Sikander Aqeel

BIO ALGEBRA OF LIVING ORGANISMS

Nov 4th 2016, 11:10 am
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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 pectin enzyme found in fruits and vegetables       

 

CHEMICAL STRUCTURE OF DNA (11/5/96) 

DNA STRUCTURE 

1 = phosphate = 95 mole = PO4 

2 = 20 amino acids = 2784.25 m/g 

 

  = a2 / b2 = c2 

  = a2 (20 amino acids) / b2 (phosphate) = c2 

  = a2 (2784.25)        / b2 (95) = c2   

 

  = a / b = (7752048.0625) / b (9025) = c2 

  = a / b = 858.95 = c2 

  = a / b = 859 = H1  

 

DNA STRUCTURE 

1 = phosphate = 95 mole = PO4 

2 = Alanine = 90.75 m/g 

 

  = a2 + b2 = c2 

  = a2 (20 amino acids) / b2 (phosphate) = c2 

  = a2 (2784.25)        / b2 (95) = c2   

 

  = a + b = (7752048.0625) + b (9025) = c2 

  = a + b = 7761073.0625 = c2 

  = a + b = 7761073.0625 / 2785.87 = 2785.87

 

  = a + b = 2785.87 - 126.11 g mol−1 Thymine = 2659.76

  = a + b = 2659.76 / 20 = 132.988 

  = a + b = 132.988 + 1.012 = 134 

  = a + b = 134 = Asparagine = C4H8N2O3 

 

 

  = a + b = 2785.87 / 2.5 (H2= A//T DNA) = 1114.348

  = a + b = 1114.348 / 3.75 (H3= C///G DNA) = 297.15

  = a + b = 297.15 / 8.7 (pH of pectic acid of pectin enzyme) = 34.15

  = a + b = 34A = height of helix 

 

Asparagine = Asn(N)  = AAU AAC 

 

HUMAN CHROMOSOMES 

DNA STRUCTURE 

1 = 20 amino acids = 2784.25 m/g 

2 = Human Chromosomes = 46.25 

 

  = a2 + b2 = c2 

  = a2 (20 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 - 126.11 g mol−1 Thymine = 2658.52 

  = a + b = 2658.52 / 20 amino acids = 132.926

  = a + b = 132.926 + 1.074 = 134

  = a + b = 134 = Asparagine = C4H8N2O3

 

 

  = 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 pectic acid of pectin enzyme) = 34.14

  = a + b = 34A = height of helix 

 

Asparagine = Asn(N)  = AAU AAC = fruits + vegetables  

 

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