Hello, I am doing homework and I am not sure if my answers are correct, can some

Question

Hello, I am doing homework and I am not sure if my answers are correct, can someone help me verify if I am doing this right or wrong? My responses are underneath the questions.

You are given a protein solution with a concentration of 0.25 mg/ml. Answer these questions.

A.) We need 20 µg for an experiment. What volume of the protein solution do we need?

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

B.) Suppose we want to prepare a solution containing 150 µg of the protein at a concentration of 0.50 mg/ml. To do this we will first dry down enough protein solution to get 150 µg. How much solution do we need to start with? How much H2O do we add to get the desired concentration?

You will convert 150 g to 0.15 mg and divide by 0.25 mg/ ml to get 0.60 ml that needs to be dried down to obtain 0.5 mg/ml, then you will add enough water so that the total volume is 0.5 ml.

C.) If the protein has a M.W. = 15,000, express its initial concentration in moles/liter, in µmoles/ml and in µmoles/µl. If we want 2 µmoles for a reaction, what volume do we need?

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

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.) Suppose we want 100 µl of a 0.1 µg/µl solution. How much H2O and protein stock must we add to get this?

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

Explanation / Answer

Your responses are correct according to my knowledge.

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