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help on Probe dilution for FISH


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#1 harry348

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Posted 15 July 2011 - 05:29 AM

Hello,

I need to dilute fluorescent probes for FISH.I need a final concentration of 200ng/ml Probe nmol = 13.8 Probe MW = 17672.00
As what I know is to how to make a 100 uM stock solution by adding 138 ul of TE or water .

i Can see this 244ug=13.8nmol written on the vial when i got from company.

Does after making it to 100uM is it equal to 244ug/ul? and if so how to make it to 200ng/ml.Serial dilution??? or any other.
there was a reply to this post earlier but could not understand.

I thank him for his reply.

Could anyone help me in this.

#2 Lapsang

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Posted 19 July 2011 - 03:28 AM

244ug = 13.8 nmol means that 244ug of your probe is the same as 13.8 nmol of probe.

The reason for this is that 17672 is the mass (in g) of one mole, and you have 13.8 nmol = 0.0000000138 mol.

0.0000000138 x 17672 = ~0.000244g = 244ug

If you have 244ug in total in the tube, and it is dissolved in 138ul of water/TE that you have added, the concentration will be 244ug/138ul = 1.768ug/ul

Sorry, the last time I replied to you I was out by x1000 as I made a stupid mistake!!! Lucky that you asked again.

#3 harry348

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Posted 21 July 2011 - 04:41 AM

244ug = 13.8 nmol means that 244ug of your probe is the same as 13.8 nmol of probe.

The reason for this is that 17672 is the mass (in g) of one mole, and you have 13.8 nmol = 0.0000000138 mol.

0.0000000138 x 17672 = ~0.000244g = 244ug

If you have 244ug in total in the tube, and it is dissolved in 138ul of water/TE that you have added, the concentration will be 244ug/138ul = 1.768ug/ul

Sorry, the last time I replied to you I was out by x1000 as I made a stupid mistake!!! Lucky that you asked again.



Can you tell me how much to dilute to make a 50ul of 200ng/ml of probe??




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