Math

Hi blog, long time no see….

I am about to receive P-32 ATP for DNA uptake experiments and I need to think in some calculations that will be necessary to be able to determine the efficiency of the labeling reaction using sheared genomic DNA as input DNA.

In order to determine the efficiency of the labelling reaction, I need:

  1. To know how hot (cpm/ng) the input fragments will be if all their (5′) ends are labelled? (Expected). I have to specify that the input I will be end-labelled T4 polynucleotide kinase. This enzyme replaces the phosphate at 5′ ends of fragments with the gamma phosphate from P32-ATP.

2.  To know how hot (cpm/ng) the input labelled fragments are?  (Observed).

 

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With these two pieces of information, I will be able to calculate the number of fragment ends that were actually labelled.

 

To calculate how hot (cpm/ng) the input fragments will be if all their (5′) ends are labelled, I first need to calculate the number of available ends to be labelled based of genomic fragments of 86-028NP Haemophilus influenzae sheared to average genomic size of 6kb and 0.25kb.

 

 

 

 

genomic size of  86-028NP Haemophilus influenzae = 1.91Mb

Molar mass of a dsDNA fragment = (# of bp) x (649 g/mol/bp)

1 mol = 6,022e+23 (avogadro number)

1.91e+6 bp * 649g/mol/bp = 1.2e+9 g/mol 

1.2e+9 g/mol =  1.2e+9 g / 6,022e+23 genomes  = 1.99e-15 g/ genome

1.91e+6 bp / genome *  1 fragment / 6000 bp (mean sheared fragment size) = 318 fragments/genome

1.99e-15g/genome * 1 ng/ 1e-9 g =  1.99e-6 ng/genome

1 genome/1.99e-6 ng * 1 ng = 5e+5 genomes/ng

5e+5 genomes/ng * 318 fragments/genome = 1.59e+8 fragments/ng 

1.59e+8 fragments/ng * 2 = 3.18e+8 labelled ends/ng

Now, I also need to calculate the amount of P32-ATP molecules that I get in a 25 µL vial

The P-32 ATP vial that I ordered had a unit size of 250 µCi or 9.25 MBq (see conversion )

specific activity is = 3000Ci /mmol

Concentration = 10Ci/L

MW of P32 – ATP = 508.18 g/mol

Molarity = 10Ci/L * mmol/3000Ci = 0.0033 mmol = 3.3 µM = 3.3e-6 M

volume = 25 µl

corrected (30/09/2015)

Avogadro-Number-CalcsII-1

25 µl * 1 L/1e+6 µl  * 3.3e-6 moles/L = 8.25e-11 moles

6.022e+23 * / 8.25e+11 moles = 4.97e+13 molecules P32-ATP

 

4.97e+13/25 µl = 1.988e+12 molecules P32-ATP/µl

Now I need to calculate the expected number of CPM/ µl from P32-ATP and the expected number of CPM/ µl from input DNA if all ends are labelled.

For this calculation, I need to convert from MBq to cpm. Unfortunately, this conversion is not straight forward since it depends on the efficiency of the equipment. To simplify calculations, I will assume that 1 dpm = 1cpm.

1 Bq = 60 dpm

1 vial P32-ATP has 9.25 MBq

9.25 MBq= 9.25e+6 Bq x 60 dpm = 5.55e+8 dpm

5.55e+8 dpm / 25 µl = 2.2e+7 dpm/µl

Is this calculation accurate? Is my assumption of 1dpm = 1cpm right?

OK lets see, I will check the notebook of a former post-doc from the lab to check the cpm/µl of fresh P32-dATP bought by him.

This former post-doc obtained 6.4e+8 cpm = 2.6e+7 cpm/ µl from a fresh P32-dATP batch (Feb 2013, experiment 49-6).

Concentration of radioactive label obtained in this experiment (2.6e+7 cpm/µl) is very similar to my calculation (2.2e+7 dpm/µl).

Finally I will calculate the expected number of CPM/ µl from input DNA if all ends are labelled, assuming the final input DNA concentration (input DNA in 20 µL end-labelled reaction) is 20 ng/µL.

3.18e+08 labelled ends/ng    

3.18e+8 molecule ends/ng  * 5.55e+8 dpm/1 P-32 ATP vial * 1 P-32 ATP vial/4.97e+13 molecules P32-ATP  = 3547.5 dpm/ ng of input DNA.

To check I will do the same calculation using the cpm/µL of P-32 dATP estimated from the former post-doc essay.

3.18e+8 labelled ends/ng  * 6.5e+08 cpm/P-32 ATP vial * 1P-32 ATP vial/4.97e+13 molecules P32-ATP = 4154 cpm/ ng of input DNA

 

Conclusions:

In summary, I have to say that for a 6kb average sheared genomic fragment, the expected cpm/ng of input DNA if all 5′ ends are labeled is between 3000 – 4000 cpm/ng. 

In progress…

I still have to calculate what is expected to happen with 0.25 kb sheared fragments.

For the 0.25Kb fragments:

1.91e+6 bp / genome *  1 fragment / 250 bp (mean sheared fragment size) =  7640 fragments/genome

1 genome/1.99e-6 ng * 1 ng = 5e+5 genomes/ng

5e+5 genomes/ng * 7640 fragments/genome = 3.82e+09 fragments/ng 

3.82e+9 fragments/ng * 2 = 7.64e+9 labelled ends/ng

7.64e+9 molecule ends/ng  * 5.55e+8 dpm/1 P-32 ATP vial * 1 P-32 ATP vial/4.97e+13 molecules P32-ATP  = 85316 dpm/ ng of input

 

 

 

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3 thoughts on “Math

  1. Rosie Redfield

    Using simpler calculations I get the same answer for the first part (ends/ng), but a much lower answer for the second part (cpm/ng).

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    Reply

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