You will be given a stock solution of BSA protein with a BSA concentration of 2 mg/mL. (Here, "mgimL is the concentration unit instead of molarity.)
1) Use the parallel dilution formula to calculate the volume of this stock solution you will need to make 100 HL of each BSA dilution shown in the table below. The desired concentrations of these dilutions are provided in the first column of the table.
2) Calculate the volume of water required so that each dilution will have a final volume of 100 μL
Transfer your calculated values to Table 1 of the Lab Procedures.
Protein concentration(mg/mL) of BSA dilution Volume of BSA stockneeded (uL) Volume of waterneeded (uL)
0.10
0.15
0.20
0.25
0.30
0.35
0.40

For the given data, index 1 represents the concentration and volume of the stock solution and index 2, the concentration and volume of the diluted solution. For example:

[tex]V_{1} = \frac{C_{2}.V_{2}}{C_{1}}[/tex]

[tex]V_{1} = \frac{0.1.100}{2}[/tex] = 5 μL

[tex]V_{1} = \frac{0.15.100}{2}[/tex] = 7.5 μL

[tex]V_{1} = \frac{0.2.100}{2}[/tex] = 10 μL

[tex]V_{1} = \frac{0.25.100}{2}[/tex] = 12.5 μL

[tex]V_{1} = \frac{0.3.100}{2}[/tex] = 15 μL

[tex]V_{1} = \frac{0.35.100}{2}[/tex] = 17.5 μL

[tex]V_{1} = \frac{0.2.100}{2}[/tex] = 20 μL

Substituing [tex]C_{2}[/tex] for each concentration given and find each volume needed.

The results is shown in the second column of the table.

2) The stock and total volume is determined. So, to calculate the volume of water needed:

[tex]V_{H_{2}O} = V_{total} - V_{stock}[/tex]

[tex]V_{H_{2}O}[/tex] = 100 - 5 = 95 μL

[tex]V_{H_{2}O}[/tex] = 100 - 7.5 = 92.5 μL

[tex]V_{H_{2}O}[/tex] = 100 - 10 = 90 μL

[tex]V_{H_{2}O}[/tex] = 100 - 12.5 = 87.5 μL

[tex]V_{H_{2}O}[/tex] = 100 - 15 = 85 μL

[tex]V_{H_{2}O}[/tex] = 100 - 17.5 82.5 μL

[tex]V_{H_{2}O}[/tex] = 100 - 20 = 80 μL

As shown in the third column of the table.