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Table of Contents

1. Title2. Introduction3. Principle4. Materials5. Procedure6. Results7. Graph8. Deduction9. Discussion10.Precautions11.Conclusion

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DETERMINE THE Km and vmax OF THE ACIDIC PHOSPHATASE

INTRODUCTION:

Catalysts are agents the increase the rate of chemical processes; those in cells are called enzymes. There are literally hundreds of enzymes in each cell. Enzymes are involved in catabolic and anabolic reactions, reactions that break down macromolecules and those that build macromolecules, respectively. The generic equation for the mechanism of enzyme action is familiar to you:

Enzyme + Substrate -- Enzyme-Substrate Complex - Enzyme + Product

There are three important features of this reaction that are important. First, enzymes display a great degree of substrate specificity. That is, they catalyze a single chemical reaction in the cell. Second, the binding of enzyme and substrate is like that of a hand and glove. This model is called the induced fit model of enzyme action and describes well the interaction of enzyme and substrate. Therefore, the enzyme will act upon only substrates that “fit” into the active site of the enzyme. Third, the enzyme is not destroyed or even modified in this reaction. Rather, it is released and is available to react with more substrate molecules. This feature allows enzymes to be very efficient in catalytic reactions and explains why small quantities of enzymes are adequate for cellular reactions.

The velocity or rate of a reaction can be determined by measuring either one of two things:

(1) substrate concentration, which should decrease as a reaction proceeds, or

(2) product formation, which should increase from an initial concentration of zero as the reaction proceeds.

PRINCIPLE:

Phosphatases are enzymes that hydrolyze phosphate monoesters and release inorganic phosphate.  Alkaline phosphatases (orthophosphoric monoester phosphohydrolases) are the phosphate hydrolases that have a maximum activity at a relatively high pH (>7.0).  This enzyme is widespread and occurs in both eukaryotic and prokaryotic cells. In Escherichia coli, its physiological role is to cleave phosphoryl groups providing the cell with a source of inorganic phosphate.

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Phosphatases catalyzes the hydrolysis of phosphate groups from phosphoesters. They are widely distributed in any organism. Many phosphatases are highly substrate-specific, like those enzymes involved in signal transduction.

A number of phosphatases, however, cleave virtually any phosphate ester. Such unspecific enzymes function mainly in the catabolic breakdown of metabolites or nutrients. Depending on the pH at which such phosphatases have optimal activity, between acidic phosphatases (also called acid phosphatases) and alkaline phosphatases can be distinguished. Acidic phosphatases are widely distributed in many organisms, including plants. They work optimally at approximately pH 5 without additional cofactors. In this experiment, we will extract an acidic phosphatase from bean sprout.

To measure its activity, we used a substrate called disodium phenylphosphate. The more activity of acidic phosphatase has, the more phenol is produced. So the content of phenol varies in proportion with the activity of the acidic phosphatase. The amount of phenol is determined by Folin-phenol method.

Figure 4-1  Acidic phosphatase catalyzes the reaction

Because of the hyperbolic shape of Vo against [S] plots, Vmax can only be determined from an extrapolation of the asymptotic approach of Vo to some limiting value as [S] increases indefinitely; and Km is derived from that value of [S] giving Vo = Vmax/2.

However, several rearrangements of the Michaelis-Menten equation transform it into a straight-line equation. The best known of these is the Lineweaver-Burk double-reciprocal plot. The Lineweaver-Burk plot is one way of visualizing the effect of inhibitors and determining the Michaelis Constant Km and the Maximum Velocity Vmax from a set of measurements of velocity at different substrate concentrations.

Vmax and Km are the two parameters which define the kinetic behavior of an enzyme as a function of [S].

If 1/Vo is plotted against 1/[S], a straight line is obtained where the slope is equal to Km/Vmax，the y-intercept is equal to 1/Vmax and x-intercept is equal to -1/ Km.

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Figure 4-2  The Lineweaver-Burk double-reciprocal plot, depicting extrapolations that allow the determination of the x- and y-intercepts and slope.

Where:

V0= initial velocity of the reaction，Vmax= maximum velocity of the reaction，

[S]= substrate concentration，

Km= the substrate concentration at which 1/2 Vmax is achieved

MATERIALS:

Apparatus: Petri dish, scissors, 5g of bean sprout, grinding pestle, mesh silk, beaker, water, Eppendorf tube, test tubes, homogenizer, centrifuge, spectrophotometer, vortex Mixer, incubator, marker, Gloves and watch. Solution : 1mL HAC(acetate) buffer (pH 5.6), 0.4mM phenol standard solution, H2O, 1 M Na2CO3, Folin-phenol reagent, 5mol/L disodium benzene phosphate, 0.2 M acetate Buffer, acid phosphatase solution.

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PROCEDURE:

A. Preparation of the acidic phosphatase

1. Weigh 5g of bean sprout.

2. Add 1mL HAC buffer (pH 5.6) and then grind the bean sprout into homogenization.

Then place it for 30min at room temperature.

3. Pour the resulting homogenization through the mesh silk into a 2mL tube then

centrifuge at 6000rpm for 20min.

4. Transfer the resulting supernatant which containing the acidic phosphatase into a fresh

tube.

5. Dilution the supernatant for 40 times with HAC buffer (pH 5.6).

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B. Determine the Km and Vmax of the acidic phosphatase

1. Plot a standard curve of phenol content

2. Select 9 tubes and mark them as 0-8, #0 tube is used as blank

Additions (mL) tube number

0 1 2 3 4 5 6 7 8

0.4mM phenol standard solution

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

H2O 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2

1M Na2CO3 2 2 2 2 2 2 2 2 2

Folin-phenol reagent 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

3. Mix well then incubate at 35℃ for 10 min

4. Measure the A680 of the samples and the blank tube is used for the zero setting.

5. A680 is plotted against phenol content.

C. Determine the Km and Vmax of the acidic phosphatase

1. Select 7 tubes and mark them as 0-6, #0 tube is used as blank

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Additions (mL) tube number

0 1 2 3 4 5 6

5mmol/L disodium benzene phosphate

0.5 0.1 0.15 0.2 0.25 0.3 0.5

0.2M acetate buffer 0 0.4 0.35 0.3 0.25 0.2 0

incubate at 35℃ for 2 min

acidic phosphatase solution

0 0.5 0.5 0.5 0.5 0.5 0.5

incubate at 35℃ for 15min

1M Na2CO3 2 2 2 2 2 2 2

Folin-phenol reagent 0.5 0.5 0.5 0.5 0.5 0.5 0.5

acidic phosphatase solution

0.5

2. Incubate at 35℃ for 10min.

3. Measure the A680 of the samples and the blank tube is used for the zero setting.

4. Calculate V0 and [S] the content of the phenol in each tube.

1/Vo is plotted versus 1/[S], a straight line is obtained where the slope is equal to Km/Vmax，the y-intercept is equal to 1/Vmax and x-intercept is equal to -1/ Km.

RESULTS:

(B) :

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Tube number 0 1 2 3 4 5 6 7 8

Absorbance / (A680)

0.00 0.14 0.247 0.346 0.449 0.562 0.657 0.767 0.876

Mass of Phenol /µmol

0.00 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.32

(C) :

From Graph (B), using y= 0.355x,

(i) When x= 0.134, y= 0.0476. Amount of phenol=0.0476 µmol.(ii) When x= 0.165, y= 0.0586. Amount of phenol=0.0586 µmol(iii) When x= 0.212, y= 0.0753. Amount of phenol=0.0753 µmol(iv) When x= 0.211, y= 0.0749. Amount of phenol=0.0749 µmol(v) When x= 0.291, y= 0.1033. Amount of phenol=0.1033 µmol(vi) When x=0.338, y=0.120. Amount of phenol=0.120 µmol

From equation: Vo = Amount of phenol Time

(i) Vo = 0.0476 1/Vo = 1 15 3.17 x 10-3

= 3.17 x 10-3 ms-1 = 315.3 m-1s

(ii) Vo = 0.0586 1/Vo = 1 15 3.91 x 10-3

= 3.91 x 10-3 ms-1 = 256.1 m-1s

(iii) Vo = 0.0753 1/Vo = 1 15 5.02 x 10-3

= 5.02 x 10-3 ms-1 = 199.3 m-1s

(iv) Vo = 0.0749 1/Vo = 1 15 4.99 x 10-3

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= 4.99 x 10-3 ms-1 = 200.3 m-1s

(v) Vo = 0.1033 1/Vo = 1 15 6.89 x 10-3

= 6.89 x 10-3 ms-1 = 145.2 m-1s

(vi) Vo = 0.120 1/Vo = 1 15 8.00 x 10-3

= 8.00 x 10-3 ms-1 = 125.0 m-1s

Tube number 0 1 2 3 4 5 6

Absorbance 0.00 0.134 0.165 0.212 0.211 0.291 0.338

[S] /mmol/L 5.0 1.0 1.5 2.0 2.5 3.0 5.0

1/ [S] /mmol-1L 0.20 1.0 0.667 0.50 0.40 0.333 0.20

Vo /(x10-3) ms-1

- 3.17 3.91 5.02 4.99 6.89 8.00

1/Vo / m-1s - 315.3 256.1 199.3 200.3 145.2 125.0

GRAPH:

(B) : Graph of A680 against phenol content

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(C) : Graph of 1/Vo against 1/[S]

DEDUCTION:(i) From graph (c), using equation y = 242.3 x +81.63,

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Where line cuts the x-axis, y=0.

Therefore, 0 = 242.3 x + 81.63,

242.3 x = - 81.63,

x = -81.63 242.3

= -0.337 Hence, x-intercept = - 0.337 mmol-1L.

As x-intercept is -1/Km,

-1/Km = -0.337

Km = 2.97 mmol/L

(ii) From graph (c), using equation y = 242.3 x +81.63, Where line cuts the y-axis, x=0.

Therefore, y = 81.63

Hence, y-intercept = 81.63 m-1s.

As y-intercept is 1/Vmax, 1/Vmax = 81.63

Vmax = 0.01225 ms-1

DISCUSSION:

There are four factors that affect enzyme activity. The first is the amount of enzyme. The initial velocity is proportional to the amount of enzyme molecules. The more enzyme, the

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greater the initial velocity will be since more product is being formed. Another factor that greatly influences enzyme activity is temperature. Proteins are usually denatured by temperatures above 50oC. Any temperature lower than that causes an increase in enzyme activity, until the freezing point is reached. Here, freezing an enzyme often denatures it and results in a loss of catalytic activity. However, for every enzyme, there is an optimal temperature that results in the greatest Vo. This temperature is usually in the range of 20oC to 40oC.

The third factor that influences enzyme activity is pH. Excesses of pH, either solutions that are too acidic or too alkaline, cause denaturation of the enzyme. Hence, all enzyme activity is lost and the Vo equals zero. Most enzymes in plants and animals operate most efficiently at near neutral pH. However the pH optimum of the enzyme you will study today is 4.5. The name of the enzyme is acid phosphatase, whose name is indicative of the pH optimum of this enzyme.

Finally, substrate concentration also influences the Vo of an enzyme-catalyzed reaction. At low substrate concentration, the active sites on the enzymes are not saturated by substrate and the enzyme is not working at maximal capacity. As the concentration of substrate increases, more and more enzyme molecules are working. At the point of saturation, no more active sites are available for substrate binding; at this point, the enzyme reaches its maximal velocity, designated Vmax. The following graph illustrates this.

The entire relationship is described by a hyperbolic curve. The initial velocity of an enzyme increases in a linear manner as the substrate concentration is increased. As substrate concentration increases beyond this point, the Vmax occurs because the enzyme is saturated with substrate. The substrate concentration required to yield half the maximal velocity (Vmax/2) can also be determined from the figure below. This number is known as the Michaelis constant and is abbreviated Km. The Km is a measure of enzyme affinity.

For example, a Km of 0.2 mole/l of substrate would indicate that the substrate-binding site would be half-saturated when the substrate is present in that concentration. Such an enzyme would have a low affinity for its substrate. In contrast, a Km of 10-7mole/l indicates that the enzyme has a high affinity for its substrate, since it is half saturated at this low concentration of substrate.

Vmax

[S]

V0

Km

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The effect of substrate concentration on the velocity of an enzyme-catalyzed reaction can be described by the Michaelis-Menten equation.

Vo = Vmax[S]

Km +[S]

Rearrangement of the equation to resemble that of a straight line (y=mx+c) by taking the inverse of both sides yields:

1 = Km 1 + 1

Vo Vmax [S] Vmax

This double-reciprocal plot is known as the Lineweaver-Burk plot and allows us to determine exactly what the Vmax and Km of a particular enzyme-catalyzed reaction are by solving for each in the equation.

PRECAUTIONS:

There may have been errors due to:- Misuse of pipettes.- Test tubes being wet or having some dirt leading to inaccurate results. - Absorbance may have been disturbed as some of the light might have been absorbed by

the glass sample of the spectrophotometer itself or by fingerprints lying on the sample glass leading to systematic errors.

Slope = Km/Vmax1/V0

1/[S]

1/Vmax

-1/Km

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- Maybe more than 2/3 of the solution was in the glass sample leading to overflow of liquid, thus affecting absorbance results

- Or perhaps less than 2/3 of the solution was in the glass sample leading to UV light passing over the solution hence affecting the absorbance results.

- The fact that different groups were using the centrifuge at the same time, the eppendorf tubes might be of different weights and they might have been arranged in the wrong position.

- Use clean glass wear and supplies- Make sure cuvettes are clean of all residues- Become familiar with spectrophotometry before proceeding- Always let the spectrophotometer warm up for 15-20 minutes before using - Know the limits of the spectrophotometer which is being used as most

spectrophotometers cannot linearly read absorbance values beyond 2.0 units. Thus, the protein sample must be diluted to work within this range

- Arrange the eppendorf tubes diagonally in opposite directions so as to keep weight balance

CONCLUSION:

From the experiment, the

Maximum velocity of the reaction, Vmax = 0.0123 ms-1

and the

Substrate concentration at which ½ Vmax is achieved, Km = 2.97 mmol/L

Enzymes increase the rate of spontaneous reactions, not changing their direction.

They do this by their ability to bind specific substrates to form transition state complexes.

Km is related to the ability of the enzyme to bind its substrate.

A higher Km means poorer binding.

The catalytic ability of an enzyme can be characterized by these two parameters.

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Virtually every chemical reaction in the cell is catalyzed by an enzyme.

The enzyme binds its substrates with great specificity and it usually aids in the chemical reaction that takes place.

However, the enzyme itself emerges unchanged.