Good Report On Practical 2: Action Of Alkaline Phosphatase

Type of paper: Report

Topic: Enzyme, Reaction, Activity, Actions, Time, Phosphatase, Business, Products

Pages: 7

Words: 1925

Published: 2020/11/01

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Introduction

Alkaline phosphatase is an enzyme that is most critical in the process of biomineralization (Whitehurst & Law 2002). The enzyme enhances the hydrolysis of substances that constitute phosphates, generates orthophosphate besides enhancing the calcium uptake. Alkaline phosphatase is a hydrolytic enzyme responsible for removing phosphate groups from DNA and proteins. A typical substrate for the enzyme is para-nitrophenol phosphate that is hydrolysed to para nitrophenol by the enzyme. This has an absorbance at 410nm and so activity of the enzyme can be measured spectrophotometrically.
In humans, the main sources of alkaline phosphatase are the liver, bile duct, kidney bone and placenta. Nonetheless, the cells of the inner lining of the small intestine also produce alkaline phosphatase in small amounts (Whitehurst & Law 2002).
This practical will entail measuring of alkaline phosphatase activity in the caco-2 cell extract that was prepared in lab 1 and comparing it to the activity of a purified form of the enzyme.

Objectives

This exercise intended to conduct an enzyme reaction, to measure the enzyme activity of the crude lysate, prepared during lab 1 and to compare this to the activity of a sample prepared by the demonstrator, to note the background (enzyme free) reaction and to appreciate the use of spectrophotometry in following the progress of enzyme reactions. The achievement of the outlined objectives demanded precise performance of the experiment together with critical evaluations appropriately.

Protocol

The protocol demanded a prior reading before beginning (Alberty 2013). It was required that time be recorded together with the spectrophotometric readings during the practical.

Method

96 wells plate was brought and labelled. It was then divided it into three sections with every section having 6 wells (rows).
In the first section, 6 wells were chosen then 100ul of buffer was added ( cell lysis buffer) and 100ul of P-nitrophenol phosphate into all 6 wells then 50ul of NaoH ( solution that stops reaction ) was again added in different times and readings taken at 0 min ( first well ) , 5 min ( second well ) , 10min (third well) , 15min ( fourth well) , 20min ( fifth well) , 25min ( sixth well ). The aim of the first section was to see the reaction without the enzyme and hence operated as a standard. In the second section, there was concentrated alkaline phosphatase, so it was diluted to 5: 95. After dilution, 100ul of alkaline phosphatase was added to 6 wells and 100ul of P-nitrophenol phosphate added again to the same 6 wells. After that, 50ul of NaoH was added in different times and read at 0 min (first well), 5 min (second well), 10min (third well), 15min (fourth well), 20min (fifth well), 25min (sixth well). This section intends to see the reaction with pure alkaline phosphatase enzyme. In the third section, 30ul of lysate (the sample) and 20ul of water were added into all the 6 wells of section 3 then again, 50ul of lysis buffer and 100ul of P-nitrophenol phosphate were added. After that, 50ul of NaoH was added in different times before a reading was taken different intervals of 0 min (first well), 5 min (second well), 10min (third well), 15min (fourth well), 20min (fifth well), 25min (sixth well). The third section intended to see the reaction with the enzyme of the sample (lysate) and see the activity of the alkaline phosphatase enzyme in the sample.
The resulting collection of substances was mixed well and placed in the spectrophotometer. The absorbance was then read at 410 nm.

Results

Write up
Assume 0.1mM solution of the reaction product produces an absorbance of 1.0.
1) Calculate the rate of the reaction in mM/min for each experiment (20 marks)
Rate of reaction = Product /Time (min)

Solution

E
If 0.1mM produces an absorbance of 1.0

In 0 minutes

0.1mM = 0.1
0.138
= 0.138 x 0.1 / 1
= 0.0138

In 25 minutes

= 0.38 x 0.1 / 1
= 0.038 mM
Product = 0.038 – 0.0138
= 0.0242 mM
Time = (25 – 0) min
0.0242 / 25
= 0.000968 mM/ min
F
If 0.1mM produces an absorbance of 1.0

In 0 minutes

0.1mM = 0.1
0.108 x 0.1 / 1
0.0108 mM

In 25 minutes

= 0.388 x 0.1 / 1.0
= 0.0388 mM
Product = 0.0388 -0.0108
= 0.028 mM
Time = 25 – 0
= 25 min
Rate of enzyme activity = 0.028 / 25
= 0.00112 mM/min
G
If 0.1mM produces an absorbance of 1.0

In 0 minutes

0.1mM = 0.1
0.043
0.043 x 0.1 / 1
0.0043 mM

In 25minutes

0.044 x 0.1 / 1
0.0044 mM
Product = 0.0044 – 0.0043
= 0.001mM
Change in time = 25 – 0
= 25 min
Rate of reaction = 0.001/ 25
= 0.00004 mM/min

Enzyme E

Rate of reaction = Enzyme Activity /Time (min)

At 0 min

0.138/ 0
= 0 nM/min

At 5 min

= 0.175/ 5
= 0.035

At 10 min

0.201 / 10
= 0.0201 nM/min

At 15 min

= 0.247 / 15
= 0.0164 nM/min

At 20 min

0.298 / 20
= 0.0149 nM/min

At 25 min

= 0.38 / 25
= 0.0152 nM/min

Enzyme F

Rate of reaction = Enzyme Activity /Time (min)

At 0 min

0.108/ 0
0 nM/min

At 5 min

0.177 / 5
0.0354 nM/min

At 10 min

0.266 / 10
0.0266 nM/min

At 15 min

0.306 / 15
0.0204 nM/min

At 20 min

= 0.346 / 20
= 0.0173 nM/min

At 25 min

= 0.388 / 25
= 0.01552

Enzyme G

At 0 min
= 0.043 / 0
= 0 nM/min

At 5 min

= 0.043 / 5
= 0.0028 nM/min

At 10 min

0.043 / 10
= 0.0043 nM/min

At 15 min

0.044/ 15
0.0029 nM/min

At 20 min

= 0.045 / 20
= 0.00225 nM/min

At 25 min

0.044 / 25
= 0.00176 nM/min
2) Calculate the enzyme catalysed rate of reaction (20 marks)

Reagents

E Lysate (sample) + Buffer + Substrate

100ul of buffer

100ul of P-nitrophenol phosphate
50ul of NaoH
Second F Enzyme + Substrate

100ul of alkaline phosphatase (95% water)

50ul of NaoH
Third
G Buffer + Substrate

30ul of lysate (the sample) and 20ul of water

50ul of lysis buffer
100ul of P-nitrophenol phosphate
50ul of NaoH
Rate of enzyme reaction = Product formed/ Unit time
= dP / dT
E
Rate of enzyme reaction = Slope
Slope = d Enzyme activity/ d Time
= (0.380 – 0.138) / (25 – 0)
= 0.245 / 5
= 0.0484 mM/min
F
Rate of enzyme reaction = Slope
Slope = d Enzyme activity/ d Time
= (0.388 – 0.266) / (25 – 10)
= 0.122 / 15
= 0.0081 mM/ min
G
Rate of enzyme reaction = Slope
Slope = d Enzyme activity/ d Time
= (0.045 – 0.043) / (20 – 10)
= 0.002 / 10
= 0.0002 mM/ min

How many times did the enzyme speed up the reaction (5 marks)

In order to get the number of times the enzyme speeded up the reaction, it is necessary to compare the reaction where the enzyme was used to that where the enzyme was never used (Whitehurst & Law 2002).

Therefore, we will compare the reaction rates G to F

G = Without Enzyme
F = With Enzyme
Rate of activity in F = 0.00112 mM/min
Rate of activity in G = 0.00004 mM/min
Speed = F / G
= 0.00112 mM/min / 0.00004 mM/min
= 28 times

Therefore, the enzyme speeded up the reaction 28 times.

Discussion
Alkaline phosphatase has similar characteristics to other enzymes that are protein in nature and specificity in reactions (Morange 2008). The enzyme is responsible for catalysing reactions, and there is a difference between the reactions that are catalysed and those that are not provided that the conditions such as temperature and pH are favourable to the enzyme (Mader 2013). A conventional expectation is that the catalysed reactions are faster unlike the reactions that are not catalysed (Morange 2008).
A reaction may continue past the anticipated limits. However, this depends greatly on the products of the reactions. That implies that, some end products of the reaction may be soluble or reactive and continue to react. As a result, there is need to stop the reaction. In the experiment explained above, the solution of NaOH was used to stop the reaction (Mader 2013).
The reactions E and F appear to be linear in their activity progressions and seem to have a definite trend. The reaction E is uniform but as the time increases, the curve appears to move inwards. This implies that the rate of reaction continues to increase faster together with the product generated as compared to time. The reaction F appears to be uniform and the increase in time is comparable to the product generation that is measured in Mn.
The reaction in G is unique as the curve maintains a stagnant state then rise steeply before sloping. Reactions are due to the number of sites where the reactants can attain successful binding sites. Without the catalyst, the reactants do not readily meet, and this slows down the reaction due to a lower activity. When more time is offered, the reactants begin to react but after the binding sites get depleted, the rate of reaction begins to reduce with time (Allemann & Scrutton 2009).
Alkaline phosphatases are abundance in their occurrence with regards to the nature hence are found in several organisms ranging from bacteria to higher organisms (Ladbury & Doyle 2005). Most of the APs are homodimeric enzymes except few, and every catalytic site has three key metal ions, such as two zinc ions and a single magnesium, needed for enzymatic action. The enzymes catalyse the process of the hydrolysis of monoesters of the phosphoric acid besides catalysing a trans-phosphorylation response in the company of greater occurrences of the phosphate acceptors (Allemann & Scrutton 2009). Whilst, the key features of the catalytic process, are reserved comparing mammalian and bacteria different APs, those of higher organisms possess greater specific activity as well as Km quantities. Imperatively, the APs have uniformity to a greater number of alternate enzymes and are, hence, constituents of the greater family of enzymes possessing particular overlapping catalytic features as well as substrate specificities (Ladbury & Doyle 2005).

Conclusion

This experiment was a success in its objectives and those that intended to evaluate the properties of the enzyme in a practical approach. The rate of a reaction was encompassing enzymes also rises with the increase in substrate concentration (Alberty, 2013). Nonetheless, the quantities of enzyme-active regions present are fewer. At lower concentrations of enzyme or else greater substrate concentrations, there is a possibility that the substrate can occupy all the sites that are active (Ladbury & Doyle 2005). Hence, augmenting the concentration of substrate more will never alter the diffusion rate. That implies that, there is a particular maximum rate of reaction (Vmax) whenever all enzyme active points are engaged. The rate of reaction will intensify with the rising quantity of substrate concentration yet must asymptotically manage the saturation rate also referred to as the Vmax is Vmax is directly comparable to the overall concentration of enzyme (Alberty 2013).

Bibliography

Alberty, R. A. (2013). Enzyme kinetics rapid-equilibrium applications of mathematica. Hoboken, N.J., Wiley. http://rbdigital.oneclickdigital.com.
Allemann, R. K., & Scrutton, N. S. (2009). Quantum tunnelling in enzyme-catalysed reactions. Cambridge, UK, Royal Society of Chemistry.
Ladbury, J. E., & Doyle, M. L. (2005). Biocalorimetry 2 Applications of Calorimetry in the Biological Sciences. Chichester, John Wiley & Sons. http://www.123library.org/book_details/?id=14439.
Mader S (2013). Biology. New York: McGraw- Hill, Print
Morange, M. (2008). A history of molecular biology. Cambridge, Mass: Harvard University Press.
Whitehurst, R. J., & Law, B. A. (2002). Enzymes in food technology. Sheffield: Sheffield Academic Press.

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