Alkaline phosphatase enzyme assay
Aim: Complete an alkaline phosphatase assay on the serum sample from CA and fill in the
Introduction [Method of King and Kind (1954)].
An enzyme is a protein molecule used as one of the most valuable tools in clinical
biochemistry. It can have many applications but the focus for this practical class is its use as
a marker of disease. The enzyme measured here is alkaline phosphatase, which is
predominantly found in bone, liver and kidney. Abnormal levels of this marker in the serum
could indicate disease in these areas.
More on Alkaline Phosphatases:
The principle of this method is as follows:
Phenol released by enzymatic hydrolysis of disodium phenyl phosphate (substrate) under defined
conditions of time, temperature and pH, is estimated photometrically by comparison with a
known phenol standard using the reaction where, in the presence of the alkaline oxidizing agent
4-aminoantipyrine, a red colour is produced with compounds containing a phenolic group.
There are several stages to this experiment:
a. A pre-incubation step – allows your substrate and buffer to reach 37 ºC
phosphate – substrate
*Red colour measured at 520 nm
b. An incubation step – time required for the enzyme ALP to convert substrate to
c. STOP reaction – At 15 min the reaction is terminated
d. Indicator step – 0.6% 4-aminoantipyrine and 2.4% potassium ferricyanide are added to
generate a colorimetric indicator.
Method (ensure to turn on your spectrophotometer)
1. Prepare the buffered substrate:
Buffered substrate = 4 mL of 0.1 M sodium carbonate/bicarbonate buffer, pH 10.0
4 mL of 0.01 M disodium phenyl phosphate mixed together.
2. You will have three samples
I) A test – this is the measurement of ALP activity in the serum sample from CA.
II) A control – This is used to control for endogenous phenol.
III) A standard – This is a known amount substrate (phenol) which will give an
equivalent absorbance and will be used to calculate the ALP activity.
3. Set up three test tubes in a rack and add the following components:
Solution Test Control Blank Standard
Buffered substrate 2.0mL 2.0mL 2.0mL 0.0mL
0.01 mg/mL phenol
0.0mL 0.0mL 0.0mL 1.0mL
0.1 M sodium
0.0mL 0.0mL 0.1mL 1.1mL
Mix well and then incubate in a 37°C water bath for 3 minutes and then add:
Serum 0.1mL 0.0mL 0.0mL 0.0mL
Mix, incubate again in the 37°C water bath for 15 minutes and then add:
0.5M NaOH 0.8 mL 0.8 mL 0.8 mL 0.8 mL
0.5 M NaHCO3 1.2 mL 1.2 mL 1.2 mL 1.2 mL
Serum 0.0 mL 0.1 mL 0.0 mL 0.0 mL
Mix well after each addition and finally add:
0.6 % 4-
1.0 mL 1.0 mL 1.0 mL 1.0 mL
2.4 % potassium
1.0 mL 1.0 mL 1.0 mL 1.0 mL
Mix after each addition and read at 520 nm
REMEMBER TO ZERO THE SPECTROPHOTOMETER WITH DISTILLED WATER BEFORE
4. Read the absorbance in the spectrophotometer (520 nm) and calculate the amount of
ALP in your test sample
One International Unit (IU) is the enzyme activity that will hydrolyse 1 µmol of substrate in one
A control should be prepared for each test sample, although one is obviously enough in this case, to guard
against the presence of endogenous (pre-existing) phenolic compounds in the test sera. However, one
standard will suffice for many test specimens.
For the purpose of this example, the absorbance of the samples from the spectrophotometer will be
• Standard = 0.219 (this is therefore equivalent to 0.01 mg or 10 µg phenol);
• Test = 1.2;
• Control for endogenous phenol = 0.010.
a. Subtract the value of control from the test (this subtracts any assay interference or background i.e.
absorbance not due to the activity of ALP in your sample).
b. To calculate the amount of product formed in your test sample, you use the standard:
Standard – Control Absorbance (0.219 – 0.010 = 0.209)
A standard solution containing 0.01 mg/mL of phenol gave a corrected absorbance of 0.209.
0.01 mg/mL = 0.209
x mg/mL = 1.2 – 0.01 (absorbance of your test – control)
Therefore, 0.01/0.209 x 1.19 = 0.057 mg
= 57 µg
This is supposed to be expressed per mL and we only added 100 µL to the assay, therefore
we need to multiply by 10 to express it per mL. Therefore the amount of phenol produced
per mL over 15 min is 570 µg/mL.
c. As mentioned this enzyme incubation was for 15 min and the law states, ’One International
Unit (IU) is the enzyme activity that will hydrolyse 1 µmol of substrate in one minute’. Therefore, the result
has to be divided by 15 to calculatethe amount of product formed per minute.
570 µg/mL / 15 min = 38 µg/mL/min.
d. The law also states that enzyme activity should be expressed in, ‘µmol of substrate formed’,
1 M phenol = 94 g/L (molecular weight of phenol)
X = 38 µg/mL/min
1 M phenol = 94 mg/mL
1 M phenol = 94 000 µg/mL
X = 38 µg/mL/min
1/94 000 x 38 = 4 x 10-4 M
= 400 µM/min
Which is the same expression as400 µmol/min/L = 400 IU/L
Thus, as one IU is the enzyme activity that will hydrolyse 1 µmol/substrate/minute, the answer in the
case of this example, is 400 lU/litre.