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A Comparison of the Amount of Catalase Enzyme in Different Plant and Animal Tissues

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A Comparison of the Amount of Catalase Enzyme in Different Plant and Animal Tissues

For this experiment I am required to compare the amount of catalase
enzyme between different plant and animal tissues.

This means that I am going to be comparing parts of animal or plant
tissues. In plants it would be the catalase content in potato, celery
or carrot samples. In animals it would be the catalase content in the
kidney, liver or heart tissue. The same parts can be compared so in
animals you can compare for example different types of potato and in
animals for example you can compare for example different animal’s

Hydrogen Peroxide is used to test for the amounts of catalase in
plants and animals because catalase breaks down hydrogen peroxide into
oxygen (O2) and water (H2O).

I am going to use a method that measures how quickly the products are
produced. I am going to measure the oxygen given off because it will
be a lot easier than water. To collect and measure the oxygen given
off I can either do it very accurately with a gas syringe or I can
measure the amount of oxygen given off by displacement in an inverted
measuring cylinder.

I have chosen to use plant tissue because it is a more stable reaction
and it is easy to regulate. One of the first things that I must do is
finding the variables that effect the reaction in different ways; I
will need to keep all of than the type of tissue, constant. I will
have to use my understanding and scientific knowledge of enzyme
activity and structure to complete this investigation and find the
best way to approach it.


Catalase is an intracellular enzyme found in the tissue of many living
things. Catalase has four active sites per molecule allowing it to
bind to four substrate molecules. It catalyses the breakdown of
hydrogen peroxide, H2O2 into oxygen and water.

Although there are large quantities of catalase in liver due to
photorespiration in mesophyll cells and lactic acid oxidation in the
cells, I am going to use plant tissue because it has a large number of
substrate molecules catalysed per minute.

I have got this information about catalase from the website:

Enzyme structure and catalysis

Enzymes are globular proteins and are defined as biological catalysts
because they speed up chemical reactions. When an enzyme and its
specific matched substrate combine bind a reaction takes place.

Enzymes are soluble in water and work in aqueous solutions in living
cells. They are sometimes described as organic catalysts. A catalyst
is a substance which affects the rate of chemical reaction and remains
unchanged in appearance and mass.

Every reaction needs activation energy to activate the reaction. If a
reaction is catalysed it will need less activation energy to get the
product, making it possible for reactions to occur at low

Uncatalysed reaction

Catalysed reaction

Lock and key mechanism

The lock and key mechanism was proposed in 1894 by Emil Fischer to
explain how enzymes and substrates interact. He stated that when the
enzyme and substrate combine they form an enzyme-substrate complex
before the products of the reaction are released.

Bond formation

Bond destruction

Hydrogen peroxide is specific to catalase’s active site so it is able
to form temporary bonds in the R-groups of catalase’s active site.
When the reaction is complete there are two products given off which
are water and oxygen. The action of the catalase and hydrogen peroxide
is shown below:

I got the information about enzyme structure and catalysis from: The
enzyme Investigation Booklet.


The rate of reaction depends on the variables below:



Surface area of tissue

Enzyme Concentration

Substrate Concentration


I will now describe these aspects and what happens if they are not


pH is a measure of acidity or alkalinity and expresses the
concentration of H+ ions in a solution. The optimum pHs for an enzyme
depends where it is e.g. Enzymes located in the stomach have an
optimum about pH2 because HCL is present. Intestinal enzymes where HCL
is not present have an optimum pH of 7.5.

The pH must remain constant throughout the experiment. If it is
changed the concentration of the H+ ions changes. The ions affect the
way that the R-groups bond with each other inside the active site by
interacting with them, and change the shape of the active site so that
the substrate molecule (Hydrogen Peroxide) cannot bind to it. This
means that the reaction cannot take place because an enzyme-substrate
complex is not formed and the substrate is not catalysed. If the pH is
changed to an extreme above or below the optimum the enzyme can be
denatured. This means that the shape of the active site will be
altered therefore causing all enzyme activity to stop and changing the
tertiary structure.


High temperatures can unfold enzymes which are made up of long,
precisely folded chains of amino acids. The optimum temperature for
human enzymes is close to 37 degrees Celsius. For most plants it is
much lower.

The temperature must remain constant because if it is decreased the
reactants will move around slower because they have less kinetic
energy so collisions between the substrate and active site will happen
less so enzyme-substrate complexes will happen less and therefore slow
down the reaction.

If the temperature is increased the reactants have more kinetic energy
so they will move around faster causing more collisions between the
substrate and active site, so there will be more enzyme-substrate
complexes happening therefore speeding up the reaction.

If the temperature is increased above the optimum the enzyme will have
too much kinetic energy and it will move around and vibrate so quickly
and vigorously that it will break the bonds between the R-groups in
its tertiary structure. This will alter the shape of the active site
causing it to denature and enzyme-substrates can no longer be created
because the substrate will no longer fit in the active site. When it
is denatured the tertiary structure has changed causing reaction to
stop completely.

I need to keep the temperature constant, so in the experiment, the
boiling tubes that I will be using will be in a water bath regulated
to a temperature of 40 degrees Celsius because this is the optimum
temperature of most biological molecules. I will keep it at this
temperature by using a thermometer. This means that I will keep the
experiment accurate and the enzyme will be operating at its optimum

Surface area of tissue

I will measure out all of the plant samples using a measuring ruler
and a sharp knife to make sure that they are all the same size so that
an equal number of enzymes and substrates react on the surface of the
plant sample. If I do not regulate the size of the sample and
therefore the amount of surface area, the samples will each have
different results because a different amount of enzyme-substrate
complexes will form because each sample will have higher or lower
catalase content because they will be different sizes.

Enzyme concentration

Enzymes catalyse reactions rapidly at very low concentrations. This is
because enzyme molecules form complexes with substrates only very
briefly. The products of the reaction are quickly released and the
enzyme is then available for further activity.

The rate which enzymes use substrates is called the turnover number.
For example a molecule of catalase can break down 40 000 molecules of
hydrogen peroxide into water and oxygen every second. The larger the
number of enzyme molecules present the greater is the amount of
substrate used in a given period of time, provided that there is an
excess of substrate available.

I need to keep this constant because if it is different each time I
will get different results when it is repeated because depending on
what the concentration is, there will be a different number of
enzyme-substrate complexes formed each time.

Substrate concentration

When there are low concentrations of substrate there is a linear
relationship between the reaction rate and substrate concentration. In
these conditions the relation of enzyme to substrate molecules is
high. A point can be reached when a further increase in substrate
concentration will not cause the reaction to go any faster. The
enzyme-to-substrate ratio is then lower, and there are more substrate
molecules present than there are free.

I need to keep this constant because if it is different each time I
will get different results when it is repeated because depending on
what the concentration is, there will be a different number of
enzyme-substrate complexes formed each time.


Many substances inhibit the activity of enzymes. There are two types
of inhibitor: reversible and non-reversible.

Reversible Inhibitors

Reversible inhibitors are substances which prevent the enzymes from
combining with the substrates. Activity of the enzyme is restored when
the inhibitor is removed.

Competitive inhibitors affect enzyme action by becoming attached to
active centres, so stopping the substrate from binding to the enzyme.
The degree of Inhibition by a competitive inhibitor is less if the
ratio of substrate –to-inhibitor molecules is high. The degree of
inhibition by non-competitive inhibitors cannot be reduced by
increasing the number of substrate molecules.

Non-reversible inhibitors

Non-reversible inhibitors become firmly bound to the active centres so
that substrate molecules cannot bind to enzymes and activity of the
enzyme is permanently stopped.

I need to keep the reaction without inhibitors because they will alter
the reaction or possibly stop all the reaction activity and will lead
to altered results.

I got the information about different variables of enzyme reactions
from: The enzyme investigation booklet.


Celery is the only plant sample that I am using which is grown above
the ground so I think it will have the largest amounts of catalase due
to photorespiration. Photorespiration is a process at least partially
opposing photosynthesis. In mitochondria develops carbon dioxide as a
product of the serine synthesis. Serine is synthesized under light
exposure and oxygen uptake from two glycerines. During the synthesis
of glyoxalate from glycolate is the heavy cytotoxin hydrogen peroxide
produced that is immediately afterwards broken down by the enzyme
catalase. As I think there will be larger quantities of catalase in
celery I predict that the reaction with celery will take place the
fastest and will produce more oxygen in the 5 minutes that I will be
timing the reaction for.

I used the following website for information for my prediction:


3 boiling tubes

2 water baths (1 is temperature regulated)


5cm3 syringe



Petri dish


I have done experiments concerning enzymes in the past so I can use my
scientific knowledge and experience to create a suitable procedure.

I am going to investigate the amount of H2O2 broken down by catalase
in three different types of plant tissue, which are carrot, potato and
celery. I have chosen these because when I do the experiment I should
get a good range of results because the amount of catalase
distribution is varied in each sample. First of all I will set up the
apparatus so that I am ready to do the experiment. I will then fill up
the water bath and add cool water until I reach the optimum
temperature of approximately 40 degrees Celsius and I will regulate it
using a thermometer and adding more boiling water if and when
necessary. I will then measure out three boiling tubes with 5cm3 of
hydrogen peroxide in each using a syringe, I will also measure out
equal sizes of the different sizes of the sample which will be 1cm3 by
1cm3. For my first sample I will place the boiling tube into the
temperature regulated water bath and check that the delivery tube is
airtight and passing through the water into the inverted measuring
cylinder. I will then take the plant sample and place it in the H2O2
solution and quickly put the bung end of the delivery tube into the
boiling tube. Then I will start the stopwatch and take down the
measurements of the volume of O2 produced at increments of 30 seconds
up to 5 minutes. I will then repeat the experiment for each sample,
and also repeat the whole experiment another 2 times so my results are
as accurate as they can be.

Risk assessment

We will be using a Hydrogen peroxide solution which is 20 volume
concentration. It is dangerous if swallowed, if this occurs plenty of
water must be taken and medical attention must be acquired. It is also
corrosive, so if a spillage occurs flood the area with large
quantities of water.

How to Cite this Page

MLA Citation:
"A Comparison of the Amount of Catalase Enzyme in Different Plant and Animal Tissues." 19 Apr 2014

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