Question

You are given a protein solution with a concentration of 0.15 mg/ml.

iii. If the molecular weight of the protein is 22,000 Da, express its initial concentration in moles/liter, micromoles/mL, and micromoles/microliters. If we want 20 micromoles of proteins for a reaction, what volume do we need to prepare?

Answer 1

We need 2.933 L of 0.15 mg /mL of protein solution.

Step-by-step explanation:

Concentration of given solution
`C_1 = 0.15 mg/mL`

1 mg = 0.001 g , 1 mL = 0.001 L

`C_1=(0.15* 0.001 mg)/(1* 0.001 L)=0.15 g/L`

Molecular weight of protein = 22,000 Da =22,000 g/mol

Initial concentration in moles/liter:

`C_1=(0.15 g/L)/(22,000 g/mol)=6.8182* 10^(-6) mol/L`

Initial concentration in micromoles/mL :

1 L = 1000 mL

`C_1=6.8182* 10^(-6) mol/L=(6.8182* 10^(-6)* 10^6 \mu mol)/(1000 mL)=6.8182* 10^(-3) \mu mole/ mL`

Initial concentration in micromoles/microLiter :

1 L = 1000,000 μL

`C_1=6.8182* 10^(-6) mol/L=(6.8182* 10^(-6)* 10^6 \mu mol)/(1000000 \mu L)=6.8182* 10^(-6)\mu mol/\mu L`

Moles of protein required = 20 μmoles

n(Moles)=C(concentration) × V(Volume of solution)

`20 \mu mol=6.8182* 10^(-6)\mu mol/\mu L* V`

`V =(20 \mu mol)/(6.8182* 10^(-6)\mu mol/\mu L)`

`V=2.933* 10^6 \mu L = 2.933 L`

We need 2.933 L of 0.15 mg /mL of protein solution.