[soph_876]

I am trying to make a calibration curve using a 20mM stock compound in DMSO.
I want to achieve the following concentrations in this order, going down the plate:
1000μM
800μM
600μM
300μM
150μM
75μM
37.5μM
18.75μM
9.375μM
4.6875μM
2.34375μM
0

I am setting this calibration curve on a Janus liquid handler. I have worked out how much compound to use in order to achieve the concentrations (i want to use as little volume of compound as possible as we are not supplied with a large volume). I also want each well to contain 5% DMSO and 20% acetonitrile.
1. aspirate 15μL of compound (20mM) into 285μL of 20% MeCN. This achieves a 1000μM conc. and 5% DMSO concentration.
2. Aspirate 5μL of compound (20mM) into well 2 which contains 120μL of 20% MeCN. This makes a conc. of 800μM and a 4% DMSO content if I am correct...
3. Aspirate 100μL of the 1000μM solution and dispense into 100μL of 5%DMSO;20%MeCN. This will give a concentration of 600μM but I am unsure as to what the overall DMSO content is now?
4. Aspirate 125μL of the 600μM solution and dispense into a well containing 125μL 5%DMSO;20%MeCN. This give a 300μM solution.  Again I am unsure what the % DMSO content would?
5. A 2 fold dilution now occurs as described in step 4 until a conc. of 2.3μM is achieved in the 11th well in the plate.

Please could someone help me work out the % of DMSO that would be in each well? I have been going round in circles for weeks now. TIA.

[chenbeier]

The calculation is as follows:

15µl of 20 mM what is equal 20 mmol/l = 0,02 µmol/µl = 0,3 µmol. This amount is dissolved in 15 µl + 285µl =300 µl. This gives a new concentration of 0,3 µmol/300µl = 0,001 µmol/µl = 1 µmol/ml = 1 mmol/l = 1 mM  = 1000 µM  It is ok!

The Acetonitrile is 20% . The dilution from 285 µl to 300 µl doesnt change much . It is 19%. DMSO is correct. 15 µl in 300 µl  is 5%.

The second gives with 5 µl only  0.1 µmol. Dissolved in 5 µl +120 µl = 125 µl. The result is 0,0008 µmol/µl = 800 µmol/l = 800 µM It is ok.

Acetonitrile 120/125 *20% =  19,2% . DMSO gives 4,16%

The third would give, if you take 100 µl of the 1000 µM and dilute with 100µl solvents a concentration of the half what means 500 µM not 600 µM.
The Acetonitrile would be 19% *20 % divided by 2 gives 19,5% and the DMSO would stay at 5%. Both solution contain 5%.

[soph_876]

Hi- thanks for the reply!

How did you work out the % Acetonitrile and % DMSO? please could you show me your working as I am still unsure.

apologies regarding the last point. I meant to type that 150uL of the 1000uM solution is added to 100uL of diluent, which would give a 600uM concentration.

Mnay thanks