Jump to content

  • Log in with Facebook Log in with Twitter Log in with Windows Live Log In with Google      Sign In   
  • Create Account

Submit your paper to J Biol Methods today!
Photo
- - - - -

Retinoic acid


  • Please log in to reply
5 replies to this topic

#1 Doraid

Doraid

    member

  • Active Members
  • Pip
  • 10 posts
0
Neutral

Posted 02 December 2014 - 04:56 PM

Hello,

Kind of embarrassing post. I'm trying to make a 10uM retinoic acid (from Sigma) http://www.sigmaaldr.../1/r2625pis.pdf

But suck on the calculations. It says "A stock solution of 0.01 M (3 mg/ml) RA in absolute ethanol was stored at -70°C for up to two weeks." I dissolve 3mg of R.A. in 1 mL of 200 proof ethanol, and then take 5 uL of that mixture and add it to 5mL of media. Are my steps correct?



#2 bob1

bob1

    Thelymitra pulchella

  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 5,831 posts
415
Excellent

Posted 03 December 2014 - 12:19 AM

Critical information missing: what final concentration are you aiming for?

 

Assuming that 3 mg/ml makes 0.01 mol/l (or 10 mmol/l) then adding 5 ul to 5 ml (1:1000 dilution) will give you 0.01 mmol/l.



#3 phage434

phage434

    Veteran

  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 2,504 posts
253
Excellent

Posted 03 December 2014 - 06:22 AM

What "calculation" is necessary? Your stock solution wants to be at 3 mg/ml.  You take 3 mg, and add it to 1 ml, and (ta-da), you have 3 mg/ml.



#4 Doraid

Doraid

    member

  • Active Members
  • Pip
  • 10 posts
0
Neutral

Posted 03 December 2014 - 10:01 AM

Critical information missing: what final concentration are you aiming for?

 

Assuming that 3 mg/ml makes 0.01 mol/l (or 10 mmol/l) then adding 5 ul to 5 ml (1:1000 dilution) will give you 0.01 mmol/l.

I want to create RPMI media with the following concentrations: 10uM, 15uM and 20uM from the 3 mg or R.A. dissolved in 1mL of 200 proof ethanol.



#5 bob1

bob1

    Thelymitra pulchella

  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 5,831 posts
415
Excellent

Posted 03 December 2014 - 11:31 AM

Ok, that makes it a bit easier.

 

Again assuming that 3 mg/ml is equivalent to 0.01 mol/l - you should check this, use n=mass/molar mass and c=n/v!

 

0.01 mol/l = 10 mmol/l or 10,000 umol/l

 

C1V1=C2V2

10000 umol/l x V1 = 20 umol/l x 5 ml

or 10000  x V1 = 100

Solve for V1

V1 =100/10000 =??? ml (note that this volume is in ml, convert to ul!)

 

repeat for the rest.



#6 Doraid

Doraid

    member

  • Active Members
  • Pip
  • 10 posts
0
Neutral

Posted 03 December 2014 - 01:05 PM

Ok, that makes it a bit easier.

 

Again assuming that 3 mg/ml is equivalent to 0.01 mol/l - you should check this, use n=mass/molar mass and c=n/v!

 

0.01 mol/l = 10 mmol/l or 10,000 umol/l

 

C1V1=C2V2

10000 umol/l x V1 = 20 umol/l x 5 ml

or 10000  x V1 = 100

Solve for V1

V1 =100/10000 =??? ml (note that this volume is in ml, convert to ul!)

 

repeat for the rest.

Thanks. I was going about it the wrong way.






Home - About - Terms of Service - Privacy - Contact Us

©1999-2013 Protocol Online, All rights reserved.