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How do you calculate the molecular weight and extinction coefficients for phosphorothioated oligos?

The molecular weigh can be calculated by using the following formula:

Anhydrous MW = ‡" ( Individual Base MW ) - 63.980 + 2.016 + (16 x number of phosphorothioate bonds)

The last portion of the formula is needed to correct for the increase in weight when sulfur is substituted for one of the oxygens in the phosphorothioate bonds. The weights of each of the bases are as follows:

dA - 313.209
dC - 289.184
dG - 329.208
dT - 304.196
dU - 290.169
dI - 314.194

Here is an example of this calculation:

Example sequence: 5'- A*T*G* TAA TGT TTG GTC* C*G*C -3' nA=3, nG=5, nC=3, nT=7, nBonds=6;

Using the formula above, the molecular weight of this oligo is (313.209 x nA) + (329.208 x nG) + (289.184 x nC) + (304.196 x nT) - 63.980 + 2.016 + (16 x nBonds) = 5536.6 g/mole

The extinction coefficient for a phosphorothioate oligo would be the same as that of an unmodified oligo.

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