ENERGY LAW IS EQUAL TO E = m2c
HEMORRHAGIC DISEASE
SYNTHESIS OF UMBILICAL CORD
ANAEROBIC BACTERIA
Placental transfer of vitamin K is very limited, and phylloquinone (vitamin K1) levels in umbilical cord blood is very low, The newborn infant’s intestinal tract is relatively sterile and takes some time to colonize with bacteria, which may have a role in synthesizing vitamin K2 (menaquinones). Because Bacteroides species are among the most common bacteria that inhabit the human intestinal tract, and because strains such as Bacteroides fragilis synthesize vitamin K, Bacteroides species are more significant in producing human vitamin K in the intestine than Escherichia coli.
1 = Important energy substrates for the small intestine (two Tryptophan) and immunocytes (glutamine)
2 = Type of genus bacteroidaceae; genus os gram-negative rodlike anaerobic bacteria producing no endospores and liveing in the gut of man and animals
1 = two Tryptophan = 414 m/g
2 = genus of gram-negative rodlike anaerobic bacteria = O4 + O4 = O8 128 free oxygen,
1 = two Tryptophan = 414 / 20.6 = 20.6 = H2O + H2
2 = an aerobic bacteria = 128 / 11.31 + 11.31 = H9
ANSWER OF DISORDER
Healthy disorders
Vitamin-K = Phylloquinone = 450.7 m/g fat,
1 = two Tryptophan = 414 / 20.6 = 20.6 = H2O + H2
2 = an aerobic bacteria = 128 / 11.31 + 11.31 = H9
A = two Tryptophan = H2O + H2
B = anaerobic bacteria = H9 absence of oxygen
A = two Tryptophan = H2O + H2 = 414
B = anaerobic bacteria = H9 = 128
ANSWER OF DISORDER
Too Sick disorders
FORMULA TO FIND OUT MEDICINE OF DISEASES
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HEMORRHAGIC DISEASE
SYNTHESIS OF UMBILICAL CORD
ANAEROBIC BACTERIA
Two Tryptophan = 414 + 414 = 828
Two Tryptophan 828 + 37 two water = 865
Anaerobic bacteria = 128 + 128 = 256
Blood = anemia
a2 + b2 = c2
a2 (Anaerobic bacteria) + b2 (Two Tryptophan + 2H2O) = c2
a2 (256) + b2 (865)
a2 (65536) + b2 (748225) = c2
a2 + b2 = 813761 = c2
a2 + b2 = 813761 / 902.08 = 902.09
a2 + b2 = 902.09 – 450.7 (Vitamin-K) = 451.39
a2 + b2 = 451.39 + 72.08 (3H2O + O = intestine) = 523.47
a2 + b2 = 523.47 = (A-T=C-G) DNA base from Gln + Glu + Asp,
So right quantity of Gln + Glu + Asp are produce Vitamin-K in intestine with water,
ENERGY IN ACT BY C1
SYNTHESIS OF UMBILICAL CORD
ANAEROBIC BACTERIA
Two Tryptophan = 414 + 414 = 828
Two Tryptophan + two water = 828 + 37 = 865
Anaerobic bacteria = 128 + 128 = 256
Blood = anemia
a2 + b2 = c2
a2 (Anaerobic bacteria) + b2 (Two Tryptophan + 2H2O) = c2
a2 (256) + b2 (865)
a2 (65536) + b2 (748225) = c2
a2 + b2 = 813761 = c2
Light speed meter seconds = 300000000 / 813761 (m2) = 368.65
= 368.65 / 24 h = 15.36
= 15.36 / 60 men = 0.256
= 0.256 * 35.15625 sec = 9
= 9 = pH of Lipoxygenase type I enzyme or
Linoleic acid a fatty acid essential for nutrition,
= 1.2 to 3.00 pH (by E = m2c)
= 3.00 pH of gastric juice
= pH + pOH = 14 / 12.477 pOH = 1.180
= 1.532 pH + 12.477 pOH = 14.00
= pH + pOH = 14.00,
= pOH = E = m2c,
HEMORRHAGIC DISEASE (E = m2c)
Anaerobic bacteria + Two Tryptophan + 2H2O = (m2)
And (m2) = 813761
For example
= 813761 (m2) / 59.1826 (3H2O + H3 = 59.25 in intestine) = 13750
= 13750 / 30.5 C(H2O) carbohydrate = 450.8
= 450.7 = (Vitamin-K) from 3H2O + H3 in intestine,
While here C(H2O) = carbohydrate as energy
So outer heat or cold is very important of body to change bio functions of inner, for example you eat m2 and for diagnose as outer c1 is very important to complete E or energy in body, supposed you eat only m2 and do not take c1 from outer side, then your diagnostic system would be damage,
SYNTHESIS OF UMBILICAL CORD
Formula of medicines
Anaerobic bacteria + Two Tryptophan + 2H2O = well colony of bacteria in intestine with vitamin-K,
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