(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 (11/17/2016)
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 = Stearic acid = 284.47 = C18H36O2
2 = 20 basic amino acids = 2784.25 m/g
= a2 / b2 = c2
= a2 (20 basic amino acids) / b2 (Stearic acid) = c2
= a2 (2784.25) / b2 (284.47) = c2
= a / b = (7752048.0625) / b (80923.1809) = c2
= a / b = 95.79 = c2
= a / b = 95.79 = PO4 = phosphate
DNA STRUCTURE
1 = Stearic acid = 284.47
2 = 20 basic amino acids = 2784.25 m/g
= a2 + b2 = c2
= a2 (20 basic amino acids) / b2 (Stearic acid) = c2
= a2 (2784.25) / b2 (284.47) = c2
= a + b = (7752048.0625) + b (80923.1809) = c2
= a + b = 7832971.2434 = c2
= a + b = 7832971.2434 / 2798.74 = 2798.75
= a + b = 2798.75 = polypeptide
= a + b = 2798.75 - 135.13 g mol Adenine = 2663.62
= a + b = 2663.62 / 40 amino acid of polypeptide = 66.5905
= a + b = 66.59 - 3.09 H3 = 63.5
= a + b = 63.5 = fructose = CH2OH
= a + b = 63.5 = fructose of smaller polypeptide
= a + b = 2798.75 / 2.5 (H2= A//T DNA) = 1119.5
= a + b = 1119.5 / 3.75 (H3= C///G DNA) = 298.53
= a + b = 298.53 / 7.4 (pH of blood plasma) = 40.34
= a + b = 40.34 + 4.91 H4 = 45.25
= a + b = 45.25 = COOH carboxylic acid of Stearic acid
Fructose + 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 = 284.479.5
= a + b = 284.479.5 / 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 Stearic acid
Fructose + carboxylic acid of Stearic acid = A..
CHEMICAL STRUCTURE OF DNA (11/17/2016)
DNA STRUCTURE
1 = Stearic acid = 284.47
2 = Adenine = 135.13 g mol
= a2 + b2 = c2
= a2 (Adenine) / b2 (Stearic acid) = c2
= a2 (135.13) / b2 (284.47) = c2
= a + b = (18260.1169) + b (80923.1809) = c2
= a + b = 99183.2978 = c2
= a + b = 99183.2978 / 314.93 = 314.93
= a + b = 314.93 - 126.11 g mol Thymine = 188.82
= a + b = 188.82 - 5.82 H5 = 183
= a + b = 183 = glucose = C6H12O6
= a + b = 183 = glucose of smaller polypeptide
= a + b = 314.93 / 2.5 (H2= A//T DNA) = 125.972
= a + b = 125.972 / 3.75 (H3= C///G DNA) = 33.59
= a + b = 33.59 / 7.4 (pH of blood plasma) = 4.53
= a + b = 5 = optimum pH of stomach
Fructose + carboxylic acid = A..T
Fructose + 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 Stearic acid from in 18 carbon atoms
Fructose + carboxylic acid of Stearic acid = A..T
Fructose + carboxylic acid of Stearic acid = A//T
CHEMICAL STRUCTURE OF DNA (11/17/2016)
DNA STRUCTURE
1 = Stearic acid = 284.47
2 = Thymine = 126.11 g mol
= a2 + b2 = c2
= a2 (Thymine) / b2 (Stearic acid) = c2
= a2 (126.11) / b2 (284.47) = c2
= a + b = (15903.7321) + b (80923.1809) = c2
= a + b = 96826.913 = c2
= a + b = 96826.913 / 311.17 = 311.17
= a + b = 311.17 - 111.1 g mol Cytosine = 200.07
= a + b = 200.07 - 151.13 g mol Guanine = 48.94
= a + b = 48.94 = COOH + H3 = carboxyl acid + hydrogen
= a + b = 48.94 = carboxyl acid + hydrogen = COOH + H3
= a + b = 48.94 = carboxyl acid + hydrogen of smaller polypeptide
= a + b = 290.32 / 2.5 (H2= A//T DNA) = 116.128
= a + b = 116.128 / 3.75 (H3= C///G DNA) = 30.96
= a + b = 30.96 / 7.4 (pH of blood plasma) = 4.18
= a + b = 5 = optimum pH of stomach
Nitrogen of pancreatic beta cells = A..T = C...G
Nitrogen of pancreatic beta cells = 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
Nitrogen + carboxylic acid of Stearic acid = A..T = C...G
Nitrogen + carboxylic acid of Stearic acid = 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,
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