Question

Quantification of Total Protein Using Bradford Protein Assay Pre-Lab Your pre-lab assignment is to plan how you will prepare your BSA standard solutions for the calibration curve. A 2 mg/mL BSA stock solution will be available. You can simply dilute with deionized water. You will need to make 25 microliters of each of the following concentrations. For the first concentration, you will need to make 160 microliters so that the BSA volume is large enough that it can be measured with a micropipettor 25 micrograms/ml 300 micrograms/mL 600 micrograms/ml 900 micrograms/ml 1200 micrograms/mL Please write up a plan for how you will prepare these standards and submit it here

Answer 1

We have stock of BSA (2mg/mL) = 2000 microgram/ mL (NOTE- 1mg = 1000 microgram)

Using formula C1 x V1=C2 x V2, we will calculate volume V1 of stock required to be added, to achieve required concentration

For 25 microgram/mL concentration

Stock: C1=2000 microgram/mL; V1=?

Required dilution: C2=25 microgram/mL; V2=160 microliter

2000 x V1 = 25 x 160

V1= 25 x 160 / 2000 = 20 microliters

Thus 20 microliters of BSA stock will be diluted by deionized water to make up a volume to 160 microliters

For 300 microgram/mL concentration

Stock: C1=2000 microgram/mL; V1=?

Required dilution: C2=300 microgram/mL; V2=25 microliter

2000 x V1 = 300 x 25

V1= 300 x 25 / 2000 = 3.75 microliters

Thus 3.75 microliters of BSA stock will be diluted by deionized water (21.25 microliter) to make up a volume to 25 microliters

For 600 microgram/mL concentration

Stock: C1=2000 microgram/mL; V1=?

Required dilution: C2=600 microgram/mL; V2=25 microliter

2000 x V1 = 600 x 25

V1= 600 x 25 / 2000 = 7.5 microliters

Thus 7.5 microliters of BSA stock will be diluted by deionized water (17.5 microliter) to make up a volume to 25 microliters

For 900 microgram/mL concentration

Stock: C1=2000 microgram/mL; V1=?

Required dilution: C2=900 microgram/mL; V2=25 microliter

2000 x V1 = 900 x 25

V1= 900 x 25 / 2000 = 11.25 microliters

Thus 11.25 microliters of BSA stock will be diluted by deionized water (13.75 microliter) to make up a volume to 25 microliters

For 1200 microgram/mL concentration

Stock: C1=2000 microgram/mL; V1=?

Required dilution: C2=1200 microgram/mL; V2=25 microliter

2000 x V1 = 1200 x 25

V1= 1200 x 25 / 2000 = 15 microliters

Thus 15 microliters of BSA stock will be diluted by deionized water (10 microliter) to make up a volume to 25 microliters

This is summarized as

Final concentration (microgram/mL)	BSA stock (2mg/mL) (microliter)	Deionized water (microliter)	Final volume (microliter)
25	20	140	160
300	3.75	21.25	25
600	7.5	17.5	25
900	11.25	13.75	25
1200	15	10	25

Plan list would be

1. Do Correct calculation for dilutions

2. Prepare protease free plastic ware (autoclaved tips, microtubes etc)

3. Label tubes properly before making dilutions

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