Calculating Vitamin B ratio as conversion to HEME (blood) base through 1/3 transfer of electrons
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Converting Alchohol To Glucose Step 1Converting Alchohol To Glucose Step 2Calculating DNA Folds
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Takes to Cell A46
Takes to Cell A88
Takes to Cell A130
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Calculating Vitamin B ratio as conversion to HEME (blood) base through 1/3 transfer of electronsLegend (e) - Electrons:
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H = 1Na = 11S = 16
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Vitamin BC63_H88_N14_O14_P_CoC = 6Mg = 12K = 19
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(e)378889811242N = 7C = 6Ag = 47
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--->
• We begin by looking for whole divides ...
O = 8Ca = 20Fe = 26
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/75414166Zn = 30
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/42228Hg = 80
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/312614
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/218944495621x
• This constitutes B as a half type (Base Type) Vitamin
Legend:Syncopations
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C = 6Ca = 20
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Converting the count to HEME base H2 N2 O2Zn = 30-- > C (5) -- Ca (5) >Fe = 26
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Zn = 30130Fe = 26
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To become 1/3 (Base Type) electron reduction must be to:30 / 5 = C130 / 26 = Fe
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Vitamin BC63_H84_N12_O12_P_Co
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(e)37884849642• This is how blood forms. P_Co is a weightWe now have 1 full syncopation (balance) for foundational atoms
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|• Which allows less frequent objects to pullZn_C(5)_Ca(5)_Fe_
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H2 N2 O2 is removed for the conversion|• Itmes off of the structure.
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This begins the base square structure seen in HEME|Where each 5 is equal to 1/4 of the whole.
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vZn_(5)_C(5)_Ca(5)_Fe(5)_
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○ Because these ratios scale
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Seeing Vitamins restructure
◙ Calculation can be done using only atom # (ratio)
Bc each has been balanced using the number (ratio) 5
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<---
○ Or only the electron #
And that is how all cells cooperate
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HEME BaseC21_H28_N4O4N
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(e)1262828247x
Through 3 we have created 4
H appears everywhere, because it is an intermediate (single electron)
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Or one HEME Base
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It
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To divide by 3 you also lose P_Co and the balance requires 7 electrons (N)Gives
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Now Fe_N4 (HEME) can attach to syncopate what it needs out of the remainderEverthing
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A
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Balance
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x2
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8x
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18
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18
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x8
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821818832x
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Converting Alchohol To Glucose Base - Step 1
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SugarC12_H22_O11_H2 N2 O2 = 32(e) - Electrons:
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(e)722288
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H = 1Na = 11S = 16
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GlucoseC6_H12_O6_C = 6Mg = 12K = 19
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(e)361248N = 7C = 6Ag = 47
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O = 8Ca = 20Fe = 26
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WaterH2OZn = 30
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(e)28Hg = 80
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AlchoholC2_H6_O_Should be seen as H6OStructured:Understanding Alchohol Effects
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(e)1268• The outside H is an intermediateC2_H5_OH_
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Alchohol (just like water) acts as a syncopator through Hydrogen
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But as a wave it looks like this:The syncopation levels are different which is why
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it causes inebriation and changes our body interactions passing through the liver.
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HCH_HCH_HOHWe become dehydrated and there are slight deteriorations because it transfers O
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Around and to other items, while also requiring 2 H to balance
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Alchohol RestructureC2_H4
O comes off very easily bc Carbon is the base
HCH_HCH
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(e)64
and H is the suspension
|
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orv
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C2_H6
Picture it as big waves (C, O, Fe)
H(HCH)_(HCH)H
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66x
Pressing out the gaps from our base H2O structure
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In a closed system H can sit outside of syncopation
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C2_H6_
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Now we have the base for Glucose and Sugar
< - - --
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C12H36O6
Multiply by 6 to see this is true
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Remember that for each C and O you need 2 H
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C12H48O12
Oxygen can act as a cap too
OHCHHO
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But this isn't a complete structure. It has too many Hydrogen. It would only stay together in space.
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We live and build in a closed system with pressure and compression so H gets pushed out
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And more dense waves (atoms) hold one another together ~~~ IE: CO, FeN
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Converting Alchohol To Glucose Base - Step 2
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.H2 N2 O2 = 32(e) - Electrons:
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.
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.H = 1Na = 11S = 16
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C = 6Mg = 12K = 19
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.N = 7C = 6Ag = 47
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.O = 8Ca = 20Fe = 26
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.Zn = 30
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It scales and the same solutions work for plants and Mg, N too (Chlorophyll)
Hg = 80
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Let's try that again...
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