Effect of substrate concentration and

enzyme inhibitor on enzyme activityDetermination of

Alkaline Phosphatase activity

IntroductionIntroduction::

Phosphatases are enzymes which Phosphatases are enzymes which catalyze the splitting off phosphoric catalyze the splitting off phosphoric acid from nonphosphoric esters.acid from nonphosphoric esters.

Two common types are estimated in Two common types are estimated in serum:serum:

Alkaline PhosphataseAlkaline Phosphatase (optimum pH 10) (optimum pH 10) Acid PhosphataseAcid Phosphatase (optimum pH 5 – 6) (optimum pH 5 – 6)

Alkaline Phosphatase

Purified forms from different sources undergo 3 types of activity:

1- Hydrolytic

R-O- P + HOH ROH + H3PO4

2- Phosphotransferase

R-O- P + R`-OH R- OH + R`- O- P 3- Pyrophosphatase: R-O- P -O- P -O- R` + HOH R-O- P + R`-O- P

Sources: Osteoblasts in the bone Bile canaliculi in liver Small intestinal epithelium Proximal tubules in the kidney. The placenta Lactating breasts

In all these sites, it seems to be involved in the transport of phosphate across membrane

ALP of normal serum is mainly derived from liver (the bone isozyme is absent)

ALP requires metal ions :ALP requires metal ions : MgMg2+2+, Zn, Zn2+ 2+ and to a lesser extent Mnand to a lesser extent Mn2+2+

Inhibitors of ALP:Inhibitors of ALP: CuCu2+2+, Hg, Hg2+2+, EDTA (chelating Mg, EDTA (chelating Mg2+2+), ),

Phosphate & some amino acids e.g. L-Phosphate & some amino acids e.g. L-phenylalanine phenylalanine

Principle of the test:Principle of the test: ALP from human serum will hydrolyze ALP from human serum will hydrolyze

the artificial substrate, the artificial substrate, disodium phenyl disodium phenyl phosphatephosphate to to phenol phenol which will react which will react with with 4-aminoantipyrine4-aminoantipyrine in the alkaline, in the alkaline, oxidizing agents giving a oxidizing agents giving a red purple red purple color color which can be measured at which can be measured at 520 nm520 nm

Objective of the test

In this experiment we will investigate the effect of changing substrate concentration on enzyme activity in presence and in absence of the inhibitor (inorganic phosphate) to identify the type of inhibition.

Blank Standard Test Test Without Inhibitor With Inhibitor

1 2 3 4 5 6 7 8

Buffer 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Substrate - - 0.25 0.50 0.75 1.0 - - - -

Substrate + Inhibitor - - - - - - 0.25 0.50 0.75 1.0

Standard - 1.0 - - - - - - - -

Dist. Water 1.1 0.1 0.75 0.50 0.25 - 0.75 0.50 0.25 -

Serum sample - - 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

Mix and incubate the tubes at 37C for exactly 15 min.

NaOH 0.5 N 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.8

Na HCO3 0.5 N 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2

4-amino-antipyrine 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Potassium ferricyanide 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Mix and read the tubes against blank at 520 nm

PROCEDURE: Into 10 test tubes, add the following (in ml):

ALP activity (v) = X 10

= mg of phenol produced/100 ml

serum in 15 minutes

= King Armistrong (KA) units/100 ml

O.D test – O.D controlO.D test – O.D control

O.D Standard – O.D blankO.D Standard – O.D blank

Calculate ALP activity (v) for each substrate concentration using the following formula:

Calculation of substrate concentration:

Substrate concentration = 0.01 MSubstrate concentration = 0.01 M

Substrate M. Wt. = 254Substrate M. Wt. = 254

Substrate concentration = 0.01 X 254 = Substrate concentration = 0.01 X 254 = 2.542.54

[S] = substrate conc. X volume[S] = substrate conc. X volume

= 2.54 X volume= 2.54 X volume

1. Calculation:1. Calculation: O.D. BO.D. B = 0.0 = 0.0 O.D. CO.D. C = 0.04 = 0.04 O.D. SO.D. S = = [S] = 2.54 X volume[S] = 2.54 X volume

Test tube

Sub.volume(ml)

[S] 1/[S] O.D.T T- CV = --------- X 10 S – B

1/V

1 0.25 0.635 1.575

2 0.50 1.27 0.787

3 0.75 1.905 0.525

4 1.00 2.54 0.394

5 0.25 0.635 1.575

6 0.50 1.27 0.787

7 0.75 1.905 0.525

8 1.00 2.54 0.394

-I+

I

2. Type of inhibition: Using 1/V and 1/[S], draw the Linweaver-Burk plot to calculate the Km and Vmax of

the reaction in presence and absence of the inhibitor

1/Vmax(-I) = Vmax (-I) = - 1/Km (-I) = Km (-I) =

1/Vmax(+I) = Vmax (+I) =

- 1/Km (+I) = Km (+I) =

Non - Competitive Inhibitor

1/vi

1/Vmax

- 1/Km 1/[S]

No Inhibitor

+ Inhibitor

1/Vmax’

1/vi

CompetitiveInhibitor

NoInhibitor

- 1/Km

Vmax is the same in

presence of a

competitive inhibitor

Competitive Inhibitor

Apparent Km is increased in presence

of a competitive Inhibitor

1/vi

1/Vmax

- 1/Km 1/[S]

No Inhibitor

+ Inhibitor

Uncompetitive Inhibitor

3. Calculation of the percentage inhibition:3. Calculation of the percentage inhibition: V - ViV - Vi % inhibition = ---------------- X 100 % inhibition = ---------------- X 100

V V

where V = rate without inhibitor, Vi = rate with inhibitor where V = rate without inhibitor, Vi = rate with inhibitor

[S] V Vi % inhibition

0.635

1.27

1.905

2.54

4. Does the % inhibition change as [S] increase?

5. Do the results of this calculation confirm your conclusions as to the type of inhibition?