3 We have a solution of protein that has a concentration of

3. We have a solution of protein that has a concentration of 0.25 mg/ml.

a.We need 20 micro-grams of the protein for an experiment. What volume of the protein solution do we need?

b.Suppose we want a solution containing 150 mircograms of the protein at a concentration of 0.50 mg/ml. To do this we will first evaporate the liquid from enough of the protein solution to get 150 mircograms. How much solution do we need to start with? How much H₂O do we add to get the desired concentration?

c.If the protein has a M.W. = 15,000, express its initial concentration in mirco moles/ml and in mircomoles/l. If we want 2 mirco moles of the protein for a reaction, how much of the original solution do we need?

d.Suppose we want 1 ml of a 10 g/ml solution. How much H₂O and protein stock must we add to get this?

e.Suppose we want 100 mircoliter of a 0.1 mircogram/l solution. How much H₂O and protein stock must we add to get this?

Solution

A) Convert 20 g into mg and you will get 0.02 mg. Then C (Concentration) = M (Mass)/V (Volume) rearrange for Volume V= M/C , V= (0.02 mg)/(0.25 mg/ ml) = 0.08 ml or 80 l.

B) Convert 150 g into 0.15 mg and then divide it by 0.25 mg/ ml which gives 0.60 ml that needs to be dried down to obtain 0.5 mg/ml, then you will add enough amount of water so that the total volume is 0.5 ml.

C) Convert 0.25 mg/mL = 0.00025 g/mL = 0.00025 g/0.001 L = 0.25 g/L If M.W. of the protein is 15,000 g/mol then 0.25 g = 0.25/15,000 moles = 1.67 x 10^-5 moles. This many moles is contained in a liter (L) therefore concentration is 1.67 x 10^-5 moles/L = 1.67moles/L = 1.67 moles/1000 mL = 1.67 x 10^-3 moles/mL = 1.67 x 10^-3 moles/1000 L = 1.67 x 10^-6 moles/L. The concentration of the solution is 1.67 x 10^-6 moles/L, 1 mole is present in 1/(1.67 x 10^-6) L and if 2 moles are present in 2/(1.67 x 10^-5) L = 1.20 x 10^-5 L = 1.20 L

D) C1V1 = C2V2 (250 g/ l)(V1) = (10 µg/ml)(1ml) V1= 0.04 ml or 40 l (Volume of protein stock for solution) Volume of water 1.0 ml- 0.04ml = 0.96 ml

E) C1V1 = C2V2 (0.25 g/ l)(V1)= (0.1 g/ l)(100 l) V1=40.0 l (Volume of protein stock for solution) Volume of water 100 l- 40.0 l = 60.0 l