DNA Quantification of PCR Products - (Jan/24/2014 )

I'm interested in sequencing a PCR product.  In order to do so, I'm told I will need a total of 550ng DNA, at a concentration of ~10 ng/uL.

 

Product length: 550 bp

Product was present after PCR, confirmed by gel electrophoresis in agarose.

Purified PCR reaction with Qiagen QIAquick PCR Purification Kit

DNA concentration post-purification - using nano-drop: 0.007 ug/uL

 

This seemed low to me, so I decided to calculate the theortical yield.  Please check my math.

 

Number of copies: 2^30 (30 cycles)

Number of bp in solution post PCR: (2^30)*(550bp) = 5.97x10^11 bp

 

Average weight of 1bp = 650 Da.

 

(5.97x10^11)*(650Da)= 3.8805x10^14 Daltons DNA in solution

 

1 Da = 1.6605x10^-24g

 

~0.64436 ng DNA in 40uL = 0.016109 ng/uL

 

I think I must be making an error - this is even lower than what was calculated by the nanodrop.

 

 

Hi There!

 

To address your main concern, which seems to be, do I have enough DNA to sequence, I would say yes.  Another way to quantitate DNA is to look at the gel you ran.  You have to have at least 20ng of DNA to visualize a band of DNA.  You can "eyeball" the intensity of the band and compare it to the intensity of the size standards you also ran (right?).  Usually the size standards will tell you something like "if you load 5uL of size standard X, there will be 100ng of DNA in the 1100bp band".

 

Most commercially available sequencing services say that you need 30 ng of DNA in an 8uL volume (for your less than 1000bp PCR product), but you can usually fudge this.  Contact the commercial service your lab uses, to see what they recommend.

 

So go ahead and try sending your product off for sampling.  You did clean it up with a PCR product purification kit, right?  If not, do so, then re-quantitate it!

 

DeeAnn Visk

I noticed some errors with my calculations.

 

My previous calculations assumed I started with only one copy of the genome, which wasn't the case.  I had 50ng of DNA in the PCR reaction, which translates to around 28 copies of the genome (dog).  That means my equation for calculating strands would look like:

 

28 x 2^n , where n is the round of replication.

 

Therefore, after 30 cycles there would be ~3.0065x10^10 copies.

 

3.0065x10^10 strands x  (550bp / 1 strand) x (1 mole / 6.022 x 10^23 bp) x (650g / 1 mole) x ( 10^9 ng / 1 g) = 18.2ng DNA.

 

Final DNA concentration = Original Concentration + Geneterated DNA

68.2 ng = 50ng + 18.2 ng

 

That's still nowhere near the 280ng that the spectrophotometer calculated but it's getting closer.

 

Also, DeeAnn, I'm not familiar with the 20ng of DNA to visualize concept.  I loaded 5uL of non-purified PCR product.  The PCR reaction was a 30uL volume.

 

(30 uL / 5 uL) * 20ng = 120 ng DNA - at the minimum.

 

How does changing contrast, brightness, etc. affect this fact?  I can freely manipulate these parameters on our camera.

 

Jake