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From the lock and key hypothesis,

1. Amylase is the lock, while starch is the key.

2. The starch with a three-dimensional shape complementary to that of the active site can fit into the amylase molecule.

3. An enzyme-substrate complex is formed.

4. The activation energy is lowered and bonds are formed/broken.

5. maltose is formed and released from the active site.

6. The amylase molecule is free to bind with more starch molecules.

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One source of error is the wind speed which effects surrounding temperature of each test tube which cannot be controlled. Different temperatures across test tubes may decrease the activity of the enzymes, in turn increasing the time recorded for the indicators to turn yellow.

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When contents were added into test tubes, enzymes were already starting to breakdown fats. It is hard to determine the exact time taken for enzymes to break down fats completely.

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Gp 1: Arielle, Silin, Ralph
Gp 2: Shirley, Shayndel, Xavier
Gp 3: Hayley, En le, Kieran
Gp 4:  Maansha, Zhe Kang, Shayne Ee
Gp 5: Adena, Jun Rui, Jia Yin, Chun Yeng
Gp 6: JunHao, Kerryn, Kripa, Efei
Gp 7: Kai Yan, Liu Yan, Jurgen

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Enzyme is the lock, while the substrate is the key. The substrate with a three-dimensional shape complementary to that of the active site can fit into the enzyme molecule. Thus, an enzyme-substrate complex is formed and the activation energy is lowered and bonds are formed/broken. Products are formed and released from the active site. The enzyme molecule is free to bind with more substrate molecules.

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1. Temperature, enzyme needs an optimum temperature, too high denature, too low inactive
2. PH value, too high or too low denatures

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Enzyme needs an optimum temperature.
When temperature is below optimum temperature for the enzyme, enzyme activity is low. When temperature is above optimum temperature, the enzyme will be denatured and substrate is not able to fit into the active site and no enzyme substrate can be formed.
pH value must be optimal for the enzymes affected.
Extreme changes in pH denature the enzymes and enzymes loses its three-dimensional shape and the substrate cannot fit into the active site causing no enzye substrate complex to be formed.

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When it is below optimum teamperature the enzymes become inactive and enzyme activity becomes low.
When it is at the optimum temperature, the enzymes are the most active and the rate of enzyme activity is the highest.
Above optimum temperature, the active site loses its 3 dimensional shape and it is denatured. The substrate cannot fit into the active site.
At extreme pHs, the enzymes become denatured and the active site loses its 3 dimensional shape and the substrate cannot fit into the active site.

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Form new products by combining(bond making) or breaking down(bond breaking) substances.

Carried out using the lock and key hypothesis:
1. Enzyme is the lock, substrate is the key.
2. They have 3 dimensional active site with the shape complementary to the substrate.
3. When the substrate and enzyme bind, they form an enzyme substrate complex.
4. Ea is lowered
5. Products are formed and released form the active site.
6. Enzymes are free to bind with other substrates.

e.g.: In the small intestine, protein is broken down by trypsin into polypeptides.

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Enzyme molecules contain an active site, which allows a specific substrate molecule to fit and bind to the enzyme. The substrate has a three-dimensional shape that complements with the active site of the enzyme molecule. As the substrate fit and binds to the enzyme, a enzyme-substrate complex is formed. The activation energy is lowered and bonds are either broken or formed at the active site. Products are formed and released from the active site. The enzyme molecule is now free to bind with more substrate molecules.

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pepsin breaks down protein in the stomach into polypeptides

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protein -> pepsin ( in stomach ) -> polypeptide

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Amylase in the saliva breaks down starch into maltose in the mouth.

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Lipase breaks down lipids into fatty acids and glycerol

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Salivary amylase:
Starch --> Maltose by salivary amylase

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An example of an enzyme is pepsin. Pepsin breaks down proteins to polypeptides

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Carbonic anhydrase in the red blood cell is the lock, while carbonic acid is the key.
Carbon dioxide and water are formed and released from the active site.

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Sodium carbonate is used to increase the pH in the syringe so the starting pH value will be 10, as the optimum pH for lipase is 10

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the sodium carbonate is used to neutralise any excess acid, which is formed from lipase breaking down fat into fatty acids and glycerol

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Test tube 1 : Start at 0 min. Contents in the test tube turned yellow at 3minutes and 21 seconds.

Test tube 2 : Start at 0 min. Contents in the test tube turned yellow at 1minute and 39 seconds.

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dilute sodium carbonate solution is used in order to increase the starting pH of the solutions in the test tube to 10. This is done as the optimal pH for lipase is 10. However, this should be exercised with caution as a pH value of over 10 could denature the lipase in the solution.

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used to increases pH in the syringe to reach the optimum pH for the enzymes to act on substrate.

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test tube 1:start at 0 min.
contents in test tube turned yellow at 1 minute and 32 seconds.

test tube 2:start at 0 min.
contents in the test tube turned yellow at 6 minutes 43 seconds.

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The time taken for the contents in test tube 1 to turn yellow is more than that of test tube 2. This is because there was bile found in test tube 2 which emulsifies fat. This increase surface area to volume ratio of lipids and increases the efficiency of lipase on lipids. Lipids are broken down into fatty acids and glycerol faster, turning the solution yellow faster. However, test tube 1 did not contain bile, hence it was slower. The contents in test tube 3 did not turn yellow as the lipase added was boiled. This denatures the lipase and active site loses its three dimensional shape. Lipids are unable to form a enzyme substrate complex with enzyme. Hence, no lipids were broken down into fatty acids and glycerol.

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test tube 1 start at 0 contents turned yellow at 4 minutes
test tube 2 started at 0 and the contents turned yellow at 3 minutes

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In test tube 1, solution took a longer time to turn yellow as compared to test tube 2 as there was no bile salt present in it. There is no emulsification of larger fat molecules into smaller fat molecule which will speed up the rate of reaction as emulsification of fat increases surface area to volume ratio of fat molecules to lipase. lipase will then breakdown fat molecule at a slower rate, solution turns acidic at a slower rate causing solution to turn yellow slower. in test tube 3, solution remained blue as boiled lipase were added. lipase becomes denature at high temperature due to boiling, there is no breakdown of fat in fatty acid and glycerols in the solution hence solution remains alkaline and remains blue

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test tube 1 started at 0 min, turned yellow at 11 minutes
test tube 2 started at 0 min, turned yellow at 7 minutes

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test tube 2. bile was added and the bile emulsified the larger fat droplets into smaller fat globules which increased the surface area to volume ratio which increased the ate of lipase breaking fats down into fatty acids and glycerol. As fatty acids are acidic the Bromothymol blue indicator turned yellow most rapidly.

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In tube 3, boiled enzymes are denatured due to the high temperature. Bile salts help to emulsify the fats in test tube 2 so fats are broken down into fatty acids and glycerol in test tube 2 the fastest. Since fatty acid is acidic, it cause the pH to be lowered causing the indicator to be lowered.

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Test tube 1 does not have bile salts but it has fresh lipase to digest fats into fatty acids and glycerol. Hence, the solution will become acidic and the time taken for bromothymol blue will turn from blue to yellow will be more than test tube 2 due to a lack of bile salts.

Test tube 3 has denatured lipase due to the high temperature caused by boiling the lipase. Therefore, the fats cannot fit into the active site of the lipase and fats cannot be digested into fatty acids and glycerol, leading to the bromothymol blue remaining blue.

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Test tube 1 contains only fresh lipase while test tube 2 contains both fresh lipase and boiled lipase and test tube 3 only contains boiled lipase. Lipase breaks down fate in the fresh milk into fatty acid and glycerol. Test tube 1, having the highest concentration of fresh lipase, the contents turns from blue to yellow the fastest as there is the fastest rate of production of fatty acids and glycerol.

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hard to determine the exact time of colour change of the solution in the test tubes which leads to an inaccurate result.

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Surrounding temperature is not optimum for the enzymes therefore it is not the most active. Therefore the rate of enzyme activity may vary and the time taken is longer.

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The temperature of the surroundings are not at 37 degrees which is the optimum temperature for the lipase to act. Therefore the rate of activity of the lipase would be decrease and the results collected would not be accurate.

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Unable to determine exactly when lipids are completely broken down into fatty acids and glycerol. The indicator turns yellow as the alkaline solution turns acidic. A small amount of fatty acids may be present after lipids are broken down, causing the solution to turn yellow. Though the solution may be yellow, all the lipids might not have been fully broken down.

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Test tube 1 took a longer time as the fat globules were not emulsified using bile salts like in test tube 2, so there is a larger surface area to volume ratio, decreasing the rate of fat being broken down.

Test tube 3 did not turn yellow as it has boiled lipase, which is denatured and cannot break down fats into fatty acids and glycerol.

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ensures that there is no excess acid present as it neutralises the acid. making the results more accurate as the change in colour will be solely due to the fatty acids present

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one source of error may be the time used to mix the mixture, as we do not know how long it takes for the contents to be fully mixed. this may result in some test tubes being undermixed, affecting the rate of breaking down of fat

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test tube 1 took 45s to turn yellow
test tube 2 tool 24s to turn yellow

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test tube 1 start 0 min
turns yellow at 1 min

test tube 2 start 0 min
turns yellow at 20 seconds

test tube 3 no change

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All test tubes stayed blue.
No change until adding more lipase.

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Dilute sodium carbonate will be neutralised and the pH will be lowered, ensuring that the change in the colour of bromothymol blue is solely due to the fatty acids

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https://docs.google.com/presentation/d/1-T9VPmnUCeMEEIBKH84NgVL0_9wNnn3aBf8PuGg00J4/edit#slide=id.g216cbee4592_0_426

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1. As a group, conduct the experiment and answer all the questions in part (a).
2. As you pen down your answer in your worksheet, update on the padlet (according to the groups stated).
3. Grps 1 to 4 pls review Grps 5 to 7 work on (ii), (iii) and (iv) and vice versa and provide suggestions/ questions you have.