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)
1 = 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 + 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
TRANSCRIPTION
= a2 + b2 = 813761 = transcription
= a2 + b2 = 813761 * 61 Carbohydrate = 49639421
= a2 + b2 = 49639421 / 112.0868 Uracil = 442865.89
= a2 + b2 = 442865.89 = Two helix of DNA
Uracil + Cytosine = Pyrimidine
Two helix of DNA = 442865.89 / 37 C human temperature = 11969.34
= 11969.34 / 183 glucose = 65.40
= 65.40 + 0.1 () = 65.5
= 65.5 = 3H2O + H8 or three acids + Anaerobic bacteria
BEGINNER TRANSCRIPTION OF RNA
= a2 + b2 = 442865.89 = beginner transcription of RNA
= a2 + b2 = 442865.89 * 347.5 Sucrose = 153895896.775
= a2 + b2 = 153895896.775 / 135.13 Adenine = 1138872.91
= a2 + b2 = 1138872.91
Adenine + Guanine = Purine
BEGINNER TRANSCRIPTION OF RNA
Transcription of RNA = 1138872.91 / 511 ATP = 2228.71
= 2228.71 - 11.46 (C) = 2217.25
= 2217.25 g/mol = (COOH)- 49 of DNA
Carboxyl group occurs too much prepared of mRNA, tRNA, rRNA,
AMINO ACIDS OF mRNA
= a2 + b2 = 1138872.91 / 530.166 (A-T=C-G + H5 = 529.72) = 2148.14
= a2 + b2 = 2148.14 = (Polymerase enzyme of RNA)
= a2 + b2 = 2148.14 * 1.2961 (H) = 2784.21
= a2 + b2 = 2784.25 = 20 amino acids of mRNA from Polymerase enzyme
So Polymerase enzyme is jointed with (A-T=C-G)
tRNA FROM CYTOPLASM
= a2 + b2 = 1138872.91 * 7.4 pH blood plasma = 8427659.534
= a2 + b2 = 8427659.534 / 40 different cytoplasm in cell = 210691.48
= a2 + b2 = 210691.48 / 157.25 (Histidine) = 1339.85(amino molecules)
= a2 + b2 = 1339.85 - 1046.94 [(A-T=C-G)2] = 292.91
= a2 + b2 = 292.91 - 132.41 (PO4 + 2H2O = 132) = 160.5
= a2 + b2 = 160.5 (AMP) energy from Anaerobic bacteria + Two Tryptophan + 2H2O,
= a2 + b2 = 210691.48 = tRNA, mRNA, rRNA from process of cytoplasm
MESSENGER mRNA
= a2 + b2 = 210691.48 / 46.25 chromosomes = 4555.49
= a2 + b2 = 4555.49 information for the ribosomal (messenger mRNA)
INFORMATION OF MESSENGER = 4555.49
= a2 + b2 = 4555.49 * 149.25 g/mol (Gln) = 679906.8825
= a2 + b2 = 679906.8825 / 244.19% nucleated amino-acids move RNA = 2784.33
= a2 + b2 = 2784.25 = 20 amino acids of polypeptide
= a2 + b2 = 2784.25 * 244.19 Mononucleotide = 679886.0075
= a2 + b2 = 679886.0075 = molecular weight from 244.19 Mononucleotide
DAILY DIET
= a2 + b2 = 679886.0075 / 5 (H3 + H2 of DNA) = 135977.2015
= a2 + b2 = 135977.2015 – 36000 pepsin = 99977.2015
= a2 + b2 = 99977.2015 – 98000 Casein = 1977.2015
= a2 + b2 = 1977.2015 = Polypeptide
5 (H3 + H2 of DNA) = (A-T=C-G)
PROTEIN
= a2 + b2 = 679886.0075 * 23 pair chromosomes cell environment = 15637378.1725
= a2 + b2 = 15637378.1725 + 462.25 (Leucine + Gln + Purine) = 4593826.01975
= a2 + b2 = 15637840.4225 m/g = protein from Anaerobic bacteria + Two Tryptophan + 2H2O,
Total protein = 15637840.4225 / 64500 main hemoglobin = 242.44
= 242.44 / 15.57 times protein = 15.57 times vitamin-k from Anaerobic bacteria + Two Tryptophan + 2H2O, and we have already 244.19 phosphoric Mononucleotide for RNA DNA;
SYNTHESIS OF UMBILICAL CORD
Formula of medicines
Anaerobic bacteria + Two Tryptophan + 2H2O = perfect DNA RNA for produce vitamin-K and protein for health of intestine,
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