**1. About the problems of the earlier view of the dark realizations of the genetic code**

Consider first the problems of the earlier views of the realization of the dark genetic codes in terms of dark proton triplets at monopole flux tubes parallel to the ordinary DNA and to the realization in terms of dark photon triplets.

- The TGD based inspired model of the dark photon genetic code (see this, this and this) assumes that the dark realization of genetic code involves 3 icosahedral Hamiltonian cycles giving rise to 20+20+20 dark DNA codons and the unique tetrahedral Hamiltonian cycle giving the remaining 4 codons.
The obvious problem of icosa-tetrahedral picture is that one must assume that icosahedron and tetrahedron are disjoint. If they have a common face, the number of faces reduces to 63 and one DNA codon is missing. This raises the question whether icosahedron and tetrahedron could be disjoint pieces of a larger structure.

- In the dark proton realization a given codon would correspond to a selected triangular face of I or T carrying dark protons at the vertices of this face. The original view was that dark 3-proton states would correspond to 64 codons. The problem was that one obtains only 8 states for dark proton triplets from spin and antisymmetrization in spin degrees of freedom would not allow any states unless the spatial wave function is totally antisymmetric and spins are in the same direction.
In the original proposal also neutrons were assumed so that the codon corresponds to a sequence of 3 nucleons with both spins. 3 nucleons would give rise to 64 states as required. Dark protons can also be effectively neutrons as far as charge is considered. This might be possible if the bonds connecting the dark protons can be both neutral and negatively charged. Weak interactions are as strong as electromagnetic interactions in a given biological scale (such as DNA scale) if the dark Compton length proportional to h

_{eff}is larger than this scale and the weak transitions change the dark protons to effective dark neutrons.This option leads to a problem with the fact that DNA nucleotides have negative unit charge. One should have protons to neutralize this charge and stabilize DNA. Also variants of the proposal in which there are flux tube connections between dark protons having 2 different neutral states analogous to neutral pion and neutral ρ meson.

The simplest proposal, which is consistent with the idea that genetic codons correspond to cyclotron transitions of dark proton triplets assignable to the triangular faces of an icosahedron or tetrahedron is as follows. Besides 2 spin states, dark protons can also have 2 states with spin +/- 1 corresponding to the analog of rotation in the discrete space defined by the vertices of the triangle. This would give 2

^{3}× 2^{3}=64 states.

- The codons identified as dark proton triplets assignable to one of the 20 triangular faces of icosahedron and tetrahedron have in quantum situations a wave function in the discrete space of the faces, which is in general delocalized. Could these wave functions in the set of faces give rise to states in 1-1 correspondence with the icosahedral and tetrahedral codons? There would be 20 wave functions for an icosahedron and 4 wave functions for a tetrahedron. The number of icosahedral states must be tripled to 60 corresponding to the 3 basic types of icosahedral Hamiltonian cycles with symmetries Z
_{n}, n=6,4,2.The 3 dark protons also have spin degrees of freedom. The dark proton triplet in the ground state(s) would be naturally spontaneously magnetized so that all spins are in the same direction. Also the states in which some dark protons are excited are allowed by Fermi statistics and are needed since these excitations could correspond to the spatial wave functions in face degrees of freedom.

- Dark photon triplets are needed for communications. The vision is that they correspond to the representation of codons as frequency triplets represented by the realization of icosahedral and tetrahedral Hamiltonian cycles as frequency triplets. The assumption has been that the 3 frequencies of dark 3-photon are associated with the cyclotron (or Larmor transitions if only spin is dynamical) of dark protons of a dark proton triplet.
Dark photon communications between identical codons would take place by 3-resonance. The de-excitation of the first codon would lead to the excitation of an identical codon: one would have a kind of flip-flop. Also dark genes as sequences of N dark codons could act as a single quantum coherent unit and 3-N resonances between identical dark genes would become possible. The mechanism is very similar to that used in the computer language LISP. The modulation of the frequency scale by modulating the thickness of the monopole flux tubes would make possible coding of the signal and it would be transformed to a sequence of resonance pulses at the receiving end.

Dark photon triplet states could correspond to wave functions in the space of icosahedral and tetrahedral faces.

- Cyclotron transitions would be needed in order to generate dark photon triplets. This would require excitations of the dark protons of the spontaneously magnetized ground state(s). If only spin matters, the cyclotron transitions reduce to Larmor transitions. The correspondence with the icosahedral Hamiltonian cycles in terms of dark photon triplets would suggest that these excitations correspond to icosahedral genetic codons as wave functions in the set of faces. The cyclotron transition would provide the energy needed to excite the wave function in the set of faces. 64 transitions would be needed.
It is important to notice that cyclotron transitions rather than cyclotron states of dark protons would correspond to codons of icosa-tetrahedral representation represented as wave functions in the set of facs.
There are however only 8 states per face if only Larmor transitions are allowed. This is much less than the number 20 20+20+4=64 for icosahedral and tetrahedral Hamiltonian cycles. An additional two-valued degree of freedom is needed. The simplest possibility is the assignment to each dark proton an analog of angular momentum eigenstate with spin +/- 1 corresponding to a discrete rotation around the triangle. This would give 8× 8=64 states per face. Could the excitations of these states correspond to 20+20+20 icosahedral states plus 4 tetrahedral states?

- Hitherto the considerations have been implicitly classical in that a localization in the set of faces has been assumed. Quantum theory allows us to give up this assumption. Icosahedral realization suggests that dark proton triplet has a icosahedral wave function delocalized to the set of 20 faces with symmetry fixed by the Hamiltonian cycle to Z
_{n}, n=6,4 or 2, and that the excitation of the dark proton triplet in the face degrees of freedom provides the energy changing the wave function in the set of faces. The same would apply to the tetrahedron with symmetry Z_{4}allowing 4 wave functions.The orbital and angular momentum degrees of freedom would be coupled. The transition from the ground state for dark proton triplet would excite wave function in the set of faces. This could imply the desired correspondence between the dark proton representations and dark photon realizations of the code.

- There is a further problem. Spontaneously magnetized states of 3 dark protons would define ground states of codons. The ground state proton triplet cannot have lower energy states and cannot emit dark photon triplets and are therefore "mute" and unable to communicate, presumably necessary for processes like transcription and translation. Note that ground states are however not deaf.

^{3}allows to realize the dark genetic code.

The icosa-tetrahedral honeycomb is the unique honeycomb, which involves only Platonic solids. This inspires the question whether genetic code could be universal and realized in all scales by induction, which means that the tessellation of H^{3} induces tessellation of 3-surface X^{3}⊂ H^{3} by restriction. Also the induction to H^{3}(a) projection of X^{4} makes sense.

The TGD view of holography indeed predicts the special role of hyperbolic 3-spaces. The space-time surfaces in H=M^{4}× CP_{2} are analogs of Bohr orbits, which go through H^{3}(a_{n})⊂ M^{4}⊂ H, where a_{n} corresponds to a root of the polynomial with integer coefficients determining to a higher degree a given region of the space-time surface by M^{8}-H duality (see this and this) .

In the sequel the detailed realization of the genetic code in terms of the icosahedral honeycomb will be discussed with an emphasis on the problems noticed above.

**2. The realization of the code in terms of icosa-tetrahedral tessellation**

The fundamental region of the icosa-tetrahedral tessellation contains 30 octahedrons, 20 tetrahedrons, and 12 icosahedrons and the cautiously proposed interpretation is that the cells meeting at each * edge* of the tessellation have either the cyclic structure TOTI or OTOI, and each vertex involve 3 O:s, 2 T:s and 1 I. Could one interpret this in terms of the dark icosahedral realization of the genetic code?

**2.1. Ideas related to the detailed realization of the genetic code**

The detailed realization of the dark genetic code is far from completely understood and one might hope that icosa-tetrahedral realization could bring in the constraints allowing us to fill in the details. It is useful to proceed by considering basic requirements on the realization of the dark code.

- There are 3 O:s per single I in vertex if 10 instead of 12 icosahedral cells are included. The reasons for this become clear from the proposed relation between DNA double strand and fundamental cell of icosahedral honeycomb. What could the role of O:S be?
Imagine that it is possible to arrange the polyhedrons for a given I to cycles as -I-O-T-O-T-O-: here cyclicity is assumed. The two tetrahedrons and I would be disjoint. This would solve the problem due to the common face of T and I (only 63 DNA codons) but give 60+4+4 faces and 68 dark DNA codons. There is however the problem posed by the mute codons. Could the presence of mute DNA codons reduce the number of DNA codons from 68 to 64. This would imply that their transcription allows only 64 dark mRNA codons. Could mute mRNA codons reduce the effective number of mRNA codons to 61 for the standard code (stop codons would be mute)? What about its variants with a smaller number of stop codons?

- Bioharmony involves 3 icosahedral Hamiltonian cycles. All the combinations of the 3 -cycles with symmetries Z
_{6}, Z_{4}and Z_{2}predict the same code. These bioharmonies are interpreted as correlates for emotional states appearing already at the basic bio-molecular level. The motivation comes from the fact that the icosa-tetrahedral harmony emerges as a geometric model for the music harmony and music indeed both creates and expresses emotions.Could icosahedral honeycomb allow us to understand the realization of these 3 icosahedral Hamiltonian cycles in terms of cyclotron frequency triplets? One must have closed magnetic monopole loops in order to have cyclotron transitions. Could these loops form triangles of form I-T-O. This would be 6 different triangles and 3 different positions of I for given T. This kind of loop would be assigned with each vertex of the face. Could the magnetic field strengths depend on the loop and for a given T give rise cyclotron frequency triplets characterizing a given icosahedral Hamiltonian cycle.

- One can criticize the assumption that there is only a single codon per single I and T. I:s could in principle carry several codons. This however gives a restriction that the codons inside given I and T are different and restricts the representative power of the code if it involves more than 2 strands. This restriction is however automatically satisfied for the base-paired codon and anticodon in the DNA double strand!

**2.2 Dark photon realization of the icosahedral part of the code**

Consider first the realization of the icosahedral part of the code in terms of dark photons.

- The 3 icosahedral Hamiltonian cycles have symmetries. The 20 codons with Z
_{6}symmetry correspond to 3 6-plets and 1 doublet of Z_{6}and for unbroken symmetry the codons inside these multiplets code for the same amino acid. This means 3+1= 4 amino acids. Z_{4}symmetry has 5 4-plets and in absence of symmetry breaking this corresponds to 5 amino-acids. Z_{2}symmetry as 10 2-plets, and also this symmetry is also almost exact and corresponds to the almost exact symmetry with respect to the third letter of the codon analogous to isospin symmetry. - Icosahedral part of the icosa-tetrahedral realization involves 3 icosahedral Hamiltonian cycles characterized by different symmetries. For Z
_{6}symmetry, there are 6+6+6+2=20 codons codons. These sets of codons can be regarded as orbits of Z_{6}and correspond to amino-acids. This if the Z_{6}symmetry is not broken. This means 3+1 amino acids in absence of symmetry breaking.Z

_{4}symmetry has 5 4-plets and in absence of symmetry breaking this corresponds to 5 amino-acids coded by 4 codons each. Z_{2}symmetry has 10 2-plets and this symmetry is also almost exact. This symmetry corresponds to the almost exact symmetry with respect to the third letter of the codon. - Dark photon codons are represented as cyclotron frequency triplets of dark photons created in 3-cyclotron transitions for dark proton triplets involving simultaneous emission of 3 dark photons made possible by quantum coherence. In the case of genes with N codons one has 3N-cyclotron transition and 3N dark proton-state represents a gene as a quantum coherent unit.

**2.3 Dark proton realization of the icosahedral part of the code**

Consider next the dark proton realization of the icosahedral part of the code.

- The basic problem of the dark proton realization of the code is that the states of dark protons triplets give only one half of the codons. One way to obtain doubling is to assume that dark codons are connected by flux tubes which can carry charges 0 or -1. It is far from clear whether this is consistent with the constant negative charge density of DNA: charge -e per DNA codon. Could 2×T degeneracy give it? For each I defining an icosahedral codon one should select either T. If both the selected and unselected T code for 4 codons, one obtains additional 4+4=8 codons.
- Icosa-tetrahedral realization should give 20+20+20+4 =64 dark proton triplets assignable to the faces of I and T. Suppose that the cells can be thought of as forming a cycle O-I-O-T-O-T with O and T ends connected. The two T:s have no common faces with O and without additional conditions give rise to 4+4 additional codons giving 68 codons. How can one reduce the number of dark DNA codons to 64?
- Dark proton codons have a ground state, or possibly several of them, which by definition cannot decay to lower energy states by emission of dark photon cyclotron triplet. Ground state codon is mute since it cannot produce dark photon triplets as 3-chords.
The natural first guess is that the ground states correspond to the 6 combinations 3 icosahedral Hamiltonian cycles and 2 tetrahedral cycles assignable to 2× T. The 3 stop codons are transcribed but not translated so that the interpretation of 3 DNA stop codons as icosahedral ground state dark codons unable to send 3-photon signals is not correct. For mRNA this interpretation could make sense if the mRNA images of DNA stop codons represent ground state codons.

- Cyclotron excitations of ground state codons are induced by dark photon triplets. Conversely cyclotron de-excitatons generate dark proton triplets except for the ground state codons with minimum total energy. Suppose that there are 6 ground state codons as combinations of 3 dark codon ground states assignable to the 3 icosahedral Hamiltonian cycles and 2 dark proton ground states assignable to tetrahedral cycles of the two T:s. This would give 8 mute states. The total number of dark DNA codons is 60+8=68. Note that the mute states are not deaf: they can receive messages.
One would obtain only 60 DNA codons, which can be transcribed to mRNA codons if the transcription involving dark photon codons. How could one get 64 as an effective number of DNA codons?

One can imagine transitions between otherwise mute codons, which generate dark photon triplets coupling to mRNA associated with DNA. Let A, B and C the ground state codons with minimal total dark cyclotron energies in an increasing order for the 3 icosahedral Hamiltonian cycles. If for a given T (two options) the cyclotron transitions are possible only between codons C and B and B and A one obtains 2 DNA-mRNA pairings for both T:s. One would have 60+2+2=64 mRNAs pairing with DNA and effectively 64 DNA codons.

Note that the transcription produces only 64 dark mRNA codons from 68 dark DNA codons.

For 64 mRNA codons it could happen that there are no transitions between the 3 icosahedral codons for both choices of T so that there are 6 mute mRNA codons. If there are transitions C → B and B → A, the number of mute icosahedral codons is 4. If there are no transitions between tetrahedral ground state codons, one has effectively 60 mRNA codons since the translation stops due to the absene of dark 3-photon signals to tRNA. If there is a transition between the 2 ground state nRNA codons associated with the two T:s, one obtains 61 effective mRNA codons of the standard realization of the code. The transitions between tetrahedral codons can increase the effective number of mRNA codons.

- What about tRNA appearing as a pair of amino-acid and single RNA codon. Could the RNA of tRNA and amino-acids correspond to the unique icosahedral honeycomb of H
^{3}and to icosahedral Hamiltonian cycles so that the number of dark codons in absence of tetrahedral degeneracy would reduce to 32, which is the minimal number of ordinary tRNA codons, which is increased by the non-uniqueness of the ordinary tRNA itself? Note that mute tRNA codons are not deaf: they can receive messages but cannot send them. Obviously, tRNA and amino-acids would correspond to the lowest evolutionary level.

The dark DNA and RNA codons are dynamical and are not fixed to be the same as ordinary codons. This is required only during the communications with ordinary DNA possibly taking place by dark photons transforming to ordinary photons and inducing resonant transitions of ordinary DNA and other basic biomolecules. This strongly suggests that dark DNA and RNA act as a kind of R&D laboratories making it possible to test variants of the genes. Actually their ground states would correspond to 3 icosahedral representations and 2 tetrahedral representations and would correspond to aminoacids via transcription and translation.

Needless to say, this picture is highly speculative and one can probably imagine variants for it. The basic idea is however clear: icosa-tetrahedral tessellation could explain the details of the standard genetic code and its modifications.

**2.4 Realization of the flux tube structures associated with dark codons**

The following represents an attempt to make the above picture more concrete.

- The selection of 1 O from 3 O:s could mean a selection of an icosahedral Hamiltonian cycle with symmetry group Z
_{6}, Z_{4}, or Z_{2}. This gives for icosahedral realization 20+20+20 =60 icosahedral codons. Tetrahedral Hamiltonian cycles associated with the two T:s should give the remaining 4 codons. One can however imagine several ways for how this could occur. - The selection of O should correspond to a choice of the icosahedral cycle. What does this mean geometrically? To each dark proton of the codon, one must assign a closed monopole flux tube. The strength of the magnetic field of the flux tube fixes the cyclotron frequency scale for each flux tube. The 20 dark-photon chords defining a given icosahedral bioharmony differ for different choices of O and T. The frequencies are fixed if the Hamiltonian cycle corresponds to a quint cycle such that the frequencies associated with the neighboring vertices of the Hamiltonian cycle differ by a scaling 3/2. This requires that the magnetic field strengths along the cycle differ by scaling 3/2.
- How to concretely realize the correlation of the bioharmony with the choice of O and T for a given I? Suppose that for a given I, the closed flux tube connects I and the selected O and T. There would be a closed I-O-T flux tube for each vertex of the face defining the codon. This kind of flux tube would define an analog of a string of a musical instrument.
These closed flux tubes would be hyperbolic analogies of closed circuits formed by Euclidian nearest neighbour lattice bonds. If makes sense to assign to each I a cycle O-I-O-T-O-T, with O and T at ends being connected, the cycle I-O-T would go through the either T, and this implies that tetrahedral codons correspond to the other face of T. One would obtain 64 dark proton codons with 3 mute dark proton codons identifiable as stop codons. In the transcription the signal as a dark photon triplet would not reach the dark RNA codon and the transcription would stop. Could this mean that dark RNA codon attaches first to dark DNA codon and the transcription of DNA to ordinary RNA occurs after that in the usual way.

- The proposed transitions between ground state codons for icosahedral Hamiltonian cycles modify the cycle geometrically since the O in cycle I-O-T changes. If the transitions for given T are only of C→ B and B→ A with energies in increasing order, one can imagine that the O is replaced by a neighboring O in the transition in the O-I-O-T-O-T.

- The symmetry breaking for the icosahedral codons with Z
_{n}, m=6,4,2 should be understood. This symmetry breaking can be assumed to occur at the level of dark mRNA and modify the frequency triplets from those for completely symmetric mRNA codons. The replacement T→ U might relate to the symmetry breaking.UUG, CUG, and the very common AUG appear as start codons. They correspond to symmetry breaking for 6-plet (Z

_{6}) coding for leu and 4-plet (Z_{4}) coding for ile. All symmetry breakings occur for start codons UUG, CUG, and for codons UAA and UAG and UGA and UGG closely related to stop codons. - Can one understand the reduction of the number of mRNA stop codons to 2 or 1 occurring for some variants of the code? In these situations, the stop codon of mRNA can code for an exotic amino acid pyrrolysine and selenocysteine. Could the transition between stop codon of dark mRNA icosahedral Hamiltonian cycle to a stop codon of another Hamiltonian cycle take place such that the dark photon triplet generated couples to tRNA involving the exotic amino acid. Situation would be almost like in the case of DNA where only two ground state codons stop the transcription.
- What can one say about the strength of the magnetic fields assignable with the monopole flux tubes? Nanometer length scale 1 nm, naturally assignable to the DNA double strand, corresponds from the formula l
_{B}= 26 nm/(B/Tesla)^{<1/2>}to 12.2 GHz. What is interesting is that the gravitational Compton frequency for Earth is 67 GHz and defines a lower bound for the gravitational quantum coherence time. If the strengths of the magnetic fields span 7 octaves, the thickness of the flux tube would vary by a factor 10 in the range about .1 nm - 1 nm. - Note that the 12-note scale can be realized using powers (3/2)
^{k}, k=1,...,12, of the fundamental and by using octave equivalence to reduce the note to the basic octave. Since the monopole flux is quantized, the realization of the scale requires variation of flux tube thickness inducing variation of magnetic field strength and therefore of that cyclotron frequency scale.There is nothing cherished in the rational quint cycle as the basis of the 12-note scale. For instance, the well-tempered scale actually replaces the Pythagorean scale with an algebraic scale coming in powers of 2

^{1/12}.

** 3. Description of the entire DNA double strand in terms of icosatetrahedral tessellation**

The most ambitious model would describe the entire DNA double strand and relate the model bioharmony to the properties of the icosa tetrahedral tessellation. There are however many questions remaining.

- Single DNA and RNA strand would correspond to a "half realization" for which the T and I cells would contain only single codon. The splitting of DNA could have a geometric interpretation as an effective replication of the induced tessellation to two tessellations to RNA type tessellations.
- There are 20 amino-acids and an icosahedron involves 20 faces. Is this a mere accident? Could icosahedral honeycomb describe amino-acid sequences geometrically. tRNA appears as a single unit. tRNA-amino-acid pairing would involve pairing of two icosahedral tessellations as also the pairing of RNA and tRNA in the translation. tRNA would naturally correspond to a single cell of icosahedral tessellation. This would also explain why the number of tRNA molecules is considerably smaller than RNA codons.
- Does RNA correspond to icosahedral or icosa-tetrahedral tessellation? Tetrahedral Hamiltonian cycles are needed, in particular the dark proton triplets associated with the tetrahedral faces. Therefore icosa-tetrahedral tessellation is the natural option also for RNA.
- It is thought that DNA and RNA nucleotides float freely in the cellular water and DNA and RNA codons are built from them in replication/transcription. This is probably the case at the biochemical level, whose dynamics is controlled by dark level (I have however considered the possibility that freely floating nucleotides could actually form loosely bound codons).
At the dark level both replication and transcription would involve replication of the induced icosa-tetrahedral tessellation: a similar process occurs for clay crystals, and is suggested to be a precursor of DNA replication. This process is a holistic quantum process occurring in a single quantum jump. This would explain the incredible accuracy of these processes, which is extremely difficult to understand in the chemical approach.

The replication would determine the outcome, be it a pair of DNA double strands or of DNA and RNA. After this the chemical processes leading to the formation of chemical codons from nucleotides and their pairing with dark codons of the induced icosa-tetrahedral tessellation would take place.

- B-form is believed to dominate in cells. From the table of the Wikipedia article one learns that for the B-form the rise per base pair (bp) is 3.32 Å, that full turn corresponds to 10.5 bps, and that the pitch of the helix per turn 33.2 Å, which corresponds to 10 bps per turn. The pitch/turn should be equal to 10.5× 3.32= 34.52 Å. There is obviously a mistake in the table.
- The solution of the puzzle is that straight DNA in solution has 10.5 bps/turn and 10 bps/turn in solid state (see this). Since DNA double helix corresponds to solid state then 10 codons correspond to 3 full turns. Therefore my earlier assumption 10 bps/turn in the double helix is correct. 10 codons would correspond 3 full turns and to the length 99.6 Å ≈ 10 nm, which in TGD framework corresponds to the p-adic length scale L(151).

- The total volume of the fundamental region is V= 20V(T)+30V(O)+12V(I)= 341.44 using 2a as length unit. Using the estimate V
_{real}= L(151)^{3}=10^{6}Å^{3}, one obtains a= L(151)/2V^{1/3}≈ 0.07× L(151). The volume fraction of single I would be 17.45/V≈ .05 and 10 I:s would take 1/2 of the volume. - The circumradius of single I would be R=(3-φ)
^{1/2}φ a/2≈ .1× L(151)=1 nm. This conforms with the assumption that there are 10 codons per length L(151)! The diameter of the B-type DNA strand is 20 Å is also consistent with the value of the circumradius. Maybe the proposed picture works! - Notice that if an icosahedral cell corresponds to 2 tetrahedral cells and 3 tetrahedral cells, then 10 codons is the maximum for the realizable DNA codon.

- For 10.5 bps/turn for a straight DNA in solution, the smallest portion of strand, which corresponds to integer numbers of turns and of codons is 6 full turns. This corresponds to 63 bps and 21 codons.
- With an inspiration coming from the notion of Combinatorial Hierarchy (see this) defined in terms of Mersenne primes M
_{n}= 2^{n}-1 defined by the recursive formula M(k)= M_{M(k-1)}= 2^{M(k-1)}-1, I proposed decades ago that ordinary genetic code could correspond to Mersenne prime M_{7}=2^{7}-1 (see this and this). The basic idea is that a system with 2^{7}-1 states corresponds to a Boolean logic with 7 bits but with one state missing: this state would correspond to empty set in the set theoretic realization or fermionic vacuum state in the realization as a basis for fermionic Fock states. Only 6 full bits can be realized and the number of realizable statements is 64, the number of genetic codons. - Memetic code corresponds to the Mersenne prime M
_{127}= M_{M7}-1= 2^{127}-1. Now the number of codons would be 2^{126}=2^{6× 21}and is realizable as sequences of 21 DNA codons! Note that higher Mersenne numbers in the hierarchy were proposed by Hilbert to correspond to Mersenne primes but for obvious reasons this has not been proven. - Could 6 full turns of straight DNA define a memetic codon? During the transcription and replication, DNA double strand opens and becomes straight. Could memetic code be established during the transcription and replication periods? A further intriguing observation is that the cell membrane involves proteins consisting of 21 amino-acids.

For a summary of earlier postings see Latest progress in TGD.

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