(Bio Mathematical Lab of Sikander Aqeel)
CHAPTER [3] PROTEINS
Proteins
(3/16/2016)
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
2 polypeptide = identical a chain = 141 residues
2 polypeptide = identical b chain = 146 residues
The Human Adrenocorticotrophin and S – S linkage
Ser His
| |
Tyr Phe
| |
Ser Arg
| |
Met Trp
| |
5- Glu Gly -5 ----first 10 amino acids sequence by Sulfur oxide
Identical a chain Identical b chain
Serine = C3H7NO3 = 106.75 m/g
Tyrosine = C9H11NO3 = 183.75 m/g
Serine = C3H7NO3 = 106.75 m/g
Methionine = C5H11NO2S = 151.75 m/g
Glutamate = C5H9NO4 = 149.25 m/g
Total = 698.25 = Identical a chain
Histidine = C6H9N3O2 = 157.25 m/g
Phenylalanine = C9H11NO2 = 167.75 m/g
Arginine = C6H14N4O2 = 177.5 m/g
Tryptophan = C11H12N2O2 = 207 m/g
Glycine = C2H5NO2 = 76.25 m/g
Total = 785.75 = Identical b chain
For example
= a2 + b2 = c2
= a2 = Identical a chain and b2 = Identical b chain = c2
= a2 = 698.25 and b2 = 785.75 = c2
= a698.25 + b785.75 = c2
= 487553.0625 + 617403.0625 = c2
= a + b = 1104956.125 = c2
= a + b = 1104956.125 / 1051.18 = 1051.18
= a + b = 1051.18 = two Polypeptide
= a + b = 1051.18 / 64 (S2) = 16.4246875 = cross-linked of Sulfur Oxide
= a + b = 16.4246875 * 6.8 (Hemoglobin Iso pH) = 111.68
= a + b = 112 = Residues
= a + b = 1051.18 / 64 (S2) = 16.4246875 = cross-linked of Sulfur Oxide
= a + b = 16.4246875 * 7 (Myoglobin Iso pH) = 114.9728125
= a + b = 115 = Residues
= a + b = 1051.18 * 61.35 (1.9CH2OH) = 64489.893
= a + b = 64489.893 + 10.107 = 64500
= a + b = 64500 = molecular weight of (Hemoglobin)
= a + b = 1051.18 = Polypeptide
= a + b = 1051.18 / 23.23 (H2O + H4) = 45.25
= a + b = 45.25 = COOH = carboxyl-group,
= a + b = 1051.18 = Polypeptide
= a + b = 1051.18 / 23.89 (H2O + H4) = 44.00
= a + b = 44.00 = carboxyl-terminal end
= a + b = 1051.18 = Polypeptide
= a + b = 1051.18 / 128 (S4) = 8.21234375
= a + b = 8.21234375 - 1.41234375 (H) = 6.8
= a + b = 6.8 Iso-electric pH of hemoglobin
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 / 1051.18 Polypeptide = 15879.46
= 15879 / 44 Carboxyl-terminal = 360.8
= 361 day biological system of earth and hemoglobin,
Supposed a cell function = 16692180 numbers / 1051.18 Polypeptide = 15879.46
= 15879 / 1752 light sec = 9.06
= 9 = age
= 9 = year age of protein by process of hemoglobin,
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