How does the solute concentration of Hydrogen Peroxide affect the enzyme rate of homogenized liver catalyze?
The Effect of Catalase1
Liver and other living tissues contain the enzyme catalase. As with all enzymes, catalase of homogenized liver is a protein, meaning that it is synthesized within the cell from building blocks called amino acids . In addition to the amino acids that make up the protein, catalase carries around a heme group. Mostly known as Hemoglobin. In the middle of the heme group sits an iron atom the catalase enzyme uses this iron atom to help it break the bonds in the two molecules of hydrogen peroxide, shifting the atoms around to release two molecules of water and a molecule of oxygen gas
Reaction : 2H2O2 –> 2H2O + O2 (fix)
The effect of Substrate Concentrations on enzyme activity2
At low substrate concentrations, enzyme activity increases steeply as substrate concentration increases. This is because random collision between substrate and active site happen more frequently with higher substrate concentrations.
Rate of Velocity
V = V(max) * Concentration of Hydrogen Peroxide
Km * (Concentration of Hydrogen Peroxide)
Dependent and Independent Variables
Dependent (in method referred as DP)
* The volume of homogenelized liver ( cm3)
Independent (in method referred as IV)
* Concentration of different hydrogen peroxide (H2O2) solutions
* (rate of enzyme activity)
Controlled (in method referred as CV)
* Volume of the substrate of Hydrogen Peroxide in each tube (5 cm3)
* Time (min)
* (Temperature- room temperature )
Method – controlling the variables
DP- the variables can be measured in different ways; in this experiment we measure the volume of homogenelized liver by using a pipette. So all the solutions will have the same amount of homogenelized liver in cm3
IV- The rate of enzyme activity can not be controlled. But it is influenced by the controlled and dependent variables.
CV – The rate of enzyme activity is affected by the amount of solution given of Hydrogen Peroxide to react. By using a measuring cylinder each solution will have the same amount of Hydrogen Peroxide to react with. Time is also an important factor, since the reactions take place over time, the longer the solution is exposed to the homogenelized liver the more it has reacted. Therefore time has to be constant with all the different solutions.
The key is to keep all the variables the same, except for one, which is in this case the concentration of hydrogen peroxide solution.
* Measuring cylinders (100 cm3) – to use as test tubes (when using less cylinders then different solutions make sure that the old solution has been washed out)
* Measuring cylinder (10cm3) for hydrogen peroxide(make sure that you wash the cylinder every time after you used it)
* Different concentrations of hydrogen peroxide solutions (%)
* Stopwatch or clock
* Pipette for homogenelized liver
* Homogenelized Liver (blended)
1. Decide which concentrations of Hydrogen Peroxide you will use, best is when you take even or uneven numbers of concentration such as 1,3,5,7, (%) et cetera.
2. Make a table for your results.
3. Take 5 cm3 of Hydrogen Peroxide in any concentration (depending which ones you will do) and put this in the measuring cylinder. At smaller concentrations you do not need a measuring cylinder of 100cm3 but you can take a smaller one e.g. 25 cm3. (The accuracy of your reading will increase with smaller measuring cylinders, do take in mind that the higher concentration the faster the rate most likely will be)
4. Add 1 cm3 of homogenelized liver, at the moment where you start the time. Before this happens make sure that you know at which time you will stop and when you will note down a result. ( In this case you could use every 30 seconds up till ï¿½ 4 minutes)
5. Now repeat this with different concentrations, every time note down the results. After 4 minutes when it the reaction still takes place, make a notice of it.
6. After you have done each uneven number or even number, you should repeat the same experiment to make it more reliable.
Observations, trends and patterns
We can see in Graphs of 1.1 show a rising volume of foam produced. When you add the polynomial line you notice that there is a pattern, namely that the volume will increase by a certain amount when the substrate concentration becomes higher. We can see more patterns in the Graph of 1.2, where you can clearly see that a higher concentration increases the reaction the most in the beginning. What is noticeable is that with each reaction (and different concentrations) the volume has risen most in the first 30 seconds. After this the reaction seems to rise slower, possible because the enzyme starts denaturizing. The higher the concentration the more volume is added in the first 30 seconds. We can see clearly the returning pattern in each concentration.
Graph 1.3 shows the rate of velocity increasing slightly when the concentration becomes higher. Even though there should be a pattern, (pre- knowledge) we can not necessarily see one clearly enough to compare it with the internet values.
We can calculate the velocity rate with the formula: (Michael Constant Equation)
V = V (max) * Concentration of Hydrogen Peroxide = Vmax
Km3 * (Concentration of Hydrogen Peroxide) Km
Vmax is the maximum reaction rate.
Km is the Michaelis-Menten constant, and for catalase, Km = 25mM (25
But first we have to find the V max for each single concentration:
V max is found by finding the highest possible gradient.
V max of Concentration 0% –> Gradient = Y2-Y1 = 0 = 0.000
X2- X1 30
V max of Concentration 1% –> Gradient = Y2-Y1 = 9 = 0.300
X2- X1 30
V max of Concentration 3% –> Gradient = Y2-Y1 = 32.5= 1.083
X2- X1 30
V max of Concentration 5% –> Gradient = Y2-Y1 = 39.75= 1.325
X2- X1 30
V max of Concentration 7% –> Gradient = Y2-Y1 = 76 = 2.533
X2- X1 30
Then we can find the velocity of each concentration:
V of concentration 0% = 0.000 * Concentration of Hydrogen Peroxide = 0.000 = 0 mM
25 * Concentration of Hydrogen Peroxide 25
V of concentration 1% = 0.300 = 0.012 mM
V of concentration 3% = 1.083 = 0.043 mM
V of concentration 5% = 1.325 = 0.053 mM
V of concentration 7% = 2.533= 0.101 mM
From this we can presume that the Velocity increases slightly, though since the concentrations are quite low we are not able to conclude since we do not have enough information.
Because 9% did not have 2 results I chose not to put this in the processed data hence also not in the calculations and graphs. It can not be known if it is reliable enough.
Improvements and Reliability
The experiment was done twice, to make the data more reliable. From these we took an average. In both occasions almost identical reactions seemed to be happening, confirming that are data is quite reliable. The trend also helps us with the reliability since most reactions have a certain pattern, and since the polynomial trend line is quite accurate I would say our data is reliable.
This does not take away that there are no improvements. At the concentration of 5% , first attempt, the measuring cylinder moved slightly causing a speeding up of the enzyme activity at that moment from 57 to 69.7cm3. . Though this only happened once, it did change the graph and the end results since only an average was taken of 2 results for each time period. The reading of the hydrogen peroxide + foam was also not always accurate. With accuracy rate of ï¿½ 1 cm3 and also having foam increasing more on one side of the measuring cylinder then the other it was hard to establish the exact volume of the foam. Next time we should be more accurate on the reading of the measuring cylinder and make sure the measuring cylinder does not move during the process. Also the time was ï¿½ the same, though since the rate of enzyme activity was fast enough to increase quite a lot in a second it was hard to be accurate since we could not always read at the exact same second.
To evaluate this experiment we need to take in consideration the improvements and the reliability of the experiment. I think our experiment went well, as we found a trend and we performed the experiment with care and tried to be as accurate as possible.
Compared values were found on the internet. Where you can not see a comparison considering our results, these graphs were found using the formula, and looking for information of ï¿½catalase effect on hydrogen peroxide reactionï¿½. It can not be taken into such extend of comparison since we only had concentrations from 1 till 9 % of Hydrogen Peroxide. We do not know till which extend other experiments carry with concentrations. When we compare the results to our graph 1.3 we can see that on both graphs the rates are rising. Our data is most possibly relevant to the first part of the graph, still before the half maximum velocity. Even if we do we can not see on the graph 1.3 that the line is rising in a steep way such that is represented on the compared value. (Possible reasons would be the accuracy of our results)
We can conclude that the rate of enzyme activity is affected by the concentration of Hydrogen Peroxide till certain extend. We know that an enzyme only works efficient up till a certain point: where the enzyme denaturizes. In our experiment we can not tell at which point the enzyme denaturizes since we do not have enough different concentrations to conclude this. The chemical reactions were at their fasted in the first 30 seconds as the activation energy5 of the catalase was at its highest. The reaction was exothermic since no heat has been given off and the energy released was greater then the activation energy. The reaction was also visible, and you could see quite clearly that a higher concentration fastened the enzyme activity rate .The enzyme rate differed between 0 and 0.101mM (milliseconds per liter).This leads up to the visible result which is the foam produced. According to graph 1.3 we can see that up till 9% the enzyme activity is still rising, and not showing any appearances of denaturation. We can therefore presume that the enzyme rate will still increase up till a certain point.
Where at 9% this was a lot higher then with 5 %, the end result was in amount of foam produced that the one with the concentration of 9% produced far more than 5% though the reaction with the 5% concentration substrate seemed to be going on longer at attempt 1 then at a concentration of 9% of hydrogen peroxide.
2 Andrew Allot – IB Study Guide Biology for Standard and Higher Level. Page 18/19
3 KM in the 10-25 mM range ( took the highest) –> http://www-saps.plantsci.cam.ac.uk/records/rec136.htm
4 Internet site : http://www.steve.gb.com/images/science/michaelis_constant_defintion.png
5 Activation energy: The energy needed to break bonds within the reactant. Later during this process energy is given off as new bonds are made. ~ Andrew Allot – IB Study Guide Biology for Standard and Higher Level. Page 69