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enzyme dilution help


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#1 claritylight

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Posted 23 March 2005 - 05:33 PM

I have to dilute DNase I 1:2500 in 1X reaction buffer using my 10X reaction buffer.

My 10X reaction buffer contains:
250uL Tris-HCl
50uL MgCl2
50uL CaCl2
4650uL water

What I did was:
(1/10)* (2500) = 250 fold dilution

My DNase weighs 1ug/uL.
250* (1uL) = 250uL DNase
If I add in 2250uL reaction buffer, then I'll have the 250+2250 = 2500 dilution required.

Is this a good way to do it?

Edited by claritylight, 23 March 2005 - 05:55 PM.


#2 jadefalcon

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Posted 24 March 2005 - 01:33 AM

hi!

I'm not quite sure if I understand you correctly, but if you're telling me that you want to dillute 250Ál DNase in 2250Ál 10x reaction buffer, you will dillute your DNaseI 1:10 and your reactionbuffer will be 9x in the final solution.

The way I see this, you should think of the volume of your to be digested DNA solution and calculate like this, for example:

1000Ál DNA-solution
125Ál 10x reaction buffer
125Ál H2O

gives you 1250Ál DNA in 1x DNase buffer

then add 0.5Ál DNaseI for a dillution 1:2500 (= 0.5:1250).

If you would be overly correct, you would only add 124.5Ál H2O, since 0.5Ál DNaseI is added lateron, but this is a very small error which can be neglected.

mike
--- He who finds typos may keep them! ---

#3 claritylight

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Posted 24 March 2005 - 04:11 AM

Hi,
No, I have 1ug/uL DNase I and want to dilute it 1:2500 in the cold reaction 1X buffer. I only have a 10X stock. (This is where I got the 1/10).

edited:
Basically, I'm trying to find out how much DNase I to use, along with how much 10X reaction buffer to use to get my DNase I to dilute 1:2500 in a 1X reaction buffer.

What I did was (1/10)* 2500 = 250 fold dilution

I think this would give me the dilution factor for 1x buffer. With this dilution factor, I'm a bit uncertain where to go from here. My DNase I weighs 1 ug/uL. I need to somehow dilute this DNase I down 1:2500 in a 1X reaction buffer.

So I have to make a 1X reaction buffer from my 10X reaction buffer, and I think I found the dilution factor for it so far. Then, I have to dilute down the DNase I with this buffer to give me a final 1:2500 dilution. Since 1ug = 1uL, then I did 1uL*250 = 250uL. Or should I divide instead (1uL/250)? Whichever way, my goal is to find how much DNase I to add into the 1X buffer.

Finally, how much of the reaction buffer do I need to add in? Do I subtract (amt of DNase I) from 2500?

I don't completely understand what your calculations are and I'd like to be exact. I hope this clears up what I'm asking.

Edited by claritylight, 24 March 2005 - 05:46 AM.


#4 claritylight

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Posted 24 March 2005 - 04:11 PM

Nevermind, I figured it out.

#5 diana

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Posted 25 March 2005 - 12:59 PM

I have to dilute DNase I 1:2500 in 1X reaction buffer using my 10X reaction buffer.

My 10X reaction buffer contains:
250uL Tris-HCl
50uL MgCl2
50uL CaCl2
4650uL water

What I did was:
(1/10)* (2500) = 250 fold dilution

My DNase weighs 1ug/uL.
250* (1uL) = 250uL DNase
If I add in 2250uL reaction buffer, then I'll have the 250+2250 = 2500 dilution required.

Is this a good way to do it?

<{POST_SNAPBACK}>


I think you'll have to know your DNAse I final concentration in your 1X reaction buffer. If you want 2500ul final volume, you add 1ul DNAseI to 2499ul 1x reaction buffer. In this case you'll have to first dilute your 10x reaction buffer to 1x to get to a volume of 2500ul (250ul 10x buffer + 2250ul water). Thus, your DNAse I final concentration will become 1ug/2500ul or 0.0004ug/ul which is very dilute. From your calculations, your DNAse I final concentration goes from 1ug/ul to 0.1ug/ul (10 fold).

I hope this last one (1:10) is the dilution you wanted for your DNAse.




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