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I was reading "A genomic code for nucleosome positioning" (by Eran Segal et al). And I am having 2 doubts. enter image description here

The figure(b) in this image from the paper shows the graph of fraction (3-bp moving average) of AA/TA/TT dinucleotides of nucleosome dna sequence they analysed statistically as far as I understand. What is 3-bp moving average here? I also don't understand how they chose 0th position (the so called dyad). Also what does it mean to have oscillations (correlation?) in this graph?

UPDATE : I am adding some supplementary information related to finding the dinucleotide fractions. Still I don't understand why is the fraction found so?

enter image description here

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    $\begingroup$ Your image is a little blurry. Can you upload a higher resolution image? $\endgroup$
    – WYSIWYG
    Commented Mar 12, 2015 at 4:58
  • $\begingroup$ @WYSIWYG : done $\endgroup$
    – dexterdev
    Commented Mar 12, 2015 at 8:24
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    $\begingroup$ Okay I edited the answer. Let me know if that is clear. $\endgroup$
    – WYSIWYG
    Commented Mar 12, 2015 at 9:03

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Dyad is the centre of the DNA that is wrapped around the nucleosome core (It basically is the centre of symmetry of the nucleosome). It is a common practice to set it at 0 thereby making incoming DNA half, negative and outgoing DNA half, positive.

By oscillations the authors mean that there is a periodic repeat of A/T dinucleotide. IMO it is actually not correct to call it oscillation which is mostly used in a time course dynamical sense.

I guess this is what is meant by the 3-nt moving average:

You have conditional dinucleotide probabilities for each position (As shown in the figure). Now you calculate the A/T dinucleotide probability which is: 

 P[A/T] = PAA + PAT + PTT + PTA

Now you find the moving average for 3 steps:

MA(n) = (1/3)×(P[A/T](n) + P[A/T](n-1) + P[A/T](n-2))

where n is the nth position of the DNA
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  • $\begingroup$ The term "oscillations" is my word, not authors'. $\endgroup$
    – dexterdev
    Commented Mar 12, 2015 at 3:22
  • $\begingroup$ I did not understood the 3-bp moving average filter argument, but now I have access to ~70 page supplementary data of the paper. I am updating my question. $\endgroup$
    – dexterdev
    Commented Mar 12, 2015 at 3:49
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    $\begingroup$ @dexterdev I think TA and AT are grouped together as TA (combination of T and A); there is no rationale for not including AT. $\endgroup$
    – WYSIWYG
    Commented Mar 12, 2015 at 9:09
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    $\begingroup$ @dexterdev Unless I read the entire paper I wont know what they have done. I know that AT is different from TA but all I am saying is that when TA is counted then it makes no sense to exclude AT. You need not modify your question because this doubt is just about what have they counted and for that you just need to look at the paper. The answer will still remain the same. $\endgroup$
    – WYSIWYG
    Commented Mar 17, 2015 at 8:22
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    $\begingroup$ @dexterdev they must have mentioned why they are doing so. You just need to read the paper carefully. You must exempt me from that effort :) $\endgroup$
    – WYSIWYG
    Commented Mar 17, 2015 at 8:25

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