Question

Calculations Needed: ANSWER ALL QUESTIONS IN ONE SINGLE POST PLEASE!!!!

1. Prepare 10ml of a 10% solution of Triton-x100 using undiluted Triton-100. Hint: Undiluted solution are always 100%

I got 1mL for this question. I am not sure if I am right though.

2. You have 300μl of reaction for a PCR experiment. You want to adjust so that there is 1.25 U of Taq polymerase per 50μl. The taq is supplied as 5U/μl. How much Taq should you add to the reaction mix?

3. You have 100X enzyme inhibitor cocktail. You want to adjust 3ml of lysing buffer to 1X inhibitor cocktail. How much of the inhibitor cocktail do you add to the buffer?

4. You want to adjust 100ml of buffer to 10 mM NaCl. You have 1M stock solution. How much stock do you need to add to the buffer to achieve 10 mM?

Answer 1

Ans. #1. Using           C1V1 (100% triton X) = C2V2 (10% triton X)

            Or, 100% x V1 = 10% x 10.0 mL

            Or, V1 = (10% x 10.0 mL) / 100%

            Hence, V1 = 1.0 mL

Therefore, required volume of 100% triton-X solution = 1.0 mL

Preparation: Mix 1.0 mL of 100% triton-X with 9.0 mL distilled water. The resultant solution in 10.0 mL of 10% triton-X.

#2. Required amount of Taq pol for 300uL PCR mix =

(1.25U / 50 uL PCR mix) x 300 uL PCR mix

= 7.5U

Now,

Required volume of stock Taq pol =

Required amount of Taq pol/ Concentration of stock Taq pol

= 7.5 U / (5U/ uL)

= 1.5 uL

#3. Using C1V1 (100X inhibitor cocktail) = C2V2 (1X lysis buffer)

            Or, 100X x V1 = 1X x 3.0 mL

            Or, V1 = (1X x 3.0 mL) / 100X

            Hence, V1 = 0.03 mL

Therefore, required volume of inhibitor cock tail = 0.03 mL = 30.0 uL

#4. Required [NaCl] in working buffer solution = 10.0 mM = 0.010 M

Using C1V1 (stock solution) = C2V2 (working solution)

            Or, 1M x V1 = 0.010M x 100.0 mL

            Or, V1 = (0.010M x 100.0 mL) / 1M

            Hence, V1 = 1.0 mL

Therefore, required volume of stock NaCl = 1.0 mL