AP Bio Lab Report

Kunal Bhattacharjee
Period 8
AP Bio

Enzyme Catalysis

Lab Report by Kunal Bhattacharjee

1

Kunal Bhattacharjee
Period 8
AP Bio
Enzyme Lab Report
INTRO:

Enzymes are a type of proteins that are formed by Amino acids and help speed up metabolic reactions. They are able to do this by interacting with the substrate
.
The substrate is what is being breaking down in the reaction. The substrate comes in to contact with the enzyme by binding to the enzyme’s unique active site
.
This is called the enzyme­substrate complex
.
The complex may cause a reduction in the activation energy for the reaction. the rates of reaction can be affected based on concentration, pH, Temperature, Activations, and Inhibitors
.
Low concentration can cause the enzyme to denature and will form an inactive precipitate while high concentration can cause new interactions between charged will occur and ordinary interactions will be blocked.
The enzyme will denature in extreme high or low pH but perform the best when pH levels are neutral. High temperatures will speed up the chemical reaction while low temperatures will slow it down. This is because temperature has a direct causation in the mobility of the molecules and rates of kinetic energy. If the temperature is raised really high, the proteins will denature because it will cause a negative effect among the enzyme molecules. This is called the “temperature optimum”. Activators are molecules other than the substrate that interact with the active site and they help increase the rate of reaction. Inhibitors bind and react with the side chains and slow the rate of reaction.
In this experiment, hydrogen peroxide (H
O
) acted as the substrate being broken down while
2
2 the catalase was the enzyme. The hydrogen peroxide formed a decomposition reaction as it was broken down to water (H
O) and oxygen (O
). The balanced equation is evident below.
2
2

2 H
O
→2 H
O + O (gas)
2
2
2
2

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Kunal Bhattacharjee
Period 8
AP Bio
RESULTS


Time (sec.) KMnO
4
(mL)

10 sec.

30 sec.

60 sec.

90 sec.

120 sec.

180 sec.

360 sec.

Baseline

11.6 mL

9.8 mL

10.6 mL

11.3 mL

9.7 mL

8.32 mL

9.6 mL

H
O
used
2
2
(Class)

3.65 mL

3.7 mL

5.9 mL

0.3 mL

7.6 mL

1.7 mL

10.3 mL

H
O
used
2
2
(Perfect)

0.9 mL

1.8 mL

2.4 mL

3.1 mL

3.3 mL

3.4 mL

3.4 mL

Timed Data Time Intervals

0­10 sec.

10­30 sec. 30­60 sec. 60­90 sec. 90­120 sec. 120­180 sec. 180­360 sec. Rates

0.9

0.045

0

0.02

0.023

0.006

0.0016

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Kunal Bhattacharjee
Period 8
AP Bio

DISCUSSION

Here is a comparison of the perfect data (right) and the class data (left). In the perfect data there is a gradual rise in the hydrogen peroxide used (rise in y­axis) as the time goes on (across the x­axis). This make sense because during the reaction, more and more hydrogen peroxide is being decomposed and broken down. Near the end of the graph, the line is constant, this is true because there is no more substrate left to break down so it will be infinitely constant unless more substrate is added. The reason why the graph turned out like because there was a constant baseline and the amount of H
0
used was
2
2 gradually increasing until it reached a constant. In the class data however, there was no consistent baseline which is part of the reason why the graph became jumbled. Also, there was plenty of human error in conducting the experiment which is why it seemed like someone interfered with the reaction. because it is not possible for a substrate to randomly increase and decrease unless it was added manually in the middle of the reaction.

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Kunal Bhattacharjee
Period 8
AP Bio
CONCLUSION (Questions 1­6) 1.

The rate of reaction table should be in the results section of this report.

2.

The reaction rate was highest in the initial ten seconds because there was more substrate to break down.

3.

Going off number two, if the initial ten seconds were the highest rate in the reaction then consequently from 180 to 360 seconds would be the lowest in the reaction because there is less substrate to break down.

4.

The sulfuric acid competes with other substrate for binding with the active site since its a competitive inhibitor and this slows down the rate of the reaction.

5.

I predict lowering the temperature would decrease the rate of enzyme activity because it decreases the mobility of the molecules therefore decreasing the kinetic energy. If it’s too low it may be out of the optimal conditions for the enzyme to function and may cause denaturation.

6.

I would perform multiple variations of the same experiment. Each time I would vary the pH, concentration, temperature, and substrate and compare the behaviors of each.

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