where [111] is the 111 length sequence and * is
the additional amino acid. Since any of 20 amino acids can go
into this * position, there are 20 +20=40 copies of any given 111
sequence in the phase space of the 112 sequence. By merely cutting
off one end, I have the increased the probability not by 10 but
by 40 to find a [111].
Consider other lengths factor increase is in {}.
[110]** + *[110]* + **[110] = {3 x 2^20=3145728
[109]*** + *[109]** + **[109]* + ***[109] {4 x 3^20=13.9 billion}
[108]****+*[108]***+**[108]**+***[108]*+****[108] {5 trillion}
etc. etc.
If each length has 10^93 functional permutations, then there are
10^12 x 10^93 = 10^105 functional [108] in the phase space of the
[112]. These [108] can be obtained by merely cutting the [112].
To cut it in the proper place is (1/112)^2=1/12544.
In support of the concept that this type of cut is to be expected
I will quote from a well known creationist. He says,
"But, according to the laws we have just been studying,
reaction time in reversible reactions will also increase still
more the possibility of degradation (randomness) of already
synthesized
molecules, that is, if their entropy is lower than that of the
starting materials. It is so easy to forget that the possibility
of decomposition in reversible reactions increases with time just
as the chance of synthetic processes does." A.E. Wilder-Smith,
Man's Origin, Man's Destiny, p. 67.
Brian, you are not dealing with the CONTAINED functional sequences.
This is a BIG hole in all probability calculations. Assuming your
10^93 usable functions for each of the sequences from 111 down to
103 in length, then the phase space of the [112] sequence has
40 x 10^93 [111]'s + 3,145,728 x 10^93 [110]'s + 13,900,000,000 x
10^93 [109]'s + 5,000,000,000,000 x 10^93 [108]'s.
The largest contributor to that sum is the last term. And there
can not be 10^93 functional sequences in short sequences which
don't have 10^93 permutations, but subject to that limitation,
there are lots of functional permutations CONTAINED in a longer
sequence.
We are not dealing with the independent production of sequences of
103,104,105... but with the independent production of 112's and
the volume of contained 103's, 104's 105's.... This is a very
different problem than what is normally presented.
Brian wrote:
>>This seems to me to be just wild speculation. Perhaps a
possible length of a protein that could perform some nontrivial
function? Is it even remotely conceivable that something as
simple as leu-ser could perform the function of cytochrome c?<<
I was using the ser-leu as an illustration of the mathematics of
the phase spaces of contained sequences. I am NOT making the
claim that ser-leu could perform that function. All I wanted to
show was that a short sequence has LOTS of copies in a longer
sequence and this I successfully showed.
If my memory does not fail me, oxytocin is only 8 residuals long.
I bet you could find a longer sequence which would perform that
function, but you might have to search a long time.
glenn