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Ok, the easiest way to do this is v=N/C (volume=number/concentration) so for sample 1 you want 50 ng (volume doesn't matter apart from needing less than 60 ul)
v=50ng/2.857
v=17.54ul make up to 60 with water (60-17.54=)... repeat for others.
#149328 do I have enough sample?
Posted
on 31 January 2013 - 11:58 AM
#100607 final concentration
Posted
on 14 February 2011 - 01:40 PM
if you mix 1ml of 20uM forward primer with 1ml of 20uM reverse primer then you will have 2ml of 20uM primer mix but 10uM each primer.
#100420 final concentration
Posted
on 12 February 2011 - 06:27 AM
claritylight, on 12 February 2011 - 06:18 AM, said:
Thank you pito. Will this be same or different for primers, if I have forward and reverse primer at 20uM each, and I mix them together, they will be at 20uM final as well?
1 M means: 1 mol per liter...
Think like with the stones..
1 stones in each liter... I have have 2 bottle like that and I put them into 1 bigger bottle, how many stones (moles) will I have then?
1 liter of 20µM + 1 liter of 20µM, how many µM will that be in the 2 liter bottle? And if you recalculate it to 1 liter after putting them together?
Always remember: concentrations dont change as long as you dont dilute it.... So if you add X times the same concentration together, the concentration stays the same...
Look it like this: if you had a tube containing 20µl with a 10ng/µl solution, how much ng would you have if you took 5µl or 15 µl ? And how much would this be in concentration...?
It goes the other way around to: adding 5 µl of a 6ng/µl concentration together with 8µl of a 6ng/µl concentration is in the end still a 6ng/µl concentration (do the math... check it).
It doenst matter wether we are talking about primers, stones, sugar, DNA, or whatever it is you have there .....
#103640 please correct
Posted
on 15 March 2011 - 12:38 AM
claritylight, on 14 March 2011 - 02:35 PM, said:
Since we fixed my number mistake now, the given is 1ug/ul of stock DNA and I have 10ul so I have 10ug DNA amount right? I was just given the numbers of 10ug and 10ul
Just tell me if the below calculations are correct and I will figure out what I want on my own and we can all move on:
If I put 10ul of the 1ug/ul DNA into a tube, add 190ul water, this = 200ul total volume, and the DNA in this tube
now has concentration of (10ul/200ul)*1000ng/ul = 50ng/ul right?
What I want to know is how much amount of DNA is in the tube. I did 50ng/ul*200ul = 10000ng = 10ug, so there's 10ug of DNA in the tube. Is that right?
Just tell me if the below calculations are correct and I will figure out what I want on my own and we can all move on:
If I put 10ul of the 1ug/ul DNA into a tube, add 190ul water, this = 200ul total volume, and the DNA in this tube
now has concentration of (10ul/200ul)*1000ng/ul = 50ng/ul right?
What I want to know is how much amount of DNA is in the tube. I did 50ng/ul*200ul = 10000ng = 10ug, so there's 10ug of DNA in the tube. Is that right?
Ok, now I get it, see: telling you have 10µl of a 1µg/µl stock is not the same as just saying: 10µl of 10ng...
10µl of 1µg/µl in a volume of 190µl means you have in total 10µg in 200µl thus 10µg/200µl = 1/20µl = 1000ng/20µl = 100ng/2µl = 50ng/µl.
this is what homebrew allready did.
So yes, if you have 50ng/µl and you have 200µl , yeah, you have 50*200 ng DNA= 10000ng = 10µg...
But I dont understand why you do this second calculation!
You started by saying you would add 10µl of a 1µg/µl solution in that tube... So you knew from the start it was 10µg ..
(10µl*1ng/µl = 10ng .. and no matter how much water you add, even its 19999999liter, it will still stay 10ng of DNA in the end... Its the concentration that changes, not the amount of DNA itself)
You calculate B , using A to calculate A in the end ...
But still: I have the feeling you dont understand the formula or that you dont know what you are doing.
#94688 pooling different pcr cycles?
Posted
on 12 December 2010 - 02:40 PM
well it depends if the plateau phase has been achieve after that many cycles.
If both PCR reaction mixes have reached the plateau phase than mixing the two could not change the concentration. Since the concentration of PCR product would be identical.
However if the plateau phase has not been achieved, then the reaction mix with the fewer number of cycles would logically has less product, and thus a lower concentration. Pooling the two would increase the amount of PCR product present but decrease the concentration.
I would add 5 cycle even 10 cycles is rather few. I use 25 cycle to amplify my PCR product.
Aside from not being as efficient as possible, I don't see a problem pooling the two.
If both PCR reaction mixes have reached the plateau phase than mixing the two could not change the concentration. Since the concentration of PCR product would be identical.
However if the plateau phase has not been achieved, then the reaction mix with the fewer number of cycles would logically has less product, and thus a lower concentration. Pooling the two would increase the amount of PCR product present but decrease the concentration.
I would add 5 cycle even 10 cycles is rather few. I use 25 cycle to amplify my PCR product.
Aside from not being as efficient as possible, I don't see a problem pooling the two.
