Biology Paper Essay Research Paper Introduction In

Biology Paper Essay, Research Paper

Introduction

In this experiment, I will be studying the activity of the enzyme catalase. Catalase breaks down of Hydrogen peroxide H2O2 into water H2O and oxygen O2.

catalase

2H2O2 2H2O + O2

catalase

Hydrogen peroxide water + oxygen

Enzymes are biological catalysts. They are made up of protein. They will only act upon one substance, is substrate. Enzymes only work at their optimum in a specific environment. There are many different factors in their environment, which affect an enzyme’s ability to decompose their substrate. Enzymes work at an optimum if you find the right balance between these factors. Some examples of these factors are temperature, pH and surface area of the potato. They are three main groups of enzymes involved in breaking down food molecules Proteases, Carbohydrases and Lipases. Proteases breakdown proteins, pepsin and trypsin are proteases. Carbohydrases breakdown carbohydrates, amylase is a Carbohydrase. Lipases breakdown fats.

Enzymes aren t just used inside the body they re also used outside the body in many different ways. Examples of these are

+ Biological detergents

+ Fruit industry

+ Photographic industry

+ Dairy industry

+ Brewing industry

+ Baby foods

+ Paper industry

+ Starch industry

+ Rubber industry

+ Baking industry

At the moment the hypothesis about how enzymes work is the Lock and Key Theory. This theory states that the active site of the enzyme and the substrate are extractly the same size. In the right environment (right pH and temperature) it is thought that the active site of the enzyme will catalysis the reaction. If one of the varibles is changed like the substrate or the pH the enzyme won t catalyse the reaction.

This is a diagram of the lock and key theory. It shows the enzyme s active site and the substrate are the exactly the same size and shape. By binding together the reaction takes place. The products of the reaction then leave the active site and the enzyme can look for another substrate molecule. In the experiment that I m doing the products of the reaction are water (H2O) and oxygen (O2). The enzyme is catalase, from potatoes, and the substrate is Hydrogen peroxide.

Problem

What factors influence the catalase s ability to decompose Hydrogen peroxide into water and oxygen.

Factors

There are many different factors that affect this reaction. They are as follows:

The factors that we can test are pH, surface area of potato containing catalase and temperature of the solution. I have chosen to keep the temperature constant because it s know that catalase will work best at body temperature. That is its natural environment. I have also chosen to keep the surface area constant. This is because the larger the surface area of potato quicker the better the reaction. This is why the body physically breaks down food so those enzymes contained in bile have a larger surface area to work on. I have chosen to vary the pH because I don t know exactly how it affects the enzymes’ ability to breakdown Hydrogen Peroxide.

Question

What will happen to the rate of the reaction catalysed by catalase if I vary the pH.

Prediction

I predict that the catalase will work at optimum at a neutral or slightly acidic environment. I say this because catalase is present in liver cells. It is in the liver because for this reason. The liver job in the digestion system is to neutralise any toxins in the bloodstream. Hydrogen peroxide is a toxic by-product of digestion so it is there are neutralise Hydrogen peroxide. The products of this reaction (water H2O and oxygen O2) are essential to the function of the body.

Catalase is present in all our liver cells. This means it s natural environment is a neutral or slightly acidic pH. This is backed up by my preliminary work (e.g. the effect of catalase on liver cells) in which I ve studied the rate of the reaction. In this case the rate of reaction is fastest at a neutral pH.

The theory behind this prediction is the Lock and Key theory. This theory states that the active site of the enzyme and the substrate are exactly the same size. In the right environment (right pH and temperature) it is thought that the active site of the enzyme will catalysis the reaction. If one of the variables is changed or at an extreme like the substrate or the pH the enzyme won t catalyse the reaction, because if difference in the environment alters the hydrogen bonds, ionic charge or whatever is making it hold its shape. This has a particular effect on the active site of the enzyme changing its shape so it and the substrate won t fit. This means that the enzyme has been denatured. Therefore, this is why I think at pH 2,4 and 10 will have a slow rate of reaction or no reaction will take place.

I predict that my results will look like this graph:

The catalase with work at optimum at pH 7 (7.07) and the rate of the reaction will drop down steadily either side of that pH.

Plan

I will collect the pieces of equipment I need and set them up as shown in the diagram. I will collect 20 milliliters of hydrogen peroxide and of each of the five pH buffers. I will measure the pH s using a pH meter m

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