The post-doc asks you prepare him 100ml of 1x TAE buffer to demonstrate your buf

Question

The post-doc asks you prepare him 100ml of 1x TAE buffer to demonstrate your buffer preparation technique. He gave you the recipe of the 1x TAE buffer, and pointed out where to obtain the stock solutions. The largest graduated cylinder you can use in the prepartion process, and store the final buffer product, is 1L.

However, due to time consumption issue and limited supplies, you are only allowed to use max of 2 graduate cylinders of any size you choose for each reagent preparation (2 dilutions max for serial dilution). Any volume lower than 1 ml is impossible to measure on any graduate cylinder. You are not allowed to use any pipettes in the preparation. Knowing that you are struggling with math, the evil post-doc hinted that to make 1mM EDTA solution, you need to mix 0.37g of EDTA with 1000 ml distillated water. Show all steps.

Recipe for 1X TAE Buffer

40 mM Tris pH 7.6

30 mM Acetic Acid

1mM EDTA sln

Stock solutions

10M Tris pH 8.0

Acetic Acid (MW = 60.05)

5% EDTA sln

Thank you in advance!

Explanation / Answer

For preparation of 40mM of Tris

M= w/Mol.wtX 1000/volume

Therefore,

40/1000 = w/121.14 X 1000/100 = 0.48 gm

For preparation of 20 mM Acetic acid:

Molarity of acetic acid in bottle = 17.4 M

So formula applied will be : C1V1= C2V2

Here, C1 = 17.4 M

C2 = 20 mM

V1= unknown

V2= 100 ml

Therefore, 17.4 X V1 = 20/1000 X 100 = 0.114 mL or 114 Microliter

EDTA has been prepared.

Now, 0.48 gram of Tris will be weighed, and mixed with 114 microliter of acetic acid, followed byadding 0.37 gm of EDTA. All will be water will be added. After adjusting pH, water volume will be made up to 1000 mL. This will result in preparation of 1000 mL TAE buffer.

But here, Stock solutions have been given.

So For 40 mMTris,

C1V1 = C2 V2 ( C1= 10M, C2= 40mM, V2= 1000 mL)

10 X V1= 2.02 X 1000 = 2 mL.

For acetic acid: Molarity will be calculated from above formula and weight will be calculated.

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