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### #1 jiajia1987

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Posted 08 January 2010 - 05:24 AM

Hi everyone,

I have been working in a lab for about 8 months and I am still really bad at calculations and seriously need some guidance! It has been driving me crazy for a day so I would appreciate any help!! So here I go...

I have a stock of 100ng/uL of DNA, and the DNA is about 100bp, so the molecular weight of it would be 66000Da, which can also be written as... 66000g/mole. I am supposed to add 10nM of the DNA into 220uL of quenching buffer. This quenching buffer will eventually be used to break of emulsions in my selection work.

I am unsure as to whether I am right about this. I am going to be very detailed in my working:

1M = 1 mole/L
10nM = 10 X 10^-9 mole/L
I need to times the concentration with the molecular weight: 10 X 10^-9 mole/L X 66000 g/mole = 6.6 X 10-4 g/L, which is also 0.66 ng/uL

If.... V1C1 = V2C2
0.66 ng/uL X 220uL = V2 X 100ng/uL
V2 = 1.452uL rounded off to 1.5uL

So this means I need 1.5uL of 100ng/uL DNA stock to get 10nM in the 220uL quenching buffer.

Am I right? Please help me to check my working as I tend to make a lot of mistakes in it. I asked a few people and they used different methods to solve this for my experiment and as a result, they all had different answers and I am so confused!

### #2 jiajia1987

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Posted 08 January 2010 - 05:28 AM

By the way.. I found this website http://forum.onlinec...ead.php?t=11182 and tried to backcalculate but could make no sense out of it..

### #3 almost a doctor

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Posted 08 January 2010 - 06:01 AM

Calculating it in a slightly different way I got the same number so.... you must be right

Hi everyone,

I have been working in a lab for about 8 months and I am still really bad at calculations and seriously need some guidance! It has been driving me crazy for a day so I would appreciate any help!! So here I go...

I have a stock of 100ng/uL of DNA, and the DNA is about 100bp, so the molecular weight of it would be 66000Da, which can also be written as... 66000g/mole. I am supposed to add 10nM of the DNA into 220uL of quenching buffer. This quenching buffer will eventually be used to break of emulsions in my selection work.

I am unsure as to whether I am right about this. I am going to be very detailed in my working:

1M = 1 mole/L
10nM = 10 X 10^-9 mole/L
I need to times the concentration with the molecular weight: 10 X 10^-9 mole/L X 66000 g/mole = 6.6 X 10-4 g/L, which is also 0.66 ng/uL

If.... V1C1 = V2C2
0.66 ng/uL X 220uL = V2 X 100ng/uL
V2 = 1.452uL rounded off to 1.5uL

So this means I need 1.5uL of 100ng/uL DNA stock to get 10nM in the 220uL quenching buffer.

Am I right? Please help me to check my working as I tend to make a lot of mistakes in it. I asked a few people and they used different methods to solve this for my experiment and as a result, they all had different answers and I am so confused!

### #4 jiajia1987

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Posted 08 January 2010 - 06:26 AM

Calculating it in a slightly different way I got the same number so.... you must be right

Hi everyone,

I have been working in a lab for about 8 months and I am still really bad at calculations and seriously need some guidance! It has been driving me crazy for a day so I would appreciate any help!! So here I go...

I have a stock of 100ng/uL of DNA, and the DNA is about 100bp, so the molecular weight of it would be 66000Da, which can also be written as... 66000g/mole. I am supposed to add 10nM of the DNA into 220uL of quenching buffer. This quenching buffer will eventually be used to break of emulsions in my selection work.

I am unsure as to whether I am right about this. I am going to be very detailed in my working:

1M = 1 mole/L
10nM = 10 X 10^-9 mole/L
I need to times the concentration with the molecular weight: 10 X 10^-9 mole/L X 66000 g/mole = 6.6 X 10-4 g/L, which is also 0.66 ng/uL

If.... V1C1 = V2C2
0.66 ng/uL X 220uL = V2 X 100ng/uL
V2 = 1.452uL rounded off to 1.5uL

So this means I need 1.5uL of 100ng/uL DNA stock to get 10nM in the 220uL quenching buffer.

Am I right? Please help me to check my working as I tend to make a lot of mistakes in it. I asked a few people and they used different methods to solve this for my experiment and as a result, they all had different answers and I am so confused!

Hey that's cool! Thanks a lot!!

But I would like to know what method you used, mind teaching me?

### #5 almost a doctor

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Posted 08 January 2010 - 06:51 AM

I did the same, just slightly different.

You need 10nM in 220ul, so from M= moles/V(L) you know you need 2.2x10^-12 moles.

now 2.2x10^-12 moles x 660000 Da = need 1.452x10^-7 grs

you have 100ng in each ul so: 1.452x10-7 / 100x10-9 = 1.452 ul

same thing really

another way will be.

100ng/ul = 1.51515x10^-6M now apply V1C1=V2C2

so V1 = 10x10^-9 x 220ul / 1.51515x10^-6 = 1.452ul

All the same, as long as you know your units and keep things consistent, you'll get to the same numbers

### #6 jiajia1987

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Posted 09 January 2010 - 06:34 AM

Great!! Now I know of a few ways to solve it. I am grateful for the guidance!! Thanks so much!!

I did the same, just slightly different.

You need 10nM in 220ul, so from M= moles/V(L) you know you need 2.2x10^-12 moles.

now 2.2x10^-12 moles x 660000 Da = need 1.452x10^-7 grs

you have 100ng in each ul so: 1.452x10-7 / 100x10-9 = 1.452 ul

same thing really

another way will be.

100ng/ul = 1.51515x10^-6M now apply V1C1=V2C2

so V1 = 10x10^-9 x 220ul / 1.51515x10^-6 = 1.452ul

All the same, as long as you know your units and keep things consistent, you'll get to the same numbers

### #7 HomeBrew

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Posted 09 January 2010 - 09:13 AM

This is correct as long as "I am supposed to add 10nM of the DNA into 220uL of quenching buffer" is correct. In other words, if you're supposed to wind up with 10 nM of DNA in a final volume of 221.5 ul.

With such a small amount added, it's not likely to make a difference in this case (the marigin of error in measuring 220 ul is likely to add as much error), but with larger volumes (or differently worded protocols) you'd need to take the additon into account if the final volume was specified. In other words, if your protocol called for you to combine 10 nm of DNA and the quenching buffer in a final volume of 220 ul, you'd need to add your 1.5 ul to 218.5 ul of the buffer.

### #8 jiajia1987

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Posted 11 January 2010 - 01:22 AM

This is correct as long as "I am supposed to add 10nM of the DNA into 220uL of quenching buffer" is correct. In other words, if you're supposed to wind up with 10 nM of DNA in a final volume of 221.5 ul.

With such a small amount added, it's not likely to make a difference in this case (the marigin of error in measuring 220 ul is likely to add as much error), but with larger volumes (or differently worded protocols) you'd need to take the additon into account if the final volume was specified. In other words, if your protocol called for you to combine 10 nm of DNA and the quenching buffer in a final volume of 220 ul, you'd need to add your 1.5 ul to 218.5 ul of the buffer.

Yes Sir! I get it now!

### #9 almost a doctor

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Posted 11 January 2010 - 01:41 AM

This is correct as long as "I am supposed to add 10nM of the DNA into 220uL of quenching buffer" is correct. In other words, if you're supposed to wind up with 10 nM of DNA in a final volume of 221.5 ul.

With such a small amount added, it's not likely to make a difference in this case (the marigin of error in measuring 220 ul is likely to add as much error), but with larger volumes (or differently worded protocols) you'd need to take the additon into account if the final volume was specified. In other words, if your protocol called for you to combine 10 nm of DNA and the quenching buffer in a final volume of 220 ul, you'd need to add your 1.5 ul to 218.5 ul of the buffer.

Indeed!! I was always considering 220ul final volume, so you have to add 220 - 1.5 = Volume of buffer (218.5ul).

In the case that you HAVE to add 10nM to 220ul of buffer you can still use C1V1=C2V2 to calculate the volume of DNA you have to add (V1) but in this case your
V2 = volumen of buffer (220ul) + volume of DNA needed (V1)

C1V1=C2(220+V1) V1= 220*C2 / (C1-C2)

### #10 jiajia1987

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Posted 12 January 2010 - 12:37 AM

This is correct as long as "I am supposed to add 10nM of the DNA into 220uL of quenching buffer" is correct. In other words, if you're supposed to wind up with 10 nM of DNA in a final volume of 221.5 ul.

With such a small amount added, it's not likely to make a difference in this case (the marigin of error in measuring 220 ul is likely to add as much error), but with larger volumes (or differently worded protocols) you'd need to take the additon into account if the final volume was specified. In other words, if your protocol called for you to combine 10 nm of DNA and the quenching buffer in a final volume of 220 ul, you'd need to add your 1.5 ul to 218.5 ul of the buffer.

Indeed!! I was always considering 220ul final volume, so you have to add 220 - 1.5 = Volume of buffer (218.5ul).

In the case that you HAVE to add 10nM to 220ul of buffer you can still use C1V1=C2V2 to calculate the volume of DNA you have to add (V1) but in this case your
V2 = volumen of buffer (220ul) + volume of DNA needed (V1)

C1V1=C2(220+V1) V1= 220*C2 / (C1-C2)

Okay great! Now, I know why sentence construction matters so much in science. My lecturer was emphasizing on how we should bring across our point once as it could mean different things!

### #11 jajell

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Posted 13 January 2010 - 12:34 PM

http://www.clearspri...af34526e5cf6e3a

This might help out a little!

Jason
Invitrogen web channel lead - have questions on cell culture or drug discovery services? Drop me a PM!

### #12 jiajia1987

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Posted 21 January 2010 - 01:30 AM

http://www.clearspri...af34526e5cf6e3a

This might help out a little!

Jason

This is awesome, thanks!!

### #13 HomeBrew

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Posted 21 January 2010 - 04:29 AM

There's another good calculator here.