The Effects of Enzyme Concentration on Rate of Reaction
To investigate how enzyme concentration affects the rate of reaction
on the decomposition of hydrogen peroxide in the presence of catalase
There are 5 possible variables that may affect the rate of the
1. Concentration of enzyme (catalase)
2. Concentration of substrate (hydrogen peroxide)
3. Pressure conditions of reaction
4. Temperature conditions of reaction
5. pH conditions of reaction
My objective is to consider the effects on the reaction rate on
altering the concentration of the enzyme. Thus, to ensure a fair test
I will aim to maintain all the other variables constant.
Hydrogen peroxide will breakdown to oxygen and water in the presence
of Catalase. I expect the time taken to evolve a set volume of gas
will decrease as the concentration increases. In other words, to
increase the concentration of catalase will increase the rate of the
reaction. I would expect that if the concentration is doubled, the
rate of the reaction will double.
Therefore: rate of reaction
Background Theory and Justification of Prediction:
Enzymes are globular proteins which act as a catalyst to metabolic
reactions. A catalyst alters the speed of a chemical reaction without
itself undergoing any permanent change. Enzyme molecules have a small
region that is functional. This is called the active site.
The basic mechanism of enzyme action
on a reaction can be understood
by the illustration below. In our case the substrate will be hydrogen
peroxide (H2O2). During the reaction the enzyme molecule and substrate
molecule form a complex and then products are formed as a result,
which in our case will be water (H2O) and oxygen(O2).
[IMAGE] 2H2O2 + Catalase 2H2O + O2 + Catalase
Oxygen is formed as a product and clearly the volume of oxygen
produced can act as an indicator to the rate of the reaction. Every
enzyme is specifically shaped to fit the substrate upon which it
works, and therefore they do not work in conditions that denature the
molecules (e.g. very high temperatures). As the reaction proceeds,
product molecules are formed and these reduce the number of collisions
between enzyme and substrate molecules. The rate of reaction therefore
slows down. So as long as there is an excess of substrate, an increase
in the amount of enzyme will lead to a proportionate increase in the
rate of the reaction. This is part of the collision theory.
Illustrative representation of effect of enzyme concentration
[Figure from Essential AS Biology, Glenn and Susan Toole, Page 50]
To ensure that only the concentration of the enzyme affects the
reaction; all other variables need to be controlled.
Temperature/Pressure - As temperature or pressure increases, molecules
move faster this is part of the kinetic theory. In an enzyme catalysed
reaction, this increases the rate at which the enzyme and substrate
molecules meet and therefore the rate at which the products are
formed. As the temperature/pressure continues to rise, however, the
hydrogen and ionic bonds, which hold the enzyme molecules in shape,
are broken. If the molecular structure is disrupted, the enzyme ceases
to function as the active site no longer accommodates the substrate.
The enzyme is denatured.
To control this variable, the temperature and pressure will be
maintained at a constant level that allows the enzyme to work
effectively, for this experiment it will be room temperature of around
230C. This will be achieved by using tongs to handle the apparatus so
that the heat from my hands
does not affect the catalase.
pH - A change in pH affects the ionic and hydrogen bonding in an
enzyme and so changes its shape. Each enzyme has an optimum pH at
which its active site best fits the substrate. Variation either side
of pH results in denaturation of the enzyme and a slower rate of
reaction. In this experiment, distilled water will be used at
different volumes to change the concentration of yeast (catalase).
Therefore the water will act as a control at around a pH of 7.
Substrate Concentration - When there is an excess of enzyme molecules,
an increase in the substrate concentration, produces a corresponding
increase in the rate of reaction. If there are sufficient substrate
molecules to occupy all of the enzymes, the rate of reaction is
unaffected by further increases in substrate concentration as the
enzymes are unable to break down the greater quantity of substrate. To
control the substrate concentration, identical quantities of the
substrate will be used for each reading.
Apparatus and chemicals
Clamp, 50ml measuring cylinder, 50ml gas syringe, stopwatch, delivery
tube, hydrogen peroxide (concentration:1 mole dm-3), 100ml distilled
water, yeast solution, conical flask, water bath, 5ml measuring
Water Bath (Keeps solution at room temperature)
1. I will set up the apparatus shown above.
2. Using the measuring syringe I will place 3ml of Hydrogen Peroxide
in the conical flask.
3. I will pour 5 ml of yeast solution using the 5ml syringe into a
4. As concurrently as possible, I shall add the yeast solution to the
hydrogen peroxide, place the bung in the neck of the conical flask,
and start the stopwatch.
5. When the volume of evolved gas reaches 30ml in the syringe, I shall
stop the watch, and record the time. I will make sure I look at the
syringe at eye level so as to reduce parallax error.
6. Starting again, I will measure 4ml of yeast solution using the
syringe, put it in a beaker and add 1ml of distilled water to it.
7. I will repeat steps 4 and 5 with different concentrations of yeast
solution and complete a table like below:
Volume of Yeast Solution (ml)
Volume of Distilled Water (ml)
Conc. Of Enzyme /
Time taken for 30ml oxygen to evolve/s
Trial No. 1
Trial No. 2
Trial No. 3
9. I shall repeat the experiment 3 times (as indicated by the trial
numbers in the table), to achieve an average time for each
To ensure a fair test, I shall only alter the concentration of the
yeast solution. All other variables shall remain constant
(temperature; volume hydrogen peroxide; concentration hydrogen
peroxide; volume enzyme solution; pressure; pH). Using 0ml of yeast
solution at one point is a control to make sure catalase in fact makes
a difference at all.
The concentration of hydrogen peroxide is 1 mole dm-3, and 1ml of this
concentration of hydrogen peroxide yields approximately 10ml of oxygen
when reacted with the enzyme catalase. This information allows me to
estimate how much gas might be evolved with the volume of hydrogen
peroxide I use.
[IMAGE] 2H2O2 2H2O + O2 (no.of moles = conc. x volume)
· No. of moles of H202 = 1 moldm-3 x 0.001 dm-3 = 0.001 moles
· Produces 0.0005 moles of O2 We know 1 mole of gas occupies 24dm-3
· Therefore 0.0005 x 24 = 0.012dm-3 which is approximately 10ml of
There are various safety precautions I will take with this experiment.
Hydrogen Peroxide is corrosive and poisonous, so it must be handled
with care. So as I do not mistake the distilled water for the Hydrogen
Peroxide, I will make sure all solutions are clearly labeled. My
college has general lab procedures which I follow. These include no
running; bags under laboratory tables etc. All of this will be
underlined in a risk assessment form.
The table below shows the precision of the instruments I will use. It
does not show how accurately each instrument can give readings; it
shows how accurately each item reading might be recorded. The
stopwatch, for example, gives readings to 0.01seconds, which is far
too accurate for human reactions.
I will expect that with greater concentrations of catalase the time
taken to evolve 30ml of oxygen will decrease. A graph of 'time taken
to evolve 30ml oxygen' versus concentration of catalase according to
my prediction will look something like this:
Graph number 1
Concentration of enzyme solution (% catalase)
Although the above graph shows that for increasing concentrations of
catalase, the time taken to produce 30ml of oxygen decreases. I
believe it will be a curve because higher concentrations of enzyme
contain more enzyme molecules than the lower concentrations. If there
are more molecules, then there are subsequently more collisions taking
place over the period of a second. This means that more reactions
between enzyme and substrate molecules take place in a second, and
therefore the product (O2 in this case) is evolved more promptly.
Again this is because of the collision theory I mentioned in the
background to my prediction. So at higher concentrations of enzyme
solution, the oxygen is given off more rapidly because there are more
enzyme molecules working on substrate molecules in a given second (or
a given period of time).
The rate of the reaction will actually be the speed at which the
products (i.e oxygen) is formed. We know that speed =
displacement/time, or in this case the volume of oxygen displaced/time.
Thus Rate = displacement of oxygen / time
But as I will be using a fixed volume of gas (30ml)
Then I can say: Rate = 1/time (units are seconds-1)
Graph number 2
Concentration of enzyme solution (% catalase)
This graph shows that the concentration of enzyme is proportional to
the rate of the reaction.
Rate µ Concentration
I would expect my results to yield this graph and if it does then
result is the one in my prediction. The collision theory was the basis
for this outcome.
Increasing the concentration of enzyme is to increase the number of
enzyme molecules present in solution. More molecules mean more
reactions taking place; twice as many molecules mean twice as many
reactions. Therefore, if the concentration is doubled, the speed of
the reaction also doubles.
In terms of the apparatus I aim to use, there are other viable
alternatives. For example I could have used a manometer to measure the
oxygen levels and use test tubes instead of the conical flask. Sources
of catalase include potato and liver. Rather than measure the time
taken to evolve a set volume of gas, I could alternatively measure the
volume of gas at various times. I picked to measure the time taken for
my plan because I feel it is easier to react to stopping a stopwatch
at a certain level of gas, than reading off a volume with my eyes at
certain at 10 seconds intervals for example.
The experiment will hope to be repeated 3 times to get average
results, this should reduce error levels. The average time taken will
be the one plotted on the graphs. If for any reason I do not have time
to complete 3 trials of this experiment I will use results of others
within my group who shall be doing the experiment.
Step 4 of the method states that the enzyme will be put into the
hydrogen peroxide at the same time as the bung will be replaced, and
the stopwatch started. All of this will be done as concurrently as
possible, but there will inevitably be some human delay involved, so
some gas will be lost. This would have slightly affect results, but as
the procedure will be repeated on each occasion, the error will be the
same throughout the experiment.
Readings shall be taken by the eye, and although I will look at the
scale as near to right-angles as possible, there may well be possible
I cannot foresee the straight line graph having all the measurement
points exactly in line. But it is hoped that these anomalies are small
enough to be discounted. I do not expect them to be distorting my
results in any way.
The study of the enzyme catalase could be further investigated by
analysing the effects of temperature upon the rate of the reaction
between the enzyme and hydrogen peroxide. Although catalase can
withstand reasonably high temperatures, it would probably denature at
extreme temperatures. It would be interesting to investigate at what
temperature it stops working. I could also look at how the substrate
concentration affects this reaction.
Going even further it could be interesting to look at other enzymes
and other reactions, this could to confirm that all enzymes work in
the same way.
O Essential AS Biology for OCR, Glenn and Susan Toole
O Advanced Level Biology A2 by Lowrie, Pauline
Enzyme Kinetics Enzymatic Decomposition of Hydrogen Peroxide