# Equal Protein Quantity - Loading onto Well (May/09/2010 )

Hi,

I am still struggling with my first Western Blot.

I made Bradford Assay for quantification but don`t know how to calclulate the amount of sample I need in order to load the same quantity of proteins.

Could, please, someone show me how to calculcate this?

I have obtained numbers from the Bradford assay, such as 0.4765 and I don`t know what to do with them. Is this protein in mg/ml, or is it absorbance? I sense that I will need my favorite formula c1v1=c2v2, but how to apply it. And how, after I have found out the protein concentration, to decide what amount of sample I need to mix with the buffer in order to load 20ul protein, for example.

Any help will be greatly appreciated!

-vigri-

the spec gives you absorbance readings that should, if you are in the linear range of the bradford, correspond directly to your concentration. To analyse your results you need to take the absorbance values from your standards (a dilution range of samples of known concentration) and then use these to generate a standard curve. From the equation of the standard curve, you can calculate your unknown sample concentrations. Remember to keep your units consistent!

-bob1-

yeah i second that... you will get absorbance values of your standards. You will have to plot a graph of your values vs concentration (preferably a 4PL graph if your standard range is very high)
then calculate the concentratuon of your unknowns according to the absorbance values obtained from the assay and then u have concentrations and so you can load equal amounts!!!
Best luck!!

-Prep!-

Well, with your help I managed to calculate the concentration.
I made a standard curve and used the formula from excel to calculate the exact concentration. Then I multiplied with the dilution and obtained my final mg/ml.

However I am still unsure how to determine the volume I have to take out from my initial sample. Let`s say I have c= 2.0769 mg/ml. So if I want to load 15ug how do I calculate this. Sorry, I am completely blocked and the number 7.24ml doesn`t seem to be right...

In fact, I have to determine a certain amount (15 ug) that can be loaded onto the gel. Does anyone has the answer?

Or is it 7.24ul? I see that I am not changing the units.

-vigri-

2.0769 mg/ml * x = 15 ug

2.0769 ug/ul * x = 15 ug

then x = 15/2.0769 ul
= 7. 24 ul.

vigri on May 10 2010, 11:31 AM said:

Well, with your help I managed to calculate the concentration.
I made a standard curve and used the formula from excel to calculate the exact concentration. Then I multiplied with the dilution and obtained my final mg/ml.

However I am still unsure how to determine the volume I have to take out from my initial sample. Let`s say I have c= 2.0769 mg/ml. So if I want to load 15ug how do I calculate this. Sorry, I am completely blocked and the number 7.24ml doesn`t seem to be right...

In fact, I have to determine a certain amount (15 ug) that can be loaded onto the gel. Does anyone has the answer?

Or is it 7.24ul? I see that I am not changing the units.

-zienpiggie-

Thank you zienpiggie!

Finger crossed for my antibodies

-vigri-

vigri on May 11 2010, 01:12 AM said:

Thank you zienpiggie!

Finger crossed for my antibodies

Best luck vigri!!

-Prep!-