(Bio Algebra of living organisms)
(Mathematical Lab of Sikander Aqeel)
CHAPTER 3 PROTEINS
Proteins = Clay
(1/21/2016)
The Size of Proteins Molecules
The molecular weight of proteins can be determined by means of various physical-chemical methods, some characteristic values, most range from about 12,000 to 1,000,000 or more, it is not possible even among proteins having the same type of function, and different enzymes, for example have molecular weight that very over a wide range,
we can calculate the approximate number of amino acid residues in a sample protein containing on prosthetic group by dividing its molecular weight by 120,
The average molecular weight of 20 different amino acids in proteins is about 138,
But since a molecule of water [mol wt 18.0 is removed to create each Peptide bond,
The average amino acid residue weight about 120, also gives the approximate number of amino acids residues for some proteins of different size and function,
Ribonuclease, Cytochrome c, and myoglobin, which are among the best known small proteins contain between 100 and 155 amino acids residues,
Arginine = C6H14N4O2 = 177.5 m/g
Isoleucine = C6H13NO2 = 134.25 m/g
Leucine = C6H13NO2 ¬= 134.25 m/g
Valine = C5H11NO2 = 119.75 m/g
Total = 565.75 m/g
A = for example
= a2 + b2 = c2
Molecular weight residues No, of chain
Insulin (bovine) ----------- 5733 ------------ 51-------------- 2 ------
Four amino acids = C23H51N4O8 = 565.75 g mol
Molecule of water = 18.0 mol
a2 + b2 = c2
a = Four amino acids = C23H51N4O8 = 565.75 m/g
b = Molecule of water = Zn1 = 18.0 mol
= a = C23H51N4O8 and b = water
= a = 565.75 and b = 65
= a = 565.75 mm height, and b = 18.0 mm bottom = c2
= a2 + b2 = c2
= a2 + b2 = c2 = C23H51N4O8 + water = c2
= a2 + b2 = c2 = a565.75 mm + b18.0 mm = c2
= a565.75 mm + b18.0 mm = c2
= a320073.0625 + b324 = c2
= a + b = 320397.0625 = c2
= a + b = 320397.0625 / 566 = 566
= c = 566 slope / 20 amino acids = 28.3 half residues
= 28.3 * 2 No, of chain = 56.6 complete residues
= c = 566 slope / 20 amino acids = 28.3 half residues
= 28.3 * 2 No, of chain = 56.6 complete residues
Small proteins contain between = 100 and 155
Small proteins contain between = 101.14 * 56.6 complete residues = 5724.524
= 5724.524 + 8.476 = 5733
= 5733 Molecular weight of Insulin (bovine)
Molecular weight residues No, of chain
Insulin (bovine) ----------- 5733 ------------ 51-------------- 2 ------
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
r = Four amino acids = C23H51N4O8 = 565.75 m/g
r = Molecule of water = Zn1 = 18.0 mol
As plus Rate Time Glycogen
n = 16.8 O r + 565.75 60 minute 60(r + 565.75)
Four amino acids dehydrogenises protein
60r + 60(r + 565.75) = 67890
60r + 60 + 33945 = 67890
120r + 33945 = 67890
120r + 33945 + 67890= 67890- 33945
120r = 33945
120r = 33945
----------- = -------------
120 = 120
r = 282.875
r = 282.875 / 16.8 Oxygen = 16.8
r = 16.8 + 1.2 = 18.0 mol water removed to create each peptide chain
Molecular weight residues No, of chain
Insulin (bovine) ----------- 5733 ------------ 51-------------- 2 ------
Always remember my words,
Enzymes = Oxygen
Bacteria = Nitrogen
Virus = Carbon +
-----------------
= life by air, fire clay, and water,
And hydrogen is a partner of these three elements,
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