Can I compare selective pressure between coding and non-coding sequences?
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6.1 years ago

I have two sequences: one is coding (A) and one is non-coding (B). I calculated selective pressure for A using Nei-Gojobori method and Wong-Nielsen for B, respectively. Let's assume I get ζ=0.3 for B and ω=0.5 for A. The question is, can I compare ζ and ω, saying that purifying selection on sequence B stronger than A's? WN method is a modification of NG so both parameters seems equal, in my opinion.

sequence SNP selection evolution • 1.4k views
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Why not measure the same thing for each?

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It's actually the same value, but ω can be calculated for coding sequences and ζ for non-coding exclusively.

Let me explain how to get these values.

For ω we calculate all theoretically possible synonymous substitutions (S) and difference in synonymous substitutions for pair of sequences (Sd). Then we'll get pS=Sd/S, which after Jukes-Cantor correction will become dS - characteristic of neutral evolution.

Same thing for non-synonymous substitutions (N - all theoretically possible non-syn substitutions, Sn - difference in non-syn substitutions for pair of sequences, pN=Sn/S, dN is JK-corrected pN value).

After this calculations ω value have been calculated: ω=Sn/Sd

ζ is almost the same, but in this method we considering neutral evolution in non-coding sequences appers with the same (to genes) speed: we calculate number of differences in non-coding sequences (SNC) and divide it to the total lenght of sequence (NC), SNC/NC ratio after correction called dNC.

Finally, ζ value: ζ=dNC/dS, where dS is value of neutral evolution for adjacent gene.

I am not strong in math neither programming, so this problem is a bit tricky to me.