BIO ALGEBRA OF LIVING ORGANISMS

(Bio Mathematical Lab of Sikander Aqeel) 

CHAPTER [3] PROTEINS  

Proteins  

(3/22/2016) 

Amino Acid Sequence of Polypeptide Chain:  

The next important landmark was the identification of the amino acid sequence of the two types polypeptide in crystalline hemoglobin, 

This was the first sequence **ysis of oligomeric or multichain proteins, 

The hemoglobin contain on 4 polypeptide chain  

Polypeptide = identical a chain = 141 residues 

Polypeptide = identical b chain = 146 residues    

        The Human Adrenocorticotrophin and S – S linkage 

1 polypeptide = Ser, Tyr, Ser, Met, Glu, His, Phe, Arg, Trp, Gly = 10 amino acids 

2 polypeptide = Lys, Pro, Val, Gly, Lys, Lys, Arg, Arg, Pro, Val = 10 amino acids 

3 polypeptide = Lys, Val, Tyr, Pro, Asp, Ala, Gly, Glu, Asp, Gln = 10 amino acids 

4 polypeptide = Ser, Ala, Glu, Ala, Phe, Pro, Leu, Glu, Phe, = 9 amino acids 

The longest polypeptide chains for which complete amino acids sequence have been deduced to date are those of bovine trypsinogen (229 residues) and bovine chymotrypsinogen (245 residues) and the polypeptide chain of glyceraldehydes 3-phosphate dehydrogenase of lobster muscles (333 residue) Although the amino acids sequence of over 50 different proteins are known,      

        The Bovine ribonuclease and S – S linkage 

1 polypeptide = Lys, Glu, Thr, Ala, Ala, Ala, Lys, Phe, Glu, Arg = 10 amino acids 

          Lys                         Ala                         

            |                            |                            

          Glu                         Lys 

            |                            | 

          Thr                         Phe 

            |                            | 

          Ala                         Glu 

            |                            | 

      5- Ala                        Arg -5 ----second 9 amino acids sequence by Sulfur oxide  

Identical a chain          Identical b chain 

                Lysine = C6H14N2O = 149.5 m/g 

            Glutamate = C5H9NO4 = 149.25 m/g 

                    Threonine = C4H9NO3 = 121.25 m/g

                Alanine = C3H7NO2 = 90.75 m/g 

                Alanine = C3H7NO2 = 90.75 m/g 

                                    Total = 601.5 = Identical a chain  

                Alanine = C3H7NO2 = 90.75 m/g 

                Lysine = C6H14N2O = 149.5 m/g 

     Phenylalanine = C9H11NO2 = 167.75 m/g 

            Glutamate = C5H9NO4 = 149.25 m/g 

                    Arginine = C6H14N4O2 = 177.5 m/g

                                    Total = 734.75 = Identical b chain   

For example 

  = a2 + b2 = c2 

  = a2 = Identical a chain and b2 = Identical b chain = c2 

  = a2 = 601.5 and b2 = 734.75 = c2 

  = a601.5 + b734.75 = c2 

  = 361802.25 + 539857.5625 = c2 

  = a + b = 901659.8125 = c2  

  = a + b = 901659.8125 / 949.5 = 949.5

  = a + b = 949.5 = Polypeptide

  = a + b = 949.5 / 64 (S2) = 14.607 = cross-linked of Sulfur oxide

  = a + b = 14.607 * 6.8 (Hemoglobin Iso pH) = 99.3276

  = a + b = 99.3276 * 2.305 (H2 hydrolyzed)) = 228.95 

  = a + b = 229 = Residues of bovine trypsinogen  

  = a + b = 949.5 / 64 (S2) = 14.835 = cross-linked of Sulfur oxide

  = a + b = 14.835 * 9.5 (trypsinogen Iso pH) = 140.9325

  = a + b = 140.9325 * 1.738 (H1hydrolyzed)) = 244.94 

  = a + b = 245 = Residues of chymotrypsinogen 

  = a + b = 949.5 / 64 (S2) = 14.835 = cross-linked of Sulfur oxide

  = a + b = 14.835 * 9.5 (trypsinogen Iso pH) = 140.9325

  = a + b = 140.9325 * 2.362 (H2 hydrolyzed) = 332.88 

  = a + b = 333 = Residues of lobster muscles 

Table: for little age and ** age

Little age = 229 amino acids for bovine trypsinogen 

Little age = 245 amino acids for chymotrypsinogen 

Little age = 333 amino acids for lobster,      

  = a + b = 949.5 * 6.037 (protons) = 5732.1315

  = a + b = 5732.1315 + 0.8685 = 5733 

  = a + b = 5733 = molecular weight of (Bovine Insulin) 

  = a + b = 949.5 * 13.312 (CH) = 12639.744

  = a + b = 12639.744 + 0.256 = 12640 

  = a + b = 12640 = molecular weight of (Bovine Pancreas) 

  = a + b = 949.5 * 353.87 (11CH2OH) = 335999.565

  = a + b = 335999.565 + 0.435 = 336000 

  = a + b = 336000 = molecular weight of (Bovine Liver)          

  = a + b = 949.5 = Polypeptide

  = a + b = 949.5 / 20.98 (H2O + H2) = 45.25

  = a + b = 45.25 = COOH = carboxyl-group, 

  = a + b = 949.5 = Polypeptide

  = a + b = 949.5 / 21.57 (H2O + H3) = 44.00

  = a + b = 44.00 = carboxyl-terminal end

  = a + b = 949.5 = Polypeptide

  = a + b = 949.5 / 128 (S4) = 7.41796875

  = a + b = 7.41796875 + 2.08203125 = 9.5

  = a + b = 9.5, Iso-electric pH of chymotrypsinogen by 4 sulfur linkage    

  Carboxyl-group = COOH = 45.25 * 12 activate carbon atom = 543 

                                      = 543 + 49 [3(OH ion)] = 592 

                                      = 592 mole Original peptide 

Supposed a cell function = 16692180 numbers / 949.5 Polypeptide = 17579.96

                                    = 17580 / 44 Carboxyl-terminal = 399.54

                                    = 400 day biological system of earth and bovine, 

Supposed a cell function = 16692180 numbers / 949.5 Polypeptide = 17579.96

                                    = 17580 / 1752 light sec = 10.03

                                    = 10 = age

                                    = 10 = year age of protein of hemoglobin, 

This feat was carried out by groups of different algebra of Sikander Aqeel,