QUANTIFYING DNA

The quantity of the oligonucleotide may be expressed in terms of A_{260} units, nanomoles, or micrograms.
The definitions of these units and conversion factors are listed below:


A_{260 }unit is that quantity of material which, when dissolved in 1.0
milliliter of water and read in a spectrophotometer at 260 nanometers in a cuvette with a 1 centimeter path, gives an absorbance of 1.0

An oligonucleotide with an evenlydistributed base composition which has
an absorbance of 1.0 at 260 nanometers has a concentration of approximately 30 microgram/milliliter.
The base composition has a dramatic effect on this approximation, since a sequence of:

All A's = 20 micrograms/milliliter when A_{260} = 1.0
All T's = 35 micrograms/ milliliter when A_{260} = 1.0
All C's = 39 micrograms/ milliliter when A_{260} = 1.0
All G's = 26 micrograms/ milliliter when A_{260} = 1.0



The number of nanomoles of an oligonucleotide is calculated from the measured
A_{260} value
using the millimolar extinction coefficient (mM E_{260}) of the specific oligonucleotide.


Nanomoles = 1000(Measured A_{260} value x dilution factor x total volume)/
(E_{260} of bases + E_{260} of modifiers)


The mM E_{260} values MIDLAND uses are:

A = 15.1  G = 11.7  T = 8.7  C = 7.4  I = 6.8


For example, a 2.0 milliliter sample with the sequence 5'ATC GAT CGG ATT3' has a
measured A_{260} value of 0.329 when measured at a 1:11 dilution. The total number of
nanomoles in the sample is calculated as follows:


Nanomoles = 1000 x 0.329 x 11 x 2.0 / 3(15.1) + 3(11.7) + 4(8.7) + 2(7.4) = 55.7 nanomoles



The number of micrograms of oligonucleotide is calculated using the molecular weight (MW)
and the nanomole value (which was calculated from the A_{260} measurement).

The molecular weight of the free acid form of DNA can be calculated as follows:

Enter the number of A's__________ x 313.22 = __________
Enter the number of C's__________ x 289.18 = __________
Enter the number of G's__________ x 329.22 = __________
Enter the number of T's__________ x 304.21 = __________
Enter the mass contribution of any modifications = __________
Subtract (there is one less phosphate than nucleoside)  61.98
The molecular weight of the oligonucleotide is = __________
Micrograms (ug) DNA = MW x nanomoles / 1000


As a general rule of thumb, for an oligonucleotide with
an evenlydistributed base composition with sequence length of 20:

1 A_{260} unit = 30 micrograms = 5 nanomoles

