How would set up a serial dilution to cover a concentration range from approximately 21 µM...

How would set up a serial dilution to cover a concentration range from approximately 21 µM to 120 µM of DHF in anassay (in the well) having a total volume of 200 µL of which 30 µL is DHF solution? You are provided a 800 µM DHFstock solution. You need to make a set of DHF dilutions in Eppendorf tubes first. From these you will take 30 µL eachand dilute into total volume of 200 µL in the wells

can you plase give me the ansewer step by step please

Expert Solution

Following is the general idea to solve the problem :

Stock solution ----> Intermediate solutions in eppendorf tubes-----> Final solutions in wells

You are given a stock solution concentration of 800 uM.

Lets assume that the eppendorf tube solutions will contain solutions of volume 50 uL, out of which we will be taking out 30 uL to prepare final solutions in wells.

This 30 uL is diluted to 200 uL, so the concentration of solution in wells is 3/20 times that of solution in eppendorf tubes.

Since we want a final concentration range of 21 uM to 120 uM, so

concentration range in eppendorf tubes is : (21*20/3)uM to (120*20/3)uM , which is equal to : 140 uM to 800 uM

Lets assume we make 7 eppendorf tubes in this range, with concentrations as 140,250,360,470,580,690,800 (unit is uM).

I am showing an example to produce one such tube, you can prepare the rest by the same calculation method as below.

Let's assume we want to make a solution of conc. 360 uM in the eppendorf tube with volume 50 uL.

Stock concentration is 800 uM. In this case you will take (360/800)*50 = 22.5 uL of stock and dilute it to 50 uL. The final concentration will be 360 uM.

queen_honey_blossom answered 1 year ago

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