Sikander Aqeel

BIO ALGEBRA OF LIVING ORGANISMS

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(Bio Mathematical Lab of Sikander Aqeel) 

CHAPTER [3] PROTEINS  

Proteins  

TECHNIQUES UTILIZED TO DETERMINE THE AMINO ACID SEQUENCE OF A PROTEIN  

Knowledge of the primary structure (amino acid sequence) of a protein is required for an understanding of the relationship of a protein’s structure to its function on a molecular level, 

In the determination of primary structure, first the number of polypeptide chain in the protein must be ascertained, to begin with the protein is denatured, and the treated with a reagent, such as Insulin, that forms a covalent bond with the NH2-termina a-amino groups of each polypeptide chain within the protein the tagged protein is then hydrolyzed to its constituent amino acids, typically condition for complete protein hydrolysis are 6 N HCl at 110 centigrade for 195 to 36 hours, in a sealed tube under vacuum, the vacuum prevents degradation of oxidation sensitive, 

 

Since most polypeptide chain in proteins contain more then 30 to 40 amino acids, they have to be hydrolysed into smaller fragments and sequenced in section, 

Both enzymatic and chemical methods are used to break polypeptide chain into smaller polypeptide fragments, for example we are making polypeptide from smaller 20 basic amino acids, or we are going in large polypeptide from small polypeptide,         

 

CHEMICAL STRUCTURE OF DNA (11/22/2016) 

DNA STRUCTURE 

For example we are making polypeptide from smaller 20 basic amino acids, or we are going in large polypeptide from small polypeptide,

 

1 = Insulin = 254 mole 

2 = 20 basic amino acids = 2784.25 m/g 

 

  = a2 / b2 = c2 

  = a2 (20 basic amino acids) / b2 (Insulin) = c2 

  = a2 (2784.25)              / b2 (254 mole) = c2   

 

  = a / b = (7752048.0625) / b (64516) = c2 

  = a / b = 120.15 = c2 

  = a / b = 120.15 = C5H4N4 = Purine

 

DNA STRUCTURE 

1 = Insulin = 254 mole 

2 = 20 basic amino acids = 2784.25 m/g 

 

  = a2 + b2 = c2 

  = a2 (20 basic amino acids) / b2 (Insulin) = c2 

  = a2 (2784.25)              / b2 (254 mole) = c2   

 

  = a + b = (7752048.0625) + b (64516) = c2 

  = a + b = 7816564.0625 = c2 

  = a + b = 7816564.0625 / 2795.81 = 2795.81

 

  = a + b = 2795.81 = polypeptide

 

 

  = a + b = 2795.81 - 135.13 g mol Adenine = 2660.68

  = a + b = 2660.68 / 51 amino acid of Insulin = 52.17

  = a + b = 52.17 = (CH2OH)1 + H2O + H = fructose + water of smaller polypeptide  

 

 

  = a + b = 2795.81 / 2.5 (H2= A//T DNA) = 1118.324

  = a + b = 1118.324 / 3.75 (H3= C///G DNA) = 298.21

  = a + b = 298.21 / 7.4 (pH of blood plasma) = 40.29

  = a + b = 40.29 + 4.96 (H4) = 45.25

  = a + b = 45.25 = (COOH) carboxylic acid of Insulin

 

Fructose + fourteen carboxylic acid = A..  

 

HUMAN CHROMOSOMES 

DNA STRUCTURE 

1 = 20 basic amino acids = 2784.25 mole

2 = Human Chromosomes = 46.25 

 

  = a2 + b2 = c2 

  = a2 (20 basic amino acids) + b2 (Chromosomes) = c2 

  = a2 (2784.25)              + b2 (46.25) = c2 

 

  = a + b = (7752048.0625) + (2139.0625) = c2 

  = a + b = 7754187.125 = c2 

  = a + b = 7754187.125 / 2784.63 = 2784.63 

 

  = a + b = 2784.63 = polypeptide

 

  = a + b = 2784.63 - 135.13 g mol Adenine = 254 mole9.5

  = a + b = 254 mole9.5 / 40 amino acids of polypeptide = 66.2375

  = a + b = 66.23 - 2.73 H2 = 63.5

  = a + b = 63.5 = fructose = CH2OH 

  = a + b = 63.5 = fructose of smaller polypeptide 

 

 

  = a + b = 2784.63 / 2.5 (H2= A//T DNA) = 1113.852

  = a / b = 1113.852 / 3.75 (H3= C///G DNA) = 297.0272

  = a / b = 297.0272 / 8.7 (pH of blood plasma) = 40.13

  = a + b = 40.13 + 5.12 H4 = 45.25 

  = a + b = 45.25 = COOH carboxylic acid of Insulin

 

Fructose + carboxylic acid of Insulin = A.. 

 

 

CHEMICAL STRUCTURE OF DNA (11/22/2016) 

DNA STRUCTURE 

1 = Insulin = 254 mole  

2 = Adenine = 135.13 g mol

 

  = a2 + b2 = c2 

  = a2 (Adenine) / b2 (Insulin) = c2 

  = a2 (135.13)  / b2 (254 mole) = c2   

 

  = a + b = (18260.1169) + b (64516) = c2 

  = a + b = 82776.1169 = c2 

  = a + b = 82776.1169 / 287.70 = 287.71

 

  = a + b = 287.71 - 126.11 g mol Thymine = 161.6

  = a + b = 161.6 + 21.4 (H2O + H2) = 183

  = a + b = 183 = C6H12O6 glucose of smaller polypeptide 

 

 

  = a + b = 287.71 / 2.5 (H2= A//T DNA) = 115.084

  = a + b = 115.084 / 3.75 (H3= C///G DNA) = 30.68

  = a + b = 30.68 / 7.4 (pH of blood plasma) = 4.14

  = a + b = 4.14 = H3 

 

Hydrogen + carboxylic acid = A..T 

Hydrogen + carboxylic acid = A//T 

 

 

HUMAN CHROMOSOMES 

DNA STRUCTURE 

1 = Adenine = 135.13 m/g 

2 = Human Chromosomes = 46.25 

 

  = a2 + b2 = c2 

  = a2 (Adenine) + b2 (Chromosomes) = c2 

  = a2 (135.13)   + b2 (46.25) = c2 

 

  = a + b = (18260.1169) + (2139.0625) = c2 

  = a + b = 20399.1794 = c2 

  = a + b = 20399.1794 / 142.82 = 142.83 

 

  = a + b = 142.83 - 126.11 g mol Thymine = 16.72

  = a + b = 16.72 + 0.53 = 17.25

  = a + b = 17.25 = alcohol of smaller polypeptide, 

 

 

  = a + b = 142.83 / 2.5 (H2= A//T DNA) = 57.132

  = a / b = 57.132 / 3.75 (H3= C///G DNA) = 15.2352

  = a / b = 15.2352 / 1.269 (pH of pepsin) = 12.00

  = a + b = 12.00 = one Carbon atom of Insulin from in 18 carbon atoms

 

Fructose + carboxylic acid of Insulin = A..T 

Fructose + carboxylic acid of Insulin = A//T 

 

 

CHEMICAL STRUCTURE OF DNA (11/22/2016) 

 

DNA STRUCTURE 

1 = Insulin = 254 mole 

2 = Thymine = 126.11 g mol

 

  = a2 + b2 = c2 

  = a2 (Thymine) / b2 (Insulin) = c2 

  = a2 (126.11) / b2 (254 mole) = c2   

 

  = a + b = (15903.7321) + b (64516) = c2 

  = a + b = 80419.7321 = c2 

  = a + b = 80419.7321 / 283.58 = 283.58

 

  = a + b = 283.58 - 111.1 g mol Cytosine = 172.48

  = a + b = 172.48 - 151.13 g mol Guanine = 21.35

  = a + b = 21.35 = (H2O + H2) hydrogen and water of smaller polypeptide 

 

 

  = a + b = 283.58 / 2.5 (H2= A//T DNA) = 113.432

  = a + b = 113.432 / 3.75 (H3= C///G DNA) = 30.24

  = a + b = 30.24 / 7.4 (pH of blood plasma) = 4.08

  = a + b = 4.08 = H3

 

carboxyl acid + hydrogen = A..T = C...G  

carboxyl acid + hydrogen = A//T = C///G 

 

HUMAN CHROMOSOMES 

DNA STRUCTURE 

1 = Guanine = 151.13 m/g 

2 = Human Chromosomes = 46.25 

 

  = a2 + b2 = c2 

  = a2 (Guanine) + b2 (Chromosomes) = c2 

  = a2 (151.13)   + b2 (46.25) = c2 

 

  = a + b = (22840.2769) + (2139.0625) = c2 

  = a + b = 24979.3394 = c2 

  = a + b = 24979.3394 / 158.04 = 158.05

 

  = a + b = 158.05 - 111.1 g mol Cytosine = 46.95

  = a + b = 46.95 - 1.7 H = 45.25 

  = a + b = 45.25 = COOH

  = a + b = 45.25 = carboxyl acid of smaller polypeptide, 

 

 

  = a + b = 158.05 / 2.5 (H2= A//T DNA) = 63.22

  = a / b = 63.22 / 3.75 (H3= C///G DNA) = 16.85

  = a / b = 16.85 / 1.20 (pH of pepsin) = 14.04

  = a + b = 14.04 = nitrogen atom of acids

 

Chromosomes + carboxyl acid + hydrogen = A..T = C...G 

Chromosomes + carboxyl acid + hydrogen = A//T = C///G 

 

The DNA structure takes less time to joint with each another in presence of fatty acids, or it is just a effect of fatty acids on the mRNA for make proteins of fat people,  

 

Lowers blood glucose level, promotes glucose storage as glycogen and fat. Fasting decreases insulin production. 

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