ALP - BLOOD TEST

(Three peptide in mathematical machine of Sikander Aqeel) 

THREE PEPTIDES AND EARTH   

CELL MEMBRANE OF HUMAN 

Cell membrane = 100 % = 40 % lipids + 55 % protein + 5 % carbohydrate   

Earth = 365 days  

    = a2 + b2 = c2  

    = a2 Earth + b2 Cell membrane = c2 

    = a2 365 + b2 100 = c2 

    = a2 133225 + b2 10000 = c2         

    = a + b = 143225 = c2 

    = a + b = 143225 / 378.45 = 378.45

    = a + b = 378.45 / 2.5 (H2= A//T DNA) = 151.38

    = a + b = 151.38 - 5.37 (H4) = 156.75

    = a + b = 156.75 = Di-peptide   

The earth atmosphere which belongs with 365 day of peptide, 

    Peptide = 85.00, = C4H4O2  

 Di-Peptide = 156.75, = C6H7N2O3  

Tri-Peptide = 224.00, = C9H8N3O4 

Total three peptide = 465.75 

                    = 465.75 – 100 Cell membrane = 365.75 

                    = 365.75 – 0.75 % use of membrane = 365 

                    = 365 biological days of a cell, 

Use of Cell membrane per minute = 0.75 %  

Use of water in cells = 18.5 / 0.75 % = 24.666666666666666666666666666667 

                      = 24.666666666666666666666666666667 * 24 hour = 592 

                      = 592 unit use of water in a cell from the Cell membrane, 

                      = 592 / 32 (O2 of atmosphere) = 18.5, 

Use of water per cell = 592 unit / 100 Cell membrane = 5.92 

                      = 5.92 liter water need of body per day, 

(7/1/2016)  

The Cell membrane absorption is 0.75 % nutrients per minute, for three peptides, and it will be called the use of cell membrane, now we have (ALP) tissues protein result to know total (ALP) tissues protein in complete body  

ALP - blood test     

Alkaline phosphates (ALP) are a protein found in all body tissues. Tissues with higher amounts of ALP include the liver, bile ducts, and bone.

A blood test can be done to measure the level of ALP.

A related test is the ALP is enzyme 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   

LOWER HEALTHY

(ALP) tissues protein = 44 / 0.75 % = 58.666666666666666666666666666667

                      = 58.666666666666666666666666666667 * 24 hour = 1408 

                      = 1408 alkaline phosphates (ALP) tissues protein from the Cell membrane, 

                      = 1408 / 32 (O2 of atmosphere) = 44, 

Use of tissues protein = 1408 unit / 100 Cell membrane = 14.08 

                       = 14.08 liter (ALP) tissues protein in body like lower healthy,    

So the 14.08 liter tissues protein in whole body, which will be counted thin body and lower healthy,  

HIGH HEALTHY

(ALP) tissues protein = 147 / 0.75 % = 196

                      = 196 * 24 hour = 4704 

                      = 4704 alkaline phosphates (ALP) tissues protein from the Cell membrane, 

                      = 4704 / 32 (O2 of atmosphere) = 147, 

Use of tissues protein = 4704 unit / 100 Cell membrane = 47.04

                       = 47.04 liter (ALP) tissues protein in body like high healthy,    

So the 47.04 liter tissues protein in whole body, which will be counted high healthy, 

DISEASES OF (ALP) (7/1/2016) 

Lower-than-normal ALP levels 

• The Hypophosphatasia is disease from lower-than-normal ALP levels, or Hypophosphatasia (HPP) is a rare genetic disorder characterized by the abnormal development of bones and teeth. These abnormalities occur due to defective mineralization, the process by which bones and teeth take up minerals such as calcium and phosphorus. 

Hypophosphatasia is caused by mutations in the tissue nonspecific alkaline phosphatase (ALPL) gene. This gene is also known as the TNSALP gene. Such mutations lead to low levels of the tissue nonspecific alkaline phosphatase enzyme. Depending on the specific form, hypophosphatasia can be inherited in an autosomal recessive or autosomal dominant manner. 

Cell membrane = 100 % = 40 % lipids + 55 % protein + 5 % carbohydrate 

Cell membrane = 55 % protein  

(ALP) tissues protein = 14.08 liter in lower condition 

    = a2 + b2 = c2  

    = a2 lower (ALP) + b2 Cell membrane = c2 

    = a2 14.08 + b2 55 % protein = c2 

    = a2 198.2464 + b2 3025 = c2         

    = a + b = 3223.2464 = c2 

    = a + b = 3223.2464 / 56.77 = 56.77

    = a + b = 56.77 / 2.5 (H2= A//T DNA) = 22.708

    = a + b = 22.708 - 0.458 = 22.25

    = a + b = 22.25 = H2O + H3 of DNA base 

DNA base and water = 22.25 / 0.75 % = 29.666666666666666666666666666667

                   = 29.666666666666666666666666666667 * 24 hour = 712 

                   = 712 

                   = 712 / 32 (O2 of atmosphere) = 22.25, 

DNA base and water = 712 unit / 100 Cell membrane = 7.12

                   = 7.12 liter water in protien, and 6.20 times enzyme alkaline phosphatase,  

                   = 7.12 * 6.20 = 44 normal results (ALP) tissues protein, 

PHOSPHATASE ENZYMES AND HYPOPHOSPHATASIA DISEASE (7/3/2016) 

2. Start the reaction. The reaction mixture is simple, made of three parts: 

a. 1.5 mL of 0.05 M Citrate Buffer, pH 4.8 

b. 1.5 mL of 1mM p-nitrophenyl phosphate (substrate) 

c. 1.5 mL of 1 mg/mL phosphatase The buffer and substrate are added to a clean test tube. Then, the enzyme is added to initiate the reaction, the tube is quickly and briefly stirred (vortex), and the stopwatch is started. Remember that the reaction starts immediately after the enzyme is added, so work quickly! 

Hydrogen Phaphate ion = H2PO4 = 97.5 m/g      

Let H = 24 hour of one day for 97.5 speed of Phaphate 

Let M = 60 minute of one hour for 52 week of a year  

Rate = Time and Phaphate --------------- Time of Phaphate with weeks 

     = h + m = 97.5--------------------24h 60m = 52(97.5)  

       = h + m = 97    

                 h = 97.5 – m  

                24h 60m = 52(97.5) 

                24(97.5 – m) + 60m = 52(97.5) 

                24(97.5 – m) + 60m = 52(97.5) # (97.5 * 24 = 2340) 

                   2340 – 24 + 60m = 52(97.5) # (97.5 * 52 = 5070) 

                        2340 + 36m = 5070 # (5070 - 2340 = 2730) 

                               36m = 2730 # (2730 / 36 = 75.83) 

                                 m = 76 # (36 / 6r = 6) 

                                 m = 76 # (76 / 6r = 12.66)  

                                 m = 12.66 * R2 = 6.33

                                   = Enzymes = (6.33)2 times 

B. Standard Phosphatase Reaction Assay You are performing a discontinuous enzyme assay. The reaction occurs for a total of 10 minutes, but an aliquot of the reaction mixture is removed every two minutes and stopped (for a total of 5 aliquots). In the end, 

you will have a test tube with a 2-minute-old reaction, 

one with a 4-minute-old reaction, 

one with a 6-minute-old reaction, and so on. Each of these test tubes can be assessed individually in the spectrophotometer. You must also consider what control reaction is important to test your hypothesis and set it up also. The directions below are for the experimental treatments. You must determine how to set up the control.

The directions below are for the experimental treatments. You must determine how to set up the control 

CONTROL

Enzymes = (6.33)2 times 

Enzymes = 12.66 * 2 minute = 25.32 

        = 25.32 * 4 minute = 101.28 

        = 101.28 * 6 minute = 607.68 

c. 1.5 mL of 1 mg/mL phosphatase 

enzymes = 607.68 * 1.5 = 911.52 

phosphatase enzymes = 911.52 / 44 Normal Results = 20.71 

                    = 20.71 – 14.08 liter = 6.63 

                    = 6.63 times enzyme of Hypophosphatasia disease,