Laboratory Calculations Problem Help (2024)

1. If 10x TBE contains 0.89 M Tris-borate, 0.89 M Boric acid, and 0.02 MEDTA, what is the Molar concentration of Tris-borate in 100 ml of 1x TBE?
   • Since molar concentrations are an expression of the molecular weight perliter, the fact that the question asks for 100 ml is irrelevant. Whether youare dealing with 1000 ml, 100 ml or 10 ml, the Molar concentration will be thesame. What is important is the change from 10x to 1x.
   • Since the concentration, 10x is divided by 10 to arrive at a 1xconcentration, then the Molar concentration is also divided by 10.
   • The concentration of Tris-borate in 100 ml of 1x TBE is 0.089 M.
2. How many ml of 0.5 M EDTA are required to make 100 ml of a 0.1 M EDTAsolution?
   • Remember: initial concentration x initial volume = final concentration xfinal volume
   • (0.5 M) (X ml) = (0.1 M) (100 ml)
   • (X ml) = (0.1 M) (100 ml) / (0.5 ml)
   • X ml = 20 ml of 0.5 M EDTA
3. If a solution of DNA is 3 mg/ml, 0.5 mg of DNA would be contained in howmany ml?
   • This can be set up as a simple proportion
   • 3 mg/ 1 ml = 0.5 mg/ X ml
   • (3 mg) (X ml)/ 1 ml = 0.5 mg
   • (3 mg) (X ml) = (0.5 mg) (1 ml)
   • X ml = (0.5 mg) (1 ml)/3 mg
   • X ml = 0.17 ml
   • 0.5 mg of DNA would be contained in 0.17 ml of the 3mg/ml solution
4. If you have a 5x stock solution, how much of this stock would be needed tomake 100 ml of a 1x solution? 10 ml of a 1x solution? 33 ml of a 1x solution?
   • initial concentration x initial volume = final concentration x final volume
   • For 100 ml:
   • (5x) (X ml) = (1x) (100 ml)
   • X ml = (1x) (100 ml)/(5x)
   • X ml = 20 ml of a 5x
   • For 10 ml:
   • (5x) (X ml) = (1x) (10 ml)
   • X ml = (1x) (10 ml)/(5x)
   • X ml = 2 ml of a 5x
   • For 33 ml:
   • (5x) (X ml) = (1x) (33 ml)
   • X ml = (1x) (33 ml)/(5x)
   • X ml = 6.6 ml of a 5x
5. How many grams of agarose are needed to make 100 ml of a 2% agarosesolution?
   • This is a weight to volume equation so remember:
   • g % means the number of grams in 100 ml.
   • 2% is 2 g in 100 ml of diluent.
6. How many ml of the detergent Nonidet P-40 would be used to make 43 ml of a7 % detergent solution?
   • This is a volume to volume conversion so use:
   • initial concentration x initial volume = final concentration x final volume
   • (100 %) (X ml) = (7 %) (43 ml)
   • X ml = (7 %) (43 ml)/ 100 %
   • X ml = 3.01 ml of Nonidet P-40
7. An extraction buffer for DNA is stored as a 15x stock solution. How muchof this stock solution would be required to make 7.5 ml of a 1x workingsolution? a 0.5x solution?
   • initial concentration x initial volume = final concentration x final volume
   • For a 1x working solution:
   • (15x) (X ml) = (1x) (7.5 ml)
   • X ml = (1x) (7.5 ml)/ 15x
   • X ml = 0.5 ml of the 15x solution
   • For a 0.5x solution:
   • (15x) (X ml) = (0.5x) (7.5 ml)
   • X ml = (0.5x) (7.5 ml)/ 15x
   • X ml = 0.25 ml of the 15x solution
8. Given the following stock solutions: 1 M Tris, 0.5 M EDTA, 4 M NaCl, dothe calculations necessary to make 1 L of a solution containing 100 mM Tris, 100mM EDTA, 250 mM NaCl.
   • The information needed to complete this problem includes calculating thevolume of each of the stocks required then the volume of diluent required toreach the final volume.
   • First, convert the units so you are working in similar units
   • Second, use
   • initial concentration x initial volume = final concentration x final volume
   • to determine each of the volumes of stocks to be used
   • For Tris:
   • (1000 mM ) (X ml) = (100 mM) (1000 ml)
   • X ml = (100 mM) (1000 ml)/ 1000 mM
   • X ml = 100 ml of 1 M Tris
   • For EDTA:
   • (500 mM) (X ml) = (100 mM) (1000 ml)
   • X ml = (100 mM) (1000 ml)/ 500 mM
   • X ml = 200 ml of 0.5 M EDTA
   • For NaCl:
   • (4000 mM) (X ml) = (250 mM) (1000 ml)
   • X ml = (250 mM) (1000 ml)/ 4000 mM
   • X ml = 62.5 ml of 4 M NaCl
   • Before calculating the amount of diluent needed, you must know what is thecombined volume of all of the initial dilutions
   • 100 ml Tris + 200 ml EDTA + 62.5 ml NaCl = 362.5 ml This is the totalamount of initial volumes to be used.
   • Finally use
   • final volume - combined initial volumes = diluent
   • to find the amount of diluent needed to bring the final volume to 1000 ml.
   • 1000 ml - 362.5 ml = 637.5 ml of diluent
9. How many ml of a 1 x 10 cells/mlsuspension should be used to plate out 500 cells/plate? 100cells/plate?
   • All that is being asked in this problem is to determine the volume whichcontains the number of cells needed. A simple proportion will work here.
   • The volume which contains 500 cells would be:
   • 1 x 10 / 1 ml =500 / X ml
   • X ml = (500) (1 ml) / 1 x 10
   • X ml = 0.0005 ml would contain 500 cells
   • The volume which contains 100 cells would be:
   • 1 x 10 / 1 ml =100 / X ml
   • X ml = (100) (1 ml) / 1 x 10
   • X ml = 0.0001 ml would contain 100 cells
10. The following stock solutions are available to make a protein extractionbuffer: 100% Nonidet P-40, 1 M Tris-Cl, and 0.5 M EDTA. What quantity of theoriginal stocks will be needed to make 250 ml of buffer with the following finalconcentrations: 0.5% Nonidet, 150 mM Tris-Cl, and 10 mM EDTA?
    • This should be done in the same manner as Problem number 8.
    • For Nonidet P-40:
    • (100 %) (X ml) = (0.5 %) (250 ml)
    • X ml = (0.5 %) (250 ml)/ 100 %
    • X ml = 1.25 ml of Nonidet P-40
    • For Tris-Cl:
    • (1000 mM) (X ml) = (150 mM) (250 ml)
    • X ml = (150 mM) (250 ml) / 1000 mM
    • X ml = 37.5 ml of Tris-Cl
    • For EDTA:
    • (500 mM) (X ml) = (10 mM) (250 ml)
    • X ml = (10 mM) (250 ml) / 500 mM
    • X ml = 5 ml of
    • 1.25 ml + 37.5 ml + 5 ml = 43.75 mlcombined initial volumes
    • 250 ml - 43.75 ml = 206.25 ml of diluent are needed

Serial Dilution Problems

Thanks are given to the MMG graduate students for suggesting these problems.