# Predicting 260/280 of DNA or RNA from sequence?

I asked this in biology, but I think they don't want to deal with the chemistry behind UV-Vis.

My question has to do with the Poly A tails on mRNA. I've been doing in vitro transcription and tailing for a while now and have noticed that the untailed RNA has a 260/280 of about 2.0, like you'd expect, but successful tailing results in ratios from 2.15 to 2.45, with higher ratios correlating to better band shifts on agarose gels, implying a longer tail. I suspect the reason for this is because the adenine has a much higher 260/280 than the other bases, so if you add a lot of adenines to the 3' end of the RNA you add much more 260 absorbance than 280 absorbance.

After seeing this I tried to come up with a mathematical method to estimate tail length based on 260/280, since gels are annoying, while I can get the ratio directly from the nanodrop.

I started by using the 260/280 for each nucleotide, A: 4.50, G: 1.15, C: 1.51, U: 4.00, T: 1.47 source here: 1. I then took the sequence for the Luciferase gene I'm using for mRNA and counted the nucleotides, 449 A, 566 C, 470 G, 374 U. I tried to get the weighted average of the 260/280 for my RNA.

A: 449 x 4.50 = 2020.50
C: 566 x 1.51 =  854.60
G: 470 x 1.15 =  540.50
U: 374 x 4.00 = 1496.00
Sum:            4911.60

Nucleotides: 449 + 566 + 470 + 374 = 1859
Sum / Nucleotides : 4911.60 / 1859 = 2.64


2.64 is clearly not 2.0. Even accounting for the potential effect of pH and ionic strength on RNA 260/280 ( acidic solutions have lower ratios ) I can't explain why the calculated ratio is so high. Even if we assumed an RNA with 25% of each base, the weighted average is still (1.15 + 4.50 + 1.51 + 4.00)/4 = 2.79, even higher than for my real sequence.

Doing the same calculation for DNA, (1.15 + 4.50 + 1.51 + 1.47)/4 = 2.16.

So why don't my calculated ratios for RNA and DNA 260/280 match the "ideal" ratios of 2.0 and 1.8? It probably has something to do with base stacking interactions, but I don't enough to compensate for it.