Biology Enzymes G.C.S.E

Is there a relationship between the quantity of enzymes and the rate of reaction with the substrate?

Hypothesis

Using my scientific knowledge my hypothesis is that as the quantity of enzymes increases so will the rate of reaction.

Reasons for this hypothesis

I am basing my hypothesis on a hypothesis known as the 'lock and key hypothesis.' It explains that when an enzyme substrate complex forms by means of a spontaneous reaction the substrate fits into the active site like a key fits a lock. It is known as a spontaneous reaction not because it happens spontaneously but because it requires initial energy in order for the reaction to take place and the products have less energy than the reactants.

An active site occurs in the enzyme because a few of the amino acids on the surface of the 3d molecule fold inwards thus creating an indention in the enzyme. The 'lock and key hypothesis' uses this analogy due to the fact that only one substrate fits one particular active site. This is because simply the substrate and active site both contain certain elements that will only product certain temporary bonds with particular other elements.

However, despite the 'lock and key hypothesis' theory being very strong it was discovered that there was a flaw in the theory. It was realised that small molecules like water could enter the active site. This caused for a refinement in the theory to be made. The refinement was the cause of the discovery that when a substrate enters an active site small changes in the active site have to occur before the spontaneous reaction can take place.

This background information is what I am using as a foundation for my hypothesis. This is because as I increase the surface area I will be exposing more enzymes, therefore more active sites are made available for the substrates to be pulled towards them. I will be able to see the rate of reaction increasing because as I increase the surface area more oxygen will be released due to the catalytic reaction. However, if I was looking at increasing the concentration of the substrate rather than the quantity of enzymes I would find that there would be a limit to the amount in which the rate of reaction could increase. Therefore on the graph after a certain amount of positive correlation the line would stay at the same level. This is because all the active sites would be being used.

To consolidate my own knowledge of this topic I used the website http://www.bbc.co.uk/education/asguru/biology

Variables to be changed

In my particular experiment I am going to change the surface area of the potato five times. To give a range of observations I will start with a surface area of 5.9 cm^3. Each observation I will get that surface area again and cut it in half and keep on doing so until I have done five experiments.

Variables to be kept the same

Variable How I will keep it the same

Mass of potato Use the same borer to extract the potato from and measure it to the nearest mm each observation.

Volume of hydrogen peroxide Accurately measure out 25ml of hydrogen peroxide each observation.

Concentration of hydrogen peroxide Look carefully at the bottle to ensure that I am using the same concentration.

Temperature Use a thermometer to test the temperature of the experiment.

Time Use a stopwatch accurately and allow the same amount of time for each experiment.

Why is it important that I control these variables?

I know that, for example, temperature affects enzymes the rate of reaction. Above the optimum temperature of about 40 degrees (where they work fastest) the rate decreases as more and more of the enzyme molecules denature. The thermal energy breaks the bonds holding the structure of the enzyme together and so the enzyme and also the active site lose its shape and looks like a random coil. At very high temperatures this is irreversible and so the enzyme has become de-natured. Below the optimum temperature the rate also slows down as the enzymes are becoming inactive however this is irreversible. This means that is important I keep the temperature the same as if I increased it by a couple of degrees the volume of oxygen produced would double and if I lowered the temperature the volume of oxygen produced would be less.

Measurements

The measurements I will use are as follows:

* Ml for the hydrogen Peroxide

* Mm for the length of the potato (as part of the surface area formula)

* Cm cubed for the surface area to 1.dp

Apparatus

Below is a list of apparatus I will use for the experiment.

* Water

* Test tubes

* Trough

* 2 potatoes

* 1 tile

* Borer with a 1cm diameter

* Delivery tube and bung

* 2 measuring cylinders

* Hydrogen Peroxide

* Stop Watch

* Water bath

* thermometer

Diagram

Potato

Bung

Test Tube

Hydrogen Peroxide

Delivery Tube

Measuring cylinder

Water

Oxygen

Water trough

Plan of procedure

* Get all required apparatus out.

* Wearing safety goggles measure out accurately 25 ml of hydrogen peroxide and pour into a test tube.

* Pour

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