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 calculate the concentration of the diluted sample.

1. Determine the concentration in the diluted sample: The diluted sample (which corresponds to the 10-to-1 dilution) contains 0.05 mg of protein in a total volume that includes both the dilution and the Bradford reagent.

   • Volume of diluted sample added = 5 µL = 0.005 mL
   • Volume of Bradford Reagent = 3 mL

   Therefore, the total volume of the mixture is:

   \[ \text{Total volume} = 0.005 , \text{mL} + 3 , \text{mL} = 3.005 , \text{mL} \]

   The concentration of protein in this total solution can be calculated as follows:

   \[ \text{Concentration in diluted sample} = \frac{0.05 , \text{mg}}{3.005 , \text{mL}} \approx 0.01666 , \text{mg/mL} \]

2. Determine the concentration of the stock solution: Since this concentration is from a 10-to-1 dilution, we need to account for this dilution factor to find the concentration in the stock solution.

   A 10-to-1 dilution means the dilution factor is 10, so the stock concentration (C_stock) can be calculated as:

   \[ C_{\text{stock}} = \text{Concentration in diluted sample} \times \text{Dilution Factor} \]

   \[ C_{\text{stock}} = 0.01666 , \text{mg/mL} \times 10 = 0.1666 , \text{mg/mL} \]

Thus, the stock concentration of the protein solution is approximately 0.167 mg/mL.