Five µl of a 10-to-1 dilution sample was added to 3ml of Bradford Reagent. The absorbance reading was 0.68. According to the standard curve (not shown) this represents 0.05mg of protein. Determine the stock concentration of the protein solution. (give your answer in mg/ml)

Answer 1

To determine the stock concentration of the protein solution, we first need to understand the relationship between the diluted sample and the original stock solution.

1. Dilution Factor Calculation: The sample was diluted in a 10-to-1 dilution, which means that for every 1 part of sample, there are 10 parts in total (1 part sample + 9 parts diluent). Therefore, the dilution factor is 10.

2. Amount of Protein in the Sample: From the absorbance reading, we know that the amount of protein in the 5 µl of the diluted sample is 0.05 mg.

3. Amount of Protein in the Original Sample: Since the protein concentration in the diluted sample is 10 times lower than in the stock solution, we can find the amount of protein in the original stock solution by multiplying the amount in the diluted sample by the dilution factor:

   \[ \text{Amount in stock} = \text{Amount in diluted sample} \times \text{Dilution Factor} = 0.05 , \text{mg} \times 10 = 0.5 , \text{mg} \]

4. Calculate the Concentration: To find the concentration in the stock solution, we need to know the total volume of the stock solution that correlates to the 0.5 mg of protein. Since we added 5 µl of the sample to the Bradford Reagent, we want to express this in a volumetric unit. The dilution factor indicates that the original solution (stock) was diluted down to meet this measurement.

   Therefore, to find the concentration in mg/ml, we consider how much volume the protein (0.5 mg) is representing in the stock solution based on how the dilution was performed.

   Assuming the stock solution is what we used proportionally in the dilution, we can designate that if 5 µl diluted equals 0.5 mg, we express the concentration based on a milliliter volume:

   Since 5 µl is \( \frac{5}{1000} = 0.005 \) ml, and we originally had 0.5 mg protein in this diluted 5 µl:

   \[ \text{Concentration of stock solution} = \frac{\text{Amount of protein}}{\text{Volume of stock sample}} = \frac{0.5 , \text{mg}}{0.005 , \text{ml}} = 100 , \text{mg/ml} \]

Thus, the concentration of the stock protein solution is 100 mg/ml.