 # Quantifying DNA

The quantity of the oligonucleotide may be expressed in terms of A260 units, nanomoles, or micrograms. The definitions of these units and conversion factors are listed below.

A260 Units

A260 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 evenly-distributed 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 A260 = 1.0
All T´s = 35 micrograms/ milliliter when A260 = 1.0
All C´s = 39 micrograms/ milliliter when A260 = 1.0
All G´s = 26 micrograms/ milliliter when A260 = 1.0

Nanomoles (nmoles)

The number of nanomoles of an oligonucleotide is calculated from the measured A260 value using the millimolar extinction coefficient (mM E260) of the specific oligonucleotide.

Nanomoles = 1000(Measured A260 value x dilution factor x total volume)/ (E260 of bases + E260 of modifiers)

The mM E260 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 ATT-3′ has a measured A260 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

Micrograms (mg)

The number of micrograms of oligonucleotide is calculated using the molecular weight (MW) and the nanomole value (which was calculated from the A260 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 evenly-distributed base composition with sequence length of 20:

1 A260 unit = 30 micrograms = 5 nanomoles ### QUESTIONS?

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