2. A comparison
of the total number of atoms in the R-chains
with number of C-atoms in R+B-chains:
Total of atoms in R-chains:
Total of C-atoms in R+B-chains: 128. Quotient 2/1.
Can this halving above in the codon groups be more than a coincidence?
How to express the "law" behind this, if any?
Which skeleton as the C-skeleton doubles its nuclear
positions in its active reactive or "radical" part ? One
example could be the doubled projection of the body in the brain?
a) His + Cys co-operate in the active centre of the enzyme
which breaks P-fructose-P into two halves, the start of the glycolyse.
C-atoms: 24 ams: 128 R+B 1
All atoms: 24 ams: 256 R 2
All atoms: 20 ams: 384 R+B 3
All atoms, 24 ams: 468 R+B
Note: number 384 of 20 ams, without double-coded, is divided
- 207 in R-chains
- 177 in B-chains, approximately the same division as of
Note: A2 + C2: 104 atoms ~ number of C+N+O+S
in all, R-chains
+ U2: 152 atoms ~ number of H in all, R-chains
Number of C-atoms in R+B-chains, 24 ams, = 128:
in A1 + U1-coded ams: 81 = entire number of C-atoms
in R-chains +1
in G1 + C1-coded ams: 47 = entire number of C-atoms
in B-chains - 1