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Topic: The same four proportional weighting factors work for each
00/01/10/11 when 0.25 is subtracted from each !!!

Replies: 506   Last Post: Nov 20, 2012 9:21 PM

 Messages: [ Previous | Next ]
 Ray Koopman Posts: 3,383 Registered: 12/7/04
Re: The relationship of UCP's to "u"
Posted: Jun 2, 2012 3:19 AM

On May 30, 6:10 pm, djh <halitsk...@att.net> wrote:
> You asked the right initial question ? how do the UCP?s relate to
> ?u?. That?s the first question that must be answered in order for you
> to understand the first table in my last post.
>
> So, let me begin with the necessary basics, which I?m going to provide
> with no mention of biology whatsoever.
>
> Imagine you work for an eccentric billionaire who for some reason
> known only to him merely wants you to continuously set type for brief
> announcements in the public notice section of a weird kind of
> newspaper he publishes.
>
> All these announcements are similar in that they must each consist of
> a string of letters drawn from the 20-letter alphabet
> {ACDEFGHIKLMNPQRSTVWY}, and for the sake of discussion, we?ll say that
> each announcement must be 24 to 124 letters long.
>
> Because of his eccentricity, your boss gives you announcements to
> typeset in a weird code that only you are privy to.
>
> In this code of his, each of the twenty letters {ACDEFGHIKLMNPQRSTVWY}
> is DEGENERATELY represented by AT LEAST one of the 64 3-letter words
> that can be formed from the four letter alphabet (t,c,a,g}, except for
> the three words taa,tag, and tga. (These three triples do not
> represent any of the letters {ACDEFGHIKLMNPQRSTVWY} because your boss
> uses them to signify ?end of announcement?, i.e. when you see one of
> them, you know that the announcement has ended.)
>
> looks like this:
>
> ttt F tct S tat Y tgt C
> ttc F tcc S tac Y tgc C
> tta L tca S taa - tga -
> ttg L tcg S tag - tgg W
>
> ctt L cct P cat H cgt R
> ctc L ccc P cac H cgc R
> cta L cca P caa Q cga R
> ctg L ccg P cag Q cgg R
>
> att I act T aat N agt S
> atc I acc T aac N agc S
> ata I aca T aaa K aga R
> atg M acg T aag K agg R
>
> gtt V gct A gat D ggg G
> gtc V gcc A gac D ggc G
> gta V gca A gaa E gga G
> gtg V gcg A gag E ggg G
>
> This code is ?degenerate? because with the exception of the letters W
> and M (which are each only encoded by one 3-letter word), the
> remaining 18 letters of the alphabet {ACDEFGHIKLMNPQRSTVWY} are
> encoded by 2, 4, or 6 different 3-letter words. For example, the
> letter F is encoded by the two 3-letter words ttt,ttc, the lettter V
> is encoded by the four 3-letter words gcX (X=t/c/a/g)m and the letter
> L is encoded by the six 3-letter words tta, ttg, ctX (X=t/c/a/g).
>
> And at first glance, your boss seems to be providing you with encoded
> announcements in which he?s chosen randomly among the 3-letter words
> that can signify the same letter. For example, sometimes, he?ll
> signify the letter N of an announcement with the code word aat, and
> sometimes with the code word aac, and there doesn?t seem to be any
> rhyme nor reason for the choices he makes.
>
> But, because your work is so boring, you decide to pay attention to
> the choices your boss is making, and you realize that they?re not
> random. In particular, you?ve noticed that in the encoded
> announcements he gives you to typeset, he uses the the two-word
> sequences att ccc, att cca, and att cct to represent the two-letter
> sequence IP far more frequently than he uses the other 9 possible two-
> word sequences that can represent IP according to the code he?s given
> you (see above.) And in fact, the following 63 2-word sequences all
> seem to be over-represented in the encoded announcements he gives you
> to typeset:
>
> ttt ttc|FF
> ttt tta|FL
> ttg ctc|LL
> ttt gcc|FA
> tta aaa|LK
> ctg ctc|LL
> ctt tcc|LS
> ctc acc|LT
> ctc tac|LY
> ctc ggc|LG
> att ccc|IP
> att cca|IP
> att cct|IP
> att gcc|IA
> atg ctc|ML
> gtg ctg|VL
> gtt tcc|VS
> gtc acc|VT
> gtc agc|VS
> gtc ggc|VG
> tcg ctg|SL
> tct cca|SP
> ccg gtg|PV
> ccc ggc|PG
> acg ctg|TL
> acg ccg|TP
> gct tta|AL
> gcg ctg|AL
> gcg gtg|AV
> gcg ccg|AP
> gct cca|AP
> gct aat|AN
> gcc agc|AS
> tat ccg|YP
> cat ccg|HP
> cag cga|QR
> aaa aat|KN
> aaa aag|KK
> gat ttg|DL
> gaa aat|EN
> gaa aaa|EK
> gag cga|ER
> gag cgg|ER
> tgg cga|WR
> tgg cgg|WR
> cgg gtg|RV
> cgg tac|RY
> cgc tgg|RW
> cgg cga|RR
> agt gga|SG
> agg cga|RR
> agg cgg|RR
> aga aga|RR
> ggt ttt|GF
> ggg ctg|GL
> ggg tat|GY
> ggc tac|GY
> ggt tat|GY
> gga aaa|GK
> ggc aag|GK
> ggc tgg|GW
> ggg cgg|GR
> ggc agc|GS
>
> For example, since there are 12 ways to represent the two letter
> sequence IP in the code he?s given you:
>
> att cct
> att ccc
> att cca
> att ccg
> atc cct
> atc ccc
> atc cca
> atc ccg
> ata cct
> ata ccc
> ata cca
> ata ccg
>
> each of the three two-word sequences att ccc, att cca, and att cct
> should occur 1/12 of the time if your boss is choosing randomly,
> whereas in the actual announcements he gives you to typeset, they each
> occur significantly more than 1/12 of the time (significantly in the
> formal statistical sense.)
>
> So just to pass the time when you?re not actively typesetting, you
> decide to determine the degree ?u? of over-representation of the above
> 63 two-word sequences in any given announcement that your boss has
> given you to typeset.
>
> And to do so, you use the following table of expected frequencies for
> the 49 2-letter sequences that are encoded by the above 63 two-word
> sequences:
>
> AL 0.0833 (the above 63 2-word sequences contain 2 out of the 12 ways
> to code AL)
> AN 0.1250 (the above 63 2-word sequences contain 1 out of the 8 ways
> to code AN)
> AP 0.1250 (the above 63 2-word sequences contain 2 out of the 16 ways
> to code AP)
> AS 0.0417 ...
> AV 0.0625
> DL 0.0833
> EK 0.2500
> EN 0.2500
> ER 0.1667
> FA 0.1250
> FF 0.2500
> FL 0.0833
> GF 0.1250
> GK 0.2500
> GL 0.0417
> GR 0.0417
> GS 0.0417
> GW 0.2500
> GY 0.3750
> HP 0.1250
> IA 0.0833
> IP 0.2500
> KK 0.2500
> KN 0.2500
> LG 0.0417
> LK 0.0833
> LL 0.0556
> LS 0.0278
> LT 0.0278
> LY 0.0833
> ML 0.1667
> PG 0.0625
> PV 0.0625
> QR 0.0833
> RR 0.1111
> RV 0.0278
> RW 0.1667
> RY 0.0833
> SG 0.0417
> SL 0.0278
> SP 0.0417
> TL 0.0417
> TP 0.0625
> VG 0.0625
> VL 0.0417
> VS 0.0833
> VT 0.0625
> WR 0.3333
> YP 0.1250
>
> In particular, you use this table as follows.
>
> Suppose your boss gives you an encoded announcement in which:
>
> a) a 2-word sequence for AP appears just once and this 2-word sequence
> is att ccc (one of the above 63 2-word sequences)
>
> b) a 2-word sequence for AL appears just once and this 2-word sequence
> is gcg ctg (one of the above 63 2-word sequences)
>
> c) no other 2-word sequence from the above 63 occurs in the
> announcement (equivalently, you will typeset an announcement in which
> none of the above 49 2-letter sequences appears other than one AP and
> one AL.)
>
> Then the degree ?u? of average over-representation of the 63 2-word
> sequences in the announcement is
>
> 1 / 0.8333 (for AL)
>
> plus
>
> 1 / 0.25 (for AP)
>
> equals
>
> 12 + 4 = 16
>
> divided by 2 (
>
> equals 8.
>
> That is, on average, the 63 2-word sequences are represented in the
> announcement 8 times more frequently than expected.
>
> Let me stop there and see if I?ve made the relationship between UCP?s
> and ?u? clear, and if not, what questions you have.
>
> If you do now understand the ?expected frequencies? table discussed
> above for S63, you should also understand why ?by definition? (as you
> said), the expected frequencies for C711 are those above subtracted
> from 1. If you don?t, then that will be the topic of my next post.

I understand things will enough that I've been able to reproduce
your table of the expected frequencies of the 49 dipeptides that
are encoded by the 63 study dicodons, except for LT and RV, for
both of which I get 1/24 but you give 1/36 -- actually, the decimal
equivalent of that: .0278 in the post to which I am replying (May
30, 6:10 pm), and .027777778 in an earlier post (May 30, 5:53 am).

I used the following inverse table (that I posted earlier):

A gct gcc gca gcg
C tgt tgc
D gat gac
E gaa gag
F ttt ttc
G ggg ggc gga ggg
H cat cac
I att atc ata
K aaa aag
L tta ttg ctt ctc cta ctg
M atg
N aat aac
P cct ccc cca ccg
Q caa cag
R cgt cgc cga cgg aga agg
S tct tcc tca tcg agt agc
T act acc aca acg
V gtt gtc gta gtg
W tgg
Y tat tac

Since both LT and RV appear only once in S63, and since both of
them can be encoded in 6*4 = 24 ways, shouldn't their expected
frequencies be 1/24 ? Is my table wrong?

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