From a protein assay, you determine that your sample of purified protein contains 3 mg protein/mL. You need to prepare a sample for SDS-PAGE. The final volume of the sample for SDS-PAGE should be 24μL and the final concentration of sample buffer in the SDS-PAGE sample should be 1X. You would like to load a total of 15μg of your purified protein on the gel. Determine the volume of the purified protein, volume of 6X sample buffer, and volume of water needed to prepare a sample for SDS-PAGE (recall final volume should be 24μL).

0.12 μL of purified protein, 4 μL of a 6X sample buffer, 19.88 μL of water.

Explanation:

The law of conservation of mass is used to solve this problem. It indicates that mass is neither created nor destroyed, it is only transformed. In a chemical reaction the sum of the mass of the reagents is equal to the sum of the mass of the products. It can be written as:

Concentration 1 x Volume 1 = Concentration 2 x Volume 2

1= initial values

2= final values

The concentration of the protein is 3mg/mL, we need to load a total of

15 μg/mL (0.015 mg/mL), and we know the final volume should be 24μL. So we calculate this as:

3mg/mL x Volume 1 = 0.015 mg/mL x 24 μL

Volume 1 = 0.12 μL.

This means, we should use 0.12 μL of te solution that contains 3 mg of proteins/mL.

The same equation is used to calculate the amount of sample buffer which is 6 times concentrated (6X)

6X x Volume 1 = 1X x 24 μL

Volume 1 = 4 μL,

This means, we should use 4 μL of a 6X sample buffer.

So far we have 0.12 μL of protein + 4 μL of sample buffer. Since the final volume is 24 μL, we must add a quantity of water in order to reach that final volume.

24 μL - 0.12 μL - 4 μL = 19.88 μL of water,

cuz he's blad

explanation: