Probe Dilution Problem = 100uM to 5ng/ul? + Final Volume for FISH - Probe dilution and FISH - Advise on volume for FISH hibridization buff (Jan/31/2011 )
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
Amigo on Mon Jan 31 23:10:51 2011 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
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