Units of Measurement:  

 Milli  (10^-3)       Micro (10^-6)    Nano (nl)    (10^-9)   Pico (10^-12)

See also Dilutions    See also cell concentrations

Question (Concentrations: stock solutions): We want to use 50 wells and put 50 ul of solution in each well, so the total volume is (50*50ul)=2500 ul * 1ml/1000=2.5ml. This stock concentration of our protein (GP160) is 100 ug/ml. How much of our 100 ug/ml stock should be add to get a final concentration of 0.75 ug/ml? Answer: 18.75 ul

one method                                                                                        Easier Method (C_I V_I = C₂ V₂)

0.75 ug/ml / 100 ug/ml = x/2.5ml                                             (10 0 ug/ml)(V₁) = (0.75 ug/ml) (2.5 ml)   

= 0.75 ug/ml * 2.5 ml = 100 ug/ml (x ml)                                   (V₁) = (1.875 ug) /(100 ug)

= 1.875 ug = 100 ug/ml (x ml)                                                  (V₁) =  0.01875 ml  or 18.75 ul

= 1.875 ug / 100 ug = x

x = 0.01875   Since this is ml   0.01875ml * 1000 ul/1ml    = 18.75 ul

Question (Concentrations: stock solutions): The stock concentration of your antibody is 0.96 mg/ml and you wish to dilute this stock concentration to an initial concentration of 500 ug/ml in dilution buffer. Prepare 350 ul of this solution.

(C_I V_I = C₂ V₂)     First, get the same units. ( 0.96 mg/ml) (1000 ug/mg) = (960 ug/ml)

(960 ug/ml) V_I =  (500 ug/ml) (350 ul)

V_I = 182.3 ul  of the antibody

The volume of the buffer that must be added to this is 350 - 182.3 = 167.7 ul

Question (Concentrations: stock solutions): The stock concentration of your antibody is 0.96 mg/ml and you wish to dilute this stock concentration to an initial concentration of 250 ug/ml in dilution buffer. Prepare 350 ul of this solution.

(C_I V_I = C₂ V₂)     First, get the same units. ( 0.96 mg/ml) (1000 ug/mg) = (960 ug/ml)

(960 ug/ml) V_I  = (250 ug/ml) (350 ul)

V_I =  91 ul  of the antibody

The volume of the buffer that must be added to this is 350 - 91 = 259 ul

Question (Concentration: stock solutions): A stock concentration of your antibody is 1.0 mg/ml. You want a final concentration to be 250 ug/ml. Make 350 ul of this concentration.

(C_I V_I = C₂ V₂)     First, get the same units. (1.0 mg/ml) (1000 ug/mg) = 1000 ug/ml

(1000 ug/ml) V_I = (250 ug/ml) 350 ul

V_I = 87.5 ul       and  for the buffer we would add 350 - 87.5 = 262.50

Question (Concentration: stock solutions): A stock concentration of your antibody is 0.5 mg/ml. You would like a final concentration of 2 ug/ml in a 1 ml volume. How many ul from the stock solution should you add to 1 ml of buffer?

(C_I V_I = C₂ V₂)     First, get the same units. (0.5 mg/ml) (1000 ug/mg) = 500 ug/ml

500 ug/ml) V_I = (2 ug/ml) 1 ml

V_I = 0.004 ml or 4 ul

Question (Concentration: stock solutions): Your stock solution of IL-4 is labelled "100 ng/ml". You want to use a concentration of 100 pg/ml. How many ul should you add to 1 ml to get the appropriate volume?

(C_I V_I = C₂ V₂)     First, get the same units (100 ng/ml)(1000 pg/ng)=100,000 pg/ml

100,000 pg/ml) V_I = (100 pg/ml) 1 ml

V_I =  .001 ml or 1 ul

Question (Concentration: stock solutions): Your stock concentration is 100 ng/ml. What volume in ul should you add for a 1000 pg/ml?

(C_I V_I = C₂ V₂)     First, get the same units (100 ng/ml)(1000 pg/ng)=100,000 pg/ml

100,000 pg/ml) V_I = (1000 pg/ml) 1 ml

V_I =  .01 ml or 100 ul

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Question (Concentrations: Molarity): On the bottle of your compound you find the number 84.01g. You want to make a 0.1M solution of this compound.

M=#moles/liter solution

84.05 g/L would be the 1M. Since your want a 0.1M you need 1/10 of this of this solution. You could take 1 ml of it and add it to 9 ml water.

Question (concentration: Molarity EGCG): The MW for EGCG is 458.4. How many ug/ml would 100 uM of EGCG be?

1M=458.4 g/L. So 458.4 ug/L = 1 uM. So 45840 ug/L=100 uM. So 45.8 ug/ml = 11 uM.

Question (1a): We have 1 mg peptide (powder) in a tube. How much DMSO would we add in order to obtain a final concentration of 5 mg/ml final concentration?

1mg/x = 5mg/1ml                                                                

1mg/5mg=x/1ml

x=0.2ml  or 200 ul

Question (1c): If we want to use 10 ml of solution for 6 plates, we will need 60 ml of the solution. The concentration of our peptide is 5 mg/ml. We want the final concentration in our 60 ml solution to be 10 mg/ml. What dilution do we need to make?

5 mg/ml * 60 ml = 300 mg        Step1

300mg/(x)ml=30 mg/ml            Step 2

300 mg = 30 mg/ml  (x) ml

x=30 ml

Question (2): We want to make serial dilutions by a factor of 10 for 12 tubes to include a dilution buffer and an antibody (ID6) which has a concentration of 1.59 mg/ml. How much buffer and how much AB should we add if we want a 500 ul total starting volume for the first tube and want a final concentration of 250 ul/ml for the starting tube?

1.59 mg * 1000 ug/1mg  = 1590 ul

1590 ul/(x) ml = 250 ug/ml

(x)=6.36 ml

Since we are using 500 ul we would take 500 ul or 0.5 ml and divide by that 6.36 ml = 0.0786 ml or 78.6 ul

As far as the amount of buffer to add, we would take 500 ul and substract that 78.6 ul = 421.4 ul

Question (3): Same question as (2) above but the antibody (IGG) you want to use has a concentration of 1.0 mg/ml.

1.0 mg/ml * 1000 ug/1 mg = 1000 ug

1000 ug/(x)ml = 250 ug/ml

(x) = 4 ml

Again since we are using 500 ul we would take 500 ul or 0.5 ml and divide it by 4 ml = 0.125 ml or 125 ul

As far as the amount of buffer, we would take the 500 ul and subtract that 125 ul = 375 ul

Question (Dilutions): Prepare a 10 fold dilution of a 50 ug/ml stock solution.

Question (Dilution): Make a 100 ul volume total using a 1:5 dilution.

1/5 of 100 is 20. So you would need to use 20 ul for every 80 ul to make a 1:5 dilution which adds up to 100 ul.

Concentrations (percentages): The solution you need to prepare has 3.0% BSA. How should you prepare it?

When you see % think of g/100 ml

So 3g/100 ml would give you the 3%.