Question

Calculate the final concentration of BSA in the problems below using the formula

C1*V1 = C2*V2,

where C1 and V1 are the concentration and volume of the starting solution, and C2 and V2 are the concentration and volume of the final diluted solution. Include proper units and show your work for full credit.

A. 300 ul of 5 mg/ml BSA is added to 700 ul TBS



B. 50 ul of 1.5 mg/ml BSA is mixed with 450 ul H2O and 500 ul Bradford Reagent

C. 10 ul of 1 mg/ml BSA is mixed with 990 ul TBS.

D. 10 ul of 0.1 mg/ml BSA is mixed with 990 ul TBS.

Answer 1

A.
`C_2=1.5(mg)/(mL)`

B.
`C_2=0.075(mg)/(mL)`

C.
`C_2=0.01(mg)/(mL)`

D.
`C_2=0.001(mg)/(mL)`

Step-by-step explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

`C_2=(C_1V_1)/(V_2)`

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

`C_2=(300\mu L*5(mg)/(mL) )/((300+700)\mu L)\\\\C_2=1.5(mg)/(mL)`

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

`C_2=(50\mu L*1.5(mg)/(mL) )/((50+450+500)\mu L)\\\\C_2=0.075(mg)/(mL)`

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

`C_2=(10\mu L*1(mg)/(mL) )/((10+990)\mu L)\\\\C_2=0.01(mg)/(mL)`

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

`C_2=(10\mu L*0.1(mg)/(mL) )/((10+990)\mu L)\\\\C_2=0.001(mg)/(mL)`

Best regards.