## Icosa-tetrahedral and icosa-dodecahedral bioharmonies as candidates for genetic codeBoth the icosa-tetrahedral (see this) and icosa-dodecahedral harmony to be discussed below can be considered as candidates for bio-harmony as also the harmony involving fusion of 2 icosahedral harmonies and toric harmony (see this). The basic reason is that the third harmony corresponds to doublets. One cannot exclude the possibility of several equivalent representations of the code.
Icosahedral harmonies can be characterized by a subgroup of icosahedral isometries A - Z
_{6}gives rise to 3 AAs coded by 6 codons each (leu,se,arg) and 2 AAs coded by 2 codons: the choice of the doublet would require additional conditions. One option is ile doublet. - Depending on whether one includes reflection or not, one can have either Z
_{4}⊂ A_{5}(60=4× 15) or Z_{4}=Z_{2,rot}× Z_{2}⊂ A_{5}× Z_{2}. I have assumed that Z_{4}=Z_{2,rot}× Z_{2}but the recent argument suggests the first option. This does not have any implications for the earlier model. Icosahedral Z_{4}gives rise to 5 AAs coded by 4 codons each (5× 4=20). This leaves 11 AAs and 3 "empty" AA formally coded by stop codons. - Icosahedral Z
_{2}gives rise to 10 dublets. These 4-plets would correspond to (phe, tyr, his, gln, asn, lys, asp, glu, cys, stop-doublet) This leaves (stop,trp) double and (ile,met) doublet with broken Z_{2}symmetry.The fusion with tetrahedral code with 4- codons and 4 AAs should explain these 4 AAs. Tetrahedral isometries form group S _{3}and reduce to group Z_{3}for tetrahedral cycle.- One could argue that ile-triplet and and met correspond to 3-element orbits with 1-element orbit. (stop,trp) would be formed by Z
_{2}symmetry breaking from trp doublet and there is no obvious mechanism for this. - If one tetrahedral face is fixed as a face shared with icosahedron, the symmetry group of tetrahedral cycle reduces to Z
_{1}. This would give 4 singlets identifiable as (ile,met) and (stop,trp) symmetry broken doubles. Since ile appears also in doublet, tetrahedral 1-orbit and icosahedral 2-orbit must have a common doubled triangle identifiable as the common face of icosahedron and tetrahedron. The doubling of the common triangle replaces ile-doublet with ile-triplet. This option looks rather reasonable.
- One could argue that ile-triplet and and met correspond to 3-element orbits with 1-element orbit. (stop,trp) would be formed by Z
Dodecadedral harmony
Dodecahedral harmony correspond to the unique Hamilton cycle at dodecahedron. Dodecahedral harmony as 20 notes and and 12 5-chords. If one assumes that the octave divides to 20 notes, this brings in mind "eastern" view about harmony. The obvious objection against dodecahedral harmony is that dodecahedral faces are pentagons so that dodecahedral chords would be 5- rather than 3-chords so that the correspondence between chords and DNA codons would be lost. The situation changes if 3 notes - 3-chord - determine the 5-chord completely and one can assign a unique 3-chord to each pentagon. This is indeed the case! - 3-edges meet in every dodecahedral vertex (this makes the dodecahedral cycle unique apart from rotations) and each edge pair in the vertex belongs to same pentagon (in the case of icosahedron there are 5 edges per vertex so that this is not true). Therefore each pentagon must contain at least 2 edges of Hamilton's cycle.
The cycle must visit all vertices of pentagon, and the visit to the vertex means that the cycle shares at least one edge with pentagon. Since all vertices of the pentagon must be visited, there are two options. For option a) given pentagon shares with the cycle disjoint 2-edge with 3 vertices and 1-edge with two vertices. For option b) the pentagon shares with the cycle 4-edge with 5 vertices. - The numbers n
_{a}of pentagons with 4-edges and n_{b}=12-n_{a}2-edge+ 1-edge (making 3 edges) can be deduced easily. Cycle has 20 edges. Pentagon of type a) shares 3 edges with the cycle and the edge is shared by 2 pentagons. This gives 3n_{a}/2 edges. Pentagon of type b) shares 4 edges with the cycle. This gives 2n_{b}= 2(12-n_{a}) edges. The total number of edges is 3n_{a}/2+2n_{b}= 20, which gives n_{a}=8 and n_{b}=4. Dodecahedral Hamilton's cycle can be found from web (see this). The structure is as deduced here.For case a) the 3-chords correspond naturally to the 3 vertices of the 2-edge shared with the cycle. Therefore it is possible to assign unique 3-chords to the dodecahedral harmony and to obtain connection with codons in this case. One however obtains also 12 2-chords: could they have some genetic counterpart? What about 5-chords for pentagons of type b)? Hamiltonian cycle can be oriented and this is induces orientation of the pentagons. One can say that the first vertex in the 4-edge is the vertex at which cycle arrives to the pentagon and identify the 3-chord as the first three vertices. It turns out that for the replacement of quint cycle this is not actually necessary.
Is icosa-dodecahedral harmony consistent with the genetic code?
One must check whether icosa-dodecahedral harmony is consistent with the degeneracies of the genetic code. - A fusion of 2 icosahedral harmonies and 2 copies of dodecahedral harmony would be in question. As in the case of icosahedral harmony already discussed, the two icosahedral harmonies would have symmetry groups Z
_{6}and Z_{4}and give the codons coding for 3 6-plets and 1 doublet+ 5 4-plets + two copies of dodecahedral harmony. - Can the model predict correctly the numbers of codons coding for AAs? It is known that dodecahedral Hamilton cycle divides dodecahedron to two congruent pieces related by Z
_{2}symmetry (see this). Also the Hamiltonian cycle defining the common boundary has Z_{2}symmetry. A good guess is that these Z_{2}:s corresponds to reflection symmetry and rotation by π but I am not able to exclude Z_{4}⊂ G_{0}, where G_{0}consists of 60 orientation preserving isometries. In this case some orbits - presumably all 3 of them - could contain 4 pentagons. This is not consistent with the condition that one has doublets and singlets.If the second symmetry corresponds to reflection, it can be excluded by simply assuming that reflections change the orientation of the cycle. - Rotation by π has two fixed points corresponding to opposite poles so that one has 5 2-orbits and 2 1-orbits giving 12 triangles for each copy. Two copies of dodecahedral harmony would give 5+5=10 doublets and 2+2=4 singlets. A possible interpretation would be as (ile,met) and (stop,trp).
- Why two copies of dodecahedral code? What distinguishes between them? If imirror symmetry leaves the cycle invariant apart from orientation the copies could be mirror images and consist of same faces. The second option is that they related by a rotation?
- The number of dodecahedral AAs is 24 rather than 20. Could the additional 4 AAs as orbits have interpretation as AAs in some sense. Could the "empty" AAs coded by stop codons be counted as AAs exceptional in some sense. In TGD framework one can consider the possibility that although AA is "empty", there is analog of AA as physical signature for the end of protein telling what stopping codon it corresponds. The magnetic body of protein is a good candidate.
Genetic code has several slightly differing variants. Could the 2 additional exotic AAs Pyl and Sec correspond in some situations to the additional AAs? - Essential for the bio-harmony as a fusion of harmonies is that one can select from each orbit single face as a representative of the AA it codes - kind of gauge choice is in question - and that the orbits corresponding to different AAs can be chosen to be disjoint. Otherwise codons belonging to the orbits of different Hamilton cycles can code for the same AA if the AA can be chosen to be in intersection. If not, the same codon can code for 2 different AAs - this can indeed occur in reality (see this)!
The condition that orbits of different cycles do not interesect seems quite stringent but has not been proven. But what if it is actually broken? Indeed, in the case of icosahedral harmony with Z _{1}symmetry tetrahedron and icosahedron could have common a doubled face the breaking of this condition would geometrically explain why ile belongs to both icosahedral and tetrahedral orbit.Ile is the problem also in the case if icosa-dodecahedral harmony. Dodecahedral singlet codes for ile as also icosahedral doublet. Could one talk about doubling of ile face so that it corresponds to a pair of triangle and pentagon (in 1-1 correspondence with triangle as chord). - The two copies of the dodecahedral code should correspond to 5 doublets and 2 singlets each. One expects that together they give rise to 10+2 +10+2 =24 faces. Do they? Mirror symmetry and rotation by π act as symmetries of the cycle so that neither can map the two cycles to each other. Dodecahedral (equivalently icosahedral) rotations give rise to new equivalent cycles. The action on pentagons corresponds to the action on vertices of icosahedron so that it easy to understand what happens.
Each symmetry corresponds to a rotation around some axis and has opposite icosahedral vertices at this axis as fixed points. Hence any two cycles obtained in this manner have 2 common pentagons. This means reduction 24→ 22 unless one interprets the situation in terms doubled faces? Could the disappearing doublet correspond to stop-doublet? What about the remaining stop of the vertebrate code pairing with trp? Why does second singlet correspond to empty AA and not something else such as exotic AA. - There is also further problem. Suppose that an intersection of orbits takes place at single triangle. Suppose that one cannot choose this triangle to be "AA" triangle for both orbits. In this case it is not clear to which AA the codon codes. This kind of phenomenon actually takes place in some cases and is known as homonymy. It is associated with the deviations of the code from the vertebrate code and involves exotic AAs Pyl and Sec. Codons can serve as a stop codon or code for an exotic AA.
See the chapter or the article Hachimoji DNA from TGD perspective. |