• Create Account

# Probe Dilution Problem = 100uM to 5ng/ul? + Final Volume for FISH

2 replies to this topic

### #1 Amigo

Amigo

member

• Members
• 4 posts
0
Neutral

Posted 31 January 2011 - 03:10 PM

Hi everyone,

I need to dilute fluorescent probes for FISH. Basically, I need a final concentration of 5 ng/ul. Probes are in dry powder. Probe nmol = 20.02. Probe MW = 6298.00 (no units provided for some reason). I know I can make a 100 uM stock solution by adding 200 ul of TE or water.

Could someone show me how to calculate this? I have never been good at these things and always get confused. Thank you in advance.

What is the final volume? I am asking that to myself too. I have to do FISH staining with these probes. This is the first time I do this and I don't know how much buffer is usually used. My samples are small so I figure 50 to 100 ul would be enough. Is that ok?  Well... I guess I don't know what the maximum amount of buffer I can make with a concentration of 5 ng/ul of probes.

Any help on this would be very much appreciated, thank you

### #2 protolder

protolder

Veteran

• Active Members
• 293 posts
9
Neutral

Posted 31 January 2011 - 11:21 PM

Amigo, on 31 January 2011 - 03:10 PM, said:

Hi everyone,

I need to dilute fluorescent probes for FISH. Basically, I need a final concentration of 5 ng/ul. Probes are in dry powder. Probe nmol = 20.02. Probe MW = 6298.00 (no units provided for some reason). I know I can make a 100 uM stock solution by adding 200 ul of TE or water.

Could someone show me how to calculate this? I have never been good at these things and always get confused. Thank you in advance.

What is the final volume? I am asking that to myself too. I have to do FISH staining with these probes. This is the first time I do this and I don't know how much buffer is usually used. My samples are small so I figure 50 to 100 ul would be enough. Is that ok?  Well... I guess I don't know what the maximum amount of buffer I can make with a concentration of 5 ng/ul of probes.

Any help on this would be very much appreciated, thank you
Hola, I´m going to explain you how I do step by step. Your 100uM solution is well made. So, you have 100umols/liter <>100nanomols /ml which in nanograms are 100x6298= 629800 ng/ml <> 629.8 ug/ml  ; in 200ul (1/5) 125.96 ug . So you have 200ul of 100uM sol having 125.96 ug. So in 1ul => 125.96/200= 0.630ug/ul <>630ng/ul.

630/5= 126 times you have to dilute your 100uM sol. to have a 5ng/ul sol.

125 of water or TE + 1ul of your 100uM and you have the 5ng/ul.

I never have done FISH, so if you think that 125 ul by sample is enought, you already know how do it. Buena suerte

### #3 Amigo

Amigo

member

• Members
• 4 posts
0
Neutral

Posted 01 February 2011 - 04:01 PM

Hola Veteran,

Muchas Gracias! Your explanation is very clear. When I was making the calculations, at one point I also got a dilution of 126 times... but I just didn't think it was right because when I talked to someone about FISH, this person told me I needed a high concentration of probes. This 1/126 (5ng/ul) just seemed to diluted. But, I am just not going to be an independent thinker and I will simply follow the protocol and see what happens.

I'll see if I can find a FISH discussion on this site to see if anyone can give me advise on how much volume to use. That is not on the protocol.

Muchas Gracias nuevamente,

Amigo