What factors affect the activity of urease? What are the kinetics of the urea-urease reaction?

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Sam Hassani U6DST 8052

Chemistry Coursework

What factors affect the activity of urease?

What are the kinetics of the urea-urease reaction?

Aim:

To investigate closely the factors that can affect the activity of urease. The effect of concentration and temperature can be studied over a period of time leading to the order and activation enthalpy of the reaction.

Introduction:

UREASE

(NH2)2CO + 3H2O ----------> 2NH3 (aq) + CO2 (g)

Urease is an enzyme found in microorganisms, invertebrates, and higher plants. It catalyses the hydrolysis of urea to ammonia and carbon dioxide. In some bacteria and invertebrates this reaction is the final step in the breakdown of complex nitrogenous compounds before ammonia is excreted.

Method:

Quench with known volume and concentration of HCl to stop the reaction. Some of the acid will react with the ammonia produced by Urease to determine the amount of HCl used up. Contents of the flask are titrated with standard Sodium Carbonate solution using methyl orange indicator. From the results of this titration, the amount of HCl that did not react with the ammonia can be found by substituting the amount of HCl used to neutralise ammonia can be found.  

Background Knowledge:

An enzyme, such as Urease, is a protein that acts as a catalyst, speeding up the rate at which a biochemical reaction proceeds but not altering the direction or nature of the reaction. An enzyme catalyses virtually every metabolic reaction that takes place within a living organism. Enzyme molecules have a special feature in that they possess an active site. The active site is a region to which another molecule or molecules can bind. This molecule is the substrate of the enzyme. The shape of the active site allows the substrate to fit perfectly and to be held in place by temporary bonds, which form between the substrate and some of the R groups of the enzyme’s amino acids. This combined structure is termed the enzyme-substrate complex. Each type of enzyme will usually act on only one type of substrate molecule. This is because the shape of the active site will only allow one shape of molecule to fit in it. The enzyme is said to be specific for this substrate.

            There are two types of enzyme inhibitors: Competitive and Non-Competitive inhibition. If an inhibitor molecule binds only briefly to the active site of the enzyme, there is competition between it and the substrate for the active site. This means it is more likely that the substrate will collide with a filled active site and so this is known as competitive inhibition. It is said to be reversible (not permanent) because it can be reversed by increasing the concentration of the substrate.

Sometimes, the inhibitor can remain permanently bonded with the active site and therefore cause a permanent block to the substrate. No competition occurs as it does not matter how much substrate is present. This is non-competitive irreversible inhibition.

Other factors affecting enzyme activity:

       

Substrate concentration

As substrate concentration increases, the initial rate of reaction also increases as collisions between enzyme molecules and reactants become more frequent. When the enzymes begin to approach the maximum rate at which they can combine with reactants and release products, the effects of increasing substrate concentration diminish. At this point the enzyme is saturated and the reaction remains at the saturation level.

pH level

A change in pH can also effect enzyme activity. Each enzyme has an optimal pH range that help maintain its normal configuration in an environment which it operates. Pepsins is a enzyme found in the stomach and functions at pH 2. At pH 2 the tertiary structure of pepsin is not altered and will catalyze the reaction. Trypsin is and enzyme found in the small intestine and prefers a pH of about 8. The tertiary structure of a protein depends on interactions such as hydrogen bonding, between R groups. A change in pH can alter the ionization of these side chains and disrupt the normal configuration and in some case denature the enzyme. A denatured protein cannot combine with a substrate.

Temperature

As temperature rises, the enzyme and substrate molecules move faster. Moreover, when they do collide, they do so with more energy. This makes it easier for bonds to be broken so that the reaction can occur. As temperature continues to increase, the speed of movement of the substrate and the enzyme molecules also continues to increase (increase in kinetic energy). However, above a certain temperature the structure of the enzyme molecules vibrates so energetically that some of the bonds holding the enzyme molecule in its precise shape begin to break. The enzyme molecule begins to lose its shape and activity and is said to be denatured. This is often irreversible.

                     The temperature, which an enzyme catalyses a reaction at the maximum rate is called the optimum temperature. Most human enzymes have an optimum temperature of around 40°C.

Enzyme concentration

Increasing enzyme concentration increases the rate of reaction by increasing the possibility of collision with substrate as more active sites are available. Therefore increasing the rate at which substrate is converted to product.

Making Standard Solution(s)

What concentration of Na2CO3 is needed?

Possible range of HCl in titration flask

  1. If urea is not reacted

  • Therefore, no NH3 is produced.

               -    Therefore, HCl is unreacted.

So, 50 cm3 of 1M HCl

= 50/1000 x 1 = 0.05 mol HCl

In Titration/

2 HCl & Na2CO3 = 2 NaCl & CO2 & H2O

Therefore, 2:1 mol ratio

Therefore, mol Na2CO3 = 0.025 moles

SAY MAX IN BURETTE = 50 cm3

→ 0.025 = 50/1000 x conc.

→ Conc. of Na2CO3 = (0.025x 1000)/50

THEREFORE, Standard solution of Na2CO3 = 0.5M

Preparing a standard solution of Sodium Carbonate

Based on my calculations, I have come to the conclusion that a 1dm3 solution of 0.5M Sodium Carbonate will have to be made up.

  • Molar Mass of Na2CO3 = (2x23) + (1x12) + (3x16) = 106g
  • Therefore, for a 1 Molar solution 106 grams of Sodium Carbonate is needed to dissolve in 1dm3.
  • Therefore, to make a 0.5 Molar solution, 53 grams of Sodium Carbonate is needed to dissolve in 1 dm3 of water.
  • A magnetic stirrer was used to try to get the Sodium Carbonate solution to dissolve.

Starting with 2M solution

How much solution is needed?

-100 cm3 at each dilution

Therefore, the Total Volume needed = 212.5 cm3 of 2M

NB/ The 2M solution was only used in my preliminary work.

Preparing my solutions of Urease

E.G. making a 1% solution of Urease

A 1% solution means 1 gram in 100cm3.

Therefore, in a 250cm3 solution, 2.50 grams of Urease needs to be measured out on electronic weighing scales.

Once this is achieved, the powder should be placed into a 250cm3 volumetric flask and filled up to the bottom of the meniscus with the addition of distilled water.  

The Urease solution is very difficult to dissolve and needs intensive stirring using a magnetic stirrer.

Calculations

       2M Urea – 80cm3

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+ 1% Urease – 20cm3

            Total – 100cm3

In 80cm3 of 2M Urea,

                                   No. of moles = 2 x 80/1000 = 0.16

Therefore, max. amount of ammonia made = 0.32 moles

Titration/

NH3 (aq) + HCl (aq) → NH4Cl (aq) + H2O (l)

Therefore, we need 0.32m of HCl.

Therefore, in 100cm3 acid we need 0.32 mol.

C = n/v = 0.32/ 100/1000

Therefore, we need to use HCl ...

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