VITAMIN-A AND 120 DAY LIFE OF RED CELLS

(Two Polynucleotide in mathematical machine of Sikander Aqeel) 

WILSON DISEASE 

Lower-than-normal ALP levels 

• 1 = Hypophosphatasia 

• 2 = Malnutrition 

• 3 = Wilson disease 

•

• Wilson disease has a range of clinical manifestations, from an asymptomatic state to fulminate hepatic failure, chronic liver disease with or without cirrhosis, neurologic, and psychiatric manifestations

• Consider hepatic Wilson disease in the differential diagnosis of any unexplained chronic liver disease, especially in individuals younger than 40 years. The condition may also manifest as acute hepatitis. Hepatic dysfunction is the presenting feature in more than half of patients. The 3 major patterns of hepatic involvement are as follows: 

• (1) chronic active hepatitis, 

• (2) cirrhosis, and 

• (3) fulminant hepatic failure. The most common initial presentation is cirrhosis.  

(1) CHRONIC ACTIVE HEPATITIS (8/9/2016) 

(2) JAUNDICE TO BE SHIFTED IN TYPHOID 

the hepatitis is called the Inflammation of liver like (Jaundice),  the red blood cells life occurs 120 day in blood by force or charge of Hydrogen, 

The Jaundice is not dangerous disease, but as Jaundice is transfer in typhoid then Jaundice may reason of death, and from this reason people quickly treatment of Jaundice, 

The Jaundice till the long time is equal to reduction of Vitamin-A or retinol, and vitamin-A is most important for whole healthy body, 

Reduction of Vitamin-A 

Causes of broken tissues, Inflammation, many kinds of infections, mostly nervous infections, too cold, reduction of appetite, to lost ability of expel moisture, and disorder of smell and taste,       

1 = Hydrogen = 5 photons force * 24 hours = 120 day life = normal condition 

2 = Hydrogen = 2 photons force * 24 hours = 48 day life of red blood cell = (Jaundice) 

3 = Hydrogen = 0 photons force * 24 hours = 24 day life = heavy (Jaundice) 

1 = VITAMIN-A AND 120 DAY LIFE OF RED CELLS 

1 = 75 % VITAMIN-A OR RETINOL     

Retinol = C20H30O = 293.5 + 120 Centigrade temperature of fiver = 413.5 

Red blood cells healthy life = 120 day 

Retinol + fiver = 413.5 / 100 % = 4.135

                = 3.0425 * 75 % = 310.125

                = 310.125 = 75 percent Retinol fiver  

    = a2 / b2 = c2 

    = a2 (75 % Retinol fiver) / b2 (Red cells life) = c2 

    = a2 (310.125) / b2 (120 day) = c2 

    = a / b = (96177.515625) / b2 (14400) = c2 

    = a / b = 6.67 + 0.23 = 6.9 pH of liver

    = a / b = 6.1 = pH of muscles 

VITAMIN-A AND JAUNDICE FIVER  

75 % Retinol = 310.125

Red blood cells healthy life = 120 day 

    = a2 + b2 = c2 

    = a2 (75 % Retinol fiver) + b2 (Red cells life) = c2 

    = a2 (310.125) + b2 (120 day) = c2 

    = a2 (96177.515625) + b2 (14400) = c2  

    = a + b = 110577.515625 = c2 

    = a + b = 110577.515625 = c2 

    = a + b = 110577.515625 / 332.53 = 332.53

    = a + b = 332.53 / 2.5 (H2= A//T DNA) = 133.012

    = a + b = 133.012 / 3.75 (H3= C///G DNA) = 35.46

    = a + b = 35.5 = Fluid of cell = (NH3)2 = 35.5  

    = a + b = 35.5 = two Ammonia    

CONDITION OF DNA    

    = a2 + b2 = c2 

    = a2 (75 % Retinol fiver) + b2 (Red cells life) = c2 

    = a2 (310.125) + b2 (120 day) = c2 

    = a2 (96177.515625) + b2 (14400) = c2  

    = a + b = 110577.515625 = c2 

    = a + b = 110577.515625 = c2 

    = a + b = 110577.515625 / 332.53 = 332.53

    = a + b = 332.53 / 2.5 (H2= A//T DNA) = 133.012

    = a + b = 133.012 / 2.939 H2 = 45.25 (H = 2.939 * 21.435(H2O + H3) = 63 k cal/mole)  

    = a + b = 45.25 = COOH = carboxylic acid,  

CONDITION OF DNA  

    = a / b = 332.53 / 2.5 (H2= A//T DNA) = 133.012

    = a / b = 133.012 / 3.75 (H3= C///G DNA) = 35.46

    = a / b = 35.46 * 50 ratio = 1773

    = a / b = 1773 / 42.10A = 42.11A

    = a / b = 42A = distance of one helix  

    = red cell = 42A / 10 mononucleotide = 4.2

    =          = 4.2A distance between two base pair, 

So here jaundice fiver depend on 4.2A height of helix and 4.2A distance between two base pair, jaundice swelling to helix from 120 fiver, 

         Normal red cells life = 120 day / 3.4A normal helix = 35 Chlorine 

         Normal red cells life = 120 day / 4.2A sick helix = 28.5 two nitrogen, one nitrogen of ammonia and one nitrogen of urea, 

1 = VITAMIN-A AND 120 DAY LIFE OF RED CELLS (8/11/2016) 

1 = 61.49 % VITAMIN-A OR RETINOL  

The less vitamin-A than the 61.49 % causes of all Wilson disease  

Retinol = C20H30O = 293.5 

Red blood cells healthy life = 120 day 

Retinol = 293.5 / 100 % = 2.935

        = 2.935 * 61.49 % = 180.47315

        = 180.5 m/g = 61.49 percent Retinol,  

    = a2 / b2 = c2 

    = a2 (61.49 % Retinol) / b2 (Red cells life) = c2 

    = a2 (180.5) / b2 (120 day) = c2 

    = a / b = (32580.25) / b2 (14400) = c2 

    = a / b = 2.26 optimum pH of pepsin of stomach

VITAMIN-A AND HEALTH  

61.49 % Retinol = 180.5

Red blood cells healthy life = 120 day 

    = a2 + b2 = c2 

    = a2 (61.49 % Retinol) + b2 (Red cells life) = c2 

    = a2 (180.5) + b2 (120 day) = c2 

    = a2 (32580.25) + b2 (14400) = c2  

    = a + b = 46980.25 = c2 

    = a + b = 46980.25 = c2 

    = a + b = 46980.25 / 216.74 = 216.75

    = a + b = 216.75 / 2.5 (H2= A//T DNA) = 86.7

    = a + b = 86.7 / 3.75 (H3= C///G DNA) = 23.12

    = a + b = 23.12 = Fluid of cell = (NH2 + H2 + H2 + H = 22.75) = 35.5  

    = a + b = 23.12 = amino group and hydrogen    

CONDITION OF DNA    

    = a2 + b2 = c2 

    = a2 (61.49 % Retinol) + b2 (Red cells life) = c2 

    = a2 (180.5) + b2 (120 day) = c2 

    = a2 (32580.25) + b2 (14400) = c2  

    = a + b = 46980.25 = c2 

    = a + b = 46980.25 = c2 

    = a + b = 46980.25 / 216.74 = 216.75

    = a + b = 216.75 / 2.5 (H2= A//T DNA) = 86.7

    = a + b = 86.7 / 1.916 H2 = 45.25 (H = 1.916 * 32.881(O2 + H) = 63 k cal/mole)  

    = a + b = 45.25 = COOH = carboxylic acid,  

CONDITION OF DNA  

    = a / b = 216.75 / 2.5 (H2= A//T DNA) = 86.7

    = a / b = 86.7 / 3.75 (H3= C///G DNA) = 23.12

    = a / b = 23.12 * 50 ratio = 1156

    = a / b = 1156 / 34A = 34A

    = a / b = 34A = distance of one helix  

    = red cell = 34A / 10 mononucleotide = 3.4

    =          = 3.4A distance between two base pair, 

So 61.49 % Retinol depend on 3.4A height of helix and 3.4A distance between two base pair, 61.49 % Retinol doing healthy of body, 

         Normal red cells life = 120 day / 3.4A normal helix = 35 Chlorine 

ALP - BLOOD TEST        

Normal Results (ALP) tissues protein

The normal range is 44 to 147 IU/L (international units per liter) 

The normal range = 147 IU/L for maximum weight and young age    

The normal range = 44 IU/L for minimum weight and over age 

NON-JAUNDICE HEALTHY LIVER

RETINOL IN ALP BLOOD TEST (LOWER) 

(ALP) tissues protein = 44 * 4.10 (Optimum pH from lipase enzyme of stomach) = 180.4

                      = 180.4 + 0.1 = 180.5

                      = 180.5 = 61.49 % Retinol in body, 

DANGER-JAUNDICE 

RETINOL IN ALP BLOOD TEST (HIGHER) 

(ALP) tissues protein = 147 * 1.227 (Optimum pH from pepsin enzyme of stomach) = 180.369

                      = 180.369 + 0.131 = 180.5

                      = 180.5 = 61.49 % Retinol in body,