 , 10,8,6,4,2 µl of the stock (2 mg/ml BSA ) completed till 20µl buffer, So the protein would be 40,20,16,12,8,4 µg/µl on my standard   I did the standard curve depending on the absorbance

For my samples I used 20,10,5,2 µl completed till 20µl buffer and get the protein amount  from the standard curve for example I obtained 64.61, 34.77 , 11.48, 0.67 for one sample

To obtain the amount mg/ml I divided on the sample volume it would be 3.23, 3.47, 2.29,  0.33

I think there are something wrong as these values should be equal for the same sample ??????

Does I have to use  the absorbance within the standard  absorbance values  only and reject the samples  absorbance values higher or lower the standard absorbance values or can use all depend on the trend character of excel??

Finally if I used the value 3.47 mg/ml and I used 200µl sample  for the  enzyme assay so  the specific activity would be divided on (200*3.47)/1000=0.694

Thanks again for help

Your standard curve goes from 4 to 40 µg BSA. Was there also a 0 µg standard? If not, you cannot make measurements for amounts of protein below 4 µg because that would be extrapolation. Even if you had the 0 µg standard, the absorbance measurements for amounts of protein below 4 µg may have been so low as to be unreliable. Also, you cannot make measurements above 40 µg for the same reason. So your 20 µl sample was off-scale and should not be used. Your 2 µl sample should have been on-scale, but wasn't, so I assume something went wrong with that one. That leaves your 10 and 5 µl samples, which gave 34.8 and 11.5 µg, or 3.47 and 2.29 mg/ml. These values are pretty far apart, suggesting a methodological problem. If you only need a rough estimate of the amount of protein, then average them. If you need a precise measurement, you should repeat the experiment using one sample volume in triplicate, at least. (Make all measurements, including the standard curve, in triplicate.) Examine your technique to see if you can improve the reproducibility of your measurements.

Can you show the absorbance data from the experiment? It might help us to spot problems.

Dear

I purpose simple formula like: Concentration of protein (mg ou ug)/ml= OD of the sample/OD of the standard X concentration of the standard 

Thank

Marius

Thanks All,

Here you are some of my absorbance data

BSA µg/ 20 µl reading 1 reading 2      reading 3       reading 4 average OD 595

40                 0.583          0.573           0.572            0.568            0.574

20                  0.373        0.359             0.344            0.336            0.353

16                  0.362         0.351            0.33               0.322           0.34125

12                  0.286          0.282           0.28               0.269           0.27925

8                    0.239          0.241          0.232                                   0.23733

4                    0.239          0.249          0.212                                 0.208 0.227

for the samples

          20 µl        10 µl       5 µl                2 µl

1      0.819         0.518      0.283       0.174

2      0.475        0.235       0.151        0.061

3       0.361      0.17          0.062        -0.019

4      0.976        0.551       0.481        0.322

5      0.646        0.454        0.168        0.115

6       0.503       0.317        0.137        0.112

I leaved the reaction for about 40 minutes then take the absorbance

I did not take zero reading for the standard

I need the protein percent to calculate the proteinase enzyme activity of my samples

Thanks Inadvance

One thing you can do is to plot your absorbance values against volume. According to Lambert-Beer's law, this should result in straight lines. In practice, you may get a curve at high protein amounts. Disregard the curved data and also any data points that are clearly outliers (nothing prevents more interpretation errors than "eyeballing" your data). Then use linear regression to calculate the slopes b of the lines, these are related to concentration c  by b1 / c1 = b2 / c2. This is a very precise method to calculate the concentration of your sample. The standard deviation of the slope of the regression line gives you an idea about the precision of your measurement, with practice you should get a relative error below 2%. You can use statistics software like SciDAVis (open source), Origin or Sigma-Plot (both closed source) for plotting and calculations.

In the long term, you should take a course in scientific data analysis.

The triplicate measurements for your standard curve are in reasonably good agreement and show that the standard curve was useful between 8 and 40 µg.

Unfortunately, the 6 replicates for each volume of the sample show a very wide range of values. There must have been some methodological problem with the assay at that point.

Thanks All,

Dear Adam the samples values were  for a different samples not replicate , I will repeat them with the volumes give absorbance  values within the standard curve values

Thanks

Also, make sure to include a 0 µg standard, and test for any signal produced by the buffer.