You have protein conc. 6.18ug/ul and you need to load 50ug protein per well. Then,

6.18ug protein is present in 1ul,

So, 50ug willbe present in (1ul x 50ug)/6.18ug = 8.09ul

Do the calculation accordingly for other concentrations.

Good luck

Dear Sandeep,

Formula is correct but unit calculation is bit different. You need 50 ug/20ul, that means 2.5 ug/ul. Your protein concentration is 6.18 ug/ul.

Then, N1=6.18ug/ul, V1=?, N2=2.5 ug/ul, V2=20ul.

V1=(2.5 X 20)/6.18= 8.09 ul + 11.91ul buffer will give you 50ug protein in 20 ul of solution.

Hope this will help you.

Hi Kumar,

If i get you right,you want to calculate equal concentration of proteins to load on SDS-PAGE.Since you have your proteins with different concentration,you can only calculate the volume to be loaded using a constant amount(equal amount) .Since Conc(ug/ul)= Amount(ug)/Volume(ul), so, you can calculate the volume by using same amount e.g 50ug of your first protein using conc. of 6.18(50/6.18=8.1ul).....and add(5ul) of 4x-SDS dye +6.9ul of DW or buffer to your desired volume(20ul)and then load.Also calculate for others .You can only load equal amount of all the proteins but not same concentration.

I hope this will help.

You have protein conc. 6.18ug/ul and you need to load 50ug protein per well. Then,

6.18ug protein is present in 1ul,

So, 50ug willbe present in (1ul x 50ug)/6.18ug = 8.09ul

Do the calculation accordingly for other concentrations.

Good luck

Hi Kumar,

According to your question and the answers that have been given by the expert, they all explain the same way how to go about your protein analysis. It is up to you to choose which way is simpler to use to calculate the amount of your protein. I hope your question has been answered and good luck with your experiment. Xiexie.

N1*V1=N2*V2 is used in titrations to calculate changes of a single solution. In your case of protein solutions for WB, it is much easier to calculate with dilution factors (which is basically what the previous answers did, the 50/6,18=8,09). This is basically the exact same thing as a N1*V1=N2*V2. However, you don't have to use the entire stock solution which is obviously much more applicable. Once you have your dilution factor, you can note it on the stock solution or in your lab book and use and modify from there, even on the fly. In short: if you want to dilute your entire stock use N1*V1=N2*V2. If you want to dilute some of your stock, use a dilution factor.

What I mean is this.

[stock_concentration]/[desired_concentration]=dilution_factor.

Then it's easy to calculate your working solutions: [desired_volume]/[dilution_factor] = ul_of_stock. Then add water/buffer to desired_volume.

Example in your case (6,18 ug/ul). dilution_factor = 6,18 / 2,5 (50 ug/20 ul) = 2,742. If you, for example, want to have enough working solution to load 6 wells. You have 6 wells * 20 ul = 120 ul (desired_volume). 120/2,742= 43,76 ul, add (120-43,76=) 76,24 ul water/buffer. Then you have your working solution of 2,5 ug/ul.

And if you persist on using N1*V1=N2*V2 and want to dilute the entire stock, you rephrase the question like this: You have, for example, a 100 ul solution of 6,18 ug/ul protein dilution and want to get a solution of 2,5 ug/ul, how much water/buffer do I add?

Then:

N1=6,18; V1=100; N2=2,5; V2=?.

Thus: V2=6,18/2,5*100=247,2. Thus add 147,2 ul to your 100 ul stock to get a solution of 2,5 ug/ul.

This is probably what you have done to get your 161,18. But you didn't provide the stock volumes. Remember, that you want a certain concentration or amount of protein. in your case 50 ug/20 ul=2,5 ug/ul. If you were to take 20 ul of the 247,2 solution you get exactly that.

Hopefully this helps. And good luck.