P-32 calculations

 

The next step in my calculations is to determine how many nanograms of perfectly labelled sheared DNA would I have to load in the gel to see the size distribution with the phosphoimager.

To calculate this, first I need to know how many dpm in a band I need to see it using a phosphoimager.

 

My methodology is based on exposing a dried agarose gel with radioactive labelled P32-DNA to a Phosphor Screen which then is scanned using a typhoon imager laser scanner. Ideally I need to find detailed information on the limit of detection (LOD) of P32 using a given exposure time and scan resolution. Unfortunately,  information on the internet is not abundant. The manufacturer only provides an LOD of 2 dmp/mm2 for 14C  scanned at 200 µm.

According to this information then if I load a ng of perfectly labelled DNA (3547.5 dmp/ng, see my last post) in a well (4mm * 5mm) I will have a band with 177 dpm/mm2. However, my DNA will not run as a band but as a smear, since it has been sheared. The size of the smear will depend on the migration time of the gel. With a rough eye calculation, I would say that the smear is 4 mm * 20 mm (after migrating it for an hour).  If that is the case, then my smear will be: 3547.5 dpm/ 80 mm2 =  44 dpm / mm2.  Of course this is a rough calculation since the average dpm/ng that I am using was calculated with an average 6Kb fragment size, and smaller fragments with more ends will be hotter and I expect them to be more intense in the scanned gel.

I am not quite convinced that this calculations are right since the  LOD presented by the manufacturers is really vague. Fortunately, I found some useful data on a presentation in figshare.

p32_1

Roy, Christian (2013): Limits of detection for P32 Phosphor Screens. figshare.

http://dx.doi.org/10.6084/m9.figshare.726151

Retrieved 15:52, Oct 06, 2015 (GMT)

This figure shows the amount of radioactivity compared with relative band volume. Each colour represent a different exposure time. The darker grey represent “the sweet spot” or the ideal visibility of the bands according to the author.

Lets start with unit conversions

1nCI = 2220 cpm

100 pCI = 222 cpm

10 pCI = 22 cpm

So if my smear has a 44 dmp / mm2 then it will be visible by exposing it for 8 hours while exposing it for only an hour will give me a barely visible band.  So, according to this figure the minimum cpm that will be visible using a 8-hour exposure is 1 pCI = 2.2 cpm. This LOD is equal to the  reported LOD by the manufacturer  in the first place.

Interestingly, the former postdoc recommended me to expose my radioactively labelled isotopes overnight. This exposure time is congruent with the data showed by the figure. Maybe he optimised the exposure in a similar way.

This data suggests that the minimum amount of perfectly labelled DNA that I need is:

X dpm/ 80 mm2 =  2.2 dpm / mm2

X = 176 dpm

1  ng/ 3547.5 dpm *176 dpm =  0.05 ng = 50 pg of gDNA sheared to a 6Kb average size.

 

 

 

 

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