You have a DNA stork solution with a concentration of 0.75 mg/ml. Describe how y

Question

You have a DNA stork solution with a concentration of 0.75 mg/ml. Describe how yon would use this stock solution to prepare 8.0 ml of DNA solution with an A_200 = 0.39. Again, write your answer as clear and concise numbered steps. A DNA stock with an A_260= 1.13 is diluted 5 fold. What is the concentration of this dilution in mug/ml and ng/ml? A protein stock solution (protein mol weight = 29.000 Da, of 4.5 mg/ml is diluted threefold at each total of 6 steps. What is the -clarity of the protein in the 5 h tube?

Explanation / Answer

1. First, you need to take into account the Beer's law and use the relation between Absorbance and concentration, which is:

A = E*b*c

Where:

E = molar absorbity (L/cm mol)

b = Lenght of the cell (cm)

c = Concentration (mol/L)

However, in this case, the beer's law change because we have concentrations in ppm, so the above expression changes as:

A = a*b*c

a = absorvity (L / g cm)

So, with the original concentration (in ppm) we change that to g/L and then, we can do 1 of 2 things:

1. Measure the absorbance of the original concentration, and then, solve for a. With this value, we can know what would be the concentration of the stock solution diluted to 8 mL, and then, we can actually know the moles dissolved in the stock solution.

2. We can assume a stock volume (say 100 mL), and assume we take a volume of solution that is diluted to 8 mL. Then, we can calculate the "a", and we can know what do we have to do.

Basing on the information provided, I'll use the second option, and I'll assume we have a 100 mL of stock solution and we are taking 1 mL of this stock solution that it will be diluted to 8 mL, the concentration of the stock would be:

c = 0.75 mg/mL * (1/8) = 0.0938 mg/mL

a = A / bc Assuming b = 1 cm

a = 0.39 / 1 * (0.0938)

a = 4.16 L / g cm

With this value, we can do part 4.

4. Assuming the same absorvity from part 3. we calculate the concentration:

c = A / ab

c = 1.13 / 4.16 * 1

c = 0.2716 g/L

If this is diluted 5 fold, is something like 1 mL to 5 mL so:

c = 0.2716 * (1/5) = 0.0543 mg/mL * 1000 ug/mg = 54.3 ug/mL * 1000 ng/ug = 54300 ng/mL

5. We convert the stock of 4.5 mg/mL to mol:

4.5 mg/mL or 4.5 g/L * 1 mol / 29000 g = 1.55x10-4 mol/L

if we are dilluting three fold at each step, it means, that we take 1 mL of stock solution, and we dillute to 3 mL for tube 1, and on and on to tube 5, so:

c5 = 1.55x10-4 * (1/3)5 = 6.38x10-7 mol/L

Now why do I elevate to 5? because we are making a dilution once and then again above the diluted solution. It's not the same thing dilute the original 15 times (which would be 3*5) that 3 times every moment, so we have to elevate.

Hope this helps

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