You are given a protein solution with a concentration of 0.25 mg/ml. 1. You need

Question

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

1. You needs 100 ug of protein for an experiment. What volume of the protein solution do you need?

2. Suppose you want to prepare a solution containing 200uh of the protien at a concentration of 1 mg/ml. You will first freeze dry enough protein solution to get 200 ug of the protein. How much solution will you need to freeze dry? How much H2O will be added to the dried protein to get the desired 1 mg/ml concentration?

3. If the protein has a MW =22,000 express its initial concentration in moles/liter, in umoles/liter and umole/uL. If you need 20 u moles for a reaction, what volume do you need?

4. If you want 1 ml of 10 ug/ml solution. How much water and protein stock do you add to get it?

5. If you want 100 ul of 0.1 ug/ul solution. How much water and protein stock do you add to get it?

Explanation / Answer

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

1. You needs 100 ug of protein for an experiment. What volume of the protein solution do you need?

Solution :- Initial concentration = 0.25 mg / ml

0.25 mg * 1000 ug / 1 mg = 250 ug .

Therefore 1 ml solution contains 250 ug of the protein

We want 100 ug protein for the reaction therefore lets calculate the volume of the 0.25 mg / ml solution of the protein needed for the reaction

1 ml = 250 ug

? ml = 100 ug

100 ug * 1 ml / 250 ug = 0.4 ml

Therefore we need 0.4 ml of the 0.25 mg/ml solution.

2. Suppose you want to prepare a solution containing 200uh of the protien at a concentration of 1 mg/ml. You will first freeze dry enough protein solution to get 200 ug of the protein. How much solution will you need to freeze dry? How much H2O will be added to the dried protein to get the desired 1 mg/ml concentration?

Solution :- lets calculate the volume of the solution needed to dry freeze to get 200 ug protein

250 ug = 1 ml

200 ug = ? ml

200 ug * 1 ml / 250 ug = 0.8 ml

Therefore 0.8 ml of the original solution needed to be dry freeze

Now we want concentration 1 mg / ml so lets calculate the volume of water needed to be added

200 ug * 1 mg / 1000 ug = 0.2 mg

We want 1 mg/ ml

So

0.2 mg / 0.2 ml = 1 mg/ ml

So we need to add 0.2 ml water

3. If the protein has a MW =22,000 express its initial concentration in moles/liter, in umoles/liter and umole/uL. If you need 20 u moles for a reaction, what volume do you need?

Solution :-

Lets first calculate moles of protein

Moles = mass / molar mass

           =(0.25 mg *1 g/1000 mg) * (1 mol / 22000 g per mol)

           = 1.14*10^-8 mol

Therefore concentration in moles per liter is as follows

Mol/ L = 1.14*10^-8 mol / 0.001 L = 1.14*10^-5 mol / L

Now lets calculate concentration in umol/L

(1.14*10^-8 mol per L * 1*10^6 umol / 1 mol )= 11.36 umol/ L

Now lets calculate umol per uL

11.36 umol per L * 1 L / 1*10^6 uL = 1.14*10^-5 umol / uL

4. If you want 1 ml of 10 ug/ml solution. How much water and protein stock do you add to get it?

Solution :-

We have 250 ug / ml solution

So to get 10 ug we have to find the volume of the solution needed to freeze

10 ug * 1 ml / 250 ug = 0.04 ml protein stock

So use 0.04 ml of original solution and dilute to 1 ml

So volume of water needed = 1 ml – 0.04 ml = 0.96 ml

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