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Show that the copper found in copper copper(II)sulphate is indeed a non-competitive inhibitor of the enzyme called catalase which is mainly found in the liver.

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Introduction

Introduction Catalase is an enzyme that consists of a protein complex with haematin groups and catalyses the decomposition of hydrogen peroxide into water and oxygen, this equation shows this. 2H O 2H O + O Free radicals are produced in most cells of the body as a byproduct of metabolism, although some cell types manufacture larger quantities for specific purposes. The most important free radicals found in aerobic cells, such as those in humans, are oxygen, superoxide, hydroxyl radical, hydrogen peroxide, and the transition metals. When free radicals form within cells they can oxidise biomolecules (molecules used inside cells, especially lipids) and thus cause cell death and injury. However, the human body has developed various mechanisms in order to protect itself from the damaging effects of free radicals. One of these mechanisms are enzymes, such as catalase which decompose peroxides and transition metals in the liver. A simple definition of an enzyme would be that they are protein molecules that act as biological catalysts. Enzyme molecules are globular proteins. The shape of an enzyme molecule is a precise 3-dimensional shape, with hydrophilic R groups on the outside of the molecule ensuring that they are soluble. The properties that are possessed in the structure of the enzyme molecule are shared with all other globular proteins. Enzymes are specialized molecules, the special feature being an active site. The active site of an enzyme is a region to which another molecule or molecule can bind to as stated in Biology 1 (Page 42). The active site is often perfectly shaped so that the substrate can fit. Each type of enzyme will usually bind with one type of substrate molecule. The reason this happens is that each type of enzyme has a different shape that is only specific to one type of substrate molecule. This concept is shown in diagram 1 below. Diagram 1 As you can see enzyme A and substrate A fit perfectly as the active site and the substrate are the same shape. ...read more.

Middle

After that calculate the average volume of oxygen collected. Table 1- A table to show which experiments are to be carried out Experiment Volume H�O� (cm�) Volume Distilled water (cm�) Volume copper (II) sulphate (cm�) Concentration copper (II) sulphate (mol) Total volume in tube (cm�) A 5 5 0 0 10 B 5 4 1 0.2 10 C 5 3 2 0.4 10 D 5 2 3 0.6 10 E 5 1 4 0.8 10 F 5 0 5 1 10 Fair Test To achieve accurate and reliable results, a fair test must be carried out. Even a slight change could alter the outcome of a test, so devising a fair test is imperative. To make the experiment a fair test, firstly the temperature at which the reaction takes place in must remain constant throughout. To do this a water bath would be used. It is vital that the temperature remains constant because enzyme activity increases or decreases depending on whether the temperature is high or low. At low temperatures the reactions take place at a low rate. The reason is that the substrate molecules and the enzyme molecule are moving relatively slowly due to the fact that they contain less energy as a direct result of the low temperature. This would mean that the enzyme molecules and substrate molecules would collide less often. Therefore the binding between enzyme and substrate reduces hence a slower reaction. If the temperature were to be increased the enzyme and substrate molecules would move faster, more collisions would take place and as they collide they would collide with more energy helping the bonds to break faster, this all adds up to a faster reaction. If one experiment was done at 35�C and another was done at 20�C then there will be more enzyme activity in the former rather than the latter. Secondly the volume of hydrogen peroxide and the total volume in the test tub must be the same in all the experiments. ...read more.

Conclusion

In four out of the six tests there was more substrate present than copper(II)sulphate, but in all four of the tests the rate of reaction was reduced, which meant that the copper was inhibiting the function of the enzyme. In the second test the there was 5cm� of hydrogen peroxide and 1 cm� of copper(II)sulphate. If copper was a competitive inhibitor there would have been no inhibition, but there was 13.5 cm� of less oxygen being released from the reaction. Theses are not the characteristics of a competitive inhibitor. Therefore copper must be a non-competitive inhibitor. . Evaluation The results obtained from the experiment was accurate enough to be able to find out that copper is a non competitive inhibitor of catalase. But I feel that the results could have been more accurate. The main reasons that I raise this matter are that there were a few flaws in the method that had been used. Firstly as the hydrogen peroxide was added in the test tube the rubber bung had to be replaced very quickly, during that split second oxygen could have escaped. Whether the amount of oxygen that escaped is significant enough to give wrong results is not known, but the possibility still remains there. Also there could have been errors made when the hydrogen peroxide, copper(II)sulphate and distilled water were being measured in the measuring cylinder. These mistakes were most likely human error. The error would most likely have been a misjudgement of just exactly how much was in the measuring cylinder as a minuscule usually forms. This confuses the mind, which leads to the wrong amount being measured. There is no definite solution to this problem, the only logical thing to do if not sure is to get a second opinion to confirm the initial opinion. I think that my results are reliable enough to support my conclusion. ?? ?? ?? ?? Page 1 of 12 ...read more.

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