**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:
256

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?

* Footnote:

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.

**Survey:**

C-atoms: 24 ams: 128 R+B 1

All atoms: 24 ams: 256 R 2
quotient

All atoms: 20 ams: 384 R+B
3

|-
84

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
number 385 among 24 amino acids.

Note: A2 + C2: 104 atoms ~ number of C+N+O+S
in all, R-chains

>
24 ams

G2
+ 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

*