Question

# Dilute the IgG protein standard $200 µg/mL$ to create a 6-point standard curve ranging from 0 – 75 µg/mL $in duplicate$. Determine the valid concentrations for each tube. The final volume is 1.0 mL. Table 1. Preparation of IgG standard curve including volumes, concentrations, and absorbance values determined by undertaking a Bradford assay $2 mark$. Tube 1 2 3 4 5 6 7 8 9 10 11 12 IgG $200 µg/mL$, mL H2O, mL Final [IgG] $µg/mL$ 0 0 75 75

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## Answer to a math question Dilute the IgG protein standard $200 µg/mL$ to create a 6-point standard curve ranging from 0 – 75 µg/mL $in duplicate$. Determine the valid concentrations for each tube. The final volume is 1.0 mL. Table 1. Preparation of IgG standard curve including volumes, concentrations, and absorbance values determined by undertaking a Bradford assay $2 mark$. Tube 1 2 3 4 5 6 7 8 9 10 11 12 IgG $200 µg/mL$, mL H2O, mL Final [IgG] $µg/mL$ 0 0 75 75

Cristian
4.7
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.}}

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