Hello!

For a serial dilution of factor 1/5, we can use a simple calculation,

C1V1 = C2V2 [Where C1 & V1 = Initial concentration and volumes respectively; C2 & V2 = Final concentration and volumes respectively]

Let your initial concentration be "c units", so the final concentration would be "0.2c units" for 5 fold dilutions. Also, let's consider initial volume be "v unit"

Using the equation,

c units * v uL = 0.2c units * 300uL. [Since you need 300uL for each final volume]

v = 60uL

So, for each step, you take 60uL of initial solution, and add it to 240uL of solvent (300-60 = 240uL), including the stock solution.

To summarise:

Fill every well with 240uL solvent and serially dilute 60uL of the previous solution.

I hope it helps you a bit.

Also, please refer to the materials I found may be interesting for you:

https://www.bio.umass.edu/micro/immunology/elisa/serial.htm

https://www.youtube.com/watch?v=NXirIFfpKdg

Have a nice day.

Subhradeep Haldar thank you so much it was very useful

That way you are performing a 1:5 dilution, but you end with 240 uL in each well.

You must calculate which volume, x, you must add to 300 uL and then remove to maintain a 1:5 dilution, while leaving 300 uL:

x/(300+x) = 1/5

Solving the equation gives you x = 75 uL

This way, you fill all wells with 300 uL solvent/buffer. Then, add 75 uL of stock solution to the first well, homogenize, take 75 uL and add to the next well. Then, repeat as required.