To prepare a 6-point standard curve ranging from 0 – 75 µg/mL using the 200 µg/mL IgG protein standard, we will dilute it with water (H2O) to create the desired concentrations in each tube.
Let's denote:
- V1 as the volume of 200 µg/mL IgG protein standard required in each tube (in mL).
- V2 as the volume of water (H2O) required in each tube (in mL).
- C1 as the initial concentration of IgG protein standard (200 µg/mL).
- C2 as the final concentration of IgG protein standard in the tubes (ranging from 0 – 75 µg/mL).
- V_total as the final volume of each tube (1.0 mL).
Given that:
- Tube 1 and 2 will have 0 µg/mL IgG.
- Tube 11 and 12 will have 75 µg/mL IgG.
We can set up a proportion to calculate the volumes of the 200 µg/mL IgG protein standard and water needed in each tube:
For Tube 1 and 2 (0 µg/mL IgG):
\frac{V1}{1.0} = \frac{0}{200} \Rightarrow V1 = 0 \, \text{mL}
V2 = 1.0 - V1 = 1.0 \, \text{mL}
For each tube from 3 to 10 (intermediate concentrations):
\frac{V1}{1.0} = \frac{C2}{C1}
V1 = \frac{C2}{C1} \times 1.0
V2 = 1.0 - V1
For Tube 11 and 12 (75 µg/mL IgG):
\frac{V1}{1.0} = \frac{75}{200} \Rightarrow V1 = 0.375 \, \text{mL}
V2 = 1.0 - V1 = 0.625 \, \text{mL}
Therefore, valid concentrations for each tube are:
- Tube 1: 0 µg/mL IgG (0 mL of 200 µg/mL IgG standard, 1.0 mL H2O)
- Tube 2: 0 µg/mL IgG (0 mL of 200 µg/mL IgG standard, 1.0 mL H2O)
- Tube 3 to 10: Concentrations ranging between 0.0 µg/mL to 75 µg/mL
- Tube 11: 75 µg/mL IgG (0.375 mL of 200 µg/mL IgG standard, 0.625 mL H2O)
- Tube 12: 75 µg/mL IgG (0.375 mL of 200 µg/mL IgG standard, 0.625 mL H2O)
\boxed{\text{Answer: Valid concentrations for each tube have been determined.}}