BIO ALGEBRA OF LIVING ORGANISMS

(Bio Mathematical Lab of Sikander Aqeel) 

CHAPTER [3] PROTEINS  

Proteins  

(3/18/2016) 

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                

          Lys                         Ala 

            |                            | 

          Val                         Gly 

            |                            | 

          Tyr                         Glu 

            |                            | 

          Pro                         Asp 

            |                            | 

      5- Asp                         Gln -5 ----second 10 amino acids sequence by Sulfur oxide  

Identical a chain          Identical b chain 

           Lysine = C6H14N2O2 = 149.5 m/g 

            Valine = C5H11NO2 = 119.75 m/g

         Tyrosine = C9H11NO3 = 183.75 m/g

            Proline = C5H9NO2 = 117.25 m/g 

    Aspartic acid = C4H7NO4 = 134.75 m/g 

                                Total = 705 = Identical a chain  

                   Alanine = C3H7NO2 = 90.75 m/g 

            Glycine = C2H5NO2 = 76.25 m/g

            Glutamate = C5H9NO4 = 149.25 m/g 

    Aspartic acid = C4H7NO4 = 134.75 m/g 

        Glutamine = C5H10N2O3 = 148.5 m/g 

                                Total = 599.5 = Identical b chain   

For example 

  = a2 + b2 = c2 

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

  = a2 = 705 and b2 = 599.5 = c2 

  = a705 + b599.5 = c2 

  = 497025 + 359400.25 = c2 

  = a + b = 856425.25 = c2  

  = a + b = 856425.25 / 925.4 = 925.4

  = a + b = 925.4 = Polypeptide

  = a + b = 925.4 / 64 (S2) = 14.459375 = cross-linked of Sulfur oxide

  = a + b = 14.459375 * 6.8 (Hemoglobin Iso pH) = 98.32375

  = a + b = 98.32375 + 13.67625 CH = 112 

  = a + b = 112 = Residues 

  = a + b = 925.4 / 64 (S2) = 14.459375 = cross-linked of Sulfur oxide

  = a + b = 14.459375 * 7 (Myoglobin Iso pH) = 101.215625

  = a + b = 101.215625 + 13.784375 CH = 115 

  = a + b = 115 = Residues 

  = a + b = 925.4 * 69.69 (2CH2OH + H5) = 64491.126

  = a + b = 64491.126 + 8.874 H7 = 64500 

  = a + b = 64500 = molecular weight of (Hemoglobin)     

  = a + b = 925.4 = Polypeptide

  = a + b = 925.4 / 20.45 (H2O + H2) = 45.27

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

  = a + b = 925.4 = Polypeptide

  = a + b = 925.4 / 21.03 (H2O + H2) = 44.00

  = a + b = 44.00 = carboxyl-terminal end

  = a + b = 925.4 = Polypeptide

  = a + b = 925.4 / 128 (S4) = 7.2296875

  = a + b = 7.2296875 - 0.4296875 (H) = 6.8 

  = a + b = 6.8 Iso-electric pH of hemoglobin 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 / 925.4 Polypeptide = 18037.79

                                    = 18038 / 44 Carboxyl-terminal = 409.95

                                    = 410 day biological system of earth and hemoglobin, 

Supposed a cell function = 16692180 numbers / 925.4 Polypeptide = 18037.79

                                    = 18038 / 1752 light sec = 10.29

                                    = 10 = age

                                    = 10 = year age of protein of hemoglobin, 

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